Bhagya – Page 296 – PSEB Solutions

PSEB 9th Class Science Notes Chapter 5 The Fundamental Unit of Life

November 19, 2021 by Bhagya

This PSEB 9th Class Science Notes Chapter 5 The Fundamental Unit of Life will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 5 The Fundamental Unit of Life

→ All living organisms around us are complex structural compartments called cells.

→ A.V. Leeuwenhoek (1674) first studied the living cell. He examined bacteria, sperms, and erythrocytes (RBC).

→ The biosphere is the highest level of organisation of living organisms.

→ Level of organisation is Atoms → Elements → Cell → Tissue → Organ → Organ System → Living Organisms.

→ Robert Hooke (1665) examined dead cells,

→ Rober Brown (1831) observed the nucleus in the centre of cell.

→ Huxley regarded protoplasm as the ‘Physical basis of life’.

→ The cytoplasm is the fluid content of cells present between the nucleus and plasma membrane.

→ It contains metabolites and organelles.

→ Organelles are special components of cells performing specific functions.

→ Cells are of two types i.e. prokaryotic cell and eukaryotic cell.

→ Organisms may be single-celled e.g. Amoeba, CMamydomonas (an algal plant), Paramecium or they are multicellular.

→ Higher plants and animals are made up of a large number of cells.

→ A true nucleus is present, It is generally single-spherical and central in position.

→ The nucleus is separated from the cytoplasm by a double-layered membrane called nuclear membrane. It controls the functioning of cells.

→ The nucleus contains chromosomes composed of DNA and protein.

→ A functional segment of DNA is called a gene.

→ All living organisms start their life cycle from a single cell.

→ Cell size varies from 0.2 – 0.5 micron to 30 micron (one micron = 1/1000 mm ). Nerve cells may be as long as a few metres.

→ A plant cell is bounded by a protective cell wall.

→ A plasma membrane is a living membrane.

→ Mitochondria are rod-shaped, double-membranous, light-microscopic, eukaryotic structures. Inner membranes have cristae and exosomes.

→ Functions: These act as powerhouses or ATP mills as they are sites for cellular respiration and release energy.

→ The centrosome is an animal structure and is formed of two microtubular centrioles, each being formed of 9 triplet microtubules showing a 9 + 0 arrangement.

→ Function: These help in cell division.

→ Basal bodies give rise to cilia or flagella centrioles form.

→ Contractile vacuoles are present in freshwater protozoans.

→ The cell is a Latin word for a little room’.

→ The electron microscope was discovered in 1942.

→ Water obeys the laws of diffusion.

→ The nucleus plays a central role in cellular reproduction, the process by which a cell divides and forms two new cells.

→ Plastids are the largest-sized eukaryotic structures of plant cells. These are of three types: Leucoplasts (colourless and store the food).

→ The primary function of the leucoplast is storage. Chloroplasts (green coloured and are sites of photosynthesis, so-called kitchens of cells)

→ Each chloroplast is a double-membranous structure having grana in its inner chamber.

→ Each granum is formed of many chlorophyll-containing thylakoids present in stacks.

→ Functions: Chloroplasts are sites for photosynthesis.

→ Chromoplasts (coloured and help in pollination of dowers and dispersal of seeds and fruits).

→ The endoplasmic reticulum is an electron-microscopic interconnected network of cisternae, vacuoles, and tubules.

→ It is of two types: RER (cisternae are studded with ribosomes and involved in protein synthesis) and SER (tubules are without ribosomes).

→ Functions:

It is a passageway for intracellular and intercellular transport of materials.
It gives internal support to the cell.
SER is involved in the synthesis of lipids and steroids.
RER is concerned with protein synthesis.

→ Lysosomes are electron-microscopic single membrane-bound vesicular structures of animal cells and contain hydrolytic enzymes.

→ Functions:

These are centres of intracellular digestion and act as both digestive bags and suicidal bags.
They destroy foreign substances.
They remove cellular debris.

→ Living organisms are composed of one or a large number of cells. The cell is the structural and functional unit of life.

→ A large number of build-up and breakdown reactions take place in the cell.

→ Life is passed on from one generation to the next generation in the form of cells.

→ Robert Hooke (1665) first discovered cells on the basis of compartments observed in a thin section of the bark of a tree.

→ Prokaryotic cells lack a well-organized nuclear membrane and membrane-bound organelles. They have 70 S type of ribosomes.

→ Eukaryotic cells have a proper nucleus and membrane-bound organelles. 80 S ribosomes are present.

→ The Golgi body is formed of stacked cisternae with swollen ends, vacuoles, and vesicles.

→ Functions:

It is involved in cell secretions such as mucous, enzymes, and hormones.
It helps in the storage of secretory products.

→ Cell inclusions include reserve food in the form of glycogen granules or lipid droplets or starch grains.

→ Vacuoles and fluid-filled membrane-bound spaces each containing cell sap within a tonoplast, wastes, gases, secretions, etc.

→ Functions: They help in the storage of food, water, and other wastes.

→ Ribosomes are composed of RNA and proteins, granular electron-microscopic particles without membrane.

→ Functions: These act as protein factories.

→ Prokaryotes: The simple organisms called Monerans without a proper nucleus. e.g. Bacteria, Blue-green algae.

→ Eukaryotes: Organisms with true nucleus (plant and animal cells).

→ Organelles: Special living components of cells each performing a definite function.

→ Leucoplasts: Colourless plastids.

→ Centriole: Star-shaped structure present near the nucleus in an animal cell. It forms a spindle during cell division.

→ Genes: They are present on chromosomes and act as carriers of characters from parents to offspring.

→ Lysosome: These are electro-microscopic structures bounded by a single membrane. They are full of digestive enzymes. They are called ‘suicidal bags.’

→ DNA (Deoxyribose nucleic acid): It controls cellular functions and also acts as genetic material.

→ RNA (Ribose nucleic acid): It plays important role in protein synthesis.

→ Cyclosis: Streaming movements of cytoplasm.

→ Autolysis: Self-digestion of the cell by its lysosomal enzymes.

→ Autophagy: Digestion of its own cell organelles or reserve food by the lysosome.

→ Cristae: Infolds of the inner mitochondrial membrane.

→ Chloroplast: Chlorophyll-containing green coloured photosynthetic plastids.

→ Chromoplast: A pigmented plastid.

→ Tonoplast: Vacuolar membrane present around cell sap.

→ Camillo Golgi discovered Golgi bodies and shared Noble Prize in 1906 with Santiago Ramony Cajal for their work on the structure of the nervous system.

→ Nucleoid: A primitive nucleus of prokaryotes, not covered by a nuclear membrane.

PSEB 9th Class Science Notes Chapter 4 Structure of the Atom

November 19, 2021 by Bhagya

This PSEB 9th Class Science Notes Chapter 4 Structure of the Atom will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 4 Structure of the Atom

→ It was known by 1900 that the atom was not a simple, indivisible particle but contains at least one sub-atomic particle i.e., electron identified by J.J. Thomson.

→ Even before the electron was identified by J.J. Thomson, E. Goldstein in 1886 discovered the presence of new radiations in a discharge tube called canal rays.

→ The canal rays were positively charged radiations which led to the discovery of the proton.

→ Proton has a charge, equal in magnitude but opposite in sign to that of electron and a mass approximately 2000 times as that of the electron.

→ Generally, the electron is represented by ‘e’ and a proton as ‘p’.

→ The mass of a proton is taken as one unit and charge +1.

→ The mass of an electron is considered to be negligible and its charge -1.

→ α-particles are doubly-charged helium ions having mass 4u \(\left({ }^{4}{ }_{2} \mathrm{He}^{2+}\right)\)

→ E. Rutherford discovered the nucleus of an atom on the basis of an α-ray scattering experiment.

→ E. Rutherford was awarded the Nobel prize in chemistry for his famous work discovery of radioactivity and the discovery of the nucleus of the atom.

→ On the basis of his experiment, Rutherford put forward the nuclear model of an atom.

→ According to Rutherford’s model, there is a positively charged centre of an atom called a nucleus and contains nearly all the mass and all the positive charge.

→ The electrons revolve around the nucleus in circular paths. The size of the nucleus is very small as compared to the size of the atom.

→ Neils Bohr’s model of the atom was more successful.

→ He suggested that only certain special orbits known as discrete orbits of electrons are allowed inside the atom. While revolving electrons do not radiate energy.

→ J. Chadwick discovered another sub-atomic particle that had no charge but mass nearly equal to a proton. This particle is called the neutron.

→ The orbits or shells in an atom are designated as K, L, M, N ………… shells starting from the nucleus side.

→ J.J. Thomson suggested that an atom is a uniform sphere of positive electricity in which electrons are embedded it.

→ The total positive charge is equal to the total negative charge and the atom, on the whole, is electrically neutral.

→ Electron is a negatively charged particle with 1.602 × 10^-19 coulomb negative charge (-1 unit) and mass 9.1089 × 10^-19 kg (negligible mass). It is represented by the symbol ‘e’. It is a fundamental particle of an atom.

→ Proton is a positively charged particle with a 1.602 × 10^-19 coulomb positive charge (+1 unit) and mass 1.672 × 10^-27 kg, it is represented by the symbol ‘p’. It is a fundamental particle of an atom.

→ Neutron is a neutral particle with no charge and mass equal to 1.678 × 10^-27 kg. It is represented by the symbol ‘n’. It is a fundamental particle of an atom.

→ The nucleus is the small, positively charged, and heavy central portion in an atom that contains in it protons and neutrons.

→ Nucleons. The neutrons and protons present in the nucleus of an atom are collectively known as nucleons.

→ An atomic number of an element represents the number of protons in an atom. It is denoted by the symbol Z.

→ Shells of an atom are designated as K, L, M, N, etc. These are also called energy levels.

→ The valence shell of an atom represents the outermost shell where electrons are present and the electrons are called valence electrons.

→ A mass number of an element is the sum of the number of protons and neutrons present in the nucleus of the atom. It is denoted by A.

→ The valence shell of an atom represents the outermost shell where electrons are present and the electrons are called valence electrons.

→ Valency. It is the combining capacity of an atom of the element.

→ Isotopes are the atoms of the same element having the same atomic number but different mass numbers.

→ Isobars are the atoms of the different elements having the same mass number but different atomic numbers.

PSEB 9th Class Science Notes Chapter 3 Atoms and Molecules

November 19, 2021 by Bhagya

This PSEB 9th Class Science Notes Chapter 3 Atoms and Molecules will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 3 Atoms and Molecules

→ The idea of divisibility of matter was considered long back in India around 500 R.C.

→ Another Indian Philosopher, Pabuda Katyayama said these particles normally exist in a combined form which gives us various forms of matter.

→ Greek Philosophers-Democritus and Leucippus suggested that if we go on the dividing matter, a stage will come when particles obtained cannot be divided further.

→ Democritus called these indivisible particles atoms.

→ By the end of the 18th century, scientists recognized the difference between elements and compounds and how and why elements combine, and what happens when they combine.

→ Antoine L.Lavoisier gave two important laws of chemical combination given below.

→ According to the law of conservation of mass, mass can neither be created nor destroyed in a chemical reaction.

→ John Dalton provided the basic theory about the nature of matter.

→ Dalton picked up the idea of divisibility of matter and it was based upon laws of chemical combination.

→ Dalton gave Dalton’s atomic theory.

→ According to Dalton’s atomic theory, all matter, whether an element, a compound, or a mixture is composed of small particles called atoms, which are indivisible.

→ Dalton’s theory provided an explanation for the law of conservation of mass and the law of definite proportions.

→ All matter is made up of very tiny particles called atoms which cannot be seen with the naked eye.

→ The elements are represented with the help of symbols.

→ The symbol represents one atom of the element.

→ Berzelius suggested that the symbols of elements be made up from one or two letters of the name of the element.

→ The atomic mass of an element is the average relative mass of one atom of the element as compared to the mass of one atom of carbon-12, taken as 12 a.m.u. (atomic mass unit).

→ According to the latest IUPAC recommendations, a.m.u. has now been replaced by U (uniformed mass).

→ A molecule is in general a group of two or more atoms that are chemically bonded together.

→ A molecule can be defined as the smallest particle of an element or a compound that can exist independently and shows all the properties of that substance.

→ The molecules of an element are constituted by the same type of atoms.

→ The number of atoms constituting a molecule is known as its atomicity.

→ Atoms of different elements join together in definite proportions to form molecules of compounds.

→ Compounds composed of metals and non-metals contain charged species called ions. An ion is a charged particle and can be negatively or positively charged.

→ A negatively charged ion is called an anion and the positively charged ion is called a cation.

→ A group of atoms carrying a charge is known as a polyatomic ion.

→ The chemical formula of a compound is a symbolic representation of its composition.

→ The chemical formulae of different compounds are based upon their valencies.

→ The combining capacity of an element is known as its valency.

→ The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance.

→ The formula unit mass of a substance is a sum of the atomic masses of all the atoms in a formula unit of a substance.

→ One mole of any species (atoms, molecules, ions, or particles) is that quantity in number having mass equal to its atomic or molecular mass in grams.

→ One mole of any substance represents 6.022 × 10²³ particles (atoms, molecules, or ions) of it.

→ This number is an experimentally obtained value and is called Avogadro’s number or constant.

→ The mass of 1 mole of a substance is equal to its relative atomic or molecular mass in grams.

1 mole = 6.022 × 10²³ particle = Relative mass in grams.

→ An Indian philosopher Maharishi Kanad, postulated that if we go on dividing matter (padarth), we shall get smaller and smaller particles.

→ Ultimately, a stage will reach when further division will not be possible. He named these particles ‘Parmanu’.

→ The radius of an atom is expressed in nanometres (nm)

1 nm = 10^-9 m.
10⁹ nm = 1 m.

→ Atom: It is the smallest or ultimate particle of an element that takes part in chemical reactions. It may or may not exist independently.

→ Molecule: It is the smallest or ultimate particle of a substance (element or compound) that can exist freely. It shows all the properties of a substance.

→ Law of conservation of mass: It states that matter (or mass) can neither be created nor destroyed during any known physical or chemical change.

→ Law of definite proportions (or Law of constant compositions): It states that a pure chemical compound is always found to be made up of the same elements combined together in the same fixed ratio by mass.

→ Symbol: It is the shorthand representation of an element. It is made up of one or two letters from the name of the element.

→ Relative atomic mass (RAM) or Atomic mass: It is the average relative mass of one atom of the element as compared to one atom of carbon-12 taken at 12 a.m.u.

→ Relative molecular mass (RMM) or Molecular mass: It is the average relative mass of one molecule of a substance as compared to one atom of carbon-12 taken at 12 a.m.u.

→ Ion: It is an atom or group of atoms carrying some charge and can exist freely in solution.

→ Polyatomic ion: It is an ion having more than one atom.

→ The chemical formula of a molecular compound is determined by the valency of each element.

→ The chemical formula of an ionic compound is determined by the charge on each ion.

→ Mole: It is the amount of substance that contains the same number of particles (atoms/ions/molecules/formula units.) etc. as there are atoms in 12 g of carbon-12.

→ Avogadro’s number or constant is the number of atoms in 12 g of carbon-12. It is denoted by N₀ and its value is 6.022 × 10²³, N₀ = 6.022 × 10²³.

→ The molar mass of a substance is the mass of one mole of a substance.

→ Variable valency: When an element shows more than one valency, it is said to have variable valency.

→ Radical: It is an atom or group of atoms having positive or negative charges and behaving as a single unit in chemical reactions. e.g. Na⁺, \(\mathrm{SO}_{4}^{2-}, \mathrm{NO}_{3}^{-}, \mathrm{NH}_{4}^{+}\) etc.

→ Simple radical: A radical which is made up of only one kind of atom is called a simple radical, e.g. Na⁺, Cl^– etc.

→ Compound radical: A radical which is made up of more than one kind of atom is called a compound radical, e.g. \(\mathrm{NH}_{4}^{+}, \mathrm{SO}_{4}^{2-}, \mathrm{CO}_{3}^{2-}\), etc.

→ Atomic mass unit (AMU or u): It is 1/12th of the mass of one atom of carbon (C^12- isotope).

→ S.T.P. stands for standard temperature (0°C or 273K) and pressure (1 atmosphere).

→ Molar volume: It is the volume occupied by one mole of a gas and its value at S.T.P./ N.T.P. is 22.4 litres.

→ Valency: It is the combining capacity of an atom of an element and is numerically equal to the number of hydrogen atoms or number of chlorine atoms or double the number of oxygen atoms with which one atom of the element can combine.

PSEB 9th Class Science Solutions Chapter 3 Atoms and Molecules 6

PSEB 9th Class Science Solutions Chapter 3 Atoms and Molecules 7

Atomic masses of some common elements in a.m.u. or u.

Element	Symbol	Atomic Mass	Element	Symbol	Atomic Mass
Aluminium	A1	27	Lead	Pb	208
Argon	Ar	40	Lithium	Li	7
Beryllium	Be	9	Magnesium	Mg	24
Boron	B	10.8	Neon	Ne	20
Bromine	Br	80	Mercury	Hg	201
Calcium	Ca	40	Nitrogen	N	14
Carbon	C	12	Oxygen	O	16
Chlorine	Cl	35.5	Phosphorus	P	31
Copper	Cu	63.5	Potassium	K	39
Fluorine	F	19	Silicon	Si	28
Helium	He	4	Silver	Ag	108
Hydrogen	H	1	Sodium	Na	23
Iodine	I	127	Sulphur	S	32
Iron	Fe	56	Zinc	Zn	65

Formulae of a few common compounds

Name of the compound	Formula	Elements Present
Hydrogen	H₂	Hydrogen
Nitrogen	N₂	Nitrogen
Ammonia	NH₃	Nitrogen, Hydrogen
Carbon dioxide	CO₂	Carbon, Oxygen
Water	H₂O	Hydrogen, Oxygen
Sulphur dioxide	SO₂	Sulphur, Oxygen
Nitric acid	HNO₃	Hydrogen, Nitrogen, Oxygen
Hydrochloric acid	HCl	Hydrogen, Chlorine
Sulphuric acid	H₂SO₄	Hydrogen, Sulphur, Oxygen
Calcium carbonate	CaC0₃	Calcium, Carbon, Oxygen
Silver nitrate	AgN0₃	Silver, Nitrogen, Oxygen
Caustic soda	NaOH	Sodium, Oxygen, Hydrogen
Caustic potash	KOH	Potassium, Oxygen, Hydrogen
Baking soda	NaHC0₃	Sodium, Hydrogen, Carbon, Oxygen
Phosphoric acid	H₃PO₄	Hydrogen, Phosphorus, Oxygen
Carbonic add	H₂CO₃	Hydrogen, Carbon, Oxygen
Nitrous acid	HNO₂	Hydrogen, Nitrogen, Oxygen
Marble	CaC0₃	Calcium, Carbon, Oxygen
Phosphine	PH₃	Phosphorus, Hydrogen
Hydrogen sulphide	H₂S	Hydrogen, Sulphur
Urea	NH₂CONH₂	Nitrogen, Hydrogen, Carbon, Oxygen
Butane	C₄H₁₀	Carbon, Hydrogen
Benzene	C₆H₆	Carbon, Hydrogen
Acetic acid	CH₃COOH	Carbon, Hydrogen, Oxygen
Methane	CH₄	Carbon, Hydrogen
Soda ash	Na₂C0₃	Sodium, Carbon, and Oxygen

Symbols or formulae of some common ions or radicals.
(A) Positive ions or Cations or Electropositive radicals:
(a) Cations having 1+ charges:

Name	Symbol or Formula
Hydrogen	H⁺
Lithium	Li⁺
Sodium	Na⁺
Potassium	K⁺
Ammonium	NH₄⁺
Silver	Ag⁺
Cuprous	Cu⁺
Aurous	Au⁺
Mercurous	Hg⁺

(b) Cations having 2+ charges:

Name	Symbol or Formula
Magnesium	Mg²⁺
Barium	Ba²⁺
Calcium	Ca²⁺
Strontium	Sr²⁺
Cobalt	Co²⁺
Nickel	Ni²⁺
Manganous	Mn²⁺
Zinc	Zn²⁺
Cadmium	Cd²⁺
Ferrous	Fe²⁺
Cupric	Cu²⁺
Mercuric	Hg²⁺
Plumbous	Pb²⁺
Stannous	Sn²⁺
Platinous	Pt²⁺

Name	Symbol or Formula
Ferric	Fe³⁺
Chromium	Cr³⁺
Aluminium	Al³⁺
Auric	Au³⁺

(d) Cations having 4+ charges:

Name	Symbol or Formula
Plumbic	Pb⁴⁺
Stannic	Sn⁴⁺
Platinic	Pt⁴⁺
Manganic	Mn⁴⁺

(B) Negative ions or Anions or Electronegative radicals:
(a) Anions having 1- charges

Name	Symbol or Formula
Flydride	H^–
Fluoride	F^–
Chloride	Cl^–
Bromide	Br^–
Iodide	I^–
Cyanide	CN^–
Hypochlorite	CIO^–
Chlorate	ClO₃^–
Perchlorate	ClO₃^–
Bicarbonate	HCO₃^–

Name	Symbol or Formula
Bisulphite	HSO₃^–
Bisulphate	HSO₄^–
Bisphide	HS^–
Hydroxide	OH^–
Meta-aluminate	AlO₂^–
Nitrite	NO₂^–
Nitrate	NO₃^–
Acetate	CH₃COO^–
Permanganate	MnO₄^–
Sulphocyanide	SCN^–

(b) Anions having 2- charges:

Name	Symbol or Formula
Carbonate	CO₃^2-
Sulphite	SO₃^2-
Sulphide	S^2-
Sulphate	SO₄^2-
Zincate	ZnO₂^2-
Oxide	O^2-
Manganate	MnO₄^2-
Chromate	CrO₄^2-
Dichromate	Cr₂O₇^2-
Oxalate	Cr₂O₄^2-
Peroxide	O₂^2-
Silicate	SiO₃^2-

Name	Symbol or Formula
Nitride	N^3-
Phosphide	P^3-
Phosphite	PO₃^3-
Phosphate	PO₄^3-
Arsenite	AsO₃^3-
Arsenate	AsO₄^3-
Ferricyanide	[Fe(CN)₆]^3-
Borate	BO₃^3-

(d) Anions having 4- charges:

Name	Symbol or Formula
Carbide	C^4-
Ferrocyanide	[Fe(CN)₆]^3-
Pyrophosphate	P₂O₇^4-

PSEB 9th Class Science Notes Chapter 2 Is Matter Around Us Pure

November 19, 2021 by Bhagya

This PSEB 9th Class Science Notes Chapter 2 Is Matter Around Us Pure will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 2 Is Matter Around Us Pure

→ A pure substance consists of only one type of particle.

→ Mixtures are constituted by more than one kind of pure form of matter known as a substance.

→ The mixture is obtained by mixing one or more pure elements and/or compounds.

→ A substance cannot be separated into other kinds of matter by any known physical process.

→ Whatever the source of the substance may be, it will always have the same characteristic properties.

→ Homogeneous mixtures may have different separate components.

→ Heterogeneous mixtures can be separated into their respective constituents by simple physical methods.

→ A solution is a homogeneous mixture of two or more substances.

→ Alloy is a solid solution and the air is a gaseous solution.

→ The component of a solution which is generally in small amounts and is dissolved into another component is called the solute.

→ The component of a solution that dissolves the other component is called the solvent.

→ This component is generally present in large amounts.

→ Air is a mixture of gas. A solution of sugar in water is solid in a liquid solution. A solution of iodine in alcohol is known as ‘tincture of iodine.

→ Aerated drinks are gas in liquid solutions.

→ A solution is a homogeneous mixture.

→ The particles of the solution are smaller than 1 nm (10^-9 m) in diameter.

→ Depending upon the amount of solute present in the solution, it can be called a dilute, concentrated or saturated solution.

→ The different substances have different solubilities in the given solvent at the same temperature.

→ Suspension is a heterogeneous mixture and the particles of a suspension can be seen with a naked eye.

→ In the suspension, the solute particles do not dissolve but remain suspended throughout the bulk of the medium.

→ The suspended particles have sizes of more than 100 nm.

→ The particles of a colloid are uniformly spread throughout the solution.

→ The scattering of a beam of light is called the Tyndall effect.

→ The size of colloidal particles lies between 1 to 100 nm and they can’t be seen with the naked eye.

→ The colloidal particles cannot be separated from the mixture by the process of filtration but can be separated by ultracentrifugation.

→ The volatile component of a solution (solvent) can be separated from the non-volatile component (solute) by the method of evaporation.

→ The cream is separated from milk by a centrifugal machine.

→ Ammonium chloride, camphor, naphthalene, and anthracene can be separated by sublimation.

→ The process of separation of coloured components of a mixture is known as chromatography.

→ The crystallization method is used to purify solids.

→ Crystallization technique is better than simple evaporation technique.

→ Colour, hardness, rigidity, fluidity, density, melting point, boiling point, etc. are the physical properties.

→ Chemical change brings a change in the chemical properties of a matter and we get new substances.

→ A chemical change is also called a chemical reaction.

→ Robert Boyle was the first scientist to use the term element in 1661.

→ An element is a basic form of matter that cannot be broken down into simple substances by chemical reactions.

→ Elements can be normally classified into metals, non-metals, and metalloids.

→ Mercury is the only metal that is liquid at room temperature.

→ Metalloids show the properties of metals as well as non-metals.

→ Pure substances can be elements.

→ The process of separation of components of a mixture containing two miscible liquids that boil without decomposition and have sufficient difference in their boiling points.

→ The different gases in the air and different components of petroleum can be separated by fractional distillation.

→ Mixtures are of two types:

Homogeneous mixtures
Heterogeneous mixtures

→ Pure Substance: It is a material containing particles of only one kind having a definite set of properties. Pure substances include elements and compounds.

→ An Element is a pure substance that is made up of only one kind of particle called atoms.

→ It can neither be built up nor broken down into two or simpler substances by any known physical or chemical methods, e.g., copper, silver, etc.

→ A compound is a pure substance that is obtained by the chemical combination of two or more elements in a fixed ratio by mass, e.g., water, ammonia, etc.

→ Mixture: It is a material obtained by mixing two or more substances in any proportion without any chemical change taking place.

→ The solution is a homogeneous mixture of two or more substances whose composition can be changed within certain fixed limits.

→ A binary solution is a solution having two components.

→ The solute is the minor component of a solution whereas solvent is the major component of a solution.

→ The concentration of a solution is the amount of solute present per unit volume or per unit mean of solvent or solution.

→ A saturated solution is one that does not dissolve any more of the solute at a given temperature and pressure.

→ Colloids are heterogeneous mixtures in which the particles have a size of more than 100 nm.

→ These particles are called colloidal particles and constitute the dispersed phase whereas the medium in which colloidal particles are dispersed constitutes the dispersion medium.

→ Suspensions: Materials that are insoluble in a solvent and have particles that are visible to naked eyes form suspensions.

→ Physical Change: It is a temporary change in which only the physical properties of substances change and can be reversed.

→ Chemical Change: It is a permanent change in which the chemical properties of substances change and there is a change in composition and cannot be reversed.

→ Filtration: The process of separation of an insoluble solid component of a mixture from a liquid component is called filtration.

→ Evaporation: It is the slow process of conversion of liquid into a gaseous state (vapour) at a temperature below its boiling point.

→ Distillation: It is the process of conversion of liquid into a gaseous state by heating it to the boiling point and condensing the vapour to get pure liquid.

→ Fractional distillation: It is the process of separating two miscible liquids having different boiling points by distillation using a fractionating column.

→ Chromatography: It is the process of separation of dissolved components of a mixture by adsorbing on a suitable substance (called adsorbent).

PSEB 9th Class Science Notes Chapter 2 Is Matter Around Us Pure 1

PSEB 9th Class Science Notes Chapter 2 Is Matter Around Us Pure 2

PSEB 9th Class Science Notes Chapter 1 Matter in Our Surroundings

November 19, 2021 by Bhagya

This PSEB 9th Class Science Notes Chapter 1 Matter in Our Surroundings will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 1 Matter in Our Surroundings

→ Matter: It is anything that occupies space, has mass, and can be judged by any one or more of the known five physical senses.

→ Early Indian philosophers classified matter into five basic elements called Panch Tatva. These are air, earth, fire, sky, and water.

→ Modern-day scientists have classified matter on the basis of its physical properties and chemical nature.

→ The matter is made up of extremely small particles and there are vacant spaces in them.

→ Particles of matter are always in motion and have kinetic energy.

→ The speed of particles increases with the increase in temperature.

→ The mixing up of particles of different substances is called diffusion.

→ All matter is made up of a large number of extremely small particles called molecules.

→ The material represents a particular kind of matter.

→ Material can be homogeneous or heterogeneous.

→ Homogeneous material is a material that has a uniform composition throughout. It consists of a single phase.

→ Heterogeneous material is a material that does not have a uniform composition throughout. It consists of two or more phases.

→ The substance is a homogeneous material that is made up of only one kind of atom or material. (Substance always refers to pure substance).

→ Based upon its physical state, there are three states of matter, i.e. solid, liquid, and gas.

→ There are two new states of matter. These are plasma and Bose-Einstein condensate.

→ Solid: It is that state of matter which has a definite mass, volume, and shape.

→ Liquid: It is that state of matter which has a definite mass and volume but has no definite shape.

→ Gas: It is that state of matter which has a definite mass but has neither definite shape nor definite volume.

→ The forces of attraction between the particles are maximum in solids, intermediate in liquids, and maximum in gases.

→ The vacant spaces between the constituent particles and kinetic energy of particles are minimum in the case of solids, intermediate in liquids, and maximum in gases.

→ The arrangement of constituent particles is most ordered in the case of solids, in the case of liquids the layers can slip over each other while in the case of gases, there is no order, particles can move randomly.

→ The different states of matter are interconverted by changing temperature or pressure or both.

→ Melting point is the temperature at which solid changes into a liquid state.

→ The boiling point is the temperature at which a liquid changes into vapour under atmospheric pressure. At boiling point, vapour pressure of the liquid is equal to atmospheric pressure.

→ Boiling is a bulk phenomenon as it involves the whole of the liquid.

→ The rate of evaporation depends upon the surface area, temperature, humidity, and wind speed.

→ The slow passing out of molecules of a liquid into a gaseous state at a temperature below its boiling point is called evaporation.

→ Evaporation causes cooling. It is a surface phenomenon.

→ Boiling is a fast process whereas evaporation is a slow process.

→ Sublimation is the process due to which a solid directly changes into the gaseous state on heating and a gaseous state directly changes into a solid-state on cooling without changing into the liquid state.

→ Vapour is a substance that exists in the gaseous state at a temperature lower than the boiling point of its liquid state.

PSEB 9th Class Science Notes Chapter 1 Matter in Our Surroundings

→ Temperature on Kelvin scale = 273 + Temperature on centigrade scale. (T K = 273 + t°C) Water freezes at 0°C or 273 K.

→ Water boils at 100°C or 373 K.

→ Units of Latent heat of fusion are kJ/kg or kcal/kg

→ Units of Latent heat of vaporisation are kJ/kg or kcal/kg

→ \(\frac{F-32}{9}=\frac{C}{5}\)
F = Temperature on Fahrenhiet scale.
C = Temperature on Centigrade scale.

Some measurable quantities and their units are:

Quantity	Unit	Symbol
Temperature	kelvin	K
Length	metre	m
Mass	kilogram	kg
Weight	newton	N
Volume	cubic metre	m3
Density	kilogram per cubic metre	Kg m^-3
Pressure	pascal	Pa

→ Matter: It is anything that occupies space, has mass, and can be judged by any one or more of the known five physical senses.

→ Panch Tatva: Indian philosophers classified matter into our basic elements called Panch Tatva. These are air, earth, fire, sky, and water.

→ Diffusion: It is the property of the mixing of particles of two or more substances.

→ Solid: It is that state of matter which has a definite mass, volume, and shape.

→ Liquid: It is that state of matter which has a definite mass and volume but has no definite shape.

→ Gas: It is that state of matter which has a definite mass but has neither definite shape nor definite volume.

→ Density: It is the mass per unit volume of a substance.

→ Melting point: It is the temperature at which solid changes into the liquid state.

→ Fusion: It is the process of conversion of a solid into a liquid.

→ Solidification or freezing: It is the process of conversion of a liquid into a solid.

→ Sublimation: It is the process due to which a solid directly changes into the gaseous state on heating and a gaseous state directly changes into the solid state on cooling without changing into a liquid state.

→ Latent heat of fusion is the amount of heat energy required to change 1 kg of solid into liquid at its melting point.

→ Latent heat of vaporisation is the amount of heat energy required to change 1 kg of a liquid to gas at atmospheric pressure and at its boiling point.

→ Boiling is a bulk phenomenon as it involves the whole of the liquid.

→ Boiling point: It is the temperature at which a liquid changes into vapour under atmospheric pressure.

→ Evaporation: The slow passing out of molecules of a liquid into a gaseous state at a temperature below its boiling point.

→ Dry Ice: Solid carbon dioxide is called dry ice.

→ Freezing point: It is the temperature at which a liquid substance changes into a solid substance.

→ Liquified Petroleum Gas (LPG): When butane is subjected to the action of high pressure, it is liquid to give LPG which is used as a fuel.

→ Compressed Natural Gas (CNG): When natural gas is subjected to the action of high pressure it gets liquified to give CNG which is used as a fuel for automobiles.

→ Vapour: It is a substance that exists in the gaseous state at a temperature lower than the boiling point of its liquid state.

PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1

November 18, 2021 by Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.1

1. In ΔABC, P is midpoint of BC, then
(i) BP = ……………..
(ii) AP is a …………….. of ΔABC
(iii) ∠ADC = ……………..
(iv) BD = BC (True/False)
(v) AD is an …………….. of ΔABC
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 1
Solution:
(i) PC
(ii) Median
(iii) 90°
(iv) False
(v) Altitude

2. (a) Draw AD, BE, CF three medians in a ΔABC.
(b) Draw an equilateral triangle and its medians. Also compare the lengths of the medians.
(c) Draw an isosceles triangle ABC in which AB = BC. Also draw its altitudes.
Solutions:
(a) We are given ΔABC D, E and F are mid points of the sides BC, CA and AB respectively. Join AD, BE and CF.
The AD, BE and CF are the required medians.
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 2

(b) Draw an equilateral triangle ABC, D,E and F are the mid points of sides BC, CA and AB respectively. On joining AD, BE and CF, we get the required medians AD, BE and CF. Measure the lengths of AD, BE and CF we observe that the three medians AD, BE and CF are equal in length.
∴ AD = BE = CF.
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 3

(c) Draw an isosceles ΔABC in which AB = BC Altitude can be drawn as below :
AD is the altitude from A to D.
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 4

3. Find the value of the unknown exterior angles.

Question (i).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 5
Answer:
In the given triangle,
By exterior angle property of a triangle
Exterior angle = sum of interior opposed angles
x = 100° + 40°
∴ x = 140°

Question (ii).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 6
Answer:
By exterior angle property of a triangle
Exterior angle = Sum of interior opp. angles
x = 20° + 30°
∴ x = 50°

Question (iii).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 7
Answer:
By exterior angle property of a triangle
Exterior angle = Sum of interior opp. angles
x = 60° + 60°
∴ x = 120°

Question (iv).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 8
Answer:
By exterior angle property of a triangle
Exterior angle = Sum of interior opp. angles
x = 90° + 30°
∴ x = 120°

4. Find the value of x in the following figures.

Question (i).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 9
Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
40° + x = 120°
x = 120° – 40°
x = 80°.

Question (ii).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 10
Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
x + 90° = 135°
x = 135° – 90°
x = 45°

Question (iii).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 11
Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
x + 80° = 130°
x = 130° – 80°
x = 50°

Question (iv).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 12
Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
x + 25° = 155°
x = 155° – 25°
x = 130°

5. Find the value of y in following figures.

Question (i).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 13
Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
y + y = 140°
2y = 140°
y = \(\frac{140^{\circ}}{2}\)
y = 70°

Question (ii).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 14
Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
y + 90° = 160°
y = 160° – 90°
y = 70°

Question (iii).
PSEB 7th Class Maths Solutions Chapter 6 Triangles Ex 6.1 15
Answer:
By exterior angle property of a triangle
exterior angle = Sum of interior opp. angles
5y = y + 80°
5y – y = 80°
4y = 80°
y = \(\frac{80^{\circ}}{4}\)
y = 20°.

PSEB 9th Class Science Solutions Chapter 12 Sound

November 19, 2021November 18, 2021 by Bhagya

Punjab State Board PSEB 9th Class Science Book Solutions Chapter 12 Sound Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Science Chapter 12 Sound

PSEB 9th Class Science Guide Sound Textbook Questions and Answers

Question 1.
What is sound and how is it produced?
Answer:
Sound: Sound is a form of energy which produces in our ears the sensation of hearing. It is produced due to vibration of a body about its mean position.

How to produce sound? We can produce sound in different bodies by plucking, by rubbing, by blowing or by giving jolt. In other words, by producing vibration in bodies sound can be produced. By vibration we mean moving a body rapidly to and fro about its mean position.

Question 2.
Describe with the help of diagram, how compressions and rarefactions are produced in air near a source of sound?
Answer:
Sound in air gets propagated in the form of longitudinal wave motion consisting of regions of compressions and rarefactions. Consider, for example, a tuning, fork in a state of vibrations. [Fig.(a)] As prong moves towards right, it compresses the layer of air in contact with it. As air has elasticity, the compressed air tends to relieve itself of its strain and moves forward to right to compress the next layer and so on.

Thus, a wave of compression moves towards the right. At the point of compression, there is an increase of pressure and is shown in form of crest C. At the point of rarefaction of concentration of particles is least and has been shown as trough R.

When the prong moves towards left, a region of reduced pressure or rarefaction is produced towards right [Fig. (b)].
PSEB 9th Class Science Solutions Chapter 12 Sound 1
Thus, a wave of rarefaction starts moving towards right. This way a series of compressions arid rarefactions move in forward direction.

Question 3.
Cite an experiment to show that sound needs a material medium for its propagation.
Or
Describe an experiment to show that sound needs a material medium for its propagation.
Or
Describe an activity to show that sound is a mechanical wave and needs a material medium for its propagation.
Answer:
Sound needs material medium for propagation: Sound is a mechanical wave which needs a material medium to travel, (propagate) It can travel through air, water, steel, etc but cannot travel through vacuum. This can be demonstrated by the following experiment.
PSEB 9th Class Science Solutions Chapter 12 Sound 2
Experiment: Take an electric bell and a glass bell-jar. Suspend the electric bell in a bell jar with the help of a cork fitted in the mouth of the jar. Connect the bell jar to a vacuum pump as shown in Fig. Press the switch of electric bell when sound is heard.

Now work the exhaust pump and remove air from the jar slowly. As air is removed the sound becomes fainter and fainter. After sometime when most of the air is removed, a feeble sound will be heard. If the whole of the air from the jar is removed no sound of electric bell will be heard. This proves that material medium is needed for the propagation of sound.

Question 4.
Why is sound wave called longitudinal wave?
Answer:
Sound waves when travel through a medium, the particles of the medium move to and fro in the same direction in which the disturbance (wave) travels. That is why, the sound waves are called longitudinal waves.

Question 5.
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room.
Answer:
On the basis of quality or timbre of sound, we can identify our friend’s voice.

Question 6.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?
Answer:
The speed of sound (344 m s^-1) is much smaller than the speed of light (3 × 10⁸ m s^-1). So thunder is heard a few seconds after the flash is seen although these are produced at the same time.

Question 7.
A person has a hearing range from 20 Hz to 20 KHz. What is the typical wavelength of sound waves in air corresponding to these frequencies? Take the speed of sound in air as 344 ms^-1.
Solution:
Given speed of sound (υ) = 344 m s^-1
Lower limit of frequency (ν₁) = 20 Hz
Upper limit of audible frequency (ν₂) = 20 KHz
= 20 × 1000 Hz.
= 20000 Hz
PSEB 9th Class Science Solutions Chapter 12 Sound 3

Question 8.
Two children are at opposite er Is of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Solution:
PSEB 9th Class Science Solutions Chapter 12 Sound 4

Question 9.
The frequency of source of sound is 100 Hz. How many times does it vibrate in a minute?
Solution:
Frequency of source sound = 100 Hz
i.e, Number of vibrations produced in 1 second = 100
Number of vibrations produced in 1 min = 60 s = 100 × 60 = 6000

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Answer:
PSEB 9th Class Science Solutions Chapter 12 Sound 6
Yes, sound follows the same laws of reflection as light does. Like light sound is reflected from solid or liquid surface.
These laws are:

First Law: The directions of incident sound and reflected sound make equal angles with the normal to the surface at the point of incidence.
i.e. \(\angle i=\angle r\)
Second Law: The incident sound wave, the reflected sound wave and normal to the reflecting surface at the point of incidence all lie in the same plane.

Question 11.
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
Answer:
On a hotter day the speed of sound increases with the increase of temperature. So on that day reflected sound returns tq source earlier than 0.1 s. Hence a clear echo sound can not be heard.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
PSEB 9th Class Science Solutions Chapter 12 Sound 7
Stethoscope is doctor’s device which is used to hear the sound produced inside heart or lungs. The speed of sound of a patient’s heart beat is guided along the tube to the doctor’s ears by multiple reflection of sound.

2. The front part of musical instruments like megaphone or loudspeaker, horn, shehnai is made open and conical so that the sound waves produced may be reflected repeatedly and may be reflected repeatedly and may be sent forward towards the listeners.
PSEB 9th Class Science Solutions Chapter 12 Sound 8

Question 13.
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s^-2 and speed of sound = 340 m s^-2.
Solution:
Here, initial velocity of sound (u) = 0
Height of the lower (i.e. distance covered) (S) = 500 m
Acceleration due to gravity (g) = 10 ms^-2
PSEB 9th Class Science Solutions Chapter 12 Sound 9

Question 14.
A sound wave travels at a speed of 339 m s^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Solution:
Speed of sound (υ) = 339 ms^-1
Wavelength of sound (λ) = 1.5 cm
= \(\frac {1.5}{100}\)
= 0.015 m
Frequency of wave (ν) =?
We know, frequency (ν) = \(\frac{υ}{\lambda}\)
= \(\frac {339}{0.015}\)
= 22600 Hz
Yes, sound waves are inaudible because these have frequency 22600 Hz which is not within the audible range 20 Hz to 20,000 Hz.

Question 15.
What is reverberation? How can it be reduced?
Answer:
Reverberation: The persistance of sound due to repeated reflection of sound is called reverberation. If sound after its production is allowed to suffer repeated reflection from walls and ceiling of big halls of concert so that it persists is called reverberation. It is unwanted sound because of which sound is not distinctly heard. To reduce reverberation effect of sound, walls and ceiling should be covered with sound absorbing materials like compressed fiber or heavy curtains having folds etc.

Question 16.
What is loudness of sound? What factors does it depend on?
Answer:
Loudness of sound is the measure of sensitivity of human ears. Like intensity it is not sound energy passing through unit area in 1 second. Two sounds can have same frequencies but still these may be heard having different loudness.

Question 17.
Explain how bats use ultrasounds to catch a prey?
Answer:
PSEB 9th Class Science Solutions Chapter 12 Sound 10
In darkness bats while flying in search of their prey emit ultrasound waves and then detect these waves after reflection. Very high-frequency ultrasonic squeaks of bat are reflected from prey and returned to bat’s ear. Amount and time delay of reflected wave helps bat in estimating the position and distance of prey.

Question 18.
How is ultrasound used for cleaning?
Answer:
Ultrasounds are used to clean parts located in hard-to-reach places e.g., complicated electronic components, watches, spiral or odd shaped parts. Appliances to be cleaned are placed in cleaning solutions and ultrasonic waves are sent through cleaning solution. Due to high frequency of ultrasounds, the dust, oil, grease and dirt get detached.

Question 19.
Explain the working and applications of SONAR.
Or
Write the full name of SONAR. How will you determine the depth of a sea using echo ranging?
Or
Write full form of SONAR. List any two purposes for which, it is used and explain its working for any one such purpose.
Answer:
SONAR: The acronym Sonar stands for Sound Navigation and Ranging. Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects.
Principle: It uses the phenomenon of echoes in determining the sea-depth and locating the presence of underwater objects.

Working: Sonar consists of a transmitter T, and a detector D, installed below a ship as shown in Fig. The wave produced by transmitter travel through water and are reflected by sea-bed of obstacle. Reflected waves are sensed by the detector. Detector converts ultrasonic waves into electric signal. These signals are interpreted by detector.
PSEB 9th Class Science Solutions Chapter 12 Sound 11
If time interval between transmission and reception is t and speed of sound in sea-water is v, then 2d = υ × t or d = υt/2, where d is the depth of the sea. This method is also called echo sounding.
Practical Applications: It is used to locate underwater submarines, icebergs, sunken ships and underwater hills etc.

Question 20.
A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from submarine is 3,625 m.
Solution:
Time between emission of sound and its collection (t) = 5 s
Depth of sea (2d) = 2 × 3625 m = 7250 m
We know, 2d = Speed of sound × Time
7250 = Speed of sound × 5
∴ Speed of sound (υ) = \(\frac {7250}{5}\)
= 1450 m s^-1

Question 21.
Explain how defects in a metal block can be detected using ultrasound?
Answer:
In industries metallic components are used in the construction of big structures like buildings, bridges, machines, scientific equipments, etc. ultrasounds (ultrasonic waves) are used to detect the internal defects or cracks in big metallic blocks which are not visible from outside.

Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back and does not reach the detector, as shown in figure. This indicates the presence of a defect.
PSEB 9th Class Science Solutions Chapter 12 Sound 12

Question 22.
Explain how the human ear works.
Answer:
Ear is very sensitive device used to hear sound. It converts compressions and rarefactions of frequency range 20 Hz to 20,000 Hz into electric signals that travel to brain via auditory nerve.

The ear consists of three sections:

the outer ear,
the middle ear and
the inner ear.

The outer ear consists of Pinna and Auditory Canal. Pinna is a cup-shaped fleshy part of the outer ear. Pinna collects and amplifies sound waves which then pass on the auditory canal. At the end of auditory canal, there is a thin membrane called tympanic membrane or eardrum. When compression reaches the eardrum, the pressure on membrane increases and the ear drum is forced inwards. When rarefaction reaches the eardrum, it moves outwards.

The vibrations are amplified by lever action of three bones called hammer, anvil and stirrup in the middle ear. In turn, the middle ear transmits the amplified pressure variations to the inner ear. The amplified pressure variations are converted into electric signals by cochlea in the inner ear. The electric signals generated are conveyed to the brain
PSEB 9th Class Science Solutions Chapter 12 Sound 13
via the auditory nerve. The brain interprets them as sound. In fact we do not hear with ear. We hear with brain through ears.

Science Guide for Class 9 PSEB Sound InText Questions and Answers

Question 1.
How does the sound produced by the vibrating object in a medium reach your ear?
Answer:
When the vibrating object (such as tuning fork or school bell) moves forward then it compresses the air particles lying just ahead of it which results in production of high pressure region. This region is called compression. This pressure moves forward in the direction in which the object is vibrating, when this vibrat ing object moves backward then a region of low pressure is produced which is known as rarefaction.
PSEB 9th Class Science Solutions Chapter 12 Sound 14
When the vibrating object rapidly moves to and fro then a series of compression and rarefaction pulses are formed i.e. sound wave is produced.
In this way the transmission of sound is caused in the form of transmission of change in density which reaches our ear and forces the tympanic membrane to vibrate. This produces the sensation of hearing in us.

Question 2.
Explain how sound is produced by your school bell?
Answer:
When the school bell is hit with a hammer, it begins to vibrate which produces sound waves. If we gently touch the bell, we feel vibrations. Wave is a disturbance which produces motion in the neighbouring particles of the medium. These particles handover the disturbance in the next particles lying close to the vibrating particles so that sound waves reach us. The particles of the medium do not move from one place to another, it is only the disturbance that travels forward.

Question 3.
Why are sound waves called mechanical waves?
Answer:
Sound is a kind of energy which cannot be produced by itself. To produce it some mechanical enery is required which may be by clapping or by striking bell with a hammer. This sound energy is transmitted in the form of waves by producing disturbance of the particles of the medium. Therefore, sound waves are called mechanical waves.

Question 4.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
For propagation of sound, air or some other material medium is required. On the surface of moon there is no such medium present as a result of which sound can not be propagated from one place to another place. So you can neither talk to your friend nor the sound produced by your friend can be heard by you.

Question 5.
Which wave property determines
(a) loudness
(b) pitch?
Answer:
(a) Loudness: The loudness of a sound wave is determined by its amplitude. The amplitude of sound wave depends upon magnitude of force. The more the force, the loud is sound produced. Loud sound traverses more distance because it has more energy in it. The more the sound is away from the source, the less is its loudness. Hence, loudness depends upon square of the amplitude.

(b) Pitch: The frequency of sound produced is called pitch. Frequency determines the aitch of a sound. The more is the vibration of the source, the higher will be its pitch.
PSEB 9th Class Science Solutions Chapter 12 Sound 15
So more the frequency, higher the pitch of sound.
In sound of high pitch the number of compressions passing through a fixed point in a unit time will be more.

Question 6.
Guess, which sound has a higher pitch: guitar or a car horn?
Answer:
Though sound of car horn is louder than that of guitar but guitar has higher pitch than car horn.

Question 7.
What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:
1. Wavelength of wave: The distance travelled by the wave during the time, the particle of the medium completes 1 vibration is called wavelength.
Or
The distance between two consecutive compressions or rarefactions in a longitudinal wave or the distance between the consecutive crests or two consecutive troughs is called wavelength. It is denoted by a greek letter ‘λ’ (lambda). SI unit of wavelength is meter (m).
PSEB 9th Class Science Solutions Chapter 12 Sound 16
2. Frequency: In any medium when wave propagates the number of vibrations made by a particle of the medium is called frequency. It is denoted by ‘ν’. S.I. unit of frequency is Hertz (Hz). It is determined by the number of compressions or rarefactions passing through a fixed point.

3. Time Period: It is the time taken by a particle to complete one vibration during the propagation of wave. It is denoted by “T”. S.I. unit of time period is second.
Or
Time taken by two nearest compressions or rarefactions of sound waves to cross a point is called time period.

4. Amplitude: The maximum displacement of a particle of the medium on either side of mean position is called amplitude, of wave. It is denoted by letter ‘A’. For sound wave its unit is same as that of pressure or density. The loudness of sound depends on its amplitude.

Question 8.
How are the wavelength and frequency of a sound wave related to its speed?
Answer:
Velocity of sound wave (υ) = Wavelength (λ) × Frequency (ν).

Question 9.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m s^-1 in a given medium.
Solution:
Velocity of sound wave (υ) = 440 ms^-1
Frequency of sound (ν) = 220 Hz
Wavelength of sound wave (λ) = ?
We know, υ = ν × λ
440 = 220 × λ
or λ = \(\frac {440}{220}\)
or wavelength (λ) = 2 m

Question 10.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of sound. What is the time interval between successive compressions from the source?
Solution:
Given frequency of sound (ν) = 500 Hz
Time taken between two successive compressions (T) = ?
We know, time period (T) = \(\frac{1}{\text { Frequency }(ν)}\)
= \(\frac {1}{500}\)
= 0.002 s

Question 11.
Distinguish between loudness and intensity of sound.
Answer:
Difference between Loudness and Intensity:

Loudness	Intensity
1. The loudness of sound is the measure of senstivity of ears.	It is the sound energy passing through a unit area in 1 second.
2. The loudness of sound can not be measured.	The intensity of sound can be measured.
3. For different observers the loudness of sound is different.	The intensity of sound is same for different persons.
4. The loudness of ultrasonic and infrasonic waves is zero because they are inaudible.	There is a possibility of intensity in ultrasonic and infrasonic sound in ultrasonic and infrasonic sound waves.

Question 12.
In which of the three media, air, water or iron sound travels the fastest at a particular temperature?
Answer:
Sound travels fastest in iron as compared to air and water. The velocity of sound in iron is 5950 m s^-1, followed by water [1500 m s^-1], air [350 m s^-1],

Question 13.
An echo returned in 3 s. What is the distance of reflecting surface from the source? Given that the speed of sound is 342 m s^-1.
Solution:
Velocity of sound (v) = 342 ms^-1
Time taken for echo to be heard (t) = 3s
Distance travelled by sound (S) = υ × t
= 342 × 3
= 1026 m
i.e. sound takes 3 s to travel from source to reflecting surface and then back to source and during this time the distance travelled is 1026 m.
Distance between source and reflecting surface = \(\frac {1026}{2}\) m
= 513 m

Question 14.
Why are the ceilings of concert halls curved?
Answer:
PSEB 9th Class Science Solutions Chapter 12 Sound 17
Ceilings of concert halls are made curved as is shown in Fig so that sound after reflection from all surfaces of hall may spread evenly to all parts and heard equally clear.

Question 15.
What is audible range of average human ear?
Answer:
For average human ear the audible range of sound is 20 Hz to 20,000 Hz.

Question 16.
What is the range of frequencies associated with
(a) infrasound
(b) ultrasound?
Answer:
(a) For infrasound the frequency range is less than 20 Hz.
(b) For ultrasound the frequency range is more than 20 KHz (i.e. 20,000 Hz)

Question 17.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m s^-1, how far away is the cliff?
Solution:
Time is taken by the sound to travel from submarine to cliff and back to the submarine
= 1.02 s
Speed of sound in saltwater = 1531 ms^-1
Distance travelled by sound (2d) = Speed of sound × Time taken
= 1531 × 1.02 [∵ d is the distance between submarine and cliff]
= 1561.62 m
or d = \(\frac {1561.62}{2}\) m
i.e. Distance between submarine and cliff (d) = 780.81 m

PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति

November 18, 2021 by Bhagya

Punjab State Board PSEB 10th Class Hindi Book Solutions Hindi Grammar prapatra poorti प्रपत्र पूर्ति Exercise Questions and Answers, Notes.

PSEB 10th Class Hindi Grammar प्रपत्र पूर्ति

प्रपत्र पाठ्यक्रम-बैंक, डाकघर तथा रेलवे से संबंधित प्रपत्र पूर्ति
‘प्रपत्र’ अंग्रेज़ी शब्द फार्म का अनुवाद है। आम बोलचाल की भाषा में प्रपत्र को फार्म भी कहते हैं। प्रपत्र शब्द पत्र के आगे ‘प्र’ उपसर्ग लगाने से बना है। बालक के जन्म से लेकर मृत्यु तक के प्रपत्र भरकर बालक का जन्म प्रमाण-पत्र तथा मृतक का मृत्यु प्रमाण-पत्र प्राप्त करना होता है। इसके अतिरिक्त स्कूल, कॉलेज में प्रवेश के लिए, परीक्षा देने के लिए, नौकरी के लिए, प्रतियोगी परीक्षाओं के लिए, राशन कार्ड, आधार कार्ड, बिजली-पानीटेलीफोन-मोबाइल, गैस कनेक्शन लेने के लिए, पासपोर्ट बनवाने, बैंक/डाकखाने में पैसे जमा कराने या निकालने के लिए, रेल-हवाई यात्रा के समय आरक्षण कराने के लिए, विवाह के पंजीकरण, वाहन पंजीकरण, व्यवसाय पंजीकरण आदि अनेक कार्यों के लिए भी प्रपत्र भरने होते हैं। कोई भी प्रपत्र भरते समय निम्नलिखित तथ्यों का ध्यान रखना चाहिए-

(क) प्रपत्र के साथ दिए गए निर्देशों को अच्छी प्रकार से पढ़ कर ही प्रपत्र भरना चाहिए।
(ख) प्रपत्र में दिए गए निर्देश के अनुसार ही प्रपत्र को पैन/बॉलपैन/पेंसिल से भरना चाहिए।
(ग) प्रपत्र में जिन स्तंभों (कॉलमों) को अंग्रेज़ी के बड़े (कैपिटल) अक्षरों में भरना हो, उन्हें बड़े अक्षरों में भरना चाहिए।
(घ) प्रपत्र पर अपना छायाचित्र (फोटो) उचित स्थान पर चिपकाएं या पिन से नत्थी कीजिए। यदि फ़ोटो को सत्यापित (अटैस्ट) कराना है अथवा अपने हस्ताक्षर करने हैं, तो वह भी कीजिए।
(ङ) प्रपत्र के साथ निर्देशानुसार स्वहस्ताक्षरित अथवा राजपत्रित अधिकारी द्वारा सत्यापित (अटैस्टिड) दस्तावेज़ लगाएँ।
(च) प्रपत्र का कोई भी स्तम्भ (कॉलम) रिक्त न छोड़ें। यदि वह स्तम्भ आप पर लागू नहीं होता तो वहां लागू नहीं लिख दीजिए।
(छ) प्रपत्र में दिए गए स्थान पर अपने हस्ताक्षर कीजिए तथा स्थान और दिनांक लिखिए।
(ज) प्रपत्र साफ़, स्पष्ट, सुंदर तथा पढ़ा जा सके-ऐसे अक्षरों में भरा जाना चाहिए।
(झ) प्रपत्र भरकर व्यक्तिगत रूप से जमा कराते समय उसकी जमा करने वाले कार्यालय से रसीद ले लें। डाक से रजिस्ट्री अथवा स्पीड-पोस्ट से प्रपत्र भरकर भेज सकते हैं। यहाँ कुछ प्रपत्रों को भरने के उदाहरण दिए जा रहे हैं।

1. उदाहरण-किसी खेल में चयन के लिए दिए गए प्रपत्र को भरना-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 1

2. उदाहरण-नौकरी के लिए दिए गए प्रपत्र को भरना
(क) व्यक्तिगत तथा पारिवारिक विवरण
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 2

(ग) व्यावसायिक अनुभव (यदि कोई हो)
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 3

दसवीं के पाठ्यक्रम में बैंक, डाकघर तथा रेलवे से संबंधित प्रपत्रों की पूर्ति निर्धारित है। इसलिए यहाँ इन्हीं तीनों विभागों से संबंधित प्रपत्रों पर विस्तार से चर्चा की जा रही है-

(क) बैंक से संबंधित प्रपत्र
आधुनिक युग में बैंकों का हमारे दैनिक जीवन में बहुत महत्त्व है। इससे हमारे लेन-देन तथा व्यवसाय में बहुत सहायता मिलती है। बैंक में खाता खुलवाने के लिए कुछ प्रमाण-पत्रों की आवश्यकता होती है, जो निम्नलिखित हैं-

आवासीय पते का प्रमाण-पत्र; जैसे-राशन कार्ड, आधार कार्ड, ड्राइविंग लाइसेंस, पासपोर्ट, वरिष्ठ नागरिक प्रमाण-पत्र आदि।
जिस बैंक में खाता खुलवाना हो वहाँ के किसी खाताधारी से गवाही।
फ़ोटो।
पैनकार्ड की प्रतिलिपि।
जन-धन योजना में शून्य राशि से खाता खुल सकता है।
विद्यार्थियों को अपने विद्यालय के पहचान-पत्र तथा प्राचार्य के हस्ताक्षरित खाता खोलने के फार्म के आधार पर बैंक में शून्य राशि से खाता खोलने की सुविधा है।
बैंक में खाता खोलने वाले को बैंक में अपने नमूने के हस्ताक्षर भी देने पड़ते हैं, जिससे कोई अन्य व्यक्ति उसके खाते से रुपए नहीं निकाल सके।
बैंक में खाता खोलने के बाद मिलने वाले ए०टी०एम० कार्ड से खाता धारक किसी भी ए०टी०एम० बूथ से कभी भी पैसे निकलवा सकता है तथा खरीददारी भी कर सकता है।

बैंक में खातों के प्रकार-बैंक में खातों के अनेक प्रकार होते हैं; जैसे-बचत खाता, चालू खाता, आवर्ती खाता, सावधि जमा खाता आदि। विद्यार्थियों को बचत खाता खुलवाना चाहिए, जिसमें उनकी छात्रवृत्ति, अपनी बचत आदि जमा हो सकती है। बचत खाता, आवर्ती खाता तथा सावधि खाता पर बैंक जमा राशि पर नियमानुसार ब्याज भी देते हैं, जो खाताधारक के खाते में जमा होता रहता है। बैंक के लेन-देन में मुख्य रूप से निम्नलिखित प्रपत्र प्रयोग में आते हैं-

बैंक में खाता खोलने का प्रपत्र
रुपए नकद जमा करने का प्रपत्र
राशि चैक द्वारा जमा करने का प्रपत्र
रुपए निकलवाने का प्रपत्र/चैक
ड्राफ्ट बनवाने का प्रपत्र।

यहां प्रमुख प्रपत्रों की रूपरेखा तथा उनके भरने के उदाहरण दिए जा रहे हैं-
1. बैंक में खाता खोलने का प्रपत्र

बैंक में खाता खोलने के प्रपत्र के प्रत्येक कॉलम को ध्यान से पढ़कर साफ़-साफ़ शब्दों में भरिए। खाता एक व्यक्ति अकेले या दो-तीन व्यक्ति मिलकर भी खोल सकते हैं तथा इस संबंध में परिचालन का विकल्प चुन सकते हैं। किसी खाता धारक से परिचय भी दिया जाता है तथा नामांकन का पूरा विवरण भी भरिए। नामिती नाबालिग हो तो उसके संरक्षक का नाम भी लिखना होता है। खाता खोलने हेतु आवेदन फार्म का एक उदाहरण यहां दिया जा रहा है-
1. खाता खोलने का प्रपत्र
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 4

प्रश्न 1.
विद्यार्थियों को बैंक में खाता खुलवाने के लिए किन-किन दस्तावेज़ों की आवश्यकता होती है?
उत्तर:
आधुनिक युग में बैंकों का हमारे दैनिक जीवन में बहुत महत्त्व है। इससे हमारे लेन-देन तथा व्यवसाय में बहुत सहायता मिलती है। बैंक में खाता खुलवाने के लिए कुछ प्रमाण-पत्रों की आवश्यकता होती है, जो अग्रलिखित हैं-

आवासीय पते का प्रमाण-पत्र ; जैसे-राशन कार्ड, आधार कार्ड, ड्राइविंग लाइसेंस, पासपोर्ट वरिष्ठ नागरिक प्रमाण-पत्र आदि।
जिस बैंक में खाता खुलवाना हो वहाँ के किसी खाताधारी से गवाही।
फ़ोटो।
पैनकार्ड की प्रतिलिपि।
जन-धन योजना में शून्यराशि से खाता खुल सकता है।
विद्यार्थियों को अपने विद्यालय के पहचान-पत्र तथा प्राचार्य के हस्ताक्षरित खाता खोलने के फार्म के आधार पर बैंक में शून्य राशि से खाता खोलने की सुविधा है।
बैंक में खाता खोलने वाले को बैंक में अपने नमूने के हस्ताक्षर भी देने पड़ते हैं, जिससे कोई अन्य व्यक्ति उसके ख़ाते से रुपए नहीं निकाल सके।

2. बैंक में पैसे जमा कराने का प्रपत्र
बैंक में नकद राशि जमा करने के लिए जो प्रपत्र भरना होता है, उसके दो हिस्से होते हैं। एक बैंक की प्रति तथा दूसरी ग्राहक की प्रति होती है। इन दोनों हिस्सों को भरना होता है, जिसमें दिनांक, खाता संख्या, खाताधारी का नाम, जमा राशि तथा जमा राशि का विवरण कि किस-किस राशि के कितने नोट हैं लिखना होता है। दिए गए स्थान पर मोबाइल नंबर लिखकर जमाकर्ता के हस्ताक्षर होते हैं। यहाँ एक उदाहरण दिया जा रहा है जिसमें विजय सिंह और सुरजीत कौर के खाते में एक हज़ार रुपए नकद जमा करने के लिए प्रपत्र भरा गया है-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 6

3. बैंक में चैक द्वारा राशि जमा कराने का प्रपत्र बैंक में चैक जमा कराने का प्रपत्र भी दो भागों में होता है। एक प्रति बैंक की तथा दूसरी ग्राहक की होती है। इसमें दिनांक, खाते का प्रकार तथा संख्या, खातेदार का नाम, चैक संख्या, चैक जारी करने वाले बैंक तथा शाखा का नाम, दिनांक तथा राशि भरनी होती है। यहाँ एक उदाहरण दिया जा रहा है जिसमें मनमोहन सिंह अपने खाते में दस हज़ार रुपए का चैक जमा कर रहे हैं-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 7

4. बैंक से रुपए निकलवाने का प्रपत्र/चैक बैंक से रुपए निकलवाने के लिए बैंक के भगतान प्रपत्र तथा बैंक द्वारा दिए गए चैक का प्रयोग किया जाता है। चैक द्वारा किसी संस्था अथवा व्यक्ति को भी भुगतान कर सकते हैं। बैंक से भुगतान प्रपत्र पर रुपए निकलवाने के लिए पासबुक साथ लगानी पड़ती है। इसमें खातेदार अपने खाते की संख्या, निकाले जाने वाली राशि, दिनांक तथा अपने हस्ताक्षर कर के राशि निकलवा सकता है। प्रीतम कौर अपने खाते से दस हज़ार रुपए भुगतान प्रपत्र पर निकाल रही है। इसका उदाहरण यहाँ दिया जा रहा है-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 8
चैक द्वारा किसी को भुगतान करने अथवा स्वयं रुपए निकालने के लिए चैक में दिनांक, खाता संख्या, किसे भुगतान करना है, राशि तथा खाताधारी के हस्ताक्षर करना आवश्यक होता है। यहाँ एक चैक राजनाथ सिंह द्वारा भारतीय जीवन बीमा निगम को आठ हजार नौ सौ बीस रुपए का जारी किया गया है उसका प्रारूप यहाँ दिया जा रहा है। चैक को रेखांकित कर देना चाहिए इस चैक को कोई अन्य नहीं भुना सकता।
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 9

5. बैंक ड्राफ्ट बनवाने का प्रपत्र
बैंक ड्राफ्ट एक मांग-प्रपत्र होता है, जिसे कोई भी व्यक्ति बैंक को निर्धारित शुल्क देकर बनवा सकता है। यह बैंक की एक शाखा द्वारा अपनी उस शाखा के नाम जारी किया जाता है जहाँ ग्राहक ने अपनी राशि भेजनी है। बैंक ड्राफ्ट पाने वाला व्यक्ति उस शाखा से ड्राफ्ट की राशि नकद अथवा अपने बैंक खाते में जमा करवा कर प्राप्त कर सकता है। बैंक ड्राफ्ट के प्रपत्र में दिनांक, आवेदन का नाम पता, किसके पक्ष में, कहाँ भेजना है, ड्राफ्ट की राशि, शुल्क आवेदक के हस्ताक्षर, मोबाइल नंबर भरने होते हैं। यहाँ राजेन्द्र सिंह द्वारा श्री रघुराज भालेराव पाठक को नागपुर पाँच हज़ार रुपए का ड्राफ्ट बनवाने का प्रपत्र उदाहरण के रूप में भरकर दिया जा रहा है। ड्राफ्ट के लिए राजेन्द्र सिंह ने दिए पाँच हजार तीस रुपए हैं परन्तु उसे ड्राफ्ट पांच हज़ार रुपए का ही मिलेगा। तीस रुपए ड्राफ्ट बनवाने का शुल्क है। इसके भी दो भाग होते हैं। एक ग्राहक के लिए तथा दूसरा बैंक के लिए-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 10

बोर्ड परीक्षाओं में पूछे गए प्रपत्र-पूर्ति सम्बन्धी प्रश्नोत्तर

प्रश्न 1.
मान लो आपका नाम निर्मला देवी है। आपको सेविंग्स बैंक खाता नं0 79684 में से दस हज़ार रुपये निकलवाने हैं। अत: आप नीचे दिए गए प्रपत्र की रूपरेखा को अपनी उत्तर-पुस्तिका पर उतार कर भरें-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 11
उत्तर:

प्रश्न 2.
मान लो आपका नाम तमन्ना है। आपका मुख्य डाकघर, अमृतसर में बचत खाता नं0 764329 है। आपके इस खाते में ₹4,500/- हैं। आपको इस खाते में ₹ 1,500/- जमा करवाने हैं। इसलिए नीचे दिए गए प्रपत्र के प्रारुप को अपनी उत्तर-पुस्तिका पर उतार कर भरें-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 13
उत्तर:

प्रश्न 3.
मान लीजिए आपका नाम विनोद कुमार है। आप पटियाला में रहते हैं। आपको अपने बचत खाता नं० 17321986 में र दो हज़ार जमा करवाने हैं। अत: आप नीचे दिए गए प्रपत्र की रूपरेखा को अपनी उत्तरपुस्तिका पर उतार कर भरें-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 10
उत्तर:

(ख) डाकघर से संबंधित प्रपत्र

डाकघर में भी बैंक की तरह अनेक प्रकार से लेन-देन होता है तथा यहाँ भी रुपए जमा करने, निकालने, मनीआर्डर से रुपए भेजने आदि से संबंधित अनेक कार्य होते हैं, जिनके लिए अनेक प्रपत्रों का प्रयोग करना पड़ता है। डाकखाने में प्रयोग में आने वाले मुख्य प्रपत्र अग्रलिखित हैं-

खाता खोलने का प्रपत्र
रुपए जमा कराने का प्रपत्र
रुपए निकलवाने का प्रपत्र
मनीआर्डर से रुपए भेजने का प्रपत्र

1. डाकघर में खाता खोलने का प्रपत्र
डाकघर में खाता खोलने के लिए दिए गए प्रपत्र को ध्यानपूर्वक पढ़कर सभी कॉलम साफ़-साफ़ शब्दों में भरने चाहिए। खाता एक व्यक्ति अथवा संयुक्त रूप से दो-तीन व्यक्ति भी खोल सकते हैं। परिचायक का नाम, पता तथा खातेदारों के हस्ताक्षर करने होते हैं। खाते में नामांकन भी किया जा सकता है। उदाहरण के लिए एक भरा हुआ प्रपंत्र यहाँ दिया जा रहा है-
डाकघर बचत-बैंक
खाता खोलने के लिए आवेदन पत्र
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 17

4. मैंहम किसी भी समय सुसंगत नियम में विनिर्दिष्ट सीमा के भीतर अपने सभी एकल या संयुक्त बचत बैंक/सा० स० ज० खातों में अतिशेष बनाए रखने का और डाकघर बचत बैंक से मांग किए जाने पर ऐसे सभी खातों की विशिष्टियां भी देने का वचन देता हूँ/देते हैं।

5. मैं/हम केंद्रीय सरकार द्वारा बनाए गए ऐसे नियमों का पालन करने के लिए सहमत हूँ/हैं जो समय-समय पर खाते को लागू हों।

6. मैं हम सरकारी बचत बैंक अधिनियम, 1873 (1873 का 5) की धारा 4 के अधीन नीचे व्यक्ति (व्यक्तियों) को अपनी मृत्यु हो जाने की दशा में खाते में जमा रकम के लिए एकमात्र प्राप्तकर्ता (प्राप्तकर्ताओं) के रूप में नाम निर्दिष्ट करता हूँ करते हैं।
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 18

2. डाकघर में पैसे जमा कराने का प्रपत्र
डाकघर में धनराशि जमा करवाने के प्रपत्र में खाताधारी को डाकघर का पता, दिनांक, खातेधारी का नाम, खाता संख्या, जमा कराने की राशि शब्दों और अंकों में, जमा के बाद कुल राशि लिख कर अपने हस्ताक्षर करने होते हैं। इसके साथ डाकघर बचत बैंक की पास बुक भी लगानी होती है, जिसमें बचत बैंक सहायक जमा की गई राशि लिखकर, डाकघर की मोहर लगाकर तथा अपने हस्ताक्षर कर जमाकर्ता को वापिस कर देता है। यहां जालंधर के डाकघर में सतीश कुमार शर्मा के खाता संख्या 1901 में एक हज़ार रुपए जमा करने के प्रपत्र को भरकर उदाहरण के रूप में दिया जा रहा है।
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 19

3. डाकघर बचत बैंक से पैसे निकलवाने का प्रपत्र
डाकघर बचत बैंक से पैसे निकलवाने के लिए प्रपत्र में जमाकर्ता को डाकघर का नाम, दिनांक, खाता संख्या, निकाली गई राशि, शेष राशि, यदि आवश्यकता हो तो संदेशवाहक का नाम, हस्ताक्षर देने होते हैं। अदायगी आदेश में राशि मिलने के हस्ताक्षर खातेदार अथवा संदेशवाहक द्वारा किए जाते हैं। यहां एक उदाहरण के रूप में भरा हुआ प्रपत्र दिया जा रहा है-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 20

प्रश्न 3.
मान लीजिए आपका नाम नरेश कुमार है। आपको लोकेश कुमार को दिनांक 10.04.2016 को ₹10,000/- का स्वहस्ताक्षरित रेखांकित किया हुआ चेक लिखकर देना है। इस अनुसार निम्नलिखित चेक के प्रपत्र को भरें-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 21
उत्तर:

4. मनीऑर्डर से रुपये भेजने का प्रपत्र
मनीऑर्डर से रुपए भेजने का काम डाकघर में वर्षों से हो रहा है। डाकघर से मनीऑर्डर प्रपत्र लेकर भरना होता है। इसके 6 भाग होते हैं। पहला भाग जिसे रुपये भेजने हैं रुपयों की संख्या शब्दों और अंकों में प्राप्तकर्ता का नाम, पता, पिनकोड, दिनांक लिखकर भेजने वाला अपने हस्ताक्षरकर्ता है। चौथा भाग में भेजने वाले का नाम पता तथा पिन कोड लिखना होता है। छठे भाग में भेजने वाला संदेश लिख सकता है। दूसरा, तीसरा और पांचवां भाग डाक विभाग ने भरना होता है। डाविभाग भेजने वाली राशि पर निर्धारित शुल्क भी लेता है। यहाँ मनीऑर्डर प्रपत्र को भर कर प्रस्तुत किया जा रहा है-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 23
(संदेश के लिए स्थान)
जालन्धर शहर प्रिय, सुजान
02-12-2000 तुम्हारे पत्र के अनुसार 800 सौ रुपए भेज रहे हैं । मिलने पर सूचना देना और रुपयों की ज़रूरत हो तो लिखना। सब को सत श्री अकाल कहना।
तुम्हारा मित्र,
सुखदेव सिंह

5. इलेक्ट्रानिक मनीआर्डर (ई०एम०ओ०) आधुनिक युग में इंटरनेट की प्रगति के साथ-साथ अपनी धनराशि किसी दूसरे को भेजने का इलेक्ट्रानिक मनीऑर्डर जल्दी से जल्दी और सरलता से भेजने का डाक विभाग द्वारा चलाया गया मनीऑर्डर का नया रूप है। इससे आम मनीआर्डर भेजने की अपेक्षा कम समय लगता है। जिस दिन ई०एम०ओ० किया जाता है, वह उसी दिन प्राप्तकर्ता को मिल जाता है। इस पर भी डाक विभाग निर्धारित शुल्क लेता है। इसके प्रपत्र की रूपरेखा यहाँ प्रस्तुत की जा रही है-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 24

6. मोबाइल मनीट्रांसफर सर्विस (इंसटेंट मनीऑर्डर) मोबाइल मनीट्रांसफर सर्विस धन भेजने का सबसे तेज़, सुविधाजनक, विश्वसनीय और सरल माध्यम है। इसे इंसटेट मनीऑर्डर अथवा तत्काल मनीआर्डर भी कहते हैं। इस माध्यम से कुछ ही मिनटों में धनराशि प्राप्तकर्ता को मिल जाती है। डाकघर से प्राप्त प्रपत्र को भर कर अधिकारी को देने पर प्रेषक को कम्प्यूटर द्वारा सोलह अंकों का सीलबंद गुप्त नम्बर दिया जाता है, जिसे प्रेषक प्राप्तकर्ता को फोन, एस०एम०एस० अथवा ई-मेल द्वारा बता देता है, जो इंसटेंट मनीऑर्डर के अपने नगर के केंद्र पर जाकर वहाँ के अधिकारी को बताता है तथा अपनी पहचान का फोटो पहचानपत्र दिखाकर प्रेषक द्वारा उसे भेजी गई राशि प्राप्त कर सकता है। इस सेवा में उन्नीस हज़ार रुपए तक की राशि नकद प्राप्त की जा सकती है जबकि बीस हज़ार रुपए अथवा उससे अधिक राशि का भुगतान चैक द्वारा किया जाता है। इस सेवा को प्राप्त करने का प्रपत्र अग्रलिखित है-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 25

प्रश्न 1.
इन्सटेंट मनीऑडर्र (आई०एम०ओ) किसे कहते हैं?
उत्तर:
मोबाइल मनीट्रांसफर सर्विस धन भेजने का सबसे तेज़, सुविधाजनक, विश्वसनीय और सरल माध्यम है। इसे इंसटेट मनीऑर्डर अथवा तत्काल मनीआर्डर भी कहते हैं। इस माध्यम से कुछ ही मिनटों में धनराशि प्राप्तकर्ता को मिल जाती है। डाकघर से प्राप्त प्रपत्र को भरकर अधिकारी को देने पर प्रेषक को कम्प्यूटर द्वारा सोलह अंकों का सीलबंद गुप्त नम्बर दिया जाता है, जिसे प्रेषक प्राप्तकर्ता को फोन, एस०एम०एस० अथवा ई०-मेल द्वारा बता देता है, जो इंसटेंट मनीऑर्डर के अपने नगर के केंद्र पर जाकर वहाँ के अधिकारी को बताता है तथा अपनी पहचान का फोटो पहचानपत्र दिखाकर प्रेषक द्वारा उसे भेजी गई राशि प्राप्त कर सकता है। इस सेवा में उन्नीस हजार रुपए तक की राशि नकद प्राप्त की जा सकती है जबकि बीस हज़ार रुपए अथवा उससे अधिक राशि का भुगतान चैक द्वारा किया जाता है।

7. अंतर्राष्ट्रीय धन अंतरण
डाकघर में विदेशों से धन मंगवाने अथवा भेजने के लिए मनीऑर्डर सेवाओं के अन्तर्गत अंतर्राष्ट्रीय धन अंतरण सेवा, मनीग्राम, इलेक्ट्रानिक क्लियरेंस सेवाएं (ई०सी०एस०) आदि भी समुचित मूल्य पर उपलब्ध हैं।

8. इंडियन पोस्टल ऑर्डर
इंडियन पोस्टल ऑर्डर डाकघर से निश्चित राशि के प्राप्त किये जा सकते हैं। इनके द्वारा एक व्यक्ति किसी दूसरे व्यक्ति अथवा संस्था को धनराशि भेज सकता है। ये भारत के सभी डाकघरों से प्राप्त किए जा सकते हैं। विभिन्न प्रतियोगी परीक्षाओं, नौकरियों के आवेदन-पत्रों आदि का शुल्क इंडियन पोस्टल ऑर्डर द्वारा मांगा जाता है। इन्हें रेखांकित भी कर सकते हैं। इसकी खरीद तिथि से वैधता दो वर्षों की होती है। इसे प्रयोग नहीं करने पर निश्चित अवधि में वापस करने पर खरीदी हुई राशि वापस भी ली जा सकती है।

इसके दो भाग फॉइल और काउंटर फॉइल होते हैं जिसके नाम इंडियन पोस्टल ऑर्डर भेजा जाता है उसे फॉइल कहते हैं और जो हिस्सा भेजने वाले के पास रहता है उसे काउंटर फॉइल कहते हैं। पोस्टल आर्डर में राशि प्राप्तकर्ता का नाम, पता, शहर, प्रेषक का नाम पता, दिनांक तथा जिस डाकघर से प्राप्तकर्ता को धनराशि प्राप्त करनी है उसका नाम लिखा जाता है। यदि इसे रेखाकिंत करना है तो वह भी किया जाता है। यहाँ इंडियन पोस्टल ऑर्डर का एक उदाहरण दिया जा रहा है, जिसमें गुरनाम सिंह, फिरोजपुर, पंजाब लोक सेवा आयोग, लुधियाना को एक रुपए का पोस्टल ऑर्डर भेज रहा है-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 26

9. रजिस्ट्री-पत्र की पावती
डाकघर से किसी को रजिस्ट्री-पत्र भेजते समय उसके साथ रजिस्ट्री-पत्र की पावती भी लगानी होती है। यह पावती रजिस्ट्री-पत्र प्राप्तकर्ता हस्ताक्षर कर प्रेषक के पास भेज देता है। इसमें भेजने वाले को अपना तथा प्राप्तकर्ता का पूरा पता लिखना होता है। इसके आगे-पीछे दो भाग होते हैं। इसका प्रारूप निम्नलिखित है-
रजिस्ट्री-पावती प्रपत्र
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 27

(ग) रेलवे आरक्षण प्रपत्र
रेलवे में आरक्षण करवाने अथवा आरक्षण रद्द करवाने के लिए प्रपत्र भरना होता है। दोनों ही स्थितियों में एक प्रकार का ही प्रपत्र प्रयोग में आता है। भरने से पहले उस पर लिखना होता है कि आरक्षण लेने के लिए प्रपत्र भर रहे हैं । अथवा आरक्षण रद्द करवाने के लिए। इस प्रपत्र में तीन भाग होते हैं । प्रथम भाग में आरक्षण लेने अथवा रद्द की जाने वाली रेलगाड़ी का नाम, क्रमांक, यात्रा करने की तिथि, श्रेणी, कहां से कहाँ तक, यात्रियों के नाम, आयु, लिंग, आवेदक का नाम, पता, मोबाइल नम्बर तथा हस्ताक्षर होते हैं। दूसरा भाग में वापसी यात्रा का विवरण तथा तीसरा भाग रेलवे के कर्मचारी भरते हैं। इसका एक उदाहरण यहाँ दिया जा रहा है-

रेलवे आरक्षण कराने के लिए प्रपत्र का नमूना
उत्तर:
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 29

आगे की यात्रा/वापसी यात्रा का विवरण
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 30

टिप्पणी : 1. अधिकतम अनुपेय यात्रियों की संख्या प्रति मांग पत्र 6 व्यक्ति।
2. एक बार एक व्यक्ति से केवल एक ही मांग पत्र स्वीकार किया जाएगा।
3. कृपया खिड़की छोड़ने से पहले अपने टिकट और शेष राशि की जांच कर लें।
4. ठीक ढंग से न भरे हुए तथा अपठनीय फार्म स्वीकार नहीं किए जाएंगे।
5. विशेष मांग पर उपलब्धता के आधार पर विचार किया जाएगा।

अब रेलवे आरक्षण/आरक्षण रद्द करने की सुविधा इंटरनेट/मोबाइल फोन द्वारा भी उपलब्ध है, जिसकी पूरी जानकारी आई०आर०सी०टी०सी० की बेबसाइट पर उपलब्ध है।

1. मान लीजिए आपका नाम तमन्ना शर्मा है।आपका गुड बैंक, शाखा गंगानगर में एक बचत खाता है। आपका खाता नम्बर-45632789 है। आपका मोबाइल नं0 1009321432 है। आपको दिनांक 22.10.2015 को अपने इस खाते में 12,000 रुपये नकद जमा करवाने हैं। आपके पास जमा करवाने के लिए एक हज़ार के 8 नोट, पांच सौ के नोट तथा पचास के 20 नोट हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:
उत्तर:
गुड बैंक, गंगानगर
बैंक में रुपये जमा करवाने के लिए प्रपत्र ( बैंक प्रति)
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 31

गुड बैंक, गंगानगर
बैंक में रुपये जमा करवाने के लिए प्रपत्र (ग्राहक प्रति)
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 32

2. मान लीजिए आपका नाम रत्नलीन कौर है। आपका मोबाइल नम्बर 1890876535 है। आपका हिमालय बैंक, शाखा सोलन में एक बचत खाता नम्बर 96237180 है। आपको दिनांक 11.08.2015 को अपने इस खाते में से 7,000 रुपये निकालवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें
उत्तर:
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 33

3. मान लीजिए आपका नाम प्रकाश सिंह है। आपका भारतीय डाक के सेक्टर 15 चंडीगढ़ में 344565 नम्बर एक बचत खाता है। आपके इस खाते में अब तक 45,000 रुपये जमा हैं। आपको दिनांक 02.01.2016 को अपने इस खाते में से 4,000 रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:
उत्तर:
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 34

4. मान लीजिए आपका नाम तुषार कपूर है। आप मकान नं० 987, पुरानी दिल्ली रोड, गुड़गाँव (हरियाणा)-122001 में रहते हैं। आपका बेटा मकान नं० 456 सिविल लाइन्स बरेली (उत्तर प्रदेश) 243001 में रहता है। आप अपने बेटे को भारतीय डाक के माध्यम से दिनांक 05.11.2015 को मनीआर्डर के द्वारा 5,000/- रुपये भेजना चाहते हैं। इस प्रपत्र में उसे संदेश भी लिखें ‘अपनी सेहत का ध्यान रखना’। इस अनुसार निम्नलिखित प्रपत्र भरें।
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 36

5. मान लीजिए आपका नाम शिखर कुमार है। आपको लोकेश कुमार को दिनांक 6.12.15 को 10,000/- रु० का स्वहस्ताक्षरित रेखांकित किया हुआ चेक लिखकर देना है। इस अनुसार निम्नलिखित चैक के प्रपत्र को भरें:
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 37

बोर्ड परीक्षा में पूछे गए प्रश्न

निम्नलिखित में से किसी एक प्रपत्र को अपनी उत्तर-पुस्तिका पर उतारकर भरें
1. (क) मान लीजिए आपका नाम सुलोचना कुमारी है। आपका हिमालय बैंक, शाखा-शिमला में एक बचत खाता है, जिसका नम्बर, 18954326781 है। आपको अपने इस खाते में से 2500 रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 38
उत्तर:

(ख) मान लीजिए आपका नाम रूप सिंह है। आपका भारतीय डाक के सेक्टर 15, करनाल में एक बचत खाता है, जिसका नम्बर 128947654431 है। आपको अपने इस खाते में से 4000/- रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 40
उत्तर:

2. (क) मान लीजिए आपका नाम पीयूष सिन्हा है। आपका कल्याण बैंक, शाखा-भुवनेश्वर में एक बचत खाता है, जिसका नम्बर 1150000234567 है। आपको अपने इस खाते में 18000/- रुपये जमा करवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 42
उत्तर:

(ख) मान लीजिए आपका नाम दीदार सिंह है। आपका भारतीय डाक के सेक्टर-25, देहरादून में एक बचत खाता है, जिसका नम्बर 18634956744 है। आपको अपने इस खाते में से 10000/- रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 44
उत्तर:

3. (क) मान लीजिए आपका नाम निर्मला कौर है। आपका भारतीय डाक के सेक्टर-42, जम्मू में एक बचत खाता है, जिसका नम्बर 62745367823 है। आपको अपने इस खाते से 8200/- रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 46
उत्तर:

(ख) मान लीजिए आपका नाम दिनेश कुमार है। आपको मुनीश कुमार को 12000/- रु० का स्वहस्ताक्षरित चेक लिखकर देना है। इस अनुसार निम्नलिखित चेक के प्रपत्र को भरें:
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 48
उत्तर:

4. (क) निम्नलिखित में से किसी एक प्रपत्र को अपनी उत्तर-पुस्तिका पर उतारकर भरें-
(i) मान लीजिए आपका नाम पूनम कुमारी है। आपका भविष्य बैंक, शाखा-शिमला में एक बचत खाता है, जिसका नम्बर 373748488 है। आपको अपने इस खाते में से ₹ 5500/- निकलवाने हैं। इस अनुसार
निम्नलिखित प्रपत्र की रूपरेखा अपनी उत्तर पुस्तिका पर उतार कर भरें-
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 50
उत्तर:

(ख) मान लीजिए आपका नाम सुप्रिया देवी है। आपका भारतीय डाक के सेक्टर-15 नोएडा में एक बचत खाता है, जिसका नम्बर 834747993 है। आपको अपने इस खाते में से दिनांक 23.05.2018 को ₹4500/- निकलवाने हैं। इस अनुसार अग्रलिखित प्रपत्र की रूपरेखा अपनी उत्तर-पुस्तिका पर उतार कर भरें:
PSEB 10th Class Hindi Vyakaran प्रपत्र पूर्ति 52
उत्तर:

PSEB 10th Class Agriculture Notes Chapter 11 Plant Clinic

November 18, 2021 by Bhagya

This PSEB 10th Class Agriculture Notes Chapter 11 Plant Clinic will help you in revision during exams.

PSEB 10th Class Agriculture Notes Chapter 11 Plant Clinic

→ A plant clinic is a place where diagnosis and remedial measures for diseased plants, nutrient deficiency, insect attack, etc. are provided to the farmers.

→ A plant clinic is a centre where diagnoses of diseased plants are done and their treatment is done.

→ Punjab Agricultural University established a central plant clinic at Ludhiana in the year 1993.

→ Plant clinics are running at 18 Krishi Vigyan Kendras (KVKs) in various districts and four at regional research stations Abohar, Bathinda, and Gurdaspur and Department of Fruit Science, PAU, Ludhiana.

→ When the number of insects increased upto a specific number or affects a specified number of plants of the crop.

→ A spray of pesticides should start after this level has reached.

→ This way crops, as well as farmers, will get some benefit. This process is called the economic threshold level.

→ Farmers can resolve their problems by dialing phone no. 0161-240-1960 with extension no. 417. Mobile number is 9463048181.

→ By sending pictures of diseased or affected plants via email to the plant clinic, the problem can be resolved and e-mail is planted [email protected]. WhatsApp can also be used to get quick solutions.

→ Some equipment is required at the plant clinic are the microscope, magnifying lenses, chemicals, incubator, scissors, knife, computer, photo camera, projector, books, etc.

→ Some chemicals which are used at plant clinics are formalin, copper acetate, acetic acid, alcohol, etc.

→ Computer, scanners, etc. are also an important part of the plant clinic.

PSEB 9th Class Maths Solutions Chapter 1 Number Systems Ex 1.1

November 18, 2021 by Bhagya

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 1 Number Systems Ex 1.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1

Question 1.
Is zero a rational number? Can you write it in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0?
Answer:
Yes. Zero is a rational number. Zero is a whole number as well as an Integer. As we know, all integers are rational numbers, zero is also a rational number.

Zero can be expressed in the \(\frac{p}{q}\) form in infinitely many ways like \(\frac{0}{5}\), \(\frac{0}{11}\), \(\frac{0}{-3}\), …….

In short, \(\frac{0}{q}\), where q is an integer other than zero, is the \(\frac{p}{q}\) form of 0.

Question 2.
Find six rational numbers between 3 and 4.
Answer:
Since six rational numbers are to be found. between rationals 3 and 4, we express both of them with denominator 7(6 + 1).
Now, 3 = \(\frac{3 \times 7}{1 \times 7}\) = \(\frac{21}{7}\) and 4 = \(\frac{4 \times 7}{1 \times 7}=\frac{28}{7}\)

Then, \(\frac{21}{7}\) < \(\frac{22}{7}\) < \(\frac{23}{7}\) < \(\frac{24}{7}\) < \(\frac{25}{7}\) < \(\frac{26}{7}\) < \(\frac{27}{7}\) < \(\frac{28}{7}\) so, \(\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}\) are required six rational numbers between 3 and 4.

By other methods, takIng 3 = 3.0 and 4 = 4.0, we can state that 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 are six rationals between 3 and 4. Still, by the method of averages, we can state \(\frac{7}{2}\), \(\frac{13}{4}, \frac{15}{4}, \frac{25}{8}, \frac{27}{8}, \frac{29}{8}\) are six rationals between 3 and 4. As there lies infinitely many rationals between any two rationals, we can give many different answers.

Question 3.
Find five rational numbers between \(\) and \(\).
Answer:
\(\frac{3}{5}=\frac{3 \times 6}{5 \times 6}=\frac{18}{30}\) and \(\frac{4}{5}=\frac{4 \times 6}{5 \times 6}=\frac{24}{30}\)
Now, \(\frac{18}{30}\) < \(\frac{19}{30}\) < \(\frac{20}{30}\) < \(\frac{21}{30}\) < \(\frac{22}{30}\) < \(\frac{23}{30}\) < \(\frac{24}{30}\)
i.e., \(\frac{3}{5}\) < \(\frac{19}{30}\) < \(\frac{2}{3}\) < \(\frac{7}{10}\) < \(\frac{11}{5}\) < \(\frac{23}{30}\) < \(\frac{4}{5}\).
Hence, \(\frac{19}{30}, \frac{2}{3}, \frac{7}{10}, \frac{11}{15}\) and \(\frac{23}{30}\) are five of the many rational numbers lying between \(\frac{3}{5}\) and \(\frac{4}{5}\).

Question 4.
State whether the following statements are true or false. Give reasons for your answers :
(i) Every natural number is a whole number.
Answer:
The given statement is true as the collection of whole numbers contain all the natural numbers.

(ii) Every integer is a whole number.
Answer:
The given statement is false as any negative integer like – 2, – 3, – 5, etc. Is not a whole number. The collection of whole numbers contains 0 and all natural numbers, but not the opposites of the natural numbers.

(iii) Every rational number is a whole number.
Answer:
The given statement is false as any rational number lying between two consecutive whole numbers is not a whole number.

e.g., \(\frac{5}{2}\) is a rational number lying between whole numbers 2 and 3, but it is not a whole number.
Skill Testing Exercise