Class 9

Haryana State Board HBSE 9th Class Science Notes Chapter 3 Atoms and Molecules Notes.

Haryana Board 9th Class Science Notes Chapter 3 Atoms and Molecules

→ Any compound is composed of two or more than two elements.

→ In a compound of water, the ratio of masses of hydrogen and oxygen is always 1: 8, it does not concern whether which source does the water belong to.

→ In ammonia (NH₃), Nitrogen and Hydrogen always exist in the ratio of 14: 3 according to their masses.

→ John Dalton was bom in 1766 in England.

→ In a compound, the relative number of atoms and its type are definite.

→ The structural unit of all the fluids is atom.

→ The atomic radius is measured in nanometre (nm). (1nm = 10^-9m)

→ Nowadays the International Union of Pure and Applied Chemistry (IUPAC) approves names of elements.

→ Each element have a name and an unmatched chemical symbol.

→ Each element has a characteristic atomic mass.

→ We assume unit of atomic mass equal to that of the mass of a carbon atom.

→ Atomic masses of all the elements are obtained in relation to single atomic mass of isotope of carbon-12

→ Generally, a molecule is a group of two or more than two atoms which are interlinked by a chemical bond.

→ Molecules of an element are composed of only one kind of atoms.

→ Molecule with negative charge is called as electron and the molecule with positive charge is called as proton.

→ The chemical formulae of molecular compound are based on by means of combination of each element.

→ In ionic compounds, chemical formulae of compound are learnt by the number of electrons on each of the ions.

→ The number of particles (atoms, molecules or ions) in a mole of a substance is constant whose mass is 6.022 × 10²³.

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

→ Law of Constant Proportions: In a pure chemical compound, elements always exist in constant proportions of masses, it is called the law of constant proportions.

→ Atom: The smallest particle of an element is atom, which can exist independently and it exhibits all its chemical properties.

→ Molecule: Molecule is that minute particle of an element or compound which can remain in a free state in normal conditions. It exhibits all properties of matter.

→ Relative Atomic Mass Unit: 1/12th part of atomic mass of an atom of carbon-12 isotopes is taken to be the relative atomic mass unit.

→ Atomicity: The number of atoms being used in structural formation of any molecule is called atomicity of its molecule.

→ Ions: Compounds are composed of metals and non-metals contain charged particles. These charged particles are called as ions e.g. sodium, magnisium.

→ Polyatomic Ions: A group of atoms that behave as an ion are called polyatomic ions.

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

→ Valency: The combining power (or capacity) of an element is the valency of that element.

→ Binary Compounds: The simplest compounds formed by combination of two different elements are known as binary compounds.

→ Molecular Mass: Molecular mass is the sum total of masses of overall combined atoms of matter. It is expressed by atomic mass units (u).

→ Formula Unit Mass: Formula unit mass of a substance is a sum total of atomic masses of its whole combined atoms.

→ Mole: Mole is that quantity of a substance in which the number of particles (atom, ion, molecule or formula unit) is equal to the existing atoms in exactly 12g of carbon =12.

→ Molar Mass: Mass of molecules of a single mole of a substance is called its molar mass.

Haryana State Board HBSE 9th Class Science Notes Chapter 2 Is Matter Around Us Pure Notes.

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

→ A pure substance is composed of only one type of molecules.

→ Sodium chloride dissolved in water can be separated from water by evaporation or distillation method.

→ Mixture contains more than one substances. .

→ Copper sulphate forms homogeneous mixture in water.

→ The mixture of water and oil is heterogeneous.

→ Lemon water and soda water are examples of solution.

→ In the solution of sugar and water, sugar is solute and water is solvent.

→ Air is a homogeneous mixture of 21% of oxygen and 78% of nitrogen.

→ The molecules of a solution are even smaller than in diameter of 1nm (1 × 10^-9m).

→ Solution is a homogeneous mixture from which the molecules of solute cannot be separated by

→ Suspended particles are larger than 100 nm (10^-7m). These particles can be seen with eyes.

→ The molecules of colloid spread uniformly in solution.

→ Milk is a colloidal solution.

→ Divergence of sunrays by colloidal molecules is called Tyndall effect.

→ The size of the molecules of colloid is between 1 nm to 100 nm.

→ Different components of colloid can be separated through centrifugation process.

→ Two insoluble liquids can be separated by separating funnel.

→ Ammonium chloride, camphor, naphthalene and anthracene are volatile solid materials.

→ Amixture of two soluble liquids whose components have more difference in their boiling point can be separated through distillation process.

→ The process of crystallization is applied to refine the solid materials.

→ Chemical change brings about change in the chemical properties of matter.

→ On the basis of chemical composition, materials can either be classified into elements or compounds.

→ Elements can generally be divided into metals, non-metals and metalloids.

→ Two elements-mercury and bromine are in liquid state at room temperature.

→ The properties already present in the elements of the compounds are different from that of the properties of the compounds. Though, the elements and compounds present in mixture shows their own properties.

→ Pure Substance: A substance composed of only one type of molecules is called as pure substance.

→ Mixture: A substance composed with the combination of one or more than one pure elements or compounds is called as mixture.

→ Solution: Homogeneous mixture of two or more than two substances is called as solution.

→ Alloys: Those homogeneous mixtures of metals that cannot be isolated through physical processes from their components are called alloys.

→ Solvent: That molecule of solution (whose quantity is more than the other one) which mixes other molecules into solution is called solvent.

→ Solute: That component of solution which dissolved in solvent is called solute.

→ Saturated solution: At a given fixed temperature if in the solution, solute is not miscible, that is called saturated solution.

→ Solubility: That quantity of solute material which at a definite temperature is present in the saturated solution, is called as solubility.

→ Unsaturated Solution: If in any solution quantity of solute material remains little to saturation, then it is called as unsaturated solution.

→ Supersaturated Solution: If in any solution the concentration of Solute is more than that of the saturated concentration, that is called as supersaturated solution.

→ Strength of Solution: Quantity (volume) of solute material dissolved into solvent’s volume is called strength of solution.

→ Suspension: Suspension is an heterogeneous mixture, in which the solute particles do not dissolve rather keep suspending in the medium.

→ Principle of Centrifugation Method: In the centrifugation method, on rotating the mixture fastly the heavier particles settle down at the bottom and the lighter particles come upward.

→ Chromatography: Chromatography is such a process which is used to isolate those solute materials which are dissolved in just one type of solvent.

→ Crystallisation: Crystallisation is that process by which pure solid substance & separated from solution in the form of crystal.

→ Element: Element is that basic form of matter which cannot be further disintegrated into smaller pieces through chemical process.

→ Metalloids: The elements showing the properties in between metals and non-metals are called metalloids, e.g., boron and silicon.

→ Compound: The substance made up of combining chemically two or more than two elements together in the same ratio is called as compound.

Haryana State Board HBSE 9th Class Maths Solutions Chapter 7 त्रिभुज Ex 7.5 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 7 त्रिभुज Exercise 7.5

प्रश्न 1.
ABC एक त्रिभुज है। इसके अभ्यंतर में एक ऐसा बिंदु ज्ञात कीजिए, जो ΔABC के तीनों शीर्षों से समदूरस्थ है।
हल :

माना एक ΔABC में, OD तथा OE, क्रमशः भुजा BC तथा CA के लंब समद्विभाजक हैं।
O रेखाखंड BC के दोनों सिरों B तथा C से समान दूरी पर है, क्योंकि O, BC के
लंब समद्विभाजक पर स्थित है।
इसी प्रकार O, C तथा A से भी समान दूरी पर है, क्योंकि O, CA के लंब
समद्विभाजक पर स्थित है।
आकृति इसीलिए, भुजा BC, CA तथा AB के लंब समद्विभाजकों प्रतिच्छेदन बिंदु O, आवश्यक बिंदु है जो ΔABC के शीर्षों से समदूरस्थ है। [इति सिद्धम]

प्रश्न 2.
किसी त्रिभुज के अभ्यंतर में एक ऐसा बिंदु ज्ञात कीजिए, जो त्रिभुज की सभी भुजाओं से समदूरस्थ है।
हल :

माना BE और CF क्रमशः ∠ABC तथा ∠ACB के कोण समद्विभाजक हैं। ये AC तथा AB को क्रमशः E तथा F पर काटते हैं।
क्योंकि O, ∠ABC के कोण समद्विभाजक BE पर स्थित है।
इसलिए O, AB तथा BC से समान दूरी पर है।
इसी प्रकार O, ∠ACB के कोण समद्विभाजक पर स्थित हैं। इसलिए O, BC और CA से समान दूरी पर है। इस प्रकार ΔABC के कोणों का समद्विभाजक अभीष्ट बिंदु है। [इति सिद्धम]

प्रश्न 3.
एक बड़े पार्क में, लोग तीन बिंदुओं (स्थानों) पर केंद्रित हैं। (देखिए आकृति)
A : जहाँ बच्चों के लिए फिसल पट्टी और झूले हैं।
B : जिसके पास मानव-निर्मित एक झील है।
C : जो एक बड़े पार्किंग स्थल और बाहर निकलने के रास्ते के निकट है।
एक आइसक्रीम का स्टॉल कहाँ लगाना चाहिए, ताकि वहाँ लोगों की अधिकतम संख्या पहुंच सके ?
हल :

आइसक्रीम का स्टॉल A, B और C से समान दूरी पर होना चाहिए, जिसके लिए AB और BC का समद्विभाजक करना पड़ेगा, जोकि O पर मिलेंगे।
इसीलिए O आवश्यक बिंदु है जोकि A, B तथा C से समान दूरी पर है,
जहाँ पर आइसक्रीम का स्टॉल स्थापित किया जा सकता है।

प्रश्न 4.
षडभुजीय और तारे के आकार की रंगोलियों [देखिए आकृति (i) और (ii)] को 1 cm भुजा वाले समबाहु त्रिभुजों से भर कर पूरा कीजिए। प्रत्येक स्थिति में, त्रिभुजों की संख्या गिनिए। किसमें अधिक त्रिभुज हैं ?

हल :

प्रत्येक आकृति को 1 cm के समबाहु त्रिभुजों के साथ भरने पर हमें प्राप्त हुआ कि आकृति
(i) में इस प्रकार के त्रिभुजों की संख्या = 150 तथा आकृति
(ii) में इस प्रकार के त्रिभुजों की संख्या 300 है।
अतः तार के आकार की रंगोलियों में त्रिभुजों की संख्या अधिक है।

Haryana State Board HBSE 9th Class Science Notes Chapter 1 Matter in Our Surroundings Notes.

Haryana Board 9th Class Science Notes Chapter 1 Matter in Our Surroundings

→ Materials found around us have different sizes, shapes and forms.

→ Matter is a substance that possesses mass and occupies space.

→ The ancient philosophers of India have classified matter mainly into five basic elements viz. air, earth, fire, water and sky (which are said to be ‘Panch Tatvas’).

→ The modem scientists have classified matter on the basis of its physical properties and chemical nature.

→ The particles of a matter have presumably sufficient space in them.

→ The purity of honey can be tested merely by pouring down just a drop of it into a glass of water as if honey falls in form of a colour line, then it is supposed to be pure.

→ The particles in the matter do always remain in motion continuously i.e., they do have kinetic energy in them.

→ The kinetic energy of the particles speeds up with the increase in temperature.

→ The particles of matter attract one another, but the capability of force of cohesion is different in every material.

→ Matter exists in three different states namely solid, liquid and gas.

→ Solids are fully incompressible, of definite shapes and fixed volume.

→ The particles of solids have the ability to rotate about accordingly to their mean positions.

→ The solids do not have the property to lose their indentity while intermixing into other solid materials.

→ Liquids are comparatively compressible to solids. They have fixed volume, but no definite shape.

→ Gases diffuse up and dissolve into water. Due to this property of them the aquatic plants and animals sustain life in water.

→ The aquatic living beings are able to breathe in water because of the dissolved oxygen in water.

→ In liquids, there is complete possibility of solubility of solids, liquids and gases.

→ There is comparatively more mixing-rate in liquids than that of solids, since in liquid state the particles have maximum space in them and they can freely move about.

→ In comparison to solids and liquids, gases are more compressible.

→ Due to random speed and maximum space, gases mix up with other gases more quickly.

→ In order to convert the degree of temperature from kelvin into Celsius, 273 should be subtracted from the given temperature and to convert from Celsius into kelvin the given temperature should be added to 273.

→ The melting point of a solid indicates the ability of force of cohesion between its particles.

→ At 0°C, the energy of the particles of water use to be more at the similar temperature than that of the energy of the particles of ice.

→ The boiling point of water is 373 K (100°C).

→ By changing in temperature we can convert matter from one state into another.

→ With the increase in pressure and decrease in temperature, gas can be converted into liquid.

→ Camphor and ammonium chloride are volatile substances.

→ Increase in the level-region, increases the rate of vaporisation.

→ With an increase in temperature, the rate of vaporisation rises.

→ Due to vaporisation, cooling occurs.

→ Diffusion: Intermixing of particles of two different materials by themselves is called diffusion.

→ LPG: By compressing butane at high pressure the fuel-gas used in the kitchen to cook is called Liquefied Petroleum Gas (LPG).

→ CNG: The natural gas after treating it with high pressure and such treated gas which is used in vehicles in the form of fuel is called as Compressed Natural Gas (CNG).

→ Density: The mass per unit volume of a substance is called density. i.e„
Density = \(\frac{\text { Mass of the substance }}{\text { Volume of the substance }}\)
or
D = \(\frac{M}{V}\)

→ Thermometer: The device used to measure temperature is called thermometer.

→ Melting Point: That fixed temperature at which a solid converts into a liquid is called melting point.

→ Melting: The process of melting or changing of a solid into liquid state is known as melting.

→ Dormant (Hidden) Heat of Melting: The heat energy that is required to convert 1 kg of a solid at atmospheric pressure at its melting point into liquid is called dormant heat of melting.

→ Boiling Point: That temperature at a certain atmospheric pressure at which a liquid starts boiling is said to be boiling point.

→ Unexposed Heat of Vaporisation: The amount of heat energy that is required to convert 1 kg of liquid at a certain atmospheric pressure at its boiling point is called as dormant/unexposed heat of vaporisation.

→ Sublimation: The process by means of which a substance converts into gaseous state directly from solid state without changing into liauid is called sublimation.

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

→ Thawing: Changing of a liquid in solid state is called thawing.

→ Freezing Point: That fixed degree of temperature at which some liquid stmts converting into solid state is called freezing point.

→ Vaporisation: The process where a liquid changes into vapour state below the temperature of its boiling point is called vaporisation.

→ Humidity: The quantity of water vapours present in the air is called humidity.

Haryana State Board HBSE 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 1 हमारे आस-पास के पदार्थ

HBSE 9th Class Science हमारे आस-पास के पदार्थ Intext Questions and Answers

(पृष्ठ संख्या -4)
प्रश्न 1.
निम्नलिखित में से कौन-से पदार्थ हैं कुर्सी, वायु, स्नेह, गंध, घृणा, बादाम, विचार, शीत, नींबू पानी, इत्र की सुगंध।
उत्तर:
कुर्सी, वायु, बादाम व नींबू पानी, पदार्थ हैं क्योंकि ये स्थान घेरते हैं तथा द्रव्यमान रखते हैं।

प्रश्न 2.
निम्नलिखित प्रेक्षण के कारण बताएँ गर्मा-गरम खाने की गंध कई मीटर दूर से ही आपके पास पहुँच जाती है लेकिन ठडे खाने की महक लेने के लिए आपको उसके पास जाना पड़ता है।
उत्तर:
क्योंकि तापमान बढ़ने से विसरण तेज हो जाता है। इसी कारण गर्मा-गरम खाने की गंध कई मीटर से ही हमारे पास पहुँच जाती है, जबकि ठंडे खाने की महक लेने के लिए हमें उसके पास जाना पड़ता है।

प्रश्न 3.
स्वीमिंग पूल में गोताखोर पानी काट पाता है। इससे पदार्थ का कौन-सा गुण प्रेक्षित होता है?
उत्तर:
जल के कणों के बीच दूरी अपेक्षाकृत अधिक होने के कारण इसमें संपीडय का गुण पाया जाता है इसी कारण स्वीमिंग पूल में गोताखोर पानी (जल) को काट पाता है।

प्रश्न 4.
पदार्थ के कणों की क्या विशेषताएँ होती हैं? ।
उत्तर:
पदार्थ के कणों की निम्नलिखित विशेषताएँ होती हैं

1. पदार्थ के कणों के बीच रिक्त स्थान होता है।
2. पदार्थ के कण निरंतर गतिशील होते हैं अर्थात् उनमें गतिज ऊर्जा होती है।
3. तापमान बढ़ने से पदार्थ के कणों की गतिज ऊर्जा बढ़ जाती है।
4. पदार्थ के कण अपने आप अंतः मिश्रित हो जाते हैं।
5. पदार्थ के कण एक-दूसरे को आकर्षित करते हैं।

(पृष्ठ संख्या -6)

प्रश्न 1.
किसी तत्त्व के द्रव्यमान प्रति इकाई आयतन को घनत्व कहते हैं। (घनत्व = द्रव्यमान/आयतन)। बढ़ते हुए घनत्व के क्रम में निम्नलिखित को व्यवस्थित करें-वायु, चिमनी का धुआँ, शहद, जल, चॉक, रुई और लोहा।
उत्तर:
दिए गए तत्त्वों को घनत्व बढ़ने के साथ आरोही क्रम में लिखने पर निम्नलिखित क्रम प्राप्त होगाचिमनी का धुआँ, वायु, रुई, चॉक, जल, शहद व लोहा।

प्रश्न 2.
(a) पदार्थ की विभिन्न अवस्थाओं के गुणों में होने वाले अंतर को सारणीबद्ध कीजिए।
(b) निम्नलिखित पर टिप्पणी कीजिए-दृढ़ता, संपीडयता, तरलता, बर्तन में गैस का भरना, आकार, गतिज ऊर्जा एवं घनत्व।
उत्तर:
(a) पदार्थ की विभिन्न अवस्थाओं के गुणों में निम्नलिखित अंतर पाए जाते हैं

(b) दृढ़ता-पदार्थ के कणों के बीच लगने वाला आकर्षण बल पदार्थ की दृढ़ता निश्चित करता है। ठोसों में आकर्षण बल अधिक होने के कारण दृढ़ता अधिक होती है। द्रवों में उससे कम तथा गैसों में सबसे कम दृढ़ता होती है।

संपीडयता – किसी पदार्थ पर बल लगाकर उसके कणों की बीच की दूरी को कम करना संपीडयता कहलाता है। गैसों में संपीडयता का गुण पाया जाता है।

तरलता – जिन पदार्थों में बहने का गुण पाया जाता है, उन्हें तरल पदार्थ कहते हैं। द्रवों में तरलता का गुण पाया जाता है।

बर्तन में गैस का भरना-किसी बर्तन में गैस कणों में अंतराणुक बल क्षीण होने के कारण गैस सारे उपलब्ध स्थान को घेर लेती है। अर्थात् गैसों का आयतन निश्चित नहीं होता। इन्हें बर्तन में उच्च दाब पर भरा जाता है।

आकार-ठोस के कण अधिक अंतराणुक बल के साथ जुड़े होने के कारण ठोस को निश्चित आकार प्रदान करते हैं, जबकि द्रवों में यह अंतराणुक बल कम होता है जिस कारण इनका आकार निश्चित नहीं होता। गैसों में भी ऐसा होता है।

गतिज ऊर्जा-कणों की गति के कारण उत्पन्न ऊर्जा को गतिज ऊर्जा कहा जाता है। तापमान बढ़ाने से कणों की गतिज ऊर्जा बढ़ जाती है। ठोस में अधिक गतिज ऊर्जा नहीं होती, द्रव में कुछ गतिज ऊर्जा होती है जबकि गैसों में उच्च गतिज ऊर्जा होती है।

घनत्व-किसी तत्त्व के द्रव्यमान प्रति इकाई आयतन को घनत्व कहते हैं अर्थात् घनत्व = द्रव्यमान । ठोस पदार्थों का घनत्व उच्च होता है, द्रव में निम्न जबकि गैसों में नगण्य होता है।

प्रश्न 3.
कारण बताएँ
(a) गैस पूरी तरह उस बर्तन को भर देती है, जिसमें इसे रखते हैं।
(b) गैस बर्तन की दीवारों पर दबाव डालती है।
(c) लकड़ी की मेज़ ठोस कहलाती है।
(d) हवा में हम आसानी से अपना हाथ चला सकते हैं, लेकिन एक ठोस लकड़ी के टुकड़े से हाथ चलाने के लिए हमें कराटे में दक्ष होना पड़ेगा।
उत्तर:
(a) गैस का आयतन निश्चित न होने के कारण गैस पूरी तरह उस बर्तन को भर देती है जिसमें इसे रखा जाता है।
(b) गैसीय अवस्था में कणों की गति अनियमित और अत्यधिक तीव्र होती है। इस अनियमित गति के कारण गैस के कण आपस में एवं बर्तन की दीवारों से टकराते हैं। बर्तन की दीवार पर गैस कणों द्वारा प्रति इकाई क्षेत्र पर लगे बल के कारण गैस का दबाव बनता है।
(c) लकड़ी की मेज का निश्चित आकार, निश्चित आयतन तथा असंपीडय होने के कारण ठोस कहलाती है।
(d) हवा के कणों के बीच अधिक दूरी होने के कारण हवा में हम हाथ को आसानी से चला सकते हैं, जबकि ठोस लकड़ी के टुकड़े के कणों के बीच की दूरी कम होने के कारण हमें हाथ चलाने के लिए कराटे में दक्ष होना पड़ेगा।

प्रश्न 4.
सामान्यतया ठोस पदार्थों की अपेक्षा द्रवों का घनत्व कम होता है, लेकिन आपने बर्फ के टुकड़े को जल में तैरते हुए देखा होगा। पता लगाइए, ऐसा क्यों होता है?
उत्तर:
सामान्यतया ठोस पदार्थों की अपेक्षा द्रवों का घनत्व कम होता है परंतु जल का घनत्व 4°C पर अधिकतम होता है। जब इसे 4°C से नीचे ठंडा किया जाता है तो 0°C पर बर्फ जम जाती है जिसका घनत्व जल के घनत्व से कम होता है इसी कारण बर्फ जल में तैरती है।

(पृष्ठ संख्या-9)

प्रश्न 1.
निम्नलिखित तापमान को सेल्सियस में बदलें
(a) 300K
(b) 573K
उत्तर:
(a) 300K = (300 – 273)°C = 27°C
(b) 573K = (573 – 273)°C = 300°C

प्रश्न 2.
निम्नलिखित तापमान पर जल की भौतिक अवस्था क्या होगी?
(a) 250°C
(b) 100°C
उत्तर:
(a) 250°C तापमान पर समस्त जल भाप बनकर उड़ जाएगा।
(b) 100°C पर जल उबलना शुरू कर देगा क्योंकि यह जल का क्वथनांक है।

प्रश्न 3.
किसी भी पदार्थ की अवस्था परिवर्तन के दौरान तापमान स्थिर क्यों रहता है?
उत्तर:
किसी भी पदार्थ की अवस्था परिवर्तन के दौरान तापमान स्थिर रहता है क्योंकि अवस्था परिवर्तन के समय पदार्थ को दी जाने वाली समस्त ऊष्मा कणों के पारस्परिक आकर्षण बल को वशीभूत करके पदार्थ की अवस्था को बदलने में उपयोग हो जाती है।

प्रश्न 4.
वायुमंडलीय गैसों को द्रव में परिवर्तन करने के लिए कोई विधि सुझाइए।
उत्तर:
वायुमंडलीय गैसों को किसी बंद बर्तन में बंद करके तथा दाब बढ़ाकर व तापमान घटाकर द्रव में परिवर्तित किया जा सकता है।

(पृष्ठ संख्या-11)

प्रश्न 1.
गर्म, शुष्क दिन में कूलर कमरे को अधिक ठंडा क्यों करता है?
उत्तर:
गर्म, शुष्क दिन में आर्द्रता कम होने के कारण वायु शुष्क होती है जब यह वायु कूलर के पंखे द्वारा खींची जाती है तो कूलर में मैट पर टपकता हुआ जल वाष्पीकृत होकर गर्म व शुष्क वायु की गर्मी को अवशोषित कर लेता है जिसके परिणामस्वरूप वायु ठंडी हो जाती है। यह ठंडी वायु कमरे को ठंडा कर देती है।

प्रश्न 2.
गर्मियों में घड़े का जल ठंडा क्यों होता है?
उत्तर:
घड़े में छोटे-छोटे छिद्र होते हैं जिनके माध्यम से गर्मियों में जल रिसकर वाष्पीकृत होता रहता है तथा वाष्पन के लिए वह ऊष्मा जल से ही लेता है जिस कारण घड़े में रखा जल ठंडा हो जाता है।

प्रश्न 3.
एसीटोन/पेट्रोल या इत्र डालने पर हमारी हथेली ठंडी क्यों हो जाती है?
उत्तर:
एसीटोन पेट्रोल या इत्र को हथेली पर डालने से इसके कण हथेली से ऊर्जा प्राप्त कर वाष्पीकृत हो जाते हैं जिससे हथेली ठंडी हो जाती है।

प्रश्न 4.
कप की अपेक्षा प्लेट से हम गर्म दूध या चाय जल्दी क्यों पी लेते हैं?
उत्तर:
प्लेट की सतह का क्षेत्रफल कप की सतह की अपेक्षा अधिक होता है जिस कारण प्लेट की सतह से वाष्पीकरण अधिक होने के कारण गर्म दूध या चाय जल्दी ठंडी हो जाती है तथा ठंडी होने के कारण जल्दी पी जाती है।

प्रश्न 5.
गर्मियों में हमें किस तरह के कपड़े पहनने चाहिएँ?
उत्तर:
गर्मियों में हमें सूती कपड़े पहनने चाहिएँ क्योंकि गर्मियों में शारीरिक प्रक्रिया के कारण अधिक पसीना आता है, जिससे हमें ठंडक मिलती है जैसा कि हम जानते हैं कि वाष्पीकरण के दौरान द्रव की सतह के कण हमारे शरीर से ऊर्जा प्राप्त करके वाष्प में बदल जाते हैं। वाष्पीकरण की प्रसुप्त ऊष्मा के बराबर ऊष्मीय ऊर्जा हमारे शरीर से अवशोषित हो जाती है, जिससे शरीर शीतल हो जाता है। सूती कपड़े पहनने से जल का अवशोषण अधिक होता है, जिस कारण पसीना इसमें अवशोषित होकर वायुमंडल में आसानी से वाष्पीकृत हो जाता है।

HBSE 9th Class Science हमारे आस-पास के पदार्थ Textbook Questions and Answers

प्रश्न 1.
निम्नलिखित तापमानों को सेल्सियस इकाई में परिवर्तित करें
(a) 300K
(b) 573K
उत्तर:
(a) 300K = (300 – 273)° = 27°C
(b) 573K = (573 — 273)°C = 300°C

प्रश्न 2.
निम्नलिखित तापमानों को केल्विन इकाई में परिवर्तित करें
(a) 25°C
(b) 373°C
उत्तर:
(a) 25°C = (25 + 273)K 298K
(b) 373°C = (373 + 273)K = 646K

प्रश्न 3.
निम्नलिखित अवलोकनों हेतु कारण लिखें
(a) नैफ्थलीन को रखा रहने देने पर यह समय के साथ कुछ भी ठोस पदार्थ छोड़े बिना अदृश्य हो जाती है।
(b) हमें इत्र की गंध बहुत दूर बैठे हुए भी पहुँच जाती है।
उत्तर:
(a) नैफ्थलीन को रखा रहने देने पर यह समय के साथ कुछ भी ठोस पदार्थ छोड़े बिना अदृश्य हो जाती है क्योंकि नैफ्थलीन एक ऊर्ध्वापातित पदार्थ है जो ठोस से सीधा ही गैसीय अवस्था में परिवर्तित हो जाता है।
(b) हमें इत्र की गंध बहुत दूर बैठे हुए भी पहुँच जाती है क्योंकि इत्र वायु में विसरित होने का गुण रखता है।

प्रश्न 4.
निम्नलिखित पदार्थों को उनके कणों के बीच बढ़ते आकर्षण के अनुसार व्यवस्थित करें
(a) जल,
(b) चीनी,
(c) ऑक्सीजन।
उत्तर:
कणों के बीच बढ़ते आकर्षण के अनुसार पदार्थों का व्यवस्थित रूप निम्नलिखित है-
ऑक्सीजन < जल < चीनी

प्रश्न 5.
निम्नलिखित तापमानों पर जल की भौतिक अवस्था क्या है
(a) 25°C,
(b) 0°C,
(c) 100°C
उत्तर:
(a) 25°C पर जल द्रव अवस्था में होगा।
(b) 0°C पर जल ठोस (बफ) अवस्था में होगा।
(c) 100°C पर जल क्वथित अवस्था (भाप) में होगा।

प्रश्न 6.
पुष्टि हेतु कारण दें
(a) जल कमरे के ताप पर द्रव है।
(b) लोहे की अलमारी कमरे के ताप पर ठोस है।
उत्तर:
(a) जल कमरे के ताप पर द्रव है, क्योंकि

1. इस ताप पर इसके अणुओं के बीच लगने वाला आकर्षण बल सामान्य होता है।
2. इसके अणुओं की गतिज सामान्य होती है।

(b) लोहे की अलमारी कमरे के ताप पर ठोस है क्योंकि

1. लोहे के अणुओं के बीच लगने वाला आकर्षण बल काफी अधिक होता है।
2. लोहे के अणुओं के बीच दूरी नगण्य होती है जिस कारण वे निश्चित दूरी तक घूम सकते हैं।

प्रश्न 7.
273K पर बर्फ को ठंडा करने पर तथा जल को इसी तापमान पर ठंडा करने पर शीतलता का प्रभाव अधिक क्यों होता है?
उत्तर:
273K तापमान पर बर्फ को ठंडा करने पर तथा जल को इसी तापमान पर ठंडा करने पर शीतलता का प्रभाव अधिक होता है, क्योंकि बर्फ में संगलन गुप्त ऊष्मा अधिक होती है।

प्रश्न 8.
उबलते हुए जल अथवा भाप में से जलने की तीव्रता किसमें अधिक महसूस होती है?
उत्तर:
उबलते हुए जल अथवा भाप में से जलने की तीव्रता भाप में अधिक महसूस होती है क्योंकि भाप में अतिरिक्त ऊष्मा होती है जिसे वाष्पन की गुप्त ऊष्मा कहते हैं।

प्रश्न 9.
निम्नांकित चित्र के लिए A, B, C, D, E तथा F की अवस्था परिवर्तन को नामांकित करें

उत्तर:
(A) संगलन,
(B) वाष्पन,
(C) संघनन,
(D) जमना,
(E) ऊर्ध्वपातन,
(F) ऊर्ध्वपातन।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.4 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Exercise 9.4

प्रश्न 1.
समांतर. चतुर्भुज ABCD और आयत ABEF एक ही आधार पर स्थित हैं और उनके क्षेत्रफल बराबर हैं। दर्शाइए कि समांतर चतुर्भुज का परिमाप आयत के परिमाप से अधिक है।
हल :

दिया है : एक समांतर चतुर्भुज ABCD और एक आयत ABEF एक ही आधार AB पर स्थित हैं और उनके क्षेत्रफल भी समान हैं।
सिद्ध करना है : समांतर चतुर्भुज ABCD का परिमाप > आयत ABEF का परिमाप।
प्रमाण : क्योंकि समांतर चतुर्भुज और आयत की सम्मुख भुजाएँ बराबर होती हैं।
∴ AB = DC [∵ ABCD एक समांतर चतुर्भुज है।]
और AB = EF [∵ ABEF एक आयत है।]
⇒ DC = EF ……(i)
या AB + DC = AB + EF ……(ii)
क्योंकि दी गई रेखा के किसी बिंदु से खींचे जा सकने वाले सभी खंड इस पर स्थित नहीं हैं, अतः लंब खंड सबसे छोटा है।
∴ BE < BC व AF < AD
या BC > BE व AD > AF
या BC + AD > BE + AF ……(iii)
समीकरण (ii) व (iii) को जोड़ने पर,
AB + DC + BC + AD > AB + EF + BE + AF
या AB + BC + CD + DA > AB + BE + EF + FA
अतः समांतर चतुर्भुज ABCD का परिमाप > आयत ABEF का परिमाप। [इति सिद्धम]

प्रश्न 2.
आकृति में, भुजा BC पर दो बिंदु D और E इस प्रकार स्थित हैं कि BD = DE = EC है। दर्शाइए कि ar (ABD) = ar (ADE) = ar (AEC) है।
क्या आप अब उस प्रश्न का उत्तर दे सकते हैं, जो आपने इस अध्याय की ‘भूमिका’ में छोड़ दिया था कि “क्या बुधिया का खेत वास्तव में बराबर क्षेत्रफलों वाले तीन भागों में विभाजित हो गया है”?

हल :
दिया है : ΔABC की भुजा BC पर दो बिंदु D व E इस प्रकार हैं कि BD = DE = EC
सिद्ध करना है : ar (ΔABD) = ar (ΔADE) = ar (ΔAEC)
रचना : A से AL ⊥ BC खींचो।
प्रमाण : आकृति अनुसार, AL, ΔABD, ΔADE व ΔAEC का शीर्षलंब है।
ar (ΔABD) = \(\frac{1}{2}\) × BD × AL …..(i)
ar (ΔADE) = \(\frac{1}{2}\) × DE × AL …(ii)
ar (ΔAEC) = \(\frac{1}{2}\) × EC × AL …(iii)
परंतु BD = DE = EC (दिया है)…(iv)
समीकरण (i), (ii), (iii) व (iv) की तुलना करने पर,
ar (ΔABD) = ar (ΔADE) = ar (ΔAEC)
हां सभी त्रिभुजों के शीर्षलंब समान हैं। बुधिया इस प्रश्न के उत्तर द्वारा अपने खेत तीन समान भागों में बांट सकती है।

प्रश्न 3.
आकृति में, ABCD, DCFE और ABFE समांतर चतुर्भुज हैं। दर्शाइए कि ar (ADE) = ar (BCF) है।

हल :
दिया है : आकृति में ABCD, DCFE व ABFE तीन समांतर चतुर्भुज हैं।
सिद्ध करना है : ar (ΔADE) = ar (ΔBCF)
प्रमाण : क्योंकि समांतर चतुर्भुजों की सम्मुख भुजाएं बराबर होती हैं।
∴ AD = BC [|| चतुर्भुज ABCD की भुजाएं]
DE = CF [|| चतुर्भुज DCFE की भुजाएं]
AE = BF [|| चतुर्भुज ABFE की भुजाएं]
अब ΔADE और ΔBCF में,
AD = BC [प्रमाणित]
DE = CF [प्रमाणित]
AE = BF [प्रमाणित]
∴ ΔADE ≅ ΔBCF [भुजा-भुजा-भुजा सर्वांगसमता]
अतः ar (ΔADE) = ar (ΔBCF) [इति सिद्धम]

प्रश्न 4.
आकृति में, ABCD एक समांतर चतुर्भुज है और BC को एक बिंदु Q तक इस प्रकार बढ़ाया गया है कि AD = CQ है। यदि AQ भुजा DC को P पर प्रतिच्छेद करती है, तो दर्शाइए कि ar (BPC) = ar (DPO) है।

हल :
दिया है : एक समांतर चतुर्भुज ABCD की भुजा BC को Q तक इस प्रकार बढ़ाया गया है कि AD = CQ, AQ भुजा DC को P पर प्रतिच्छेद करती है।
सिद्ध करना है : ar (ΔBPC) = ar (ΔDPQ)
रचना : A व C को मिलाओ।
प्रमाण : क्योंकि ΔADC और ΔADQ एक ही आधार AD और एक ही समांतर रेखाओं AD व CQ के मध्य स्थित है।
∴ ar (ΔADC) = ar (ΔADQ)
दोनों ओर से ΔADP का क्षेत्रफल घटाने पर,
ar (ΔADC) – ar (ΔADP) = ar (ΔADQ) – ar (ΔADP)
⇒ ar (ΔAPC) = ar (ΔDPQ) …(i)
इसी प्रकार ΔAPC और ΔPCB एक ही आधार PC तथा एक ही समांतर रेखाओं PC व AB के मध्य स्थित हैं।
∴ ar (ΔAPC) = ar (ΔPCB)
समीकरण (i) व (ii) की तुलना से,
ar (ΔPCB) = ar (ΔDPQ)
या ar (ΔBPC) = ar (ΔDPQ) [इति सिद्धम]

प्रश्न 5.
आकृति में, ABC और BDE दो समबाहु त्रिभुज इस प्रकार हैं कि D भुजा BC का मध्य-बिंदु है। यदि AE भुजा BC को F पर प्रतिच्छेद करती है, तो दर्शाइए कि
(i) ar (BDE) = \(\frac{1}{4}\)ar (ABC)
(ii) ar (BDE) = \(\frac{1}{2}\)ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = \(\frac{1}{8}\)ar (AFC)

हल :
दिया है : ΔABC और ΔBDE दो समबाहु त्रिभुज इस प्रकार हैं कि D भुजा BC का मध्य-बिंदु है। AE भुजा BC को F पर प्रतिच्छेद करती है।

सिद्ध करना है : (i) ar (ΔBDE) = \(\frac{1}{4}\)ar (ΔABC)
(ii) ar (ΔBDE) = \(\frac{1}{2}\)ar (ΔBAE)
(iii) ar (ΔABC) = 2 ar (ΔBEC)
(iv) ar (ΔBEF) = ar (ΔAFD)
(v) ar (ΔBFE) = 2 ar (ΔFED)
(vi) ar (ΔFED) = \(\frac{1}{8}\)ar (ΔAFC)
रचना : EC व AD को मिलाओं तथा EL ⊥ BC खींचो।
प्रमाण : माना समबाहु ΔABC की प्रत्येक भुजा = a मात्रक
(i) ar (ΔABC) = \(\frac{\sqrt{3}}{4}\)a² …(1) [∵ समबाहु त्रिभुज का क्षेत्रफल = \(\frac{\sqrt{3}}{4}\)(भुजा)²]
ar (ΔBDE) = \(\frac{\sqrt{3}}{4}\left(\frac{a}{2}\right)^2\)
= \(\frac{\sqrt{3}}{4}\left(\frac{a^2}{4}\right)=\frac{\sqrt{3} a^2}{16}\)
= \(\frac{1}{4} \times \frac{\sqrt{3}}{4} a^2\)
= \(\frac{1}{4}\)ar (ΔABC) [समीकरण (1) से]
अतः ar (ΔBDE) = \(\frac{1}{4}\)ar (ΔABC) [इति सिद्धम]

(ii) ar (ΔBDE) = \(\frac{1}{2}\)ar (ΔBEC) …..(2)
[∵ ED, ΔBEC की माध्यिका है तथा माध्यिका A के दो समान भाग करती है।]
अब ∠EBC = ∠ACB [प्रत्येक = 60%]
परंतु यह एकांतर कोण हैं।
∴ BE || AC
ΔBEC तथा ΔBEA एक ही आधार BE तथा एक ही समांतर रेखाओं BE व AC के मध्य स्थित हैं।
∴ ar (ΔBEC) = ar (ΔBEA) …..(3)
\(\frac{1}{2}\)ar (ΔBEC) = \(\frac{1}{2}\)ar (ΔBEA) …..(3)
समीकरण (2) व (3) की तुलना से,
ar (ΔBDE) = \(\frac{1}{2}\)ar (ΔBAE) [इति सिद्धम]

(iii) ar (ΔBDE) = \(\frac{1}{4}\)ar (ΔABC) ….(4) [प्रमाणित]
ar (ΔBDE) = \(\frac{1}{2}\)ar (ΔBEC) ….(5) [प्रमाणित]
समीकरण (4) तथा (5) की तुलना से,
\(\frac{1}{4}\)ar (ΔABC) = \(\frac{1}{2}\)ar (ΔBEC)
ar (ΔABC) = 2 ar (ΔBEC) [इति सिद्धम]

(iv) ∠ABD = ∠BDE [प्रत्येक = 60%]
परन्तु यह एकांतर कोण हैं।
∴ AB || DE
अब ΔBED व ΔAED एक ही आधार ED तथा एक ही समांतर रेखाओं ED और AB के मध्य स्थित हैं।
∴ ar (ΔBED) = ar (ΔAED)
दोनों ओर से ΔEDF का क्षेत्रफल घटाने पर,
ar (ΔBED) – ar (ΔEDF) = ar (ΔAED) – ar (ΔEDF)
∴ ar (ΔBFE) = ar (ΔAFD) …..(6) [इति सिद्धम]

(v) ΔABD में,
AD² = AB² – BD²
= (a²) – (\(\frac{a}{2}\))²
= \(a^2-\frac{a^2}{4}=\frac{3 a^2}{4}\)
AD = \(\frac{\sqrt{3} a}{2}\)
ΔLED में,
EL² = DE² – DL²

समीकरण (7) व (8) से,
ar (ΔAFD) = 2 ar (ΔEFD)
समीकरण (6) व (9) से,
ar (ΔBEF) = 2 ar (ΔEFD)

(vi) अब ar (ΔBDE) = \(\frac{1}{4}\)ar (ΔABC) [प्रमाणित]
ar (ΔBEF) + ar (ΔFED) = \(\frac{1}{4}\) × 2 ar (ΔADC)
2 ar (ΔFED) + ar (ΔFED) = \(\frac{1}{2}\)ar (ΔADC) {∵ ar (ΔBEF) = 2ar (ΔFED) प्रमाणित}
3 ar (ΔFED) = \(\frac{1}{2}\)[ar (ΔAFC) – ar (ΔAFD)]
3 ar (ΔFED) = \(\frac{1}{2}\)ar (ΔAFC) – \(\frac{1}{2}\) × 2ar (ΔFED)
{∵ ar (ΔAFD) = 2 ar (ΔFED) प्रमाणित}
4 ar (ΔFED) = \(\frac{1}{2}\)ar (ΔAFC)
ar (ΔFED) = \(\frac{1}{8}\)ar (ΔAFC) [इति सिद्धम]

प्रश्न 6.
चतुर्भुज ABCD के विकर्ण AC और BD परस्पर बिंदु P पर प्रतिच्छेद D करते हैं। दर्शाइए कि ar (APB) × ar (CPD) = ar (APD) × ar (BPC) है।
हल :

दिया है : चतुर्भुज ABCD के विकर्ण AC और BD परस्पर बिंदु P पर प्रतिच्छेद करते हैं।
सिद्ध करना है : ar (ΔAPB) × ar (ΔCPD) = ar (ΔAPD) × ar (ΔBPC)
रचना : AM ⊥ BD व CN ⊥ BD खींचिए।
प्रमाण: ar (ΔAPB) = \(\frac{1}{2}\) × PB × AM …(i)
ar (ΔCPD) = \(\frac{1}{2}\) × DP × CN …(ii)
ar (ΔAPD) = \(\frac{1}{2}\) × DP × AM….(iii)
ar (ΔBPC) = \(\frac{1}{2}\) × PB × CN …(iv)
समीकरण (i) व (ii) से,
ar (ΔAPB) × ar (ΔCPD) = \(\frac{1}{2}\) × PB × AM × \(\frac{1}{2}\) × DP × CN …(v)
समीकरण (iii) व (iv) से,
ar (ΔAPD) × ar (ΔBPC) = \(\frac{1}{2}\) × DP × AM × \(\frac{1}{2}\) × PB × CN …(vi)
समीकरण (v) व (vi) से,
ar (ΔAPB) × ar (ΔCPD) = ar (ΔAPD) × ar (ΔBPC) [इति सिद्धम]

प्रश्न 7.
Pऔर Q क्रमशः त्रिभुज ABC की भुजाओं AB और BC के मध्य-बिंदु हैं तथा R रेखाखंड AP का मध्य-बिंदु है। दर्शाइए कि:
(i) ar (PRQ) = \(\frac{1}{2}\)ar (ARC)
(ii) ar (RQC) = \(\frac{3}{8}\)ar (ABC)
(iii) ar (PBQ)= ar (ARC)
हल :
दिया है : ΔABC की भुजाओं AB और BC के मध्य-बिंदु क्रमशः P व Q है। R रेखाखंड AP का मध्य-बिंदु है।
सिद्ध करना है :
(i) ar (ΔPRQ) = \(\frac{1}{2}\)ar (ΔARC)
(ii) ar (ΔRQC) = \(\frac{3}{8}\)ar (ΔABC)
(iii) ar (ΔPBQ) = ar (ΔARC)
रचना : AQ तथा PC को मिलाओ।

प्रमाण : (i) ∵ ar (ΔPQR) = \(\frac{1}{2}\)ar (ΔAPQ) [∵ QR त्रिभुज APQ की माध्यिका है जो त्रिभुज को दो समान क्षेत्रफलों वाले त्रिभुजों में बांटती है]
= \(\frac{1}{2} \times \frac{1}{2}\)ar (ΔABQ) [∵ QP त्रिभुज ABQ की माध्यिका है]
= \(\frac{1}{4}\)ar (ΔABQ)
= \(\frac{1}{4} \times \frac{1}{2}\)ar (ΔABC) [∵ AQ त्रिभुज ABC की माध्यिका है।]
= \(\frac{1}{8}\)ar (ΔABC) …..(i)
अब ar (ΔARC) = \(\frac{1}{2}\)ar (ΔAPC) [∵ CR त्रिभुज APC की माध्यिका है।]
= \(\frac{1}{2} \times \frac{1}{2}\)ar (ΔABC) [∵ CP त्रिभुज ABC की माध्यिका है।]
= \(\frac{1}{4}\)ar (ΔABC) ……(ii)
समीकरण (i) व (ii) की तुलना से,
ar (ΔPQR) = \(\frac{1}{8}\)ar (ΔABC)
= \(\frac{1}{2} \times \frac{1}{4}\)ar (ΔABC)
= \(\frac{1}{2}\)ar (ΔARC) [इति सिद्धम]

(ii) ar (ΔRQC) = ar (ΔRQA) + ar (ΔAQC) – ar (ΔARC) ….(iii)
अब ar (ΔRQA) = \(\frac{1}{2}\)ar (ΔPQA) [∵ RQ, ΔPQA की माध्यिका है।]
= \(\frac{1}{2} \times \frac{1}{2}\)ar (AQB) [∵ PQ, ΔAQB की माध्यिका है।]
= \(\frac{1}{4}\)ar (AQB)
= \(\frac{1}{4} \times \frac{1}{2}\)ar(ABC) [∵ AQ, ΔABC की माध्यिका है।]
= \(\frac{1}{8}\)ar (ΔABC) …..(iv)
अतः ar (ΔAQC) = \(\frac{1}{2}\)ar (ΔABC) …..(v)
[∵ AQ, ΔABC की माध्यिका है।]
या ar (ΔARC) = \(\frac{1}{2}\)ar(ΔAPC) [∵ CR, ΔAPC की माध्यिका है।]
= \(\frac{1}{2} \times \frac{1}{2}\)ar (ΔABC) [∵ CP, ΔABC की माध्यिका है।
= \(\frac{1}{4}\)ar (ΔABC) …..(vi)
समीकरण (iii), (iv), (v) व (vi) की तुलना से,
ar (ΔRQC) = \(\frac{1}{8}\)ar (ΔABC) + \(\frac{1}{2}\)ar(ΔABC) – \(\frac{1}{4}\)ar(ΔABC)
= \(\frac{3}{8}\)ar (ΔABC) [इति सिद्धम]

(iii) ∵ ar (ΔPBQ) = \(\frac{1}{2}\)ar (ΔABQ) [∵ PQ, ΔABQ की माध्यिका है।]
= \(\frac{1}{2} \times \frac{1}{2}\)ar (ΔABC) [∵ AQ, ΔABC की माध्यिका है।]
= \(\frac{1}{4}\)ar (ΔABC)
= ar (ΔARC) [समीकरण (iv) से]
[इति सिद्धम]

प्रश्न 8.
आकृति में, ABC एक समकोण त्रिभुज है जिसका कोण A समकोण है। BCED, ACFG और ABMN क्रमशः भुजाओं BC, CA और AB पर बने वर्ग हैं। रेखाखंड AX ⊥ DE भुजा BC को बिंदु Y पर मिलता है। दर्शाइए किः

(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar (CYXE) = 2 ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
हल :
दिया है : समकोण ΔABC में ∠A समकोण है। BCED, ACFG व ABMN क्रमशः भुजाओं BC, CA व AB बने तीन वर्ग हैं। रेखाखंड AX ⊥ DE भुजा BC को बिंदु Y पर मिलता है।
सिद्ध करना है :
(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar (CYXE) = 2 ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
प्रमाण : (i) ∠BAC + ∠CAG = 180° (प्रत्येक = 90°)
∴ BAG एक सरल रेखा है।
इसी प्रकार CAN एक सरल रेखा है।
∠MBA = ∠CBD (प्रत्येक = 90°)
दोनों ओर ∠ABC जोड़ने पर,
∠MBA + ∠ABC = ∠CBD + ∠ABC
⇒ ∠MBC = ∠ABD
ΔMBC तथा ΔABD में,
BC = BD [वर्ग BCDE की भुजाएं]
MB = AB [वर्ग ABMN की भुजाएं]
∠MBC = ∠ABD [प्रमाणित]
∴ ΔMBC ≅ ΔABD [भुजा-कोण-भुजा सर्वांगसमता]
[इति सिद्धम]

(ii) ΔABD तथा आयत BYXD एक ही आधार BD तथा एक ही समांतर रेखाओं BD एवं AX के मध्य स्थित हैं।
∴ ar (आयत BYXD) = 2 ar (ΔABD)
∴ ar (आयत BYXD) = 2 ar (ΔMBC) …..(i)
[∵ ΔABD = ΔMBC प्रमाणित] [इति सिद्धम]

(iii) ΔMBC तथा वर्ग ABMN एक ही आधार MB तथा एक ही समांतर रेखाओं MB तथा NC के मध्य स्थित हैं।
∴ ar (वर्ग ABMN) = 2 (ar AMBC) …..(ii)
समीकरण (i) व (ii) से,
ar (आयत BYXD) = ar (वर्ग ABMN) [इति सिद्धम]

(iv) ΔFCB तथा ΔACE में,
CB = CE [वर्ग BCED की भुजाएं]
FC = AC [वर्ग ACFG की भुजाएं]
∠FCB = ∠ACE [प्रत्येक = ∠ACB+ 90°]
∴ ΔFCB ≅ ΔACE [भुजा-कोण-भुजा सर्वांगसमता] [इति सिद्धम]

(v) ΔACE तथा आयत CYXE एक ही आधार CE तथा एक ही समांतर रेखाओं के मध्य स्थित हैं।
∴ ar (आयत CYXE) = 2 ar (ΔACE)
∴ ar (आयत CYXE) = 2 ar (ΔFCB) ….(iii) [इति सिद्धम]
[∵ ΔACE ≅ ΔFCB प्रमाणित] [इति सिद्धम]

(vi) ΔFCB तथा वर्ग ΔCFG एक ही आधार CF तथा एक ही समांतर रेखाओं के मध्य स्थित हैं।
∴ ar (वर्ग ΔCFG) = 2 ar (ΔFCB) ….(iv)
समीकरण (iii) व (iv) की तुलना से,
ar (आयत CYXE) = ar (वर्ग ACFG)

(vii) ar (वर्ग BCED) = ar (आयत BYXD) + ar (आयत CYXE)
⇒ ar (वर्ग BCED) = ar (वर्ग ABMN) + ar (वर्ग ACFG)
{∵ ar (आयत BYXD) = ar (वर्ग ABMN) तथा ar (आयत CYXE) = ar (वर्ग ACFG)}
[इति सिद्धम]

Haryana State Board HBSE 9th Class Maths Solutions Chapter 7 त्रिभुज Ex 7.4 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 7 त्रिभुज Exercise 7.4

प्रश्न 1.
दर्शाइए कि समकोण त्रिभुज में कर्ण सबसे लंबी भुजा होती है।
हल :

माना ΔPQR एक समकोण त्रिभुज है जिसमें,
∠PQR = 90°
हम जानते हैं कि ΔPQR में,
∠PQR + ∠QRP + ∠RPQ = 180°
या 90° + ∠QRP + ∠RPQ = 180°
या ∠QRP + ∠RPQ = 180° – 90° = 90°
⇒ ∠QRP = ∠RPQ = न्यून कोण
अतः ∠QRP < 90° तथा ∠RPQ < 90°
इस प्रकार ΔPQR में ∠PQR सबसे बड़ा कोण है तथा सबसे बड़े कोण के सामने की भुजा सबसे बड़ी होती है।
∴ PR सबसे बड़ी भुजा है।
अतः समकोण त्रिभुज में कर्ण सबसे लंबी भुजा होती है। [इति सिद्धम]

प्रश्न 2.
आकृति में, ΔABC की भुजाओं AB और AC को क्रमशः बिंदुओं Pऔर Q तक बढ़ाया गया है। साथ ही, ∠PRC < ∠QCB है। दर्शाइए कि AC > AB है।

हल :
यहाँ पर दिया गया है,
∠PBC < ∠QCB या – ∠PBC > – ∠QCB
या 180° – ∠PBC > 180° – ∠QCB
या ∠ABC > ∠ACB
या AC > AB [∵ बड़े कोण की सम्मुख भुजा बड़ी होती है।] [इति सिद्धम]

प्रश्न 3.
आकृति में, ∠B < ∠A और ∠C < ∠D है। दर्शाइए कि AD < BC है।

हल :
यहाँ पर दिया गया है,
∠B < ∠A और ∠C < ∠D
⇒ AO < BO तथा OD < OC [∵ बड़े कोण की सम्मुख भुजा बड़ी होती है]
दोनों को जोड़ने पर,
AO + OD < BO + OC
⇒ AD < BC [इति सिद्धम]

प्रश्न 4.
AB और CD क्रमशः एक चतुर्भुज ABCD की सबसे छोटी और सबसे बड़ी भुजाएं हैं (देखिए आकृति)। दर्शाइए कि ∠A > ∠C और ∠B > ∠D है।

हल :
यहाँ पर, ABCD एक चतुर्भुज है, जिसमें AB सबसे छोटी भुजा है और CD सबसे बड़ी भुजा है।
AC तथा BD को मिलाया।
क्योंकि AB, चतुर्भुज ABCD की सबसे छोटी भुजा है।
∴ ΔABC में, BC > AB
∠8 > ∠3 ………..(i)
[∵ बड़ी भुजा के सामने का कोण बड़ा होता है]
इसी प्रकार CD चतुर्भुज ABCD की सबसे बड़ी भुजा है।
∴ ΔACD में,
CD > AD
∠7 > ∠4 …..(ii) [∵ बड़ी भुजा के सामने का कोण बड़ा होता है]

समीकरण (i) व (ii) से,
∠8 + ∠7 > ∠3 + ∠4
या ∠A > ∠C [इति सिद्धम]
अब ΔABD में,
AD > AB [∵ AB सबसे छोटी भुजा है]
या ∠1 > ∠6 ….(iii)
ΔBDC में,
CD > BC [∵ CD सबसे बड़ी भुजा है]
∴ ∠2 > ∠5 ……..(iv)
समीकरण (iii) व (iv) से
∠1 + ∠2 > ∠5 + ∠6
या ∠B > ∠D [इति सिद्धम]

प्रश्न 5.
आकृति में, PR > PQ है और PS कोण OPR को समद्विभाजित करता है। सिद्ध कीजिए कि ∠PSR > ∠PSQहै।
हल :

यहाँ पर ΔPQR में, PR > PQ [दिया है]
⇒ ∠PQR > ∠PRQ [∵ बड़ी भुजा के सामने का कोण बड़ा होता है]
दोनों ओर ∠1 जोड़ने पर,
∠PQR + ∠1 > ∠PRQ + ∠1
या ∠PQR + ∠1 > ∠PRQ + ∠2 ……..(i)
[∵ PS, ∠P का कोण समद्विभाजक है, ∴ ∠1 = ∠2]
अब ΔPQS में,
∠PQR + ∠1 + ∠PSQ = 180°
∠PQR + ∠1 = 180° – ∠PSQ …(ii)
इसी प्रकार ΔPSR में, ∠PRQ + ∠2 = 180° – ∠PSR ……(iii)
समीकरण (i), (ii) व (iii) से,
∴ 180° – ∠PSQ > 180° – ∠PSR
या – ∠PSQ > -∠PSR
या ∠PSQ < ∠PSR अर्थात ∠PSR > ∠PSQ [इति सिद्धम]

प्रश्न 6.
दर्शाइए कि एक रेखा पर एक दिए हुए बिंदु से, जो उस रेखा पर स्थित नहीं है, जितने रेखाखंड खींचे जा सकते हैं, उनमें लंब रेखाखंड सबसे छोटा होता है।

हल :
माना P कोई एक बिंदु है, जो रेखा l पर नहीं है,
तथा PM ⊥ l,M के अतिरिक्त रेखा l पर N कोई बिंदु है।
अब ΔPMN में,
∠M = 90°
⇒ ∠N < 90° [∵ ∠MPN + ∠PNM = 90° ⇒ ∠N < 90°]
या ∠N < ∠M
या PM < PN [∵ बड़े कोण के सामने की भुजा बड़ी होती है]
अतः P से l तक रेखाखंडों में से PM लंब सबसे छोटा रेखाखंड है। [इति सिद्धम]

Haryana State Board HBSE 9th Class Maths Solutions Chapter 7 त्रिभुज Ex 7.3 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 7 त्रिभुज Exercise 7.3

प्रश्न 1.
ΔABC और ΔDBC एक ही आधार BC पर बने दो समद्विबाहु त्रिभुज इस प्रकार हैं कि A और D भुजा BC के एक ही ओर स्थित हैं (देखिए आकृति)।
यदि AD बढ़ाने पर BC को P पर प्रतिच्छेद करे, तो दर्शाइए कि
(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP कोण A और कोण D दोनों को समद्विभाजित करता है।
(iv) AP रेखाखंड BC का लंब समद्विभाजक है।

हल :
(i) ΔABD और ΔACD में,
AB = AC [दिया है]
BD = DC [दिया है]
तथा AD = AD [उभयनिष्ठ]
∴ ΔABD = ΔACD [भुजा-भुजा-भुजा सर्वांगसमता]

(ii) ΔABP और ΔACP में,
AB = AC [दिया है]
∠BAP = ∠CAP [सर्वांगसम ΔABD और ΔACD के संगत भाग]
AP = AP [उभयनिष्ठ]
∴ ΔABP ≅ ΔACP [भुजा-कोण-भुजा सर्वांगसमता]

(iii) क्योंकि ΔABD ≅ ΔACD [प्रमाणित]
∴ ∠BAD = ∠CAD
अर्थात AD, ∠A को समद्विभाजित करता है।
अतः AP, ∠A को समद्विभाजित करता है।
अब ΔBDP और ΔCDP में,
BD = CD [दिया है]
BP = CP [सर्वागसम ΔABP और ΔACP के संगत भाग]
DP = DP [उभयनिष्ठ]
∴ ΔBDP ≅ ΔCPD [भुजा-भुजा-भुजा सर्वांगसमता]
⇒ ∠BDP = ∠CDP [सर्वांगसम त्रिभुजों के संगत भाग]
अर्थात DP, ∠D को समद्विभाजित करता है।
या AP, ∠D को समद्विभाजित करता है। ……….(ii)
समीकरण (i) व (ii) से,
AP, ∠A व ∠D को समद्विभाजित करता है। [इति सिद्धम]

(iv) क्योंकि AP भुजा BC पर स्थित है।
∴ ∠APB + ∠APC = 180°
ΔBAP ≅ ΔCAP [प्रमाणित]
∠APB = ∠APC [सर्वांगसम त्रिभुजों के संगत भाग]
अतः ∠APB + ∠APB = 180° [∵ ∠APB = ∠APC]
या 2∠APB = 180°
या ∠APB = \(\frac {180°}{2}\) = 90°
अतः ∠APB = ∠APC = 90° तथा BP = PC [प्रमाणित]
इस प्रकार AP रेखाखंड BC का लंब समद्विभाजक है। [इति सिद्धम]

प्रश्न 2.
AD एक समद्विबाहु त्रिभुज ABC का एक शीर्षलंब है, जिसमें AB = AC है। दर्शाइए कि-
(i) AD रेखाखंड BC को समद्विभाजित करता है।
(ii) AD कोण A को समद्विभाजित करता है।

हल :
यहाँ पर दिया गया है, AD, शीर्ष A से डाला हुआ लंब है, जोकि समद्विबाहु ΔABC के आधार BC के सम्मुख है। AB = AC, ∠ADC = ∠ADB = 90°
अब ΔADB और ΔADC में,
कर्ण AB = कर्ण AC [दिया है]
AD = AD
∠ADC = ∠ADB [∵ प्रत्येक = 90°]
∴ ΔADB ≅ ΔADC [समकोण-कर्ण-भुजा सर्वांगसमता]
⇒ BD = DC तथा ∠BAD = ∠DAC [सर्वांगसमता त्रिभुजों के संगत भाग]
अतः (i) AD, BC को समद्विभाजित करता है। [इति सिद्धम]
तथा (ii) AD, ∠A को समद्विभाजित करता है। [इति सिद्धमा]

प्रश्न 3.
एक त्रिभुज ABC की दो भुजाएं AB और BC तथा माध्यिका AM क्रमशः एक-दूसरे त्रिभुज की भुजाओं PQ और QR तथा माध्यिका PN के बराबर हैं (देखिए आकृति) दर्शाइए कि
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR

हल :
(i) यहाँ पर दिया गया है, ΔABC और ΔPQR में,
AB = PQ
BC = QR
तथा AM = PN
क्योंकि AM तथा PN क्रमशः ΔARC तथा ΔPQR की माध्यिकाएँ हैं।
अब BC = QR [दिया है]
या \(\frac {1}{2}\)BC = \(\frac {1}{2}\)QR
या BM = QN
अब, ΔABM तथा ΔPQN में,
AB = PQ [दिया है]
BM = QN [प्रमाणित]
तथा AM = PN [दिया है]
∴ ΔΑΒΜ ≅ ΔΡQΝ [भुजा-भुजा-भुजा सर्वांगसमता] [इति सिद्धम]

(ii) ∠B = ∠Q
सर्वांगसम त्रिभुजों के संगत भाग
अब ΔABC तथा ΔPQR में,
AB = PQ [दिया है]
∠B = ∠Q [प्रमाणित]
BC = QR [दिया है]
∴ ΔABC ≅ ΔPQR [भुजा-कोण-भुजा सर्वांगसमता] [इति सिद्धम]

प्रश्न 4.
BE और CF एक त्रिभुज ABC के दो बराबर शीर्षलंब हैं। RHS सर्वांगसमता नियम का प्रयोग करके सिद्ध कीजिए कि ΔABC एक समद्विबाहु त्रिभुज है।
हल :

ΔBCF और ΔCBE में,
∠BFC = ∠CEB [प्रत्येक = 90°]
कर्ण BC = कर्ण BC [उभयनिष्ठ]
FC = EB [दिया है]
ΔBCF ≅ ΔCBE [समकोण-कर्ण-भुजा सर्वांगसमता]
⇒ ∠FBC = ∠ECB [सर्वागसम त्रिभुजों के संगत भाग]
अतः ΔABC एक समद्विबाहु त्रिभुज है। [इति सिद्धम]

प्रश्न 5.
ABC एक समद्विबाहु त्रिभुज है, जिसमें AB = AC है। AP ⊥ BC खींचकर दर्शाइए कि ∠B = ∠C है।
हल :

यहाँ पर, ΔABP और ΔACP में,
AB = AC [दिया है]
AP = AP [उभयनिष्ठ]
तथा ∠APB = ∠APC [प्रत्येक = 90°]
∴ ΔABP ≅ ΔACP [समकोण-कर्ण-भुजा सर्वांगसमता]
⇒ ∠B = ∠C [सर्वांगसम त्रिभुजों के संगत भाग] [इति सिद्धम|

Haryana State Board HBSE 9th Class Science Important Questions Chapter 11 Work and Energy Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 11 Work and Energy

Very Short-Answer Type Questions

Question 1.
What is work ?
Answer:
When force is applied on an object and it moves in the direction of force, it is said to be work done.

Question 2.
On which the quantity of work depends ?
Answer:
The quantity of work depends upon the applied force & distance moved.

Question 3.
What is the formula of work ?
Answer:
Work = Force x Displacement.

Question 4.
What is the unit of work ?
Answer:
Newton-metre (Joule).

Question 5.
What is energy ?
Answer:
The capacity of doing work is called energy.

Question 6.
What are the types of mechanical energy ?
Answer:
Mechanical energy is of two types:
(1) Kinetic energy,
(2) Potential energy.

Question 7.
What is potential energy ?
Answer:
The energy possessed by a body due to its position or change in shape is called potential energy.

Question 8.
Which type of energies are in aircraft flying on height ?
Answer:
Both type of energies-kinetic and potential.

Question 9.
Which type of energy is there in a stone lying on a hill ?
Answer:
Potential energy.

Question 10.
What is mechanical Energy ?
Answer:
The sum of kinetic energy and potential energy of a body is called mechanical energy.

Question 11.
Which type of energy is there when a bullet is fired from a gun ?
Answer:
Kinetic energy.

Question 12.
Write the formula for kinetic energy.
Answer:
Kinetic energy = \(\frac {1}{2}\) mv², where m → mass, v → velocity.

Question 13.
Write the formula for potential energy.
Answer:
Potential energy = mgh, where m → mass, g → acceleration due to gravity and h → height.

Question 14.
What is the law of conservation of energy ?
Answer:
According to this law, energy can neither be created nor be destroyed but can only be transferred from one form to another.

Question 15.
What is the effect on energy when work is done ?
Answer:
When work is done, energy decrease in some extent.

Question 16.
What is the work done by a man when he lifts a luggage of mass 10 kg on his head and moves 50 metres on horizontal road ?
Answer:
zero.

Question 17.
From where, our body gets energy ?
Answer:
From food taken by us.

Question 18.
Which type of energy is used by plants ?
Answer:
Solar energy.

Question 19.
Write an important use of wind energy.
Answer:
Wind energy is used to grind the flour by wind mills.

Question 20.
Write the name of any two natural sources of energy.
Answer:
(i) Sun
(ii) Water.

Question 21.
Write any four types of energy.
Answer:
(i) Mechanical Energy
(ii) Chemical Energy
(iii) Light Energy
(iv) Sound Energy.

Question 22.
What is gravitational potential energy ?
Answer:
The energy stored in a body when work is done against force of gravity is called gravitational potential energy.

Question 23.
What is mean by elastic potential energy ?
Answer:
The energy possessed by an object due to change in its shape is called elastic potential energy.

Question 24.
Why spark is produces when two stones collides with each other ?
Answer:
Because kinetic energy of stone is converted into light energy.

Question 25.
How arrow gets its kinetic energy ?
Answer:
The arrow converts the potential energy from bow into kinetic energy.

Question 26.
Out of energy and acceleration, which is vector ?
Answer:
Acceleration is a vector quantity.

Question 27.
The kinetic energy goes on decreasing when it is thrown vertically upward. What will be the motion at highest point ?
Answer:
Zero.

Question 28.
Write the name of that device which converts electric energy into mechanical energy.
Answer:
Electric motor.

Question 29.
Which energy is converted in other in a hydroelectric power centre ?
Answer:
The kinetic energy of water is converted into electric energy.

Question 30.
What is power ? What is its unit ?
Answer:
The rate of doing work is power. Its unit is watt.

Question 31.
If the velocity of an object is increased three times, what will be its kinetic energy ?
Answer:
The kinetic energy will increase nine times.

Question 32.
Which type of energy is there in a expanded spring ?
Answer:
Potential energy.

Question 33.
Which type of energy is there in a dam ?
Answer:
Potential energy.

Question 34.
Which type of energy is there in the spring of watch ?
Answer:
Potential energy which gives the kinetic energy to the needles of watch. ,

Question 35.
Which type of energy is there in running train ?
Answer:
Kinetic energy.

Question 36.
What will be potential energy of that body whose height is doubled ?
Answer:
Its potential energy will be doubled.

Question 37.
What is the relation between 1 kilowatt and watt ?
Answer:
1 kilowatt = 1000 watt.

Question 38.
Why the palms become hot when rubbed ?
Answer:
Because kinetic energy of palms is converted into heat energy.

Question 39.
How much energy is required to lift a stone of mass 1 kg to a height of 1 metre ?
Answer:
9.8 Joules [∵ W = mgh = 1 x 9.8 x 1 = 9.8 joules]

Question 40.
A person and his son are running with uniform velocity. If the mass of person become two times more than of his son, then what will be the ratio between kinetic energy of both ?
Answer:
2 : 1 [ ∵ K.E. = \(\frac {1}{2}\) mv² and v is equal]

Question 41.
Which type of energy is required to move for a sailboat ?
Answer:
The sailboat moves by the kinetic energy of wind.

Question 42.
What will be the energy a body has a capacity to do a work of 10 joules ?
Answer:
10 Joules.

Question 43.
Write the name of two units of power used in daily life.
Answer:
Kilowatt and megawatt.

Question 44.
How many watts are there in one horse power ?
Answer:
1 horse power = 746 watt.

Question 45.
What is meant by 1 kilowatt hour energy ?
Answer:
The energy consumed by a source of 1 kilowatt power in 1 hour.

Question 46.
What is the relation between kilowatt hour and joule ?
Answer:
1 kilowatt hour = 36,00,000 joules.

Question 47.
Which type of energy is used when a person climbs stairs ?
Answer:
Muscular energy.

Question 48.
Which type of energy is there in our body ?
Answer:
Chemical energy.

Question 49.
The person requires the least amount of energy in which of the following activity ?
(i) In swimming
(ii) In movement
(iii) In sleeping
(iv) In running foot.
Answer:
(iii) In sleeping.

Question 50.
What is simple pendulum ?
Answer:
Simple pendulum is a metalic bob which is hang by weightless string.

Question 51.
Write the name of energy when simple pendulum is at its extreme point.
Answer:
Potential energy.

Question 52.
Write the name of energy when pendulum is in middle position.
Answer:
Kinetic energy.

Short-Answer Type Questions

Question 1.
What is meant by work ? How is it measured ? What is its unit ?
Answer:
Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the
applied force.
Thus, Work done = Force x The distance moved in the direction of force
W = F x s
Work is a vector quantity. Unit of work is newton-metre or joule (J).

Question 2.
What is energy ? Write its unit and definition. What are its various types ?
Answer:
The capacity of doing work is energy. Its unit is also joule.
Joule: Work done is said to be one joule when a force of 1 newton acting on an object and the object is displaced through 1 m in the direction of force.

Various types of energy:
(i) Mechanical energy
(ii) Heat energy
(iii) Chemical energy
(iv) Electric energy
(v) Nuclear energy’
(vi) Light energy
(vii) Atomic energy
(viii) Magnetic energy
(ix) Sound energy,
(x) Solar energy.

Question 3.
Justify with example that moving bodies have capacity to do work ?
Answer:
The examples are following :
(i) A moving marble has a capacity to displace that marble which is in rest.
(ii) The blowing wind works by rotating the wheels of windmill.
(iii) The moving water also takes the objects from one place to another.
(iv) The blowing wind is helpful for sailing boat.
Thus, we can say that moving bodies have capacity to do work.

Question 4.
What is meant by potential energy and kinetic energy ? Give two examples for each.
Answer:
Potential energy’: The energy possessed by a body due to its change in position or shape is called the potential energy.
Examples:
(1) The book lying on the table has potential energy due to its position.
(2) A binded watch has potential energy.

Kinetic energy:
The energy possessed by a body due to its motion is called the kinetic energy.

Examples:
(1) Moving air has kinetic energy.
(2) The running water has kinetic energy.

Question 5.
Is there energy in following substances ? If yes, then what is the type-potential energy or kinetic energy or both ?
(i) The switched off ceiling fan.
(ii) A man climbing up mountain.
(iii) Flying bird.
(iv) The water in a dam.
(v) The stretched bow more than its shape.
(vi) The rubber locket on a table.
(vii) Stretched locket of rubber.
Answer:
Yes, all substances have energy’.
(i) A switched off ceiling fan has potential energy only.
(ii) A man climbing up mountain has both potential and kinetic energy.
(iii) A flying bird has both kinetic and potential energy.
(iv) The water in the dam has potential energy.
(v) A stretched bow has potential energy.
(vi) The rubber locket on the table has potential energy.
(vii) Stretched locket of rubber has potential energy.

Question 6.
Write the difference between potential energy and kinetic energy.
Answer:
The differences between potential energy and kinetic energy are as follows:
Potential Energy (P.E.):
1. The energy possessed by an object due to its position is called potential energy.
2. Potential energy = mgh
3. The potential energy of an object depends upon the height or depth from the surface of earth.

Kinetic Energy (K.E.):
1. The energy possessed by an object due to its motion is called kinetic energy.
2. Kinetic energy = \(\frac {1}{2}\) mv²
3. Kinetic energy depends upon the motion of the object.

Question 7.
Which type of energy is in following ?
(1) Coiled spring of watch.
(2) Running buffallo.
(3) The arrow released from bow.
(4) A table lying on roof.
(5) Compressed air.
(6) A mango on tree.
Answer:
(1) Potential energy
(2) Kinetic energy
(3) Kinetic energy
(4) Potential energy
(5) Potential energy
(6) Potential energy.

Question 8.
When the capacity of doing work by human body decreases ?
Answer:
The capacity of doing work by human body decreases in following circumstances :
(1) When a person is ill.
(2) When a person grows old.
(3) When a person takes food less than required, his muscular energy decreases.

Question 9.
What is the relation between work and energy ? Write their units.
Answer:
The capacity of doing work is called energy. The magnitude of energy is always equal to that work which has a capacity to do that. Energy is measured in joule and the unit of work is also joule.

For example:
To lift a piece of stone of mass 1 kg upto a height of 1 metre requires 9.8 joules energy. Opposite to it, when stone is dropped from one metre, the stone does the same amount of work (9.8 J). Thus, if an object has an energy of 100 joules. It has a capacity of doing work of 100 joules.

Question 10.
How the energy is transferred in key toy ?
Answer:
There is a spring in key toy or key watch which contracts while binding and stored potential energy in it. After some moments, the spring loses slowly and the toy comes in motion. Thus, potential energy is transferred in kinetic energy.

Question 11.
A ball is thrown upward and it returns back. What is the change in kinetic and potential energy of the ball ? Explain.
Answer:
When a ball is thrown upward its kinetic energy changes continuously into potential energy. At highest point its whole kinetic energy changes into potential energy. When the ball returns back its potential energy changes continuously into kinetic energy and whole of its potential energy changes into kinetic energy near the earth but on reaching the earth, whole of its kinetic energy changes into potential energy.

Question 12.
A boy is standing by holding some weight in his expanded hand. In this stage, there seems no change in the position of both boy and weight. Is he doing any work ? If yes, then justify.
Answer:
Yet, there is no change in the position of boy and weight externally, but internally the boy is working. The muscles of boy are stretched due to weight on palm. Thus, their shape get changes. The heart pumps more blood into muscles. The energy of boy wastes in all these chemical activities and he tired soon. If the boy stand more enough in same position, the external change will be observed as sweating.

Question 13.
How much work is done when an artificial satellite revolve round the earth ? Explain in brief. –
Answer:
The artificial satellite revolves round the earth in almost circular track. The gravitational force between the earth and the satellite works along the line joining the earth and satellite and the satallite moves perpendicular to the direction of force of gravity. Thus, the magnitude of displacement of satellite in the direction of force will be zero and the work done by the earth on the satellite will also be zero.

Question 14.
How the kinetic energy of wind is utilised ? Write its important uses.
Answer:
The electric energy can be produced by the energy in the blowing air. It has much waste. Thus, wind mill is used in grinding flour and drawing out water from ponds. A small amount of electric energy is produced from wind mill. Those places where the wind blows throughout the years, wind mill are used to grind flour and drawing out water.
Important uses of mechanical energy of wind :
(1) The fields can be irrigated by drawing water from ponds.
(2) It can be used in grinding the flour.
(3) The mechanical energy of wind can be converted into electric energy by joining the various wind wheels together.

Question 15.
What is water energy ? Write its two important uses.
Answer:
The running water has a capacity to do work due to its kinetic energy, which is known as water energy.

Uses of water energy: Following are the uses of water energy :
(1) Water energy is converted into electric energy and it is used in houses and factories.
(2) Water energy is used to grind flour by windmills.
(3) The heavier wooden pieces is taken from one place to another due to water energy.

Question 16.
How water energy is used to produce electricity ?
Answer:
The kinetic energy of running water is stored as potential energy in dam. The stored water is dropped on the larger blades of water turbines. The shaft attached with running turbine rotates the generator or dyanmo quickly. The rotating coil of dyanmo produces electricity which is called ‘hydroelectric power’. The electricity is produced on Bhakhra dam by this method. This is also known as the principle of hydroelectricity.

Question 17.
Why the nail becomes hot when it is hammered in the sheet of wood ?
Answer:
The nail becomes hot when it is hammered in the sheet of wood because the lifted hammer has potential energy due to its position. When the hammer is dropped on the nail, it comes in rest position, but transfers, its whole energy in the nail. As a result of this, the nail gets some kinetic energy and moves inside the wood. When the nail enters completely and we continuous hammering the nail becomes hot because the mechanical energy of the hammer is transferred in nail and changes into heat energy.

Question 18.
What are the benefits of the solar energy ?
Answer:
The benefits of the solar energy are as follows :
(1) The solar energy is used in solar cooker to prepare food,
(2) Solar energy is used to warm the water by solar heater in hotels, hospitals and industries.
(3) Solar energy is used in the production of electricity.
(4) It is used to melt the metals in solar furnances.

Question 19.
Differentiate among work, energy and power. Write the SI unit for each. Or What is Power? Write its S.I. unit.
Answer:
Work:
When force is applied on a body and it moves in the direction of force, it is said to be work done.
Thus, Work (W) = Force (F) x Distance (S)
SI unit of work is joule.

Energy: t he capacity ot doing work is called energy. SI unit of energy is joule.

Power: The rate of doing work is called power.
Thus,

SI unit of power is watt.

Question 20.
What do you mean by transfer of energy ? Write two examples of transfer of energy.
Answer:
According to the law of conservation of the energy, The energy can neither be created nor destroyed. It can be transferred from one form to another. It is called the transfer of energy. Thus, the total magnitude of energy is never changed.

Examples:
(1) Ice on mountains melts and becomes water. This is the transfer of potential energy into kinetic energy.
(2) The electric energy in bulb changes into light energy and glows.

Numerical Questions

Question 1.
A boy of mass 40 kg is running with a ve’ocity of 0.5 ms. What is his kinetic energy ?
Solution:
Here,
m = 40 k
v = 0.5 ms^-1
We know that,
Kinetic Energy (K.E.) = \(\frac {1}{2}\) mv²
= \(\frac {1}{2}\) x 40 x 0.5 x 0.5 = 5 Joule Answer:

Question 2.
An object of mass 15 kg is moving with a uniform velocity (if 4 ms. What is tue kincth energy possessed by the object?
Solution:
Here
Mass of the body (m) = 1 5 kg
Velocity of the body (v) = 4 rn/s
We know that
Kinetic Energy (K.E) = \(\frac {1}{2}\) mv²
= \(\frac {1}{2}\) x 15 x 4(2)²
= \(\frac {1}{2}\) x 15 x 16
= 120 Jule

Question 3.
A stone of mass 100 gram is kept at a height ofS metre. What kind of energy does it possess and find out its value.
Solution:
The stone has potential energy because of height.
Here. m = 100g = \(\frac {100}{1000}\) = 0.1 kg
g = 9.8 ms^-2
h = 5m
We know that.
Potential Energy (P.E.) m x g x h = 0.1 x 9.8 x 5 = 4.9 Joule

Question 4.
To what height can an object of mass 1 kg be raised by giving it an energy of 980 joule?
Solution:
Here.
m = 1 kg
g = 9.8 ms^-2
h = ?
Potential Energy (RE.) = 980 J
We know that,
Potential Energy (RE.) = m.g.h.
or 980 = 1 x 9.8 x h
or h = \(\frac{980}{1 \times 9.8}\)
= 100 m

Question 5.
What is the increase in kinetic energy of a particle, if its velocity is increased to four times?
Solution:
Suppose the initial velocity of the particle = v
and its final velocity = 4 v
Suppose the mass of the particle = m
∴Initial kinetic energy (K₁) of the particle = \(\frac {1}{2}\) mv² ………….(i)
Final kinetic energy (K₂) of the particle = \(\frac {1}{2}\)m (4v)²
= \(\frac {1}{2}\)m x 16 v² …………..(ii)
From equations (i) and (ii),
\(\frac{\mathrm{K}_1}{\mathrm{~K}_2}=\frac{\frac{1}{2} \mathrm{mv}^2}{\frac{1}{2} \mathrm{~m} \times 16 \mathrm{v}^2}=\frac{1}{16}\)
or K₂ = 16K₁
Thus, the increase in kinetic energy is 16 times the initial kinetic energy.

Question 6.
The kinetic energy of a ball of mass loo g is 20 joule. Calculate its velocity.
Solution:
Here,
m = 100 g = \(\frac {100}{1000}\)kg = 0.1 kg
v = ?
Kinetic Energy (K.E.) = 20 J
We know that, Kinetic Energy (K.E) = \(\frac {1}{2}\) mv²
Or 20 = \(\frac {1}{2}\) x 0.1 x v²
v² = \(\frac {20 x 2}{0.1}\) = 400
or v = \(\sqrt{400}\) = 20 ms^-1

Question 7.
A coolie lifts an object of mass 20 kg from the earth to 1.5 m and keep It on his head. Calculate the amount of work done by him on the object.
Solution:
Here,
Mass of the object (m) = 20kg
Displacement (s) = 1 .5m
Work done (W) = F x s = mg x s
= 2okg x 10 ms^-2 x 15 m
= 300kgms^-2 m
= 300Nm= 300J
The amount of work done by the coolie on the object is 30 joules.

Question 8.
A man of mass 50 kg jumps to a height of 1.2 m. What is its potential energy at the highest point?
Solution:
Here,
m = 50kg
g = 9.8 ms^-2
h = 12m
We know that,
Potential Energy (P.E.) m x g x h
= 50 x 9.8 x 1.2
= 588 J
Thus, at the highest point the potential energy of the man is 588 J.

Question 9.
With what velocity should a man of mass 50 kg run so that his kinetic energy becomes equal to 625 J.
SolutIon:
Here,
m = 50kg
v = ?
Kinetic Energy (K.E.) = 625 J
We know that,
Kinetic Energy (K.E) = \(\frac {1}{2}\) mv²
Or 625 = \(\frac {1}{2}\) x 50 x v²
v² = \(\frac{625 \times 2}{50}\) = 25
or v = \(\sqrt{25}\) 5ms

Question 10.
A man throws a stone of mass 10 kg from a ladder of height 5m. When ¡t reaches the ground, what will be its kinetic energy ? What will be its velocity at the nearest point to the ground?
Solution:
Here,
m = 10kg
h = 5m
v = ?
Kinetic Energy (K.E.) = ?

We know that,
The kinetic energy (K.E.) of the stone = The loss in potential energy due to falling of the stone
= mgh
= 10 kg x 9.8 ms^-2 x 5m = 490J ……………(i)
Because kinetic energy = \(\frac {1}{2}\) mv² .(ii)
From eqns. (i) and (ii),
\(\frac {1}{2}\) mv^-2 = 490J
or \(\frac {1}{2}\) x 10kg x v² = 490J
or v² =\(\frac{490 \times 2}{10}=\frac{980}{10}\) = 98
or v = \(\sqrt{98}\) = 9.9 ms^-1 (approx.)

Question 11.
A car of mass 1000 kg is moving with a velocity of 30 ms^-1. After applying brakes it stops with uniform acceleration at a distance of 50m. Calculate the force and the work done by the brakes?
Solution:
Here,
m = 1000kg
u = 30 ms^-1
v = 0 ms^-1
s = 50m
F = ?
W = ?
We know that, v² – u² = 2as
a = \(\frac{v^2-u^2}{2 s}=\frac{(0)^2-(30)^2}{2 \times 50}=\frac{-900}{100}\) ms^-2
= – 9 ms^-2 So retardation = 9ms^-2
F = m.a.l000kg x 9ms² = 9000 N
The force applied by the brakes on the car (F) = 9000 N
Work done(W) = F x s = 9000 N x 50 m = 450000 J = 450 kJ

Question 12.
A man of mass 50kg climbs a vertical height of 10m on a mountain in 20 seconds. What ¡s his power ? (Given that g 9.8 ms-2)
Solution:
Here,
F = m x g
= 50 x 9.8 = 490N
Now, W = F x S = 490N x 10m = 4,900 Nm
Time (t) = 20 seconds
Power (P) = \(\frac{\text { Work done }}{\text { Time taken }}=\frac{4,900}{20}\) = 245 watt

Question 13.
A boy of mass 40 kg runs up a staircase of 45 steps in 9 sec. If the height of each step is 15 cm, Calculate his power ? (The value of g is 10 ms²)
Solution:
Weight of the boy (mg) 40kg x 10ms² = 400N
Total height of the steps (h) = 45 x 15/100m = 675m
Time taken to climb the steps (t) = 9s
We now that
Power (P) = Time taken = \(\frac{\mathrm{mgh}}{t}\) = \(\frac{400 \mathrm{~N} \times 6.75 \mathrm{~m}}{9 \mathrm{~s}}\) = 300 W
The power of the boy is 300 W.

Question 14.
A man of mass 50 kg climbs 30 steps in 30 sec. If height of each step is 20 cm. Then calculate the power to climb the total steps.
Solution:
Mass of the man (m) = 50 kg
Total number of steps = 30
height of each step = 20 cm = \(\frac {20}{100}\) m = 0.2 m
Total height covered by the man = 30 x 0.2 m = 6m
Acceleration due to gravity (g) = 10 ms^-2
Thus, work done by the man = m.g.h = 50 x 10 x 6J = 3000J
Time taken to climb the steps = 30 s

= \(\frac{3000 \mathrm{~J}}{30 \mathrm{~s}}\)
= 100 Js^-1 = 100 W(watt)

Question 15.
An electric bulb of 100 Wis used for 2 hours, how much electrical energy is consumed?
Solution:
Power of the bulb = 100W
Time = 2 hours 2 x 3600 = 7200 sec.
Energy consumed = Power x Time
= 100 x 7200Ws = 720000 joule

Question 16.
A rocket has a mass of 3 x 10⁶ kg. Calculate (a) its potential energy (b) its kinetic energy, at a height of 25 km moving with a velocity of 1 km/s. (Take g = 10 ms^-2)
Solution:
Here,
Mass of the rocket (m) 3 x 10⁶ kg
Velocity of the rocket (v) = 1 k ms^-1 = 1000 ms^-1
Height of the rocket (h) = 25 km = 25000 m
Acceleration due to gravity (g) = 10 ms^-2

(a) Potential Energy = m.g.h.
3 x 10⁶ x 10 x 25000 J
= 7.5 x 10¹¹

(b) Kinetic Energy = \(\frac {1}{2}\) mv²
= \(\frac {1}{2}\) x 3 x 10⁶ x (1000)² J
= 7.5 x 10¹¹J

Question 17.
A horse is pulling a cart moving with a velocity of 18 kmh^-1 with a force of 300 N. Find out the power of the horse in watt and also in horse power?
Solution:
Here,
Velocity of the cart(v) = 18 kmh^-1
= \(\frac{18 \times 1000}{3600}\) ms^-1 = 5 ms^-1
The force exerted by the horse on cart (F) = 300 N
m.g. = 300N
Work done = m.g.h
= 300 x 5 = 1500J

= \(\frac{15000}{1s}\) = 1500 w
We know that, 746 W = 1 horsepower
1 waft = \(\frac {1}{746}\) horsepower
1500 watt = \(\frac {1}{746}\) x 1500 horsepower
= 2 horsepower

Do Yourself

(1) A car is moving with a uniform velocity of 54 kmh^-1 What will be the kinetic energy of a child of mass 40 kg sitting inside it?
Answer:
[4500J]

(2) Calculate the power of that pump which stores the water of loo kg in 25 seconds into a tank kept at a height of 19m. (Let g = 10 ms²)

(3) What will be the energy of an object of mass 10 kg at a height of 40 m from the ground level?
Answer:
[Take g 100 ms²] [40003]

(4) A force of 5 N is acting on a body. The body displaces to a distance of 2m in the direction of the force. Find out the work done on the object.
Answer:
[103)

(5) An object of mass 15 kg is moving uniformly with a velocity of 4 ms². Calculate the kinetic energy of the object.
Answer:
[120J]

(6) What will be the height from the ground level where an object of mass 1 kg will have a potential energy of 1 joule if g= 10 ms²?
Answer:
[0.1 m]

(7) If 4900 joule of energy is utilised in lifting a body of mass 50 kg to a certain height, then what will be the height of the object if g = 9.8 ms²?
Answer:
[10m]

(8) A body of mass 3 kg starts falling down from rest position. After 3 sec., what will be its kinetic energyifgl0m.c2?
Answer:
[1350m]

(9) An object of mass 10 kg is lifted to a height of 6m from the ground level. Calculate the energy present in the object. The value of g is
9.8 ms².
Answer:
[588 J]

(10) An object of mass 12 kg is kept at a certain height. 1f its potential energy is 480 J; calculate its height from the ground level.
Answer:
[Take g = 10 ms²] [4 m]

(11) A man of mass 50kg takes a packet of mass 20kg upto the terrace of a building of height 14m, what is the work done by the man ?
Answer:
[Take g = 10 ms²] [9604 J]

(12) A vehicle of mass 1800 kg is moving with a velocity of 40 ms. After applying brakes it stops with a uniform acceleration at a distance of 200 m. Calculate the force applied and work done by the brakes.
Answer:
[1440 kJ]

(13) What will be the increase in kinetic energy of a particle if its velocity is doubled?
Answer:
[4 times ]

(14) How much amount of work is done on lifting an object of mass 2 kg to a height of 2m if the value of g is 9.8 ms²?
Answer:
[39.2J1

(15) The height of’Kutubminar’ is 70m. What is the work done by a man of mass 50 kg on reaching the top?
Answer:
[Take g 9.8 ms^-2] [3430011

(16) A stone of mass 10g is kept at a height of 5m. What kind of energy is present in it and calculate its value?
Answer:
[Take g 9.8 ms^-2] [0.49J]

(17) Two girls A and B of mass 400 N each, climbs a rope upto a height of 8m. IfA takes 20 sec. to do this work while B takes 50 sec.,
then calculate the individual power of both the girls.
Answer:
[A 160 W, B = 64W]

(18) Find the energy in KWh consumed inS hours by four devices of power 500 W each.
Answer:
[10Kwh]

Essay Type Questions

Question 1.
Define potential energy. Derive an expression for the potential energy of a body.
Answer:
Potential Energy:
The energy possessed by a body due to its position or change in shape is known as potential energy.

Expression for potential energy:
Consider a body of mass m kept on the.ground level at point ‘A’. it is raised to a point ‘B’ by applying a force F on it. The distance between A and B is ‘h’.
As we know that, the force F acting on the body to lift it, is equal and opposite to the gravitational force
F = \(\frac{G M_e m}{\mathrm{R}_e{ }^2}\) …………(i)

Where, M_e = Mass of the earth
R_e = Radius of the earth
But, g = \(\frac{G M_e m}{\mathrm{R}_e{ }^2}\) (g = acceleration due to gravity) ………..(ii)
From equations (i) and (ii),
F = mg
The work done in raising the body to a height ‘h’
W = F x S (∵ s = h) = mgh
Because work done is same as the energy. Therefore, at a height ‘h’ the potential energy of the object is,
Potential Energy (P.E.) = mgh

Question 2.
Define kinetic energy. Obtain an expression for it.
Answer:
Kinetic Energy:
The energy possessed by a body due to its motion is called kinetic energy.

Expression:
Let us suppose a body of mass m is dropped from a height h and its velocity becomes V.
on reaching the ground level Then –
Initial velocity of the body (u) = 0
Acceleration (a) = g (acceleration due to gravity)
Final velocity = v
Distance (s) = Height (h)
We know that, v² – u² = 2as
v² – 0 = 2gh
v² = 2gh
h = \(\frac{v^2}{2 g}\) …………. (i)
As the object comes down, its potential energy decreases and kinetic energy increases. When the body strikes to the ground level, its hoIe potential energy is converted into kinetic energy. According to the law of conservation of energy, Energy can neither be increased nr be decreased.”
Thus, kinetic energy at ground level = Potential energy at height h.
Kineiic Energy = mgh
or K.E. = mg x \(\frac{v^2}{2 g}\) (∵ h = \(\frac{v^2}{2 g}\))
K.E. = \(\frac {1}{2}\) mv²

Question 3.
Define Law of conservation of energy. Give an example of conservation of energy.
Answer:
Law of conservation of energy-According to this law, energy can neither be created nor be destroyed. ¡t can only be transfèrred from one form to another. When the energy is released in any form, the same amount of energy appears in other forms. Thus, total energy of the universe remains constant. Suppose, an object of mass m starts falling from a height h, on the ground level.

(a)EnergyatpointA, PotentialEnergy = mgh
Kinetic Energy = 0
∴ Total energy = mgh + 0 = mgh ……………(i)

(b)Energy at point B:
Suppose the object falls from A to B at a distance s, then its height from ground level will be (h-S).
∴ Potential energy = mg (h-s)
and v²-u² = 2gs
v² = 2gs [∵ u = 0]
Kinetic Energy = \(\frac {1}{2}\) mv² = \(\frac {1}{2}\)m x 2gs = mgs
Total Energy = mg (h-s) + mgh = mgs – mgs + mgs = mgh …………(2)

(c) Energy at point C; v²-u² = 2gh
v² = 2gh [∵u = 0]
Kinetic Energy = \(\frac {1}{2}\) mv² = \(\frac {1}{2}\) m x 2gh = mgh
and Potential Energy = 0
Total Energy = mgh + 0 = mgh ……………(iii)
in this way we observe from eQuestion (i), (ii) and (iii) that the energy of the object remains constant at every point. There is potential energy in the object at highest point. When it falls litle downs same part of potential energy is converted into kinetic energy and on ground level whole potential energy is converted into kinetic energy.

Practical Work

Experiment 1.
Describe by an experiment that the wark done by a force can either be positive or negative.

Procedure:
A child is pulling a toy car parallel to the ground as shown in figure. The child has exerted a force in the direction of displacement of car. In this situation, the work done will be equal to the product of the force and displacement. In this situation the work done by the force is taken as positive. Consider a situation in which an object is being displaced by the action of forces and we identit one of the forces, F acting Opposite to the direction of the displacement s, thus, the angle between the two directions is 1800.

In such a situation, the work ‘ done by the force F is taken as negative and denoted by the minus sign. The work done by the force is F x (-s) or (-F x s). When the force acts opposite to the direction of the displacement, the work done is negative. When the force acts in the direction ofdisplacernen, the work done is positive.

Experimenf 2.
Describe by an activity in which both positive and negative forces are working.

Procedure:
When an object is lifted up, both positive and negative forces work, because the force applied by us, displaces the object upward and called positive force, while the force of gravity applied by earth pulls the object downward which is called negative force.

Experiment 3.
Describe by an experiment that the object at more height has more energy.

Procedure:
(1) Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from height of about 25 cm. The ball creates a depression.
(2) Repeat this activity from heights of 50 cm, I m and 1.5 m.
(3) Ensure that all the depressions are distinctly visible.
(4) Mark the depressions to indicate the height from which the ball was dropped.
(5) Compare their depths.

We will observe that when the ball was dropped from the height of 1.5 m, the depression was the deepest; which proves that the object at more height has more energy which makes the deepest depression.

Expereiient 4.
Describe by an experiment that the kinetic energy of an object increases by increasing its mass.

Procedure:
(1) Set up the apparatus as shown in figure
(2) Place a wooden block of known mass in front of the trolley at a convenient fixed distance.
(3) Place a known mass on the pan so that the trolley starts moving.
(4) The trolley moves forward and hits the wooden block and displaces the block.
(5) Note down the displacement of the block. It is, clear that the work is done on the block by the trolley as the block has gained energy.
(6) Repeat this activity by increasing the mass on the pan.

You will observe that the displacement will increase with the increase in mass, which proves that the kinetic energy increases by increasing the mass.

Quick Review of the Chapter

1. What is required for doing a work?
(A) displacement in object
(B) force should be applied on the object
(C) both (A) and (B)
(D) no displacement in object
Answer:
(C) both (A) and (B)

2. What type of energy is there inside a body kept at height?
(A) potential
(B) kinetic
(C) both (A) and (B)
(D) no relation with energy
Answer:
(A) potential

3. The device that converts the electrical energy In heat energy:
(A) Electrical iron
(B) Electrical bulb
(C) Radio
(D) Electrical motor
Answer:
(A) Electrical iron

4. Which type of energy is there in a flying bird?
(A) potential
(B) kinetic
(C) both (A) and (B)
(D) neither of the two
Answer:
(C) both (A) and (B)

5. Rado transfer the electrical enegry in:
(A) light energy
(B) sound energy
(C) mechanical energy
(D) heat energy
Answer:
(B) sound energy

6. Which formula is correct for work?
(A) W = \(\frac {F}{s}\)
(B) W = \(\frac {s}{F}\)
(C) W= Fs
(D) W = \(\frac {1}{Fs}\)
Answer:
(C) W= Fs

7. The unit of force Is:
(A) Newton
(B) Newton metre
(C) Joule
(D) both (B) and (C)
Answer:
(A) Newton

8. If 5n force acts on an object, then displacement is 2m, what will be the work done?
(A) lONm
(B) IOJ
(C) lOJm
(D) both(A)and(B)
Answer:
(D) both (A) and (B)

9. A coolie lifts an object of mass iS kg from the earth to 1.5 m keep it on his head. Calculate the
amount of work done by the coolie.
(A) 22.5J
(B) 2.25 J .
(C) 225J
(D) 225N
Answer:
(C) 225 J

10. The potential energy of an object of mass 1kg kept at the height &f 2m will be:
(A) 9.8J
(B) 19.6J
(C) 29.4 J
(D) 39.2 J
Answer:
(B) 19.6 J

11. Whose unit is Newton metre (Nm)?
(A) work
(B) force
(C) acceleration
(D) power
Answer:
(A) work

12. How many watts are in 1 Kilowatt?
(A) 10³
(B) 10⁴
(C) 10⁵
(D) 10⁶
Answer:
(A) 10³

13. How many joules are in 1 Kilowatt?
(A) 36 lakhs
(B) 36 thousands
(C) 36 hundreds
(D) 36 crores
Answer:
(A) 36 lakhs

14. What is another name of newton metre?
(A) Erg
(B) Coulomb
(C) Joule
(D) Hertz
Answer:
(C) Joule

15. ……….. is vector.
(A) potential energy
(B) kinetic energy
(C) work
(D) none of these
Answer:
(D) none of these

16. A man stands for 10 minutes by carrying a weight of 20 kg the work done s ill be:
(A) 200joule
(B) onejoule
(C) zero
(D) 400joule
Answer:
(C) zero

17. The unit of energy is:
(A) Nm
(B) Joule
(C) both (A) and (B)
(D) no unit
Answer:
(C) both (A) and (B)

18. The example of kinetic energy is not:
(A) stone kept on mountain
(B) revolving wheel
(C) blowing air
(D) bullet shot out of a gun
Answer:
(A) stone kept on mountain

19. The example of potential energy is not :
(A) an arrow kept in the bow
(B) a stone on a mountain
(C) falling coconut
(D) a bullet in a gun
Answer:
(C) falling coconut

20. An object of 10 kg mass is carried to a height of 6 m. Calculate the energy of the object. (g = 9.8 mc²)
(A) 98 J
(B) 58.8 J
(C) 588 J
(D) 5880 J
Answer:
(C) 588 J

21. The law of conservation of mass is:
(A) all works require energy
(B) energy can be produced
(C) energy can be destroyed
(D) energy neither can be produced nor destroyed
Answer:
(D) energy neither can be produced nor destroyed

22. Which statement is not correct?
(A) Potential energy + kinetic energy constant
(B) Kinetic energy + potential energy = mechanical energy
(C) As much decrease is there in potential energy at a point so is in kinetic energy
(D) As much decrease is there in potential energy so much increase takes place in kinetic energy.
Answer:
(C) As much decrease is there in potential energy at a point so is in kinetic energy

23. Which formula is correct?
(A) P = W x t
(B) P = \(\frac {W}{t}\)
(C) P = \(\frac {t}{W}\)
(D) P = \(\frac{1}{W \times t}\)
Answer:
(B) P = \(\frac {W}{t}\)

24. What is the unit of power?
(A) Joule
(B) Nm
(C) erg
(D) Watt
Answer:
(D) Watt

25. How much energy will be consumed by a person of 600 newton weight for climbing a height of 10m?
(A) 60 joule
(B) 600 jouie
(C) ó000joule
(D) 588 joule
Answer:
(C) 6000 joule

26. When will be the minimum amount of potential energy in your body ? When you:
(A) are standing
(B) are sitting in a chair
(C) ire lying on the earth
(D) are sitting on the earth
Answer:
(C) are lying on the earth

27. When ¡s work not done?
(A) on fixing a nail in a wood
(B) on sliding a box on the floor
(C) when there is no constituent of parallel force in the direction of motion
(D) a weight hanging on a peg
Answer:
(C) when there is no constituent of parallel force in the direction of motion

28.
10 seconds are taken to fetch a bucket of 5 kg mass from a well of 10 m depth. The power used in the power will be:
(A) 5ONm
(B) 50)
(C) 50 W
(D) 500 W
Answer:
(C) 50 W

29. How much energy will be consumed by 5 fans of 100 watt each in 4 hours?
(A) 20 kwh
(B) 10 kwh
(C) 2 kwh
(D) 1 kwh
Answer:
(C)2 kwh

30. The rate of doing work is called:
(A) Power
(B) Force
(C) Energy
(D) None of these
Answer:
(A) Power

31. What is the momentum of an object of mass m, moving with a velocity u?
(A) (mυ)²
(B) mυ²
(C) \(\frac {1}{2}\) mυ²
(D) mυ²
Answer:
(C) \(\frac {1}{2}\) mυ²

Haryana State Board HBSE 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 13 Why Do We Fall Ill

Very Short-Answer Type Questions

Question 1.
What is the meaning of health?
Answer:
The proper shape of the entire organs and systems of the body, their location and functioning is called as health. Health is the normal position of the ability of physical, mental and social life.

Question 2.
What is called an ailment or disease?
Answer:
Any kind of disruption developing in good health is called as a disease. Or in other words, occurring of any disorder in the physical system is called a disease.

Question 3.
Give the name of a factor that influences health.
Answer:
Personal hygiene.

Question 4.
Give a reason that affects the community’s cleanliness.
Answer:
Failure of disposing of garbage scientifically.

Question 5.
How does food affect our health ?
Answer:
Excess of deficiency of nutritious elements in diet both leave ill effects on our health.

Question 6.
How does our environment harm us ?
Answer:
Unwanted materials found in the environment proved to be the cause of diseases for us.

Question 7.
Is it called health that the body is disease-free ?
Answer:
‘No, even if we are disease-free, we cannot consider ourselves healthy.

Question 8.
How is a disease diagnosed ?
Answer:
From the symptoms of the disease.

Question 9.
What are the symptoms of diseases ?
Answer:
The signs that indicate any disorder in our body, are called as the symptoms of disease.

Question 10.
What do the physicians do to confirm a certain disease ?
Answer:
They refer the laboratory’ test,

Question 11.
How many types of diseases are there (on the basis of time duration) ?
Answer:
Two types: Acute diseases, Chronic diseases.

Question 12.
What are called as acute diseases ?
Answer:
Ailments that are short lived are called as acute diseases.

Question 13.
Write down names of two acute diseases.
Answer:
Cough-cold, headache.

Question 14.
What are called as chronic diseases?
Answer:
Chronic diseases-the sustain for long time or throughout life.

Question 15.
Write down the name of chronic disease.
Answer:
Filariasis (elephantiasis).

Question 16.
Can a person fall ill in the inadequacy of diet?
Answer:
Yes, in the inadequacy diet a person can fall ill.

Question 17.
The antibiotic penicillin blocks the processes that build in bacteria.
Answer:
cell wall.

Question 18.
What are called as communicable diseases ?
Answer:
Diseases that spread through contact are called as communicable diseases.

Question 19.
Write down the names of two communicable diseases.
Answer:
Cholera, jaundice.

Question 20.
What are called as non-communicable diseases ?
Answer:
Diseases that do not spread through infection or pathogens are called as non-communicable diseases.

Question 21.
Write down names of two non-communicable diseases.
Answer:
Goitre and Diabetes.

Question 22.
Write names of two diseases caused due to virus.
Answer:
Cough-cold, Influenza, Polio.

Question 23.
What type of disease is cancer ?
Answer:
Non-communicable disease.

Question 24.
Which diseases are caused due to bacteria ?
Answer:
Cholera and Tuberculosis.

Question 25.
Give two examples of the diseases caused due to fungi.
Answer:
Skin diseases (dermatosis)-Ringworm, eczema etc.

Question 26.
Write names of two diseases caused by protozo
Answer:
Malaria and kala-azar.

Question 27.
What are the names of diseases caused by a worm?
Answer:
Elephantiasis (Filariasis), ascariasis.

Question 28.
Write down the name of the factor of acne or pimple.
Answer:
Staphylococci bacteria.

Question 29.
What is the cause of sleeping sickness?
Answer:
Trypanosoma protozoa.

Question 30.
Which disease is caused due to lachmania protozoan?
Answer:
Kala-azar.

Short-Answer Type Questions

Question 1.
What is health ? Write the advantages of good health.
Answer:
Health is a state of being well enough for a person to function well physically, mentally and socially. The structure and functions of all the body parts of a healthy person are normal. He is free from all types of psychological, mental and social tensions. A person free from only physical diseases is not frilly healthy.
Advantages: Following are the advantages of good health:
(1) The personal life of a healthy person is full of happiness.
(2) A healthy person utilises his potential fully as an individual, in the family, in society and in the country and does not become a burden on anyone.

Question 2.
What are the conditions necessary to maintain good health?
Answer:
The following conditions are necessary to maintain good health:
(1) Personal and family hygiene.
(2) Proper exercise and relaxation.
(3) Balanced diet.
(4) Community hygiene.
(5) Good habits and abstinence from intoxicants/drugs.

Question 3.
How does the environment affect health?
Answer:
The health of all persons is based on the surrounding environment. Physical factors constitute the environment. A person can fulfil his basic requirements only from the environment. This is possible only in a hygienic environment. A polluted environment has a bad effect on our health.

Question 4.
What things should be kept in mind while cleaning OUT homes 2
Answer:
Apart from dusting, cleaning cobwebs and disposing the wastes, we should also regularly keep a check on insects like cockroaches, mosquitoes, houseflies, spiders etc. Apart from this, it is necessary to white wash the house to keep it clean and beautiful. We should also keep our immediate surroundings neat and clean so that we don’t catch a disease.

Question 5.
What is a disease? How does disease occur?
Answer:
Any variation in the normal structure or function of a tissue or organ of a body. Or a change in psychological condition is called a disease. A diseased person does not feel comfortable.

Causes of disease:
The disease can occur because of any of the following three causes:
(1) Diseases caused due to deficiency or accumulation of some nutritive components e.g. Beri-beri, Obesity etc.
(2) Through infection (communicable) e.g., cholera, tuberculosis etc.
(3) Non-communicable diseases, like heart diseases, stones in the kidney or gall bladder etc.

Question 6.
What are non-communicable diseases ? Name any two.
Answer:
The diseases which are not caused by infection (and remain confined to the bearer only) are called non-communicable diseases, e.g. heart diseases, non-functioning of kidney, gall bladder or kidney stones etc.

Question 7.
What are hereditary (congenita!) diseases ? Give examples.
Answer:
The diseases which are carried by the parents to the offspring are called hereditary diseases. One such disease is haemophilia, in which the patient’s blood does not clot easily. Generally, the time taken for the clotting of blood increases from 2-8 minutes to 30 minutes-24 hours. A haemophilic person can die because of a continuous flow of blood even from a minor injury. Another such type of disease in sickle cell anaemia. The patient’s RBCs (Red Blood Cells) have defective haemoglobin, which reduces the oxygen-carrying capacity of the RBCs and the person goes on to become weak.

Question 8.
What are antibiotics? Why is their excessive use harmful?
Answer:
Antibiotics are the drugs which are derived from bacteria and are fatal for other bacteria. They do not harm the protoplasm of humans For example; penicillin, Terramycin, streptomycin etc. These save us by stopping the growth of pathogens or by. destroying them. If we use them excessively, then the useful micro-organisms in our alimentary canal will start getting destroyed which will affect our digestive system. .

Question 9.
What are the simple and easy precautions to remain free from the diseases caused by microorganisms ?
Answer:
The simple and easy precautions are:
(1) To keep the infected person away from a healthy one.
(2) By vaccination.
(3) To boil water before drinking.
(4) By keeping our surroundings neat and clean.
(5) By keeping check (on the growth) of mosquitoes and houseflies.
(6) By wearing properly washed clothes.
(7) By dumping the waste products in covered places. .
(8) By spraying insecticides so as to keep the surroundings free from harmful insects and pathogens.

Question 10.
What is the difference between bacteria and virus ?
Answer:
Given is the difference between bacteria and virus:
Bacteria:
1. They are living organisms.
2. They have a cellular structure.
3. They can be seen under a microscope.
4. Bacteria are bom from the same type of bacteria and multiply on their own.
5. Bacteria can be harmful as well as useful.

Virus:
1. They possess the features of both living as well as non-living.
2. Their structure is not cellular.
3. They can be seen only with the help of an electron microscope.
4. Viruses can be multiplied only inside a jost cell.
5. Viruses are only harmful.

Question 11.
How do antibiotics act against bacterial infection? Why are they not effective against virus? Explain with reason.
Answer:
Bacteria create a cell wall for their protection. But the antibiotic substances hinder the process of creation of the cell soil due to which bacteria cannot protect themselves and the antibiotics destroy them. Pathogens are destroyed by the same process. Because virus do not create any cell wall so, that antibiotics are not able to effect virus.

Question 12.
How does infection through air take place ?
Answer:
When an infected person exhales, sneezes or coughs, the pathogens spread in the air. When another person inhales the same air, these pathogens enter his body. Common cold, pneumonia and tuberculosis spread through this mode only. The occurence of common cold in crowded places is again due to this.

Question 13.
How does infection take place through water ?
Answer:
When the pathogens from the infected person enter the drinking water, then a healthy person also gets infected when he drinks this water. There is more incidence of water-borne diseases in rainy season. Cholera Spreads through this method.

Question 14.
Why are V.D. and gonorrhoea considered a social evil ?
Answer:
They are both sexual diseases. Their pathogens enter a healthy person’s body through cuts, wounds or genitals. In such a condition, these diseases are spread from a woman to a man or from a man to a woman during sexual contact. As these diseases also spread through sexual activity, so these diseases are considered a social evil because the incidence of these diseases is more in people who establish illegitimate sexual relations. Prostitution is the main cause of this. Social awareness can reduce the incidence of these diseases.

Question 15.
What is the sign inflammation ?
Answer:
During infection, immune system recruits many cells to the affected tissue to kill the disease causing microbes, this process is called imflammation. As a part of this process, there are local effects such as swelling, pain and general effects such as fever.

Question 16.
Why is vaccination of children and expectant mothers done ?
Answer:
Polio, whooping cough, measles diphtheria, typhoid and tetanus are some infective diseases common in the children. To prevent these diseases newly bom babies and children must be got vaccinated or immunised. To immunise all the children and expectant mothers under a plan, sufficient quantity of vaccines have been manufactured and distributed in India. If the expectant mothers are treated with the tetanus vaccine before she delivers the baby, the newly bom baby will have the ability of immunity. Therefore, lacs of lives of children can be saved by vaccination at proper time.

Essay Type Questions

Question 1.
How do pathogens spread diseases ?
Answer:
Those pathogens, that spread diseases are called as pathogens. With the entrance of pathogens into the human body and their reaction a man falls ill.

Pathogens can spread diseases in the following ways:
1. Through Contact: Some pathogens spread diseases through direct contact with the patient or the thing used by the patient; for example, if a healthy person puts on the clothes of a patient or uses his bed, the pathogens get entered into his body and make him ill, like pathogens causing skin diseases.

2. Through eatables: Pathogens of typhoid, cholera etc. the remaining food of the patient enter into our body and spread diseases.

3. Through air: Pathogens of common cold, influenza etc. enter the body of a healthy person through breathing.

4. Through vectors: Houseflies, mosquitoes, mice, bugs etc. are some insects that spread diseases; like malaria spreads through female anopheles mosquito bite. Plague is spread by mice.

5. Through water: Pathogen on mixing into drinking water reach to the healthy person and diseases
like cholera, diarrhoea, typhoid etc. surround the human beings. Therefore, specially during rainy season it is suggested to boil drinking water and then cool it down to drink there after, so that the pathogens might destroy.

Question 2.
Give a brief introduction of the infectious diseases.
Answer:
Infectious diseases are caused due to micro-organisms like:
1. Cholera: The cause of qholera is vibriocoli. It occurs due to drinking dirty water or infected eatables. Pathogens of cholera are spread through housefly. The patient suffers from painless water loss of due to acute vomitting and diarrhoea.

2. Tuberculosis: It’s cause is micro bacterium tuberculosis. This disease affects the lungs. The patient suffers from constant fever, blood in sputum, bodyache, diarrhoea and the neck ribs get swollow.

3. Typhoid: It spreads through salmonela bacterium typall. Its main symptoms are fever, headache, bleeding nose, black circles on the skin, reddish blisters on the chest.

4. Cough and Cold: It is caused by rhino works. Constant flowing of liquid through nostrils, redness in the eyes, irritation and headache are its main symptoms.

5. Chickenpox: It is caused by the pathogen berisola joster. Here, big circles on the patients body
get formed, severe backache, high fever and headache are its other symptoms. Its a effect lasts long for 14 days. .

6. Polio: It is an infectious disease among the children. It spreads through the pathogen of polio mylitis.
Its symptoms are fever, vomiting, bodyache, stiffness and difficulty in controlling muscles. The children develop paralysis. .

7. Rabies: It is caused by the Rabies pathogen. It is caused due to the bite of dog, cat, bat, jackal, fox etc. Patient suffering from Rabies scares water (hydrophobia). It affects the nervous system. Due to this the patient meets a painful death.

8. AIDS: It comes under the category of sexually transmitted diseases which is caused due to virus. A person suffering from HIV-AIDS, Immunisation system becomes weaker, with the result the body fails to face even the minor infection. The patient of HIV runs with pneumonia even with minor cough and cold. Presently, AIDS is an incurabled disease.

Question 3.
Which programmes have been adopted the Govt, of control communicable diseases ?
Answer:
Following programme have been adopted by the Govt, of India to control communicable diseases :

1. National Malaria Eradiction Programme:
This started as a control measure in 1953. In 1975, it was converted from control of the disease to total eradication programme. The changed programmed was started in 1977 which was effectively enforced. The programme include inspection of areas affected by malaria educated the people using suitable medicines, spraying of insecticides to kill mosquitoes, identifying the malaria patients, distribution of antimalarial drugs to keep their accounts etc.

Proper Vaccination List:

2. National Tuberculosis Eradication Programme:

Following stages are there in this programme:
(1) To identify the person suffering from tuberculosis in the very first stage and giving him treatment.
(2) To prevent the diseases by inoculating the children at an early age by BCG vaccine.
(3) Many TB clinics, district centres and education centres started in country.

3. National Cholera Eradication Programme: It includes cleaning the cholera effected area, making fresh drinking water availability and proper disposal of urban faecol waste. To educate people and for immediate treatment, mobile treatment units have been established for it. To inoculate people with cholera injections has been arranged in these.

4. National Leprosy Eradication Programme : A survey was conducted for thistdiseases in India and many treatment centres have been setup. Houses have been constructed for rehabilitation of the infected persons. Tamilnadu Govt, has opened a Leprosy Education and Research Centres, where there is an advisory committee for this diseases. Sulpha drugs are used for the treatment of the disease.

Practical Work

Experiment 1.
To study the effect of possible calamities on the health of people.

Procedure:
There are two effects of natural calamities on the health. Prepare its table with the help of science-teacher.

Experiment 2.
To study the provisions made by local municipal corporation for the supply of clear drinking water.

Procedure:
Prepare a list of provisions made by the local municipal corporation for the supply of clear drinking water.
(1) The cholorination of drinking water is done.
(2) Water storage tank is cleaned regulary.
(3) Tubewells are used to get underground drinking water.
(4) Water treatment plant is setup for canal water.
(5) Water supplies by pipes to avoid pollution.

Experiment 3.
To study the provisions made by local authority to manage garbage.

Procedure:
Prepare a list of provisions related to it from officer related to local conservator.
(1) The garbage is cleaned by landfill.
(2) Organic garbage is treated by roasting.
(3) Covered vehicles are used to throw garbage.
(4) Composte prepared from organic garbage is selling to farmers.
(5) Local water is planned to set up solid waste treatment plant.

Quick Review of the Chapter

1. Cholera is a ………… disease.
(A) bacterial
(B) viral
(C) protozoan
(D) fungal
Answer:
(A) bacterial

2. Malaria spreads by:
(A) mouse
(B) fly
(C) mosquito
(D) bug
Answer:
(C) mosquito

3. Cholera spreads by :
(A) fly
(B) mouse
(C) mosquito
(D) none of these
Answer:
(A) fly

4. Respiratory diseases spread by :
(A) food
(B) water
(C) air
(D) animal
Answer:
(C) air

5. Communicable disease is :
(A) goitre
(B) beri-beri
(C) obesity
(D) tuberculosis
Answer:
(D) tuberculosis

6. Example of chronic disease is :
(A) cough
(B) cold
(C) elephantiasis
(D) headache
Answer:
(C) elephantiasis

7. Acute disease is:
(A) lungs tuberculosis
(B) Dengue fever
(C) blindness
(D) disentary
Answer:
(D) disentary

8. Which of the following disease is caused by virus ?
(A) Anthrax
(B) Dengue fever
(C) Malaria
(D) Cholera
Answer:
(B) Dengue fever

9. The reason of AIDS is :
(A) bacteria
(B) protozoa
(C) virus
(D) fungi
Answer:
(C) virus.

10. The prostitution in society is the reason which of the following disease
(A) pneumonia
(B) gonorrhoea
(C) giardiasis
(D) influenza
Answer:
(B) gonorrhoea

11. AIDS can spread by:
(A) kiss
(B) embracing
(C) combined used blade of shave
(D) hug
Answer:
(C) combined used blade of shave

12. AIDS cannot spread by:
(A) sexual relations
(B) blood transfusion
(C) from placenta of mother
(D) by shaking hands
Answer:
(D) by shaking hands

13. Which disease can be caused by attack on lever?
(A) diabetes
(B) malaria
(C) jaundice
(D) goitre
Answer:
(C) jaundice

14. Which disease can be caused by fungi?
(A) kala-azar
(B) cholera
(C) malaria
(D) skin disease
Answer:
(D) skin disease

15. Vomiting is related to :
(A) mind
(B) liver
(C) lungs
(D) kidney
Answer:
(A) mind

16. Which fight against microbes :
(A) RBC
(B) WBC
(C) platelets
(D) lymphocytes
Answer:
(B) WBC

17. Which disease happens first and last times ?
(A) rabies
(B) polio
(C) chicken pox
(D) small pox
Answer:
(D) small pox

18. The main symptom of AIDS is:
(A) the loss in resistance power
(B) fever
(C) arthritus
(D) high cough ‘
Answer:
(A) the loss in resistance power

19. Rabies is a …………… disease.
(A) bacterial
(B) viral
(C) protozoan
(D) fungal
Answer:
(B) viral

20. Before going to fair, there is vaccination of:
(A) tuberculosis
(B) hepatitis
(C) typhoid
(D) cholera
Answer:
(D) cholera

21. Deficiency disease is not:
(A) diarrohea
(B) scurvy
(C) pellagra
(D) xeropthalmia
Answer:
(A) diarrohea

22. Communicable disease is :
(A) intluenza
(B) hemophilia
(C) goitre
(D) anaemia
Answer:
(A) influenza

23. The disease in mother’s placenta is:
(A) polio
(B) AIDS
(C) goitre
(D) scurvy
Answer:
(B) AIDS

24. Vaccination of which of the following is not available?
(A) whooping cough
(B) polio
(C) diphtheria
(D) AIDS
Answer:
(D) AIDS

25. Which disease can be caused by throwing solid waste in open?
(A) elephantiasis
(B) goitre
(C) cholera
(D) AIDS
Answer:
(C) cholera

26. Match the following:

(A) a-iv, b-i,c-ii,d-iii
(B) a-i,b-ii,c-iii,d-iv
(C) a-ii,, b-ic—iii,d-ii,
(D) a-iii. b-Ii, c-i, d’-iv
Answer:
(A) a-iv b-i, c-ii d-III

27. Which disease is caused by viruses?
(A) Common cold
(B) Dengue fever
(C) Influenza
(D) All of these
Answer:
(D) All of these

Haryana State Board HBSE 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Ex 6.2 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Exercise 6.2

प्रश्न 1.
आकृति में, x और y के मान ज्ञात कीजिए और फिर दर्शाइए कि AB || CD है।

हल :
यहाँ पर PQ एक रेखा है।

∴ ∠PRA + ∠ARQ = 180° [रैखिक युग्म]
50° + x = 180°
या x = 180° – 50° = 130° उत्तर …………..(i)
अब CD और PQ दो प्रतिच्छेद रेखाएँ S पर प्रतिच्छेद करती हैं।
∴ ∠QSC = ∠DSR [शीर्षाभिमुख कोण]
⇒ 130° = y
या y = 130° उत्तर ………………(ii)
समीकरण (i) व (ii) से
∠x = ∠y [प्रत्येक 130°]
परंतु ये एकांतर कोण हैं
∴ AB || CD [इति सिद्धम]

प्रश्न 2.
आकृति में, यदि AB || CD, CD || EF और y : z = 3 : 7 है, तो x का मान ज्ञात कीजिए। [B.S.E.H. March, 2017]

हल :
क्योंकि CD || EF तथा तिर्यक रेखा PQ उनको क्रमशः S तथा T पर काटती है।

∴ ∠CST = ∠STF [एकांतर कोण]
या 180°- y = z [∵ y + ∠CST = 180° रैखिक युग्म]
या y + z = 180°
परंतु y : z = 3 : 7
अनुपाती योग = 3 + 7 = 10
∴ ∠y = \(\frac {3}{10}\) × 180° = 54°
∠z = \(\frac {7}{10}\) × 180° = 126°
अब क्योंकि AB || CD तथा तिर्यक रेखा PQ उनको क्रमशः R तथा S पर काटती है।
∴ ∠ARS + ∠RSC = 180°
[∵ अभ्यंतर कोण संपूरक होते हैं।]
या x + y = 180°
या x = 180° – y = 180° – 54° = 126°
[∵ y = 54°]
अतः x = 126° उत्तर

प्रश्न 3.
आकृति में, यदि AB || CD, EF ⊥ CD और ∠GED = 126° है, तो ∠AGE, ∠GEF और ∠FGE जात कीजिए।

हल :
क्योंकि AB || CD तथा तिर्यक रेखा GE इन्हें क्रमशः G तथा E पर प्रतिच्छेद करती है ∠AGE = ∠GED [एकांतर कोण]
परंतु ∠GED = 126°
∴ ∠AGE = 126°
आकृति अनुसार
∠GEF = ∠GED – ∠FED
= 126°- 90° = 36°
या ∠FGE = ∠GEC [एकांतर कोण]
⇒ ∠FGE = 90° – ∠GEF [त्रिभुज GEF का तीसरा कोण]
= 90° – 36° = 54°
अतः
∠AGE = 126°, ∠GEF = 36°
∠FGE = 54° उत्तर

प्रश्न 4.
आकृति में, यदि PQ || ST, ∠PQR = 110° और ∠RST = 130° है, तो ∠QRS ज्ञात कीजिए।

हल :
यहाँ पर PQ को M बिंदु पर काटने के लिए SR तक बढ़ाया।

अब PM || ST तथा तिर्यक रेखा SM क्रमशः उनको M तथा S पर काटती है।
∠SMQ = ∠TSM [एकांतर कोण]
परंतु ∠TSM = 130°
अतः
∠SMQ = 130°
⇒ ∠QMR = 180° – 130° = 50°
[∵ SR एक रेखा है]
क्योंकि किरण RQ, PM को Q पर काटती है।
∴ ∠PQR + ∠RQM = 180°
या 110° + ∠RQM = 180°
या ∠RQM = 180° – 110° = 70°
अब ΔQRM में,
∠QRM + ∠RQM + ∠QMR = 180°
या ∠QRS + 70° + 50° = 180°
या ∠QRS + 120° = 180°
या ∠QRS = 180° – 120° = 60° उत्तर

प्रश्न 5.
आकृति में, यदि AB || CD, ∠ARQ = 50° और ∠PRD = 127° है, तो x और y ज्ञात कीजिए।

हल :
आकृति अनुसार AB || CD तथा तिर्यक रेखा PQ इनको क्रमशः P और Q पर प्रतिच्छेद करती है।
∠PQR = ∠APQ [एकांतर कोण]
परंतु ∠PQR = x व ∠APQ = 50°
∴ x = 50°
अब क्योंकि AB || CD तथा तिर्यक रेखा PR इनको क्रमशः P तथा R पर प्रतिच्छेद करती है।
∴ ∠APR = ∠PRD [एकांतर कोण]
या ∠APQ + ∠QPR = 127° [∵ ∠PRD = 127°]
या 50° + y = 127° [∵ ∠APQ = 50°]
या y = 127° – 50° = 77°
इस प्रकार x = 50° तथा y = 77° उत्तर

प्रश्न 6.
आकृति में, PQ और RS दो दर्पण हैं जो एक-दूसरे के समांतर रखे गए हैं। एक आपतन किरण (incident ray) AB, दर्पण PQ से B पर टकराती है और परावर्तित किरण (reflected ray) पथ BC पर चलकर दर्पण Rs से C पर टकराती है तथा पुनः CD के अनुदिश परावर्तित हो जाती है। सिद्ध कीजिए कि AB || CD है।

हल :
यहाँ पर दो समतल दर्पण PQ तथा RS एक-दूसरे के समांतर रखे गए हैं।
∴ PQ || RS [दिया है।

दी गई आपतित किरण AB परावर्तित होने के बाद पथ BC और CD से होकर जाती है।
BN तथा CM समतल दर्पण PQ तथा Rs के अभिलंब खींचे गए हैं।
क्योंकि BN ⊥ PQ, CM ⊥ RS तथा PQ || RS
⇒ BN || CM
इसलिए BN तथा CM दो समांतर रेखाएँ हैं तथा तिर्यक रेखा BC इन्हें क्रमशः B तथा C पर काटती है।
∴ ∠2 = ∠3 [एकांतर कोण]
परंतु परावर्तन के नियम से
∠1 = ∠2 और ∠3 = ∠4
⇒ ∠1 + ∠2 = ∠2 + ∠2
और ∠3 + ∠3 = ∠3 + ∠4
⇒ ∠1 + ∠2 = 2∠2
और ∠3 + ∠4 = 2∠3
परंतु ∠2 = ∠3 या 2∠2 = 2∠3
⇒ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠ABC = ∠BCD
परंतु ये एकांतर अंतः कोण हैं।
इसलिए AB || CD