WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Comprehensive WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter can help students make connections between concepts.

Matter: Atomic Structure, Physical and Chemical Properties of Matter Class 9 WBBSE Notes

Indian Philosopher Kanad named the basic smallest particles of matter as Kana.

Greek Philosopher Demokritos called the smallest particle as Adornos’ meaning indivisible.

John Dalton developed the idea of ultimate unit of matter which is known as atom.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Cathode Rays : It consists of negatively charged material particles called electrons. Cathode rays are produced from cathode surface when a gas at low pressure is subjected to electric discharge.

Electrons : Fundamental sub-atomic particles carrying negative charge (1.602 × 10-19 coulombs) and having mass 91 × 10-31 kg, discovered by J. J. Thomson.

Proton : A sub-atomic positively charged particle, having charge 1.602 × 10-19 coulombs and mass 1.67 × 10-27 kg. Mass of a proton is nearly 1837 times higher than the mass of an electron.

Alpha particle : He2+ ions or helium nuclei.

Rutherford’s Experiment : Led to the discovery of nucleus. Radius of nucleus (~ 10-15 m) is very small as compared with radius of an atom (~ 100-10m).

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Neutrons : Sub-atomic neutral particles having mass 1.675 x 10-27 kg, discovered by James Chadwick.

Nucleons : The constituent particles of the nucleus i.e., protons and neutrons together are called nucleons.

The mass of an electron is \(\frac{1}{1837}\) times the mass of a proton.

The mass of an electron is \(\frac{1}{1839}\)  times the mass of a neutron.

Rutherford’s model of the atom proposed that a very tiny nucleus is present inside the atom and electrons revolve round this nucleus. The stability of the atom could not be explained by this model.

Neils Bohr proposed that electrons revolve round the nucleus in a limited number of orbits, called permissible orbits. The orbits are also called energy level.

Energy levels are designated by letters K, L, M, N, O, P, Q.

Elements with their outermost shell completely filled are chemically inert.

The outermost shell of an atom is called valence shell and the electrons present in the valence shell are called valence electrons.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

The maximum number of electrons that can be accommodated in a shell is given by 2n2, where
n = shell number.

Atomic number (Z)= Number of protons
= Number of electrons
– Mass number (A) = Number of protons + Number of neutrons

Isotopes : Atoms of same element having same atomic number but different mass number.

Isobars : Atoms of different elements having same mass number but different atomic number.

Isotones: Atoms of different elements having same neutron number but different atomic and mass number.

Nuclear force : In the nucleus, continuous transformation of the closely packed neutrons and protons take place i.e. protons transform into neutrons and vice versa. As a result of this transformation a strong attractive force acts within the nucleus. This force is known as nuclear force.

Ion : A charged atom or radical is called an ion.

Cation : Positively charged atom or radical is called a cation.

Anion : Negatively charged atom or radical is called an anion.

Frequency (v) of the radiation emitted when an electron jumps from an orbit having energy E1 to an orbit having energy E1 is given as \(v=\frac{E_2-E_1}{h}\)

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Electronic arrangement of elements:

WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 1
Each element has its own characteristic emission spectrum.

Series of lines in the line spectrum of hydrogen atom :

WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 2

Orbital: Region of space around the nucleus where probability of finding an electron is maximum.

Quantum Mechanics: It is a theoretical science that takes into account the dual nature of electron.

Degenerate orbitals: The orbitals having equal energy.

Ground state: The lowest energy state of an atom.

Excited state: An atom is said to be in excited state if some lower energy orbital is vacant and the electron is present in higher energy orbital.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Electronic configuration : Distribution of electrons among various orbitals in an atom.

Quantum numbers : A set of four number used to specify energy, size, shape, orientation of the electron orbital and spin of the electron.

Principal Quantum number (n) : Tells us about energy of the electron and size of the orbital.

Azimuthai Quantum number (1): Tells us about the shape of the orbital and orbital angular momentum.

Magnetic Quantum number (m): Tells us about orientations of the electron cloud in a sub-shell.

Spin Quantum number (s) : Tells us about spin of the electron.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

s-orbitais are spherically symmetrical.

Shape of s-orbitais:
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 3
p-orbitals are dumb-bell shaped:
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 4

The five d-orbitais are : dxy dyz dxz dx2 – y2 and dz2

WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 5

Angular momentum \(m v r=\frac{n h}{2 \pi}\)
[m mass of the electron, u = velocity of the electron, r = radius of the orbit,
h = Planck’s constant, n = Principal Quantum number]

de-Brogue equation:
\(\lambda=\frac{h}{m v}\) A = [λ = wavelength, m = mass of the electron, υ = velocity of the electron, h = Planck’s constant]

Heisenberg’s uncertainty principle:
\(\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}\)
[Δx = uncertainty in position,
Δp = uncertainty in momentum
h = Plancks constant]

Schrodinger’s wave equation:
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 6

[m = mass of an electron, V potential energy, E = total energy, Ψ = wave function of electron (eigen function

Aufbau Prinuiple : The orbitais are filled in the order of increasing energy,starting with the orbital of lowest energy.

Pauli’s Exclusion Principle: No two electrons in an atom can have same set of ail the four quantum numbers.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Hund’s Rule of Maximum Spin Multiplicity: The pairing of electrons in orbitais of a particular sub-shell cannot take place until all the orbitais of the sub-shell are singly occupied.

Nuclear isomer : Atoms (nuclides) which have the same atomic number and mass number but have different radioactive properties are called nuclear isomers.
Example:
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Isodiapher : Atoms in which the difference between the number of neutrons and the number of protons is saine are termed as isodiapher.

Example:
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 8

Isoster: Molecules or ions with same number of atoms and also the same number of electrons are said to form isosteric group or more simply isosters.
Example : in N2O and CO2, number of atoms 3 and numbers of electrons = 20. So, these are isosters.

Gay Lussac’. Law : Under the same conditions of temperature and pressure, when two or more gases combine together. they do so in simple ratio by volumes, and the volumes of the products if gaseous also bear a simple ratio to the volume of the reacting gases.

Berzelius hypothesis: Under the same conditions of temperature and pressure, equal volume of all gases contain the same number of atoms.

Avogadros hypothesis : Under the same conditions of temperature and pressure equal volume of all gases (both elementary and compound) contain equal number of molecules.

Molecule : The smallest particle of an element or compound which can exist in the free state is known as molecule.

Elementary molecule : The molecule which is formed by one or more atoms of an element is known as elementary molecule. Example : Hydrogen (H2), Oxygen (O2) etc.

Compound molecule : The molecule which is formed by atoms of more than one element is known as compound molecule. Example: Water (H2O), Carbon dioxide (CO2) etc.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Vapour density: Vapour density of a gas is the ratio of the weight of a certain volume of the gas to the weight of same volume of hydrogen under similar conditions of temperature and pressure.

Molecular mass =
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 9

Atomic mass = \(\frac{\text { mass of } 1 \text { atom of an element }}{\frac{1}{12} \times \text { mass of } 1 \text { carbon }\left({ }^{12} \mathrm{C}\right) \text { atom }}\)

Gram atomic mass : When the atomic mass of an element is expressed in gram then the amount of element in gram is known as gram atomic mass of the element.

Gram molecular mass : When the molecular mass of an element or compound is expressed in gram then that amount of element or the compound in gram is known as gram molecular mass of the element or the compound.

Molar volume : The volume of a gaseous substance (element or compound) under a fixed temperature and pressure is known as molar volume of the gas.

Avogadro’s number: One gram-molecule of any element or compound containing equal number of molecules (NA 6.022 × 1023) is known as Avogadro’s number.

R. A. Millikan determined the value of Avogadro’s number by ‘oil drop experiment’ in 1913.

Modern definition of Avogadro’s number : The number of atoms present in exactly 12g of carbon (12C) is designated as Avogadro’s number.

Avogadro’s constant : ‘Avogadro’s number/mole’ is called Avogadro’s constant
i.e. 6.022 × 1023 Avogadro’s number has no unit. But Avogadro’s constant has the unit mol’. It is a universal constant.

Atomicity : It is the number of atoms by which an elementary molecule is composed of.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Atomic mass unit (amu) : It is an unit used for measusing the atomic mass of an atom.
1 amu = 16606 ×10-24g

Average atomic mass
\(\frac{\sum \text { natural abundance of isotope }(\%) \times \text { its atomic mass }}{100}\)

Mole (SI System) : A mole is the amount of a substance which contains as many entities (atoms, molecules, ion or any other particle) as there are atoms in exactly 0012 kg (or 12g) of carbon-12 isotope.

Solution : A solution (or true solution) is a uniformly homogeneous mixture of two or more substances whose relative proportions may be varied upto a certain limit.

Solute : The substances which are present in smaller quantities and gets dissolved are called solutes.

Solvent : The medium in which the solutes are uniformly dispersed through dissolution is known as the solvent.  solute + solvent = solution

Homogeneous mixture : A homogeneous mixture is one in which the constituent substances are so intimately mixed that even a very close examination cannot distinguish any surface of separation between them.

Heterogeneous mixture : A heterogenous mixture on the other hand is one in which the particles of the constituent substances of a mixture are distinctly distinguishable.

Some useful solvents : Water is a versatile liquid solvent of organic and inorganic substances, carbon disuiphide is the solvent of sulphur, oxalic acid is the solvent of rust, benzene is the solvent of rubber etc.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Types of solution:

(a) Solution of solid in solids : Solution of one metal in another gives the alloys like brass, bronze etc.

(b) Solution of liquids in solids : Mercury (liquid metal) can remain dissolved in solid metal like gold, silver, sodium etc. to form amalgams.

(c) Solution of gases in solids : Hydrogen gets dissolved in spongy palladium and in some similar metals. This phenomenon is called occlusion.

(d) Solution of solids less in liquids : Aqueous solution of common salt (NaCl), sugar, copper sulphate are examples.

(e) Solution of liquids in liquids : A mixture of water and alcohol, a mixture of benzene and xylene are examples.

(f) Solution of gases in liquids: Ordinary water contains air dissolved in it. Soda water contains carbon dioxide under pressure. Dissolved oxygen in water allows aquatic life to survive.

(g) Solution of solids in gases : e.g. smoke, where solvent is air and solute is carbon and dust particles dissolved in air.

(h) Solution of liquids in gases: e.g. fog.

(j) Solution of gases in gases : e.g. air, mainly a mixture of nitrogen and oxygen.

Classification based on concentration:

  • saturated solution
  • unsaturated solution
  • supersaturated solution

Suspension : The particle in suspension state is 10-4cm in diameter.

True solution : The particle in true solution is 10-8 cm in diameter.

Sol : It is a colloidal solution in which a solid is dispersed in a liquid (e.g.paints).

Emulsion : It is a colloidal solution in which a liquid is dispersed in another liquid (e.g. milk).

Gel : It is a colloidal solution in which a liquid is dispersed in a solid (e.g.fruit jellies, cheese etc.)

Aerosol : It is a colloidal solution in which a solid or a liquid is dispersed in a gas (e.g. fog. smoke, clouds etc.)

Lyophilic colloids : These are the substances which pass readily into the colloidal state whenever mixed with a suitable solvent (e.g. protein, starch etc.)

Lyophobic colloids : These are the substances which do not yield colloidal solutions on mere shaking with a liquid (e.g. gold, silver. Fe(OH)4, As).

Positive sols : These are the sols which carry positive charge on the dispersed phase particles. (e.g. sols of Fe(OH), Al(OH)3 etc.)

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Negative sols : These are the sols which carry negative charge on their particles (e.g. sols of Cu, Ag, Au, As2S3 etc.)

Multi-molecular colloids :These are the colloids in which the individual particles consist of aggregates of atoms of small molecules having molecular sizeless than 10-7cm in diameter. (e.g. sols. of gold atoms, platinum sol.)

Macro-molecular colloids : These are the colloids in which the size of the particles of the dispersed phase are of the order of colloidal dimensions (e.g. sol of starch, cellulose)

Peptisation : It is the process of converting precipitates into colloidal state by adding small amount of a suitable electrolyte.

Dialysis : It is the process of separating substances in colloidal state from those present in ionic states with the help of a semipermeable membrane

Tyndal I effect : It is the process of scattering of light from the surface of colloidal particles and in this process a beam of light passed through a colloidalsolution becomes visible as a bright streak.

Brownian motion : This is ceaaeless erratic irregular and random motion of colloidal particles suspended in a dispersion medium.

Elcctrophoresis : It is the movement of charged colloidal particles, under the influence of an electric field, towards the oppositely charged electrodes.

Coagulation : It is the phenomenon of change of colloidal state to suspended state. It can be brought about by adding an electrolyte to a colloidal solution.

Hardy schuize rule : It states that greater the valency of the coagulating ion added, the greater is its power to coagulate.

Gold number : ¡t is the weight in milligrams of a protective colloid which prevents the coagulation of 10 mL of a gven gold sol on adding 1 mL of 10% solution of sodium chloride.

Smaller the value of gold number, of a lyophilic colloid, the greater is its protective power.

Colloids in everyday life :
(i) Foods, medicines and pharmaceutical preparation.
(ii) Soaps, synthetic detergents,
(iii) Paints, varnishes, enamels, resins, gums, glues, rubber.
(iv) Industrial processes such as tanning, dyeing, rubber plating and removal of smoke from air are based on the colloidal nature of particles.

Suspended particulate matter (SPM) : Many partículate matter, such as fine particles of dust, sand, coal dust, pollen grains, cement dust, metal dust, fly-ash etc. remain suspended in normal air; these are termed as pollutants which cause air pollution.

Solubility : Solubility of a given solute in a solvent is defined as the weight in grams of the solute dissolved in 100 grams of the solvent so as to saturate the solution at a given temperature.

Solubility is just a number, it has no unit,

Effect of temperature and pressure on soluhility of gas in liquid:
(a) Effect of tempe rature : The solubility of gases in liquids decreases with the increase of temperature and at lower temperature, solubility increases.

(b) Effect of pressure : At a definite temperature the solubility of a gas in a particular solvent is proportional to the pressure applied on the gas. This is known as Henry’s law.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Effect of temperature Of solubility of solid in liquid :

  • The solubility of KNO3 in water, increases tremendously with the increase of temperature.
  • In case of NaCl, the increase of solubility in water is much less, almost same.
  • The solubility of CaSO4in water decreases with the increase of temperature.
  • The solubility of Glauber’s salt (Na2SO4.104O) in water first increases upto a certain temperature (324°C) and then sharply decreases with temperature.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 10

Solubility curve :

Crystals : Crystals are homogeneous solid particles (big or small) with definite geometric shape, and are bounded by plane surfaces which meet at sharp edges.

Crystallisation : Crystallisation is a process by which crystals of a substance are obtained from its solution.

Water of crystallisation : When crystal is formed from the aqueous solution of it then one or more fixed number of water molecules associated with each molecules of that substance by chemical bonding are called water of crystallisation.

Hydrated crystals: Crystals having the water molecules associated with them are called hydrated crystals.

Example:
MgSO4. 7H2O (Epsome salt)
FeSO4. 7H2O (Green vitriol)
CuSO4. 5H2O(Blue vitriol)
H2C2O4. 2H<2O (Oxalic acid)

Efflorescent substance and Efflorescence: Certain hydrated crystals when exposed to air at ordinary temperature lose their water of crystallisation partially or fully and are transformed into amorphous varieties. Such a substance is called efflorescent substance and the phenomenon is known as efflorescence.

Example:
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Cause of efflorescence: Efflorescence is happened only when the vapour pressure within the hydrated crystal at ordinary temperature is greater than the vapour pressure of the atmosphere.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Deliquescent substance and Deliquescence: Certain water soluble substances when exposed to the atmosphere at ordinary temperature absorb water vapour from the air and are soluble in it to make saturated solution, such substances are called deliquescent substances and the phenomenon is known as deliquescence.

Example:
(i) MgCl2 (Magnesium chloride)
(ii) CaCl2(Calcium chloride)
(iii) FeCl3 (Ferric chloride)
(iv) NaOH (Caustic soda)
N.B. : Edible Salt (NaCl) becomes moist in the rainy season due to the presence of deliquescent MgCl2. 6H2O as impurity in it.

Cause of deliquescence: Deliquescence occurs when the vapour pressure of water in the delequescent substance is less than the vapour pressure of the atirosphere of ordinary temperature.

Hygroscopic substances: There are certain substances which when exposed to air absorb moisture from air but are not dissolved in the absorbed water. These substances are called hygroscopic substances. Example: Conc. HSO4, CaO etc.

Drying agent: There are certain substances which absorb water or water vapour molecule from other compounds without forming any chemical bond. These substances are called drying agent.
Example: Anhydrous CaCl2, Anhydrous MgSO4, Conc. HSO4 etc.

The concentration of a solution can be expressed in several different ways:
(a)Percentage by weight (% w/w): It is the amount of solute in grams present in 100 grams of the solution.

(b) Weightivolume percentage (% w/V): It is the amount of solute in grams present in 100 mL of the solution.

(c) Volume/volume percentage (% V/V): It is the volume in mL of the solute present in 100 mL of the solution.

(d) Strength: It is the number of grams of the solute dissolved per litre of the solution.

(e) Normality (N): It is the number of gram equivalent of solute dissolved per litre of a solution.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

(f) Molarity (M): It is the number of moles of solute dissolved per litre of the solution.

(g) Molality (m): It is the number of moles of solute dissloved in 1000 gram of the solvent.

(h) Formality (F): Formality of a solution is the number of gram formula weight of the ionic solute (e.g. NaCl) dissolved per litre of the solution.

(j) Parts per million (ppm) : It is defined as the number of parts of a component per million parts of the solution.

Vapour pressure of liquid : The pressure developed above the liquid at a given temperature at the equilibrium point.

Lowering of vapour pressure : It is the difference in the vapour pressure of the pure solvent and that of the solution.

Raoult’s Law : The vapour pressure of a solution is equal to the product of the mole fraction of the solvent and its vapour pressure in pure state.

Ideal solution: The solution which obey Raoult’s law at all concentration and follow the conditions
\(\Delta \mathrm{H}_{\text {mix }}=0 ; \Delta \mathrm{V}_{\text {mix }}=0 \)

Non-ideal solution : The solution which show positive or negative deviations from Raoults law.

Azeotrope : The mixture of liquids which boils at constant temperature like pure liquid and has same composition of components in liquid as well as vapour phase.

Colligative properties : The properties of the solution which are independent of nature of solute but depend upon the concentration of solute particles.

Osmosis : The passage of solvent from pure solvent or solution of low concentration to the solution of higher concentration through the semi permeable membrane.

Isotonic solutions : The solutions of same molar concentration and same osmotic pressure at given temperature.

Acids:
(j) Arrhenius concept : A substance which furnishes H ions in aqueous solution.
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 12

(ii) Bronsted concept : A substance which is proton donor.

(iii) Lewis concept : A substance which is acceptor of electron pair.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Types of acids : According to their structure:

(i) Hydracids : Those acids which have hydrogen atoms and other non-metal or radicals but no oxygen atom are called hydracids. Example : Hydrochloric acid (HCl), Hydrocyanic acid (HCN).

(ii) Oxyacids : Those acids which have one or more oxygen atoms with hydrogen and other non-metal atoms are called oxyacids. Example : Nitric acid (HNO3), Sulphuric acid (H2SO4).

Types of oxyacids:
(a) Ous-acids : Those oxiacids which have lesser number of oxygen atom and the valency of the main element naming the acid is less are called otis-acids.

Example: HNO2 (Nitrous acid)
[valency of N = 3]
H2SO3 (Sulphurous acid)
[valency of S = 4]

(b) Ic-acids : Those oxyacids which have greater number of oxygen atom and the valency of the main element naming the acid is high. are called ic-acids.

Example: HNO3 (Nitric acid)
[valency of N= 5]
H2SO4 (Sulphuric acid)
[valency of S = 6]

(c) Hypoacids : Those acids which have less oxygen atom than ous-acids are called hypoacids.
Example : HOCl (Hypochiorous acid)

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

(d) Per-acids : Those acids which have greater number of oxygen atoms than ic-acids are called per-acids. Example : HClO4 Perchloric acid)

Basicity of acid : The number of H ions produced in aqueous solution of a molecule of acid, is called basicity of acid.

Types of acids according to basicity of acid:

(a) Monobasic acids : Those acids which ionise in aqueous solution to produce one He ion are called monobasic acids. Example : HCl, HNO3, HBr etc.

(b) Diba sic acids : Those acids which ionise in aqueous solution to produce inherit; two He ions are called dihasic acids. Example : H2SO4, H2CO3, H2SO3 etc.

(c) Tribasic acids : Those acids which ionise in aqueous solution to produce three He ions are called trihasic acids. Example : H3PO4

• Types of acids according to their ionisation :

Strong acids : The acids which ionise mostly in aqueous solution and small number of molecules exist in molecular state, are called strong acids.
Example : HCl, HNO3, H2SO4 etc.

Weak acids : The acids whose most of the molecules exist in molecular state and ionise a little number of molecules, are called weak acids.
Example : CH3COOH, H2CO3, H2S etc.

Types of acids according to their source :
Mineral acids : The acids which are produced from minerals are called mineral acids.

Example :
(i) Hydrochloric acid (HCl)
[Prepared from NaCl]

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

(ii) Nitric acid (HNO3)
[Prepared from KNO3]

Organic acids : The acids containing carbon atom and prepared from animals or plants are known as organic acids.
Example :
(i) Formic acid (HCOOH) [Source : Ant]
(ii) Lactic acid [CH3CH(OH)COOH] [Source : milk]

Properties of acids :
(i) An acid produces H  ion in aqueous solution.
(ii) Acids are generally soluble in water and the solution is acidic or sour in taste.
(iii) The aqueous solution of acids turns blue litmus into red, orange coloured methyl orange into pinkish red. On the other hand, phenol phthalein remains colourless in aqueous solution of an acid.
(iv) Electropositive metals like Zn, Mg, Fe (which belong above hydrogen in electrochemical series) react with dilute acid to produce hydrogen.
Example :
Zn + H2SO4 = ZnSO4 + H2
2Al + 6HCl – 2AlCl3 + 3H2

Acids form salt and water reacting with metallic oxides and hydroxides.
Example :
2NaOH + H2SO4 = Ha2SO4 + 2H2O
CaO + 2HCl = CaCl2 + H2O

Acids form carbon dioxide reacting with metallic carbonate and bicarbonate.
Example :
Na2CO3 + 2HCl = 2NaCl + CO2 + H2O
NaHCO3 + HCl = NaCl + CO2 + H2O

Fuming nitric acid : It is a product produced by dissolving excess amount of nitrogen dioxide in nitric acid.

Acid rain : Dissolved NO2 and SO2 of atmosphere produce nitric acid (HNO3) and sulphuric acid
(H2SO4)which come down along with rain. This phenomenon is known as acid rain.

Stone cancer : Corrosion of marble or stone due to acid rain is called stone cancer.

Aqua regia : It is a mixture of 3 volumes concentrated hydrochloric acid and 1 volume concentrated nitric acid which can dissolve metals like gold and silver.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Royal water : Aqua regia is called royal water.

Fuming sulphuric acid : Fuming sulphuric acid or oleum (H2S2O7) is obtained when sulphur trioxide is passed over 98% Sulphuric acid.

Muriatic acid : Hydrochloric acid.

Aqua fortis : Nitric acid.

King of chemicals : Sulphuric acid.

Laughing gas : Nitrous oxide (N2O).

Preparation of hydrogen chloride in laboratory :
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Preparation of nitric acid in laboratory :
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Manufacture of sulphuric acid by contact process :
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Identification of acids :

(a) Identification of Hydrochloric acid :
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(b) Identification of Sulphuric acid :
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(c) Identification of Nitric acid (Ring test) :
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Formation of Hydrogen from nitric acid : Magnesium liberates hydrogen from cold and very dilute nitric acid.
Mg + 2HNO3 = Mg(NO3)2+ H2

Passive iron : Cold and concentrated nitric acid or fuming nitric acid when comes in contact with iron produces passive iron. Passive iron is chemically inactive.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Uses of hydrochloric acid :

  • It is used in dyeing and calicoprinting.
  • It is used as a cleaning agent in galvanising and tin plating.
  • It is used in the manufacture of glucose, glue and many useful metallic chlorides.

Uses of nitric acid :

  • It is used in the production of explosives such as dynamite, gun cotton, nitro-glycerine, trinitro toluene (TNT).
  • It is used to prepare rayon, artificial silk and dyes.

Uses of sulphuric acid :

  • It is used for the preparation of different chemical compounds like HCl, HNO3, ether, alcohol etc.
  • It is used as an important raw material for dyes, medicine, plastic etc.
  • Oxides : Binary compound of oxygen with any element (metallic and non-metallic) are called oxides.

Types of Oxides :

(a) Acidic oxide : An acidic oxide is an oxide which reacts with a base to form salt and water.

  • Non-metallic oxide : CO2, SO2, P2O5, SiO2
  • Metallic oxide : CrO3 (Chromium trioxide), Mn2O7 (Manganese heptoxide) etc.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

(b) Basic oxide: A basic oxide is an oxide which reacts with an acid to form salt and water. These are generally the oxides of metals.
Example :
CaO (Calcium oxide)
MgO (Magnesium oxide)
FeO, CuO etc.

(c) Neutral oxide : The oxides which neither react with acid nor with base are called neutral oxide.
Example : CO (Carbon monoxide)
H2O (Water), N2O (Nitrous oxide)
NO (Nitric oxide)

(d) Amphoteric oxide : The oxides which react with both acids and bases to produce salt and water are called amphoteric oxide.
Example :
Al2O3 (Aluminium oxide)
ZnO (Zinc oxide)
SnO (Stanous oxide)
PbO (Lead monoxide) etc.

Peroxide : The oxides which react with cold and dilute mineral acid to produce hydrogen peroxide and have peroxylinkage (— O — O —) are called peroxide.
Example :
Na2O2 (Sodium peroxide)
BaO2 (Barium peroxide)

Mixed oxide : An oxide which is formed by oxides of more than one oxide of a metal having variable valency is called mixed oxide.
Example :
Fe3O4 [FeO + Fe2O3] ; Mn3O4 [2MnO + MnO];
Pb3O4 [PbO2 + 2PbO] etc.

Poly-oxide : The oxides which have oxygen atoms more than an ordinary oxide and also do not react with acid to produce hydrogen peroxide are called poly-oxides.
Example :
Mn2O7 (Manganese heptoxide)
PbO2 (Lead dioxide)

Sub-oxide : An oxide which has less oxygen atoms rather than required for oxidation state of the element is called sub-oxide. Example : Carbon sub-oxide (C3O2)

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Super-oxide : Metals like Na, Li, Ca form super-oxide and have negative ion (O-O) in it.
Example : LiO2, KO2 etc.

Alkali: Those metallic oxides and hydroxides react with acids to produce salts and water are known as alkali. Example : CaO, Na2O, NaOH, Al(OH)a etc.

Base (According to Arrhenius concept): The metallic oxides or hydroxides after dissolving in water produce negatively charged hydroxyl ions are known as base.
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 19

Properties of bases :

  • The aqueous solution of bases turn red litmus into blue, orange coloured methyl orange into yellow.
  • Bases produce salts and water when react with acids.
  • Concentrated solutions of some strong bases are slippery to touch.
  • The aqueous solution of bases can conduct electricity.
  • Strong bases when heated with electro-positive elements (like Al, Zn etc.) produce hydrogen.

Example : 2NaOH + Al + 2H2O = 2NaAlO2 + 3H2
Strong bases absorb carbon dioxide from atmosphere to form carbonate salt and water.
Example : 2NaOH + CO2 = Na2CO3 + H2O

Difference between alkali and bases : The oxides and hydroxides of metal and metal like radicals are called alkalis. But those alkalis which dissolve in water to form OH are called bases.

Example :
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 20

  • Acidity of bases : The number of OH ions produced in aqueous solution of one molecule of bases are called the acidity of bases.

Types of bases according to acidity of bases :

(a) Monoacidic bases : The bases which produce one OK ion in aqueous solution from one molecule are called monoacidic bases.
Example:
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 21

(b) Diacidic bases : The bases which produce two OH” ion in aqueous solution from one molecule are called diacidic bases.
Example :
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 22

Types of bases according to their ionisation :

Strong bases : The bases which ionise mostly in aqueous solution and small number of molelcules exist in molecular state are called strong bases.
Example : NaOH, KOH, Ca(OH)2 etc.

Weak bases : The bases whose most of the molecules exist in molecular state and ionise a little number of molecules are called weak bases.
Example : NH4OH, Cu(OH)2 etc.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Types of bases according to their source :

Mineral bases : The bases which are produced from minerals are called mineral bases.
Example : NaOH, KOH etc.

Organic bases : The bases cortaining nitrogen atom and whose sources are animals and plants are called Organic bases.
Example:
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 23

Salts : The replaceable hydrogen atoms in an acid when replaced by metal or basic radical partially or fully then the compound so produced is called salt.

Definition of salts according to Arrhenius concept: The compound formed by cation other than hydrogen ion (H)+ and anion other than hydroxyl ion (OH~) is known as salt.

Types of salt :

  • Normal salts : Normal salts are formed by the complete replacement of the hydrogen atoms in an acid by metals or basic redicals.

Example :
(i) NaCl [Sodium chloride] [NaOH + HCl = NaCl + H2O]
(ii) Na2SO4 [Sodium Sulphate] [2NaOH + H2SO4 = NA,SO4 + H2O]

Acid salt: An acid salt is one which is formed only by partial replacement of hydrogen atom in a molecule of an acid (di-basic, tri-basic) by metal or basic radical.
Example :
NaHSO4 (Sodium bisulphate)
NaHCO3 (Sodium bicarbonate)
Na2HPO4 (Di-sodium hydrogen phosphate)

Basic Salt: When the hydroxyl (OH) or, oxide (O) radicals are partially neutralised by acid then basic salts are produced.
Example :
(i) Pb(OH)Cl [Basic lead chloride]
(ii) Pb(OH)NO3 [Basic lead nitrate]

Double salts : When two normal salts are mixed with each other with their molecular weight ratio to form a solution and cooled the solution to make a combined crystal which is stable in solid state but ionises into different constituent ions in solution are called double salts.
Example :
K2SO4 Al2(SO4)3 . 24H2O
(NH4)2SO4FeSO4 . 6H2O

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Complex salts : When the solution of two mixed salts is concentrated and cooled then a crystal of new salt is produced. The constituent salts lose their identity in new salt. In aqueous solution they produce new complex ions. This type of salts are called complex salts.
Example : K4[Fe(CN)6] ; [Cu(NH3)] SO4

Neutralisation : The reaction in which equivalent amount of an acid reacts with equivalent amount of a base and thus the properties of acid and bases are completely lost forming salt and water, is called neutralisation reaction.
Example :
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 25

Titration : The process by which bases are neutralised with the help of acids or vice-versa is called titration.

Indicators : The chemicals which are able to determine the end point of neutralisation reaction by changing their colours are called indicators.

Indicators used in neutralisation reaction :

Name of the indicators and their actual colour In acid solution In alkali solution In neutral solution
1. Litmus (violet) red blue violet
2. Phenolphthalein colourless pink colourless
3. Methyl-orange (orange) red yellow orange

Choice of indicators :
(a) Strong acid and weak base — Methyl orange
(b) Weak acid and strong base — Phenolphthalein
(c) Strong acid and strong base — Any indicator
(d) Weak acid and weak base — No such indicator.

Vanishing colour : Vanishing colour is a coloured solution which becomes colourless on exposure to air after a lapse of time.
Example :
The aqueous solution of ammonium hydroxide (NH4OH) mixed with phenolphthalein turns pink colour. Now if this solution is allowed to spray on cloth, at once a pink colour is developed upon cloth. But since ammonia in NH4OH is highly volatile, after lapse of time NH3 disappears from the cloth and the solution becomes neutral. As a result pink colour of phenolphthalein vanishes.

Vanishing colour cannot be prepared with dilute NaOH solution, due to the fact that NaOH does not evaporate.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

pH : The symbol pH is derived from ‘Poteuz,’ the Danish word for power, pH refers to potency of hydronium ion.  pH = -log [H] = – log[OH]

pH range :

  • The pH of acidic solution is less than 7. [pH < 7, the solution will be acidic]
  • The pH of basic solution is greater than 7. [ pH > 7, the solution will be basic]
  • The pH of neutral solution is 7. [pH = 7, the solution will be neutral] So, the range of pH is from 0 to 14.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 26
Colour changes for universal indicator at different pH values.

  • The approximate pH of a solution can be determined with the help of pH papers.
  • pH -papers have different colours in solutions of different pH.
    A pH paper can determine pH of a solution with an accuracy of about 0 -5.

pH-metres : For accurate measurement of pH (upto accuracy 0.001 units), pH metres are used.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Universal indicator : A mixture of organic dyes which gives different colours with solutions of different

pH values is called a universal indicator.

Name of the Fluid pH
1. 0 1(M) NaOH solution 13
2. Milk of magnesia 10
 3. Egg white, sea water 7-8
4. Human blood  7-4
5. Tears 7-4
6. Milk 6-8
7. Human Saliva 64
8. Tomato juice 4-2
9. Soft drinks and Vinegar 30
10. Lemon juice 2-2
 11. Gastric juice 1.2
12. 1(M) HCl solution 0

Importance of pH :

Field Utility
1. Biotechnology Biochemical and organic reactions at controlled pH value give best results.
2. Agriculture Some crops such as citrus fruits grow better in alkaline soils. Sugarcane grows better in neutral soil, whereas rice grow better in slightly acidic soil. The pH of soil is tested before growing particular crop.
3. Medicine Determination of pH value of blood and urine helps to diagnose certain disease.
4. Cosmetics Soaps, shampoos, face creams are prepared for the consumers having different pH value of skin secretions.
5. Milk plants pH of milk is rigorously controlled at pH 6 8 as otherwise it turns sour.
6. Breweries Controlling pH value of wine to obtain a desired flavour.

Antacids : Medicines which can remove the excess acid from the stomach and raise the pH to appropriate level are called antacids.

Types of antacids:

Cimetidine (Tegamet): It binds to the receptors that trigger the release of hydrochloric acid into the stomach. This results in release of lesser amount of acid.
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 27
Cimetidine remained the largest selling drug in the world for a long time.

(ii) Ranitidine (Zantac): After cimetidine, ranitidine (Zantac) was introduced for treatment of hyperacidity.
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 28

Separation of mixtures:

Fractional distillation : The process by which two or more miscible liquids are separated by distillation using the difference in their boiling point is called fractional distillation.

When the boiling points of two or more miscible liquids differ by about 15°C — 20°C these are separated by simple fraction distillation.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

fractionating column : If the boiling points of two liquids are very close i.e. difference is small in that case a fractionating column is used to separate the liquid mixtures.

Separation of immiscible liquids using separating funnel : Immiscible liquids are easily separated by using a separating funnel. Each liquid forms its own liquid layer due to difference in density.

Chromatography : It is a method by which mixture of different substances are separated. In this method due to difference in adsorption of different substances in solid phase (adsorbent) and difference in migration of the substances in the mobile phase (liquid or gas), various substances are separated.

Paper Chromatography : It is a very easy technique to separate the various organic dyes present in the writing ink or printing ink.

Advantages of chromatography :

  • A small amount of the compound present in the mixture can be separated.
  • The properties of the individuals present in the mixture do not alter.
  • It is a very easier method to separate the different components present in the mixture.

Sublimation : There are some solids which on heating directly transform into vapour phase without transforming through the intermediate liquid state. This phenomenon is called sublimation.
Example : NH4Cl (ammonium chloride) and SiO2 (sand) are separated by this method.

Natural water : Natural water, except rain water, is generally impure and contains dissolved metals, apart from other impurities.

Sources of natural water :

  • Rain water
  • River water
  • Spring and well-water
  • Mineral water
  • Sea water

Soft water : Water in which lather is easily formed with soap is called soft water.

Hard water : The class of water that does not easily form lather with soap, is called hard water.

Hardness of water : The property of natural water which does not allow it to form lather is known as hardness of water.

Cause of hardness of water : The hardness of water is caused by the presence of bicarbonate, chloride and sulphate of calcium, magnesium and iron in water.

WBBSE Class 9 Physical Science Notes Chapter 4 Matter: Atomic Structure, Physical and Chemical Properties of Matter

Types of hardness :

  • Temporary hardness : It is due to the presence of soluble bicarbonate of calcium, magnesium and iron in water.
  • Permanent hardness : This is caused by the presence of chloride and sulphate of calcium, magnesium and iron in water.

Removal of temporary hardness :
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 29

Removal of permanent hardness :
Lime-soda process :
CaSO4 + Na2CO3 = CaCO3↓ + Na2SO4
CaCl2 + Na2CO3 = CaCO3↓ + 2NaCl

Permutit process :
2Na (Permutit) + Ca(HCO3)2 = Ca(Permutit)2 ↓+ 2NaHCO3

[Zeolite, a natural mineral is a giant molecule of sodium aluminium silicate (NaAlSO4 3H2O). Artificially prepared mineral called permutit is used to remove permanent hardness]

Calgon process :
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 30

Ion exchange resin process :
WBBSE Class 9 Physical Science Notes Chapter 4 Matter Atomic Structure, Physical and Chemical Properties of Matter 32
De-ionised water : Very soft and pure water which is obtained by treating hard water through the ion-exchange process, is free from all types of ions i.e., cations and anions. Such type of water is known as de-ionised water.

Water is an universal and versatile solvent :

  • Water is a polar compound. Most of the ionic compounds are soluble in polar solution and thus water is, therefore, a good ionising solvent for acids, bases and salts.
  • Water is an amphiprotic solvent i.e. it can act both as an acid (a proton donar) and a base (a proton acceptor) in the reaction.
  • Many covalent compounds like sugar, urea, alcohol, organic acid, organic bases are soluble in water through hydrogen bonding. Hence, we may conclude that water is an universal as well as versatile solvent.

Water pollution : Some soluble and insoluble matters in water which produce some harmful effect to men and aquatic animals are known as water pollution.

Pollution due to natural source :

Pollution due to arsenic : Ground water is sometime contaminated with arsenic acid and arsenius acid as a deprotonated form. Consumption of these arsenic contaminated water for a long period can cause dangerous poisonous effects in the form of black spots on palms and feet and skin become rough. This is known as black foot disease. From scientific researches it can be assumed that arsenic poison may be a cause of skin cancer.

Pollution due to fluoride : Sea-water, river-water contain fluoride from natural source. Excess fluoride in water is responsible for fluorosis disease which is a cause of dental decay and bone decay.

Pollution due to artificial or human activity : Fertilizers, insectisides, pestisides, are regularly used for increasing the production of food grains. These substances when washed with water are mixed in ponds, river etc. As a result of this, water is polluted.

Removal of arsenic from water :

  • Arsenic is removed from water if water is allowed to pass through a bag made of cloth containing alumina or ferric hydroxide replacing the candle used in the filter. Arsenic or arsenius ions are absorbed in the alumina bed or ferric hydroxide bed and thus are removed.
  • Take 20 litre of water, \(\frac{1}{4}\)th amount of tea spoon of bleaching powder and the same amount of ferrous sulphate in a bucket. After stirring well, arsenic will be removed.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 1 Concerns About Our Environment offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 1 Question Answer – Concerns About Our Environment

Very Short Answer Type Questions :

Question 1.
In which of the atmospheric layers does the increase in altitude result in increase in temperature?
Answer:
Stratosphere and Thermosphere.

Question 2.
Which one is more harmful to the human body – CO or CO2 ?
Answer:
CO is more harmful to the human body.

Question 3.
In which of the layers of atmosphere does temperature decrease with increase in altitude?
Answer:
Troposphere and Mesosphere.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 4.
Mention the range of temperature of the atmosphere ?
Answer:
The range of temperature of the atmosphere is from -92° C to + 1200° C.

Question 5.
What is the main source of carbon monoxide in the atmosphere ?
Answer:
Coal, Petrol and incomplete combustion of other fossil fuels.

Question 6.
Which of the atmospheric layers contains maximum ozone gas?
Answer:
Stratosphere.

Question 7.
Name some gases which are responsible for green house effect.
Answer:
Carbon dioxide (CO2), Methane (CH4), Water vapour (H2O), Chlorofluoro carbon (CFC), Nitrous Oxide (N2O).

Question 8.
Name the sources of dissolved oxygen in water ?
Answer:
The sources of Oxygen in water are the atmospheric oxygen or the photosynthesis carried out by the plants in water during daylight.

Question 9.
Name of the main compounds causing damage to the ozone layer?
Answer:
Nitric oxide (NO) and Freons.

Question 10.
What compound is formed when CO gas combines with blood ?
Answer:
Carboxy haemoglobin.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 11.
Which disease is caused due to the formation of hole in the ozone layer and why ?
Answer:
Ultraviolet rays coming from the sun reach the earth passing through the ozone hole cause skin cancer.

Question 12.
Name some methods of waste managment ?
Answer:
B urning and incineration, recycling process, sewage treatment and dumping method.

Question 13.
Name some methods used in green chemistry ?
Answer:
The use of sunlight, micro-waves, sound waves and the use of enzymes.

Question 14.
What is COD?
Answer:
COD means chemical Oxygen Demand.

Question 15.
Name some methods used in green chemistry.
Answer:
The use of sunlight, microwaves, sound waves and the use of enzymes.

Question 16.
Name two sinks of carbondioxide.
Answer:
Ocean which dissolves it and plants which use it for photosynthesis.

Question 17.
What type of pollution effects the sea birds ?
Answer:
Oil pollution.

Question 18.
What are the main sources of thermal pollution ?
Answer:
Nuclear plants and Thermal power plants.

Question 19.
What is siltation ?
Answer:
The mixing of soil or rock particles into water is known as siltation.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 20.
Which gas was responsible for so many deaths in Bhupal tragedy?
Answer:
The gas methyl isocyanate (MIC) was responcihle to many deaths in Bhupal.

Question 21.
What type of aromatic compounds are present as particulates in the air?
Answer:
Polycyclic aromatic hydrocarbons (PAH).

Question 22.
What are silicosis and asbestosis ?
Answer:
Lung disease caused by the particulates of silica is called silicosis and that caused by particulates as asbestos is called asbestosis.

Question 23.
What class of people are generally susceptible to attack of black lung and white lung diseases.
Answer:
Generally workers of coal mines and those of cotton mills become victims of black lung and white lung diseases respectively.

Question 24.
What is PCB ?
Answer:
PCB is polychlorinated biphenyl. It is a highly toxic compound.

Question 25.
Why does the population of fish get hindered in thermally polluted water ?
Answer:
Thermal pollution increases the temperature which in turn decreases the DO level of the water. Thus it affects the fishes badly and their growth gets retarded.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 26.
What is chlorosis ?
Answer:
The presence of SOx in the atmosphere as pollutant slows down the production of chlorophyll in the leaves of the plants and the green colour disappears. The phenomenon is called chlorosis.

Question 27.
Why high and low tides occur in sea ?
Answer:
Due to the gravitational pull of sun and moon on the spinning earth, the level of water in the sea rises and falls, these are referred to as the “high tide’ and ‘low tide’.

Question 28.
Thal energy is converted to what form of energy for use ?
Answer:
Electrical energy.

Question 29.
Where in the world geothermal energy is used to generate electricity ?
Answer:
Due to geological changes, some molten hot rocks are formed in the deeper region’s of the earth’s crust and are pushed upward and traped in some regions. In some places heat comes from the fission of radioactive material naturally present in the rocks.

Question 30.
Name a petro-crops.
Answer:
At present Euphorbias plants contain latex of high carbon content, thermal cracking of this latex produce bio-petrol or diesel. These plants are called petro-crops.

Question 31.
What is the name of anaerobic bacteria used for the fermentation of blomass ?
Answer:
Methanogenic bacteria.

Short Answer Type Questions :

Question 1.
What are combustible and non-combustible materials ?
Answer:
Combustible materials : Materials that can be burnt are combustible materials. Non-combustible materials : Materials that cannot be burnt are noncombustible materials.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 2.
What are the types of sources of energy?
Answer:
Types of sources of energy: There are two types of sources of energy. These are : (i) Renewable sources of energy. (ii) Non-renewable sources of energy.

Question 3.
What is nuclear fission?
Answer:
Nuclear fission : It is a process in which a heavier nucleus splits into lighter nuclei with release of huge amounts of energy.

Question 4.
What is a petroleum? What is natural gas?
Answer:
(i) Petroleum: An oil derived from rocks. A complex mixture of hydrocarbons. (ii) Natural gas : Natural gas is a fossil fuel that is found on the top of petroleum under the earth.

Question 5.
What is destructive distillation.
Answer:
Destructive distillation : It is a process of burning wood in limited supply of oxygen.

Question 6.
What is hydroelectric power plant ?
Answer:
Hydroelectric power plant : In this power plant energy of flowing water is utilized to produce electricity on a large scale.

Question 7.
What is wind mill?
Answer:
Wind mill : It is a device that converts wind energy to mechanical or electrical energy.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 8.
What is a solar cell?
Answer:
Solar Cell : It is a device that directly converts solar energy into electrical energy.

Question 9.
What is good source of energy ?
Answer:
Good source of energy : A good source of energy should be renewable and environment friendly.

Question 10.
What are the disadvantages of fossil fuels ?
Answer:
Disadvantages of fossil fuels: Following are the disadvantages of fossil fuels: (i) These are non-renewable (ii) The produce air pollutants on burning.

Question 11.
What is geothermal energy?
Answer:
Geothermal energy : The energy which can be hamessed from the heat of inside of the earth is called geothermal energy.

Question 12.
If you could use any source of energy for heating your food, which one would you use and why?
Answer:
We use LPG (liquefied Petroleum Gas) for heating our food; because it is easily available in our kitchen, gives on smoke on burning and takes less time to do the job.

Question 13.
Name two energy sources that you would consider to be renewable. Give reasons for your choice.
Answer:
Solar energy and wind energy are examples of renewable energy sources because they get quickly replenished.

Question 14.
Give the names of two energy sources that you would consider to be exhaustible. Give reasons for your choice.
Answer:
Coal and petroleum are exhaustible energy sources because their stock is going to be finished in the near future and it takes millions of years for their formation.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 15.
Fish do not grow as well in warm water as in cold water-explain why.
Answer:
Explanation: The solubility gas in water depends on the temperature. Consequently the amount of soluble oxygen in warm water becomes less than that of cold water. As oxygen is very essential for the respiration and growth of fish in water, so the growth of fish becomes less in warm water.

Question 16.
What is chlorosis ?
Answer:
Chlorosis: The process of sulpherdioxide gas in the atmosphere retards the production of chlorophyll in the leaves of the plants. In consequence of which the green colour of leaves is lost. This phenomenon of disappearance of green colour of leaves is known as chlorosis.

Question 17.
What is siltation ?
Answer:
Siltation: In this process of mixing of soil particles with water. The soil particles present in water procedure turbidity. Due to the turbidity of water, the free movements of the equatic organisms is hindered and as a result their growth and productivity also get reduced.

Question 18.
In which season the depletion of ozone layer on Antarctica take place and when is it filled up again ?
Answer:
During the spring time in Antarctica i.e. in the months of September and October, the depletion of ozone layer of the stratosphere takes place. But after the spring in the months of November and December the produced ozone hole is filled up.

Question 19.
What is meant by inversion temperature in different regions of atmosphere?
Answer:
Inversion temperature: As we proceed from one region of atmosphere to the next higher region, the trend of temperature changes either from increase to decrease or from decreaseto increese. This trend of either increase or decrease is called inversion temperature.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 20.
How does carbon monoxide molecule link to haemoglobin molecule?
Answer:
Explanation: In haemoglobin molecule Fe(II) atom is linked to five groups by coordinate bonds and the sixth site remains free. In normal condition aerial oxygen O2 links to Fe (II) of this vacant coordination site. Now carbon monoxide (CO) being a stronger legand than oxygen molecule in case of the presence of both CO and O2, CO occupies the vacant coordiriction site in haemoglobin instead of oxygen.

Broad Answer Type Questions

Question 1.
What is a good fuel ?
Answer:
Good fuel: A good fuel should have many desirable characteristics. Some of these are as follows:

  1. It should produce a high amount of energy for each unit of mass or volume.
  2. It should be easily available and should be easy to transport.
  3. It should be economical.

Question 2.
Why are we looking for alternate source of energy ?
Answer:
Because of growing population, the energy demand is rising. Fossil fuels are going to be exhausted in the near future and burning them is causing air pollution. Hence we need to find an alternate source of energy which renewable and environment-friendly.

Question 3.
How has traditional use of wind and water energy been modified for our convenience?
Answer:
Before the beginning of the Industrial revolution, wind and water energy were used for serving many purposes but the ways of their use were not efficient. For example windmills were used to power smaller mills like flour mills or saw mills. The flow of water was used for transporting wooden logs. Now-adays, both wind energy and water energy are being widely used to produce electricity, which means a better and more efficient use of these forms of energy.

Question 4.
What kind of mirror, concave or convex or plain would be best suited for use in a solar cooker? Why ?
Answer:
Explantation : A concave mirror is best suited for use in a solar cooker. The reason for this is the ability of a concave mirror to converge the solar energy at a point. This enables the concave mirror to produce a larger amount of heat compared to other types of mirror.

Question 5.
What are the limitations of the energy that can be obtained from the oceans?
Answer:
Limitations of energy which can be obtained from the oceans :
These forms of energy can be used only in coastal areas, which would leave a vast portion of the human habitation. The technologies for using these energies are still at experimental stage and hence are very costly and less efficient.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 6.
What are the advantages of nuclear energy ?
Answer:
Advantages of nuclear energy are as follows: A small amount of radioactive material can generate a huge amount energy. It does not produce air pollution. A nuclear power plant is more efficient than other power plants.

Question 7.
Can any source of energy be pollution-free? Why and why not ?
Answer:
Many sources of energy can be pollution-free when burning of biomass is not involved in production of energy, then there is no chance of pollution. For example, wind energy, hydel energy, solar energy etc. are pollution-free.

Question 8.
Hydrogen has been used as a rocket fuel would you consider it a cleaner fuel than CNG ? Why and why not?
Answer:
Hydrogen is a much cleaner energy source than CNG. CNG is derived from biomass and hence burning the CNG causes air pollution, although on a much smaller scale than coal and petroleum. Use of hydrogen as energy source does not polluting gases.

Brain Storming Questions :
Very Short Answer Type Questions :

Question 1.
In which layer of atmosphere least temperature is found?
Answer:
Mesopause, the upper part of mesosphere, is the coldest region having an average temperature around -85° C.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 2.
Why ionosphere is so called ?
Answer:
Ionosphere has ions and electrons produced due to different types of ionization reactions occuring there in presence of high energy solar uv radiation. For this reason it is called so.

Question 3.
What is a ‘moisture trap’ and why is it called so ?
Answer:
Tropopause is called ‘moisture trap’. Most of water vapour of stratosphere enters through the tropopause by virtue of its low temperature leaving stratosphere dry. For this reason tropopause is called so.

Question 4.
What is an ozone hole?
Answer:
Ozone hole is a region of expectionally depleted ozone in the stratosphere of the ozone layer in high altitudes i.e., the Antartic, chiefly in winter, attributed to chemical action of CFCs and other atmospheric pollutants resulting in an increase in ultraviolet rays at ground level which gives rise to an increased rink of skin cancer.

Question 5.
What do you mean by ODS ?
Answer:
The chemical substances which catalyse and accelerate the depletion of ozone are called ozone-depleting substances (ODS).
Example :
Chlorofluorocarbons (CFCs), Halons etc.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 6.
What is Dobson unit (DU) ?
Answer:
The Dobson unit (DU) is a unit of measurement of the columner density of a trace gas in the earth’s atmosphere.

Question 7.
Global warming is a consequence of which phenomenon ?
Answer:
Global warming is a direct consequence of greenhouse effect.

Question 8.
Name one compound which has approximately 20 times more green house effect than carbondloxide.
Answer:
Methane (CH4) has approximately 20 times more green house effect than CO2.

Question 9.
What is a greenhouse ?
Answer:
Greenhouse is a glass house used for raising plants, vegetables, fruits and flowers within it in cold temperature regions where growing of crops in the open field is not possible due to low temperature or cold.

Question 10.
Is the greenhouse effect applicable for earth only ?
Answer:
No, in the solar system, the atmospheres of Venus, Mars and Titan also contain gases that cause a greenhouse effect.

Question 11.
Name any two radiations emitted by the sun that are not visible to the human eye.
Answer:
Infrared radiation and ultraviolet radiation are not visible to the human eye.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 12.
Which one is the principle greenhouse gas ?
Answer:
When ranked by their direct contribution to the greenhouse effect, apart from water vapour, Carbon dioxide stands as the principal greenhouse gas.

Question 13.
Name a non-renewable source of energy other than fossil fuels.
Answer:
Nuclear energy is another category of non-renewable source of energy other than fossil fuels.

Question 14.
What is the empirical formula of methane hydrate?
Answer:
The empirical formula of methane hydrate is (CH4)4 .(H2 O)23.

Question 15.
Which process is the basis for the formation of biomass ?
Answer:
Photosynthesis is the basic process for the formation of biomass.

Question 16.
Name the country which generates maximum electricity using wind energy ?
Answer:
Germany produces maximum electricity using wind energy.

Question 17.
What should be the maximum wind speed to operate wind turbine?
Answer:
The maximum wind speed shall be 15 Km hr-1.

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Question 18.
How much power does a single photo voltaic cell produce ?
Answer:
The power output of a single solar photo voltaic cell varies between 1.0 to 1.5 w.

Question 19.
Name two semi conducting materials which are used to make solar cells.
Answer:
Solar cells are made of Silican (Si) and Gallium (Ga).

Question 20.
Between Petrol and Kerosene, which has a higher calortic value?
Answer:
Petrol (50 Kj g-1) has a higher calorific value than Kereosene (48 Kj g-1).

Question 21.
Name two devices in which solar energy is harnessed directly.
Answer:
Solar energy is harnessed directly in solar cookers and solar water heaters.

Numerical Problems :

Working fomula :

Calorific value of fuel :
Amount of heat energy liberated on burning of a fuel = (mass of fuel) × (Calorific value of fuel)

Solar constant :
Heat energy produced by incident solar radiation =(Solar constant) × (Area) × (time)
Where solar constant =1.4 kjm-2 s-1 = 14 kwm-2

WBBSE Class 10 Physical Science Solutions Chapter 1 Concerns About Our Environment

Examples 1: Calculate the energy released by a piece of wood of mass 10 kg when it is completely burnt. (The calorific value of wood is 17 Kjg-1)
Answer :
Amount of heat energy liberated on burning a fuel (piece of wood)
= (mass of fuel) × (Calorific value of fuel)
= 17 × 10000 kj
[Mass of fuel, m = 10 kg
= 170000 kj
= 10000 g
Calorific value of fuel = 17 Kjg-1 ]

Example 2 : The calorific value of a gas is 40 Kjg-1. If a furnace contains 40 g gas in 40 second, calculate the power of the furnace.
Answer :
Gas consumed by the furnace is 40 second = 40 g
∴ Energy consumed by the furnace is 40 second
= 40 × 40 kj=1600 kj
∴ Power of the furnace = \(\frac{Energy}{ Time}\) = \(\frac{40 \times 40}{40}\) kjs-1
= 40 kw

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

Comprehensive WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties can help students make connections between concepts.

Matter: Structure and Properties Class 9 WBBSE Notes

Fluid : The word fluid comes from a Latin word ‘fluere’ meaning to flow.

Hydrostatics : The branch of Physics which deals with fluid at rest is called hydrostatics.

Hydrodynamics: The branch which deals with fluid in motion is called hydrodynamics.

Density : The density of a substance is its mass per unit volume.

Specific gravity : Specific gravity of a substance is the ratio of the density of a substance to the density of water at 4°C.

Relative density : The specific gravity being ratio of two densities, it is also called relative density.
In C.G.S. system, the density of a substance is numerically equal to its specific gravity.

Hydrostatic pressure : The normal force exerted by the fluid at rest per unit area of the surface in contact with it is called the pressure of fluid or hydrostatic pressure.

Mathematical expression of pressure : If F be the normal force acting on the very small area A of the surface in contact due to the fluid, then the pressure ‘ p ‘ is given by,

p = \(\frac{F}{A}\) t or F = p × A
Thrust : The total normal force exerted by a fluid at rest on a surface in contact with it is thrust.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

Unit of pressure :

  • CGS : dyne cm-2
  • SI : Nm-2 or pascal (Pa)

Dimension of pressure : [ML-1T-2]
Pressure is a scalar quantity as pressure at any point inside a fluid is same in all directions.

Unit of Thrust :

  • CGS : dyne
  • SI : newton (N)

Pressure at a point in a liquid : Consider a liquid of density ‘d’ contained in a vessel in equilibrium of rest. We are to find the pressure at a point ‘O’ at a depth ‘h’ below the liquid surface. Consider a cylindrical vertical column of liquid of height ‘h’, the area of cross-section of the cylinder being ‘A’. Area ‘A’ is around the point ‘O’.
So, volume of liquid column = hA
mass of the liquid column = h Ad
Weight of the liquid column = h Adg
∴ Force exerted by the liquid column on area A = hAdg
∴ Pressure ‘P’ on A = \(\frac{hAdg}{A}\)
∴ P = hdg

WBBSE Class 9 Physical Science Notes Chapter 3 Matter Structure and Properties 1

Important points regarding liquid pressure :

  1. The liquid at rest exerts equal pressure at a point inside the liquid in all directions.
  2. The liquid at rest exerts equal pressure at all points at the same horizontal level in the liquid.
  3. The liquid pressure is independent of shape of the liquid surface or the area of the liquid surface, but depends upon the height of liquid column.
  4. Total pressure at a depth ‘h’ below the liquid surface is equal to (p + hdg), where ‘p’ is the atmospheric pressure and ‘d’ is the density of the liquid.
  5. Average pressure on the side walls of a containers having liquid of density ‘d’ upto a height ‘h’ is \(\frac{1}{2}\) hdg.
  6. The liquid pressure is a scalar quantity.
  7. The thrust exerted by a liquid on the walls of vessel in contact with the liquid acts normally to the surface of the vessel.
  8. The free surface of a liquid at rest is horizontal.
  9. The liquid always finds its own level.
  10. The gauge pressure at a point in a liquid is the difference of total pressure at that point and atmospheric pressure.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

Pascal’s Law : The pressure applied at any place to an enclosed fluid at rest is transmitted with undiminished magnitude to every portion of the fluid and acts normally to the surface in contact with the fluid.

Archimedes’ principle : Archimedes’ principle states that a body immersed wholly or partly in a fluid at rest, appears to lose a part of its weight, which is equal to the weight of the fluid displaced.

Buoyancy : The upward thrust which any fluid exerts upon a body partly or wholly submerged in it is called its Buoyancy.

Conditions for Archimedes’ principle and buoyancy :

As Archimedes’ principle involves the weight of a body, so the principle does not hold good when the body is in weightless condition like the state of free fall of the body in an artificial satellite moving in a circular orbit.

The buoyancy does not depend on the depth of the liquid to which the body is immersed or on the shape, mass or material of the body. It depends only on the immersed volume of the body and the density of the displaced liquid at a particular place.

Conditions for floatation of a body :

  1. When the weight of the body(w) is greater than the weight of the liquid displaced (w), then the body sinks, i.e. w>w.
  2. When the weight of the body (w) is equal to the weight of the liquid displaced (w), then the body floats completely immersed anywhere within the liquid, i.e. w = w’.
  3. When the weight of the body is less than the weight of the liquid displaced, then the body floats partly immersed on the surface of the liquid. i.e. w < w

Atmospheric pressure : The weight of a column of air of unit crosssection and which extends from the point under consideration to the top of the atmosphere is called atmospheric pressure.

Standard or Normal temperature and pressure : The standard or normal atmospheric pressure is defined as the pressure exerted by a column of mercury 76 cm high at 0°C at 45° latitude and mean sea-level.

Siphon : A siphon is a simple device for transferring liquid from one vessel to the other without disturbing the whole volume of the liquid or pouring it.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter Structure and Properties 2

Arrangement of siphon : It consists of a bent tube with one of the arms BD longer than the other AC. The tube is first filled with the liquid to be transferred. The two ends of the tube are then closed with fingers and the shorter limb is placed in the vessel to be emptied below the level of the liquid and two ends are opened. The liquid begins to flow.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

Working principle of siphon :

Let ‘p’ be the atmospheric pressure, ‘ d ‘ the density of the liquid and h1 and h2 be the vertical heights of A and B above the liquid surfaces.
The pressure P1 at A = P-h1 dg and the pressure P2 at B = P-h2 dg.
Now as h2>h1, so P1>P2,
i.e. Pressure at A > Pressure at B.
Hence the liquid at A is pushed to B. As the liquid moves from A to B the vacancy so produced is filled up by atmospheric pressure forcing more liquid into the tube from the vessel. From the point ‘B’ liquid comes out of longer tube due to gravity and a continuous flow through the siphon takes place.

Conditions for working of a siphon are :

  1. At first the whole tube must be filled with liquid.
  2. The end of the longer tube must be below the level of the liquid in the other vessel.
  3. The height h1 must be less than the height of the corresponding liquid barometer, otherwise the atmospheric pressure will not be able to raise the liquid to the top of the tube.
  4. The siphon will not work in vacuum due to the absence of atmospheric pressure.

Example : 1. We cannot siphon out water from a boat floating on a river to the river. Because the boat is floating in the river and water leaks into the boat from the river, so level of water inside the boat would be same as that of the river outside.

2. There being no atmosphere in the moon, atmospheric pressure is absent there. So normally siphon does not work on the moon. But in the case of pure liquid, the siphon action may be possible even in the moon due to cohesive force of the liquid.

Surface Tension : Surface tension is the property of a liquid by virtue of which the free surface of liquid at rest tends to have minimum area and so it behaves as if covered with a stretched membrane.

Measurement of surface tension : Surface tension is measured as the force acting on unit length of an imaginary line drawn tangentially anywhere on the free surface of the liquid at rest. It acts at right angles to both sides of the line and tangentially to the liquid surface.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

Mathematical expression of surface tension :

If ‘F’ be the total force acting on an imaginary line of length l, drawn tangentially on the liquid surface at rest, the force of surface tension ” T ” is given by:
T = \(\frac{F}{l}\)

Units of surface tension :

  • CGS : dyne cm-1
  • SI : Nm-1

Dimensional formula of surface tension : [MT-2]
Surface tension is a scalar quantity as it has no specific direction.

Examples of surface tension :

Raindrops attain spherical shape to acquire minimum area (as we know for a given volume of sphere has minimum surface area) due to surface tension, so that it has minimum surface energy.

An iron needle sinks in water but a greased iron needle placed gently on the water surface at rest is found to float on it. A slight depression on the surface of water can be seen just below the needle, showing that the water surface behaves like a stretched membrane.

A wire ring is dipped in soap solution to form a soap film on it and a loop of cotton thread is gently placed on it. The loop takes an irregular shape. Now on pricking the soap film in the middle position of the loop with a sharp needle, the loop would be found to attain circular shape. It is because the remaining soap film tries to have minimum surface area due to surface tension and for a given perimeter, the area of the circle is maximum.

A little kerosene oil dropped on water is found to spread over the water surface. This is because the surface tension of oil is much less than that of water and so the greater tension of the water stretches the oil in all directions.

Surface energy : It is the amount of work done against the force of surface tension, in forming the liquid surface of a given area at a constant temperature.

Excess of pressure inside a soap bubble or liquid bubble : A soap bubble or a liquid bubble has a very thin film having two surfaces and has air both inside and outside the bubble. So, in this case, excess of pressure ‘ P ‘ inside the bubble is given by :

P = \(\frac{4T}{R}\)

Excess of pressure inside an air bubble in a liquid :
An air bubble inside a liquid has only one surface and so excess of pressure ‘P’ inside it is given by,

P = \(\frac{2 T}{R}\)

If the air bubble is formed at a depth ‘h’ below the free surface of a liquid of density ‘dhr’, then excess of pressure inside the bubble is given by :

P = \(\frac{2T}{R}\) + hdg

Angle of contact: The angle which is tangent to the liquid surface at the point of contact, makes with the solid surface inside the liquid is called angle of contact.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

The value of angle of contact :

  1. the value depends on the nature of the liquid and the solid in contact.
  2. depends on the medium above the free surface of the liquid.
  3. is independent of the inclination of the solid with liquid surface.
  4. is fixed for a given pair of solid and liquid and surrounding medium.

Variation of surface tension :

The presence of impurities in the liquid surface or dissolved in it, considerably affect force of surface tension and depends on the degree of contamination.

The surface tension of a liquid decreases with rise of temperature. The surface tension of a liquid becomes zero at boiling point of the liquid and it vanishes at critical temperature.

Applications of surface tension :

  1. Surface tension of soap solution is low, and thus it is used for washing. Hot soap solution is still better as surface tension decreases with rise in temperature.
  2. Surface tension of lubricating oils and paints are kept low, so that they can spread easily.
  3. Antiseptics have very low surface tensions so that they can spread quickly.
  4. Oil spreads over the surface of water because the surface tension of oil is less than that of water.
  5. Rough sea can be calmed by pouring oil on sea-water.
  6. In soldering, flux is added to reduce the surface tension of molten tin, so that it can spread.

Viscosity : Viscosity is the property of a fluid by virtue of which an internal frictional force comes into play when the fluid is in motion and opposes the relative motion of its different layers.

Co-efficient of viscosity (η) : Co-efficient of a liquid is the tangential force required to maintain a unit velocity gradient between two layers of unit area.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

Dimensional formula of :
η = ML-1 T-1

Unit of co-efficient of viscosity :

  • CGS : Poise
  • SI : Pascal-second (PaS) or decapoise

1 decapoise = 1 NSm-2

Decapoise : If 1 newton tangential force is required to maintain a velocity gradient of 1 ms-1 m between two layers each of area 1 m2 is called 1 decapoise.

Poise : If 1 dyne tangential force is required to maintain a velocity gradient of 1 cms-1 / cm between two layers each of area 1 cm2 is called 1 poise.
1 decapoise =10 poise

Terminal velocity : It is the constant maximum velocity acquired by a body while falling through a viscous fluid.

The expression of terminal velocity :

v = \(\frac{2 r^2\left(d-d^{\prime}\right) g}{9 \eta}\)

[ Consider a small sphere of radius r and density d falling under gravity in a viscous fluid of density d and coefficient of viscosity η and terminal velocity is r]

Terminal velocity depends upon :

  1. The terminal velocity of a sphere varies directly as the square of the radius of it.
  2. The terminal velocity of a sphere is inversely proportional to the co-efficient of viscosity.
  3. If the density of the material of the sphere be less than the density of the fluid, then the sphere would move upwards to attain terminal velocity.

Variation of viscosity :

  1. The viscosity of a liquid decreases with increase in temperature.
  2. The viscosity of all gases increases with increase in temperature.
  3. With increase in pressure, the viscosity of liquids increases, but viscosity of water decreases.
  4. With increase in pressure, the viscosity of gases remains unchanged.

Streamline Flow : The streamline or orderly flow of a fluid is that flow in which every particle of the fluid follows exactly the path of its preceding particle and has the same velocity in magnitude and direction as that of its preceding particle while crossing that point.

Laminar Flow : If fluid flowing over a horizontal surface with a steady flow, moves in the form of layers of different velocities which do not mix with each other, then the flow of liquid is called laminar flow.

Turbulent Flow : When a fluid moves with a velocity greater than its critical velocity the motion of the particles of fluid becomes disorderly or irregular and such a flow is called turbulent flow.

Critical velocity : The critical velocity is that velocity of liquid flow, upto which its flow is streamlined and above which its flow becomes turbulent.

Reynold’s number (R): It is the ratio of the inertial force per unit area to viscous force per unit area for a flowing fluid.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

Significance of Reynold’s number (R) :

  1. R < 2000, the flow of a liquid is streamline or laminar.
  2. R > 3000, the flow of a liquid is turbulent.
  3. 2000 < R < 3000, the flow is streamline or turbulent.

Energy of a liquid in motion : A liquid in motion possesses three types of energy. These are :

Pressure energy : The energy possessed by a liquid by virtue of its pressure and is measured by the work done in pushing the liquid in the vessel against the pressure without imparting any velocity to it is called the pressure energy of the liquid.

Potential energy : The energy possessed by a liquid by virtue of its height or position above the surface of the earth or any reference level taken as zero level is called the potential energy of the liquid.

Kinetic energy : The energy possessed by a liquid by virtue of its motion or velocity is called the kinetic energy of the liquid.

Bernoulli’s Principle (1738) : For the streamline flow of an ideal fluid (non-viscous and incompressible) the sum of the pressure energy, kinetic energy and potential energy per unit mass is always constant.

Mathematical form of bernoulli’s principle :

For streamline flow of an ideal fluid of density d and passing through any cross-section at a height h with a velocity ‘v’ at pressure p we have
\(\frac{p}{d}\) = Pressure energy per unit mass
gh = Potential energy per unit mass
\(\frac{1}{2}\) vhr2 = Kinetic energy per unit mass
According to Bernoulli’s theorem,
\(\frac{p}{d}\) + gh + \(\frac{1}{2}\)v2 = const.
or, \(\frac{p}{dg}\) + h + \(\frac{v^2}{2 g}\) = const.
or, p + hdg + \(\frac{1}{2}\)dv2 = const.

For the streamline flow of an ideal fluid, the sum of pressure energy per unit volume, potential energy per unit volume and kinetic energy per unit volume is always constant at all cross-sections of the liquid.

Applications of bernoulli’s theorem :

Atomiser or sprayer : It is Based on Bernoulli’s principle. It is used for spraying liquid like perfumes over the body etc.

Blowing off the roofs during storm : During storms or cyclones, sometimes roofs of a hut are found to blow off without causing any damage to the side walls of the house.

Motion of two parallel boats : The boats sailing in the same direction parallel to each other, small distance apart are found to come closer to each other.

Lift on an aeroplane wings: The shape of an aeroplane wings i.e. aerofoil is made in such a way that its upper surface is curved more than its lower surface.

Spinning of a ball (Magnus effect) : A Cricket ball is thrown straight without any spin. The streamlines of air passing are seen evenly passing above and below the ball and the ball continues to move along the original straight direction. But if the ball is released with a spin, the streamlines of air in the same direction as the direction of spinning of the ball, while these are oppositely directed.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

Limitations of bernoulli’s theorem :

  • It is assumed that velocity of every particle of liquid passing through any particular cross-section of the tube is uniform.
  • But acutally, praticles of liquid in the central layer have maximum velocity and those closer to the tubewell have minimum velocity.
  • Thus, mean velocity of the particles should be taken.
  • The liquid in motion experiences a viscous drag, which should be considered.
  • It is assumed here that there is no loss of energy, when the liquid is in motion. But there is always some loss of
  • energy as some kinetic energy is lost as heat.
  • When the liquid is flowing on a curved path, the energy due to centrifugal force should be considered.

Elasticity : Elasticity is the property by which a body is able to resist deformation, either in shape or in volume or both, and recovers its original configuration when the deforming force is removed.

Types of bodies according to elastic property :

Perfectly rigid body : A body is said to be perfectly rigid, when no relative displacement between its parts occurs under the action of a deforming force, however large it may be.

Perfectly elastic body : A body is said to be perfectly elastic, when it regains its original configuration immediately and completely after the removal of the deforming force, however large it may be.

Perfectly inelastic body : A body is said to be perfectly inelastic or plastic, when it does not regain its original configuration at all on the removal of deforming force, however small it may be.

Partly elastic body : A body is said to be partly elastic when after the removal of the deforming force, it regains only partly its original configuration.

Elastic Limit : Elastic limit is the upper limit of deforming force upto which the body regains its original shape or size completely on removal of deforming force and beyond which on increasing the deforming force, the body loses, its property of elasticity and gets permanently deformed.

Stress : Whenever a deforming force is applied to a body, internal force of reaction comes into play, which tends to resist the deforming force and maintains the original configuration of the body.
This reaction force developed per unit area of the body is called stress.

Types of Stress :

  • Normal stress
  • Tangential stress.

Normal stress : When the deforming force acts normally over an area of a body, then the internal restoring force set up per unit area of cross-section is called normal stress.

Types of normal stress :

Tensile stress : If there be increase in length or extension of a body in the direction of the applied force, the stress developed is called tensile stress.

Compression stress : If there be decrease in length or compression of a body due to applied force, the stress developed is called compression stress.

Tangential stress : When the deforming force acts tangentially to the surface of a body to produce a change in the shape of the body, then the stress developed in the body is called tangential stress.

Strain : When a deforming force is applied on a body, there is a change in the configuration of the body and the body is said to be strained.

Measurement of strain : The fractional change in the length, volume or shape of a body relative to that in its original configuration is a measure of the strain.

Types of strain :

  1. Longitudinal strain : If the deforming force produces a change in length alone, the strain produced in the body is called longitudinal strain.
  2. Volumetric strain : If the deforming force produces a change in volume only, the strain produced in the body is called volumetric strain.
  3. Shearing strain : If the deforming force produces a change in the shape of the body alone without changing its volume, the strain produced is called shearing strain.

Hooke’s Law : The deformation of an elastic body is directly proportional to the applied force within elastic limit.

Generalised Hooke’s Law : (Modified Hooke’s Law by Scientist Thomas young)
Within elastic limit, stress is proportional to strain.
∴ stress ∝ strain
or, \(\frac{t { stress }}{t { strain }}\) = Constant
This proportionality constant is known as coefficient of elasticity or modulus of elasticity of a body which is independent of magnitude of stress and strain but depends upon the nature of material of the body and the way in which the body is deformed.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter: Structure and Properties

Types of modulus of elasticity :

  • Young’s modulus of elasticity : It is the ratio of the longitudinal stress to the longitudinal strain within the elastic limit.
  • Bulk Modulus of Elasticity : Bulk modulus of elasticity is the volume stress (normal stress) to the volume strain within elastic limit.
  • Modulus of rigidity : Modulus of rigidity or shear modulus of a material is the ratio of the shearing stress to the shearing strain within the elastic limit.

Poisson’s Ratio: The ratio of the lateral strain to longitudinal strain of the material of a wire or bar under tension is called its Poisson’s ratio.

Elastic property of Material : A stressstrain curve for a wire of elastic material shown in fig. 3.3. From the figure it is found that OA portion of the curve is a straight line, i.e., stress is linearly related to strain. Hence, the material obey’s Hooke’s law. This is the region of perfect elasticity and ‘A’ represents the elastic limit of the material.

WBBSE Class 9 Physical Science Notes Chapter 3 Matter Structure and Properties 3

When the stress on the wire is increased beyond ‘A’ i.e., of elastic limit. Hooke’s law is no longer capable of explaining the elastic behaviour. On removal of stress the wire does not regain its original length completely and the wire is said to have acquired permanent set. The wire behaves like a plastic body.

If the process of stretching continued further, a stage is reached when the wire is found to undergo a relatively large strain with almost no increase of stress. This occurs at point ‘B’ called yield point. Beyond this point, metals are called ductile.

Further increase in stress ultimately results in the breaking of the wire and corresponding point ‘C’ on the curve is called breaking point. Breaking stress is the maximum stress that can be applied to a material before it ruptures. If a solid breaks soon after crossing the elastic limit, it is called brittle.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Comprehensive WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion can help students make connections between concepts.

Force and Motion Class 9 WBBSE Notes

Rest : If a body does not change its position with respect to time, the body is said to be at rest and that body is called static body.

Motion : If a body changes its position with respect to time, the body is said to be in motion and that body is called kinetic body.

Rest and motion is always a relative term : An object in one situation may be at rest, while in another situation the same object may appear to be in motion.

For example, a man sitting in a running train finds a man standing on the platform or in a field or a tree or a building all moving backwards, i.e. in motion. But actually, to the man on the field, the tree, the building etc. appear to be stationary, while the man in the train is moving forward.

Classification of motion :

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 1

Linear motion : If any particle moves along any line, then the motion of that particle is called linear motion.

Straight linear motion : If any particle moves along any straight line and any particle of the body travels the same distance at same speed, then the motion is called straight linear motion.

Curve linear motion : If any particle does not move in a definite direction but moves in curvilinear way and changes its direction in that way, then the motion is called curve linear motion.

Circular motion : If any particle moves around the particular point or axis, then this motion is called circular motion.
Example: The motion of electric fan.

Mixed motion : If any motion is a combined form of linear motion and circular motion, then that motion is called mixed motion.

Example: The moving wheel car. Here the circular motion of wheel produces linear motion.
Concept of a particle: A particle is a geometrical mass point or a material body of neglígible dimensions.

Reference body : For determining whether a body is at rest or in motion we must compare the motion of the body with some other body, called the reference body.

Reference frame : To locate the position of body with respect to the reference body, a system of co-ordinates fixed to the reference body is constructed. This is called reference frame.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Types of reference frames :

Inertial frame of reference : A frame of reference which is either at rest or moving with uniform velocity is called an inertial frame of reference.
In this frame of reference, Newton’s first law of motion is valid.

Example :

  • A frame of reference remaining fixed with respect to stars is an inertial frame of reference.
  • A space ship moving in space without spinning and with its engine shut off is an inertial frame of reference.
  • A train moving on a horizontal track with constant velocity is an inertial frame of reference.

Non-inertial frame of reference : A frame of reference which is accelerated is called non-inertial frame of reference.
In this frame of reference, Newton’s first law of motion is not valid.
Example : A ball is lying on the floor of a stationary bus. When the bus is accelerated or retarded, the ball starts moving. The bus is said to an example of non-inertial or accelerated frame of reference.

Motions in one, two or three dimensions :

One dimensional motion : If the motion of a particle be along a straight line, the motion is one dimensional.
Example :

  • A train moving along a straight track.
  • A body falling freely under gravity.

Two dimensional motion : If the motion of a particle be in a plane, i.e. when two of the three co-ordinates are required to specify the position of the particle in motion, the motion is called two dimensional.

Example :

  • The earth revolving round the sun.
  • An insect crawling on the floor.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Three dimensional motion : If the motion of a particle be in a space, i.e. when all the three co-ordinates are required to specify the position of the particle in motion, the motion is called three dimensional.

Example :

  • A flying bird
  • A flying aeroplane
  • A flying kite

Position of a particle moving along a straight line : The position of a particle moving along a straight line is described by a quantity x. The numerical value of x, together with the unit, gives the distance of the particle from the origin, and the sign of x denotes whether the particle is in the positive direction or in the negative direction.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 14

Fig. Positions of a moving particle on a straight line path
To describe the position of the particle at a given time, we have to specify :

  • its distance from the origin
  • whether it is in the positive direction or in the negative direction as seen from the origin.

Distance traversed : The length of the actual path traversed by a particle during motion in a given interval of time is called distance traversed by the particle.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 15

Suppose a man goes from P to Q, a distance of 4 m. Then he moves a distance QR of length 3 m at right angle to PQ.

Thus starting from P, he reaches finally at R travelling a total distance PQ + QR = (4 + 3) m = 7 m.
Unit (SI) of distance traversed : meter (m).
Distance traversed : Scalar quantity

Displacement : The displacement of a moving particle in a given interval of time is the shortest distance between the two positions of the moving particle in a particular direction and is given by the vector drawn from the initial position of the particle to its final position.

Units of displacement :

  • CGS : cm
  • SI : m

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Dimensions of displacement : [M0 L1 T0]

The characteristics of displacement are :

  1. Displacement has the unit of length.
  2. The displacement may be positive, negative or zero.
  3. The magnitude of the displacement of a moving particle between two points gives the shortest distance between them.
  4. The displacement of a moving particle between two points does not give us any idea about the path followed by the moving particle.
  5. The displacement of a moving particle between two points has a unique value.
  6. The actual distance travelled by the moving particle may be greater than or equal to the magnitude of the displacement.
  7. The displacement remains unaltered, even if the origin of the co-ordinate axis be shifted.
  8. As a particle cannot be at two different positions at the same time, so the displacement is a single valued function of time.
  9. The displacement changes if its direction be changed.

Diffference between displacement and distance

Displacement Distance
i. The distance moved by a particle in any particular direction is called the displacement of the particle. i. Total path covered by a particle in a straight or curved path is called distance traversed by the particle.
ii. It is a vector quantity. ii. It is a scalar quantity.

Speed : Rate of change of position of a body with respect to time is known as its speed.
It is measured by the distance covered by body in unit time.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 2

Characteristics of speed :

  1. Speed does not give any idea about the direction of motion.
  2. Speed can be positive, but never negative or zero.
  3. Speed is a scalar quantity.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Units of speed :

  • CGS: cmsM1
  • SI: ms-1

The dimensions of speed are : M0 L1 T-1]

Types of Speed :

Uniform speed : The speed of a body is said to be uniform, if it covers equal distance in equal intervals of time, however small these intervals may be.

Non-uniform speed or variable speed : If an object does not cover equal distances in equal time-intervals, its speed is called non-uniform or variable speed.

Average speed: When a body is moving with variable speed, the constant speed with which the body covers the same distance in a given time as it does while moving with variable speed during the given time, is called the average speed.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 3

If a body moves distances S1, S2, S3, …. etc. in times t1, t2, t3 …. etc. respectively, then

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 4

Instantaneous speed: The instantaneous speed at a point is defined as the limit of the average speed over a path length that tends to become zero but always includes that point.
Thus, if the moving particle covers a path △S in time interval △t, then

Whenever we mention about the speed of a particle, it always means the instantaneous speed, if not mentioned otherwise. In case of uniform motion of a particle, the instantaneous speed is equal to the uniform speed.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Typical speeds of some objects :

Average speed of a tortoise 05 km/h
Average walking speed of man 6 km/h
Maximum speed of a bee 16 km/h
Maximum speed of a cheetah 96 km/h
Maximum speed of a falcon 152 km/h
Maximum speed of a typical fast train 225 km/h
Wind speed in a light breeze 32 km/h
Wind speed in a hurricane 320 km/h
The speed of the earth moving round the sun 1.072 × 105 km/h

If not mentioned otherwise, the term velocity means instantaneous velocity.

Difference between speed and velocity :

Speed Velocity
i. Speed is the time rate of change of position of a body in any direction. i. Velocity of a body is the time rate of change of its displacement.
ii. Speed is a scalar quantity, i.e. it has only magnitude but no definite direction. ii. Velocity is a vector quantity, i.e. it has both magnitude and also a particular direction.
iii. Instantaneous speed is the magnitude of instantaneous velocity. iii. With a proper direction attached to the instantaneous speed, instantaneous velocity is obtained.
iv. A body moving with uniform speed may not have uniform velocity. iv. A body moving with uniform velocity always possesses uniform speed.
v. Speed can be positive but never negative or zero. v. Velocity can be positive, negative or zero.

Acceleration : The rate of increase of velocity of a body with respect to time is called its acceleration.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 6

Characteristics of velocity :

  1. When the velocity increases, the acceleration is called positive and when it decreases, it is called negative or retardation or deceleration.
  2. Acceleration has both magnitude and direction, so it is a vector quantity.

Units of acceleration :

  • CGS: cms2
  • SI: ms2

Dimensions of acceleration : [M0 L1 T2]

Types of acceleration :

Uniform acceleration : The acceleration of a moving body is said to be uniform if its velocity changes be equal amounts in equal intervals of time, however small the intervals may be.

Non-uniform or variable acceleration : The acceleration of a moving body is said to be variable, if it does not have equal changes in velocity in equal intervals of time, however small these intervals may be.

Average acceleration : When a body is moving with variable acceleration, then the average acceleration of the body is defined as the ratio of the total change in velocity of the body to the total time taken.

Instantaneous acceleration : The instantaneous acceleration of a particle at a point at a certain instant of time is the limit of the average acceleration over a displacement that tends to be zero, but always includes the given point.

Thus, if the moving particle undergoes a change of velocity of \(\overrightarrow{\Delta \mathrm{V}}\) in time interval Δt, then

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 7

If not mentioned otherwise, the term acceleration means instantaneous acceleration.
Retardation : The rate of decrease of velocity of a body with respect to time is called retardation.

Characteristics of retardation :

  1. Retardation is opposite to the acceleration and thus it is considered as negative acceleration.
  2. Retardation has both magnitude and direction, and hence it is vector quantity.

Units of retardation :

  • CGS : cms2
  • SI: ms2

Dimensions of retardation : [M0L1T2]

Types of retardation :

  1. Uniform retardation : The retardation of a moving body is said to be uniform if its velocity changes by equal amounts in equal intervals of time, however small the intervals may be.
  2. Non-uniform retardation: The retardation of a moving body is said to be variable, if it does not have equal changes in velocity in equal intervals of time.
  3. Instantaneous retardation : The instantaneous retardation of a particle at a point at a certain instant of time is the limit of average retardation over a displacement that tends to be zero but always includes the given point.

Graphical Representation of Motion : A graph is a very powerful method of representing information.

To describe the motion of an object, we can use line graphs. In this case, line graphs show dependence of one such as distance or velocity, on another quantity, such as time.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Plotting a graph : A graph is plotted to display the relation between two quantities. Generally, one of the two quantities changes independently and the other quantity depends on it.

Choosing the axes: Draw two perpendicular lines, crossing each other at a point. Each line represents one of the two quantities to be plotted. Generally, a horizontal line from left to right is drawn to represent the independent quantity, and a perpendicular line is drawn to represent the dependent quantity. These lines are called the x-axis and the y-axis, or the horizontal axis and the vertical axis respectively.

Choosing the scale : The size of the paper on which a graph is drawn is limited. On the available length of the axes, values are marked at equal distances. This is done in such a way that all the values of the quantity represented on the axis can be accommodated in the available length.

Plotting the points : Each set of values of the two quantities is represented by a point on the graph.

Joining the points: Once all the points corresponding to the available sets of values of the quantities are plotted, they are joined by a smooth curve to get the graph.

Velocity-time graph : Velocity-time graphs under different conditions are shown in the Fig. 1 to Fig. 4. In Fig. 1, velocity-time graph for a particle moving with uniform velocity is shown.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 8

As the velocity V is uniform, so the plot of the velocity against time would be a straight line parallel to the time axis. The distance moved in time t is equal to v t, which is seen from the graph as the area of the rectangle OABC. Thus the distance travelled by a particle in uniform motion is given by the area under the velocity-time graph.

In Fig. 2, velocity-time graph for a particle moving in straight line with uniform acceleration and starting with initial velocity u = 0 is shown. Same motion starting with initial velocity u is shown in Fig.3. In both the cases velocity-time curves are straight line. In each case the constant slope of the straight line gives the uniform acceleration.

In Fig. 4, velocity-time graph for a particle moving with uniform retardation in a straight line is shown. The curve is again a straight line but the slope is negative. In all the cases, the distance moved by the particle is given by the area under the velocity-time graph.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Position-time graph :

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 9

In Fig. 5 position-time graph of a particle moving with uniform velocity is shown. It is a straight line and the slope of the curve gives the uniform velocity of the particle.

In Fig. 6. position-time graph for a particle moving with uniform acceleration in a straight line is given. The graph is parabolic in nature. Here the slope of the curve gives the average velocity over certain interval of time.
Equations for motion with uniform acceleration: Consider a particle
moving with acceleration ‘a’ along a straight line, starting with initial velocity ‘u’ attains a velocity ‘V’ in interval of time ‘t’ and traverses a distance ‘S’ in this interval.

In Fig. 7, the velocity-time graph for the motion is shown. Here OD = u and AC = v and OA = t.
From the graph we have

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 10

AC = AB+BC
or, v = u + a t

Again the distance ‘S’ covered by the particle in the time interval ‘t’ is given by the area of the trapezium OACD.

or, s = area of the trapezium OACD
= area of the rectangle OABD + area of the △BCD
= OA.AB + \(\frac{1}{2}\) BD.DC
= ut + \(\frac{1}{2}\) t . at
∴ s = ut + \(\frac{1}{2}\) at2
Again, s = area of the trapezium OACD
= \(\frac{1}{2}\)( Sum of parallel sides ) × perpendicular distance between them
= \(\frac{1}{2}\)(OD + AC) × BD
= \(\frac{1}{2}\)(u + v) × t = \(\frac{1}{2}\)(u + v) a /a
or, 2 as = (u + v) at = (u + v)(u – v) [∵ at = v-u]
∴ 2 as = v2-u2
Or, v2 = u2+2 a s

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Newton’s first law of motion :

Every material body continues in its state of rest or of uniform motion in a straight line if and only if it is not compelled by any external force to change that state.

Newton’s first law of motion consists of three parts :

  • A body at rest continues to be at rest until some external force acts on it.
  • A body in uniform motion in a straight line continues to move uniform unless it is compelled to change that state by the application of some external force.
  • A body cannot change its direction of motion without the application of an external force. The body continues to move along the same straight line.

Newton’s first law of motion gives two things :

  • Concepts of inertia of a body
  • Definition of force

Definition of force : Force is that which changes or tends to change the state of rest or of uniform motion of a body along a straight line.

Inertia of the body : The inability of a body to change by itself its state of rest or of uniform motion along a straight line is called inertia of the body.

Types of inertia of a body :

  • Inertia of rest
  • Inertia of motion

Inertia of rest : Inertia of rest of a body is its inability to change its state of rest by itself.
Example : When a bus starts suddenly from rest, the passengers sitting inside tend to fall backwards. This is because the lower parts of their bodies start moving with the bus being in contact with it, but upper parts of their bodies try to remain at rest due to inertia of rest and fall back.

A coin placed on a piece of cardboard covering a hollow glass tumbler, falls into it when the cardboard is suddenly struck from side by fingers. The cardboard slips off but the coin tends to the remain in place due to inertia of rest and falls into the tumbler due to gravity when the support is removed.

Inertia of motion : Inertia of motion of a body is in inability to change by itself its state of uniform motion in a straight line.

Example :
An electric fan revolves a few rounds even after the switch is put off. It is due to inertia of motion.

When a running bus stops suddenly, passengers sitting inside tend to fall forward. This is because the lower parts of their bodies come to rest as soon as the bus stops, but the upper parts of their bodies try to remain in motion due to inertia of motion and fall forward.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Newton’s second law of Motion :

The rate of change of momentum of a body is proportional to the impressed force and takes place in the direction in which the force acts.

Newton’s second law of motion gives two things :

  • Concept of inertial mass [Inertial mass of a body is the force required to produce unit acceleration in the body.]
  • Magnitude and direction of the force.

Linear momentum : It is the property of a body possessed by virtue of its mass and velocity and its magnitude is given by the product of its mass and velocity.

p = m v (p = Linear momentum,
m = mass of the body,
v = velocity of the body)

Units of momentum :

  • CGS : g.cm s-1
  • SI: kg.ms-1

Momentum has both magnitude and direction and so it is a vector quantity.
Measurement of force : Suppose a body of mass m moves uniformly in a certain direction with the velocity u and after t time, due to application an of external force p in the same direction, its final velocity becomes v and acceleration of the body will be f.
Initial momentum of the body = mu.
After time t, final momentum of the body = mv.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 11

So, the change of momentum after time t = (mv-mu).
Rate of change of momentum

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 12

Now, according to Newton’s second law of motion, rate of change of momentum

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion 13

Now, according to Newton’s second law of motion, rate of change of momentum is directly proportional to the applied external force.

∴ p α m f
or, p = k m f

[k is a constant of proportionality]
If we now choose the unit of the force in such a way that unit force is that force which when acting on a body of a unit mass produces unit acceleration on it, then we have p = 1, m = 1, f = 1, i.e. k = 1

∴ p = m f or Force = mass × acceleration

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Dimensions of force :
F = [M’ L’ T2]
The force is a vector quantity.

Types of units of force :

  • Absolute units
  • Gravitational units

Absolute units : These units are called absolute as their values are independent of the place where the measurements are made.

CGS unit Dyne: It is the force which when acting on a body of mass one gram produces in it an acceleration of one centimetre per second every second.
(b) SI unit : Newton: It is a force which when acting on a body of mass one kilogram produces in it an acceleration of one metre per second every second.

Relation between dyne and Newton :

1 Newton = 1 kg × 1 ms2
= 1000 g × 100 cms2
= 105 g.cms2
∴ 1 Newton = 105 dyne

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Gravitational units of force : The unit of force expressed in terms of gravity is known as gravitational unit.

CGS system : gram-weight (g-wt) : The force with which the earth attracts a body of mass one gram towards its centre is called one gram-weight.

SI system : kilogram-weight (kg-wt) : The force with which the earth attracts a body of mass one kilogram towards its centre is called one kilogramweight.
1 g – wt = 981 dyne
1 kg – wt = 9.81 Newton

Impulse : The total effect of a force is called impulse and is measured by the product of the force and the time for which the force acts on the body.

Impulsive force : Impulsive force is a force of very large magnitude but of extremely short duration, which acting on a body produces a finite change of its momentum, displacement of the body during the action of the force being negligible.

The effect of the force is measured by its impulse.
Example of impulsive force : A player lowers his hands while catching a cricket ball. In doing so he increases the time of catch to reduce the momentum of the ball to zero. We know,
Impulse = force × time = change in linear momentum

So, to produce a particular change in linear momentum, if one takes longer time, then the force required would be less. Thus in catching a cricket ball, the player increases the time of catch by lowering his hands and thereby decreases the force and smaller reaction on the hands.
Newton’s third law of motion : To every action there is an equal and opposite reaction.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Newton’s third law of motion gives :

  1. From Newton’s third law we find that forces always occur in pairs, i.e. a single isolated force is not possible.
  2. Action and reaction [Action means the force exerted by a body on another body and reaction means the force exerted by the second body on the first] can never cancel each other, as they occur on different bodies.
  3. Each force produces its own effect.
  4. Newton’s third law is applicable to bodies both at rest or in motion.
  5. The action and reaction forces may appear due to actual contact of the two bodies or even from a distance.
  6. Newton’s third law is applicable to all types of forces e.g., gravitational, electric or magnetic etc.

Examples of action and reaction forces : If action and reaction act on the same system of particles, the action and reaction forces balance each other. So there is no displacement of the system of particle.

A chair cannot be raised sitting on it. Because in this case, action and reaction act on the same system.
Firing from a gun : When a gun is fired, the shot moves forward (action) and the gun recoils back (reaction).

Newton’s second law is the real law of motion : Both Newton’s first and third laws of motion are found to be followed from Newton’s second law of motion. So Newton’s second law can be said to be the real law of motion.

The principle of conservation of linear momentum : In an isolated system of bodies, i.e. when no external force acts on the system, the vector sum of the linear momenta of all the bodies of the system is conserved and is not affected due to their mutual action and reaction.

Characteristics of the principle of conservation of linear momentum :

  1. The principle of conservation of linear momentum is universal, i.e. it is applicable to both the microscopic and macroscopic system.
  2. The principle of conservation of linear momentum is independent of the frame of reference.
  3. The principle of conservation of linear momentum is more fundamental than Newton’s laws of motion from which it is derived.

WBBSE Class 9 Physical Science Notes Chapter 2 Force and Motion

Application of the principle of conservation of linear momentum :

A man jumps to the shore from a boat: When a man jumps to the shore from a boat, the boat is pushed away from the shore. This is because the momentum of the boat is equal and opposite to that of the man in accordance with the principle of conservation of linear momentum.

When a gas is released from a gas-filled balloon, it moves in direction opposite to that of the gas. Here also following the principle of conservation of momentum, the momentum with which the gas is coming out would impart an equal and opposite momentum to the balloon and hence the balloon would move in the opposite direction.

Motion of a rocket is based on the principle of conservation of momentum. When the fuel is burnt inside a rocket the gas produced comes out through a nozzle at the bottom of the rocket at high speed. To equalise this momentum of the issuing gas, the rocket moves upward.

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy

Comprehensive WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy can help students make connections between concepts.

Work, Power and Energy Class 9 WBBSE Notes

The temperature of the deeper layers of the water in the pond remains nearly at 4°C and falls gradually to 0°C upwards till the layer of ice is reached. The marine life in water is thus saved.

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy 1

Work = Force x displacement of the point of application in the direction of the force

WF x d (W Work done;
F Force acting on body
d = displacement of the body from the point of application of the force)

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy

Sometimes the direction of force is placed as an angle to the horizontal vector will be taken suppose F is acting as θ angle with horizontal plane. So, the horizontal vector Fcosθ.
∴ work done,
W = Fcosθ.d = Fdcosθ ………………………… (i)

Note :
(i) If θ = 0°, cosθ = 1, then W = Fd
(ii) If θ = 90°, cosθ 1, then W = 0
(iii) If θ = 180°, cosθ = -1, then I W = – Fd

Work done by the force : If the displacement of the point of application of the force takes place in the direction of the force then the force is said to have done the work.

Example : If a body is dropped vertically, the gravitational force attracts it downward. The displacement of the body takes place in the direction of gravitational force. Here, work is done by the gravitational force.

Work done against the force: When the displacement of the point of application of the force takes place in the direction opposite to the applied force, then work is said to be done against the force.
Example: If a body is lifted vertically upward, displacement of the body is in the direction opposite to the gravitational force and work is done against the gravitational force.

No work force : When a force acts perpendicularly to the direction of displacement of the particle, then the force does not do any work and is known as no work force.

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy

Example : If a man moves along a horizontal place road with a suitcase in his hand, force of gravity does no work on the suitcase. Because in this case weight of suitcase and its displacement are at right angles to each other.

Unit of work:
Absolute unit :
(i) CGS : erg
1 erg 1 dyne x 1cm

(ii) SI : Joule
1 Joule = 1 Newton x 1 meter
= 105 dyne × 10
2 cm
= 107 dyne × cm
∴ Joule = 107erg

Gavitational unit of work :
(i) CGS : g-cm
1g-cm = 1 g-wt × 1 cm
= 981 dyne × 1 cm
1 g-cm = 981 erg

(ii) SI : kg-m
1kg-m = 1 kg-wt x 1m
= 9.8 Newton x 1m
1 kg-m = 9.8 Joule

Practical unit of work : Joule
1 Joule 10 erg

Power : Rate of work done by an agent is called its power, i.e. work done in unit time is known as power. If any work W done by applying ‘p’ force in ‘p’ time then
WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy 3
N.B : Some work done is less time by an agent means it has greater power. Power does not depend on the total amount of work done but on the time required to do the work. Power is a scalar quantity.

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy

Unit of power:

Absolute unit:

  • CGS : erg s-1
  • SI: Watt or Js-1

Gravitational unit:

  • CGS : g-cms-1
  • SI: kg-ms-1

Practical unit : The practical unit of power in both CGS and SI units is Watt.

Watt : The power of doing 1 erg of work in 1 second is known as 1 watt.

1 Watt 1 Joule second-1  = 10-7erg s-1
1 Kilowatt 10 watt
1 Megawatt 10watt
Horsepower: In FPS system, practical unit of power.

Definition of Horse Power (hp) : The power to work at the rate of 550ft-lb per second is called 1 horse power 1 horse power = 746 Watt

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy

Energy : The capacity of an object to do work is called the energy of the object.

N.B. Since by energy we mean some amount of work, so we express energy in the unit of work.

Energy is a scalar quantity.

Types of energy :
(1) Potential energy : The energy of an object because of its position or shape is called its potential energy. Examples of potential energy :

  • When a weight is raised through a certain height, work is done against gravity, which is retained in the body in this new position as potential energy and is now capable of doing some work if allowed to fall down to its former position.
  • When a spring wound, stretched or compressed it gains potential energy, as it is now capable to do work while it comes back to its normal shape.

(2) Kinetic energy : It is the energy acquired by a body by virtue of its motion.
Examples of Kinetic energy :

  • Kinetic energy of a bullet fired from a gun helps it to pierce a target.
  • Kinetic energy of blowing air and running water are used to run a windmill and a water mill respectively.

Kinetic energy of a hammer is used in driving a nail into a block of wood.
Calculation of kinetic energy: If a body of mass m be moving with velocity v, then it can be shown that,
i.e. Kinetic energy = 1/2 x mass x (velocity)2
∴ Kinetic energy = 1/2 mv2

Mechanical energy: The sum of the kinetic energy and the potential energy of an object is called its mechanical energy.

Law of conservation of energy: Energy cannot be created or destroyed. Only one form of energy is transformed to other type of energy. The total amount of energy is constant in the universe.

Examples of conservation of energy :

(i) Snow deposited at high altitudes melts and the water so formed flows down to the seas. In the process, the potential energy of the water is converted into kinetic energy. We convert this kinetic energy of water to electrical energy in hydroelectric power plants.

(ii) The food we eat comes from plants and animals that eat plants or plant products. When we eat food, several chemical processes take place, and energy in the form needed by the body is produced from the energy contained in the food. When we walk, run, exercise, this energy is used to provide kinetic energy.

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy

m Conservation of energy in a freely falling body : Suppose a ball of mass ‘m’ is kept at rest at a point A, which is at a height ‘h’ above the ground. We take the potential energy at the ground to be zero. The potential energy of the ball at A’ is then,
UA = mgh

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy 4

The ball is at rest at A, so its kinetic energy here is
KA= o
The total energy is, UA + KA = mgh (1)

The ball is now allowed to fall. The only force on it during the fall is gravity (free fall), and hence, its acceleration is a = g downwards. Consider any point B in its path.

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy

Suppose its distance from A is x. The height of the point B from the ground is (h – x). When the ball reaches point B, its potential energy is,
UA = mg (h- x).
The velocity of the ball at point B is given by
v2 = u2 + 2gx = 2gx.
The kinetic energy of the ball at point ‘B’ is :
KB – \(\frac{1}{2}\) mv2 = \(\frac{1}{2}\) m(2gx) = mgx
The total energy at point B is
UB + KB = mg(h-x) + mgx = mgh …………..(2)
Comparing (1) and (2), we see that
UA + KA = UB + KB
The point B was chosen to be any point in the path of the ball. The total energy here turned out to be equal to the initial energy. Thus, the energy at all points during the fall is the same, that is, remains conserved as the ball falls.

WBBSE Class 9 Physical Science Notes Chapter 5 Work, Power and Energy

The potential energy is gradually converted into kinetic energy.
This result is also true if a ball is thrown upwards or at some angle with the vertical. During the upward movement of a ball, its kinetic energy gets gradually converted to potential energy.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Comprehensive WBBSE Class 9 Physical Science Notes Chapter 6 Heat can help students make connections between concepts.

Heat Class 9 WBBSE Notes

Heat : It is an external agency whose absorption turns a body hot and extraction turns a body cold.

Sensible heat : It is the type of heat that is perceptible by senses. This causes rise of temperature of a body that can be measured with a thermometer.

Latent heat : It is the quantity of heat absorbed or released when a substance of unit mass undergoes a physical change of state at a constant temperature and pressure.

Radiant heat : It transmits from a source of heat in all directions with the speed of light and in the form of waves.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Temperature : The temperature of a body is the thermal condition which determines whether it will absorb heat from other body or release heat to that body when they are kept in contact.

Lower fixed point of a thermometer : At normal atmospheric pressure, the temperature at which ice melts into water or pure water freezes into ice is called the lower fixed point of a thermometer.

Upper fixed point of a thermometer : It is the temperature at which pure water boils and transforms into steam under normal atmospheric pressure.

Fundamental interval : The range of temperature between the upper and lower fixed points is called fundamental interval.

Celsius scale of temperature : In this scale, the lower fixed point is 0°C and the upper fixed point is 100°C. The fundamental difference is divided into hundred equal parts and each part represents 1°C.

Fahrenheit scale of temperature : In this scale, the lower fixed point is 32°F and the upper fixed point is 212°F. The fundamental difference is divided into 180 equal parts and each part represents 1°F.

Kelvin scale of temperature : In this scale the lower fixed point is 273K and the upper fixed point is 373K, the fundamental interval is divided into 100 equal divisions like that in the Celsius scale. Each division represents 1K.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Calorie : It is the amount of heat required to raise the temperature of one gram of pure water through 1°C.

Specific heat : It is defined as the quantity of heat required to raise the temperature of unit mass of it by 1 degree.

Specific heat capacity of a substance : Specific heat capacity of a substance is the quantity of heat required to raise the temperature of 1kg of the substance through 1K.

Units of specific heat :

  • CGS : calg-10 C-1
  • SI : Jkg-1K-1

Thermal capacity or heat capacity of a body : It is defined as the quantity of heat needed to raise the temperature of the body for unit degree rise of temperature.

Units of heat capacity :

  • CGS : calg-10 C-1
  • SI : Jk-1

Water equivalent : The water equivalent of a body is the mass of water which will be heated through one degree by the amount of heat required to raise the temperature of the body through one degree.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Units of water equivalent:

  • CGS: gram (g)
  • SI: kilogram (kg)

Fundamental principle of calorimetry :

Conditions :

  • During the process of heat transfer, there is no heat exchange with the surrounding.
  • No chemical reaction takes place between the bodies.

Principle :
Heat lost by the hotter body = Heat gained by the colder body.

First law of thermodynamics : Whenever mechanical work is converted into heat energy, the heat so obtained is directly proportional to the work done.
If, ‘W’ be the mechanical work done and ‘H’ be the corresponding heat energy produced, then from first law of thermodynamics,
W α H or W = JH
where J is the constant of proportionality and is known as mechanical equivalent of heat.

Mechanical equivalent of heat : The mechanical equivalent of heat is numerically equal to the work done to produce unit quantity of heat.

Unit of mechanical equivalent of heat :

  • CGS: 4.8 x 107 erg cal-1
  • SI : In SI, both heat and work are measured in Joule, so mechanical equivalent of heat has no unit and its value is 1.

Change of state : There are three states of matter – solid, liquid and gas. Substances can be changed from one state to other by application or removal of heat from them. This process is known as change of state.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Mathematical expression of Latent heat : Q = mL
where Q is the amount of heat supplied (or extracted) to change the state of ‘m’ mass of substance and L is latent heat

Unit of Latent heat :

  • CGS : cal g-1
  • SI : J kg-1

Types of Latent heat:

  • Latent heat of fusion
  • Latent heat of solidification
  • Latent heat of vaporisation
  • Latent heat of condensation

Latent heat of fusion of ice and latent heat of solidification of water are both equal:

  • CGS : 80 cal g-1
  • SI: 336 × 105 Kg-1

Latent heat of vaporisation of water and latent heat of condensation of steam are both equal:

  • CGS: 537cal g-1
  • SI : 226 × 106 Jkg-1

Melting point : The melting point of a solid is the definite temperature at which it melts on heating under normal atmospheric pressure and the temperature remains constant until the melting is
completed.

Boiling point : The boiling point of a liquid is the temperature at which a liquid boils under normal atmospheric pressure. The boiling point of a liquid depends upon:

  • The nature of the liquid
  • Presence of dissolved substance
  • The superincumbent pressure

pressure: Whenever a liquid evaporates at any temperature, the vapour exerts a definite pressure on everything in contact. This pressure is called vapour pressure of the liquid at that temperature.

Saturated vapour : If a liquid is allowed to evaporate in a closed space, after some time it is found that evaporation stops i.e. at a given temperature. there is a maximum limit to the amount of
vapour the space can hold. The space is said to be then saturated with vapour and the vapour is then called saturated vapour.

Unsaturated vapour : If at a given temperature, a space contains a smaller amount of vapour than the maximum possible amount that the space can hold, the space is then said to be unsaturated and the vapour is called unsaturated vapour.

Saturated vapour does not obey Charles’ law or Boyle’s law, while unsaturated vapour obeys charles’ and Boyle’s law.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Critical temperature : For every substance in gaseous state there is a certain temperature such that if the substance is below this temperature, it can be liquefied by the application of suitable pressure,
and above this temperature, it cannot be liquefied, however large the pressure may be, then the temperature is called the critical temperature for that substance.

Gas : A substance at a temperature above critical temperature is called gas.

Vapour : When the substance is at a temperature below its critical temperature it is called vapour.

Anomalous expansion of water: Usually liquids expand on heating. But in the case of water we find deviation from this general behaviour of the liquids within a certain range of temperature. It is found that if certain mass of water at an temperature about 10°C be taken and allowed to cool, its volume is observed to fall gradually till it reaches the temperature of 4°C, when with further cooling the volume of water will increase instead of decreasing.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat 1
The behaviour of water is some how different from other liquids. The volume of water is minimum at 4°C and its density is maximum at 4°C. This phenomenon is called the anomalous expansion of water. This phenomenon is called the Ienipentturi anomalous expansion of water.

Effect on marine life : Anomalous expansion of water has great practical importance on the marine life in the cold countries. In these countries when air above the surface of water of pond falls below 0°C, the water on the surface on cooling, become denser than that below and gradually sinks downwards.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat 2

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

This continues until the temperature of water falls to 4°C. As the surface temperature falls further it becomes less dense than the water below, which is at 4°C and is the densest. It thus remains at the top, cooling C more and more and ultimately a layer of ice is formed there and remains there as ice is lighter than water. The layer of ice forms a poor conducting layer on the top, which prevents the heat from water below to pass into colder atmosphere above.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat 3

The temperature of the deeper layers of the water in the pond remains nearly at 4°C and falls gradually to 0°C upwards till the layer of ice is reached. The marine life in water is thus saved.

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

Well structured WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound can serve as a valuable review tool before exams.

Sound Class 9 WBBSE MCQ Questions

Multiple Choice Questions :

Question 1.
What should be the minimum frequency of a vibrating body for producing sound?
(i) 2000 Hz
(ii) 100 Hz
(iii) 20 Hz
(iv) 150 Hz
Answer:
20 Hz

Question 2.
How does pitch of sound depend on frequency ?
(i) pitch increases with decrease of frequency
(ii) pitch decreases with increase of frequency
(iii) pitch increases or decreases accordingly with increase or decrease of frequency
(iv) none of these.
Answer:
Pitch increases or decreases accordingly with increase or decrease of frequency.

Question 3.
The unit of frequency is :
(i) ohm
(ii) mho
(iii) Hz
(iv) torr
Answer:
Hz

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

Question 4.
Migraine is caused by –
(i) sound pollution
(ii) air pollution
(iii) water pollution
(iv) soil pollution.
Answer:
sound pollution

Question 5.
When one end of a long iron pipe is struck –
(i) two
(ii) three
(iii) four sounds are heard.
Answer:
Two

Question 6.
The relation between velocity (v) of a sound with its density (d) is :
(i) v ∝ \(\frac{1}{d}\)
(ii) v ∝ \(\frac{1}{\sqrt{d}}\)
(iii) v = \(\frac{1}{d}\)
(iv) v = \(\frac{1}{\sqrt{d}}\)
Answer:
v ∝ \(\frac{1}{\sqrt{d}}\)

Question 7.
The relation between velocity (v) of a sound with the absolute temperature (T) is –
(i) v ∝ √T
(ii) v ∝ T
(iii) v = T
(iv) v = √T
Answer:
v ∝ √T

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

Question 8.
If the size of the sounding body be bigger then intensity of sound will be :
(i) increased
(ii) decreased
(iii) remains same
Answer:
increased

Question 9.
The velocity of sound through iron is –
(i) 4 times
(ii) 10 times
(iii) 15 times
(iv) 16 times than the velocity of sound through air.
Answer:
15 times

Question 10.
The sensation due to short sound remains for about –
(i) \(\frac{1}{20}\) second
(ii) \(\frac{1}{10}\) second
(iii) \(\frac{1}{30}\)
(iv) \(\frac{1}{5}\) second and is known as persistence of hearing.
Answer:
\(\frac{1}{10}\) second.

Question 11.
Relation among velocity (v), wavelength (λ) and frequency (v) is :
(i) v = v λ
(ii) v = v λ
(iii) λ = v\v
(iv) v = vλ
Answer:
v = v\λ

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

Question 12.
If the velocity of a wave havingwave length 1.7 m is 340 m / s then what will be the frequency of the wave?-
(i) 20 Hz
(ii) 200 Hz
(iii) 2000 Hz
(iv) 100 Hz
Answer:
200 Hz

Question 13.
If the tones present in a note are of frequencies 100,150,200,280, 300,450 and 500 , what is the fundamental tone ?-
(i) 100
(ii) 280
(iii) 500
(iv) 450
Answer:
100

Question 14.
The sound of lightning is heard after 6 sec. it is seen. If the velocity of sound is 330 m/s at that time what will be the distance of the cloud?
(i) 1980 m
(ii) 2000 m
(iii) 330 m
(iv) 350 m
Answer:
1980 m

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

Question 15.
Wave motion transfers –
(i) matter
(ii) energy
(iii) momentum
(iv) all of these
Answer:
energy

Question 16.
Distance between two successive compressions is –
(i) \(\frac{\lambda}{2}\)
(ii) \(\frac{\lambda}{4}\)
(iii) λ
(iv) 2 λ
Answer:
λ

Question 17.
Sound wave can travel –
(i) only in solid
(ii) only in liquid
(iii) only in gas
(iv) in all of these
Answer:
in all of these

Question 18.
The speed of sound in air-
(i) decreases with the increase in temperature
(ii) remains the same with the increase in temperature
(iii) remains the same with the decrease in temperature
(iv) increases with the increase in temperature.
Answer:
decreases with the increase in temperature.

Question 19.
Sound wave is –
(i) transverse in nature
(ii) longitudinal in nature
(iii) both transverse and longitudinal in nature
(iv) none of these
Answer:
longitudinal in nature

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

Question 20.
Distance between a compression and a rarefaction is
(i) \(\frac{\lambda}{4}\)
(ii) \(\frac{\lambda}{2}\)
(iii) λ
(iv) 2λ
Answer:
\(\frac{\lambda}{2}\)

Question 21.
Which of the following waves is a mechanical wave ?
(i) light wave
(ii) Sound wave
(iii) X-rays
(iv) Ultra-violet ray
Answer:
Sound wave

Question 22.
The speed of sound is maximum in –
(i) air
(ii) hydrogen
(iii) water
(iv) iron
Answer:
iron

Question 23.
In SONAR, we use –
(i) infrasonic
(ii) radio waves
(iii) audible sound
(iv) ultrasonic
Answer:
ultrasonic

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

Question 24.
Infra sound can be heard by –
(i) bat
(ii) rhinoceros
(iii) dolphin
(iv) human being
Answer:
rhinoceros.

Fill in the blanks :

1. Sound moves ______ through solid than through liquid.
Answer:
faster

2. Sound of frequency above is known as ultrasonic ______ sound.
Answer:
20,000 Hz

3. The sound of frequency below 20 Hz is called ______ sound.
Answer:
subsonic

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

4. Sound originates from a mechanically ______ body.
Answer:
vibrating

5. We can hear sound only if the frequency of a vibrating body lies within the limits of ______ to 20,000 Hz.
Answer:
20

6. The maximum displacement of a particle in a medium on either side of its mean position is known as ______.
Answer:
amplitude.

7. Time period is the time required by a particle to make ______ complete oscillation.
Answer:
one

8. The number of complete oscillations of a body per second is known as its ______ .
Answer:
frequency

9. To perceive the initial and the reflected sounds clearly, the minimum time interval between these should be at least ______ second.
Answer:
\(\frac{1}{10}\)

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

10. Pitch is the characteristic of a musical sound that distinguishes a sharp sound from a ______ sound.
Answer:
flat

11. Pitch increases corresponding to ______ in wavelength.
Answer:
decrease

12. Pitch of sound produced by a vibrating air column increases with the ______ of the air column.
Answer:
length

13. Note is analogous to white or polychromatic light and ______ analogous to light of a single colour.
Answer:
tone

14. In a note, the sound of least frequency is called the ______ tone.
Answer:
fundamental

15. A ______ continuous noise of level may cause tension, headache, migraine, neurological diseases.
Answer:
high

16. A ______ is necessary for the propagation of sound.
Answer:
medium

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

17. If the frequency of sound is high, pitch of the sound will also be ______.
Answer:
high

18. The particular harmonic whose frequency is double the fundamental frequency is called ______ of the fundamental.
Answer:
octave

19. Sound is a form of ______ .
Answer:
energy

20. A ______ medium is required for the propagation of sound.
Answer:
material

21. The velocity of sound in a gas is proportional to the square root of the _______ of the gas.
Answer:
density

22. The source of sound in laboratory is _______.
Answer:
sonometer

23. _______ is the safe intensity of sound according to WHO.
Answer:
45 decibel (or 45 dB)

24. Sound originates from a body that ______
Answer:
vibrates

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

25. The maximum displacement of particle in vibratory motion is called ______.
Answer:
amplitude

26. Light moves as _____ wave
Answer:
transverse

27. Sound travels through gases as ______ waves
Answer:
longitudinal

28. A wave transmits ______ through the medium
Answer:
energy

29. the unit of frequency is ______.
Answer:
Hertz.

WBBSE Class 9 Physical Science MCQ Questions Chapter 7 Sound

30. For a echo to be heard the minimum distance between the source and the reflector should be ______ m
Answer:
16.6

31. ultrasonic waves have frequency greater than ______ Hz.
Answer:
20,000

32. Loudness, pitch and quality are the _____ of musical sound
Answer:
characteristics.

WBBSE Class 9 Physical Science Notes Chapter 7 Sound

Comprehensive WBBSE Class 9 Physical Science Notes Chapter 7 Sound can help students make connections between concepts.

Sound Class 9 WBBSE Notes

Wave motion and wave : When a disturbance produced in one portion of space travels to another portion, without involving the transfer of any material with it, the motion of the disturbance is called wave motion.

The disturbance itself is called a wave.

  • Longitudinal wave : If the particles of a medium move along the direction of the motion of the wave, the wave is called a longitudinal wave.
  • Transverse wave : If the particles of a medium move perpendicular to the direction of the motion of the wave, the wave is called a transverse wave.

WBBSE Class 9 Physical Science Notes Chapter 7 Sound

Characteristics of sound :

  • Sound is produced by a vibrating source.
  • Sound travels as a longitudinal wave through a material medium.
  • A vibrating source produces compression and rarefaction pulses, one after the other in the medium. These pulses travel one behind the other as a sound wave.
  • In sound propagation, it is the energy of the sound that travels and not the particles of the medium.
  • Sound cannot travel in vacuum.

Oscillation : The change in density from one maximum value to the minimum value and again to the maximum value makes one complete oscillation.

Periodic motion : Any motion which repeats itself after a fixed interval of time is called periodic motion.

All oscillatory motions are periodic motions but all periodic motions are not oscillatory motions.

Examples of oscillatory motion :

  • Motion of a pendulum
  • To and fro motion of a loaded spring.
  • Vibrations of the wire of a stringed musical instrument.
  • Motion of liquid contained in U-tube when it is compressed once in one limb and left to itself etc.

Compression pulses correspond to regions of high density and pressure, whereas rarefaction pulses correspond to regions of low density and pressure in the medium.

Some definitions :
(a) Displacement : The displacement of a particle in oscillatory motion at any instant is the distance of the particle from its equilibrium position at that instant.

(b) Amplitude : The maximum displacement of a particle in oscillatory motion an either side of its equilibrium position is called amplitude of its motion.

(c) Time period : The time required for a complete oscillation by a particle in oscillatory motion is called its time period. It is usually denoted by T and is measured in second (s).

(d) Wavelength : The distance between two consecutive compressions or two consecutive rarefactions is called the wavelength (A,). It is measured either in centimetre (cm) or in metre (m).

(e) Frequency : The number of oscillations of the density of the medium at a place per unit time is called the frequency (v).
\(v=\frac{1}{T}\) (V = frequency, T = time period)

WBBSE Class 9 Physical Science Notes Chapter 7 Sound

Frequency is measured in per second (s-1), which is also known as hertz (Hz).
Relation among wavelength, frequency and velocity of a wave : In a medium, the distance travelled by the wave in unit time is called the velocity of the wave in the medium.

The number of complete vibrations made per unit time by a particle in the path of wave is called the frequency of the wave.
Velocity of sound = Frequency × wave length
i..e v = vγ

Reflection of sound : When sound waves hit on the boundary separating two homogeneous media, a portion of sound changes its direction from the surface of separation and returns to the first medium. This is known as reflection of sound.

Laws of reflection of sound :

  • The incident sound, reflected sound and the normal on the surface of separation through the point of incidence remain on the same plane.
  • The angle of incidence is equal to the angle of reflection.

Mach number : The ratio of speed of a body to the speed of sound is called mach number of the body. If the mach number of a body is greater than 1, then the velocity of the body is supersonic velocity.

Echo : When a sound after reflection from some reflector is again heard separated from the original sound, then this reflected sound is known as an echo.

Condition for echo : For occurrence of an echo the minimum distance between the source of sound and a reflector of sound producing echo should be around 16.6m.

Reverberation : The persistence of sound due to repeated reflection and its gradual fading away is called reverberation of sound.

Subsonic sound : We cannot hear the sound generated from the sources having frequency less than 20Hz. These sounds are known as subsonic sound or infrasonic sound.

Ultrasonic sound : We can not hear sound generated from the sources having frequency 20,000 Hz (20 kHz) or more. These sounds are known as ultrasonic sound.

Audible sound : We can hear only the sounds of the sources producing about 20 Hz to 20,000 Hz in frequencies. The sounds are known as audible sound.

Applications of ultrasound :

  • Ultrasound is generally used to clean parts located in hard-to- reach places, for example spiral tube, odd shaped parts, electronic components etc.
  • Ultrasound is used for drilling holes or making cuts of desired shape.
  • Ultrasounds can be used to detect cracks and flaws in megtal blocks.
  • Ultrasonic waves have given doctors powerful and safe tools for imaging human organs.

(a) Echocardiography is a technique in which ultrasonic waves, reflected from various part of the heart, form an image of the heart.

(b) Ultrasonography is routinely used to show image of patient’s organs such as the liver, gall bladder, uterus etc. to doctors.

(c) A technique called phacoemulsification (phaco stands for the lens of the eye) uses the power of ultrasound to break the hardened gel into tiny pieces.

(d) Ultrasound is also employed to break small ‘stones’ that form in the kidneys into fine grains.

  • Sonar : The acronym SONAR stands for SOund Navigation AND Ranging. This is a method for detecting and finding the distance of objects under water by means of reflected ultrasonic waves.
  • Note : A sound with a number of frequencies is called a note.
  • Tone : A sound with a single frequency is called a tone.

WBBSE Class 9 Physical Science Notes Chapter 7 Sound

Characteristics of Musical Sound :

  • Loudness or intensity : Loudness is the measure of how intense or loud a sound is, loudness of a sound is measured by energy contained per unit volume of the medium through which sound passes.
  • Pitch : The pitch is that physical cause by which we can distinguish a shrill sound from a flat one of same intensity. It depends on the frequency of sound.
  • Quality or timbre : Quality is that characteristic of a musical sound by which we can distinguish a note sounded on a musical instrument from a note of same pitch and intensity produced.

Structure of human ear : The structure of human ear can be divided into three parts :

(i) Outer ear : The outer ear consists of pina and ear canal. Pina is the part which is visible from the outside. The function of the outer ear is to guide sound waves to the middle ear.

(ii) Middle ear : The middle ear is separated from the ear canal by a tightly stretched membrane called eardrum. Beyond it, three interconnected bones called the hammer, anvil and stirrup are situated.
The function of the middle ear is to pick up, amplify and transmit sound waves to inner ear.

(iii) Inner ear : The inner ear consists of a liquid-filled coiled tube called cochlea, which is shaped like a snail. This tube is connected to the stirrup. The cochlea has special sensory cells called hair cells, which are so called because of the hairlike structures that stick out of them. The hair cells are connected to the auditory nerve, which is connected to the brain.

Sound pollution : Any unpleasant and unwanted sound is noise. Sound pollution means creation of discomfort, disturbance and irritation which result to ill effects to mental and physical health.

Sources of sound pollution :

  • The sound vehicle causes sound pollution.
  • The machinery used in industry creates sound of intensity more than tolerance limit and thereby causes sound pollution.
  • Loud-speakers, amplifiers, air horn played at top volume on the roads, are major sources of sound pollution.
  • The sound crackers cracked in different functions, particularly marriages and Puja Festivals causes sound pollution.

Harmful effects of sound pollution :

  • Sound pollution may cause hindrance in the growth of nervous system of a child in the womb, That is yet to be born.
  • One may become deaf due to sound pollution.
  • The sound of high intensity and pitch may cause increase of blood pressure, nervous breakdown, heart disease, loss of memory etc.

WBBSE Class 9 Physical Science Notes Chapter 7 Sound

Remedial measures of sound pollution :

  • Factories, smaller or big, should be prohibited in residential areas.
  • Sound resistant measures may be taken in industry.
  • Unnecessary use of horn in the vicinity of hospital, school etc. should be restricted.
  • Airports should be away from residential area.
  • Sound pollution awareness programme should be taken up at regular interval.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Detailed explanations in West Bengal Board Class 9 Physical Science Book Solutions Chapter 4.2 Mole Concept offer valuable context and analysis.

WBBSE Class 9 Physical Science Chapter 4.2 Question Answer – Mole Concept

Very short answer type questions

Question 1.
What is mole of a substance ?
Answer:
Molecular weight of a substance expressed in gram is called mole of the substance.

Question 2.
What is molar volume of a substance ?
Answer:
Molar volume of a substance is the volume of gram molecular weight of any substance in gaseous state.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 3.
What is vapour density of a gas ?
Answer:
Vapour density is the ratio of weight of some volume of a gas or vapour to the weight of same volume of hydrogen at same temperature and pressure.

Question 4.
What is Avogadro’s number ?
Answer:
Avogadro’s number is the number of molecules present in 1 gram molecule of a substance.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 5.
What is the value of Avogadro’s number ?
Answer:
The value of Avogadro’s number is 6.023 × 1023

Question 6.
How many molecules are present in 44g CO2?
Answer:
6.023 x 1023

Question 7.
What is the gram molecular weight of nitrogen ?
Answer:
28g

Question 8.
What is the atomicity of helium ?
Answer:
Atomicity of helium is 1.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 9.
What will be the volume of 22g of CO2 at NTP ?
Answer:
\(\frac{22 \cdot 4}{2}\) lit = 11.2 lit

Question 10.
Who postulated the atomic theory ?
Answer:
John Dalton postulated the atomic theory.

Question 11.
What is the molar volume of a gas at NTP ?
Answer:
22.4 lit

Question 12.
Give the name of a gas which has same atom and molecule ?
Answer:
Argon (Ar)

Question 13.
Who gave the first concept of molecule ?
Answer:
idea about the molecules was put forward first by an Italian physicist, Amedeo Avogadro (1811).

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 14.
Which element has atomicity 4 ?
Answer:
Phosphorus

Question 15.
What is the volume of 1 mole CO2 at NTP ?
Answer:
22.4 lit

Question 16.
What is the number of 1 mole electron ?
Answer:
6.023 x 1023

Question 17.
What is the mass of 1 gram-mole of water.
Answer:
18g

Question 18.
If a gas has vapour density 14, what will be its molecular weight ?
Answer:
Molecular weight of the gas = 2 × 14 = 28

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 19.
How many number of molecules are present in 18g water ?
Answer:
18g water contains 6.023 × 1023 molecules of water.

Question 20.
What is the atomicity of Argon gas ?
Answer:
The atomicity of Argon is 1.

Question 21.
What will be the volume of gas of 64g SO2 at NTP ?
Answer:
22.4 lit

Question 22.
Calculate the number of molecules in 4.5 g water.
Answer:
The number of molecules present in 4.5 g water
\(=\frac{6.023 \times 10^{23} \times 4.5}{18}=1.505 \times 10^{23}\)

Question 23.
What will be the weight of 44.8 lit CO2 at NTP ?
Answer:
The weight of 44.8 lit CO2 at NTP

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 24.
What are the number of electrons in 2 moles electron ?
Answer:
The number of electrons present in 2 moles electron
= 2 x 6.023 x 1023 = 12.046 x 1023

Question 25.
Which one is heavier : 2 gram-molecule ammonia and 2 gram molecule carbon dioxide ?
Answer:
2 gram-molecule ammonia = (2 x 17) g = 34g NH3
2 gram-molecule carbon dioxide = (2 x 44)g = 88g CO2
So, 2 gram-molecule carbon dioxide is heavier than 2 gram-molecule ammonia.

Question 26.
If the vapour density of any gas is 35.5, what will be its molecular weight ?
Answer:
The molecular weight of the gas will be = 2 x 35.5 = 71

Question 27.
One atomic mass unit (amu) is how much grams ?
Answer:
1.6603 x 10-24g

Question 28.
What is the volume of 7g nitrogen at STP ?
Answer:
5.6L

Short Answer Type Question

Question 1.
State Avogadro’s hypothesis.
Answer:
Avogadro’s hypothesis : Under the same conditions of temperature and pressure, equal volume of all gases (both elementary and compound) contain equal number of molecules.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 2.
What is Gay Lussac’s law ?
Answer:
Gay Lussac’s law : If different gases react chemically under the same condition of temperature and pressure and if the product is gaseous then the reactants and the products maintain a simple ratio in their volumes.

Question 3.
What is Berzelius hypothesis ?
Answer:
Berzelius hypothesis : Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of atoms.

Question 4.
What is vapour density ?
Answer:
Vapour density : Vapour density of a gas is the ratio of the weight of a certain volume of the gas to the weight of the same volume of hydrogen under similar conditions of temperature and pressure.

Question 5.
What is molecule ?
Answer:
Molecule : The smallest particle of an element or compound which can exist in the free state is known as molecule.

Question 6.
What are the types of molecules ?
Answer:
There are two types of molecules :

  • Elementary molecule
  • Compound molecule

Elementary molecule : The molecule which is formed by one or more atoms of an element is known as elementary molecule, e.g. Na, K, B, C etc.
Compound molecule : The molecule which is formed by atoms of more than one element is known as compound molecule, e.g. H2O, NH3, HCl etc

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 7.
What is molecular weight ?
Answer:
Molecular weight : The molecular weight of a substance is a number which represents how many times a molecule of the substance is heavier than \(\frac{1}{12}\)part of the weight of a C-12 isotope.

Question 8.
What is atomic weight ?
Answer:
Atomic weight : Atomic weight of an element
\(=\frac{\text { Weight of } 1 \text { atom of the element }}{\frac{1}{12} \text { of the weight of } 1 \text { atom of } \mathrm{C}^{12}}\)

Question 9.
What is gram atomic mass ?
Answer:
Gram atomic mass : When the atomic mass of an element is expressed in gram then the amount of element in gram is known as gram atomic mass or gram-atom of the element.

Question 10.
What is gram molecular mass ?
Answer:
Gram molecular mass : When the molecular mass of an element or compound in expressed in gram then that amount of element or the compound in gram is known as gram molecular mass or gram molecule or gram-mole of the element or the compound.

Question 11.
What is Molar volume ?
Answer:
Molar volume or gram-molar volume : The volume of a gaseous substance (element or compound) under a fixed temperature and pressure is known as molar volume or gram molar volume.

Question 12.
What is Avogadro’s number ?
Answer:
Avogadro’s number : One gram-molecule of any element or compound contains equal number of molecules. This number is known as Avogadro’s number.
It is denoted by N and its value is 6.023 × 1023

Question 13.
What is atomicity ?
Answer:
Atomicity : Atomicity is. the number of atoms by which an elementary molecule is composed of.
e.g. atomicity of He = 1; atomicity of oxygen = 2; atomicity of ozone = 3; atomicity of phosphorous = 4

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 14.
What do you mean by mole ?
Answer:
Mole : The ‘mole’ is regarded as the amount of the substance in grams which contains Avogadro’s number of elementary entities (such as molecules, atoms, ions) constituting the substance under investigation.

Question 15.
What is atomic mass unit (amu) ?
Answer:
Atomic mass unit (amu) : It is the quantity of mass equal to \(\frac{1}{12}th\) of the mass of carbon atom (c12).
1 amu = 1.6606 x 10-24 g

Question 16.
What is the difference between molecular weight and actual weight of a molecule ?
Answer:
Distinction between molecular weight and actual weight of a molecule : Gram molecular weight is the weight of 6.023 × 1023 molecules, while the actual weight is the weight of one molecule.
Weight of molecule = \(\frac{\text { Gram-molecular weight }}{6.023 \times 10^{23}}\)

Question 17.
Give statement of two important deductions of Avogadro’s law
Answer:
Two important deductions of Avogadro’s law :

  • Molecules of common elementary gases like hydrogen, oxygen, nitrogen etc. are diatomic.
  • The gram-molecular (or molar) volume of all gases under the same condition of temperature and pressure is the same and at STP, it is 22.4 litBroad answer type questions

Broad answer type questions 

Question 1.
Explain Berzelius hypothesis on the basis of Dalton’s atomic theory.
Answer:
Explanation of Berzelius hypothesis on the basis of Dalton’s atomic theory : Under the same conditions of temperature and pressure one volume of hydrogen and one volume of chlorine combine to form two volumes of hydrogen chloride gas. According to Berzelius hypothesis, suppose that under the same conditions of temperature and pressure each volume of a gas contains
n-number of atoms.
We can say,
n-number of hydrogen atoms + n-number of chlorine atoms
= 2n number of hydrogen chloride atoms
∴ 1 atom of hydrogen + 1 atom of chlorine
= 2 atoms of hydrogen chloride
∴ \(\frac{1}{2}\) atom of hydrogen + \(\frac{1}{2}\) atom of chlorine
= 1 atom of hydrogen chloride
Now, existence of \(\frac{1}{2}\) atom of hydrogen and chlorine is impossible according to Dalton’s atomic theory, so the hypothesis clashed with the basic concept of atomic theory. Hence, it was rejected by Dalton.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 2.
Explain Gay-Lussac’s law with the help of Avogadro’s hypothesis.
Answer:
Explanation of Gay-Lussac’s law with the help of Avogadro’s hypothesis :
From experimental result it has been found that under the same conditions of temperature and pressure one volume of hydrogen combines with one volume of chlorine to produce two volumes of hydrogen chloride gas.

Suppose, under the same conditions of temperature and pressure each volume of gas contains
n-number of molecules.
So,
n-number of hydrogen molecules + n-number of chlorine molecules
= 2n-number of hydrogen chloride molecules.
∴ 1 molecule of hydrogen + 1 molecule of chlorine
= 2 molecules of hydrogen chloride gas.
∴ \(\frac{1}{2}\) molecule of hydrogen + \(\frac{1}{2}\) molecule of chlorine
= 1 molecule of hydrogen chloride gas
This is to say,
1 molecule of hydrogen chloride must contain \(\frac{1}{2}\) molecule of hydrogen and \(\frac{1}{2}\) molecule of chlorine.
This fact does not go against Dalton’s atomic theory. Later on, it is known from Avogadro’s hypothesis that each molecule of hydrogen and chlorine contains two atoms.
So, 1 atom of hydrogen + 1 atom of chlorine
= 1 molecule of hydrogen chloride gas. in this way Dalton’s atomic theory and Gay-Lussac’s law were harmonized with the help of Avogadro’s hypothesis.

Question 3.
What are the application of Avogadro’s hypothesis ?
Answer:
Application of Avogadro’s hypothesis :

  • Excepting the inert gases, all other active elementary gases are diatomic i.e. molecule containing two atoms.
  • Molecular weight of any gas is twice its vapour density.
  • Volume of a gram-mole of all elementary or compound gases is 22.4 lit at NTP.
  • Number of atoms or molecules of all elementary or compound gases at their gram-atomic or gram-molecular weight is 6.023 x 1023 at NTP.
  • Molecular formula of a gaseous molecule can be deducted from its volumetric composition.
  • Atomic weight of an element can be found out by the application of this hypothesis.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 4.
Prove that hydrogen molecule is diatomic in nature.
Answer:
Profit of diatomic nature of hydrogen molecule : From experimental result we know that under the same conditions of temperature and pressure one volume of hydrogen and one volume of chlorine combine chemically to form two volumes of hydrogen chloride gas.

Let us suppose that under the conditions of pressure and temperature at the time of experiment, one volume of hydrogen gas contains re-molecules. So, according to Avogadro’s hypothesis, under the same conditions of temperature and pressure one volume of chlorine contains remolecules and two volumes of hydrogen chloride gas contain 2n molecules.

We can write,
re-molecules of hydrogen + re-molecules of chlorine
= 2re molecules of hydrogen chloride gas.
∴ 1 molecule of hydrogen + 1-molecule of chlorine = 2 molecules of hydrogen chloride gas.
So, 1 molecule of hydrogen chloride
= \(\frac{1}{2}\) molecule of hydrogen + \(\frac{1}{2}\) molecule of chlorine
Now, according to Dalton’s atomic theory, an atom is indivisible.
So, one molecule of hydrogen chloride contains at least one atom of hydrogen and one atom of chlorine. This one atom of hydrogen comes from \(\frac{1}{2}\) molecule of hydrogen.
So one molecule of hydrogen must contain two atoms.

Question 5.
Deduce the relation of density and vapour density of a gas on the basis of hydrogen.
Answer:
Relation of density and vapour density of a gas on the basis of hydrogen
We know,
WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept 3
∴ Density of gas (d) = D x 0.089
(Density of hydrogen gas = 0 089)
∴ d = D x 0.089

Question 6.
Deduce the relation between vapour density and the molecular weight of a gas.
Answer:
Relation between vapour density and the molecular weight of a gas : From the definition of vapour density
WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept 1
Suppose, V volume of the gas contains ‘n’ molecules. Then by Avogadro’s hypotheis, V volume of hydrogen will also contain ‘n’ molecules of hydrogen.
WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept 2
i.e. Molecular weight of a gas 2 x its vapour density.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 7.
Gram-molecular volume of all gases at NTP is 224 litres – prove it.
Answer:
If the molecular mass of a gas is M, then we know that the limiting density of the gas 0.0898 × \(\frac{M}{2}\) g/litre
From the definition of limiting density it can be written that mass of 1 litre of a gas at NTP = 00898 x \(\frac{M}{2}\)g
So. at NTP, volume of 0.0898 × g gram of the gas = 1 litre
∴ At NTP volume of 1 gram of the gas = \(\frac{2}{M \times 0.0898}\) litre
∴ At NTP. volume of M g or 1 gram molecule of the gas
\(=\frac{2 \times M}{M \times 0.00898}=\frac{2}{0.0898}\) = 2227 litre
Atomic mass of hydrogen in oxygen scale is 2 016. in that case calculating in a similar way, we can obtain,
Volume of I gram-molecule gas at NTP\(=\frac{2.016}{0.0898}=22.4 \text { litre }\)
So, at NTP, volume of 1 gram-molecule of any gas is 224 litre.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Question 8.
What is the actual weight of a molecule ? What is the actual weight of an atom?
Answer:
Actual weight of a molecule:
Suppose, the molecular mass of any substance M
So, gram-molecular weight of the substance = Mg
We know that there are present 6.023 × 102 molecules in gram-
molecular weight.
∴ Weight of 6.023 x 1023 molecules Mg
∴ Weight of 1 molecule = \(=\frac{M}{6.023 \times 10^{23}} \mathrm{~g}\)
Actual weight of an atom:
Suppose, the atomic mass of an element = A
So, gram-atomic weight of the element = Ag
We know that there are 6.023 x 1023 atoms present in gram-atomic weight.
∴ Weight of 6.023 × 1023 atoms = Ag
∴ Weight of 1 atom \(=\frac{A}{6.023 \times 10^{23}} \mathrm{~g}\)

Question 9.
What are the difference between molecular weight and acutal weight (mass) of a molecule?
Answer:
Difference between molecular weight and actual weight (mass) of a molecule :
WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept 4

Numerical Problem:
WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept 5

Problem 1.
If 200 ml of a gas at NTP weighs 0.27g and the normal density of hydrogen be 0.09 glut, calculate vapour density of the gas.
Answer:
At NTP, vapour density × 0.09 = Normal density
Now, 200 ml of the gas weighs 0.27 g
WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept 6

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Problem 2.
Find the actual weight of 1 molecule water. (H = 1, O = 16)
Answer:
Molecular weight of water = (1 × 2 + 16) = 18
gram molecular weight of water = 18 g
So, 18g water contains = 6.023 × 1023 molecules
Hence, actual weight of 1 water molecule
= \(\frac{18}{6.023 \times 10^{23}}\) = 2.988 × 10-23g (approx)

Problem 3.
How many gram-moles of chlorine are present in 213 g chlorine and what is the volume of 6.023 x 1030 hydrogen molecules at STP ? Given, atomic weight of chlorine is 35.5.
Answer:
Molecular weight of chlorine 355 × 2 = 71
∴ 71g chlorine = 1 gram-mole chlorine
∴ 213 g chlorine = \(\frac{1 \times 213}{71}\) gram-mole = 3 gram mole
At STP, 6.023 × 1023 molecules occupy 224 lit
∴ At STP, 6.023 × 1030 molecules occupy
\(=\frac{22.4 \times 6.023 \times 10^{30}}{6.023 \times 10^{23}} \mathrm{lit}\)
= 22.4 ×107 lit

Problem 4.
Find the number of hydrogen and oxygen atoms in 425.6 lit of water vapour at NTP.
Answer:
Number of molecules in 22.4 lit water vapour at NTP
6.023 x 1023
∴ Number of molecules in 425.6 lit water vapour at NTP
\(=\frac{6.023 \times 10^{23} \times 425 \cdot 6}{22 \cdot 4}=114437 \times 10^{20}\)
Now, 1 molecule of water contains number of hydrogen atom = 2
∴ In 114437 x 1020 molecules of water contains number of hydrogen atom
= 2288874 × 1020
Again, 1 water molecule contains 1 oxygen atom
∴ 114437 x 1020 water molecules contain 114437× 1020 oxygen atom.

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Problem 5.
What is the weight of one gram oxygen and how many number of oxygen atoms are there ?
Answer:
One gram-atom of oxygen = 16 g of oxygen
From Avogadro’s hypothesis,
32 g of oxygen contains = 6.023 × 1023 molecules
∴ 16 g of oxygen contains  \(=\frac{6023 \times 10^{23}}{32} \times 16\) molecules
= 3.0115 × 1023 molecules
Again oxygen is a diatomic element,
Hence, it contains 2 × 3.0115 x 1023 atoms = 6.023 × 1023 atoms.

Problem 6.
Calculate the number of mole in 9 grams of water and calculate the number of hydrogen atoms present in it.
Answer:
18 g water contain = 6.023 × 1023 molecules
9 g water contain \(\frac{6.023 \times 10^{23}}{2}\) = 3.0115 x 1023 molecules of water
One molecule of water contains 2 hydrogen atoms. So the number of hydrogen atoms =
(3.0115 × 1023) x 2 = 6.023 × 1023

Problem 7.
250 cm3 of a gas at NTP weigh 0.7924. What is the molecular weight of the gas ?
Answer:
250 cm3 of the gas at NTP weighs = 0.7924 g
∴ 22400 of the gas at NTP weighs
= \(\frac{0.7924 \times 22400}{250} \mathrm{~g}\)
= 70 999 g
∴ Gram-molecule of the gas = 70.999 g
∴ Molecular weight of the gas = 70.999

Problem 8.
The molecular weight of a gaseous substance is 200. What is the volume of 5 g of the substance at NTP?
Answer:
The molecular weight of the substance = 200
∴ 1 gram-molecule of the substance = 200 g
Volume of 1 gram-molecule or 200 g of the substance at NTP = 224 lit.
Volume of 5 g of the substance at NTP = \(\frac{22.4 \times 5}{200}=0.56 \mathrm{lit}\)

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Problem 9.
Calculate the number of molecules in a drop of water weighing 0.05 g (H = 1, 0 = 16)
Answer:
Molecular mass of water = (2 × 1 + 1 × 16) = 18
∴ Gram-molecular mass of water = 18 g
Now, 18 g of water contains 6.023 × 1023 molecule
∴ 0.05 g of water contains
\(\frac{6.023 \times 10^{23}}{18} \times 0.05\) molecule
= 1.672 × 1021

Problem 10.
Calculate the total number of electrons present in 1.6g of methane (CH4). [H = 1, C = 12]
Answer:
Molecular mass of methane = 16 g
16 g of methane 1 mole
∴ 16 g of methane = 0.1 mole
Now, 1 mole of methane = 6.023 × 1023 molecules
∴ 0.1 mole of methane = 6.023 × 1022 molecules
We know that one atom of carbon contains six electrons and one atom of hydrogen contains one electron.
∴ One molecule of CH4 contains = 6 + 1 × 4 = 10 electrons
∴ 6.023 × 1022 molecules contain = 10 x 6.023 x 1022
= 6.023 × 1023 electrons

Problem 11.
A tetra-atomic gas occupies 1.4 lit at 0°C and 76cm pressure. Find the number of atoms in the gas.
Answer:
At 0°C and 760 cm pressure, number of molecules present
= 224 lit of the gas.
= 6.023 × 1023
= 6.023 × 1023 x 4 atoms
Number of atoms present in 1.4 lit
= \(\frac{1 \cdot 4}{22 \cdot 4}\) x 6.023× 1023 × 4 = 1.505 x1023

Problem 12.
Calclulate the molecular mass of CO2
Answer:
CO2 molecule contains one C-atom and two oxygen atoms.
Molecular mass of CO2= (1 x 12 + 2 x 6) = 44U

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Problem 13.
Calculate the formula of unit mass of MgCl2
Answer:
MgCl2 molecule contains one Mg-atom and two Cl-atoms.
∴ Formula of unit mass of MgCl2
= 24+2 x 35.5 = 24+71=95U

Problem 14.
Calculate the following :
(i) 1 milimole of NH3
(ii) 3.011 x 1023 number of \({ }_8^{16} \mathrm{O}\)
Answer:
(i) 1 milimole = 10 mole.
Mass = molar mass x no. of moles
= 17 × 10-3= 0.017g

(ii) Mass = number of moles × atomic mass
\(=\frac{3.011 \times 10^{23}}{6.022 \times 110^{23}} \times 16=8 \mathrm{~g}\)

Example 15 :
Calculate the number of particles in each of the following :
(i) 35.5g of Na atoms.
(ii) 11g of CO2
(iii) 0.2 mole of 2C atoms.
Answer:
(i) the number of atoms
\(=\frac{\text { given mass }}{\text { molar mass }}\)×Avogadro’s number
\(\frac{35 \cdot 5}{23}\)

(ii) Number of Melecules of CO2

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept 7

(iii) No of 12C atoms
= No of moles of particle x Avogadro’s number.
= 0.2 × 6.022 × 1023 = 1.2044 × 1023

Problem 16.
Calculate the mass of 250 molecules of sodium chloride.
Answer:
Molecular mass of NaCl
= 1 × 23 + 1 × 35.5 = 58.5
1 mole of NaCl = 58.5g NaCl
6.022 × 1023 molecules of NaCl have mass 58.5g
250 molecules of NaCl have mass = \(\frac{58 \cdot 5 \times 250}{6.022 \times 10^{23}}\) = 2.428 ×10-23g.

Problem 17.
What is the actual weight of one molecule of H2 and one atom of hydrogen ? The gram molecule of H2 is 2.016g.
Answer:
6.022 x 10-23 molecules of hydrogen weigh 2.016g.
1 molecule of hydrogen weigh \(\frac{2 \cdot 016}{6 \cdot 022 \times 10^{23}} \mathrm{~g}\) = 3.34 × 10-24g
Hydrogen is diatomic i.e. one molecule of hydrogen consists of two atoms. Therefore, one atom of hydrogen weighs
\(=\frac{3.34 \times 10^{-24}}{2}\) = 1.67 × 10-24g

WBBSE Class 9 Physical Science Solutions Chapter 4.2 Mole Concept

Problem 18.
How many moles or milimoles are present in 0.49 g of H2SO4? [H = 1, S=32, O=16]
Answer:
Molecular mass of H2SO4
=2 x 1+1 x 32+4  x 16=98
1g mole of H2SO4 is equal to 1 mole of H2SO4
∴ 0.49g of H2SO4 is equal to \(\frac{0.49}{98}\) = 0.005 mole
So. no of millimoles
= 0.005 × 103 = 5 millimole
[Since 1 mole 103 millimole]

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Well structured WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat can serve as a valuable review tool before exams.

Heat Class 9 WBBSE MCQ Questions

Multiple Choice Questions :

Question 1.
What is the unit of heat in SI system ?
(i) calorie
(ii) Joule
(iii) erg
(iv) Newton
Answer:
Joule

Question 2.
Who designed absolute scale of temperature ?
(i) Newton
(ii) Lord Kelvin
(iii) Celsius
(iv) Boyle
Answer:
Lord Kelvin

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 3.
Which substance has the highest specific heat ?
(i) water
(ii) iron
(iii) gold
(iv) ice
Answer:
Water

Question 4.
The fundamental interval of celsius scale is divided into –
(i) 100
(ii) 180
(iii) 120
(iv) 102 division
Answer:
100

Question 5.
The fundamental interval of Fahrenheit scale is divided into
(i) 180
(ii) 100
(iii) 32
(iv) 212 division
Answer:
180

Question 6.
If C and F represent a particular temperature in Celsius and Fahrenheit scales respectively, then their relation is :
(i) \(\frac{C}{4}\) = \(\frac{F-32}{9}\)
(ii) \(\frac{C}{5}\) = \(\frac{F-32}{9}\)
(iii) \(\frac{C}{3}\) = \(\frac{F-32}{5}\)
(iv) \(\frac{C}{9}\) = \(\frac{F-32}{5}\)
Answer:
\(\frac{C}{5}\) = \(\frac{F-32}{9}\)

Question 7.
The temperature of a body is measured by the instrument :
(i) hydrometer
(ii) thermometer
(iii) voltameter
(iv) ammeter
Answer:
thermometer

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 8.
How much calorie of heat would be produced by converting 42 joule of work completely into heat ?
(i) 100 calorie
(ii) 10 calorie
(iii) 1000 calorie
(iv) 90 calorie
Answer:
10 calorie

Question 9.
Normal body temperature of a man is –
(i) 98.4° C
(ii) 98.4° F
(iii) 98.4 K
(iv) 100° F
Answer:
98.4° F

Question 10.
Which substance has the highest specific heat?
(i) ethanol
(ii) water
(iii) acetone
(iv) ether
Answer:
water

Question 11.
If the specific heat of copper be 0.09, what will be the water equivalent of a block of copper of mass 50 g ?
(i) 45 g
(ii) 4.5 g
(iii) 450 g
(iv) 44 g
Answer:
4.5 g

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 12.
What is the melting point of ice in Fahrenheit scale ?
(i) 0° F
(ii) 32° F
(iii) 180° F
(iv) 121° F
Answer:
32° F

Question 13.
What is the specific heat of water in SI system ?
(i) 4 2 joule
(ii) 4200 joule/kg
(iii) 4200 joule/kg/K
(iv) 42 erg
Answer:
4200 joule / kg / K

Question 14.
The dimension of heat is –
(i) M L T
(ii) M L2 T-2
(iii) M L-1 T-2
(iv) M L-1 T-1
Answer:
ML2 T-2

Question 15.
Generally body temperature of a man is measured in :
(i) Celsius scale
(ii) Fahrenheit Scale
(iii) Kelvin scale
(iv) none of these
Answer:
Fahrenheit

Question 16.
Heat reaches the earth from the sun by the process of –
(i) conduction
(ii) convection
(iii) radiation
(iv) expansion
Answer:
Radiation

Question 17.
The value of mechanical equivalent of heat in CGS system is :
(i) 4.18 × 107 erg / calorie
(ii) 4.8 erg / calorie
(iii) 5.18 × 107 erg / calorie
(iv) 5.18 erg / calorie
Answer:
4.18 × 107 erg / calorie

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 18.
How much work is to be done to produce 50 calorie of heat ?
(i) 210 joule
(ii) 200 joule
(iii) 201 joule
(iv) 310 joule
Answer:
210 joule

Question 19.
The water equivalent of a body is equal to :
(i) mass of the body x specific heat
(ii) mass of the body / specific heat
(iii) mass of the body + specific heat
(iv) mass of the body – specific heat
Answer:
Mass of the body × specific heat

Question 20.
Mercury remains in liquid state for long range of temperature :
(i) -39° C to 357° C
(ii) -57° C to 457° C
(iii) -10° C to 257° C
(iv) -12° C to 122° C
Answer:
-39° C to 357° C

Question 21.
-40° C is equal to :
(i) -50° F
(ii) -40° F
(iii) -60° F
(iv) -80° F
Answer:
-40° F

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 22.
1 Calorie is equal to –
(i) 4.18 Joule
(ii) 5.18 Joule
(iii) 3.18 Joule
(iv) 4.18 erg
Answer:
4-18 Joule.

Question 23.
If a temperature in Kelvin scale is TK and the temperature in Celsius scale is t° C, the relation is :
(i) TK = (273 + t°C)
(ii) TK = (273 – t°C)
(iii) TK = (273 / t°C)
(iv) none of these
Answer:
T K = (273 + t°C)

Question 24.
Calorimetry relates to the measurement of –
(i) heat
(ii) temperature
(iii) mechanical energy
(iv) none of these
Answer:
heat

Question 25.
Quantity of heat of a body depends on its –
(i) temperature
(ii) mass
(iii) material of the body
(iv) all of them
Answer:
all of them

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 26.
Quantity of heat is given by –
(i) \(\frac{m s}{t}\)
(ii) mst
(iii) \(\frac{m t}{s}\)
(iv) \(\frac{m s t}{w}\)
Answer:
mst

Question 27.
Work done W and heat produced H are related to each other as(where J is the mechanical equivalent of heat)
(i) WH = J
(ii) H = JW
(iii) W = J H
(iv) W J H = 1
Answer:
W = JH

Question 28.
The value of mechanical equivalent of heat (in erg/calorie) is –
(i) 4.2 × 10-7
(ii) 4.2 × 107
(iii) \(\frac{1}{4.2}\)
Answer:
4.2 × 107

Question 29.
The latent heat of fusion of ice is –
(i) 80 cal g-1
(ii) 3.36 × 105 cal g-1
(iii) 80 joule kg-1
(iv) 3.36 cal g-1
Answer:
80 cal g-1

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

Question 30.
Saturated vapours obeys –
(i) Boyle’s law
(ii) Charles’ law
(iii) pressure law
(iv) none of these
Answer:
none of these

Question 31.
Dews are formed at a temperature –
(i) greater than dew point
(ii) less than dew point
(iii) equal to dew point
(iv) at critical temperature.
Answer:
equal to dew point

Question 32.
The volume of water is minimum at a temperature –
(i) -4°C
(ii) 4°C
(iii) 0°C
(iv) 1°C
Answer:
4°C

Question 33.
Thermal capacity of a body of mass m and specific heat s is –
(i) \(\frac{\mathrm{m}}{\mathrm{s}}\)
(ii) \(\frac{\mathrm{m}}{\mathrm{s}}\)
(iii) \(\frac{\mathrm{m}}{\mathrm{s}}\)
(iv) ms
Answer:
ms

Fill in the blanks :

1. Heat is the cause and temperature is the _____.
Answer:
effect

2. Heat is a form of _____ a body possesses.
Answer:
energy

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

3. The range of temperature between the upper and lower fixed point is known as _____ interval.
Answer:
fundamental

4. Kelvin scale of temperature starts from the temperature corresponding to _____, taken as zero Kelvin.
Answer:
-273° C

5. Mean calorie is _____ of the heat necessary to raise the temperature of 1 gram pure water from 0° C} to 100° C.
Answer:
(\(\frac{1}{100}\)) th

6. 1 calorie = 4.18 or approximately 4.2 _____.
Answer:
joule

7. Specific heat of _____ in SI system is 4200 J/kg/K.
Answer:
water

8. Thermal capacity or heat capacity of a body is defined as the ratio of the heat supplied to a body to its consequent rise in _____.
Answer:
temperature.

9. In _____ process heat is transmitted in wave form.
Answer:
radiation

10. In radiation process _____ is not required.
Answer:
medium

11. Heat reaches the earth from the sun by _____ process.
Answer:
radiation

12. Normal body temperature of a man is _____ degree celsius.
Answer:
36.9

13. 94° F = _____ °C.
Answer:
34.4

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

14. The water equivalent of a body is = mass x _____.
Answer:
specific heat

15. A body becomes hot by _____ heat and becomes cold on giving away heat.
Answer:
taking

16. Temperature is the condition of a body.
Answer:
thermal

17. Melting point of _____ under normal atmospheric pressure is taken as the lower fixed point of a scale of temperature.
Answer:
ice

18. The quantity of heat required to raise the _____ of unit mass of a substance through one degree is called the specific heat of the substance.
Answer:
temperature

19. When work is completely converted into heat or heat is compietely converted into work, one is _____ to the other.
Answer:
equivalent

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

20. Work done to produce unit heat is known as _____ equivalent of heat.
Answer:
mechanical

21. Heat is a _____ quantity.
Answer:
scalar

22. In CGS system the unit of specific heat is _____
Answer:
Calorie/g/°C

23. The unit of specific heat in SI system is _____
Answer:
J/kg/K

24. -40°C = _____ °F
Answer:
-40

25. What is the dimension of heat ?
Answer:
The dimension of heat is ML2 T-2.

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

26. 1°C = ______ °F
Answer:
\(\frac{9}{5}\)

27. in CGS system heat is measured by the unit called _____.
Answer:
Calorie

28. Temperature of a body depends on the _____ contained in it.
Answer:
heat

29. In SI system the unit of thermal capacity is _____.
Answer:
JK-1

30. 1 kcal = _____ cal
Answer:
1000

31. In case of change of state, heat is _____ or liberated.
Answer:
absorbed

32. Latent heat of fusion of ice is _____ Jkg-1
Answer:
3.36 × 105

WBBSE Class 9 Physical Science MCQ Questions Chapter 6 Heat

33. Specific heat of _____ is highest than any other common substance.
Answer:
Water

34. The temperature at which dews are formed called _____point.
Answer:
dew

35. The density of water at 4°C is _____.
Answer:
maximum

36. In CGS system, latent heat of vapourization of water is _____.
Answer:
540 cal/g/°C