Lakshmi

Acids, Bases And Salts Very Short Answer Type:

Question 1. The aqueous solution of HCI gives the acid character. Which ion is responsible for it?
Answer: Hydronium ion (H₃O⁺).

Question 2. Give the name and formula of the anion present in aqueous solution of bases.
Answer: Hydroxyl ion (OH^–).

Question 3. What is the nature of aqueous solution of carbon dioxide?
Answer: Acidic

Question 4. What type of salt is this — NaHCO₃?
Answer: Acidic.

Read and Learn all WBBSE Solutions for Class 9 Physical Science And Environment

Wbbse Madhyamik Physical Science And Environment Class 9 Question 5.

What is the use of methyl orange?
Answer: It is used as an indicator.

Question 6. Give an example of 2 neutral oxides.
Answer: Carbon monoxide (CO)

Question 7. What is the nature of ZnO?
Answer: Amphoteric oxide.

Question 8. Give an example of an acid which has oxidising properties.
Answer: Nitric acid (HNO₃).

Question 9. What will be the colour of the solution if a few drops of phenolphthalein are added in the aqueous solution of sodium hydrogen carbonate?
Answer: Colourless.

Question 10. Which one is used in vanishing colour NH₄OH or NaOH?
Answer: NH₄OH.

Wb Class 9 Physical Science Question Answers

Question 11. What is the basicity of CH₃COOH?
Answer: The basicity of CH₃COOH is 1.

Question 12. Give an example of an oxide of metal which is not basic.
Answer: Mn₃O₄

Question 13. Give an example of an organic base.
Answer: Methyl amine (CH₃NH₂).

Question 14. Give an example of an inorganic gaseous compound which has basic properties.
Answer: Ammonia (NH₃).

Question 15. Name a normal salt and an acid salt of the same acid.
Answer: Normal salt-sodium sulphate (Na₂SO₄) and acid salt-sodium bisulphate (NaHSO₄) of acid H₂SO₄.

Question 16. Name a normal salt whose aqueous solution is basic.
Answer: Aqueous solution of sodium carbonate (Na₂CO₃) is basic.

Question 17. What is an oxide?
Answer: An oxide is a compound of oxygen formed with another element.

Question 18. Name 2 different classes of oxides. 
Answer: Basic oxide and acidic oxide are two different classes of oxides.

Question 19. Name a binary compound of oxygen which is not an oxide.
Answer: Oxygen fluoride (OF₂).

Question 20. What is an indicator?
Answer: An indicator is some weak organic acid or base which indicates distinctive colours in acid, alkali and neutral solutions.

Question 21. Which element is present in all acids?
Answer: Hydrogen is present in all acids.

Question 22. Which ion does an acid generate in an aqueous solution?
Answer: An acid generates hydrogen ion (H⁺) in aqueous solution.

Question 23. Why are CaH₂ and CH_4, not acids?
Answer: CaH₂ and CH₄ do not produce H⁺ ions in an aqueous solution, so they are not acids, for every acid produces H⁺ ions in an aqueous solution.

Question 24. Which ions disappear in neutralization?
Answer: H⁺ ions generated by acid and OH ions generated by alkali disappear in a neutralization process.

Question 25. Name an acid salt and write its formula.
Answer: Sodium bisulphate is an acid salt. Its formula is NaHSO₄.

Question 26. What is meant by dissociation?
Answer: The process of breaking a substance (electrolyte) into its constituent ions in solution is called dissociation. or ionisation.

Question 27. Which element must an acid contain?
Answer: Hydrogen.

Question 28. Give an example of normal salt.
Answer: Sodium chloride (NaCl).

Question 29. Which indicator is used in the titration of strong acid and weak base?
Answer: Methyl Orange.

Question 30. Name an acid salt.
Answer: Sodium bisulphate (NaHSO₄).

Question 31. Which ions disappear in neutralization?
Answer: H⁺ ions generated by acid OH^– ions generated by alkali disappear in a neutralization process.

Question 32. What is an oxide?
Answer: An oxide is a compound of oxygen formed with another element.

Question 33. Name a tribasic acid.
Answer: Phosphoric acid (H₂PO₄).

Question 34. Give an example of an organic acid.
Answer: Formic acid (HCOOH).

Question 35. What is the neutralisation reaction?
Answer: The reaction in which an equivalent amount of an acid reacts with a base is called a neutralisation reaction.

Question 36. Give an example of a neutral oxide.
Answer: Carbon monoxide (CO)

Question 37. What is a vanishing colour?
Answer: The aqueous solution of ammonium hydroxide mixed with phenolphthalein is called vanishing colour.

Question 38. What will be the colour of the solution if a few drops of phenolphthalein are added to in aqueous solution of sodium carbonate?
Answer: Pink colour.

Question 39. Which indicator is used in the titration of a weak acid and a strong base?
Answer: Phenolphthalein.

Question 40. What is the nature of ZnO?
Answer: Amphoteric oxide.

Question 41. How many replaceable hydrogen atoms are present in sulphuric acid?
Answer: Two replaceable hydrogen atoms are present in sulphuric acid.

Question 42. Give an example of a double salt.
Answer: K₂SO₄.Al₂(SO₄)₃ , 24H₂O.

Question 43. Give an example of an oxide of metal which is not basic.
Answer: Mn₃O₄.

Question 44. What is the acidity of NaOH?
Answer: The acidity of NaOH is 1.

Question 45. What is the basicity of  CH₃COOH?
Answer: The basicity of CH₃COOH is 1.

Acids, Bases And Salts 2 Marks Questions And Answers:

Question 1. What are mineral acids?
Answer:

Acids HCl, H₂ SO₄, HNO₃ etc. are called mineral acids because all are derived from minerals. Mineral acids are strong acids also called inorganic acids.

Wbbse Class 9 Physical Science

Question 2. What is a universal indicator?
Answer:

A universal indicator is a mixture of indicators, often sold ready-made as a solution, which can indicate pH values, usually in a range of 3 – 11. When this universal indicator is added in a solution, its colour changes and is then matched with the given indicator colour.

Question 3. Can a vanishing colour be prepared with a dilute NaOH solution?
Answer:

Vanishing colour cannot be prepared with dilute NaOH solution because sodium hydroxide (NaOH) does not evaporate.

Question 4. Why is an acid called a proton donor?
Answer:

Acid is called a proton donor: Acids produce cation (H⁺) in aqueous solution. H⁺ is nothing but a proton. So acid is called a proton donor.

Question 5. Why is alkali called a proton acceptor?
Answer:

Alkali is called a proton acceptor: Alkali produces hydroxyl ion (OH^–) in an aqueous solution, and hydroxyl accepts a proton (H⁺) coming from an aqueous solution of acid to produce neutral water. So, alkali is called a proton acceptor.

Question 6. Aqueous solution of HCl turns blue litmus red but HCl vapour does not, why?
Answer:

Reason: HCl vapour is a covalent compound, so it does not ionise in the vapour state. In an aqueous solution, HCI produces H₃O⁺ (hydronium ion) and acts as an electrovalent compound which turns blue litmus red.

Wbbse Madhyamik Physical Science And Environment Class 9 Question 7. Sodium carbonate is a neutral salt, but its aqueous solution is alkaline in nature, why?
Answer:

Reason: Sodium carbonate reacts with water and produces caustic soda, a strong alkali, and a weak acid, carbonic acid. Due to strong alkali, after neutralisation of H*, there is excess OH (hydroxyl ion) in the solution. So the solution will be alkaline.

Equation : Na₂CO₃ + H₂O \(\rightleftharpoons\)(Na⁺ + OH) +H₂CO₃

Question 8. What do you mean by the acidity of a base?
Answer:

The acidity of a base is expressed by the number of hydroxyl groups present in each molecule of the base.

For example :

Base	Acidity
NaOH, KOH	1. (Monoacidic base)
Ca(OH)l2, Zn(OH)l2, Mg(OH)l2	2. (Diacidic base)
AI(OH)3	3. (Triacidic base)

The acidity of a base (metallic oxide) which does not contain a (OH) group is measured by the number of molecules of a monobasic acid required to react completely with one molecule of it to produce salt and water.

CaO + 2HCI = CaCl₂ + H₂O (Acidity of CaO is 2)

MgO + 2HNO₃ = Mg(NO₃)₂ + H₂O (Acidity of Mg is 2)

Question 9. Point out the differences between bases and alkalis.
Answer:

Differences between bases and alkalis: A base may or may not be soluble in water but all alkalis are soluble in water. So all alkalis are bases but all bases are not alkalis. NaOH is an alkali because it is soluble in water. Al(OH)₃ is water insoluble. So it is a base but not an alkali.

Question 10. Which ions disappear in a neutralization reaction? Explain with an example.
Answer:

H⁺ ions generated by acid and OH^– ions generated by alkali disappear in a neutralization reaction.

Example : 2HCI + Ca(OH)₂ = CaCl₂ + 2H₂O. In this reaction, ions form as Ca(OH)₂ \(\rightleftharpoons\)Ca²⁺ + 2OH^–

2HCI\(\rightleftharpoons\) 2H⁺ +  2Cl.

Here, H⁺ ions generated from HCI combine with OH^– ions generated from Ca(OH) producing water molecules.

Question 11. What colours are produced when a drop of methyl orange is added separately to the aqueous solution of sodium carbonate and ammonium sulphate?
Answer:

Yellow in the first case and pink-red in the latter, for, the aqueous solution of the former is alkaline and that of the latter is acidic. (Na₂CO₃ + 2H₂O = 2NaOH + H₂CO₃) NaOH is a strong alkali while H₂CO₃ is a weak acid.

Again, (NH₄)₂SO₄+ 2H₂O = 2NH,OH + H₂SO₄  here H₂SO₄  is a strong acid but NH₄OH is a weak alkali.

Wbbse Class 9 Physical Science

Question 12. Why does the reddish violet colour of a phenolphthalein-mixed KOH solution gradually decolourise when air is blown into it with the help of a pipe?
Answer:

When air is blown, CO₂ gas mixes with the water of the alkaline KOH-solution producing carbonic acid (H₂CO₃), which in turn, neutralizes the given solution for which phenolphthalein turns colourless.

Question 13. What are oxides? What are the types of oxides?
Answer:

Oxides: Binary compounds of oxygen with any element (metallic and non-metallic) are called oxides.

Types of oxides :
(1) Acidic oxiae
(2) Basic oxide
(3) Neutral oxide
(4) Amphoteric oxide
(5) Peroxide
(6) Mixed oxide
(7) Polyoxide
(8) Sub-oxide
(9) Superoxide.

Question 14. Define acidic oxide.
Answer:

Acidic oxide: An acidic oxide is an oxide of a non-metal usually. It reacts with an alkali or a base to form salt and water.
Examples : CO₂, SO₂, P₂O₅ , etc.

Question 15. Define basic oxide.
Answer:

Basic oxide: A basic oxide is an oxide of a metal usually. It reacts with an acid to produce salt and water.
Examples: CaO, MgO, CuO, etc.

Question 16. Define neutral oxide.
Answer:

Neutral oxide: The oxides of certain non-metals which neither react with acid nor with base are called neutral oxides.
Examples: CO, H₂O, NO, etc.

Question 17. Define amphoteric oxide.
Answer:

Amphoteric oxide: The oxides of certain weak electropositive metals which react with both acids and bases to form salt and water are called’ amphoteric oxides.
Examples: Al₂O₃, ZnO, PbO, etc.

Wbbse Class 9 Physical Science

Question 18. Define normal salts.
Answer:

Normal salts: The salts produced by complete displacement of all the replaceable hydrogen atom or atoms present in the molecule of an acid by a metal or.a group equivalent to a metal, are called normal salts.

Question 19. What are alkalis?
Answer:

Alkalis: Water-soluble hydroxides of metals are called alkalis.
Examples: NaOH, KOH, etc.

Question 20. What are salts? What are the types of salts?
Answer:

Salts: The replaceable hydrogen atoms in acid, when replaced by meta! or basic radical partially or fully then the compound so produced is called salt.
Types of salts :
(1) Normal salt
(2) Acid salt
(3) Basic salt
(4) Double salt
(5) Complex salt.

Question 21. What are bases?
Answer:

Bases: A base in general is an oxide or hydroxide of a metal and when reacts with an acid produces salt and water.
Examples: Al₂O₃, Al(OH)₃, CaO, etc.

Wbbse Class 9 Physical Science Solutions

Question 22. What are acids?
Answer:

Acids: Ordinarily, an acid is a compound, the molecule of which contains one or more hydrogen atoms replaceable partially or completely, directly or indirectly by a metal or a group of elements behaving like a metal to form salt.

Examples: HCl, H₂SO₄, HNO₃, etc.

Question 23. Define acid salts ‘or bi-salts.
Answer:

Acid-salts or bi-salts: The salts produced by the partial displacement of the replaceable hydrogen atoms present in the molecule of an acid by a metal or a radical acting like a metal, are known as acid-salts or bi-salts.

Examples : NaHSO₄, NaHCO₃, Na₂HPO₄ etc.

Question 24. How does vanishing colour act 
Answer:

Action of vanishing colour: When the pink colour solution of vanishing colour is spread on a white linen, the linen turns pink. On allowing the pink-coloured wet linen to dryin air, the ammonia of the solution evaporates and the linen assumes its original white colour.

Wbbse Class 9 Physical Science Solutions

Question 25. Define basic salt.
Answer:

Basic salt: Salts formed by the partial displacement of oxide or hydroxide of alkalis by acid are known as basic salt.

Examples : Pb(OH)CI, Pb(OH)NO₃, etc.

Question 26. What is a vanishing colour? How is it prepared?
Answer:

Vanishing colour: It is a coloured solution which becomes colourless on exposure to air for a length of time.Preparation of vanishing colour: It is a dilute solution of ammonia in water with a few drops of phenolpthalein. Its colour is deep pink.

Question 27. Define double salt.
Answer:

Double salt: It is formed by the association of two or more normal salts.

Examples: K₂SO₄.Al,(SO₄)₃, 24H₂O; (NH₄)₂ SO₄.FeSO₄, 6H₂O, etc.

Question 28. Which ions disappear in neutralisation?
Answer:

lons disappear in neutralisation: H+ ions generated by acid and OH- ions generated by alkali disappear in a neutralisation process.

Question 29. Define complex salt.
Answer:

Complex salt: Complex salts contain complex ions which seem to be a complete salt.

Example : K₄[Fe(CN)₆], [Cu(NH₃)₄]SI₄, [Ag(NH₃)₂Cl], etc.

Question 30. What is an indicator?
Answer:

Indicator: It is some weak organic acid or base which indicates distinctive colours in acid, alkali and neutral solutions,

Question 31. What is neutralisation ?
Answer:

Neutralisation: It is the process in which acids and alkalis in equivalent quantities in their aqueous solutions react to produce salt and the neutral substance water.

Question 32. What do you mean by titration?
Answer:

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

Question 33. Why are CaH₂ and CH_4, not acids?
Answer:

CaH₂ and CH₄ are not acids: Calcium hydride (CaH₂) and methane (CH₄) do not produce H⁺ ions in aqueous solution, so these are not acids, for every acid produces H⁺ ions in aqueous solution.

Acids, Bases And Salts 3 Marks Questions And Answers:

Question 1. Classify acids with definitions of the types of acids and their respective examples.
Answer:

Acids are classified in many ways :

(1) based on composition :

Based on composition, acids can be classified into major two types :
(1) Hydracids and
(2) Oxy acids.

(1) Hydracids: The acids which contain hydrogen and a non-metallic element (other than oxygen) or a radical are known as hydracids.
For example:
HCI (Hydrochloric acid)
HI (Hydroiodic acid),
HCN (Hydrocyanic acid), etc.

(2) Oxyacids: The acids which contain oxygen besides hydrogen and a non-metallic element or a radical are called oxy-acids.

Examples of oxy-acids are :
HNO₃ (Nitric acid),
H₂SO₄ (Sulphuric acid),
H₃PO₄ (Phosphoric acid), etc.

Wbbse Class 9 Physical Science Solutions

(2) According to sources :

According to sources, acids can be classified into:
(1) Inorganic acids and
(2) Organic acids.

(1) Inorganic acids: Those acids which are derived from inorganic sources (e.g. – minerals) are known as inorganic acids. For example – Hydrochloric acid (obtained from common salt NaCl), nitric acid (obtained from sodium or potassium nitrate), sulphuric acid (from sulphur), phosphoric acid (obtained from calcium phosphate), etc.

(2) Organic acids: The acids which contain carbon as an essential element and can be obtained from either plants or animals are designated as organic acids. Formic acid(HCOOH), acetic acid (CH₃COOH), etc. are some of the organic acids.

(3) According to oxidising power, acids are also classified as:
(1) oxidising acid and
(2) non-oxidising acid.

(1) Oxidising acid: Acids which easily dissociate in normal conditions and produce nascent oxygen are known as oxidising acids:

Examples:

(1) Dissociation of concentrated sulphuric acid: H₂SO₄ H₂O + SO₂ + [O].
(2) Dissociation of concentrated nitric acid: 2HNO₃ H₂O + 2NO₂ + [O].

The nascent oxygen atoms produced from these acids oxidise other substances, for this reason, they are called oxidising acids.

(2) Non-oxidising acids: Non-oxidising acids are those which do not give off nascent oxygen on dissociation.
Examples: Dilute sulphuric acid (H₂SO₄), hydrochloric acid (HCI), hydrocyanic. acid (HCN), organic acids, like formic acid (HCOOH), acetic acid (CH₂COOR), etc.

(4) based on concentration: According to concentration, acids are of two types:
(1) Concentrated acid and
(2) Dilute acid.

(1) Concentrated acid: The acids which have less quantity of water in them are called concentrated acids.
Examples: conc. H₂SO₄, conc. HCl, etc.

(2) Dilute acids: Those acids which have more quantity of water mixed with them are called ditute acids.
Examples: dil. H₂SO₄, dil. HCl, etc.

(5) based on Basicity :

Definition of Basicity of acid: Basicity of an acid is its power of neutralising a base and it is measured by the number of replaceable hydrogen atoms present in one molecule of it.

Wbbse Class 9 Physical Science Solutions

According to basicity, acids are classified into three types:
(1) Monobasic acids
(2) Dibasic acids and
(3) Tribasic acids.

(1) Monobasic acids: Those acids which have only one replaceable hydrogen atom in their one molecule are called monobasic acids.

Examples: Hydrochloric acid (HCI), nitric acid (HNO), hydrocyanic acid (HCN), perchloric acid (HCIO₄), formic acid (HCOOH), acetic acid (CH₃COOH), etc.

(2) Dibasic acids: Those acids which have only two replaceable hydrogen atoms in their one molecule are called dibasic acids.

Examples: Sulphuric acid (H₂SO₄), Carbonic acid (H₂CO₃), hypophosphorous acid (H₃PO₂), etc.

(3) Tribasic acids: Those acids which have only three replaceable hydrogen atoms in their one molecule are known as tribasic acids. Phosphoric acid (H₃PO₄), citric acid (C₆H₈O₇), etc.

Question 2. What is meant by ionisation? Explain with an example.
Answer:

ionisation: The process of breaking up any compound into positively charged and negatively charged atoms or groups of atoms. by the mere process of dissolution is called ionisation. The positively charged particles are called cations and negatively charged particles are called anions.

Example: When common salt, i.e., sodium chloride (NaCl) is dissolved in water, it breaks up into positively charged sodium ions (Na⁺, cation) and negatively charged chloride ions (Cl^–, anions) by the solvating action of water. This process of ionisation may be expressed by the equation.

NaCl \(\rightleftharpoons\) Na⁺+ Cl^–.

The process of ionisation is a reversible one (indicated by = sign), and consequently, an equilibrium is established between the ions and undissociated molecules.

Wbbse Madhyamik Physical Science And Environment Class 9

Question 3. Calcium, sodium, sulphur, phosphorus, and carbon are burnt completely in separate oxygen jars. Each product is treated with water and then every solution is tested with blue and red litmus papers. Now complete the table.

Answer:

Elements	Change of colour (1)    Equation of reaction with oxygen. (2)    Equation of reaction of product with water.	Blue litmus	Red litmus
1. Calcium	(1)    2Ca+O2= 2CaO (2)    CaO + H2O= Ca(OH)2		blue
2. Sodium	(1)    2Na+O2= Na2O2 (2)    2Na2O2+2H2O=4NaOH +O2		blue
3. Sulphur	(1)    S+O2 = SO2 (2)    SO2 + H2 O = H2SO3	red
4. Phosphorus	(1)    4P + 5O2 = 2P2 O5 (2)    P2O5+3H2O= 2H3PO4	red
5. Carbon	(1)    C +O2 = CO2 (2)    CO2 +H2 O = H2 CO3	red

Question 4. What is Arrhenius’s concept of acid?
Answer:

Arrhenius’s concept of acid: According to Arrhenius, a Swedish scientist, an acid is a chemical compound that dissociates in water, producing H⁺ ions (cations) as the only Positive ions. H* ion is called proton. So, an acid is known as a proton donor. But H⁺ (proton) does not remain in a free state in solution; it attains stability being attached to a molecule of the solvent. In water solution proton combines with a water molecule to produce hydronium ion H₃O⁺.

Examples : HCl + H₂O \(\rightleftharpoons\)H₃O⁺ + CI^–

HNO₃ + H₂O \(\rightleftharpoons\) H₃O⁺ + NO^3–

Question 5. All acids are hydrogen compounds but all hydrogen compounds are not acids. Explain.
Answer:

Explanation: It is to be noted that any acid contains hydrogen but any compound containing hydrogen is not an acid. A compound containing one or more than one hydrogen atom in its molecule will be termed as acid only when its hydrogen atom or atoms are replaceable by metal and the product thus obtained must be a salt.

Example: None of the four hydrogen atoms of methane (CH₄) can be replaced by a metal. Metals like sodium, and potassium displace hydrogen from water (H₂O) but the product obtained in each case is not a salt. So, methane and water, though they contain hydrogen atoms in their molecules, are not acids.

Question 6. What are the important properties of acids?
Answer:

Important properties of acids :

(1) Taste: Generally, the aqueous solutions of acids have a sour taste.

(2) Colour changes of certain substances called indicators: Acids turn blue litmus solutions red. :

(3) In aqueous solution, the acids conduct electricity.

Wbbse Madhyamik Physical Science And Environment Class 9

(4) Reactions with metals: The solutions of acids in water react with many metals like zinc, magnesium, iron, etc. which are more electropositive than hydrogen with the liberation of hydrogen gas and formation of corresponding salts.

(5) Reactions with oxides and hydroxides of metals: Acids react with metallic oxides or hydroxides producing salts and water.

(6) Reactions with carbonates and bicarbonates: Acids are also characterised by their tendency to react with metal carbonates and bicarbonates evolving carbon dioxide.

(7) According to Arrhenius’s theory of electrolytic dissociation, the acids in aqueous solutions produce hydrogen ions as the positive ions or cations.

Question 7. What is Arrhenius’s concept of base?
Answer:

Arrhenius’s concept of base: According to Arrhenius, a base is a chemical compound which produces OH- ions (hydroxyl ions) as the only anions when these are dissolved in water.

Examples : CaO + H₂O \(\rightleftharpoons\) Ca(OH)₂  ; Ca(OH)₂ \(\rightleftharpoons\) Ca²⁺ + 2OH^–

NaOH = Na⁺ + OH^–

Wbbse Madhyamik Physical Science And Environment Class 9 Question 8.

Discuss the role of indicators in the case of titration.
Answer:

Uses of indicators in neutralisation reaction: Indicators which can determine the end-point of the neutralisation reaction are given below

Indicator	Colour changes in
	Neutral Solution	Acidic Solution	Alkaline Solution
1. Litmus	Violet	Red	Blue
2. Methyl-orange	Orange	Red	Yellow
3. Phenolphthalein	Colourless	Colourless	Pink

The selection of a suitable indicator to determine the correct end-point of a reaction :

(1) Strong acid and weak base: Methyl orange

(2) Weak acid and strong base: Phenolphthalein

(3) Strong acid and strong base: Any indicator

(4) Weak acid and weak base: No suitable indicator.

Question 9. State the properties of acids and bases about indicators.
Answer:

Properties of acids about indicators: The aqueous solution of acids turn :
(1) Blue litmus to red
(2) Orange-coloured methyl orange to pinkish red
(3) Phenolphthalein remains colourless in aqueous solutions of an acid.

Properties of bases about indicators: The aqueous solution of bases turn:
(1) Red litmus to blue
(2) Orange-coloured methyl! orange to yellow
(3) Colourless. phenolphthalein solution to pink.

Question 10. What is the basicity of an acid and what is the acidity of a base?
Answer:

Basicity of an acid: Basicity of an acid is expressed by the number of replaceable hydrogen atoms present in each molecule of the acid.

Example :

Monobasic acid : (i.e., basicity is 1): HCl, HNO₃

Dibasic acid (i.e., basicity is 2): H₂ SO₄, H₂ CO₃.

Tribasic acid : (i.e., basicity is 3): In H₃PO₄

Acidity of a base: Acidity of a base is expressed by the number of hydroxyl groups present in each molecule of the base.

Example :

Monoacidic base : NaOH, KOH.

Diacidic base : Ca(OH)₂ , Mg(OH)₂

Triacidic base : Al(OH)₃.

Question 11. All alkalis are bases but all bases are not alkalis —Explain.
Answer:

Explanation: Bases are the oxides or hydroxides of metals and react with acids to Produce salts and water. The bases which are soluble in water are known as alkalis. Ferric hydroxide Fe(OH)₃, Zinc hydroxide Zn(OH)₂, Aluminium hydroxide Al(OH)₃, Sodium hydroxide NaOH, and Potassium hydroxide KOH all are bases but not alkalis. Among them only water-soluble metallic hydroxides NaOH, and KOH are alkalis. From the above examples of bases and alkalis, it is clear that all alkalis are bases but all bases are not alkalis.

Question 12. What are the properties of acids?
Answer:

Properties of acids :

(1) An acid produces H⁺ ion in aqueous solution.

(2) Generally acids are water-soluble and the solution has an acidic or sour taste.

(3) The aqueous solution of acids turns blue litmus to red, orange-coloured methyl orange to pinkish red and phenolphthalein remains colourless in the aqueous solution of an acid.

(4) Electropositive metals like Zn, Mg, and Fe (which belong above hydrogen in the electrochemical series) react with dilute acid to produce hydrogen.

Example : Zn + H₂SO₄ = ZnSO₄ + H₂ ↑

2Al + 6HCl = 2AICl₃ + 3H₂↑

(5) Acids form salt and water reacting with metallic oxides and hydroxides.

Example : 2NaOH + H₂SO₄ = H₂SO₄ + 2H₂O

CaO + 2HCI = CaCl + H₂O

(6) Acids form carbon dioxide by reacting with metallic carbonate and bicarbonate.

Example : Na₂CO₃ + 2HCI = 2NaCl + CO₂ + H₂O

NaHCO + HCI = NaCl + CO₂ + H₂O

Question 13. Classify and define the different types of oxides.
Answer:

Types of Oxides :

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

Non-metallic oxide : CO₂ , SO₂ , P₂O₅ , SiO₂, etc.

Metallic oxide : CrO₃ (Chromium trioxide), Mn₂O₇ (Manganese heptoxide), etc.

(2) Basic oxide: A basic oxide is an oxide which reacts with an acid to form salt and water. These are generally the oxides of metals.

Examples: CaO (Calcium oxide), MgO (magnesium oxide), FeO, CuO, etc.

(3) Neutral oxide: The oxides which neither react with acid nor with base are called neutral oxide.

Examples: CO (Carbon monoxide), H₂O (Water), N₂O (Nitrous oxide), NO (Nitric oxide), etc.

(4) Amphoteric oxide: The oxides which react with both acids and bases to produce salt and water are called amphoteric oxide.

Examples: Al₂O₃(Aluminium oxide), ZnO (Zinc oxide), SnO (Tanous oxide), PbO (Lead monoxide), etc.

(5) Peroxide: The oxides which react with a cold and dilute mineral acid to produce hydrogen peroxide and have peroxy linkage (— O — O -) are called peroxide.

Examples: Na₂O₂ (Sodium peroxide), BaO, (Barium peroxide), etc.

(6) Mixed oxide: An oxide which is formed by oxides of more than one oxide of a metal having variable valency is called mixed oxide.

Examples : Fe₃ O₄ [FeO + Fe₂O₃ ]; Mn₃ O₄ [2MnO + MnO₂] ; Pb₃ O₄ [PbO₂ + 2PbO], etc.

(7) Poly-oxide: The oxides which have oxygen atoms more than ordinary oxide and also do not react with an acid to produce hydrogen peroxide are called poly-oxides.

Examples: Mn₂O₇ (Manganese heptoxide), PbO₂ (Lead dioxide), etc.

(8) Sub-oxide: An oxide which has fewer oxygen atoms rather than its oxygen atoms permitted for oxidation state: of the element is called sub-oxide.

Example: Carbon sub-oxide (C₃O₂).

(9) Super-oxide: Metals like Na, Li, and Ca form super-oxide and have negative ions (O — O)  in them.

Examples : LiO₂, KO₂, etc.

Question 14. What are the properties of bases?
Answer:

Properties of bases :

(1) The aqueous solution of bases changes the colour of litmus from red to blue, and the orange methyl orange to yellow.

(2) Reacting with acid, bases produce salt and water.

(3) Concentrated solutions of some strong bases are slippery to the touch.

(4) The aqueous solution of bases conducts electricity.

(5) Strong bases when heated with electropositive elements (like Al, Zn, etc.) produce hydrogen.

Example : 2NaOH + Al + 2H₂O = 2NaAlO₂ + 3H₂↑

Strong bases absorb carbon dioxide from the atmosphere to form carbonate salt and water.

Example : 2NaOH + CO₂ = Na₂ CO₃ + H₂O.

Question 15. Classify and define the different types of bases.
Answer:

Types of bases according to the acidity of bases :

(1) Monoacidic bases: These bases which produce one OH- ion in aqueous solution from one molecule are Called monoacidic bases.

Example : NaOH \(\rightleftharpoons\) Na⁺ + OH^–  ; KOH \(\rightleftharpoons\) K⁺ + OH^–

(2) Diacidic bases: The bases which produce two OH- ions in aqueous solution from one molecule are called diacidic bases.

Example : Ca(OH)₂\(\rightleftharpoons\)   Ca²⁺+ 2OH^–

Types of bases according to their ionisation :

(1) Strong bases: The bases which ionise mostly in an aqueous solution and a small number of molecules exist in a molecular state are called strong bases.

Examples: NaOH, KOH, Ca(OH)₂, etc.

Types of bases according to their source :

(1) Mineral bases: The bases which are produced from minerals are called mineral bases.

Examples: NaOH, KOH, etc.

(2) Organic bases: The bases containing nitrogen atoms and whose sources are animals and plants are called bases.

Example:

 

Question 16. Classify salt and define its different types.
Answer:

Types of salts :

(1) Normal salts: Normal salts are formed by the complete replacement of the hydrogen atoms in acid by metals or basic radicals.

Examples :

(1) NaCl [NaOH + HCI = NaCl + H₂O].

(2) Na₂SO₄ [2NaOH + H₂SO₄ = Na₂SO₄ + H₂O].

(2) Acid salt: An acid salt is formed only by partial replacement of hydrogen atoms in a molecule of an acid (di-basic, tri-basic) by metal or basic radical.

Examples :

(1) NaHSO₄ (Sodium bisulphate).

(2) NaHCO₃ (Sodium bicarbonate).

(3) Na₂HPO₄ (Di-sodium hydrogen phosphate).

(3) Basic salt: When the hydroxyl  (OH^–) or, oxide (O) radicals are partially neutralised by acid then basic salts are produced.

Examples :

(1) Pb(OH)CI (Basic lead chloride).

(2) Pb(OH)NO₃ (Basic lead nitrate).

(4) Double salt: When two normal salts are mixed with their molecular weight ratio to form a solution and cooled the solution to make a combined crystal which is stable in solid state but ionises into different constituent ions in solution, the salts are called double salts.

Examples : K₂SO₄.Al₂(SO₄)₃ .24H₂O

(NH₄)₂ SO.FeSO₄.6H₂O

(5) Complex salts: When the solution of two mixed salts is concentrated and cooled then crystals of a new salt are produced. The constituent salts lose their identity in new salt. In an aqueous solution,, they produce new complex ions. These types of salts are called complex salts.

Examples : K₄(Fe(CN₆) , (Cu(NH₃)₄) SO₄ .

Question 17. Write a short note on vanishing colours.
Answer:

Vanishing colours: Vanishing colour is a coloured solution which becomes colourless on exposure to air for a length of time.

Example: The aqueous solution of ammonium hydroxide (NH₄OH) mixed with phenolphthalein turns pink in colour. Now if this solution is allowed to spray on cloth, all at once a pink colour is developed upon cloth. But since ammonia in NH₄OH is highly volatile, after the lapse of time, NH₃ volatilises and the solution becomes neutral. As a result, the pink colour of Phenolphthalein vanishes. Vanishing colour cannot be prepared with dilute NaOH solution, because NaOH does not evaporate.

Wbbse Madhyamik Physical Science And Environment Class 9 Question 18. Write about the importance of pH.
Answer:

Importance of pH:

Field	Utility
(1) Biotechnology	Biochemical and organic reactions at controlled pH value give best results.
(2) Agriculture	Alkaline soils are preferred for some crops such as citrus fruits grow better in alkaline soils. Sugarcane grows better in neutral soil, whereas rice. The pH of soil is tested before growing a particular crop.
(3) Medicine	Determination of pH value of blood and urine helps to diagnose certain diseases.
(4) Cosmetics	Special soaps, shampoos, face creams are prepared for consumers having different pH value of skin secretions.
(5) Milk plants	pH of milk is rigorously controlled at pH 6.8, otherwise, it turns sour.
(6) Breweries	Controlling pH value of wine to obtain a desired flavour.

Question 19. Write a short note on the chemical properties of sulphuric acid.
Answer:

Chemical Properties of sulphuric acid :

(1) Acid Property: Sulphuric acid is a strong dibasic acid. In its aqueous solution, it is completely ionised. Its ionisation takes place in two steps

H₂SO₄ \(\rightleftharpoons\) H⁺ + SHO^–₄  ; H₂SO₄ \(\rightleftharpoons\) 2H + SO^2-₄

Its aqueous solution turns blue litmus red and reacts with bases and alkalis to produce salts and water.

(1) At ordinary temperature Sulphuric acid reacts with carbonates and bicarbonates with the formation of sulphates or bisulphates liberating carbon dioxide.

NaHCO₃ + H₂SO₄→NaHSO₄ + CO₂ + H₂O ; Na₂CO₃ + H₂SO₄→Na₂SO₄ + CO₂ + H₂O

(2) Since sulphuric acid has two replaceable hydrogen atoms, it forms two series of salts — normal salt (sulphates) and acid salts (bisulphates).

MgO + H₂SO₄→MgSO₄ + H₂O          CaO + 2H₂SO₄→Ca(HSO₄)₂ + 2H₂O

NaOH + H₂SO₄→NaHSO₄+ H₂O       2NaOH + H₂SO₄→ Na₂SO₄ + 2H₂O

(3) Reaction with metals :

With zine :

(1) At ordinary temperature, dilute H₂SO₄ reacts with zinc to produce zinc sulphate and hydrogen gas. Zn + H₂SO₄→ ZnSO₄ + H₂

(2) Hot, concentrated H₂SO₄ reacts with zinc to produce zinc sulphate, water and sulphur dioxide gas. Zn + 2H₂SO₄→ZnSO₄ + SO₄ + 2H₂O

With magnesium :

(1) At ordinary temperature, dilute H₂SO₄ reacts with magnesium to produce magnesium sulphate and hydrogen gas. Mg + H₂SO→ MgSO₄+ H₂↑

(2) Hot, concentrated H₂SO₄ reacts with magnesium to produce magnesium sulphate, water and sulphur dioxide gas. Mg + 2H₂SO₄ →MgSO₄ + SO₂ + 2H₂O

With copper :

(1) Cold, dilute H₂SO₄ does not react with copper.

(2) Hot, concentrated H₂SO₄ reacts with copper to produce copper sulphate, water and sulphur dioxide gas. Cu + 2H₂SO₄ → CuSO₄ + SO₂+ 2H₂O

(2) Test of sulphuring acid: When barium chloride solution is added to dilute or concentrated H₂SO₄, a white precipitate of barium sulphate is formed which is insoluble in

HCl₃. H₂SO₄ + BaCl₂→ BaSO₄ ↓+ 2HCI.

Wbbse Madhyamik Physical Science And Environment Class 9 Question 20. Write a short note on the chemical properties of hydrochloric acid.
Answer:

Chemical Properties of hydrochloric acid :

(1) Acid Property: Hydrogen chloride in the gaseous state is not acidic, but the aqueous solution of the gas is highly acidic and one of the strongest acids. Hydrochloric acid is < monobasic acid and turns blue litmus red. It ionises almost completely in its aqueous solution.

HCl\(\rightleftharpoons\) H⁺+Cl^–

(1) Action on bases: Hydrochloric acid reacts with metallic oxides and hydroxide forming the corresponding chlorides and water.
CaO + 2HCI → CaCl₂+H₂O             NaOH + HCI→ NaCl + H₂O

MgO + 2HCIl → MgCl₂ + H₂O        Fe(OH)₃ + 3HCl →FeCl₃ + 3H₂O.

(2) Action on metallic carbonates and bicarbonates: Hydrochloric acid reacts with metallic carbonates and bicarbonates with the evolution of CO₂.

Na₂CO₃+ 2HCI 2NaCl + CO₂ + H₂O ;     NaHCO₃ + HCl→ NaCl + CO₂+ H₂O

(3) Action on metals: The metals which are above hydrogen in the electrochemical series (Mg, Zn, Al, Fe, Sn, etc.) react with hydrochloric acid at ordinary temperature wit! the evolution of hydrogen and the formation of corresponding chloride.

Zn + 2HCI ZnCl₂ + H₂    ;  Fe + 2HCl→ FeCl₂ + H₂ ;  2Al + 6HCl →2AICI₃ + 3H₂

Copper dissolves slowly on boiling with concentrated HCI in the presence of air or oxygen producing cupric chloride and water. 2Cu + 4HCl + O₂ → 2CuCl₂ + 2H₂O

Question 21. Write a short note on the chemical properties of nitric acid.
Answer: Nitric acid is a strong monobasic acid. lt ionises almost completely from hydrogen ions and nitrate ions. HNO₃ \(\rightleftharpoons\) H⁺+ NO^–

As nitric acid is a monobasic acid, it forms normal salts only. It turns blue litmus red.

(1) Action on alkalis: It reacts with oxides and hydroxides producing nitrate salts and water.

CaO + 2HNO₃ →Ca(NO₃)₂+ H₂O ;  MgO + 2HNO₃→Mg(NO₃)₂+ H₂O

NaOH + HNO₃ → NaNO₃ + CO₂ + H₂O

(2) Action on metallic carbonates and bicarbonates: Nitric acid reacts with metallic carbonates and bicarbonates producing nitrates and liberating CO₂.

Na₂CO₃ + 2HCl→ 2NaCl + CO₂ + H₂O  ; NaHCO₃+ HC] → NaCl + CO₂ + H₂O

(3) Action on metals: The metals which are above hydrogen in the electrochemical series (Mg, Zn, Al, Fe, Sn, etc.) react with hydrochloric acid at ordinary temperature with the evolution of hydrogen and the formation of corresponding chloride.

Zn + 2HCl —> ZnCl₂ + H₂   ; Fe + 2HCI →  FeCl₂ + H₂ ; 2Al + 6HCl → 2AlCl₃ + 2H₂O

(4) Test of hydrochloric acid: An aqueous solution of HCl gives a curdy white precipitate of silver chloride with silver nitrate solution. The precipitate is insoluble in nitric acid but is

readily soluble more than the ammonium hydroxide solution. AgNO₃ + HCl → AgCl + HNO₃ .

Wbbse Madhyamik Physical Science And Environment Class 9 Question 22. Write a short note on the chemical properties of sodium hydroxide.
Answer:

Sodium hydroxide is popularly known as Caustic Soda.

Physical properties :
(1) NaOH has a sharp, bitter taste.
(2) It changes the colour of indicators turns red litmus to blue and turns colourless phenolphthalein to pink.
(3) It is a soapy substance, i.e., it is slippery to touch.
(4) NaQH is a strong electrolyte.

To note that: NaOH has a mild corrosive action which causes slight alkali burn on skin.

Chemical properties 

(1) NaOH absorbs carbon dioxide from the air to form carbonates.

CO₂+ 2NaOH→  Na₂CO₃ + H₂O

(2) It produces ammonia gas when warmed with ammonium salt.

NH₄Cl + NaOH →  NaCl+ H₂O + NH₃ (g)

(3) NaOH, being an alkali or base, neutralizes acids to form salt and water.

NaOH + HCI → NaCl + H₂O  ;  NaOH = HNO₃→NaNO₃ + H₂O

2NaOH + H₂SO₄→ Na₂SO₄ + 2H₂O

(4) Action on metals :

(1) With Aluminium: When powdered aluminium is heated with concentrated solution 4of NaOH, sodium aluminate and hydrogen gas are produced.

2Al + 2NaOH + 2H₂O → 2NaAlO₂+ 2H₂

(2) With Zinc: When powdered zinc is heated with a solution of NaOH, sodium zincates 2nd hydrogen gas are produced.

Zn =2NaOH → Na₂ZnO₂ +H₂

Question 23. How does acidity affect human life?
Answer:

(1) Decay of teeth-enamel: It has been inferred by researchers that in areas where drinking water has fluoride levels of 1 ppm (part per million) or above, children have 60% fewer cavities than do children living in areas with lesser fluoride concentrations. Dentists often suggest fluoride treatments and prescribe toothpaste containing fluoride. The enamel on teeth (and the bones of the body) are made of a compound called “hydroxyapatite [Ca₁₀(PO₄ )₆(OH)₂]. This is basic and it reacts with acids and thus decays.

Fluorides present in the body produce fluorapatite [Ca₁₀(PO₄ )₆F₂], which is more resistant to decay. Tooth cavities are caused when acids dissolve tooth enamel.

Ca₁₀(PO₄ )₆(OH)₂ + 8H⁺ (aq) → 10Ca²⁺ (aq) + 6HPO₄ ^2-, (aq) + 2H₂O (l)

(2) antacids: Drugs for stomach upset: Digestion is accompanied by the production of hydrochloric acid (HCI) in the stomach. Sometimes the stomach produces too much HCl. This may be due to emotional stress or overeating. Stomach upset and heartburn are the result. Antacids are the solution or care for such a problem.

Functioning of an antacid: The various antacid medications work by neutralizing the excess acid. The reaction is simply an acid-base reaction :

OH^– +H⁺    H₂O [for antacids that contain OH^– ions]

CO₃^2- + 2H⁺ → H₂O+ CO₂ [for antacides containing carbonate ions]

Mangesium hydroxide or magnesium carbonate are such compounds used in antacid.

Mg(OH)₂ + 2HCI MgCl₂+ 2H₂O  ;  MgCO₃ + 2HCI MgCl₂ + H₂O + CO₂

These two compounds, in small doses, act as antacids and in large doses, they act as laxatives.

(3)In fishery agriculture:

(1) Acidification of inland waters induced by acid rain or other acidic pollutants causes various harmful physiological and behavioural effects on fish.

(2) At pH ≤ 5 it induces failure of immune and reproductive functions.

(3) pH around 4 efflux of Na⁺ and Cl^– ions leading to mortality.

(4) Many populations of fish have vanished particularly due to the damage to aquatic ecosystems by acid rain.

Question 24. Write a short note on pH and soil quantity.
Answer:

pH and soil quality :

(1) pH of soil is the masier variable as it controls many chemical processes that take place. It specifically affects plant nutrients. The optimum pH range for most plants is 5.5→ 7.0.

(2) Acidity in soils comes from H⁺ and Al ³⁺  ions, which at pH = 4 →6, reacts with water to form Al(OH)₃ and releases 3H⁺  ions.

(3) Many other processes may contibute to the formation of acid soils including rainfall, fertilizer use, plant root activity, acid rains and mine spoilings.

(4) Alkaline soils are characterized by the presence of carbonates.

The effects :

(1) Below pH = 4, H⁺  ions themselves damage root cell membranes.
(2) Al³⁺ and Mn²⁺ are more soluble at lower pH and dissolved Al³⁺is toxic to plants.
(3) If pH > 7.5 (alkaline), phosphorus (P) starts forming insoluble compounds with calcium (Ca).

Note that: The common and easy way to increase soil pH is to add lime (CaCO₃ or MgCO₃ ), wood ash or burnt lime CaO: CO^2-₃ + 2H⁺→ CO₂ +H₂O

Question 25. Write a short note on acid rain.
Answer:

Under normal conditions, rain is slightly acidic, with a pH = 5.6, due to dissolved CO₂ ie., H₂CO₃. Recently, the acidity of rainwater, in many industrialized and populated areas, has increased to more than 100 times, to a PH 3 to 3.5. The major pollutants in acid rain are strong acids that arise from human activities, particularly automotive activities.

(2) The Chemistry behind acid rain   In ‘the automobile internal combustion engines and electrical power stations NO₂ is formed and gets further oxidized in air to form which in turn, forms nitric acid (and nitric oxide)

N₂(g) + O₂ (g) →  2NO (g) [at high temperature]

2NO (g) + O₂ (g) → 2NO₂ (g)

3NO₂ (g) + 3H₂O (l) → 2H₃ O⁺(aq) + 2NO^–₃  (aq) + NO (g)

Sulphur dioxide (SO₂) is produced as a by-product of the burning of fossil fuels. It reacts with water directly forming sulphurous acid (a weak acid) but produces sulphur trioxide (SO₃) in the presence of particulate matter and aerosols. This SO₃ combines with water forming sulphuric acid (H₂SO₄) (a strong acid)

SO₂ (g) + 2H₂O (l)→H₂O (l) + H₂SO₄

2SO₂(g) + O₂ (g) → 2SO₃  (g)

SO₃ (g) + 2H₂O (l) → H₂O + H₂SO₄ (or H₃O⁺ + HSO^–₄).

The effect of acid rain: In many processes, Nature likes and requires a balance of pH. Any drastic shift in the pH of rain, thus, is bound to affect it in many ways.

(1) Lakes may become so acidic that all fish life may disappear.
(2) Massive trees die off as acid rain has lowered the pH of the soil and has leached nutrients from leaves.
(3) Valuable and heritage marble statues have been slowly dissolved away as the calcium carbonate of marble has been attacked by acid rain :

CaCO₃(s) + 2H ⁺ (aq) Ca²⁺ (aq) + H₂O (l) + CO₂ (g)

The problem will grow more serious as sources of low-sulphur coal are nearly exhausted and power plants are forced to rely on more abundant sources of high-sulphur coal.

What to do :
(1) Much more work on methods to control acidic emissions is to be done.
(2) To reduce and optimize the consumption of fossil fuel, and the consumption of energy in general.
(3) To change our attitude from ‘mastering’ nature (and thus exploiting and destroying to the “sustainable coexistence” with mother nature.

Wbbse Madhyamik Physical Science And Environment Class 9 Question 26. Write a short note on neutralisation.
Answer:

Wherever acids and bases combine in the ratio of their equivalent weights, Salts and water are formed quantitatively. The resulting solution is neutral and this type of reaction is called neutralisation. Both acids and bases lose their identities completely. According to logic theory, neutralisation amounts to the formation of undissociated weak electrolyte H₂O by thé interaction of H* ions of the acid and OH^– ions of the alkalis.

HCI \(\rightleftharpoons\)  H⁺ + Cl^–

NaOH \(\rightleftharpoons\)  Na⁺ + OH^–

HCI + NaOH \(\rightleftharpoons\)  Na⁺ + Cl^– + H₂ O

It is to be remembered that all indicators cannot be used for all types of acid-base titration. The choice of an indicator depends on the nature of acid and base.

Selection of Indicators :

Titration of		Indicator to be used	Examples of neutralisation
Acid	Base
1. Strong	Strong	Any indicator	HCl + NaOH
2. Strong	Weak	Methyl orange Methyl red	HCI + NH4OH HCl+Na2CO3
3. Weak	Strong	Phenolphthalein	CH3COOH + NaOH
4. Weak	Weak	No suitable indicator	CH3COOH + NH4OH

Question 27. Write a short note on antacid.
Answer:

Antacid: Our stomach normally produces acids to help to acid in disgesting food and to kill germs. The pH of our stomach juice is 2-3. Secretion of hydrochloric acid increases when we consume too much of spicy food. Then we suffer from acidity, gas, etc.

Antacids work by counteracting (neutralising) the acid in our stomach. They do this because the chemicals in antacids are bases (alkalis). This neutralization reduces the acidity of the stomach juice. Antacids are the alkaline substances which can reduce the excess acid of stomach. They can maintain the proper pH of the stomach juice.

Some common antacids which are available in the market are gelusil, Digene, polyol, milk of magnesia, etc. All these antacids contain aluminium hydroxide and magnesium hydroxide. When we consume antacids, these alkaline substances react with the hydrochloric acid secreted in the stomach. Thus the acidity is reduced.

Nowadays, zinetac, acidic, pan-40, etc. are sold in the market as antacids. These compounds are to be taken in empty stomach. Actually these compounds can control the acid secretion from stomach.

Wbbse Madhyamik Physical Science And Environment Class 9

Question 28. How do acids and bases react? 

Answer:

According to Arrhenius’s theory, in an aqueous solution acids and bases produce H⁺ and OH^– ions respecitvely. When acid-base are mixed together in aqueous solution, they produce water molecules.

H⁺ (aq) + OH^– (aq) H₂ O (l)

The other parts of the acid and base combine to give salt.

Ca(OH)₂ + H₂ SO₄ → CaSO₄ + 2H₂O

(Base)    +   ( Acid )  →  (Salt )  + (water)

2H⁺  ions form H₂SO₄ and 2OH^– ions from Ca(OH)₂ form 2H₂O molecules. The Ca²⁺ ion and SO₄^2- ions from the salt CaSO₄. So acids and bases react to give salt and wate (some bases react with acids, given only salt no water is formed). e.g. 2NH₃ + H₂SO₄= (NH₄)₂ SO₄.

Question 29. The idea of nascent hydrogen (or oxygen) is at present considered unnecessary. Justify the statement.
Answer:

The nascent state of an element is that state just at the time of its production from a compound. So, hydrogen just at the time of its production from its compound is called the nascent hydrogen. The idea of nascent hydrogen and nascent oxygen as reducing and oxidising agents respectively at present is unnecessary, because the actual reducing agents produce hydrogen in acidic condition. These hydrogen-producing substances are stronger

reducing agents than hydrogen. On the other hand, oxidising agents are not producing
nascent oxygen, they are only oxiding agents, e.g.

(reduced)

Zn + 2FeCl₃ (yellow) FeCl₂(colourless) + ZnCl₂

(oxidised)

Zn is a reducing agent which itself is oxidised and FeCl₃ is an oxidising agent which itself is reduced. No participation of nascent hydrogen is required in the reduction process.

(oxidised)

Zn + 4HNO₃ → Zn(NO₃)₂+ 2NO₂↑+ 2H₂O

(reduced)

In this reaction the idea of nascent oxygen as the oxidising agent is baseless. Here HNO₃ itself is an oxidising agent, which is itself reduced to nitrogen dioxide (NO₂) and oxidises zinc. Here zinc is a reducing agent.

Wbbse Madhyamik Physical Science And Environment Class 9 Question 30.

Write the limitations of Arrhenius’s concept.
Answer:

Arrhenius’s concept has the following limitations :

(1) It recognises the dissociation of acids and acid-base behaviour in aqueous medium only. It cannot explain the acid-base behaviour in the absence of water and in non-aqueous medium.

(2) Arrhenius concept restricts acids to merely hydrogen-containing compounds and bases to merely hydroxyl-containing compounds. The substances such as CO₂, SO₂, SO₃, etc. are not regarded as acids and the substances such as NH₃, Na₂O, CaO, MgO, and Na₂CO₃  are not regarded as bases though they exhibit properties of acids and bases, respectively.

CaO (s) + SO₃(g) →CaSO₃(s)
(Base)   +    (Acid)

NH₃(g) + HCI (aq)→ NH₄C (s) .
(Base)  +   (Acid)

Question 31. Write a short noie on indicator.
Answer: Indicator: Indicators are complex.organic compounds which undergo a change of colour with a change of acidic, neutral or basic medium.

Litmus (a dye from plants), methyl orance (organic pigment)and phenolphthalein (an organic acid)are three commonly used indicators to determine the acidic or basic nature of a solution by a characteristic in colour. An indicator is a substance which change as to a characteristic colour in the presence of a particular concentration of ions such as H⁺ and OH^– ions.

Some useful indicators

Indicator	Colour in acid solution	Colour in alkaline solution	Colour in neutral soloution
Litmus	red	blue	purple
Phenolphthalein	colourless	pinkish red	colourless
Methyl orange	Pink	yellow	orange
Methyl red	red	yellow	light

When litmus is dissolved in distilled water which is neither acidic nor basic; the colour of litmus solution is purple. In acidic medium the colour of litmus solution is red and in basic medium the colour of litmus solution is blue. Litmus solution is called an acid-base indicator. Examples of other acid-base indicators are phenolphthalein, methyl orange, and methyl red. The natural acid-base indicator is turmeric powder, which is used in cooking vegetables in almost all Indian homes. Its colour in acidic solution is yellow and in basic solution is brown.

Wbbse Madhyamik Physical Science And Environment Class 9 Question 32.

Write the chemical reactions of H₂SO₄, HCI & HNO, with carbonate and bicarbonate compounds.
Answer:

Chemical reactions of H₂SO₄, HCl & HNO₃, with carbonate and bicarbonate compounds :
All mineral acids like H₂SO₄, HCl and HNO₃, in reaction with metallic carbonates and bicarbonates, give corresponding metallic sulphate. metallic chloride or metallic nitrate, along with carbon dioxide. The gas (CO₂) is liberated with brisk effervescence.

(1) Metallic carbonate + Hydrochloric acid Metallic chloride + water + carbon dioxide
Metallic carbonate + Sulphuric acid → Metallic sulphate + water + carbon dioxide
Metallic carbonate + Nitric acid → Metallic nitrate + water + carbon dioxide

Examples :

CaCO₃ + 2HCI = CaCl₂ + H₂O + CO₂
Na₂CO₃ + H₂SO₄ = Na₂SO₄ + H₂O + CO₂
K₂CO₃ + 2HNO₃ = 2KNO₃ + H₂O + CO₂

(2) Metallic bicarbonate + Hydrochloric acid —> Metallic chloride + water + carbon dioxide Metallic bicarbonate + sulphuric acid —> Metallic sulphate + water + carbon dioxide

Metallic bicarbonate + nitric acid —> Metallic nitrate + water + carbon dioxide

Example :

Ca(HCO₃)₂ + 2HCI = CaCl₂ + H₂O + CO₂
2NaHCO + H₂SO₄= Na₂SO₄ +H₂O + CO₂
KHCO₃ + HNO₃ = KNO₃+ H₂O+CO₂

Question 33. Write the chemical reaction of H₂ SO₄, HCl and HNO₃ with alkali or base.
Answer:

The chemical reaction of H₂SO_4, HCI and HN₃  with alkali or base ” Bases are the metallic.oxides and alkalis are water-soluble hydroxides of metals. When acids like sulphuric acid (H₂SO₄), hydrochloric acid (HCI) and nitric acid (HNO₃) react with these, they form salt and water. This is the main basic reaction for the neutralisation of acid or base.

Base + Sulphuric acid   →   Sulphate salt + water

Alkali + Sulphuric acid  →   Sulphate salt + water

Base + Hydrochloric acid  →   Chloride salt + water

Alkali + Hydrochloric acid →  Chloride salt + water

Base + nitric acid → nitrate salt + water

Alkali + nitric acid → nitrate salt + water

Examples :

Na₂O + H₂SO₄ = Na₂SO₄ + H₂O
2NaOH + H₂SO₄ = Na₂SO₄ + H₂O

[Sometime bisulphate salt can be formed by partial neutralisation of H₂SO₄,
NaOH + H₂SO₄ = NaHSO₄+ H₂O Sodium bisulphate]

CaO + 2HCI = CaCl₂+ H₂O

Ca(OH)₂ + 2HCI = CaCl₂+ 2H₂O

MgO + 2HNO₃ = Mg(NO₃)₂ + H₂O

Mg(OH)₂+ 2HNO₃ = Mg(NO₃)₂ + 2H₂O

Wbbse Madhyamik Physical Science And Environment Class 9 Question 34.

Write the chemical reaction of H₂SO₄, HCI & HNO₃ on metals.
Answer:

The chemical reaction of H₂SO₄, HCI & HNO₃ with metals (like Cu, Zn, Mg): All the mineral acids like H₂SO₄(sulphuric acid), HCI (hydrochloric acid) and HNO, (nitric acid) react differently with the metals like Cu, Zn, Mg. Generally with H₂SO₄, they form sulphate salts, with HCI they form chloride salts and with HNO₃ they form nitrate salts, under different conditions.

(1) With sulphuric acid :

(1) When copper metal is strongly heated with concentrated sulphuric acid, it forms blue-coloured copper sulphate, along with a gas having the pungent odour of burnt sulphur and water.

Cu + 2H₂SO₄= CuSO₄+ SO₂+ 2H₂O

However copper does not react with cold and dilute sulphuric acid properly.

 

(2) When zinc granules are added to dilute sulphuric acid, it form hydrogen gas at room temperature.

Zn + H₂SO₄ = ZnSO₄ +H₂
(dilute)

H₂ + H₂SO₄ = SO₄+ H₂O
(string)

Magnesium reacts with both dilute sulphuric acid, forming magnesium sulphate and hydrogen gas.

Mg + H₂SO₄ = MgSO₄ + H₂

(2) With hydrochloric acid: Dilute hydrochloric acid reacts with metals like zinc (Zn) and magnesium (Mg) forming corresponding metallic chlorides and liberating hydrogen gas, whereas concentrated hydrochloric acid reacts with copper in the presence of oxygen presence in the air, it forms cupric chloride and water. Here hydrogen gas is not produced.
Mg + 2HCI = MgCl₂ + H₂
Zn + 2HCI = ZnCl₂ + H₂
2Cu + 4HCI + O₂ = 2CuCl₂ + 2H₂O
Copper does not react with dilute hydrochloric acid.

(3) With nitric acid: Very dilute nitric acid (1%) reacts with magnesium at low temperatures.
Mg + 2HNO₃ = Mg(NO₃)₂ + H₂

Strong or concentrated nitric acid in reaction with magnesium produces magnesium nitrate and nitric oxide.
3Mg + 8HNO₃ = 3Mg (NO₃)₂ + 2NO + 4H₂O

Zinc in reaction with dilute nitric acid at ordinary temperature forms zinc nitrate, nitric oxide (NO) and water.

Zn + 8HNO₃ = 3Zn(NO₃)₂+ 2NO + 4H₂O

When zinc reacts with concentrated nitric acid at high temperatures, it forms zinc, nitrate, nitrogen dioxide and water.

Zn + 4HNO₃ = Zn(NO₃ )₂ + 2NO₂ + 2H₂O

Copper in reaction with dilute nitric acid at ordinary Temperature forms copper nitrate, nitric oxide (NO) and water.
3Cu + 8HNO₃ = 3Cu(NO₃ )₂ + 2NO₂ + 4H₂O

Copper in reaction with concentrated nitric acid at high Gitte yey Temperature forms copper nitrate, nitrogen dioxide and water. of AgCl
Cu + 4HNO₃= Cu(NO₃)₂ + 2NO₂ + 2H₂O

When nitric acid vapour is passed over red hot copper, copper immediately oxidises into cupric oxide (CuO) along with nitrogen gas and water vapour. This reaction proves the presence of nitrogen in nitric acid.

5Cu + 2HNO₃ = 5CuO + N₂+ H₂O.

Wbbse Madhyamik Physical Science And Environment Class 9

Question 35.  Write about the identification test for HNO₃, H₂SO₄ and HCl.
Answer:

Identification test for HNO₃, H₂SO₄ and HCI (Wet Test) :

(1) Presence of nitric acid (HNO,) by ring test: A little amount of given dilute acid is taken in a test tube. Freshly prepared ferrous sulphate solution is added to it in same volume. Now concentrated sulphuric acid is taken in another test tube. Concentrated sulphuric acid taken in a tube is carefully poured into the liquid mixture of the first tube by the side of the test tube, so as to form a heavy layer at the bottom of the test tube. A brown ring is formed at the junction of the two liquids, which proves the presence of nitric acid in the test tube.

6FeSO₄ + 2HNO₃+ 3H₂SO₄= 3Fe₂(SO₄)₃ + 4H₂O + NO

FeSO₄ + NO + 5H₂O = Fe[(H₂O)NO] SO₄
(brown penta aqua nitroso ferrous sulphate)

(2) Presence of sulphuric acid: In a test tube the given dilute solution of acid is taken. In that test tube, barium chloride solution is added. The test tube is well shaked. A white precipitate is formed at the bottom of the test tube which is insoluble in dilute HCI or dilute HNO₃.

The precipitate is due to the formation of barium sulphate, which proves the presence of H₂SO₄.

BaCl₂+ H₂SO₄ = BaSO₄↓ + 2HCI
(white ppt )

(3) Presence of Hydrochloric acid: In a test tube the given solution of acid is taken. In that test tube, silver nitrate solution is added. The test tube is well-shaken. A dense curdy white precipitate is formed at the bottom of the test tube which is insoluble in nitric acid but dissolves in ammonium hydroxide due to the formation of a complex salt, diamine silver chloride, which proves the presence of hydrochloric acid.

AgNO₃ + HCI = AgCl + HNO₃
( curdy white ppt)

AgCl + NH₄OH = Ag(NH₃)₂ Cl + 2H₂O
(diamine silver chloride)

Question 36. Write about the safe use of acids and alkalis.
Answer:

(1) A solution of hydrogen chloride is highly corrosive and blisters are caused on the skin by its touch.

(2) Pure and concentrated sulphuric acid is highly corrosive and burns the skin, turning it black due to the absorption of moisture from the lower layer of the skin.

(3) Dilute nitric acid reacts with the proteins of the skin and forms a yellow compound called xanthoproteic acid. Hence, the skin becomes yellow. Concentrated nitric acid causes blisters on the skin.

(4) Due to strong alkalinity, sodium hydroxide (NaOH) solution is soapy in touch and it is highly corrosive to the skin and makes blisters on the skin.

Wbbse Madhyamik Physical Science And Environment Class 9 Question 37.

Write about the safety measures for acid & alkali burns.
Answer:

Safety measures for acid and alkali burns :

(1) In case of acid burns, wash the place with plenty of water, then wash with sodium bicarbonate and again wash with water.
(2) In case of alkali burns, wash the place with plenty of water; then with 1% per cent acetic acid and again with water. Finally, dry the skin and apply any medicated skin ointment.

(3) In case of eye accidents we proceed as follows :

(1) Acid in the eye: At once the eye must be washed with water a number of times and then with 1% sodium bicarbonate solution. Finally, add a dilute solution of any medicated eye drop.

(2) Alkali in the eye: At once the eye must be washed with water repeatedly and then it should be washed with 1% boric acid solution. Finally, add a dilute solution of any medicated eye drop or add a drop of liquid paraffin in the eye.

(3) If acid or alkali is swallowed: Immediately take plenty of drinking water to dilute it, followed by taking a dilute solution of milk of magnesia to destroy the effect of acid. Immediately take plenty of drinking water to dilute it, followed by lemon or orange juice to destroy the effect of alkali.

In all these cases a skilled medical attendant is necessary, which is very important in case of safety measures.

Question 38. Write about the importance of aqueous medium in the properties of acid and alkali.
Answer:

Importance of aqueous medium in the properties of acids and alkalis: According to the concept of Arrhenius, acids exhibit their acidic properties only in aqueous solution. In fact, in the absence of water or benzene and other non-aqueous solvents anhydrous HCl cannot turn blue litmus red. The anhydrous vapour of HNO₃ cannot liberate CO₂  from CaCO₃. But such a reaction is a characteristic property of all acids. Aqueous solution of acids imparts hydronium (H₃O⁺) ions.

Hydrogen chloride is a covalent compound. It does not ionise to form an H⁺ ion in a pure state. Since it does not give H+ ions, it is not acidic in a pure state. But in contact with water, HCI forms H₃O⁺ and Cl^– ions. The acid character of HCl is developed due to the formation of H₃O⁺ ions in aqueous medium.

H₂O+HCl → H₃O+Cl^–

Similarly, when a base like Na₂O is dissolved in water it forms an alkali and in an aqueous medium, it generates OH^– and shows alkalinity.

Na₂O + H₂O \(\rightleftharpoons\) 2NaOH \(\rightleftharpoons\) 2Na⁺ + 2OH^–

So we observe that the aqueous medium plays an important role in case of acidity or alkalinity. Now, we carry out the test of dry baking soda (NaHCO₃) and tartaric acid crystals. In a test tube dry NaHCO₃ (or baking soda) and tartaric acid crystals are mixed and the mixture is kept in a test tube we observe that no; bubble of gas is evolved.

But when we add water in this mixture of test tube, i-e., in baking soda and tartaric acid, immediately bubbles of carbon dioxide gas come out. Here water is a liquid that behaves as a solvent, which separates the ions and molecules in the reaction.

At first, on dissolution in water, tartaric acid gives H⁺ ion which immediately reacts with NaHCO₃, forming carbon dioxide gas and thus bubbles of CO₂ comes out. But no bubble is evolved if a mixture of dry NaHCO₃ and tartaric acid crystals is shaken with benzene or kerosene, which is no aqueous medium, solvents like kerosene, benzene can not dissolve ionic compounds. So, we observe, some reactions always occur in aqueous medium and water has an important role in case of acidity and alkalinity of those reactions.

Question 39. Write a short note on pH.
Answer:

The acidity of a solution is expressed in terms of a parameter called pH, where pH stands for potenz of hydrogen (hydrogen power).

The quantity pH is defined as the logarithm of the reciprocal of hydrogen ion concentration (H+) and is equal to the logarithm of the hydrogen ion concentration with negative sign. Mathematically,

pH = − log₁₀ [H⁺]

The scale indicating the hydrogen ion concentration of solutions in terms of pH values is called pH scale. The pH scale is expressed by a series of positive numbers between 0 and

14. The pH scale can be explained as a series of numbers which indicates the acidic or basic nature of a solution. Thus a neutral solution has the pH = 7, an acidic solution has a pH < 7 and an alkaline solution has a pH >7. The greater the hydrogen ion concentration, the smaller is the pH.

A mixture of several indicators which can show different colours at different concentrations on ‘hydrogen ions in a solution is called universal indicator. A paper impregnated with universal indicator which is used to measure the pH of a solution, is called pH paper.

The higher the concentration of H⁺ ions, the lower is the pH of a solution. For example, among two solutions A and B having pH 2 and_5 respectively, the solution A has higher hydrogen ion concentration.

Wbbse Madhyamik Physical Science And Environment Class 9 Question 40.

Write about the acidity of water due to CO₂, SO₂ and NO₂.
Answer:

Acidity of water due to CO₂, SO₂, NO₂: When we final pH less than 5.1 in the rain water, that type of rain is called acid rain. The acidic gaseous oxides, such as SO₂, NO₂, and CO₂ in the atmosphere cause acid rain. Such rainwater is acidic.

These oxides dissolve in rain water to form acids. Sometimes SO₂ and NO₂ undergo aerial oxidation catalysed by carbon particle or ozone gas to form SO₃ & NO₅ which form H₂SO₄ & HNO₃. Carbon dioxide forms carbonic acid. These acids remain dissolved in rain water and so it becomes acidic. Due to the presence of CO₂ in air, and other oxides like SO₂ and NO₂, these gases form acids with water by photochemical reaction in the atmosphere.

CO₂ +H₂O \(\rightleftharpoons\) H₂CO₃ \(\rightleftharpoons\)H⁺ +HCO^–₃

\(\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \frac{\text { carbon particles }}{\text { light }} \mathrm{SO}_3 \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{H}_2 \mathrm{SO}_4\)

2NO₂ + O₃ → O₂+ N₂O₅

N₂O, + H₂O→ 2HNO₃

Acid rain affects plantation and agriculture by washing out the soil nutrients. It may even destory aquatic lives like fish. Exact pH value of acid rain is 5.1.

When acid rain reacts with limestone or slate, the following reaction shows the effect :

CaCO₃ + H₂SO₄ = CaSO₄ + CO₂ + H₂O

CaCO₃ + 2HNO₃ = Ca(NO₃)₂ + CO₂ + H₂O

Due to the presence of CaSO₄ Ca(NO₃)₂, the marble stone of all historical monuments slowly becomes dull and the brightness of marble stone of historical monuments is slowly lost. The top layer of stone is corroded, small holes are created and bright layer of marble stone is totally lost. Such type of destruction of stone, due to air pollution, is called stone cancer.

Question 41. Write three points of the practical application of neutralisation.
Answer:

Practical Application of neutralisation :

(1) Farmers add slaked lime (calcium hydroxide) to reduce the acidity of soil.

(2) Persons suffering from acidity are given antacid tablets, containing magnesium hydroxide mainly, which neutralises excess HCI produced in stomach.

(3) Sting of ants and bees contains formic acid. This can be neutralised by rubbing soaps, which contains dissolved base.

Question 42. Explain the concept of antacid and classify it.
Answer:

Concept of antacid: The different foodstuffs which we eat are digested in the stomach where one of the important chemicals is hydrochloric acid having pH range 2 to 3. Sometimes due to the nature of food stuff and due to the different diseases, the production of this acid is uncontrolled and this excess secretion of acid produces disturbance in the
digestion and acidity is formed, which can produce blisters in the stomach in the long run.

So we can take antacid, which regulates. the acidity and pH level in the digestion. Generally antacids are taken after eating, as antacid does not decrease the production of acid in the stomach but controls the pH of the digestive system.

Antacids are of two types:

(1) systemic and
(2) non-systemic.

(1) Systemic antacids: They function in a regular manner. They are soluble in water and can be easily absorbed. But due to the absorption of this antacid the acid-base equilibirium of our body is disturbed.

Examples of: Such type is-sodium bicarbonate (NaHCO₃), sodium citrate (Na₃C₆H₅O₇).

(2) Non-systemic antacids: They do not function in a regular manner. They are not soluble in water easily and they cannot be absorbed. Hence they do not disturb the acid base equilibrium of our body.

Examples of such type of antacids:

Magnesium compound—magnesium hydroxide

[Mg(OH)₂], magnesium carbonate (MgCO₃), aluminium compound—aluminium hydroxide

[Al(OH)₃] and calcium compound-calcium carbonate (CaCO₃).

The common composition of some antacids available in the market are:
(1) Magnesium trisilicate and aluminium hydroxide mixture.
(2) Aluminium hydroxide, magnesium hydroxide, and magnesium carbonate mixture.
(3) It is the gel of pure aluminium hydroxide, sodium carboxymethyl cellulose and magnesium hydroxide mixture.

WBBSE Solutions for Class 9 Physical Science And Environment

• Chapter 1 Measurement
• Chapter 2 Force And Motion
• Chapter 3 Matter: Structure And Properties
• Chapter 4 Matter: Atomic Structure; Physical & Chemical Properties of Matter
  • 4.1 Atomic Structure
  • 4.2 Concept Of Mole
  • 4.3 Solution
  • 4.4 Acids, Bases & Salts
  • 4.5 Separation Of Components Of Mixtures
  • 4.6 Water
• Chapter 5. Energy In Action: Work, Power & Energy
• Chapter 6 Heat
• Chapter 7 Sound

 

Chapter 2 Forces And Motion Very Short Answer Type:

Question 1. Give an example of motion when there is no change of speed but there is only change in direction.
Answer:

A runner running with uniform speed across a circular track will have to change his direction of motion at every instant of his journey.

Question 2. A body goes round the sun with constant speed in a circular orbit. Is the motion uniform or accelerated?
Answer:

Though the body moves with uniform speed, the direction of its velocity changes continuously. Hence, the motion is accelerated.

Speed And Velocity Difference Class 9

Question Question 3. Give an example where acceleration is produced due to change in the direction of velocity although the magnitude of the velocity remains unchanged.
Answer:

The revolution of earth around the sun is an accelerated motion though its velocity is uniform.

Read and Learn all WBBSE Solutions for Class 9 Physical Science And Environment

Question 4. What is meant by “motion in one direction”?
Answer:

When a body moves in a straight line, its motion is said to be in one direction (one dimension) only.

Question 5. What do you mean by the distance between two locations?
Answer:

Distance is the total path traveled from one location to another. It is a scalar quantity.

Question 6. What is a reference frame?
Answer:

To locate the position of a body with respect to the reference body or observer, a system of coordinates fixed on the observer or reference body is constructed and it is known as reference frame.

Question 7. What is a body?
Answer:

Body: A natural object possessing mass, weight and volume is called a body.

Question 8. What is a particle?
Answer:

Particle: It is a hypothetical minute object having mass and existence. So it is called point mass, it has only translatory motion, i.e., motion along a line. It has no volume, so it has no rotational motion about its own axis or center.

Speed And Velocity Difference Class 9

Question 9. What is a rigid body?
Answer:

Rigid body: Theoretically, a rigid body is one, such that under no circumstances an external applied force, however large may it be, can increase or decrease the distance between two of its constituents parts.

Question 10. What is the role of frame of reference?
Answer:

To determine relative motion or position of a body we compare the state of motion or position of it with respect to a fixed frame of reference. Without a frame of reference we cannot understand or measure the quantity of motion or distance of position of body from another body.

Question 11. What is rest?
Answer:

Rest: A body is said to be at rest if it always remains at a fixed or unchanged position from a frame of reference. : Or more simply, a body is said to be at rest if it does not change its position with time from a fixed neighboring object.

Question 12. What is motion?
Answer:

Motion: A body is in motion if its position from a fixed neighboring obj ect or reference point changes with respect to time.

Question 13. What is vibratory motion?
Answer:

Vibratory motion: Repetition of the to-and-fro motion of a body about a mean position in a horizontal plane or a vertical plane is called vibratory motion. Oscillation of a tuning fork is a vibratory motion.

Question 14. What is periodic motion?
Answer:

Periodic motion: A motion which repeats at regular intervals of time is called periodic motion.

Example: When the earth moves around the sun once in 365 days, itis a periodic motion.

Speed And Velocity Difference Class 9

Question 15. What is average speed?
Answer:

If an object traverses different distances in equal time intervals, its speed is said to be non-uniform. For a body moving with non-uniform speed, its average speed is calculated by dividing the total distance covered with the total time required to cover the distance.

What Is Uniform Velocity Class 9

Question 16. What is uniform velocity?
Answer:

Uniform velocity of an object occurs if it traverses equal distances in equal inter- vals of time in a given direction.

Question 17. What is non-uniform velocity?
Answer:

Non-uniform velocity: A body is said to move with non-uniform velocity if it covers unequal distances in a certain direction in equal time intervals or if it changes its direction of motion at different points on its path, even covering equal distances in equal time intervals.

Question 18. What is acceleration?
Answer:

Acceleration: Increase or decrease of velocity per unit time is the acceleration of a body. It is also defined as the rate of change of velocity with time.

∴\(\frac{\text { unit of displacement }}{\text { unit of time }} \times \frac{1}{\text { unit of time }} \)

Question 19. What is the principle of physical independence of force?
Answer:

The principle of physical independence of a force states that if a body is under simultaneous action of a number of forces, each force independently changes velocity or thus, the momentum of the body in its own direction, any other member of tne applied forces cannot influence it.

Question 20. Define inertia
Answer:

Inertia of a material body is its intrinsic property due to which the body tends to preserve its state of rest or uniform motion in a straight line.

What Is Uniform Velocity Class 9

Question 21. A motor car moving with uniform velocity goes 60 km in 3 hours. Find the velocity of the car.
Answer:

Velocity of the car\(=\frac{\text { displacement }}{\text { time }}\)

⇒ \(=\frac{60 \mathrm{~km}}{3 \mathrm{hrs}}\)

= 20 km/hr from north to south.

Question 22. A man is running at a velocity of 5 m/s. How far will the man goin15s?
Answer:

Distance moved = velocity x time = 5 m/s x 15s = 75 m.

Question 23. A body starts from rest with uniform Osacecly of 2 m/\(s^2\) What will be its velocity after 10 s ?
Answer:

Given, u = 0; t= 10s; f =2 m/s

Then, v=u+ft=0+2×10=20 mis.

Question 24. Same force acting on two bodies of masses 5g and 2g produces an acceleration of 2 cm/\(s^2\) in the former. What is the acceleration of the second body?
Answer:

Mass of the first body = 5 g

Acceleration produced in it = 2 cm/\(s^2\)

there fore Applied force = mass x acceleration = 5 x 2 = 10 dyne.

∴ Same force, i.e., a force of 10 dyne acts also on the second body of mass 2 g.

So, acceleration produced in it \(=\frac{\text { force }}{\text { mass }}\) =\( \frac{10}{2}\) = 5 cm/\(s^2\)

Question 25. A body initially at rest starts moving with an acceleration of 20 cm/s\(s^2\) After what time would its velocity be 2 m/s ?
Answer:

Given, a = 20 cm/\(s^2\) u = 0; v = 2 m/s = 200 cm/s.

Now, v=u+at or, 200=0+2x t

∴ Required time=\( \frac{200}{20}\)  = 10s.

Question 26. Calculate the amount of force required to produce an acceleration of 4 m/\(s^2\) in a body of mass 0.8 kg.
Answer:

Given mass (m) = 0.8 kg, acceleration (a) = 4 m/\(s^2\)

Hence, force, F = ma = 0.8 kg x 4 m/\(s^2\) = 3.2 N.

Question 27. Find the acceleration produced by a force of 20 N acting ona body of mass 5 kg.
Answer:

Given mass (m) = 5 kg, force (F) = 20 N,

acceleration (a) =?

∴F=ma

∴ a \(=\frac{\text { force }}{\text { mass }}\)

=\( \frac{20N}{5Kg}\)

=4m/\(s^2\)

Chapter 2 Forces And Motion 2 Marks Questions And Answers:

Question 1. When is a body said to be at rest?
Answer:

If the position of a body is not changing with time with respect to a given frame of rest, motionless, immobile, stationary or to have constant position.

Question 2. What do you mean by the statement that there is nothing at absolute rest?
Answer:

In reality, there is nothing at absolute rest. For example, earth’s gravitation constantly pulls objects towards its surface, while earth is one of the objects the sun constantly pulls towards itself, causing it to orbit the sun: the sun, in turn, orbits the center of the Milky Way; and so on.

Wbbse Class 9 Physical Science What is scalar

Question 3. When is a body said to be in motion?
Answer:

Motion is a change in the position of an object with respect to time. Motion is typically described in terms of displacement, distance (scalar), velocity, acceleration, time and speed. Attaching a frame of reference to an observer and measuring the change in position of the body relative to that frame, we observe motion of a body.

Question 4. What is absolute motion?
Answer:

When there is no absolute frame of reference, absolute motion cannot be determined. Thus, everything in the universe can be considered to be moving. More generally, notion is a concept that applies to objects, bodies and matter particles, to radiation, radiaion fields and radiation particles, and to space, its curvature and space-time.

Question 5. Define Displacement.
Answer:

Displacement is the distance between two locations measured along the shortest path connecting them, in specified location. It is a vector quantity. The SI unit of distance and displacement is meter (m).

Wbbse Physical Science And Environment Class 9 Solutions

Question 6. Define Speed.
Answer:

Speed is the distance travelled per unit time or the rate of change of distance. Speed = total distance traveled / time taken.

Question 7. Define Velocity.
Answer:

Velocity is the speed in a given direction or the rate of change of displacement. Average velocity = displacement/ time taken.

Question 8. Define Acceleration and Deceleration.
Answer:

Acceleration is the rate of change of velocity.

Acceleration = change of velocity/time taken, where change of velocity = final velocity (v) – initial velocity (u)

Hence, Acceleration = (final velocity – initial velocity) / time taken =(v-u)/t

Negative Acceleration is called Deceleration.

Question 9. State Newton’s first law of motion.
Answer:

Newton’s first law of motion: An object either remains at rest or continues to move at a constant velocity unless acted upon by an external force. Newton’s first law is often referred to as the law of inertia.

Question 10. State Newton’s second law of motion.
Answer:

Newton’s second law of motion: The vector sum of the external forces F on an object is equal to the mass m of that object multiplied by the acceleration vector a of the object. F=mxa

Question 11. State Newton’s third law of motion.
Answer:

Newton’s third law of motion: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

Question 12. State some applications of Newton’s first law of motion.
Answer:

A person in motion stays in motion with the same speed and in the same direction unless acted upon by the unbalanced force of a seat belt. Seat belts are used to provide safety for passengers whose motion is governed by Newton’s laws. The seat belt provides the unbalanced force that brings you from a state of motion to a state of rest.

Question 13. State some applications of Newton’s second law of motion.
Answer:

A car speeding down the road at a constant speed hits the brakes (unbalanced force), so the car slows down (negative acceleration).

Question 14. State some applications of Newton’s third law of motion.
Answer:

When a model rocket is ignited, there is a force from the engine that thrusts downward (all that cool fire & smoke). In response, the model rocket accelerates up.

Question 15. What is Friction?
Answer:

Friction is a force that opposes motion between two surfaces that are in contact with each other. The amount of friction depends on two things: the type of surface and how hard they press against each other. When we consider “how hard” two surfaces are pressing against each other, we are really talking about the force of contact between the surfaces.

There are three types of friction; they are sliding friction, rolling friction etc. fluid friction.

Question 16. Define Momentum.
Answer:

Momentum is defined as the mass multiplied by the velocity: p= m x v.

Since the momentum depends on the velocity, it is also a vector quantity. The units for momentum are [kg.m/s].

Question 17. Define Impulse.
Answer:

The force multiplied by the time has a special name, the impulse. The impulse i is the change in the momentum. Impulse is a vector quantity and is equivalent to a Newton [N], the units of impulse are [N.s].

Impulse is the force multiplied by time the force acts on an object.

I=F×t

Question 18. What is Conservation of Momentum?
Answer:

Conservation is an important principle in physics that allows us to determine That happens in collisions or explosions. Ifa body A of mass \(\mathrm{m}_1\)and velocity\(\mathrm{v}_1\) collides another body B of mass \(\mathrm{m}_2\) and velocity \(\mathrm{v}_2\) moving in the same direction.

If A exerts a force F to the right on B for a time t, then by Newton’s third law, body B will exert an equal and opposite force to the left on A also for atime t but to the left. Thus, the bodies receive equal but opposite impulses “Ft”. The changes of momentum must be equal and opposite.

Therefore, the total momentum change Is zero. In other words, the total momentum of A and B together remains constant.

Question 19. What do you mean by physical independence of a force?
Answer:

Physical independence of a force states that if a body is under simultaneous action of a number of forces, each force independently changes or thus the momentum of the body in its original direction, any other member of the applied force the single existence of it.

Question 20. Argue against the single existence of a force.
Answer:

When one body exerts a force on the other, the force is called action, the second body also offers at the same time an equal force on the first body on the opposite direction and this force is called reaction. Now, since every force of action is associated with an opposite force of reaction, no force can exist singly in nature, that is, in nature forces always appear in pairs.

Question 21. What is thrust?
Answer:

When the weight of a body acts on another body, the force the first body offers is called thrust. This is the action, the second body simultaneously exerts an equal and opposite force on the first, this is the reaction.

When a book rests on the table, its weight on the table is a thrust which is the action, the tabletop also exerts an upward force on the book as the reaction. Push is also a kind of thrust. If the leaf of a door or a window is pushed (action) by hand, the leaf also exerts equal and opposite force on the Hepa as reaction.

Question 22. What are ‘pull’ and ‘tension’?
Answer:

If some weight is attached to the lower end of a wire Supported from a beam, the weight tries to elongate the wire downward, which is the pull; it is the action. The wire also at the same time pulls the weight upward, which |is called the tension and this force of tension is the reaction.

Question 23. What are ‘attraction’ and ‘repulsion’?
Answer:

Two bodies may attract or repel each other when they are in contact or without contact. The mutual forces of ‘attraction’ and ‘reaction’ in opposite directions between the earth and a body on or near it, the force acting between two unlike magnetic poles or two unlike electric charges of opposite kind form action-reaction pairs when they are in contact or away from each other within a limit. The forces of repulsion between a pair of similar magnetic poles or electric charges also form action-reaction pairs.

Question 24. Why is retardation called negative acceleration?
Answer:

Retardation is called negative acceleration as its magnitude is same as that of acceleration but its direction is opposite to that of acceleration.

Question 25. Can a body have velocity in one direction and acceleration in the opposite direction? Give an example.
Answer:

When a stone is thrown vertically upwards, the velocity is in upward direction while it is subjected to an acceleration due to gravity in vertically downward direction.

Question 26. Is it possible for a body to have average speed with average velocity of zero? Explain with an example.
Answer:

Yes, a body can have average speed with an average velocity of zero. When a body moves in a circle, the displacement after one complete rotation is zero. So its average velocity is zero. But it covers a certain distance in a certain interval of time. So its speed is not zero.

Question 27. Two particles are moving at the same constant speed of 50 meters per minute. Which one of them has a constant velocity if Gils moves in a straight line while the other in a circle? Explain your answer.
Answer:

The particle moving in a straight line has constant velocity because both the direction and magnitude of its speed remain unchanged. The direction of motion of the particle moving in a circle constantly changes. So its velocity also changes.

Question 28. What is a rigid body?
Answer:

Rigid body: A rigid body is one such that under no circumstances an externally applied force, however large it may be, can increase or decrease the separation between two of its constituent parts. ;

Wbbse Physical Science And Environment Class 9 Solutions

Question 29. What is a frame of reference and what is its necessity?
Answer:

The frame of reference is any nearby or distant fixed object from which the distance and position of a body are observed as time elapses. Frame of reference is necessary since the state of rest or motion of a body can be observed or measured with respect to it.

Question 30. Explain that the state of rest and state of motion are relative terms.
Answer:

The state of rest and state of motion of a body are relative rather than absolute terms. An object may be in motion with regard to another object but may be in rest with respect to a third object.

For example, suppose you are travelling in a train and you pass a person standing alongside the tracks. The person standing along the tracks will see you and everyone else on the train as being in motion. But the person sitting next to you on the train will be at rest with respect to you.

Question 31. Can a body be called at absolute rest? Explain the statement.
Answer:

Nobody is present in the universe which is said to be in absolute rest. All the bodies in the universe are in the state of motion. When we speak of bodies to be at rest o. in motion, we speak of their apprent state of rest or apparent state of motion.

When we compare the position of a tree with a running train, we say that the tree is in rest. But the tree is situated on earth and the Earth is always in a state of motion.

Question 32. What is relative velocity? Give a practical example.
Answer:

Relative velocity: As in this universe all the points with respect to which motion is studied are also in motion, the motion is termed as relative motion and the corresponding velocity is known as relative velocity. Relative velocity of a body ‘B’ with respect to a body ‘A’ when both are in motion, is the velocity with which ‘B’ appears to move towards or away from ‘A’.

Example: Let us consider a flat train moving on earth on a straight track with velocity\(\mathrm{V}_{\mathrm{TE}}\) (velocity of train with respect to earth). Let us consider again a car moving on a roac parallel to the train in the same direction with a velocity \(\mathrm{V}_{\mathrm{CE}}\) (velocity of car w.r.t earth).

Now the relative velocity of the train w.r.t to the car will be \((\mathrm{V}_{\mathrm{TE}}+ \mathrm{V}_{\mathrm{CE}}\)). If the car moves in the opposite direction, the relative velocity will be \((\mathrm{V}_{\mathrm{TE}}+ \mathrm{V}_{\mathrm{CE}}\)).

Question 33. State the conditions under which a train will appear to be at rest to a person sitting in a nearby train.
Answer:

To a person sitting in a train, another train will appear to be at rest when the following three criteria are fulfilled :

(1) The two trains are moving parallel to each other
(2) The two trains are moving in the same direction
(3) The two trains are moving with the same velocity.

In other words, if the relative velocity of the trains relative to the person is always the same, the train will appear to be at rest.

Question 34. Why do both mass and velocity of a body pertain to its momentum?
Answer:

The momentum of a moving body is its quantity of motion. When two bodies of unequal masses move with same velocity, momentum or quantity of motion of the body of large mass is greater than that of the body of smaller mass. Thus momentum depends on mass of a moving body.

Again, of two bodies of equal mass moving with unequal velocities. momentum of the body moving with larger velocity is more than that of the body moving with smaller velocity. So, momentum also depends on velocity. Hence, momentum of a body depends on both its mass and velocity. Infact, momentum = mass x velocity.

Question 35. What is meant by.a unit force? Why is an absolute unit of force ‘absolute’?
Answer:

A unit force means the quantity of force that when acts on a body of unit mass produce unit acceleration. Absolute units are so called because they are universally valid without admitting any exception, they are complete and independent of restrictions like altitude, position, etc.

Question 36. What can you say regarding the acceleration of a body having a constant kinetic energy?
Answer:

The kinetic energy of a body \(=\frac{1}{2} m v^2\),where ‘m’ is the mass and ‘v’ is its velocity. Now

If Kinetic energy, i.e., \(=\frac{1}{2} m v^2\)  is constant, then the velocity ‘v’ is constant. Again, if velocity Is constant, there is no acceleration.

Wbbse Physical Science And Environment Class 9 Solutions

Question 37. Deduce Newton’s first law of motion from the second law.
Answer:

From the relation, force = mass x acceleration (or retardation), if the applied force is zero, i.e., in absence of any external force, acceleration or retardation is zero, since mass cannot be zero.

Now, that acceleration or retardation is zero means, either the body is at rest or it is in uniform motion; thus, in absence of an external force, a body is either at rest or in uniform motion, this is Newton’s first law of motion.

Question 38. How can a body have uniform speed but variable velocity? Is the converse possible?
Answer:

While a body moves along a circular path or a curved path, its direction of motion at an instant, at any point on its route is along the tangent at that point; obviously, the direction of motion of the body changes from point to point.

So, even if the body covers equal distances in equal intervals of time, i.e., it possesses uniform speed, its velocity is non-uniform due to change of direction of motion.

Thus a body may move with uniform speed but its velocity may be non-uniform. But the converse is not true. A body moving with uniform velocity covers equal distances in equal time, so its speed is uniform.

Question 39. Is it possible for a body to have acceleration without velocity? Explain with an example.
Answer:

Yes, it is possible for a body to have acceleration without velocity. We consider the instance of a continuously swinging body like the swinging bob of a pendulum. At the extreme end of the path described, on either side of the mean position, the bob stops momentarily and at that moment it has no velocity but has acceleration directed towards the mean position.

Question 40. The velocity of a body changes from 14 cm/s to 12 cm/s. How can we conclude that a force is acting on it? Is it helping or opposing the motion?
Answer:

Since the velocity of the body gradually decreases, an external force must be acting on it, because Newton’s first law of motion says that an external force only can change the motion of a body. Since, the motion of the body gradually decreases, the force opposes the motion of the body.

Question 51. Why is it that a freely falling body is in weightless condition? 
Answer:

We can feel our body weight because of a reaction force. Due to gravity, the weight of one’s body acts downward on the ground where he stands, this is the action. The ground exerts an equal upward force on his body, that is the reaction.

Due to this reaction force we can feel the weight of our body. A freely falling body has no other body on which it can exert its weight, so no reaction force acts on it. Hence, the body is in a weightless condition.

Question 52. To catch a flying cricket ball why does a fielder draw back his hands along with the ball?
Answer:

While the fielder goes for the catch, he puts his palms against the motion of the cricket ball, so the ball strikes his palms as an action. Due to reaction from the palms, the ball moves away in the opposite direction and tends to spill before the fielder clutches it.

The fielder draws back his hands at the moment of holding the ball, allowing it to move a bit more in its direction of motion so that its velocity decreases. If the ball strikes the palms of the fielder with less velocity, the reaction force due to which the ball rebounds also lessens. The fielder gets enough time to grasp the ball that rebounds now a bit slowly.

Wbbse Physical Science And Environment Class 9 Solutions

Question 53. Why is it more painful to drop on a concrete surface than on a sandy bed?
Answer:

A person when drops on a concrete surface, which is a continuous body, the thrust exerted by him is the action. The concrete surface exerts equal and opposite reaction force on the person’s feet that causes appreciable pain. But, if the person drops on a sandy bed.

Which is a discontinuous body, major part of the exerted force transmits down inside the sand layers. This force gradually weakens, for, it is utilised to compress the sand layers which have some entrapped air pockets. The reaction of the remaining force on the surface of sand does not produce any painful force on the person’s feet.

Question 54. Is the motion of a body from rest to rest uniform motion? Give reason.
Answer:

The motion of a body from rest to rest is an instance of non-uniform motion. While a body initially at rest begins to move, and after moving through some distance again comes to rest, its motion cannot be uniform. As it starts moving from rest, its velocity (or speed) gradually increases and then its velocity gradually decreases before coming to rest.

Question 55. Two identical lorries move with same velocity—one of them is loaded and the other is not, which of the two may be stopped with less effort ? Why?
Answer:

The quantity of motion of a body depends on both the mass and the velocity of the body. Here, the loaded lorry has larger quantity of motion due to greater mass than the unloaded lorry. So, the unloaded lorry of less quantity of motion can be stopped with less effort.

Question 56. Why is it easier to accelerate a moving vehicle by push than one at rest?
Answer:

When the car or a roller is at rest, to set it into motion, a large force is necessary to overcome the inertia of rest of its large mass and also to overcome its frictional force with the ground. On the other hand, when it is already in motion, since it itself tries to maintain its forward motion due to inertia of motion and also has overcome opposing frictional force, a smaller effort can accelerate it.

57. Why does an electric fan continue its motion for some time even after the switch is off?
Answer:

By switching off, the supply of electricity to the motor of the fan is stopped, so the motor no longer works, but due to inertia of motion the blades of the fan continue to move for some time. Finally, the blades stop due to opposing frictional forces of air, ball bearings, etc.

Question 58. Will a boat move when air from an electric fan situated in it strikes the sails?
Answer: T

he electric fan on the boat and the sails constitute a single system. So, the force to blow of air from the fan is not an external force, that is why the boat will not move.

Wbbse Physical Science And Environment Class 9 Solutions

Question 59. Distinguish between rotation and circular motion.
Answer:

Difference between rotation and circular motion :

(1 ) When a body moves along a circular path about a point as center, then the motion of that body. is called circular motion. But the rotation is the characteristic motion of a rigid body in which each constituent particle on it undergoes such circular motion in which the centers of all the circles lie on a single straight line perpendicular to the plane of each circular path.

(2) Circular motion is the motion of a point along a circular path but rotation is the change of orientation of an extended object without suffering change in position. For example, orbital motion of the Earth and other planets around the sun is circular motion but daily rotation of the Earth about its own axis is a rotational motion.

Question 60. Define displacement. Is it vector or scalar?
Answer:

Displacement: The change of position of a moving body in a particular direction is called the displacement of the body and is measured by the linear distance between its initial and final positions. Thus displacement has both magnitude and direction. So displacement is a vector quantity.

Question 61. Define speed. It is vector or scalar?
Answer:

Speed: The rate of change of position of a body with respect to time is known as its speed. In other words, the distance traversed by a body in unit time is its speed. In the case of speed we have not mentioned about any direction. Thus, speed has only magnitude but no direction. So speed is a scalar quantity.

Wbbse Class 9 Difference Between Speed And Velocity

Question 62. State two differences between speed and velocity.
Answer:

Question 63. Define balanced forces.
Answer:

Definition of balanced forces: When two forces are of equal magnitude but acting in opposite directions on an object simultaneously, the object continues to be in its state of rest or of uniform motion in a straight line. Such forces acting on the body are said to form a system of balanced forces.

Question 64. Define unbalanced forces.
Answer:

Definition of unbalanced forces: When two forces of unequal magnitudes act in opposite directions on an object simultaneously, then the object moves in the direction of the larger force. Such forces acting on the body are said to form a system of unbalanced forces.

Question 65. State Newton’s first law of motion and define force from it.
Answer:

Statement of Newton’s first law of motion: Everybody continues in its state of rest or of uniform motion in a straight line unless it is acted on by some unbalanced force to change that state.

Definition of Force: From Newton’s first law of motion we get the definition of force. Force is that, which acting on a body, changes or tends to change the state of rest or uniform motion of the body in a straight line.

Wbbse Physical Science Class 9

Question 66. When a running bus stops suddenly, the passengers inside it lean forward. Why?
Answer:

When a running bus stops suddenly, a passenger sitting inside it leans forward. Here also as soon as the bus stops, the lower portion of the body of the passenger being in contact with the bust stops immediately. But the the upper: portion wants to maintain its motion due to inertia of motion and hence leans forward.

Again, when a bus suddenly starts, a passenger sitting in it leans backward. As soon as the bus starts, the lower portion of the body begins to move along with it. But upper portion of the body due to inertia of rest tends to continue in that state and thus leans backward.

Question 67. State Newton’s second law of motion and hence define the unit of force in CGS system.
Answer:

Newton’s second law of motion gives the relationship: between the force applied or the body and the acceleration produced in the body. When an unbalanced force acts on a body then the applied force is proportional to the acceleration produced in it.

Wbbse Physical Science Class 9

Question 68. Why do we have to make a great effort to hold a hose pipe to throw a stream of water through it?
Answer:

Water flowing through a hose pipe: We have to make a great effort to hold a hose pipe to throw a stream of water through it. The stream of water coming out through the hose pipe in the forward direction with great speed exerts a large force on the hose pipe in the backward direction. Thus we have to make great effort to hold the hose pipe strongly to keep it at rest.

Question 69. The moving blades of an electric fan do not stop as soon as the switch is off. Why?
Answer:

The moving blades of an electric fan do not stop as soon as the switch is off.

Explanation: By switching off the supply of electricity to the motor of the fan is stopped, so the motor no longer works, but due to inertia of motion the blades of the fan continue to move for some time.

Question 70. Why should a person slightly lean back before alighting from a moving vehicle?
Answer:

Before alighting from a moving vehicle, bus or tram, one should slightly lean back.

Reason: As long as the person remains inside the moving vehicle, his whole body moves with the speed of the vehicle. As soon as he descends on ground his feet suddenly come to stop but the upper part of his body still possesses forward speed due to inertia of motion. If he has no support to grab he stumbles on ground.

But if he leans back a bit beforehand, it makes up the short span he moves through forward when he descends and ultimately he remains erect. This practice saves one from stumbling down when he unwisely alights from a moving vehicle.

Question 71. Deduce Newton’s second law of motion from his first law.
Answer:

Newton’s first law of motion from the second law: From the relation, force = mass x acceleration (or retardation), if the applied force be zero, i.e., in absence of any external force, acceleration or retardation is zero, since mass cannot be zero.

Now, acceleration or retardation is zero means either the body is at rest or it is in uniform motion; thus, in absence of an external force, a body is either at rest or in uniform motion, that is Newton’s first law of motion.

Wbbse Class 9 Physical Science Chapter 2 Question Answer

Question 72. The velocity of a train in a particular direction increases from 30 km/hr to 60 km/ hr in 1 minute. If the train be moving with uniform acceleration, what is its acceleration?
Answer:

Initial velocity of the train = 30 km/hr.

Final velocity of the train = 60 km/hr.

Time required = 1 minute\(=\frac{1}{60}\)  hr.

Change in velocity = 60 — 30 = 30 km/hr.

∴acceleration \(=\frac{\text { change in velocity }}{\text { time }}\)=\(\frac{30 \mathrm{~km} / \mathrm{hr}}{1 / 60 \mathrm{hr}}\)

= 1800 km/hr)\(^2\).

Question 73. A train starting from rest attains a velocity of 72 km\(\mathrm{h}^{-1}\)in 5 minutes. Assuming that the acceleration is uniform, find
(1) the acceleration and
(2) the distance travelled by the train for attaining this velocity.
Answer:

Given,u=0,v=72km\(\mathrm{h}^{-1}\)

⇒ \(=\frac{72 \times 1000}{60 \times 60} \mathrm{~ms}^{-1}\)

⇒ \(=20 \mathrm{~ms}^{-1}\)

t=5 min= 300s.

(1)Fromv=u+at 0r, a\(=\frac{v-u}{t}\)=\(\frac{20-0}{300}\)=\(\frac{1}{15} \mathrm{~ms}^{-2}\)

(2) Again,\(v^2\)=\(u^2+2 a s\) or, s \(=\frac{v^2-u^2}{2 a}\)=\(\frac{20^2-0^2}{2 \times \frac{1}{15}}\)=3000m=3km.

Question 74. A force of 1000 kg-wt is applied on a body. Express this force in dyne and Newton.
Answer:

1000 kg-wt = 1000 x 9.81 newton = 9810 N

Again, 1N \(=10^5\) dyne

therefore 9810 N = 9810 x \(10^5\) dyne = 9.81 \(10^5\) dyne.

Question 75. When a force acts on a body of mass 10 g initially at rest, it acquires a velocity of 15 cm/s in 5 seconds. Find the acceleration and final momentum of the body and the magnitude of the force.
Answer:

Final velocity c = 15 cm/s; mass of the body = 10 g. therefore Final momentum of the body = mass x speed = 10 x 15 = 150 g-cm/s

Again,acceleration a\(=\frac{v-u}{t}\)=\(\frac{15-0}{5}\)=3cm/\(s^2\) and force = mass x acceleration = 10 x 3 = 30 dyne.

Wbbse Class 9 Physical Science Chapter 2 Question Answer

Question 76. The initial velocity of a body of mass 20 g is 10 cm/s and after 5 s its velocity becomes 40 cm/s. Find the magnitude of the force acting on the body.
Answer:

Given, mass of the body m = 20 g; initial velocity u = 10 cm/s

final velocity v = 40 cm/s; time t = 5 s.

Now, v=u+at

or,acceleration a\(=\frac{v-u}{t}\)=\(\frac{40-10}{5}\)=\(\frac{30}{5}\)=6 cm/\(s^2\)

∴ Required force = m.a = 20 x 6 = 120 dyne.

Question 77. A body of mass 147 g at rest is subjected to a force of 15 g-wt. What is the acceleration of the body ? What will be the velocity of the body after 2s?
Answer:

Given, m = 147g; u=0

p= 15 g-wt = 15 x 980 dyne

∴acceleration a\( =\frac{p}{m}\)=\(\frac{15 \times 980}{147}\) =100cm/\(s^2\)

Again, v=u+at=0+ 100 x 2=200cm/s=2 m/s.

Question 78. A bullet of mass 100 g is fired from a gun of mass 20 kg with a velocity of 100. m\(s^{-2}\). Calculate the velocity of recoil of the gun.
Answer:

Given mass of the bullet, \(\left(m_1\right)\)= 100 g = 0.1 kg

velocity of the bullet, v = 100\(s^{-1}\); mass of the gun, \(\left(m_2\right)\) = 20 kg.

Let the recoil velocity of the gun = V.

Initial momentum of the gun-bullet system\(=\left(m_1+m_2\right)\)x 0 =0.

Final momentum of the system\(=m_1 v+m_2 v\) 0.1 x 100 + 20x V=10+ 20V.

Applying the law of conservation of momentum,

10+20 V=0 V=-0.5\(s^{-1}\)

The -ve sign indicates that the gun moves backward.

Wbbse Class 9 Physical Science Chapter 2 Question Answer

Question 79. Velocity-time graph for a body moving with retardation is shown. Calculate the
(1) retardation and (2) displacement of the body.
Answer:

(1) Here, retatdation \(=\frac{\text { Decrease in velocity }}{\text { time taken }}\)

⇒ \(=\frac{(10-0) \mathrm{m} / \mathrm{s}}{(10-0) \mathrm{s}}\)=\(\frac{10}{10} \mathrm{~m} / \mathrm{s}^2\)=2m/\(s^2\)

(2) Displacement = Area of velocity-time graph 3 5

= Area of ΔAOB B

=½×OB×OA=(½×10×10)m=50\(m^2\).

Question 80. Calculate the force required to impart to a car a velocity of 40 m/s in 10 seconds starting from rest. The mass of the car is 1500 kg.
Answer:

Here, mass (m) = 1500 kg,

Initial velocity (u) = 0, Final velocity (v) = 40 m/s, Time taken (t)= 10s.

∴Force required, F=ma

⇒ \(=m\left(\frac{v-u}{t}\right)\)=\(1500\left(\frac{40-0}{10}\right) N\)\(=\left(1500 \times \frac{40}{10}\right) \mathrm{N}\)=6000N.

Wbbse Class 9 Physical Science Chapter 2 Question Answer

Question 81. A cricket ball of mass 200 g moving with a speed of 30 m/sis brought to rest by a player in 0.03 seconds. Find
(1) The change in the momentum of the ball and
(2) The average force applied by the player.
Answer:

Here, mass (m) = 200g = 0.2 kg, Initial velocity (u) = 30 m/s,

Final velocity (v) = 0, Time taken (t) = 0.03 s.

(1) Change in momentum = (mv — mu) = 0.2 (0 — 30) =— 6 kg m/s.
(2) Average force applied, F = Rate of change of momentum

⇒ \(=\frac{\text { change of momentum }}{\text { time }}\)

⇒ \(=\frac{6}{0.03} \frac{\mathrm{kg} \mathrm{m} / \mathrm{s}}{\mathrm{s}}\)=200N.

Question 82. A force of 4 kg-wt acts on a body of mass 9.8 kg. Calculate the acceleration. [Take g = 9.8 m/\(s^2\) ]
Answer:

4 kg-wt = 4 kg. x 9.8 m/\(s^2\) = 39.2 N.

Now, force = mass x acceleration, ∴ 39.2 N = 9.8 kg x acceleration

∴ acceleration \(=\frac{39.2 \mathrm{~N}}{9.8 \mathrm{~kg}}=\)=4m/\(s^2\)

Question 83. Two bodies A and B are of equal mass and are at rest. A is acted upon with 1. kg-wt force and B with 1N. Will the acceleration generated in them be equal? force
Answer:

Using the relation, acceleration \(=\frac{\text { force }}{\text { mass }}\),Acceleration of A

⇒ \(=\frac{1 \mathrm{~kg} \mathrm{wt}}{\mathrm{m}}\)=\(\frac{9.8 \mathrm{~N}}{\mathrm{~m}}\),where ‘m’ is the mass of A or B.

acceleration of B \(\frac{1\mathrm{~N}}{\mathrm{~m}}\). ∴acceleration of A is 9.8 times more than that of

Question 84. A constant force of 30 dyne acts on a body, initially at rest, for 4 seconds and then ceases to act. The body moves 48 cm in the next 8 seconds. What is the mass of the body?
Answer:

On withdrawal of the force, the body moves with a constant velocity with which it covers 48 cm in 8 s.

∴Velocity acquired by the body after 4 seconds \(=\frac{48 \mathrm{~cm}}{8 \mathrm{~s}}\) = 6 cm/s.

Now, acceleration\(=\frac{\text { Change of velocity }}{\text { time }}\)\(=\frac{(6-0) \mathrm{cm} / \mathrm{s}}{4}\)

(∴ initial velocity = 0) = 1.5 cm/\(\mathbf{s}^2\)

From the relation, force = mass x acceleration,

30 dyne = mass x 1.4 cm/\(s^2\)

∴  mass \(=\frac{30 \mathrm{dyn}}{1.5 \mathrm{~cm} / \mathrm{s}^2}\)=20 g.

Wbbse Class 9 Physical Science Chapter 2 Question Answer

Question 85. A cricket ball of mass 100 g moves with a velocity 20 m/s. A batsman stops the ball in 0.05 seconds. Find the force applied.
Answer:

Retardation \(=\frac{(0-20) \mathrm{m} / \mathrm{s}}{0.05 \mathrm{~s}}\)=\(400 \mathrm{~m} / \mathrm{s}^2\)

Mass   of the ball \(=\frac{100}{1000} \mathrm{~kg}\)=o.1 kg.

Hence, force applied on the ball = 0.1 kg x 400 m/\(s^2\) = 40 N.

Question 86. A force acting on a body of mass 10 kg changes the velocity of the body from 5m/s to 10 m/s in 5 seconds. Find the value of the force in dyne and Newton.
Answer:

10 kg = 10,000g, 5 m/s = 500 cm/s, 10 m/s = 1000 cm/s.

Now, acceleration=\(\frac{(1000-500) \mathrm{cm} / \mathrm{s}}{5 \mathrm{~s}}\)=100 cm/\(s^2\)

So, the applied force = 10,000 g x 100 cm/\(s^2\)  \(=10^6\) dyne = 10 N.

Question 87. What is straight linear motion?
Answer:

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

Question 88. What is curvilinear motion?
Answer:

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

Question 89. What is circular motion?
Answer:

Circular motion: If any particle moves around a particular point or axis, then this motion is called circular motion.

Question 90. What is mixed motion?
Answer:

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

Wbbse Physical Science And Environment Class 9 Solutions

Question 91. What is inertial frame of reference?
Answer:

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

Wbbse Physical Science And Environment Class 9 Solutions

Question 92. What is non-inertial frame of reference?
Answer:

Non-inertial frame of reference: A frame of reference which is accelerated is called non-inertial frame of reference. In this frame of reference Newton’s first law of motion is not valid.

Question 93. What is one-dimensional motion?
Answer:

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

Example :
(1) A train moving along a straight track.
(2) A body falling freely under gravity.

Question 94. What is two dimensional motion? 
Answer:

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

Example :

(1) The earth revolving around the sun.
(2) An insect crawling on the floor.

Wbbse Physical Science And Environment Class 9 Solutions

Question 95. What is three dimensional motion?
Answer:

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

Example :

(1) A flying bird.
(2) A flying aeroplane.
(3) A flying kite.

Question 96. State the units of displacement in CGS and SI systems.
Answer:

Units of displacement :

(1) CGS: cm
(2) SI: m.

Question 97. State two characteristics of displacement.
Answer:

The characteristics of displacement are:

(1) Displacement has the unit of length.
(2) The displacement may be positive, negative, or zero.

Question 98. State two differences between displacement and distance.
Answer:

Difference between displacement and distance :

Question 99. State two characteristics of speed.
Answer:

Characteristics of speed :

(1) Speed does not give any idea about the direction of motion.
(2) Speed can be positive, but never negative or zero.

Wbbse Physical Science And Environment Class 9 Solutions

Question 100. State the units of speed in CGS and SI systems.
Answer:

Units of speed :
(1) CGS : cm/\(s^{-1}\)
(2) SI: m\(s^{-1}\).

Question 101. What is average speed?
Answer:
Average speed: When a body is moving with variable speed, the constant speed with which the body covers the same distance in a given time as it does while moving with variable speed during the given time, is called average speed.

Average speed \(=v_{a v} \)= \(\frac{\text { total dis tance covered }}{\text { total time taken }} \)

If a body moves distances \(\mathrm{S}_1, \mathrm{~S}_2, \mathrm{~S}_3\)……..etc. in times \(\mathrm{t}_1, \mathrm{~t}_2, \mathrm{~t}_3\) ………etc. respectively, then

Average speed \(=v_{a v} \)\(=\frac{S_1+S_2+S_3+\ldots \ldots}{t_1+t_2+t_3+\ldots \ldots}\)

Question 102. State two characteristics of velocity.
Answer:

Characteristics of velocity :

(1) The velocity can be positive, negative or zero.
(2) Velocity has both magnitude and direction, and hence, it is a vector quantity.

Wbbse Physical Science And Environment Class 9 Solutions

Question 103. State the units of velocity in CGS and SI systems.
Answer:

Units of velocity (same as speed) :

(1) CGS : cm/\(s^{-1}\)
(2) SI: m/\(s^{-1}\).

Question 104. What is average velocity ?
Answer:

Average Velocity: When a body is moving with variable velocity, the average velocity is the value of that constant velocity at which the same body could have under- gone the same displacement in the same interval of time as the body has actually gone.

Average velocity \(=\frac{\text { Total displacement }}{\text { Total time }}\)

Question 105. State two characteristics of acceleration.
Answer:

Characteristics of acceleration :

(1) When the velocity is increasing, the acceleration is called positive and
(2) when it is decreasing, it is called negative or retardation or deceleration.

Question 106. State the units of acceleration in CGS and SI systems
Answer:

Units of acceleration :

(1) CGS : cm/\(s^{-2}\)
(2) SI: m\(s^{-2}\)

Question 107. What is average acceleration?
Answer:

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

Wbbse Physical Science And Environment Class 9 Solutions

Question 108. State two characteristics of retardation.
Answer:

Characteristics of retardation :

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

Question 109. What is uniform retardation?
Answer:

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

Question 110. What is non-uniform retardation ?
Answer:

Non-uniform retardation: The retardation of a moving body is said to be variable if it does not have equal changes in velocity in equal intervals of time.

Chapter 2 Forces And Motion 3 Marks Questions And Answers:

Question 1. State the different types of motion
Answer:

The different types of motion are :

(1)Simple harmonic motion.

(2)Anharmonic motion

(3)Periodic motion

(4)Rectilinear motion — motion that follows a straight linear path, and whose displacement is exactly the same as its trajectory

(5) Reciprocal motion (e.g. vibration)

(6) Random motion (e.g. vibration)

(7) Brownian motion (i.e., the random movement of particles)

(8)Circular motion (e.g. the orbits of planets)

(9)Rotary motion – a motion about a fixed point. (e.g. Ferris wheel).

(10)Curvilinear motion – It is defined as the motion along a curved path that may be planar or in three dimensions.

(11)Rotational motion – (as of the wheel of a bicycle)

(12)Oscillation

(13)Combination (or simultaneous) motions – Combination of two or more above-listed motions

(14)Projectile motion – uniform horizontal motion + vertical accelerated motion.

Question 2. What are Balanced and Unbalanced Forces?
Answer:

A force is defined as a push or a pull and more than one force can and usually does act on an object at the same time. Every force has a certain strength or magnitude. Forces also have direction.

The net force on an object is the sum (in both magnitude and direction) of all the forces acting on it. If the net force is zero, the forces are balanced. Balanced forces produce a change in motion of an object.

When the net force is greater than zero, the forces acting on the object are not balanced. Unbalanced forces cause an object at rest to move and an object in motion to change speed and/or direction.

Question 3. How is measurment of force obtained from the second law of motion?
Answer:

Measurement of force: From the second law of motion, measurement of force is possible. Say, a body of mass ‘m’ moves uniformly in a certain direction with the velocity ‘u’ and after ‘t’ seconds, due to the application of an external force F, in the same direction, its final velocity becomes ‘v’.

∴ the initial momentum of the body = mu

final momentum of the body = mv

So, change of momentum after ‘t’ seconds = mv – mu

∴ Change of momentum after 1 second \(=\frac{m v-m u}{t}\)\(=\frac{m(v-u)}{t}\)

i.e., rate of change of momentum \(=\frac{ v- u}{t}\)

=ma(∴\(=\frac{ v- u}{t}\)=acceleration,a)

Now, according to Newton’s second law of motion,

∴F α ma or F = k ma where k is a constant.

Now, the force that acting on a body of unit mass produces unit acceleration in the body is called unit force.

So, when m = 1 and a= 1 then F = 1; hence, the relation F = kmf changes to 1 = k.1.1. So, k = 1. Therefore F = ma, i.e., Force = mass x acceleration.

Question 4. Establish the relation between absolute and gravitational units of force.
Answer:

Relation between gravitational and absolute units of force :

The CGS absolute unit of force 1dyne = 1 g x 1 cm/s?, the CGS gravitational unit of force

1 gram weight = 1 gram x g (acceleration due to gravity)= 1 gram x 981 m/\(s^{2}\)

= (1 gram x 1 cm/\(s^{2}\)) x 981

= 1 dyne x 981

= 981 dynes

= g dynes.

Hence, 1 gram weight = g dynes. This is the relation between gravitational unit (1 gram weight) and absolute unit of force (1 dyne).

Also, the MKS absolute unit of force = 1 kg x 1 m/\(s^{2}\)? and the MKS gravitational unit of

force = 1 kg x g (g = acceleration due to gravity) = 1 kg x 9.81 m/\(s^{2}\)?

= (1kg x 1m/\(s^{2}\)?) x 9.81 = 1N x 9.81 = 9.81 N = g Newton.

Hence, in general, it can be written as

Gravitational unit of force = Absolute unit of force x g.

Wbbse Physical Science And Environment Class 9 Solutions

Question 5. How does a horse cart move forward, though the force with which the horse pulls the cart is equal and opposite to the force with which the cart pulls the horse?
Answer:

It may be noticed that just at the moment of start, the horse bends its head and body to exert a force on ground in an inclined way with its legs [The same thing does a rickshaw-puller; athletes also do so at the start of a race.]. The ground also exerts an equal and opposite reaction force on the horse in an inclined direction.

This reaction may be resolved in the vertical and the horizontal directions. The vertical component reduces the weight of the horse and the horizontal component helps the horse and the cart to move forward. If the horizontal component can exceed the frictional force between the ground and wheels of the cart, then the whole system, horse and cart, moves forward.

Question 6. Taking force (F) length (L) and time (T) as fundamental quantities, express (1) density, (2) pressure, and (3) momentum.
Answer:

Acceleration  \(=\frac{L}{T^2}\)  mass \(=\frac{\text { force }}{\text { acceleration }}\)=\(\frac{F}{L}\)=\(\frac{F T^2}{L}\)

∴(1)\(\frac{\text { force }}{\text { acceleration } \times L^3}\)=\(\frac{F T^2}{L^4}\)

(2)\(\frac{F}{L^2}\)

(3)\(\frac{\mathrm{FT}^2}{\mathrm{~L}} \times \frac{\mathrm{L}}{\mathrm{T}}\)=\(\mathrm{FT} \text {. }\)

Question 7. Define translation. Classify it.
Answer:

If the different particles constituting a body travel in a definite direction so that the line joining any two points of it remains parallel to itself, then the motion of the body is called translation. Translational motion, again, is divided into two kinds:

(1) Rectilinear motion : When the translational motion of a body takes place along a straight line, it is called Rectilinear motion. (e.g. motion of a freely falling body under gravity).
(2) Curvilinear motion: When the translational motion of a body takes place along a curved path, it is called curvilinear motion. (e.g. motion of a seat of vertical merry-go-round).

Question 8. Wrtie a short note on circular motion.
Answer:

Circular motion: If a body moves along a circular path around a fixed point or about a fixed axis, then the motion of the body is called circular motion. Examples include the Supvllingsametion motion of the earth about the sun, the motion of a satellite around the earth, the motion of an electron around the nucleus of an atom, a stone tied at one end of a string and revolving along a circular path, etc.

In translation a body moves along a straight line while in circular motion a body moves along a circular path etc.. is around a fixed point. The direction of motion does not Cha Rotational change in case of translation but the direction of motion axis changes continuously in circular motion.

Moreover, in ’ translation, if a body moves with constant speed, it has i. no acceleration; but in circular motion, a body, even if travels with uniform speed possesses acceleration.

Again, motion may occur in one dimension, in two dimensions, and also in three dimensions. When a body moves along a straight line it is known as one-dimensional motion. Motion of a freely falling body under gravity, oscillatory motion of a body motion.

If a body moves along a curved path in a plane, then its motion is known as two-dimensional motion. Motion of an ant crawling on a globe, revolution of a planet around the sun, etc. are examples of two-dimensional motion.

Again, when a body moves in space, then its motion is known as three-dimensional motion. This kind of motion is the most general type of motion. Its examples are a bee flying in air or a kite flying on a windy day, etc.
Question 9. Write a short note on rotational motion
Answer:

Rotational motion: To describe rotational motion, the concepts adopted in circular motion are not sufficient; because in rotational motion change of orientation rather than position of an extended object is to be considered. An extended body is made up of a continuous distribution of particles of fixed masses with an external force.

Let us consider three constituent particles P, Q, and R of an extended rigid body and for a pure rotation of this extended body; each and every particle undergoes circular motion and centers of such motion for those particles lie on the same straight line or axis, known as axis of rotation.

Axis of rotation for an extended rigid body may pass through the body or it may lie outside the body.

Wbbse Physical Science And Environment Class 9 Solutions

Now, for an extended rigid body three kinds of rotation are possible, viz. orbital motion, spin motion and precessional motion. When a body is rotated around a distant axis of rotation then the motion of the body is called orbital motion.

Example: Motion of different planets around the sun.

When an extended rigid body rotates around its axis of symmetry then the motion of the body is called spin motion. Motion of a spinning top, daily rotation of the Earth about its axis passing through its two poles are examples of spin motion. Again, the spin axis of a rotating rigid body wobbles around another fixed axis to describe a cone.

Then that kind of motion is called precessional motion. At the last stage of a spinning top just before it comes to stop, the motion is such that the axis of the top rotates like a cone around a definite vertical axis. It is an example of precessional motion.

Question 10. Distinguish between translation and rotation
Answer:

We get the following differences between translation and rotation :

(1) In pure translation, the body travels along a straight line but in pure rotation, a body describes circular motion in a plane.

(2) In a pure translation, the direction of motion of the body remains fixed whereas in pure rotation axis of rotation remains fixed.

(3) In a pure translation, the displacement of all the particles constituting a rigid body occurs in the same direction and in pure rotation, the orientation of a rigid body changes but the relative distance between the constituent particles remains unaltered.

Wbbse Class 9 Physical Science Distance And Displacement

Question 11. Distinguish between distance and displacement.
Answer:

Difference between distance and displacement :

12. State and explain the meaning of resolution of forces.
Answer:

Resolution of forces: Resolution of force means splitting up of a force into two components. When a force is resloved into mutually perpendicular components then those components are called rectangular components of the force.Resolution of a force is utilised in towing of a boat by a simple rope. If a boat is pulled

Along the direction \(\overrightarrow{O C}\)with the help of a rope then with OC as diagonal and a parallelogram

OACB is drawn, then according to parallelogram law of vectors, the vector \(\overrightarrow{O C}\) can be resolved into two mutually perpendicular components \(\overrightarrow{O A}\)and \(\overrightarrow{O B}\).

The component \(\overrightarrow{O B}\) being normal to the river tries the boat to move across the river but its effect is cancelled by the rudder and as a result, the boat moves in the direction of the component\(\overrightarrow{O A}\)

Question 13. Write a short note on balanced and effective forces.
Answer:

Effective force and balanced force : If the resultant of a number of forces acting on a body is not zero, then the forces are called unbalanced forces. The net force acting on the body under the influence of a number of unbalanced forces is called the effective force. If the rider of a bicycle stops pedalling it then it will slow down. This is because. when the

Rider stops pedaling, the force of friction acting between the tires of the bicycle and the road opposes the motion of the bicycle.

An effective force has to be applied on the bicycle to keep it in motion. When the bicycle is moving with some non-uniform velocity, then the net force or unbalanced force acting on the bicycle becomes equal to the force applied by the rider while pedaling the bicycle minus force of friction between the tires and the road.

Unbalanced forces:

(1) may increase the speed,
(2) may decrease the speed or
(3) may

change the direction of motion of a body. This means that unbalanced forces or an effective force can produce acceleration in a body. Here, effective force is the cause and acceleration (change of motion) is the effect. On the other hand, if the resultant of all the forces acting on a body is zero, the forces are called balanced force.

Question 14. State and explain the concept of inertial mass.
Answer:

Concept of inertial mass: Newton’s second law of motion leads us to the concept of inertia and the concept of inertial mass.
According to Newton’s second law of motion, we know that for a given body, the acceleration is directly proportional to the net force applied.

Thus, force,\(\overrightarrow{F}\)α acceleration \(\overrightarrow{a}\)Or, \(\overrightarrow{F}\) = m.\(\overrightarrow{a}\), where m is a constant of proportionality,.called mass of the body.

Let an equal amount of force \(\overrightarrow{F}\)be applied separately on two bodies of different masses \(\mathrm{m}_1\) and\(\mathrm{m}_2\) (\(\mathrm{m}_1\), > \(\mathrm{m}_2\))

and if the accelerations produced in the two bodies are \(\overrightarrow{a_1}\) and a\(\overrightarrow{a_2}\) respectively, then according to the relation \(\overrightarrow{F}\)=m. \(\overrightarrow{a}\) we get,

⇒ \(\overrightarrow{F}\)=\(\mathrm{m}_1\) \(\overrightarrow{a_1}\)=\(\mathrm{m}_2\)\(\overrightarrow{a_2}\)

∴\(\frac{\overrightarrow{a_1}}{\overrightarrow{a_2}}=\frac{m_2}{m_1}\)

∴(\(\mathrm{m}_1\), > \(\mathrm{m}_2\))

∴\(\overrightarrow{a_1}\) › a\(\overrightarrow{a_2}\)

Hence, acceleration produced in the heavier body is less than that in the lighter body. Hence, mass of a body may be considered as the measure of the opposition for the produced in it. It leads to give us the concept of ‘intertial mass’.

Question 15. Is Newton’s second law of motion consistent with his first law?
Answer:

Consistency of Newton’s second law of motion with the first law of motion: According to Newton’s second law of motion, \(\overrightarrow{F}\) = (\(\overrightarrow{m}_a\) where F is the force applied on a body, m is the mass of the body and a is the acceleration produced in the body.

Now, in absence of any external force Bing on the body, i.e., when \(\overrightarrow{F}\) =0, \(\overrightarrow{m}_a\) =0.

∴ m can not be zero ,so \(\overrightarrow{a}\) =0.

Now, acceleration (1) of the body zero indicates that if the body is initially at rest, it will remain at rest and if it is in motion, then it will continue its motion with the same velocity which is nothing but Newton’s first law of motion. Therefore, Newton’s second law of motion is consistent with the first law.

Question 16. Write a short note on contact forces.
Answer:

Contact forces: The forces which exist because of actual contact between the sources and the body are called contact forces.

Different types of contact forces are

(1) Normal reaction: When a body appliés a force on another, the latter body also exerts a reactionary force which acts normal to its surface and this is known as normal reaction. For example when a book is placed on a table, the table exerts normal reaction on the book against its weight in the upward direction.

(2) Elastic force: It is the force exerted by.a deformed elastic body by virtue of its elasticity. Force exerted by a compressed string is an example of elastic force.

(3) Tension: The force that comes into play in a suspension string is called tension.

(4) Frictional force: Force that acts tangentially to the two surfaces due to the interlocking of undulations present on the surfaces is called frictional force. Force of friction exerted between the tyre of a car and the road surface is an example of such force.

(5) Air resistance: When the motion of a body takes place through air then the retarding force which comes into play is called air resistance. This force is exerted on an aeroplane when it flies through air at high speeds.

Wbbse Physical Science And Environment Class 9 Solutions

Question 17. Write a short note on non-contact forces
Answer:

Non-contact forces: The forces which exist without any physical contact between the bodies but are due to some intrinsic characteristics of the bodies, are called non-contact forces.

Some non-contact forces are:

(1) Gravitational force: The force of attraction acting between. any two bodies in the universe due to gravitational attraction between them.

Example: force of attraction acting on a body nearer to the Earth by virtue of which the body falls downwards towards the center of the earth.

(2) Electrical force: The force of attraction or repulsion acting between two unlike or like charges is called electrical force. Force of attraction acting between the nucleus of an atom and its orbiting electrons is an example of such kind of force.

(3) Magnetic force: The force that comes into play between two bodies due to their magnetic behavior is called magnetic force. Force of attraction acting between two unlike

Question 18. What are the four basic forces in nature?
Answer:

Basic forces in nature : There are four basic forces in nature. These are:

(1) Gravitational force: It is always attractive, mass-dependent but change independent force which obeys inverse-square law; gravitational is a long-range, the weakest and conservative central force. :

(2) The weak force: It is a short range but 10% times stronger than the gravitational force. In the process of β-decay, this force plays an important role and it explains the interaction between leptons, mesons and baryons.

(3) The strong force : It is a very short Tange, attractive, charge-independent but spin-dependent force; gravitational force is about 100 times stronger than electromotive force and is operative in between different nucleons inside a nucleus.

(4) Electromagnetic forces: It is a long-range, mass-independent, attractive or repulsive force that obeys the inverse-square law; electromagnetic force is a central force and is about 10*5 times stronger than the weak force. It is also a conservative force.

Question 19. Write a short note on the principle of conservation of linear momentum.
Answer:

The principle of conservation of linear momentum states that — “If no net external unbalanced force acts on the system of isolated bodies then the total linear momentum of the system in a given direction remains constant.”

According to Newton’s second law of motion, the time rate of change of linear momentum is equal to the applied force.

Thus,\(\vec{F}\)=\(\frac{\Delta \vec{p}}{\Delta t}\)

If the system is isolated, then \(\vec{F}\) =0

∴\(\frac{\Delta \vec{p}}{\Delta t}\)=0

∴\(\vec{P}\)=Constant.

Collisions of two bodies moving in same straight line: Let us consider an isolated system consisting of two bodies of masses m, and m, respectively. Let the two bodies be moving along a straight line in the same direction with velocities \(\overrightarrow{u_1}\) and \(\overrightarrow{u_2}\) ( \(\overrightarrow{u_1}\)>\(\overrightarrow{u_2}\))respectively.

The two bodies will collide after some time and let their final velocities after collision be \(\overrightarrow{v_1}\) and \(\overrightarrow{v_2}\) respectively.

Prior to collision, total linear momentum of the system\(=m_1 \overrightarrow{u_1}+m_2 \overrightarrow{u_2}\)

After collision takes place, total linear momentum of the system \(=m_1 \overrightarrow{v_1}+m_2 \overrightarrow{v_2}\)

Now, at the time of collision the first body of mass \(m_2\), exerts of force \(\vec{F}\)on the second body of mass \(m_2\) , and the second body also exerts an equal but opposite reactionary force ( \(\vec{-F}\) ) on the first body.

Now, according to Newton’s second law of motion, force acting on the body of mass \(m_2\) is

⇒ \(\vec{F}\)= Rate of change of momentum of the body of mass \(m_2\)

⇒  \(=m_2\left(\frac{\overrightarrow{v_2}-\overrightarrow{u_2}}{t}\right)\)……………..(1) (t=time during which collision occurs)

Again, force acting on the body of mass \(m_2\) is pars}

⇒  \(\vec{-F}\)= Rate of change of mementum of the body of mass\(\overrightarrow{m_1}\)

⇒  \(=m_1\left(\frac{\overrightarrow{v_1}-\overrightarrow{u_1}}{t}\right)\)………………(2)

From equation (1)and (2), we get

⇒  \(=m_2\left(\frac{\overrightarrow{v_2}-\overrightarrow{u_2}}{t}\right)\)–\(=m_1\left(\frac{\overrightarrow{v_1}-\overrightarrow{u_1}}{t}\right)\)

or,\(m_2 \vec{v}_2-m_2 \vec{u}_2=-m_1 \vec{v}_1+m_1 \vec{u}_1\)

or,\(m_2 \vec{v}_2+m_1 \vec{v}_1=m_2 \vec{u}_2+m_1 \vec{u}_1\)

∴\(m_1 \vec{u}_1+m_2 \vec{u}_2=-m_1 \vec{v}_1+m_2 \vec{v}_2\)

i.e., total momentum of the system before collision is equal to total momentum of the system after collision this is the law of conservation of linear momentum.

Wbbse Physical Science And Environment Class 9 Solutions

Question 20. Derive the equations of motion by graphical method.
Answer:

Equations of motion by graphical method : An object moving along a straight line with initial velocity u and uniform acceleration a can be related to its velocity v and distance s covered in a certain time interval t by the following equations :

(1) v=u+at

(2) s = ut + ½a\(t^2\)

(3) \(v^2\)=\(u^2+2 a s\)

These equations are called equations of motion and can be derived from the velocity-time graph.

(1) Derivation of v = u + at: The velocity-time graph of an object moving under uniform acceleration a and initial velocity u is shown in Fig. It is seen from the graph, the initial velocity is u at A which increases to v at B in time t. If BC and BE be the perpendiculars drawn from B on time and velocity axes respectively, then u = OA, v = BC and t = OC.

So, BD = BC − CD = change in velocity in time interval t.

If AD be drawn parallel to OC, then from the graph,

BC = BD + DC=BD+OA

or, v=BD+u

or, BD=v-u

Now, acceleration, a \(=\frac{\text { change in velocity }}{\text { time }}\)=\(\frac{\mathrm{BD}}{\mathrm{AD}}\)=\(\frac{\mathrm{BD}}{\mathrm{OC}}\)

Putting OC=t, \(a \)= \(\frac{B D}{t}\)  or, BD = at=v-u

∴ V=u+at

(2) Derivation of s = ut + ½a\(t^2[\): Let the object moves a distance s in time t under uniform acceleration a. The distance traveled by the object is given by area OABC enclosed under the velocity-time graph A

So, distance traveled,

s = area OABC

= area of rectangle OADC + area of the AABD

= OA x OC + ½ (AD x BD) =u t+ ½(t xat)

∴ s=ut+ ½at.

(3) Derivation of \(v^2\)=\(u^2+2 a s\): Let the object travels a distance s in time t under uniform acceleration a. It is seen from the.
s = area of trapezium OABC \(=\frac{(O A+B C) \times O C}{2}\)=\(\frac{(u+v) t}{2}\)

Further, from the relation v = u + at, we have t \(=\frac{v-u}{a} .\)

so,s\(=\frac{(v+u)(v-u)}{2 a}\)

V \(=u^2+2 a s\).

Question 21. Show the displacement-time graphs of an object moving with uniform velocity and uniform acceleration.
Answer:

Displacement-Time graphs: The displacement of an object moving with uniform velocity along a straight path. The hypothetical time-displacement data of a particle are

Show in the following table.

Time (t) (second)             0  10  20  30  40
Displacement (s) (meter) 0  9    19  28  38

The displacement-time graph is shown in fig. 1. It is a straight-line graph passing through origin. From the graph it is seen that in time interval AC \(=\left(t_2-t_1\right)\) the corresponding displacement BC is equal to (\(\left(s_2-s_1\right)\)) =s.

Thus, v \(=\frac{s_2-s_1}{t_2-t_1}\)=\(\frac{s}{t}\)

Again when the object is moving with uniform acceleration along a straight path, let the corresponding data are shown in the following table.

Time (t) in s             0  1  2  3  4    5   6
Displacement in m  0  1  4  9  16  22  36

The displacement-time graph is shown in the

Question 22. Write a short note on the parallelogram law of forces.
Answer:

The parallelogram law of forces : If simultaneously two forces act on a particle, the forces being represented in magnitude and direction by two adjacent sides of a resulting force, represented in magnitude and direction by the diagonal of the parallelogram passing through the same point.

If P and Q be two forces acting at an acute angle θ with each other, and be represented in magnitude and direction by R.

The two adjacent sides of a parallelogram, Q then their resultant force R is given in magnitude and direction by the diagonal of the parallelogram.

And if α be angle, the resultant R makes with the force P, then its value can be determined by measuring this angle.

In our daily life we find applications of this law. Two applications are given as below :

(1) A boat pulled by two tugs in two different directions, then the boat does not move in either of these two directions, but moves along direction which is the diagonal of the parallelogram formed by two tugs as the sides of the parallelogram.

(2) If a man walks across the floor of a compartment of a train moving along a straight railway line, then the resultant velocity of the man can be found by the diagonal of the rectangle so formed.

Wbbse Physical Science And Environment Class 9 Solutions

Question 23. Give some examples of action and reaction forces.
Answer:

The following examples explain the action and reaction forces:

(1) When we sit on a chair, their is a downward gravitational force on us (i.e., our weight) which acts on the chair. This.is action. The chair also applies an equal reaction force on us in the opposite (i.e., upward) direction.

Because of these two equal and opposite forces acting on us we keep steady and do not fall to the ground. It is to be noted that equilibrium is restored not between the action and reaction forces but between our weight and the reaction of the chair, both acting on our bodies.

(2) While flying, a bird applies a force on the air by its wings. The air also exerts an equal and opposite reaction force to this action force of the wings which helps the bird to fly floating on the air. For this reason birds cannot fly in vacuum.

(3) Consider two magnets separated by a distance. When opposite poles of the magnets face each other, they attract each other and similar poles are facing each other they repel. Of the two magnets, weaker or lighter one is drawn or repelied from the other.

Here both the magnets attract or repel each other with some force but the lighter one moves towards or away from the heavier one. If we hold one of them hard on the table, the other would move towards or away, when the force of attraction or repulsion between the magnets are sufficient to pull each other.

(4) The Earth orbits around the sun. The only force that acts on the Earth by the sun is the gravitational force exerted by the sun on the earth, which provides the centripetal force required by the earth for its revolution.

Now where is the corresponding reaction force? The answer is the gravitational attractive force of the Earth on the sun. It is known that these gravitational forces are equal in magnitude and opposite in sign. Thus it is seen that action and reaction forces are equal and opposite to each other.

(5) Let us take two spring balances whose hooks be connected to each other. Holding them firmly, they are drawn in the opposite directions. The readings of the spring balances are found to be always the same.

Now, if one of the balances is fixed to a firm support like wall and the other is pulled with a certain force, it will be found that each of the spring balances shows same reading. Thus it proves conclusively that action and reaction forces are equal and opposite.

Question 24. What are the ways in which a force may be exerted?
Answer:

There are four different ways in which a force may be exerted. These are :

(1) Pull: It is usually applied along some length of a substance.

Example: the force is applied along a string, a chain, etc. and the string is said to be in a state of tension.

The pull substance like a rod. A rope tied to the ceiling through a hook. When pull the rope from the lower end, a force is exerted on the hook. The hook, in turn, exerts a reaction through the rope.

A pull is thus exerted along the length of a substance. Thus the effect of the pair of action and reaction forces exerted through a material medium is to bring the two interacting bodies together orto prevent them from separating.

(2) Push: When a man presses a wall with his hands, he applies a force on the wall, which in turn, applies a reaction on the man. As a result, the man and the wall tend to separate from each other.

So, by push it is meant a pair of action and reaction between two bodies which tries to keep the bodies apart from each other. A push distributed over an area is often called thrust and can be applied through rigid bodies only and not by flexible bodies. :

(3) Attraction and repulsion: When a pair of actions between two bodies separated by a distance tries to bring them together, attraction between them is said to exist and when it tends to keep apart, repulsion between them occurs.

Gravitational, magnetic, electrical forces are examples of this type of force. These forces exist always. It does not matter whether there is the presence of a material medium between them or not, though the magnitude of the forces may depend upon them.

(4) Friction: When a body tends to slide over another, the second body exerts a reaction on the first acting opposite to the direction of motion and trying to prevent the motion is called friction.

Wbbse Physical Science And Environment Class 9 Solutions

Question 25. “Rest and motion are relative”. Explain.
Answer:

Rest and motion are relative: A body seems to be at rest with respect to one neighboring fixed object but the same body may appear to be in motion with respect to some other fixed neighbouring object. So, rest and motion are relative terms.

Examples :

(1) In a moving train, a passenger finds his co-passengers at rest, but to an observer standing outside the train the same passengers appear to be in motion.

(2) A person standing on a road finds a car moving away from him. So, the car is in motion with respect to him. Now, if another car near it moves with the same speed and direction, thento a person in the second car, the first car seems to be at rest. In fact, every object in the universe is in motion with respect to one or the other object.

Example: A passenger in a moving train finds trees, buildings, etc. in motion, but. those things appear to be at rest to a person standing somewhere outside the train. ~ Thus, absolute rest and absolute motion are meaningless terms.

Question 26. Write a short note on displacement.
Answer:

Displacement: When an object changes its position as time passes on, the distance measured in the direction from its initial to final position along a straight line is known as its displacement.

Displacement is a vector quantity, it is directed from the initial position of a body to its final position. The measurement of displacement of a body solely depends upon its initial and final positions, not on the paths followed by it in reaching the final position from its initial position.

As in the adjacent figure, an object moves from its initial position A to its final position B. It may do this in two ways.

It may either follow the path ACDEB or the path AOB, but in each instant the displacement is the straight line AB that joins the initial position C, D, E in the latter case and at O in the former case. A

lso the distances of ACDEB, and AOB are of different magnitudes but displacement AB is same in both some cases. One thing is common to both. It is that both distance and displacement denote some length of path covered by a moving body.

But measurement of displacement is always directed from the initial position of a body to its final position. If the initial position of a moving body coincides with the final position occupied by it after moving through some length of path, the displacement of the body is zero.

Thus a body while moving along a circular or a curved path, if returns to its starting position, its displacement and the distance covered are different in magnitude but if it travels along a straight line, both the distance and displacement are equal in magnitude.

Wbbse Physical Science And Environment Class 9 Solutions

Question 27. Write a short note on inertia of rest.
Answer:

Inertia of rest: It is that property of a body by virtue of which the body tends to remain at rest and itself does not change its state.

Definition: Inertia of rest is the inherent property of a material body by virtue of which it continues in its state of rest until and unless no external unbalanced force compels it to move.

Example: A book kept on a table remains at rest as long as it is not disturbed by any externally applied force.

For disturbing from its state of rest, it can be raised or displaced.to other positions only by the application of a force by way of pulling up or pushing through Of course, the body at rest will not respond to an external force of any magnitude, the force to which the body responds, depends on the mass of the body; greater the mass of the body, greater external force is required.

A smaller force can disturb the rest condition of a body of a smaller mass. Another definition of the inertia of rest may also be derived : it is the ability of a body to stistain its initial state of rest. This property of sustaining ability of a body depends on its mass.

For example, a blackboard duster lying at rest on a table can be easily set into motion by pushing it gently with a finger, but such a small force cannot move a table of larger mass. From this we can infer that the ability of a body to sustain its initial state of rest or motion increases with increase of its mass.

But it is true that the force exerted on the table with finger brings forth a tendency of its motion, for, if the magnitude of the force is increased, the table starts moving. Thus, the ability of a body to sustain its intial state of rest or motion is limited.

Examples of inertia of rest :

(1) To remove dusts from a blanket or a cloth it is struck by a stick or jerked vigorously. When struck by a stick or jerked suddenly, fibres of the article are just made to move away, but the dust particles, due to their inertia of rest, remain at their initial positions and they fall off in absence of any support.

(2) A person sitting or standing in a bus at rest leans backward as the bus starts suddenly. While the bus is at rest, the person’s body in contact with it also remains at rest. As soon as the bus starts, the lower part of the person’s body in contact of the bus moves forward but his upper part tends to remain in the former position due to inertia of rest, the result is nothing but leaning back.

Question 28. Write a short note on the inertia of motion.
Answer:

Inertia of motion: Definition: The tendency of a moving body to maintain its uniform motion in a straight line in absence of any effective external force is called inertia of motion.

Another meaning is that a body moving uniformly tries to defy the effect of an external. force acting on it. The ‘effect’ of the force here refers to the disturbing actions over the uniformity of motion of the body.

But this ability is limited upto certain magnitudes of the said force. A body of large mass can maintain this property upto a large value of this force while a smaller mass yields to a much smaller force.

Example :

A person runs or walks uniformly along a road. His uniform motion may be changed simply by pushing or pulling him with hand, but a similar force cannot affect the uniform motion of a running locomotive engine that possesses a huge mass. But, a vigorous pull or push executed with the help of another engine can definitely change its uniform motion of it.

An external force changes the uniformity of motion of a body in either of these ways — if the force is directed in the direction of initial motion of the body, velocity of it increases.If it is in opposite direction, velocity decreases and if it is tangential to the body, the direction of motion changes.

Also, everywhere the inertia of a body and an external force acting on it are in opposition to each other. Thus it is also true that inertia of rest or inertia of motion of a body is a measure of its ability of sustaining its state of rest or uniform motion.

Wbbse Physical Science And Environment Class 9 Solutions

Question 29. “Mass is a measure of inertia.”  Explain.
Answer:

Mass is a measure of Inertia: This means, both types of inertia (of rest and of motion) are large for a body of large mass and they are small for a body of small mass. That is, greater the mass, greater is the ability of a body to sustain its initial state of rest or motion even under the effects of an external force and smaller the mass, smaller is its ability to do so.

Thus, a small force of push offered with hand can easily displace a book at rest. But to displace a large chunk of rock, a force of enormous magnitude is necessary. Here, the book of small mass has a large inertia of rest, soa very large force can move it.

Similarly, a small force can change the motion of a cyclist but a very large force is necessary to do so in case of a moving truck, for the inertia of motion of the truck of large mass is greater than that of the cyclist of smaller mass.

Question 30. State and explain Newton’s third law of motion.
Answer:

Third law of motion: To every action there is an equal and opposite reaction. Discussions on the third law of motion: If someone pushes with his hand a fixed massive body, like a wall or a tabletop, he feels pain on his fist; if he offers blow gently or violently he will feel the effect accordingly.

This means whenever a body applies a force on another, the second body also returns a force of the same magnitude on the former.

For application of force, always two different bodies or two different parts of a single body are necessary. When one body exerts a force on the other, the force is called the action, the second body also offers at the same time an equal force on the first body in the opposite direction and this force is called reaction.

Due to this fact Newton’s third law of motion Says, action and reaction are equal in magnitude but opposite in direction.

Since every force of action is associated with an opposite force of reaction, no force can exist singly in nature, that is, in nature forces always appear in pairs.

Any force acting at some point in an inclined direction may be considered to consist of two parts or two components in two suitable directions such that the result of the combined effect of the components on the point is exactly same as that of the initial force. So the components are equivalent to the single force.

Question 31. What are the characteristics of displacement’?
Answer:

Displacement has the following characteristics :

(1) Displacement has the unit of length.

(2) The displacement may be positive, negative or zero.

(3) The magnitude of the displacement of a moving particle between two points gives the shortest distance between them.

(4) From the displacement of a moving particle between two points we do not get any idea about the path followed by the moving particle.

(5) The displacement of a moving particle between two points has a unique value.

(6) The actual distance traveled by the moving particle may be greater than or equal to the magnitude of the displacement. ;

(7) Altering the origin of the coordinate axis has no effect on displacement.

(8) As a particle cannot be at two different positions at the same time, so displacement is a single-valued function of time.

(9) The displacement changes if its direction be changed.

Wbbse Class 9 Difference Between Speed And Velocity

Question 32. Distinguish between speed and velocity.
Answer:

Difference between Speed and Velocity :

Question 33. Write a short note on velocity-time graphs.
Answer:

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

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

In, the velocity-time graph for a particle moving in a straight line with uniform acceleration and starting with initial velocity u = 0 is shown. All these conditions being same, motion starting with initial velocity u is shown in .

In both cases, velocity-time curves are straight lines. In each case the constant slope of the straight line gives the uniform accleration.

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

Question 34. How is a graph plotted?
Answer:

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

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

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

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

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

Question 35. Write a short note on impulse.
Answer:

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

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

The effect of the force is measured by its impulse.

Example of impulsive force: A player lowers his hands while catching a cricket ball. In doing so he increases the time of catch to reduce the momentum of the ball to zero. We know,

Impulse = force x time = change in linear momentum.

So, to produce a particular change in linear momentum, if one takes longer time, then the force required would be less. Thus in catching a cricket ball, the player increases the time of catch by lowering his hands and thereby decreases the force and smaller reaction occurs on the hands.

Question 36. How does the value of ‘g’ vary from place to place?
Answer:

The value of ‘g’ varies from place to place :

(1)Both as we go up from earth’s surface or go down inside below the surface of the earth, The value of acceleration due to gravity decreases.

(2) At the center of the earth the sete of acceleration due to gravity is found to be zero.

(3) Moreover, acceleration due to gravity changes due to earth’s spin about its axis also. Standard value of acceleration due to gravity is taken at 45° latitude sea level.

(4) Again, on the surface of earth it is maximum at the poles. Because, due to orange shape of the earth, the distance of the poles is shorter than that of the equator from the center of the earth.

Question 37. An object of mass 20 kg is accelerated uniformly from a velocity of 36 km\(h^{-1}\)“ to 54 km\(h^{-1}\)” in 20 s. Calculate
(1) the initial momentum of the body
(2) final momentum of the body and
(3) the force acting on the body.
Answer:

m = 20 kg; initial velocity, u = 36 km\(h^{-1}\) = 36 x 18 =10m\(s^{-1}\)

final velocity, v = 54 km\(h^{-1}\)\(=54 \times \frac{5}{18}\) =15 m\(s^{-1}\); time, t= 25s

(1) Initial momentum of the body, \(\mathrm{p}_1\) = mu = 20 x 10 = 200 kg m\(s^{-1}\)
(2) Final momentum of the body, \(\mathrm{p}_2\) = mv = 20 x 15 = 300 kgm\(s^{-1}\)

(3) Force \(=\frac{\text { change in momentum }}{\text { time taken }}\)=\(\frac{p_2-p_1}{t}\)\(=\frac{300-200}{25}\)=4N.

Question 38. A revolver of mass 500 g fires a bullet of mass 10 g with a speed of 100 ms. Find
(1) momentum of the bullet
(2) the initial momentum of the revolver and bullet as a system and
(3) recoil velocity of the revolver.
Answer:

Given, mass of the revolver, \(\mathrm{m}_1\) = 500 g = 0.5 kg.

mass of the bullet, \(\mathrm{m}_2\) = 10g = 0.01 kg

initial velocity of (revolver + bullet) system, u = 0

final velocity of bullet, v = 100 m\(s^{-1}\)

Let the recoil velocity of the revolver be V m\(s^{-1}\)

(1) momentum of the bullet = \(\mathrm{m}_2\)v = 0.01 x 100= 1 kgm\(s^{-1}\)
(2) initial momentum of (revolver + bullet) = (\(\mathrm{m}_1+ \mathrm{m}_2\))u= 0
(3) momentum of the revolver after firing = \(\mathrm{m}_1\)v = 0.5 x V kgm\(s^{-1}\)
momentum of (revolver + bullet) after firing = initial momentum of (revolver + bullet)

or,0.5v+1=0  or, V= -2m\(s^{-1}\)

∴ recoil velocity of the revolver = 2 m\(s^{-1}\).

39. Velocity-time graph for a body is shown. Find its (1) A acceleration, (2) velocity and (3) distance covered in 20 
Answer:

(1) Acceleration \(=\frac{\text { Change in velocity }}{\text { Time }}\)=\(\frac{0}{t}\)=0

(2) Velocity of the body = 15 m/s (from the graph).

(3) Distance covered in 20 seconds

= Area under v — t graph.
= OA x AB = (20 x 15) m = 300 m.

Question 40. Velocity-time graph for the motion of a particle is shown:
(1) Which part of the graph indicates accelerated motion? Calculate the acceleration
(2) Which part of the graph shows retarded motion?
(3) Calculate the distance traversed by the body in the first 20 seconds of its journey.

Answer:

(1) In the graph OA part indicates accelerated motion as velocity goes on increasing from O to A.

Acceleration (a) \(=\frac{v-u}{t}\)=\(\frac{150-0}{20}\)=7.5m/\(s^{2}\).

(2) The part BC of the graph indicates retarded motion as velocity goes on decreasing from B to C.

Retardation(-a)\(=\frac{v-u}{t}\)=\(\frac{0-150}{10}\)=-15m/\(s^{2}\).

(3) Distance traveled by the body in the first 4 s of its journey Area under OA part of the graph = Area of OAD

=½×OD×AD=½×4×4=8cm.

Question 41. The driver of a train A travelling at a speed of 54 km/h applies brakes and retards the train uniformly.
The train stops in 5 seconds. Another train B is travelling on the parallel track with a speed of 36 km/h. His driver applies the brakes and the train retards uniformly. The train B stops in 10 seconds. Draw speed-time graphs for both the trains on the same graph and calculate the F 2 distance travelled by each train after the brakes were Ot>34567 8910 applied.

Answer:

In case of train A:

Initial velocity (u) = 54 km/h\(=\left(54 \times \frac{5}{18}\right)\)m/s = 15 m/s

Final velocity (v) = 0, time (t) =5s

Velocity-time graph for a is shown by straight line EF

∴ Distance travelled by A = Area of ΔEOF

=½×OF×OE=(½×5×15)m=3.75m.

In case of train B :

Initial velocity (u) = 36 km/h =\(=\left(36 \times \frac{5}{18}\right)\)m/s =10m/s

Final velocity (v) = 0

Time taken (t) = 10s

Velocity-time graph for (B) is shown by the straight line CD.

Distance traveled by B = Area of ΔCOD

= ½×OD×OC=(½×10×10)m=50m.

Question 42. For a moving object distance-time graph is shown beside. From the graph,
(1) Find the speed of the object during first 4 s of its motion.
(2) State how long it was stationary.
(3) State whether this situation is real or not.

Answer:

(1) Speed of the object during first 4s

= slope of OP part of the graph Time (s}

⇒ \(=\frac{P S}{O S}\)=\(\frac{75 m}{4 s}\)=18.75m/s.

(2) From 4 seconds to 14 seconds, distance does not change with time and hence it remains stationary for (14−4) = 10 seconds.

(3) This situation is impractical because the distance traversed can not be decreased with time. Moreover, time can not go backward as is shown by QR part of the graph. Hence it is not a real situation.

Question 43. Velocity-time graph for a moving body is shown.
(1) What is the initial velocity?
(2) What is the velocity of the body at the point C?
(3) Find the acceleration of the body between A and B.
(4) Find the acceleration acting on the body between B and C.
Answer:

(1) Initial velocity of the body = 20 m/s.
(2) Velocity of the body at the point C = 40 m/s.
(3) Acceleration between A and B

⇒ \(=\frac{v-u}{t}\)\(=\frac{(40-20) m / s}{3 s}\)=\(\frac{20}{3} \mathrm{~m} / \mathrm{s}^2\)

=6.67m/\({s}^2\)(approx).

(4) Acceleration between B and C\(=\frac{(v-u)}{t}\) =\(\frac{0}{t}\)

(∴ No change in velocity occurs between B and C) = 0

44. A force of 5 N gives a mass m, an acceleration of 8 m/s? and a mass m, an acceleration of 24 m/s. What acceleration would it give if both the masses are tied together?
Answer:

In the first case: force (F) = 5 N, mass (m) =\(\mathrm{m}_1\)  acceleration (a) = 8 m/\({s}^2\)

∴F=ma ∴ 5=\(\mathrm{m}_1\) 8 or,m=\(\frac{5}{8}\) kg

∴ The mass \(\mathrm{m}_1\)  is \(\frac{5}{8}\) kg.

In the second case: Force (F) = 5 N, mass (m) =\(\mathrm{m}_2\), acceleration (a) = 24 m\({s}^2\)

∴F=ma  ∴5=\(\mathrm{m}_2\)  x 24  ∴\({m}^2\)  =\(\frac{5}{24}\) kg

∴The mass\(\mathrm{m}_2\)  is\(\frac{5}{24}\)kg.

In the third case: mass (m) = \(\mathrm{m}_1\) +\(\mathrm{m}_2\)=\(\left(\frac{5}{8}+\frac{5}{24}\right) \mathrm{kg}\)=\(\frac{20}{24} \mathrm{~kg}\)=\(\frac{5}{6} \mathrm{~kg}\)

Force (F) =5N

Acceleration (a) =?

∴F=ma      ∴5\(=\frac{5}{6}\)a      ∴ a=5×\(\frac{6}{5}\)=6m/\({s}^2\)

∴Hence, the required acceleration is 6 m/\({s}^2\).

Question 45. From the given velocity-time graph, answer the following :
(1) What type of motion does OA represent?
(2) What type of motion does AB represent?
(3) Find the acceleration of the body. ‘ Y
(4) Find the distance traveled by the body from O to C.
Answer:

(1) The graph is a straight line having upward slope, it represents uniform acceleration.

(2) Since it is a straight line parallel to time axis, it represents uniform speed.

(3) We know acceleration is the slope of the time-velocity graph. So, \(\frac{A D}{O D}\)=\(\frac{10 m}{5 s}\) = 2 m/s.

(4) Distance is the area enclosed between velocity time graph and time axis = area of

trapezium OABC = ½ (AB + OC) x BE = ½{(25−5) + 30) x 10 m/s = 250 m.

Question 46. From the given velocity-time graph of a body :
(1) Calculate the distance traveled by it in 4 s.
(2) Calculate the distance traveled by it in 8 s.
Answer:

(1) We know that the area between ~ velocity time graph and time axis gives the distance traveled by the body. The distance traveled in 4 sec is the area of
triangle OAC = ½ x OC x AC= ½ x4s x 10 m/s = 20m.

The distance traveled in 4 s is 20 m. Time per second

(2) Distance traveled in 8 s is given by the area of the trapezium OABD.

Area of trapezium OABD = ½ (AB + OD) x BD

Now, AB = CD = OD-OC=8-4=4s

BD = 10 m/s. OD = 8s.

Area of trapezium OABD = ½(4 + 8) x 10 =60m.

Thus, distance traveled in 8 s = 60 m.

Question 47. A particle moves in a straight line with a retardation of 10 m/s\(s^2\) If its initial velocity is 50 m/s, when would it stop?
Answer:

(1) The deceleration is 10 m/ \(s^2\)means that in each second velocity of the particle diminishes by 10 m. So, at this rate, the time in which its initial velocity diminishes.
To zero, then it stops. Hence, the required time \(=\frac{50 \mathrm{~m} / \mathrm{s}}{10 \mathrm{~m} / \mathrm{s}^2}\)=5s.

(2) By using the formula, v = u + at, where symbols are of usual significance, when the particle stops, its final velocity, v = 0, and ‘a’ should be (-ve) for deceleration.

∴V=u-at    or, 0=50−10.t   or, 10t=50      ∴ t=5.

So, the required time is 5 seconds. 

Question 48. Equal forces act on two masses ‘m’ and ‘2m’. If the acceleration acquired by ‘m’ be ‘f’, what is the acceleration acquired by 2 m?
Answer:

From the relation, force = mass x acceleration, the force acting on the mass‘m’ is mf.

Now the same force mf acts on the mass 2 m, and since, acceleration \(=\frac{\text { force }}{\text { mass }}\),
acceleration of the second body \(=\frac{m f}{2 m} \)= \(\frac{f}{2}\).

Otherwise: since, acceleration α\(\frac{1}{\text { mass }}\) , if the mass of one body is double the other, for the same force, the acceleration of the former body is half that of the latter and it is \(\frac{f}{2}\).

WBBSE Solutions for Class 9 Physical Science And Environment

• Chapter 1 Measurement
• Chapter 2 Force And Motion
• Chapter 3 Matter: Structure And Properties
• Chapter 4 Matter: Atomic Structure; Physical & Chemical Properties of Matter
  • 4.1 Atomic Structure
  • 4.2 Concept Of Mole
  • 4.3 Solution
  • 4.4 Acids, Bases & Salts
  • 4.5 Separation Of Components Of Mixtures
  • 4.6 Water
• Chapter 5. Energy In Action: Work, Power & Energy
• Chapter 6 Heat
• Chapter 7 Sound

Chapter 6 Heat Very Short Answer Type:

Question 1. Which instrument is used to measure the temperature of a body?
Answer: Thermometer.

Question 2. Which instrument is used to measure heat gained or lost by a body?
Answer: Calorimeter.

Wbbse class 9 physical science chapter 6

Question 3. How many Fahrenheit degrees is equal to (1) Celsius degree?
Answer: \(\frac{9^{\circ}}{5} \mathrm{~F}\)

Question 4. What is the value of 100°C in the Fahrenheit scale?
Answer: In Fahrenheit scale the temperature is 212°F.

Read and Learn all WBBSE Solutions for Class 9 Physical Science And Environment

Question 5. Is heat a vector quantity?
Answer: No, heat is a scalar quantity.

Question 6. Is there any change of temperature during the change of state?
Answer: There is no change in temperature during the change of state.

Question 7. What is the value of the specific heat capacity of pure water?
Answer: The specific heat capacity of pure water is 4200 J/kg K.

Question 8. Is the specific heat of solids greater than that of liquids?
Answer: Generally specific heat of solids has a lower value than that of liquids.

Question 9. Express 45°C in Fahrenheit scale.
Answer: 45°C=111°F.

Question 10. What is a thermometer?
Answer: The instrument with the help of which the temperature of a body can be measured accurately is called a thermometer.

Question 11. What will be the direction of heat flow when two bodies of different temperatures are kept in contact?
Answer: In thermal contact, heat will flow from the body with a higher temperature to the body at a lower temperature.

Question 12. What are the upper and lower fixed points in the Fahrenheit scale?
Answer: In the Fahrenheit scale the upper fixed point is 212°F and the lower fixed point is 32°F.

Wbbse class 9 physical science chapter

Question 13. How many divisions are there in the fundamental interval of a Celsius scale?
Answer: The fundamental interval of a Celsius scale is divided into 100 equal divisions.

Question 14. How many divisions are there in the fundamental interval of a Fahrenheit scale?
Answer: The fundamental interval of a Fahrenheit scale is divided into 180 equal parts.

Question 15. Of 0°C and 0°F – which one is less?
Answer: Of 0°C and 0°F, 0°F is less.

Question 16. In which scale is a clinical thermometer graduated?
Answer: A clinical thermometer is usually graduated in Fahrenheit scale. But, nowadays, a clinical thermometer is also being graduated in a Celsius scale.

Question 17. At what temperature are the readings in Celsius and Fahrenheit scales equal?
Answer: At -40 temperature the readings in Celsius and Fahrenheit scales are equal.

Question 18. What is the normal body temperature of a human being in a Celsius scale?
Answer: The normal body temperature of a human being in a Celsius scale = 36°9 °C.

Question 19. What is the normal body temperature of a human being?
Answer: The normal body temperature of a human being is 98°4°F.

Question 20. What is the relation between Celsius and Kelvin scales of temperature?
Answer: lf any temperature in the Kelvin scale be T K and in the Celsius scale it is t°C, then T K = (273 + t)°C.

Wbbse class 9 physical science chapter 6

Question 21. What is the relation between calories and joules?
Answer: (1)cal = 42 J.

Question 22. Which liquid has the highest specific heat?
Answer: The specific heat of water is the highest.

Question 23. What is the specific heat of pure water?
Answer: Specific heat of pure water in CGS system is 1cal \(g^{-1}\)°\( C^{-1}\) and in SI system it is 4200 J.K \(g^{-1}\)\(K^{-1}\)

Question 24. The water equivalent of a body is 20 g. What will be its thermal capacity?
Answer: As the water equivalent of the body is 20. g, its thermal capacity will be 20 cal °\( C^{-1}\).

Question 25. What is the SI unit of heat?
Answer: Joule (j) is the SI unit of heat.

Question 26. What is the SI unit of thermal capacity?
Answer: The SI unit of thermal capacity is Joule/Kelvin or Joule/°C.

Question 27. What is a clinical thermometer? 
Answer: The thermometer used for measuring the temperature of the human body is called a clinical thermometer.

Question 28. What is the thermometric liquid in a clinical thermometer?
Answer: Mercury.

Question 29. Name the liquid which has the lowest specific heat. State one use of this property
Answer: Mercury. That is why, it is used as a thermometric liquid.

Question 30. What are the two special features of a clinical thermometer?
Answer:

(1) A very short temperature range from 35°C to 42°C.
(2) A constriction in the glass tube just above the mercury bulb.

Wbbse class 9 physical science chapter 6

Question 31. What is the value of 273K temperature in celsius scale?
Answer: Zero degrees centigrade (0°C).

Question 32. What is the unit of specific heat in the C.G.S. system?
Answer: Calorie per gram per degree Celsius (cal \(g^{-1}\)°\( C^{-1}\)).

Question 33. State the S.I. unit of specific heat (capacity).
Answer: The S.1. unit of specific heat capacity is J K \(g^{-1}\)\(K^{-1}\).

Question 34. What is the value of specific heat of the water in S.I. system?
Answer: The sp. heat of water in S.1. system is 4200 JK \(g^{-1}\)\(K^{-1}\).

Question 35. What is the value of specific heat of water in M.K.S. system?
Answer: In M.K.S. system the value of specific heat of water is 4200 J/kg°\(C^{-1}\).

Question 36. Equal masses of mercury and water are given equal amounts of heat. In which case will the temperature rise more and why?
Answer: Mercury, as it has lower specific heat than water.

Question 37. Mention two uses of high specific heat of water.
Answer: For heating purpose (hot water bag) and cooling purpose (car radiator).

Question 38. Mention the conditions under which the principle of Calorie a is valid.
Answer: No loss or gain of heat by the calorimeter.

Question 39. What is the unit of heat in CGS system?
Answer: Calorie is the unit of heat in the CGS system.

Question 40. What is the main principle of calorimetry?
Answer: Heat lost by the hot body = Heat gained by cold body.

Question 41. What will be the temperatures of the bodies at the state of thermal equilibrium?
Answer: The temperatures of the bodies at thermal equilibrium will be the same.

Question 42. What are the different kinds of heat?
Answer: Sensible heat, Latent heat and Radiant heat.

Wbbse class 9 physical science chapter 6

Question 43. What is steam point?
Answer: Steam point is the boiling point of water under normal pressure.

Question 44. What is (1)calorie?
Answer: 1 calorie is defined as the heat required to raise the temperature of 1g of water through one degree celsius.

Question 45. What is fundamental interval?
Answer: It is the range of temperature between the upper and lower fixed points in the scale of temperature.

Question 46. What are the factors for determining the quantities of heat in a body?
Answer:

The quantity of heat depends upon the following factors :

(1) Temperature
(2) Mass
(3) Material of the body.

Question 47. What are the lower fixed point and upper fixed point of the thermometer in Celsius scale?
Answer: Lower fixed point is 0°C and upper fixed point is 100°C.

Question 48. What is the fundamental interval of Celsius scale?
Answer: The fundamental interval is divided into 100 equal parts in celsius scale.

Question 49. What is the melting point of ice in Kelvin scale?
Answer: Melting point of ice in Kelvin scale is 273 K.

Question 50. Is specific heat of solid greater than that of liquid?
Answer: Generally specific heat of solid has. lower value than that of liquid.

Question 51. Give an advantage of use of mercury in thermometer.
Answer: Equal volume of mercury increases or decreases for each degree increase or decrease of temperature in a regular manner.

Wbbse class 9 physical science chapter 6

Question 52.  What is radiant heat?
Answer: Heat which reaches us from a source by the process of radiation is called radiant heat.

Question 53. What are the different scales of thermometer?
Answer: Gelsius, Fahrenheit and Kelvin or Absolute scale of temperature.

Question 54. What is specific heat capacity?
Answer: Specific heat capacity is defined as the quantity of heat required to raise the temperature of unit mass of any substance through one degree.

Question 55. What do you mean by the statement that “S.H.C. of aluminium is 880 J k\(g^{-1}\)\(k^{-1}\)
Answer: The statement means 1kg of aluminum requires 880 J of heat for 1° K rise of temperature.

Question 56.The specific heat of copper is 0.09.’ — What does the statement mean?
Answer: ‘Specific heat of copper is 0.09’ means that 0.09 calorie of heat is required to raise the temperature of 1g of copper through 1°C.

Question 57. What is the temperature of 30°C in Kelvin scale?
Answer: The temperature of 30°C is 303 K in Kelvin scale.

Question 58. What is the boiling point of water in Kelvin scale?
Answer: The boiling point of water in Kelvin scale is 373 K.

Question 59. By which process heat reaches the earth from the sun?
Answer: Radiation.

Question 60. What is relation between work done and heat produced?
Answer: W =J x H.

W = work done
H = Heat produced
J = Mechanical equivalent of heat.

Question 61. Why do we feel warmer with two thin blankets than one blanket twice as thick?
Answer:

Air is a bad conductor of heat. When two blankets are used a layer of air is trapped in between. It does not allow the body temperature to go out. Hence, we feel warmer with two thin blankets than one blanket twice as thick.

Question 62. Why is it more convenient to cook when the cooking pot has a bottom made of copper?
Answer: Copper is an excellent conductor of heat. When the bottom of the cooking pot is made of copper, the heat is distributed very fast and equally. Hence, it is convenient to cook.

Question 63. Two coins, one of copper and the other of silver, are heated from room temperature to 100°C. Do they absorb same amount of heat? Explain your answer.
Answer: No. Since the specific heat of copper is more than that of silver, the copper coin will absorb more amount of heat.

Question 64. What is the melting point of ice in Kelvin scale?
Answer: Melting point of ice in Kelvin scale is 273K.

Question 65. The partial pressure of water vapor in the atmosphere is 10 mm of Hg at 20°C. Find the relative humidity. (Vapour pressure of water at 20°C = 17.5 mm of Hg)
Answer:

The vapour pressure of water at 20°C is 17.5 mm of Hg.

Hence, relative humidity \(=\frac{10}{17.5} \times 100\)= 57%.

Question 66. The rise in temperature of a body is 60° in celsius scale. What is the corresponding rise in temperature in Fahrenheit scale?
Answer:

We know, 1°C = \(=\frac{9}{5}^{\circ} \mathrm{F}\)

Question 67. How much heat would be produced by converting 42 Joule or work completely into heat?
Answer:

We know, W = JH

or, H\(=\frac{W}{J}\)=\(\frac{42 \text { Joule }}{4.2 \text { Joule } / \mathrm{Cal}}\)= 72 Joule/Cal = 10 Cal.

Question 68. Mass of a body is 150 g and its specific heat is 0.09. How much heat is required to raise its temperature by 10°C?
Answer:

Required heat = mass of the body x specific heat x rise in temperature = 150 x 0.09 x 10 = 135 cal.

Question 69. How much heat is required to raise the temperature of 50 g of lead from 20°C to 100°C? Specific heat of lead is 0.03. 
Answer:

Heat gained by lead \(=m s\left(t_2-t_1\right)\)

= 50 x 0.03 x (100 −20) = 120 cal.

Question 70. What quantity of heat will be taken by 50 g of water at 20°C to reach the boiling point at normal pressure?
Answer:

Boiling point of water at normal pressure = 100°C.

So, heat taken by water = 50 x 1 x (100 −20) = 4000 cal.

Question 71. How many calories of heat are needed to melt 60 g of ice at 0°C?
Answer:

Given, the mass of ice,  m = 60 g;  temperature of ice = 0°C.

H = mL = 60 x 80 = 4800 cal.

Chapter 6 Heat 2 Marks Questions And Answers:

Wbbse class 9 physical science chapter 6

Question 1. Two bodies have the same heat capacity. If they are combined to form a single composite body, show that the equivalent specific heat of this composite body is independent of the masses of the individual bodies.
Answer:

Let the two bodies have masses\(\mathrm{m}_1\) ,\(\mathrm{s}_1\) and specific heats\(\mathrm{s}_2\) and,,

then \(\mathrm{m}_1\)\(\mathrm{s}_1\)=\(\mathrm{m}_2\)\(\mathrm{s}_2\)

Let s = specific heat of the composite body.

Then,(\(\mathrm{m}_1\)+\(\mathrm{m}_2\)) S =\(\mathrm{m}_1\)\(\mathrm{s}_1\)+\(\mathrm{m}_2\)\(\mathrm{s}_2\)

Question 2. Define thermal capacity or heat capacity.
Answer:

Thermal capacity, also referred to as heat capacity, is the amount of heat required to change the temperature of an object by a certain degree.

Question 3. Define specific heat capacity.
Answer:

The specific heat capacity, also referred to as the specific heat of a material, is the amount of heat needed to raise the temperature of an object per unit mass of that object.

Question 4. Define Latent Heat.
Answer

: Latent heat is the energy absorbed by or released from a substance during a phase change from a gas to a liquid or a solid or vice versa.

If a substance is changing from a solid to a liquid, for example, the substance needs to absorb energy from the surrounding environment in order to spread out the molecules into a larger, more fluid volume.

If the substance is changing from something with lower density, like a gas, to a phase with higher density like a liquid, the substance gives off energy as the molecules come closer together and lose energy from motion and vibration.

Question 5. Define Sensible Heat.
Answer:

Sensible heat is the energy required to change the temperature of a substance with no phase change. The temperature change can come from the absorption of sunlight by the soil or the air itself.

Or it can come from contact with the warmer air caused by the release of latent heat (by direct conduction). Energy moves through the atmosphere using both latent and sensible heats acting on the atmosphere to drive the movement of air molecules which create wind and vertical motions.

Question 6. Define Latent Heat of Fusion and Vaporisation.
Answer:

The energy required to change the phase of a substance is known as latent heat. The word latent means hidden. When the phase change is from solid to liquid we must use the term the latent heat of fusion, and when the phase change is from liquid to gas, we must use the term the latent heat of vaporisation.

Question 7. What is the mechanical equivalent of heat energy?
Answer: The amount of heat energy or thermal energy created is proportional to the work done. This relationship is called the mechanical equivalent of heat and can be expressed by the equation

W=JH

where,

W is the work done in joules (J),

J is the relationship constant 4.18 joules/calorie (J/c),

H is the thermal energy created from the work in calories (c).

Question 8. What is ‘Anomalous Expansion of Water’?
Answer:

Water shows unusual expansion when it is cooled from four degree centigrade to zero degree centigrade, this expansion is known as “anomalous expansion of water.” The unusual behaviour of water, when it expands below 4° celsius to 0°, is called anomalous expansion of water.

Question 9. How does the anomalous expansion of water help to preserve aquatic life?
Answer:

The anomalous expansion of water helps to preserve aquatic life during very cold weather. When temperature falls, the top layer of water in a pond contracts, becomes denser and sinks to the bottom.

A circulation is thus set up until the entire water in the pond reaches its maximum density at 4°C. If the temperature falls further, the top layer expands and remains on the top till it freezes. Thus, even though the upper layer is frozen, the water near the bottom is at 4°C and the fishes, etc. can survive in it easily.

Question 10. Give an example of the anomalous expansion of water.
Answer:

If we take a cube of ice at -5°C and heat it, it expands till ice starts melting. During melting its temperature remains 0°C but its volume decreases. If heat is continuously supplied to water at 0°C, it further contracts upto 4°C and then it starts expanding. Thus, water has its minimum volume and maximum density at 4°C.

Question 11. What is absolute scale or Kelvin scale of temperature?
Answer:

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

Question 12. What is specific heat capacity of a substance?
Answer:

Specific heat capacity of a substance: The specific heat in SI system is known as specific heat capacity.

Definition: Specific heat capacity of a substance is the quantity of heat required to raise the temperature of 1 kilogram of the substance through 1 K as 1°C.

Question 13. What does the first law of thermodynamics state?
Answer:

First law of thermodynamics: When. work is completely converted into heat or heat is completely converted into work, one is equivalent to the other.

Mathematical expression of the first law of thermodynamics: If by converting work W completely heat H is obtained then Wα H

Or, W=JH     (J = mechanical equivalent of heat)

Question 14. What is mechanical equivalent of heat? What are the units of mechanical equivalent of heat in CGS and SI systems?
Answer:

Mechanical equivalent of heat: Work done to produce unit heat is called mechanical equivalent of heat.

Units of mechanical equivalent of heat :

CGS system : 4.18 x \(10^7\)erg/cal

SI system:     4.18 J/Cal

Question 15. Which indicates higher temperature, one division of Celsius scale or one division of Fahrenheit scale?
Answer:

One division of Celsius scale indicates higher temperature than one division of

Fahrenheit scale, for, 1°C \(=\frac{9}{5} \ °F\)

Question 16. Explain why a clinical thermometer should not be dipped into boiling water.
Answer: A clinical thermometer should not be dipped into boiling water. This is because the Temperature of boiling water is 212°F, much higher than the maximum limit (110°F) of the Clinical thermometer. So, if the thermometer is dipped in boiling water, the expanding mercury column will exert a great force which may break the thermometer tube.

Question 17. Why is the bulb of the thermometer made Cylindrical and not spherical?
Answer:

A thermometer becomes quick-acting if its bulb is made cylindrical in shape instead of spherical. Heat absorbed by a body due to conduction is directly proportional to its surface area in contact.

Volume being the same, the elongated cylindrical bulb has a larger contact area than a spherical bulb. The cylindrical bulb will, therefore, absorb more heat in the same time than the spherical bulb, and hence, will respond quickly.

Wbbse class 9 physical science chapter 6

Question 18. Two coins, one made of copper and another made of silver, are heated to 100°C at room temperature. Do they absorb same amount of heat? Explain.
Answer: The two coins will not absorb same quantity of heat.

Explanation : Heat absorbed = mass of the body x its specific heat x rise in temperature. Though the rise in temperature of both the copper and silver coins is the same, their masses and specific heats are different. So, the two coins will absorb different quantities of heat.

Question 19. A kettle containing water is cooled from 80°C to 40°C. Do they (the kettle and the water in it) release same amount of heat? Explain.
Answer: The kettle and the water inside it will not lose same amount of heat.

Explanation : Heat given up = mass of the body x its specific heat x fall in temperature.

Though the fall in temperature of both the kettle and the water in it is the same (80 – 40 = 40°C), their masses and specific heats are different and hence, the kettle and water in it will not release an equal amount of heat.

Question 20. Does milk get hot more quickly than water?
Answer:

The specific heat of milk is less than that of water. Hence, if the same amount of water and milk are heated separately at the same rate, the rate of increase in temperature is more for milk than for water. So, milk gets hot more. quickly than water.

Question 21. What is the specific heat of pure water?
Answer: S

pecific heat of pure water in CGS system is 1 cal\(g^{-1}\)\(C^{-1}\)and in SI system it is 4200 J. k\(g^{-1}\)

Question 22. What is the upper fixed point of a thermometer?
Answer: The upper fixed point or the steam point corresponds to the temperature of steam from water under normal atmospheric pressure (76 cm of mercury).

Wbbse class 9 physical science chapter 6

Question 23. Express 45°C of temperature in Fahrenheit scale.
Answer:

We know that, \(\frac{C}{5}\)=\(\frac{F-32}{9}\)

Or, \(\frac{45^0}{5}\)=\(\frac{F-32}{9}\)

Or, \(9\)=\(\frac{F-32}{9}\)

Or, F − 32 = 81

Or, F.= 81+ 32 = 113.

∴  In Fahrenheit scale 45°C is 113°F.

Question 24. A piece of copper of mass 100 gram is cooled from 60°C to 40°C. How much heat will be lost? (Specific heat of copper = 0.09 cal\(g^{-1}\)\(C^{-1}\)).
Answer:

Heat lost = mass x specific heat x fall of temperature

= 100.g x 0.09 cal\(g^{-1}\)\(C^{-1}\) x (60 – 40)° C = 180 calories.

Question 25. The mass of a piece of copper is 3 kilograms. The specific heat of copper is 0.09 cal \(g^{-1}\)°\(C^{-1}\). How much heat will be required to raise its temperature by 10°C?
Answer:

Here, mass (m) = 3 kg = 3000 gm.

Heat required to raise the temperature by 10°C

= mass x specific heat x rise in temperature

= 3000 x 0.09 x 10 calories = 2700 calories.

Wbbse class 9 physical science chapter 6

Question 26. Explain the term Heat.
Answer:

Heat is a form of energy. It can be experienced by our senses. If we touch a body, we can say whether it is hot or cold. A body appears hot when our hand receives heat from it, while it appears cold when heat passes from our hand to that body.

Heat: The external cause due to which a cold body becomes hot directly or indirectly, or a warm body becomes cold, is known as heat.

Since heat is a form of energy, so it cannot be-created or destroyed, but can be transformed into another form or a number of forms of energy. When heat is applied to a body, it gradually gets heated and when heat is taken out of it, the body becomes cold.

Question 27. What do you mean by the temperature of a body?
Answer:

Temperature : Temperature is a quantity which defines the thermal state of a body. It determines the direction of flow of heat when two bodies of coi temperatures are kept in contact.

Thus, temperature is the condition of a body which determines the direction along which heat will flow. Heat is the cause and temperature is the effect. Temperature indicates the
degree of hotness or coldness of the body. It does not tell us the quantity of heat energy contained in the body. Hence, temperature is the thermal state of a body.

Question 28. State and define the units of heat.
Answer:

The S.I. units of heat is 1 Joule (J).

The C.G.S. unit of heat is 1 Calorie (Cal).

Calorie:nOne calorie heat is the quantity of heat required to raise the temperature of (1)gm of water through 1°C (from 14. 5° C to 15.5°. C).

1 calorie (1  cal) = 4°2 Joules (J) nearly.

1 Kilo calorie = 1000 calorie

= 4200 Joules. 

Question 29. How is temperature measured? Name the instrument to measure it.
Answer:

Temperature of a body is measured by an instrument called thermometer. The common type of thermometers used in the laboratory or to measure the temperature of a human body are mercury thermometers.

The most commonly used thermometers using the scale are:
(1) Celsius or centigrade scale and
(2) Fahrenheit scale.

WBBSE class 9 question and answers

Question 30. What are the different scales of temperature used in thermometers?
Answer:

To measure the temperature of a body accurately a standard temperature scale is essential.

A temperature scale is prepared by using fixed points on thermometers:
(1) Celsius scale of temperature, and
(2) Fahrenheit scale of temperature.

Question 31. What do you mean by calibration of thermometers?
Answer:

The calibration of thermometers is a process of fixing two points on it and then dividing the interval between them into convenient number of equal parts. The number of equal parts is called the scale of temperature. The fixed points are chosen in such a way that they can be easily obtained. These two fixed points are called lower fixed point and upper fixed point.

Question 32. Establish the relation between the two scales of temperature — Fahrenheit and Celsius.
Answer:

The given figures shows the two scales of temperature, the Celsius scale and the Fahrenheit scale of temperature.

The interval between the upper and lower fixed points is the same in both the scales. Thus, 100 centigrade (div.) degree = 180 Fahrenheit (div.). degree.

∴\(\frac{C-0}{100}\)=\(\frac{F-32}{180}\)

Or, \(\frac{C}{5}\)=\(\frac{F-32}{9}\)

This the relation between the two scales of temperature.

Question 33. Explain Celsius scale (or centigrade scale) and Fahrenheit scale of temperature.
Answer:

Celsius scale (or centigrade scale): On this scale, the lower fixed point or ice point is marked as 0° C. The upper fixed point or steam point is marked as 100° C. The interval between the ice point and the steam point is divided into 100 equal parts. Each of these divisions is called one degree Celsius and is written as °C.

WBBSE class 9  question and answers

Fahrenheit scale of temperature: On the Fahrenheit scale of temperature, the ice point is marked as 32°F and the steam point is marked as 212°F. The interval between the two fixed points is divided into 180 equal parts.-Each division is called one degree Fahrenheit and is written as °F.

Question 34. Show by an experiment that two balls of different materials of same mass and temperature absorb different quantities of heat.
Answer:

Two balls of same mass but of different materials, say, lead and aluminum, are taken. They are immersed in water and heated for some time. After some time both balls are taken out of water and placed on a block of ice. It will be seen that the aluminum ball will melt more ice. This shows that the quantity of heat taken depends upon the material of the body if other conditions are constant. This property is known as the heat capacity of the body.

Question 35. Define thermal or heat capacity of a body.
Answer:

Thermal capacity: The thermal capacity (or heat capacity) is the amount of heat required to raise its temperature by 1° C.

Hence, thermal capacity =\(=\frac{\text { Amount of heat }}{\text { Rise in temperature }}\)

The unit of thermal capacity in S.I. system is Joule per degree.

The unit of thermal capacity in C.G.S. system is calorie per degree.

Question 36. The specific heat of lead is 0°03” – What is its meaning?
Answer:

The specific heat of lead is 0°03” means that one gram of lead will require 0°03 calories of heat to raise the temperature of lead through 1° C.

Question 37. Do one gram of iron and one gram of copper contain equal quantities of heat at same temperature? Explain.
Answer:

The amount of heat required to raise the temperature of a body The specific heat of copper is less than that of iron. Hence the amount of heat required to raise the temperature of both metals of same mass will be different.

WBBSE class 9  question and answers

Question 38. Define water equivalent of a body.
Answer:

The amount of water whose temperature is raised through 1° C by the same quantity of heat which is required to raise the temperature of a body through 1° C, is known as the water equivalent of a body.

If the mass of a body is m gm and its specific heat be s, then the water equivalent (W) is W = ms gm.

The unit of water equivalent measured in gm or kg.

Question 39. State the difference between water equivalent and aennal capacity.
Answer:

Water equivalent and thermal capacity of a body are numerically equal, but their units are different. Water equivalent is an amount of water and it is measured in gram, but thermal capacity is an amount of heat and is expressed in calories.

Question 40. State the principle of caiorimetry. 
Answer:

The principle of calorimetry is based on the law of conservation of energy. When a hot body is mixed (or kept in contact) with a cold body, heat passes from hot body to the cold body till both bodies attain the same temperature. If no heat is lost with the surrounding,then

Heat lost by the hot body = Heat gained by the cold body.

Question 41. Why is water used in hot water bottles?
Answer: Water has a high specific heat in comparison to other liquids. Hence for the same mass of water and other liquids, our body can get greater amount of heat from water.

Question 42. Why do farmers fill their fields with water to protect their crops from frost?
Answer:

Waiter having high specific heat does not allow the temperature in the surrounding area of plants to fall below 0° C. In the absence of water, if on a cold night, temperature falls below 0° C, the water in the fine capillaries of plants will freeze and veins will burst due to increase in volume of water on freezing. As a result the crops willbe destroyed.

Question 43. Why should the bore of a thermometer be uniform and also narrow?
Answer:

The bore should be uniform, otherwise, unequal expansion of mercury thread will take place at different parts of the bore for equal changes in temperature. The bore of thermometer tube should be narrow so that the expansion of mercury thread is long enough for a small change of temperature.

WBBSE class 9  question and answers

Question 44. What do you mean by a quick-acting thermometer? What are the requisites for making a thermometer quick-active?
Answer:

The thermometer with which the temperature of a body can be measured very quickly is called a quick-acting thermometer

To make a quick-acting thermometer:
(1) Volume of its bulb should be small
(2) Glass wall of the bulb should be thin
(3) The bulb should be made cylindrical and
(4) The thermo metric liquid should be of high thermal conductivity.

Question 45. On what factors does the total quantity of heat in a body depend?
Answer:

The quantity of heat in a body depends upon its
(1) Mass
(2) Specific heat and
(3) Temperature.

Question 46. The body temperature of an ill person is 104° F. What would be its temperature in Celsius scale and Kelvin scale?
Answer:

By relation,\(\frac{C}{5}\)=\(\frac{F-32}{9}\) given F = 104° F.

Then,\(\frac{C}{5}\)=\(\frac{104-32}{9}\)  Or, C\(=\frac{5 \times 72}{9}\)=40°.

The required temperature in Celsius scale is 40° C.

∴ Reading in Kelvin Scale = 273 + 40 = 313K.

Question 47. The mass of a piece of copper is 3 kilogram. The specific heat of copper is 0°09 cal/g °\(C^{-1}\). How much heat will required to raise its temperature by 10° C?
Answer:

The mass of copper (m) = 3 Kg = 3000 Kg. Specific heat (s) = 0°09 cal/g\(C^{-1}\)  and rise in temperature (1) = 10° C. Heat required = mass x specific heat x rise in temperature

= 3000 x 0°09 x 10 cal = 2700 caI.

Question 48. Calculate the amount of heat required to raise the temperature of 150 gm of water from 0°C to its boiling point.
Answer:

The rise in the température of water from 30° C to 100° C (boiling point)

= (100 – 30)°C = 70°C. 4

Heat taken by water = mass of water x specific heat x rise in temperature

= 150 x1 x 70 calories = 10500 calories.

Question 49. What quantity of heat will be taken by 50 gm of water at 20°C to reach its boiling point at normal pressure ?
Answer:

Mass m = 50 gm.

Temperature difference t = (100 − 20)°C = 80°C. ~ Specific heat of water s = 1.

We know that, H = mst

= 50 x (1)x 80

= 4000 cal

∴ Heat taken = 4000 cal.

Question 50. A thermometer having the corresponding upper and lower fixed points 95° and 15°, reads a temperature 65°, what will this temperature be in Celsius scale?
Answer:

The relation between temperature of Celsius and the given scale can be written from the general relation between any two scales as, \(\frac{65-15}{95-15}\)=\(\frac{C-0}{100-0}\)

⇒ \(\text { or, } \frac{50}{80}\)=\(\frac{C}{100}\)

80C= 50×100

∴C\(=\frac{50 \times 100}{80}\)=6.25.

The required temperature is 62.5°C.

WBBSE class 9  question and answers

Question 51. In Fahrenheit scale the temperature of a body increases by 36°. What is the corresponding risé of temperature in Celsius scale?
Answer:

1°F\(=\left(\frac{5}{9}\right)^{\circ} \mathrm{C}\)

∴ 36° F\(=\left(36 \times \frac{5}{9}\right)^{\circ} \mathrm{C}\)

=20° C.

Question 52. What are the types of heat?
Answer:

Types of heat: There are three types of heat.These are :
(1) Sensible heat
(2) Latent heat
(3) Radiant heat.

Question 53. What is radiant heat?
Answer:

Radiant heat: It is that thermal condition of the body which determines whether it will absorb heat from other body or release heat to that body.

Question 54. What is lower fixed point of a thermometer?
Answer:

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

Question 55. What is upper fixed point of a thermometer?
Answer:

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

Question 56. What do you mean by fundamental interval?
Answer:
Fundamental interval: The range of temperature between the upper and the lower fixed points is called Fundamental interval.

Question 57. What is absolute scale or Kelvin scale of temperature?
Answer:

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

Question 58. What are the units of heat in CGS and SI systems?
Answer:

Unit of heat in CGS system is calorie: It is the amount of heat required to raise the temperature of one gram of pure water through 1°C. Unit of heat in SI system is joule : As heat is a form of energy, so now-a-days, heat is measured in the unit of energy. That is heat is expressed in joule. It is the SI unit of heat.
1 calorie = 4%8 joule.

Question 59. What do you mean specific heat of a substance?
Answer:
Specific Jeeat of a substance: It is defined as the quantity of heat required to raise the temperature of unit mass of it by 1 degree.

Question 60. What are the units of specific heat in CGS and SI systems?
Answer:

Unit of specific heat in CGS system is cal/g/°C.

Unit of specific heat in SI system is J/kg K.

Question 61. What are the units of heat capacity in CGS system and SI systems?
Answer:

Unit of heat capacity in CGS system is Calorie/°C.

Unit of heat capacity in SI system is Joule/kelvin.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 62. State the relation between thermal capacity and specific heat of a body.
Answer:

Relation between thermal capacity and specific heat of a body: Quantity of heat required to raise the temperature of a body of mass ‘m’ and specific heat is by 1° = m.s.1= m.s.

i.e., Thermal capacity = mass x specific heat.

Question 63. Show that specific heat is the heat capacity of unit mass.
Answer:

Thermal capacity of a body

= mass x specific heat of material of the body.

Hence, thermal capacity = ms.

Putting, m = 1, in the above relation,

Thermal capacity = 1 x specific heat.

So, it can be concluded, specific heat is the thermal capacity per unit mass.

Question 64. What are the units of water equivalent in CGS system and SI system?
Answer:

Unit of water equivalent in CGS system is gram (g).

Unit of water equivalent in SI system is Kilogram (kg).

Question 65. What are the differences between thermal capacity and water equivalent?
Answer:

Differences between thermal capacity and water equivalent :

Thermal capacity	Water equivalent
(1)    Thermal capacity represents some quantity of heat.	(1)    Water equivalent represents some quantity of water.
(2)    Units of thermal capacity in CGS and SI systems are calorie/°C and J/K respectively.	(2)    Units of water equivalent in CGS and SI systems are gram and kg respectively.

 

Question 66. What is the fundamental principle of calorimetry?
Answer:
Fundamental principle of calorimetry: If no heat flows to outside from the system of bodies and if no chemical reaction takes place between these, >, then from the principle of conservation of energy, it can be said, heat lost by the hotter bodies = heat gained by the colder bodies. This is the principle of calorimetry.

Question 67. What is the relation between heat and molecular velocity?
Answer:
Relation between heat and molecular velocity: The molecules in a hotter body have slower motions. Also, if heat is supplied to a body, its molecules move more rapidly. If heat is withdrawn from a body, the motion of its molecules becomes slower. Thus, increase or decrease of molecular velocity corresponds to higher or lower quantity of it.

Question 68. Why should the bore of a thermometer be uniform?
Answer:

Reason: The bore should be uniform, otherwise, unequal expansion of mercury thread will take place at different parts of the bore for equal change of temperature.

Question 69. Why is meant by sensitivity of a thermometer?
Answer:

Sensitivity of a thermometer: A thermometer is considered sensitive if it :

(1) Quickly picks up the temperature of the body with which it is kept in contact
(2) Can detect small changes of temperature.

Question 70. Why is water preferred to other liquids in a hot water bag?
Answer:

Reason: Since water has the highest specific heat, a certain mass of water heated to a certain temperature contains more heat than the same mass of any other liquid heated to the same temperature. So, hot water will take longer time to cool than any other liquid heated to the same temperature. So, for hot compression, it can be used for a sufficiently long time.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 71. When a mass m of hot water was added to a mass 3 m of cold water (initially at 10°C), the mixture attained a final temperature of 20°C. Find the initial temperature of hot water.
Answer:
Solution: Let s be the specific heat capacity of water and t be the initial temperature of hot water. Then we have,

Heat lost by hot water = ms(t −20)

and heat gained by cold water = 3ms (20 −10) = 30 ms.

Since heat lost = heat gained. :

We have, ms (t- 20) = 30ms_ or, t-20=30 or,  t= 50°C

The initial temperature of the hot water was, therefore, 50°C.

Question 72. One joule of work expended in winding a 30 g steel watch and the watch is wrapped in a heat insulator. How much warmer would the watch be by the time it has run down? Specific heat of steel = 0.(1)cal/g°c.
Answer:

W = JH or, H=W/J \(=\frac{1.0 \mathrm{~J}}{4.18 \mathrm{~J} / \mathrm{cal}}\)  = 0.24 cal.

Now, H ( or Q ) = ms ΔT

∴ ΔT=\(\frac{\mathrm{H}}{\mathrm{ms}}\)=\(\frac{0.24 \mathrm{cal}}{30 \mathrm{~g} \times 0.1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}}\)=0.08 C°.

Question 73. The temperature on a day was 94°F. What is it in Celsius sca
Answer:

We know that, \(\frac{C}{5}\)=\(\frac{F-32}{9}\)        F=94°

or,\(\frac{C}{5}\)=\(\frac{94-32}{9}\)                                          C=?

or, C\(=\frac{62 \times 5}{9}\) =34.4°.

Question 74. The lowest attainable temperature so far is — 270° Celsius. What is it in Fahrenheit scale?
Answer:

We know,  C =− 270°

⇒ \(\frac{C}{5}\)=\(\frac{F-32}{9}[/latex, F=?

or, [latex]\frac{-270}{5}\)=\(\frac{F-32}{9}\)

or,−54 \(=\frac{F-32}{9}\)

or, F − 31=−54×9=−486

or, F = −486+32=− 454

∴ f=−454°5.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 75. Convert 50°C to Fahrenheit scale.
Answer:

We know,  C=50°

⇒ \(\frac{C}{5}\)=\(\frac{F-32}{9}\)   F=?

or, \(\frac{-50}{5}\)=\(\frac{F-32}{9}\)

or, 90=F−32

or, F  = 90+32=122

Hence, 50°C =122°F

Question 76. The difference in the readings of a Fahrenheit and a Celsius Thermometer is  48°. What are the actual readings shown in the thermometers?
Answer:

Let the actual readings in the Rahsenhelt and the Celsius thermometers a to F and C.

∴ F−C =  or,48°C  F=C+48°

Again ,from the relation, \(\frac{C}{5}\)=\(\frac{F-32}{9}\)

or, \(\frac{C}{5}\)\(=\frac{F+48-32}{9}\)=\(\frac{C+16}{9}\)

or, 9C=5C+90

or, AC=80

or, C=20°

Now,F= C + 48° = 20° + 48° = 68°.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 77. A certain temperature is 104°F, what is its value in Kelvin scale?
Answer:

We know the relation between Fahrenheit and Kelvin scales is

⇒ \(\frac{F-32}{9}\)=\(\frac{K-273}{5}\), F=104°

or, \(\frac{104-32}{9}\)=\(\frac{K-273}{5}\), K=?

or, \(\frac{72}{9}\)=\(\frac{K-273}{5}\)

or, 8×5=K−273

or, K=313

∴ 104°F=313K

Question 78. The normal temperature of human body is 98.4°. What is the value of this temperature in centigrade scale?
Answer:

We know, F = 98.4°

⇒ \(\frac{C}{5}\)=\(\frac{F-32}{9}\)

or, \(\frac{C}{5}\)=\(\frac{98.4-32}{9}\)

or,\(\frac{C}{5}\)=\(\frac{66.4}{9}\)

or, 9C=6.64×5

or, \(C\)=\(\frac{332.0}{9}\)=36.89

So, the required temperature = 36.89°C,

Question 79. Which temperature is same in both centigrade scale and Fahrenheit scale?
Answer:

Let the temperature be X°.

According to question, C = F = X.

We know, \(\frac{C}{5}\)=\(\frac{F-32}{9}\)

or, \(\frac{x}{5}\)=\(\frac{x-32}{9}\)

or, 9x=5x-160

or, 9x-5x= -160

or, 4x=-160

or,x\(=\frac{-160}{4}\)=−40°

So, (−)40° temperature will read same both in Fahrenheit and centigrade scales.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 80. Mass and specific heat of a substance 100 g and 0.09 respectively. Find the Thermal capacity and water equivalent of the body.
Answer:

We know,  m= 100g;     s = 0.09

Thermal capacity = ms

= 100 x 0.09 = 9 Cal/°C

Water equivalent = ms

= 100 x 0.09 =9g.

Question 81. Calculate the amount of heat required to raise the temperature of 60 g of water from 20°C to its boiling point.
Answer:

We know,   m = 50g

Absorbed heat = ms\(\left(t_2-t_1\right)\) \(t_1\)= 20°C

=50 x 1x(100−20) ⇒ \(t_2\)  = 200°C

=50×80,   s=1

= 4000 cal.

Wb Class 9 Physical Science Question Answers

Question 82. A body of iron having mass 60 g is cooled from 200°C to 100°C. Calculate the amount of heat given out. Specific heat of iron = 0.12.
Answer:

We know, m = 60g

The amount of heat given out by iron    \(t_1\)= 100°C

= ms\(\left(t_2-t_1\right)\)       \(t_2\) = 200°C

= 60 x 0.12(200 − 100) s= 0.12

= 60 x 0.12 x 100 = 720 cal.

Question 83. Find the quantity of heat required to raise the temperature of 1 kg of water from 10°C to 90°C.
Answer:

Mass of water = 1 kg = 1000 g

Rise in temperature of water = 90 − 10 = 80°C

Specific heat of water = 1cal \(g^{-1}\) \( C^{-1}\)

Thus, the required heat = mst = 1000 x 1 x 80 = 80,000 cal.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 84. A kettle and the water in it are cooled down from 80°C to 40°C. Do the kettle and water give up the same amounts of heat? Explain

your answer.
Answer:

Heat lost by a body = mass of the body x specific heat of the body x fall in temperature. Although fall in temperature of water or kettle = 80 −40 = 40°C is same, their masses and specific heats being different, they would not give up same amounts of heat.

Question 85. If the specific heat of copper be 0.09, calculate the thermal capacity and water equivalent of a block of copper of mass 50 g.
Answer:

Thermal capacity of the piece of copper

= mass x specific heat = 50 x 0.09 = 4.5 cal/°C.

Water-equivalent of the piece of copper

= mass x specific heat = 50 x 0.09 = 4.5 g.

Question 86. What quantity of heat will be taken by 50 g of water at 20°C to reach its boiling point at normal pressure?
Answer:

Boiling point of water at normal pressure = 100°C.

So, heat taken by water = 50 x 1x (100 −20) = 4000 cal.

Question 87. The quantity of heat required to raise the temperature of one keane ofa metal is 2300 cal. Find the specific heat and water equivalent of the metal! piece.
Answer:

Given,

mass of the metal piece m = 1 kg = 1000 g;

rise in temperature = 20°C; total amount of heat Q = 2300 cal.

If specific heat of the metal s, then

Q=mst_ or, 2300 = 1000 x s x 20

∴ s\(=\frac{2300}{1000 \times 20}\)=0.115

Again, water-equivalent of the metal piece = ms = 1000 x 0.115 = 115 g.

Wbbse Madhyamik Class 9 Physical Science And Environment 

Question 88. 300g of water at 15°C is mixed with 500 g of water at 30°C. What is the resultine temperature?
Answer:

Given, \(\mathrm{m}_1\)= 300g; \(\mathrm{m}_2\)  = 500g; \(\theta_1\) = 15°C; \(\theta_1\) = 30°C.

Let the resulting temperature be θ°C.

Heat gained by cold water =\(\mathrm{m}_1\)xs (θ −\(\theta_1\)) = 300 x1x (θ−15) calorie

Heat lost by hot water = \(\mathrm{m}_2\)x s x (\(\theta_2\)− θ) = 500 x (1)x (30 − θ) calorie.

From the relation, Heat gained = Heat lost.

300(θ −  15) = 500(300 − θ )

or, (300 + 500)θ = 500 x 30 + 300 x 15

∴θ \(=\frac{15000+4500}{800}\)

=24.38°C.

Question 89. Calculate the amount of work to be done to convert 100g of water at 0°C to water at 100°C. Mechanical equivalent of heat = 4.2 Joule.
Answer:

Heat required, H = mst = 100 x 1x (100 −0) = 10,000 cal.

Work required, W = JH = 4.2 x 10,000 = 42,000 J.

Question 90. A steel ball of specific heat 0.12 has its temperature increased by 0.5°C on falling through a certain height. If J = 4.2 x\(10^7\)erg\(\mathrm{cal}^{-1}\) , calculate the height.
Answer:

Given, specific heat, s = 0.12

Increase in temperature, θ = 0.5°C; J = 4.2x \(10^7\)erg\(\mathrm{cal}^{-1}\)

Let mass of the ball be m gram and the height be h cm. Then

PE of the ball = mgh.

Heat produced = msθ = m x 0.12 xθ

mgh=Jxmx0.12xθ

h\(=\frac{\mathrm{J} \times 0.12 \times 0.5}{g}\)=\(\frac{4.2 \times 107 \times 0.12 \times 0.5}{980}\)

=2.571×\(10^3\)cm=25.71m.

Question 91. A kilogram of water at 20°C is placed in a refrigerator. How much heat must be extracted before all the water turns to ice? Latent heat of ice = 80 Cal \(g^{-1}\).
Answer:

Mass of water m = (1)kg = 1,000 g; L = 80\(\text { calg }^{-1}\); \(t_1\) = 20°C; \(t_2\) = 0°C

Heat extracted from water, \(\mathrm{H}_1\) = ms(\(t_1\) −\(t_2\) ) = 1000 x (20 −0) = 20,000 cal.

Heat extracted from water to freeze into ice, \(\mathrm{H}_2\)= mL = 1000 x 80 = 80,000 cal.

Total heat extracted = \(\mathrm{H}_1\)+\(\mathrm{H}_2\)= 20,000 + 80,000 = 1,00,000 cal.

Question 92. After a few pieces of dry ice had been added to 120 g of water at 46°C, the final temperature of the mixture was found to be 0°C. What is the mass of ice added? Latent heat of ice = 80 cal\(g^{-1}\).
Answer:

Given, \(\mathrm{m}_1\)= mass of water = 120g \(t_1\) = temperture of water = 46°C.

Final temperature of the mixture,,\(t_2\)= 0°C; L = 80 cal\(g^{-1}\)

Let mass of the ice added be \(\mathrm{m}_2\)g.

Heat lost by water at 46°C to be water at 0°C = \(\mathrm{m}_1\)s(\(t_1\)−\(t_2\))

= 120 x (1)x (46 – 0)

= 120 x 46 cal.

Heat required by m,g ice at 0°C to water at 0°C.= \(\mathrm{m}_2\)L =\(\mathrm{m}_2\)x 80 cal.

Heat Lost = Heat gained; 120 x 46 = \(\mathrm{m}_2\) x 80

∴ \(\mathrm{m}_2\)

Wb Class 9 Physical Science Question Answers

Question 93. How much heat is released in converting 5g of steam at 100°C to water at 60°C ?Latent heat of steam = 540 calg”.
Answer:

Heat released in converting 5g steam at 100°C to 5g of water at 100°C = 5 x 540 cal.

Heat released in converting 5g of water at 100°C to 5g of water at 60° =5x 1(100-60) =5 x 40 cal.

∴ Total heat released = 5 x 540 + 5 x 40 = 5 x 580 = 2900 cal.

Question 94. A piece of copper heated to 35°C is dropped into 140 g of water at 15°C. The mass of copper is 150 g and the specific heat of copper is 400 J/kgK. Find the final temperature. (sp. heat of water = 4200 J/kgK)
Answer:

Here, the hot piece of copper will give up heat and cold water will absorb it. Let the final temperature be t°C.

Now, the heat lost by copper = its mass x sp. heat fall of temperature

= 0.15 x 400 x (35 – t) = 60 x (85-1) J.

And heat gained by water = its mass x sp. heat x rise of temperature

= 0.14 x 4200 x (t − 15) = 588(t − 15) J. Now, heat lost = heat gained

or, 60 x (35 − t) = 588 (t− 15) or, 175 – St = 49t – 735

∴ t= 16.85°C.

Question 95. A lump of copper of mass 100 g is cooled from 60°C to 40°C. How much heat will be rejected by the lump ? Specific heat of copper = 0.09 cal \(g^{-1}\) \( C^{-1}\)
Answer:

Amount of heat rejected = mass x sp. heat x fall in temperature

= 100 x 0.09 x (60 − 40) Cal = 180 Cal.

Question 96. A brass sphere of mass 0.4 kg at 100°C is dropped into 1 kg of water at 20°C. The final temperature of the mixture is 23°C. Find the $.H.C. of brass. Given, S.H.C. of water = 4200 J/kgK.
Answer:

Heat lost by brass sphere = m x s x (\(t_1\)−\(t_2\))

= 0.4 x s x (100 — 23) = 30.8 x sJ.

Heat gained by water

=mxsx (\(t_1\)−\(t_2\)) =1 x 4200 x (23 − 20) = 12600 J.

Now, Heat lost = Heat gained

30.8 x s = 12600

∴ s  = 409.1 J/kgK.

Question 97. How many calories of heat will be evolved on the expenditure of 84 joules of energy?
Answer:

4.2 joule will produce 1 calorie heat.

∴  84 joules will produce\(\frac{1 \times 84}{4.2}\) calorie = 20 calorie.

Wb Class 9 Physical Science Question Answers

Question 98. How much work is to be completely converted to heat to melt 50 g ice at orc?
Answer:

Heat required to melt 1  g ice at 0°C to water at 0°C = 80 calories.

∴ Heat required to melt 50 g ice at 0°C to water at 0°C = 80 x 50 calories = 4000 calories.

Now, for 1   calorie, 4.2-joule work is to be converted to heat.

∴ For 4000 calories, 4.2 x 4000 joule work is to be converted to heat = 16800 Joule.

Wb Class 9 Physical Science Question Answers

Question 99. A mass 10 kg falls a height 1 kilometer to the ground. If all the energy is converted into heat, find the amount of heat developed (given 4.2 joule = 1 calorie).
Answer:

The potential energy mgh stored in the body when it was 1  km above the ground converts to heat when it falls to the ground.

∴ Potential energy = mgh = 10 kg x 9.8 m/\( s^2\) x 1000 m = 98000 J.

∴  The required heat developed = 98000 J.

Question 100. How much heat is necessary to convert 10 g ice at 0°C to 10 g water at the same temperature? 
Answer:

Latent heat of fusion of ice is 80 calories per gram or 336 J per gram.

∴  Heat required to convert (1)g ice at 0°C to water at 0°C = 80 calories or 336 J.

∴  Heat is required to convert 10 g ice at 0°C to water at 0°C.

= (80 x 10) calorie or (336 J x 10) = 800 calorie or 3360 J.

Question 101. Find the quantity of heat required to convert 15 g water at 100°C to 15 g steam at 100°C. (Specific latent heat of vapourization of water = 22.5 x 105 J/kg)
Answer:

15 g = 0.015 kg.

∴ Heat required to convert 1 kg water at 100°C to steam at 100°C = 22.5 x 10.

∴  Heat required to convert 0.015 kg water at 100°C to steam at 100°C

= 22.5 x \(10^5\) x 0.015 J = 33750 J.

Chapter 6 Heat 3 Marks Questions And Answers:

Question 1. State and explain the principle of calorimetry.
Answer:

When two bodies (one being solid and the other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from the body at a higher temperature to the body at a lower temperature till both acquire the same temperature. The body at higher temperatures releases heat while the body at lower temperatures absorbs it, so that

Heat lost = Heat gained,

i.e., the principle of calorimetry represents the law of conservation of heat energy While using this principle always keep in mind that

(1) Temperature of the mixture (T) is always > lower temperature (\(T_1\)) and < higher temperature(\(T_2\)),i.e.,\(T_1\)<T<\(T_2\),

ie., the temperature of the mixture can never be lesser than lower temperature (as a body cannot be cooled below the temperature of the cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of the heating body when there is no chemical reaction).

(2) When the temperature of a body changes, the body releases heat if its temperature falls and absorbs heat when its temperature rises. The heat released or absorbed by a body of mass m is given by:

Q=mc ΔT

Where c is the specific heat of the body and AT change in its temperature in °C or K.

(3) When the state of a body changes, change of state takes place at constant temperature [m.pt. or b.pt.] and heat released or absorbed is given by

Q=mL

Where L is latent heat. Heat is absorbed if solid converts into liquid (at m.pt.) or liquid converts into vapor (at b.pt.) and is released if liquid converts into solid or vapor converts into liquid.

Wb Class 9 Physical Science Question Answers

Question 2.20 gm steam at 100°C is let into a closed calorimeter of water equivalent to 10 gm containing 100 gm ice at — 10°C. Find the final temperature of the calorimeter and its contents. The latent heat of steam is 540 cal/gm, the latent heat of fusion of ice = 80 cal/gm, and the specific heat of ice = 0.5 cal/°C gm.
Answer:

Heat lost by steam = mL + ms (100 − t)

where, t is the equilibrium temperature

Heat lost by steam = 20 x 540 + 20 x (1)(100 − t) =10800 + 2000 – 20t.

Heat gained by (ice + calorimeter) = 100 x 80 + 100 x 0.5 x 10+ 100 x t.

Now, Heat lost = Heat gained :

or, t 10800 + 2000 − 20 t = 8000 + 500 + 100%

or, 120 = 4300 or t = 4300/120 = 35.83°C.

Question 3. A tube leads from a flask in which water is. boiling under atmospheric pressure to a calorimeter. The mass of the calorimeter is 150 gm, its specific heat capacity is 0.1 cal/g/°C, and it contains originally 340 gm of water at 15 °C. Steam is allowed to condense in the calorimeter until its temperature increases to 71°C, after which the total mass of the calorimeter and contents are found to be 525 gm. Compute the heat of condensation of steam.
Answer:

Mass of calorimeter and contents before passing steam = (150 + 340) = 490 gm.

Mass after passing steam = 525 gm.

=> mass of steam which condenses = (525 – 490) gm = 35 gm.

= 286

Let L = latent heat of steam.

Heat lost by steam = heat gained by water + heat gained by calorimeter.

35L +35 x 1 (100-71) =340 x 1 x (71-15) + 150 x 0.1x (71- 15), Thus, L=539 cal/gm.

Question 4. What is a Calorimeter?
Answer:

A calorimeter is a device used to measure the heat of a reaction. It can be sophisticated and expensive or simple and cheap A styrofoam cup is used as a calorimeter because it is a container with good insulated walls to prevent heat exchange with the environment.

In order to measure heat of reactions, we often enclose reactants in a calorimeter, initiate the reaction, and measure the temperature difference before and after the reaction. The temperature difference enables us to evaluate the heat released in the reaction. A calorimeter may be operated under constant (atmosphere) pressure or constant volume.

Wb Class 9 Physical Science Question Answers

Question 5. State the laws of thermodynamics.
Answer:

There are two major laws concerning thermodynamics:

First Law of Thermodynamics: The First Law of Thermodynamics is the law of Conservation of Energy. It states that energy cannot be created or destroyed. Instead, it is converted from one form to another, such as from mechanical work to heat, from heat to light, from chemical to heat or such. One example of that is how the kinetic energy of a moving car is converted into heat energy at the brakes and tire surfaces. Another example is when chemical energy is released in burning and is converted into light and heat energy.

Second Law of Thermodynamics: The Second Law of Thermodynamics has several variations which will be explained below. Some heat is wasted in conversion  One version of the Second Law of Thermodynamics states that some heat is wasted when converting heat into mechanical energy. In other words, in a car engine, not all of the heat created from the exploding gasoline is used in turning the engine or moving the car. Some of the heat simply heats the engine. The percentage of heat turned to work is called the ‘thermal efficiency’ of the engine.

Question 6. Explain the terms saturation vapor pressure (s.v.p.) and unsaturated vapor.
Answer:

Let us suppose that some liquid is poured into a bottle which is then corked up. Owing to evaporation, the space above the liquid begins to fill with vapor. The vapor molecules move about in all directions and exert pressure when they bounce off the walls of the bottle.

They also strike the surface of the liquid and many re-enter it. Eventually, a state of dynamic equilibrium is reached in which the rate at which molecules leave the liquid is equal to the rate at which others return to it.

Under these conditions, the space above the liquid is said to be saturated with vapor, and the pressure exerted is called the saturation vapor pressure ( s.v.p.).

For a given temperature the saturation vapor pressure of a liquid is always the same whether there is air in the space above it or not. Before equilibrium has been reached in the manner described the vapour is said to be unsaturated.

Question 7. What are the kinds of heat?
Answer:

Heat is mainly of three types:

(1) Sensible heat: The kind of heat perceptible by senses or can be detected by a thermometer is known as Sensible heat.

(2) Latent heat: It is the quantity of heat absorbed or released when a substance of unit mass undergoes a physical change of state at constant temperature and pressure. This heat can not be measured by a thermometer.

(3) Radiant heat: The heat which radiates from its source to all directions is called Radiant heat. E.g. Sunlight.

Question 8. What are the conditions required to fulfill the principle of calorimetry?
Answer:

The conditions are:

(1) There should not be a transfer of heat from the bodies to the surroundings.
(2) There should not occur a change of state of any body.
(3) Chemical reaction must not take place between the components.
(4) The bodies should not be soluble in each other.

Question 9. Equal masses of milk and water are separately taken in two identical vessels. The vessels are heated simultaneously in the same oven. Which one will have the quicker rise of temperature, milk or water, and why?
Answer:

The specific heat of milk is less than that of water, so milk will require a smaller quantity of heat for each degree rise in temperature than that needed for the same mass of water. Hence, the supply of heat being the same to the equal masses of milk and water, the rise in the temperature of milk will be quicker than that of water.

(Mathematically, let H quantity heat be supplied to a body of mass m and specific, heat

S. Then if the rise of temperature is t, H = mst. Or, t\(=\frac{\mathrm{H}}{\mathrm{m}} \cdot \frac{1}{\mathrm{~S}}\)

Now if the same quantity in heat H be supplied to equal mass of different substance, then t ∝\(\frac{1}{s}\)

(∴ H, m are constants).

This means a substance of lower specific heat will have a greater rise in temperature. So, in this case, milk will have a quicker rise in temperature.)

Question 10. What do you mean by the sensitivity of a thermometer? Write down the characteristics of a sensitive thermometer.
Answer:

The efficiency of a thermometer to measure even a very little difference of temperature is called: its sensitivity. A thermometer becomes more sensitive if it can measure the slight differences in temperature more accurately.

Conditions for sensitivity :

Bulb:

(1) Should be of large volume (cylindrical)

Glass capillary tube:   

(1) The bore of the capillary tube should be fine
(2) The area of the cross-section should be the same everywhere
(3) Lengthy

Thermometric liquid:

(1) Should be of high coefficient of volume expansion
(2) Must be of low thermal capacity

Wbbse Physical Science And Environment Class 9 Solutions

Question 11. Will the temperature of a body containing more heat will always be higher than that of another body containing less heat? Answer with an explanation.
Answer: No.

The quantity of heat in a body depends on three factors:

(1) Mass of the body,
(2) Specific heat of the body and
(3) Change in the temperature of the body. The flow of heat depends on the temperature of the bodies. Heat always flows from the body of high temperature to the body of low temperature.

Question 12. What are the effects of heat on a body?
Answer:

When heat is supplied to a body, the following changes are found to occur:

(1)  Change in temperature
(2)  Cange in volume
(3)  Change of state
(4)  Change in physical properties
(5)  Change in chemical properties
(6)  Creation of light energy, electric energy, burning, etc.

Question 13. Why is mercury used in thermometers?
Answer:

Mercury is an ideal substance for commonly used thermometers. It has the following advantages:

(1) Mercury is a liquid metal and a good conductor of heat, hence it easily acquires the temperature of the body with which it is kept in contact.
(2) The volume of mercury increases sufficiently for a small increase in temperature.
(3) Mercury is opaque and bright like silver, so it can be easily seen from outside the glass tube.
(4) While rising or falling, mercury does not stick to the walls of the glass tube.
(5) The freezing point of mercury is – 30°C and its boiling point is 357° C. Any temperature between this range is measurable by it:
(6)Pure mercury is available easily.

Wbbse Physical Science And Environment Class 9 Solutions

Question 14. Write three differences between heat and temperature.
Answer:

The main differences between heat and temperature are the following :

Heat	Temperature
1.    Heal is a form of energy.	1.    Temperature is a thermal condition of the body, which shows how hot or cold the body is.
2.    Heat is the cause. If a body is heated without a change of states, its temperature increases.	2.    Temperature is the effect.
3.    The total quantity of heat contained in two bodies may not be equal but their temperatures may be same.	3.    The temperatures of two bodies may be the same, whether their quantity of heat contained be not equal
4.    The quantity of heat contained in two bodies may be same but temperatures may be different.	4.    The temperature of two bodies may be different but may have the same quantity of heat.
5.    The direction of flow of heat from one body to another does not depend upon the quantity of heat	5.    The direction of flow of heat from one body to another body depends upon
possessed by them.	their temperature difference.
6.    Heat is measured in Joules or in Calories	6.    Temperature is measured in degrees.
7.    During the change of state, the amount of heat in a body changes	7.    During the change of state, the temperature of the body remains constant.

Question 15. Explain the factors on which the quantity of heat of a substance depends.
Answer:

If the change of state of a substance does not occur due to the application of heat, then the quantity of heat of the substance depends on its mass, material, and temperature.

(1) On Mass: The heat absorbed by a body is proportional to its mass. Let two balls of the same material, having a mass of one ball double the other, be immersed in boiling water at the same time. The heat gained by the first ball (whose mass is double) will be more than that of the second ball.

(2) On material: The quantity of heat taken by the body depends on the material of the body.

(3) On temperature: Heat taken by a body is directly proportional to the rise in temperature.

Question 16. Define specific heat (or, specific heat capacity). State its mathematical expression. State its units.
Answer:

Specific Heat: The heat required to raise the temperature of the unit mass of a substance through t°C is known as specific heat or specific heat capacity of the substance.

If Q is the quantity of heat supplied to a body of mass m so that its temperature rises through t? C, then specific heat (S) is given by

⇒ \(S\)=\(\frac{Q}{m t}\), Quantity of heat Q=mst.

Units of specific heat: The S.I. unit of specific heat is Joule per kilogram per degree C ( Joule/kg/ \(C^{-1}\)).

The C.G.S. unit of specific heat is calorie per gram per degree C cal/g °\(C^{-1}\).

Wbbse Physical Science And Environment Class 9 Solutions

Question 17. Explain mathematically that heat lost by the hot body is equal to the heat gained by the cold body if there is no loss of heat due to exchange with the surroundings.
Answer:

Let a substance A of mass \(m_1\), specific heat \(s_1\), and a higher temperature \(t_1\) is mixed with another substance B of lower temperature. Let the mass of the substance B is \(m_2\), specific heat be \(s_2\) and final temperature be \(t_2\)

Let the final temperature of the mixture be t.

Then, fall of temperature of A =\(t_1\)– \(t_2\) and gain of temperature of B =t−\(t_2\),

Heat lost by A = \(m_1\).\(s_1\). (\(t_1\) − t).

Heat gained by B =\(m_2\) \(s_1\)(t – \(t_2\)). If no heat is lost in surroundings then, heat lost by A = Heat gained by B

or,\(m_1\)\(s_1\)(\(t_1\)−t)=\(m_2\) \(s_1\)(t−\(t_2\))

Question 18. What will be the reading in Fahrenheit scale of temperature scale corresponding to 100 C? Sites Careak 392
Answer:

From the relation, \(\frac{C}{5}\)=\(\frac{F-32}{9}\), C=100° C

⇒ \(\frac{100}{5}\)=\(\frac{F-32}{9}\)   or, F−32=20×9

or, F = 180 + 32 = 212° F.

Question 19. At what temperature is the reading on the Fahrenheit scale five times the reading on the centigrade scale?
Answer:

Let the reading on the centigrade scale be x°, then the reading on the Fahrenheit scale = 5x°.

From the relation,\(\frac{C}{5}\)=\(\frac{F-32}{9}\) or, \(\frac{x}{5}\)=\(\frac{5 x-32}{9}\)

or, 25 x − 160 = 9x or, 16 x = 160, or, x =10.

Therefore, reading on a centigrade scale is 10° C and Fahrenheit scale is = 10 x 5 = 50° F.

Question 20. A body of mass 500 gm requires 3000 joules of heat in order to raise its temperature from 35° C to 45° C. Calculate the specific heat of the body.
Answer:

Given, the mass of water (m) = 500 gm = 0.5 Kg

Heat supplied (Q) = 3000 J, and rise in temperature (t) = 45-35 = 10°C.

Let the specific heat of the body =s

Heat supplied = m.s.t

3000 =0°5x s x 10,

or, 5S = 3000, or s = 600.

Hence, specific heat of body = 600 J/Kg/° C

Question 21. An iron ball of mass of 50 gm is warmed in a furnace and then dropped in a vessel containing 240 gm of water at 30° C. The water equivalent of the vessel is 10 g. If the temperature of water rises to 50° C, find the temperature of the furnace. The specific heat of iron = 0.1.
Answer:

Let the required temperature of the furnace be t° C.

Heat gained by water = mass x specific heat x rise in temperature

= 240 x (1)x (50 – 30) = 4800 cal.

Heat gained by vessel = 10 x (50 – 30) = 200 cal.

Total heat gained by water and vessel = (4800 + 200 ) cal = 5000 cal.

Heat lost by iron ball = mass x specific heat x fall in temperature

= 50 x 0.1 x ( t −50) cal.

By the principle of calorimetry,

Heat lost by iron ball = Heat gained by water and vessel

50 x 0.1 x ( t− 50) = 5000

or, t= 1000 + 50 = 1050° C

Hence the temperature of the furnace = 1050° C.

Wbbse Physical Science And Environment Class 9 Solutions

Question 22. If a clinical thermometer reads 98°4°F as the temperature of a healthy person then what would be the reading of the temperature of this person, if measured in Celsius scale?
Answer:

From the relation, \(\frac{C}{5}\)=\(\frac{F-32}{9}\)

C=\(=\frac{5}{9}(F-32)\)

⇒ \(=\frac{5}{9}(98 \cdot 4-32)\)

⇒ \(=\frac{5}{9} \times 66.4\)

= 36.8°C.

Question 23. Calculate the amount of heat given out when the temperature of a piece of iron of mass 60 grams is lowered from 200°C to 100°C. (Specific heat of iron = 0°12)
Answer:

We know that H = mst

Mass of iron = 60 gm.  Where t = rise in temperature

Fall in temperature = 200°C − 100°C   and S = Specific heat = 100°C.

Specific heat of iron = 0.12

∴ H = 60 x 0.12 x 100:

= 720 cal.

∴ The amount of heat given out

= 720 cal.

Question 24. The difference in the readings of a Fahrenheit and a Celsius thermometer is 48°, What are the actual readings shown in the thermometers?
Answer:

Let the actual readings in the Fahrenheit and the Celsius thermometers correspond to F and C.

∴F−C = 48° or, F=C + 48°

Again, from the relation,  \(\frac{C}{5}\)=\(\frac{F-32}{9}\)

⇒ \(\frac{C}{5}\)=\(\frac{C+48-32}{9}\)=\(\frac{C+16}{9}\)

∴ S =20°

Now, F = C + 48° = 20° + 48° = 68°.

∴ The reading in the Fahrenheit thermometer is 68° and that in the Celsius thermometer is 20°.

Question 25. What are the advantages of mercury as a thermometric substance?
Answer:

Advantages of mercury as a thermometric substance :

(1) Mercury being a metal, it absorbs very little heat from the source and thus there is almost no change in the temperature of the source.
(2) Mercury is a shining liquid, so it can be easily seen through glass.
(3) Mercury remains in the liquid state for a long range of temperatures (— 39°C to 357°C). So temperature can be measured over a long range.
(4) Mercury does not wet the glass wall of the tube, so it can move up and down easily through the glass tube.

Question 26. What steps should be taken to make a thermometer sensitive?
Answer:

Requisites of a sensitive thermometer :

(1) The thermometer bulb should have thin walls, so that heat from a body in its contact may easily reach the mercury inside it.
(2) The bore of the thermometer tube should be narrow so that the expansion of the mercury thread is long enough for a small change in temperature.
(3) The bore should be uniform, otherwise, unequal expansion of mercury thread will take place at different parts of the bore for equal changes in temperature.
(4) The thermometric liquid should be good conducting so that it quickly assumes the temperature of the body in contact of the thermometer.

Wbbse Physical Science And Environment Class 9 Solutions

Question 27. Obtain an expression for the heat absorbed or given out by a body.
Answer:

Calculation of the amount of heat absorbed or given out by a body: If S be the specific heat of a substance, then we can write from the definition of specific heat that,

1 unit mass of a substance, for 1° rise or fall of temperature, will absorb or give out ‘S’ units of heat.

m unit mass of a substance, for t° the rise or fall of temperature, will absorb or give out ‘mst’ units of heat.

That is,
Quantity of heat (absorbed or given out)

= mass of the body x specific heat x rise or fall of temperature.

Question 28. What is the relation between Celsius, Fahrenheit, and Kelvin scales?
Answer:

Relation between Celsius, Fahrenheit, and Kelvin scales: Since the range of temperature from lower fixed point to upper fixed point is equal in all three scales, 100 centigrade degrees = (212 — 32) or 180 Fahrenheit degrees = (373 − 273) or 100 absolute degrees.

We consider that the three thermometers in the above three scales are dipped simultaneously in a liquid of a certain temperature. Let the temperatures recorded in the Celsius, Fahrenheit, and Kelvin thermometers respectively be C, F, and K. Now, it can be proved that

⇒ \(\frac{C-0}{(100-0)}\)=\(\frac{F-32}{(212-32)}\)=\(\frac{K-273}{(373-273)}\)

or,\(\frac{C}{100}\)=\(\frac{F-32}{180}\)=\(\frac{K-273}{100}\)

or,\(\frac{C}{5}\)=\(\frac{F-32}{9}\)=\(\frac{K-273}{5}\)

Question 29. State and explain the equivalence of work and heat.
Answer:

From various phenomena, it is possible to consider that there exists a relation between heat and work. In some phenomena, heat transforms to mechanical work and in some, the reverse happens.

In 1798 Rumford noticed the intense heat generated in a steel body that was being hollowed for making barrels of cannon. He concluded that the mechanical work done in the following process was converted into heat energy.

When a football bladder or a bicycle is pumped, the body of the pump is heated. Here, due to the push of the piston, the motion of the air particles inside the pump increases, i.e., their kinetic energy increases.

The kinetic energy of the air particles converts to heat energy. When heat is supplied to a gas confined in a vessel at constant pressure, the volume of the gas increases, which means, the gas particles do work against the pressure. Here, heat is converted to work.

Here, to cause expansion, i.e., to increase the space between the gas particles some heat energy from the gas particles is expanded, the loss of heat causes cooling. Thus, heat is used to do the mechanical work in increasing the separation between the gas particles.

From the above example, it is clear that heat and mechanical work are related to each other. So, if W amount of work produces H quantity of heat or H quantity of heat produces W mechanical work, according to Joule, W ∝ H or W = JH, where J is a constant, called Joule’s constant. J is also called the Mechanical equivalent of heat.

The above said relation, W=JH, J\(=\frac{W}{H}\)

lf H = 1, ie., when unit quantity of heat is produced, W = J. This means J, the mechanical equivalent of heat is equal to the mechanical work that produces a unit quantity of heat. So, the definition of the mechanical equivalent of heat is it is the quantity of mechanical work that can produce a unit quantity of heat.

If the unit of work chosen is erg, and that of heat, calorie, then in the C.G.S. system, the value of J is 4.2 Joule/calorie. Thus, the mechanical equivalent of heat is 4.2 Joule; it means that, if 4.2 Joule work is fully converted to heat, (1)calorie heat is produced.

This value of J has also been confirmed by many experiments.

Question 30. What is latent heat? Classify it and define its different types.
Answer:

The quantity of heat given out or absorbed per unit mass of a substance during the change from one state to another keeping the temperature constant at the temperature at which the change of state occurs is called the latent heat of the substance.

There are different types of change of states and the following are the corresponding latent heats:

Latent heat of fusion: The quantity of heat required to convert the unit mass of a solid substance at its melting point to its liquid state keeping the temperature constant is known as its latent heat of fusion.

The values of latent heat of fusion of ice in C.G.S. and SI systems are 80 calorie/g and 3.36 x \(10^5[l/latex] J/kg respectively.

Latent heat of vaporization: The quantity of heat given out to convert the unit mass of a liquid at its freezing point to its solid state keeping the temperature constant is known as its latent heat of solidification.

Latent heat of condensation: The quantity of heat given out to convert the unit mass of a vapor at its normal boiling point to its liquid state keeping the temperature constant is known as its latent heat of condensation.

Question 31. Distinguish between. saturated and unsaturated vapor.
Answer:

Difference between saturated and unsaturated vapor:

(1) A vapor in a closed enclosure at a given temperature is said to -be saturated if the pressure exerted by it is maximum at that temperature and the pressure exerted by it is called saturation vapor pressure at the temperature. A vapor in a closed enclosure at a given temperature is said to be unsaturated if the pressure exerted by it is less than the maximum value at that temperature.

(2) Saturated vapor does not obey Boyle’s law, while unsaturated vapor obeys Boyle’s law.

(3) Saturated vapor does not obey the pressure law, while unsaturated vapor obeys the pressure law.

(4) A certain quantity of unsaturated vapor can become saturated if the pressure is increased or the temperature is decreased at constant volume.

Question 32. Write a short note on the anomalous expansion of water on marine life.
Answer:

Generally, liquids are found to expand on heating, i.e., their volumes increase with the rise in temperature. But in the case of water, this phenomenon does not hold true within a certain range of temperature.

When water is cooled from room temperature, it contracts uniformly till the temperature reaches 4°C. Below this temperature instead of contraction, water expands with fall in temperature till 0°C, when water freezes to form ice. This phenomenon is called the anomalous expansion of water.

This peculiar behavior of expansion of water has great importance on marine life in cold countries. In these countries when air above the surface of the water of a pond falls below 0°C, the water on the surface, cooling, becomes denser than that below and gradually sinks downwards. It keeps ‘on till the water temperature falls to 4°C. As the surface temperature falls further, it

Becomes less dense than the water below, which is at 4°C, and is the densest. It thus remains at the top, cooling more and more and ultimately a layer of ice is formed there and remains there, as ice is lighter than water.

The layer of ice forms as a poor conducting layer on the top, which prevents the heat from the water below from passing into the colder atmosphere
above.

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

Wbbse Physical Science And Environment Class 9 Solutions

Question 33. When 0.15 kg of ice at 0°C mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Calculate the heat of the fusion of ice. ([latex]S_{\text {water }}\)=4186J\(\mathrm{kg}^{-1} \mathrm{~K}^{-1}\))
Answer:

Let mw = mass of water = 0.30 kg

⇒ \(m_i\)= mass of ice = 0.15 kg

⇒ \(\mathrm{T}_1\)= initial temperature of water.= 50°C

⇒ \(\mathrm{T}_2\)= final temperature of water = 6.7°C

⇒ \(\mathrm{L}_f\)= Heat of fusion of ice

⇒ \(\mathrm{S}_w\) = Specific heat of water = 4186 J\(

⇒ [latex]\mathrm{kg}^{-1} \mathrm{~K}^{-1}\)

Heat lost by hot water \(=m_w s_w\)  \(\left(T_1-T_2\right)\) = (0.30kg) (4186 Jk\(g^{-1}\) (50 – 6.7)°C = 5437614 J

Heat needed to melt ice =\(m_i\)\(\mathrm{L}_f\)= (0.15 kg) (\(\mathrm{L}_f\))

Heat is needed to raise the temperature of ice water

=\(m_i\)\(\mathrm{S}_w\) (\(\mathrm{T}_2\)−0°C) = (0.15 kg) (4186 J\(\mathrm{kg}^{-1} \mathrm{~K}^{-1}\)) (6.7 −0)°C = 4206.93 J

Heat lost = Heat gained

∴ 54376.14 J = (0.15 kg)\(\mathrm{L}_f\)  + 4206.93 J

or, \(\mathrm{L}_f\) = 3.34\(10^5\) J k\(g^{-1}\)

Question 34. 300g lead at 99°C is dropped in 100 g waster at 18°C. The final temperature of the mixture is 27°C. Find the specific heat capacity of lead if specific heat capacity of water is 4200 J k\(g^{-1}\)\(K^{-1}\)
Answer:

Let the specific heat capacity of lead be ‘s’ J/kg K.

Heat lost by 300g lead to cool to 27°C

⇒ \(=\frac{300}{1000} \times 5 \times(99-27) \text { Joules }\)

= 0.3 x s x 72 joules = 21.6s Joules

Heat gained by 100 g water when the temperature rises from 18°C to 27°C

⇒  \(=\frac{100}{1000} \times 42,000 \times(27-18) \text { joules }\)

= 3780 joules.

So, from the principle of calorimetry, heat lost = heat gained

∴21.68 = 3780

or, s\(=\frac{3780}{21.6}\)=\(175 \mathrm{JKg}^{-1} \mathrm{~K}^{-1}\).

Question 35. A piece of copper heated to 35°C is dropped into 140 g of water at 15°C. The mass of copper is 150 g and the specific heat of copper is 400 J/kg K. Find the final temperature. (Specific heat of water = 4200 J/kg.K)
Answer:

Let the final temperature be t°C.

Now, the heat lost by copper.

= 0.15 x 400 x (35 − t) = 60(35 – t) joule.

Heat gained by water

= 0.14 x 4200 x (t − 15) = 588 (t−15) joule.

So, from the principle of calorimetry,

heat lost = heat gained

60(36 − t) = 588 (t −15)

or, 60 x 35 −60t = 588t 588 x 15

or, 60 x 35 + 588 x 15 = 188t + 60t

∴ 648t = 60  35 +588  15

or, t\(=\frac{60 \times 35+588 \times 15}{648}\)

=16.85° C

Question 36. How much heat is required to convert fully 1 kg of ice at 0°C to steam at 100°C? (S.L.H. of fusion of ice = 3.36 x \(10^5\)J/kg, S.L.H. of vapourisation of water = 22.6 x \(10^5\) J/kg and specific heat capacity of water = 4.2 x \(10^3\) J/kg)
Answer:

Heat required to convert 1kg ice at 0°C to 1 kg water at 0°C = 3.36 × \(10^5\) J……….(1)

∴The heat required to raise the temperature of 1 kg water at 0°C to 100°C =1x 4.2 x\(10^3\) x 100 J = 4.2 x\(10^5\) J……….(2)

Heat required to convert (1)kg water at 100°C to 1 kg steam at 100°C = 22.6 x  x\(10^5\) J………… (3)

∴ The total quantity of heat required to convert 1 kg ice at 0°C to (1)kg steam at 100°C

= (1) + (2) + (3) = (8.36 + 4.2 + 22.6) x \(10^5\) J

= 3016 x \(10^5\)  J.

Question 37. 300 g lead at 99°C is dropped in 100 g water at 18°C. The final temperature of the mixture is 27°C. Find the specific heat capacity of lead if the specific heat capacity of water is 4200 J K\(g^{-1}\)\(K^{-1}\)
Answer:

Let the specific heat capacity of lead be ‘s’ J/kgK.

Heat lost by 300 g leads to cool from 99°C to 27°C

= \(\frac{300}{1000}\) xs x (99-27) J=0.3 xS x 72 J= 21.6 SJ.

Heat gained by 100 g water when its temperature rises from 18°C to 27°C

=\(\frac{100}{1000}\) x 42000 x (27 −18) J = 3780 J.

From the principle of calorimetry, heat lost = heat gained

or, 21.6 s= 3780

∴ S= \(\frac{3780}{21.6}\)  Jkg-1K-(1)= 175 Jk\(g^{-1}\)\(K^{-1}\).

WBBSE Solutions for Class 9 Physical Science And Environment

• Chapter 1 Measurement
• Chapter 2 Force And Motion
• Chapter 3 Matter: Structure And Properties
• Chapter 4 Matter: Atomic Structure; Physical & Chemical Properties of Matter
  • 4.1 Atomic Structure
  • 4.2 Concept Of Mole
  • 4.3 Solution
  • 4.4 Acids, Bases & Salts
  • 4.5 Separation Of Components Of Mixtures
  • 4.6 Water
• Chapter 5. Energy In Action: Work, Power & Energy
• Chapter 6 Heat
• Chapter 7 Sound

 

Physical Science Class 9 WBBSE Chapter 1 Measurement Very Short Answer Type :

Question 1. What do you mean by mass?
Answer:

Mass : The mass of a body is the quantity of matter contained in it.

Question 2. What is the function of ordinary scale?
Answer:

Function of ordinary scale: We generally use an ordinary scale for the measurement of length.

Question 3. What is solar day?
Answer:

Solar Day: The interval of time between two consecutive apparent transits of the sun across the meridian at any place is called a solar day.

Read and Learn all WBBSE Solutions for Class 9 Physical Science And Environment

Question 4. What is the definition of second in the SI system?
Answer:

Definition of a second in SI system: The time taken for 9, 192, 631, and 770 vibrations of a radioactive atom CS133 in a standard magnetic field is called one second.

Question 5. What do you mean by volume?
Answer:

Volume: It is defined to be the space occupied by a substance. Solids have three dimensions, i.e., length, breadth, and height.

Wbbse Class 9 Physical Science Solutions

Question 6. What is liter?
Answer:

Litre: A volume of 1 kg of pure water at 4°C or 277 K is called a liter.

Question 7. What is spring balance?
Answer:

Spring balance: With the help of this instrument, the weight of a body can be measured.

Question 8. Is the density of a material dependent on its mass or volume?
Answer:

The density of a substance is independent of its mass or volume.

Question 9. In which unit is the wavelength of X-rays measured?
Answer:

The wavelength of x-rays is measured in angstrom (A) unit.

Question 10. The yellow light of sodium has a wavelength of 589 nanometres. Express it in meters and Angstrom.
Answer:

589 nanometre = 0.000000589 metre = 0.0000000589 Angstrom.

Question 11. Give an example of a physical quantity that has no unit. Write the dimension of velocity.
Answer:

A physical quantity that ‘has no unit is the Mechanical Advantage of a machine. The dimension of velocity is \(\mathrm{LT}^{-1} .\)

Question 12. Name two quantities having the same unit.
Answer:

Velocity and speed have the same unit, e.g. m/s in the SI system.

Wbbse Class 9 Physical Science Solutions

Question 13.1 A=__cm,1 micron =___ cm. [Fill in the blanks]
Answer:

⇒ \(10^{-8}\), \(10^{-4}\)

Question 14. What is the unit of measuring the distance between the stars?
Answer:

Light year is the unit used for measuring the distance between the stars.

Question 15. Which physical quantity is normally measured with a common balance?
Answer:

Mass is normally measured with a common balance.

Question 16. Name the suitable units that should be chosen to measure

(1) Distance between two cities
(2) Mass of a small grain
(3) Distance between two stars
(4) Atomic radius
(5) Volume of water contained in an overhead tank. (Each part of the question carries )
Answer:

(1) Kilometre
(2) Milligram
(3) Light year
(4) Fermi
(5) Kilolitre.

Question 17. Why is the temperature 4°C mentioned in defining the density of water in the CGS system?
Answer:

Because at 4°C, water has maximum density, which is 1g/cc.

Wbbse Class 9 Physical Science Solutions

Question 18. What is the necessity of measurement of a physical quantity?
Answer:

Measurement of a physical quantity gives a clear idea about the largeness or smallness of the quantity. Simple eye estimation does not give any exact magnitude of the physical quantity, so measurement of a physical quantity is very important.

Question 19. Define light year.
Answer:

A light year is the distance traveled by light in one year at the rate of 3 x \(10^8\) meters per second.

Question 20. Name a scalar quantity which is the product of two vector quantities.
Answer:

The scalar quantity is work which is the product of two vector quantities force and displacement in the direction of force.

Question 21. Name a physical quantity that has both magnitude and direction but it is not a vector quantity.
Answer:

The quantity is electric current.

Question 22. Name a physical quantity that has a unit but not a dimension.
Answer:

Plane geometric angle.

Question 23. What is a cubic foot?
Answer:

It is the volume of a cube on the edge of one foot.

Question 24. What is the relation between a cubic foot and a gallon?
Answer:

One cubic foot = 6.25 gallons.

Wbbse Class 9 Physical Science Solutions

Question 25. What is the physical quantity?
Answer:

Physical Quantity: A physical quantity is any measurable quantity of an object or an event.

Example: mass, length, time, weight, density, etc.

Question 26. Define scalar quantity.
Answer:

Scalar quantity: The quantity which has only magnitude but no direction is called a scalar quantity.

Question 27. Define vector quantity.
Answer:

Vector Quantity: The quantity which has both magnitude and direction is called a vector quantity

Wbbse Physical Science And Environment Class 9 Solutions

Question 28. What is the unit?
Answer:

Unit: In measuring any physical quantity, some convenient and definite quantity of it is taken as the standard and in terms of this standard the physical quantity is measured. This standard is called a unit.

Question 29. Mention one importance of units.
Answer:

Importance of units: Measurement of any physical quantity without a unit is meaningless because we cannot have any idea about a physical quantity with its magnitude only.

Question 30. What is fundamental unit?
Answer:

Fundamental Unit: The units of physical quantities which are independent of each other and from which other units can be derived are called fundamental units.

Question 31. What is a derived unit?
Answer:

Derived unit: The units of physical quantity which are derived with the help of one or more than one fundamental unit are called derived units.

Example: The unit of area is obtained by using the unit of length twice. Similarly, units of speed, force, work, etc. are all derived units.

Question 32. What is accuracy?
Answer:

Accuracy: The closeness of the measured value to the true value of the physical quantity. is known as the accuracy of measurement.

Wbbse Physical Science And Environment Class 9 Solutions

Question 33. What is precision?
Answer:

Precision: It means the extent or limit to which the measurement of a physical quantity is done

Physical Science Class 9 WBBSE Chapter 1 Measurement 2 Marks Questions And Answers:

Question 1. Is every physical quantity a unit?
Answer:

No, some physical quantities are measured by the ratio of two identical quantities, e.g. atomic weight, specific heat, specific gravity, etc. These physical quantities have no units, they are pure numbers.

Question 2. What are the characteristics of fundamental units?
Answer:

Characteristics of fundamental units :

(1) They do not depend on each other.
(2) They can form the units of other physical quantities.
(3) They are only three in number.

Question 3. What are the types of different systems of units? y
Answer:

Different systems of units: At present two different systems of units are used in science throughout the world.

(1) CGS or Metric system.
(2) SI or international systems of units (1960).

Question 4. What is the CGS system?
Answer:

CGS or French or Metric system: This system originated in France but now it is used in scientific measurements and is widely used throughout the world including India. From the initial letters of the words centimeter, gram, and second, this system is called the CGS system. This is a decimal system.

Question 5. What is the SI system of units?
Answer:

SI or International System of Units: In 1960, an international system of units of measurement known as the SI system was recommended by the General Conference on Weights and Measures to have a consistent system of units and to simplify communications among scientists.

Question 6. Define meter.
Answer:

Metre: The distance between the two marks made on a platinum-iridium (90: 10) bar maintained at 0°C temperature preserved at the International Bureau of Weights and Measures at Sevres near Paris, is considered as one meter (symbol: m).

Question 7. What is the modern concept of meter?
Answer:

Modern concept of meter (October 1983): The distance traveled by light in;1/3x \(10^8\)sec in vacuum is called 1 meter.

Wbbse Physical Science And Environment Class 9 Solutions

Question 8. What do you mean by mass?
Answer:

Mass: The mass of a body is the quantity of matter contained in it.

Question 9. State the bigger units of mass.
Answer:

Bigger units of mass:

(1) Quintal (= 100 kg)
(2) Metric ton or tonne (= 1000 kg = 10 Quintal)
(3) CSL (= 1.4 x mass of the sun).

Question 10. State the smaller units of mass.
Answer:

Smaller units of mass :

(1) a.m.u. (= 1:6603 x \(10^{-24}\)g)
(2) 1 carat = 200 mg

Question 11. What is density? State its CGS and SI units.
Answer:

Density: The density of the material of a body is the quantity of its mass per unit volume.

Unit of density in CGS system: g \(\mathrm{cm}^{-3}\)

Unit of density in SI system: kg \(\mathrm{m}^{-3}\)

Wbbse Physical Science And Environment Class 9 Solutions

Question 12. What do you mean by solar day?
Answer:

Mean solar day: The length of the solar day varies from day to day owing to various reasons and the mean value of the actual solar days averaged over a full year is called the mean solar day.

Question 13. What are the advantages of the metric system?
Answer:

Advantages of the metric system:

(1) In the metric system, multiples and sub-multiples of the units of length and mass are always ten times the preceding ones.

(2) In the metric system, units of length, mass, and volume are related by the same relation.

(3) In the metric system, the same prefixes (such as milli, centi, deci, etc.) are used in the units of length and mass.

Question 14. Define dimension.
Answer:

Definition of dimension: Dimensions of a derived unit are the powers to which the fundamental units of mass, length, and time must be raised to represent that unit.

Wbbse Physical Science And Environment Class 9 Solutions

Question 15. Define dimensional expression.
Answer:

Definition of dimensional expression: Dimensional expression is a product or quotient of the symbols of fundamental physical quantities involved in a derived physical quantity raised to appropriate powers.

Example: Dimensional expression of force : \(\left[\mathrm{MLT}^{-2}\right]\)

Question 16. State physical quantities without dimension.
Answer:

Physical quantities without dimension: A physical quantity is expressed as a ratio between two physical quantities of the same unit, it becomes a number only, which will not have dimension.

Example: Mechanical advantage of a machine, specific gravity, refractive index, etc. are dimensionless. Again, angle, although has a unit (radian), it has no dimension.

Question 17. How can you measure the length of a curved line?
Answer:

Measurement of the length of a curved line: If the length of a curved line is to be measured with the help of a scale, we make use of a thread and place it along the curved line and measure the length of the straightened thread with the help of a scale.

Question 18. Why are scales made usually with wood or plastic?
Answer:

Scabies are made usually with wood or plastic: In making an ordinary scale wood or plastic is usually used instead of metal because, with the change in atmospheric temperature, metal scales change in length.

A metal scale gives correct readings only at a temperature at which it has been graduated. At any other temperature, it would give incorrect measurements. But in the case of wood or plastic, these changes are negligible. So the scales are made usually with wood or plastic.

Question 19. What is common balance?
Answer:

Common balance: In the laboratory, we measure the mass of a body with a common balance. The weighing balance used in our daily lives is a simple version of this common balance.

Question 20. What is the working principle of measurement of mass?
Answer:

Working principle of measurement of mass: We find the mass of a body by comparing its mass with some standard weights. Generally, the measurable object is kept on the left pan of the balance and the standard weights are placed on the right pan. When the balance beam is horizontal, then.

Mass of the objective= Mass of the standard weights

Question 21. What is the working principle of the measurement of weight?
Answer:

Working principle of the measurement of weight: The spring balance should be kept vertical and the pointer should coincide with the ‘0’ mark of the scale. When a weight is suspended from the hook, the spring elongates, elongation being proportional to the weight.

So, the pointer attached to the spring moves over the graduated scale. By noting its position against the scale, we can directly measure the weight of a body.

Question 22. What is a measuring cylinder?
Answer:

Measuring cylinder: The instrument by which the volume of a liquid is measured is known as measuring cylinder.

Question 23. What points should be noted during measuring the volume of a liquid
Answer:

Points to be noted while measuring the volume of a liquid

(1) The volume of a liquid cannot be measured by measuring a cylinder that reacts with glass.

(2) The volume of a liquid cannot be measured by a cylinder which is very volatile.

Question 24. How can you measure the density of an irregularly shaped body by using a common balance and measuring cylinder?
Answer:

Density: The density of a substance is defined as mass per unit volume of the substance.

If M is the mass of the substance, V is the volume, then the density D is given by D = M/V…..(1)

The mass of the irregularly shaped body is determined by a common balance and volume is measured by a measuring cylinder. From equation (1) density of the body can be determined.

Wbbse Physical Science Class 9 Chapter 1

Question 25. How will you determine the volume of a solid by a measuring cylinder?
Answer:

Determination of the volume of a solid by a measuring cylinder: To find the volume of the solid, some liquid in which the solid sinks but does not dissolve is taken in a dry measuring cylinder. The reading of the upper meniscus of the liquid in the cylinder is noted, let it be \(\left(V_1\right)\) ml.

The solid substance tied with a waxed thread is now slowly immersed in the liquid. Let the reading of the upper meniscus of the liquid in a measuring cylinder be\(\left(V_2\right)\)ml. So, the volume of the given solid is \(\left(V_2-V_1\right)\)ml.

Question 26. What is measurement?
Answer:

Measurement: Measurement is the determination of the size or magnitude of something. , The comparison of an unknown quantity with some standard quantity of the same type is known as measurement.

Question 27. What are the basic units to measure the physical quantities?
Answer:

The basic units are shown in the following table.

Question 28. Define Astronomical Unit.
Answer:

Astronomical unit (AU): It is the mean distance between the sun and the earth. It is used in measuring astronomical distances.

AU = 1.5 x \(10^{13}\) m

Question 29. Why is temperature mentioned in the unit of length but not in the unit of mass?
Answer:

The length is measured by a scale made of metals, plastic, wood gataparcha, etc. The length of these substances varies with temperature difference. So in determining the unit of length, temperature is to be specified to get the exact length. On the other hand, mass is an intrinsic property of matter, it does not vary with the change of temperature. So, it is not necessary to specify the temperature.

Wbbse Physical Science Class 9 Chapter 1

Question 30. If the mass and volume of a piece of iron be 760 g and 100 cm, then what would be its density?
Answer:

Density of iron = mass of iron/volume of iron

= \(\frac{760 \mathrm{~g}}{100 \mathrm{~cm}^3}\)=7.6g/\(\mathrm{cm}^3\)

Question 31. State two rules for writing the S.I. units.
Answer:

The following guidelines are followed while writing the units of physical quantities.

(1) The symbol for a unit that is not named in honor of any scientist is written in small letters.

Example: The symbol! for meter is ‘m’.

(2) The symbol for a unit that is named in honor of any scientist is written with an initial capital letter.

Example: The symbol for a unit of force (newton) is N.

Question 32. When is a common balance called defective?
Answer:
A defective common balance: A common balance is called defective if

(1) the arms of the balance are of unequal length.

(2) the scale pans are of unequal weight.

(3) the center of gravity is not exactly below the fulcrum of the balance when the beam is horizontal.

Question 33. Why are the masses of weights in a weight box taken in the ratio of 5: 2 : 2: 1?
Answer:

The masses of the standard weights in a weight box in gm are 100g, 50g, 20g, 20, 10g, 5g, 2g, 2g, and 1 gm and in milligrams are 500, 200, 200, 100, 50,20, 20 and, 10, i.e in the ratio of 5 : 2: 2: This is because, with this ratio of the masses, the mass of any body within 211.11 gm can be measured with a common balance without the need of any larger weight.

For example, if the mass of a body is 56.86 gm., its mass can be measured by placing standard weights of (50 + 5 + 1) gm and (500 + 200 + 100 + 50 + 10) mg in the right-hand pan of the balance.

Question 34. How will you measure the volume of tea in a teacup?
Answer:

The volume of tea in a teacup can be conveniently measured with the help of a narrow measuring cylinder whose volume is graduated in centimeters and decimals of centimeters.

The total amount of tea is poured inside the cylinder and the reading of the upper level of tea in the cylinder gives the volume of tea.

Question 35. If the bus fare per mile is changed into per kilometer, find whether a Passenger pays more or less than the actual fare if he travels a distance of 10 miles and pays at the rate of a kilometer.
Answer:

As the rate of fare is equal both in the mile and kilometer and 10 miles being equal to 10 x 1.60 km = 16 km, the Passenger will pay more if he pays at the rate of kilometer.

Question 36. In a ration shop, one metric ton of rice is sold per week. How many kilograms of rice are sold per week?
Answer:

1 metric ton = 1000 kilograms.

So, 1000 kilograms of rice are sold per week.

Wbbse Physical Science Class 9 Chapter 1

Question 37. Which is greater in length a table of 2 meters in length or a table of 2 yards in length? Give a reason for your answer.
Answer:

1 (one) metre = 39.37 inches
2 metres     = 39.37 x 2 inches
                     = 39.37 x 2/12 x 3 yards
                     = 2.186 yards.
As 2 meters = 2.186 yards, hence a table of the length of 2 meters is greater than a table of the length of 2 yards.

Question 38. Which is cheaper—rice at Rs. 2 per kg or rice at Rs. 1 per pound?
Answer:

1 (one) kg = 2.2056 Ibs.

Therefore, the cost of 2.2046 Ibs of rice at the rate of Rs. one per pound = Rs. 2.204. Again, the cost of 1 kg of rice at a rate of Rs. 2.00 per kg = Rs. 2.00. Hence, rice of 2 rupees per kg is cheaper than that of one rupee per pound.

Question 39. Which one will be profitable to you—kerosene costing rupees forty per gallon or kerosene costing ten rupees per liter? Give a reason for your answer.
Answer:

We know,

1 gallon = 4.536 litres.

Now if a gallon of kerosene is purchased at the rate of liter, it costs rupees 45.36 (Rs.4.536 x 10), but if it is purchased at a gallon rate, it costs rupees forty only. Hence, it is profitable to purchase at a gallon rate.

Question 40. What is the mass of 100\(\mathrm{cm}^3\)? of a liquid whose density is 1.2 gm/cc?
Answer:

We know, density = mass/volume
or, Mass = volume x density.
Hence, volume = 100 cubic centimeters.
Density = 1.2 gm/c.c.
Mass = 100 x 1.2 = 120 gm.

Question 41. The mass and volume of a piece of iron are 510 gm and 60 c.c. respectively. What is the density of the iron piece? Express the value also in S.I. unit.
Answer:

Solin. Density (d) =\(\frac{\text { mass }(M)}{\text { volume }(V)}\)=\(\frac{510 \mathrm{~g}}{60 \mathrm{v} \cdot \mathrm{c}}\)=8.5/c.c

Density of iron in S.I. unit = 8.5\(\times 10^3 \mathrm{~kg} / \mathrm{m}^3\)

Question 42. How is the volume of a regularly shaped body determined?
Answer:

To find the volume of a regularly shaped body its dimensions such as length, breadth, height, radius, etc. are measured.

Then use the following relations:

Volume of cuboid = length x breadth x height

Volume of a cube =\((\text { side })^3\)

Volume of sphere =\( \frac{4}{3} \pi \text { (radius) }^3\)

Question 43. Is light year a fundamental unit or a derived unit? Arrange in descending order of magnitude—angstrom, femtometre, micron, nanometre.
Answer:

A light year is a fundamental unit since it denotes length. Arranged in descending order: micron, nanometer, angstrom, femtometre.

Question 44. Why does one not express his age in seconds?
Answer:

A person does not express his age in seconds because, with such a small unit of time, the number expressing his age will be very large which does not give one a clear idea about the age.

Wbbse Physical Science Class 9 Chapter 1

Moreover, in calculating age in seconds and then expressing it with a large number, some more seconds will pass off which do not come into the calculation. Thus, the correct age of one cannot be expressed in seconds.

Question 45. Is the density of a material dependent on its mass or volume?
Answer:

The density of a substance is independent of its mass or volume. If the mass of a substance is increased or decreased, its volume correspondingly increases or decreases proportionately and vice versa, so density, the ratio of mass to volume, remains unchanged.

Question 46. What precautions should be taken while determining the volume and density of a solid body with the help of a common balance?
Answer:

To find the volume and thus density of a body, the following points should be kept in mind :

(1) The body must sink completely in the liquid taken in the measuring cylinder.

(2) The body must not react chemically with the liquid and also it must not dissolve in the liquid.

(3) While dipping the body in the liquid taken in the measuring cylinder, it should be done carefully so that no liquid splashes out of the cylinder.

Question 47. The mass of a proton is 1.67 x 10 Kg. How many protons would be required to make 1 kg?
Answer: Number of protons = \(\frac{1 \mathrm{~kg}}{1.67 \times 10^{-27} \mathrm{~kg}}    =\frac{10^{27}}{1.67} \)

=\(6 \times 10^{27}\) approx.

Question 48. Light of a certain color is composed of a train of waves each of 0.3048 microns long. How many of these are there in a meter?
Answer:

Number of waves    =\(\frac{1 \mathrm{~m}}{0.3048 \times 10^{-6} \mathrm{~m}}\)=\(\frac{1 \times 10^6}{0.3048}\)

=\(33 \times 10^5\) approx.

Wbbse Physical Science Class 9 Chapter 1

Question 49. Mankind has existed for about 10° years, whereas the universe is about 10 years old. if the age of the universe is taken to be one day, for how many seconds has mankind existed?
Answer:

⇒ \(10^{10}\) years is taken as equivalent to 24×60×60 s.

so, yrs is equivalent to \(\frac{24 \times 60 \times 60 \times 10^6}{10^{10}}\)s.

⇒\(\frac{24 \times 9}{25} s\)

∴ \(8.64 \mathrm{~s}\)

Question 50. What are the characteristics of fundamental units?
Answer:

Characteristics of fundamental units:

(1) They do not depend on each other.
(2) They can form the units of other physical quantities.
(3) They. are only three in number.

Question 51. What is the working principle of measurement of weight?
Answer:

Working principle of measurement of weight: The spring balance should be kept vertical and the pointer should coincide with the ‘0’ mark of the scale. When a weight is suspended from the hook, the spring elongates, elongation being proportional to the weight.

So, the pointer attached to the spring moves over the graduated scale. By noting its position against the scale, we can directly measure the weight of a body.

Wbbse Class 10 Physical Science 

Question 52. Why is the unit of area not called a fundamental unit?
Answer:

The area is the product of two lengths (length and breadth) whose unit is a fundamental unit. So unit of area has been obtained by multiplying the two fundamental units.

Question 53. Which will be of greater density—a kilogram of iron or a kilogram of cotton? Give a reason for your answer.
Answer:

If the volume of a kilogram of cotton is larger than that of a kilogram of iron, the density of cotton will be less than that of iron.

Question 54. How can a physical quantity be classified? :
Answer:

A physical quantity can be classified into two broad categories:

(1) Scalar quantity
(2) Vector quantity.

Question 55. Express a light year in a megametre.
Answer:

Light year is the distance traveled by light in one year at the speed of 3 x \(10^8\) VMS.
One megametre = \(10^6\) m.
Distance traveled by light in one year (865 days) in megametre (Mm)
\(\frac{3 \times 10^8 \times 60 \times 60 \times 24 \times 365}{10^6}\)Mm=9.4608×\(10^9\)  Mm

Question 56. Arrange in ascending order of magnitude; 1 micron, 1 angstrom, 1 fermi, 1 millimeter, 1 nanometre.
Answer:

1 micron     = \(10^{-6}\)m
1 angstrom = \(10^{-10}\)m
1 fermi (or femtometre)= \(10^{-15}\)m
1  millimetre = \(10^{-3}\)m
1 nanometer = \(10^{-9}\)m.

The required ascending order is 1 fermi, 1 angstrom, 1 nanometre, 1 micron, 1 millimeter.

Question 57. Light of a certain color is composed of a train of waves, each of wavelength 6355 angstrom. How many such waves are present in an inch or 2.54 centimeters?
Answer:

1 angstrom = \(10^{-8}\)mcm

∴ 6355 angstrom.= 6355 x \(10^{-8}\)cm. Since, 1 inch = 2.54, cm.

∴  The required number of waves=\(\frac{2.54}{6355 \times 10^{-8}}\)=\(40 \times 10^3\)

Question 58. In the CGS system density of water is 1 g/cc. Find the density of water in SI.
Answer:

Density of water =1 g/cc=1g/1\(\mathrm{cm}^3\)
=\(\frac{1 g}{(1 \mathrm{~cm})^3}\)=\(\frac{\frac{1}{1000} \mathrm{~kg}}{\left(\frac{1}{100} \mathrm{~m}\right)^3}\)=\(\left(\frac{1}{1000} \times \frac{1000000}{1}\right) \mathrm{kg} / \mathrm{m}^3\)

=\(1000 \mathrm{~kg} / \mathrm{m}^3\)

 Therefore,Density of water in the SI system is \(1000 \mathrm{~kg} / \mathrm{m}^3\)

Wbbse Class 10 Physical Science 

Question 59. In a faulty balance, the lengths of the left and right arms are respectively 11 cm and 10.5 cm. The apparent mass of a body determined by the balance is found to be 22 g; find the true mass of the body.
Answer:

The standard mass of 22 g is kept in the right pan. So, from the principle of balance, mass on the left pan x length of left arm = mass on the tight pan x length of right arm, or m X 11 = 22x 10.5
m=\(\frac{22 \times 10.5}{11}\)=21

So, the true mass is 21 g.

Question 60. The mass of an empty bucket of 10 I Capacity is 1 kg. Find its mass when it is completely filled with a liquid of density 800 \(\mathrm{kg} / \mathrm{m}^3\)
Answer:

1 =I \(\mathrm{dm}^3\)=8kg;

Mass of 10 I of the given liquid=\(\frac{800 \mathrm{~kg}}{(10 \mathrm{dm})^3}\times 10 \mathrm{dm}^3\)

So, the mass of the bucket with the liquid (8+1) or 9 kg.

Question 61. The dimension of a physical quantity is given as [M][L]?(T}2. Name the quantity. What is its SI unit? Give reasons for your answer.
Answer:

\([\mathrm{M}][\mathrm{L}]^2[T]^{-2}\)=M\(\left[\frac{\mathrm{L}}{\mathrm{T}}\right]^2 \)  = mass x velocity?.

=2x½× mass ×velocity? = 2 x kinetic energy. So, the given quantity is twice the kinetic energy. The SI unit of K.E. is Joule, for a unit of kinetic energy is the same as the unit of work.

Question 62. Deduce the dimension of power.
Answer:

Power=\(\frac{\text { work }}{\text { time }}\)=\(\frac{\text { force } \times \text { distance }}{\text { time }}\)=force x velocity

=mass x acceleration \(\times \frac{\text { distance }}{\text { time }}\)

=Mx Lx \(\mathrm{T}^{-2} \times  \mathrm{LT}^{-1}\)=\(\mathrm{ML}^2 \mathrm{~T}^{-3}\)

Question 63. Find the value of 1 unified atomic mass unit (a.m.u.) in kg.
Answer:

1 a.m.u=\(\frac{1}{12} \text { of a } C_{12} \text { isotope. }\)

\(1 \mathrm{ml}^{12} \mathrm{C}\)=\(12 \mathrm{~g}^{12} \mathrm{C}\)

\(\text { i.e., } \mathrm{N}_{\mathrm{A}} \text { number of }{ }^{12} \mathrm{C}\)=12g\(\left[N_A=\text { Avogadro number }\right]\)

Therefore, 1 a.m.u =\(\frac{1}{12} \times \frac{1}{N_A} \times 12 g\)=\(\frac{1}{N_A} g\)=\(\frac{1}{1000 N_A} \mathrm{~kg}\)=\(1.66 \times 10^{-27} \mathrm{~kg}\)

Question 64. Calculate the angle of (a) 1° (degree) (b) 1° (minute of arc or arc minute) (c) 1” (second of arc or arc second) in radians
Answer:

360°=2π rad, 1°=60′, 1′=60″.

(1)360°=2π rad=
∴ 1°=\(\frac{2 \pi}{360} \times \mathrm{rad}\)=\(1.745 \times 10^{-2}\)rad

(2) 1°=60°=\(1.745 \times 10^{-2}\)rad
∴ 1′=\(\frac{1.745}{60} \times 10^{-2} \mathrm{rad} \simeq 2.91 \times\)rad

(3) 1′=60″=2.91\(2.91 \times 10^{-4}\)rad
∴ 1″=\(\frac{2.91}{60} \times 10^{-4}\)rad=\(4.85 \times 10^{-6}\)rad

Question 65. Express the numbers in exponential notation :
(a) 563
(b) 0.0012
(c) 43,900,000
(d) 0.0000007190
Answer:

(a) 5.63 x \(10^2\)
(b) 1.20 x \(10^{-3}\)
(c) 4.39 x \(10^7\)
(d) 7.19 x \(10^{-7}\)

Question 66. How many protons would be required to make 1 kg? m, = 1.67 107 kg.
Answer:

Number of protons(n)× \(m_p\) =1 kg
∴n=\(\frac{1}{1.67 \times 10^{-27}}\) =\(5.99 \times 10^{26}\)

Question 67.
(1) How many millimeters (mm) are there in a kilometer (km)?
(2) How much larger is a millisecond (ms) than a nanosecond (ns)?
(3) A microwatt (UW) is what fraction of a kilowatt (kW)?
(4) A gigaparsec (GPsc) is how many kiloparsecs (kpsc)?
(5) What fraction of a kilovolt (kV) is a millivolt (mV)?
Answer:

(a) \(10^6\)
(b)\(10^6\)
(c)\(10^{-9}\)
(d) \(10^6\)
(e) \(10^6\)

Question 68. If 325 sheets of paper make a stack 2.54 cm high, what is the thickness of a sheet of a single paper?
Answer:

The thickness of a single sheet of paper \(=\frac{2.54 \mathrm{~cm}}{325}\)=\(7.8 \times 10^{-3}\)cm.

Question 69. There are 120 small divisions on a stopwatch dial. What is the least count of the watch?
Answer:

1 Complete rotation =60 s
i.e., 120 division =60 s
∴ divisions= \(\frac{60}{120}\) =0.5 s
∴Lest count=0.5 s.

Wbbse Class 10 Physical Science 

Question 70. The length of an object was measured to be 30 cm by a linear scale of least count 0.1 cm. What is the % error?
Answer:

 % error=\(\frac{\text { least count }}{\text { measured value }} \times 100\)=\(\frac{0.1}{30} \times 100\)=0.33 \%

Question 71. A piece of iron has a volume of 500 mi and a mass is 4kg. Find the density of the object.
Answer:

Density =\(\frac{\text { mass }}{\text { volume }}\)=\(\frac{4 \mathrm{~kg}}{500 \mathrm{ml}}\)\(=\frac{4000 \mathrm{~g}}{500 \mathrm{ml}}\)=\(8 \mathrm{~g} / \mathrm{mL}\)

Question 72. 100 g of water is added to a 100 cc solution of sugar of density 1.5 g/cc. What is the density of the new mixture?
Answer:

Mass of sugar = density x volume = 1.5 g/cc x 100 cc = 150 g.
Assume 100 g water = 100 cc water (density of water = 1 g/cc).
∴ Total volume = 100 + 100 = 200 cc, mass of solution = 150 g + 100 g = 250 g.
∴ Density \(=\frac{250 \mathrm{~g}}{200 \mathrm{cc}}\)=1.25g/cc

Question 73. Find the dimension of force.
Answer:

Force = mass x acceleration =\(\text { mass } \times \frac{\text { change in velocity }}{\text { time }}\)

\(=\frac{\text { mass } \times(\text { final velocity }-\text { initial velocity )}}{\text { time }}\)

\(=\frac{m \times \frac{L}{T}}{T}\)=\(\frac{M L}{T^2}\)=\(\left[M L^2 T^{-3}\right]\)

Question 74. Power [P] is the rate of doing work. Find the dimensional equation of power.
Answer:

power\(=\frac{\text { work }}{\text { time }}\)\(=\frac{\text { force } \times \text { displacement }}{\text { time }}\)

\([P]=\frac{\left[M L T^{-2}\right] \times[l]}{[T]}\)=\(\left[\mathrm{MLT}^{-2}\right]\)

Question 75. Specific gravity is the ratio of two densities. Show that it has no dimension.
Answer:

Specific gravity of a substance \(=\frac{\text { Density of the substance }}{\text { Density of water at } 4^{\circ} \mathrm{C}}\)

\(=\frac{M L^{-3} T^0}{M L^{-3} T^0}\)=\(\left[M^0 L^0 T^0\right]\)

So, Specific gravity is dimensionless.

Wbbse Class 10 Physical Science 

Question 76. Kinetic energy is given by amv’ and potential energy by mgh. Show that both of them have the same dimensions.
Answer:

Kinetic energy \(=\frac{1}{2} m v^2\)=\((M) \times\left(L T^{-1}\right)^2\)\(=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)

Potential energy = mgh =(M)\(×\left(\mathrm{LT}^{-1}\right)^2\)=\(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)
So, kinetic and potential energies have the same dimensions.

WBBSE Class 9 Physical Science Solutions Chapter 1 Measurement 3 Marks Questions And Answers:

Question 1. State different smaller units of length in the SI system.
Answer:

Smaller units of length in the SI system :

(1) Femtometre or Fermi (\(=10^{-15}\) m) (fm)

(2) Picometre ( \(=10^{-12}\) m) (pm)

(3) X-ray unit (  \(=10^{-13}\) m) (Xu)

(4) Angstrom ([late]=10^{-10}[/latex] m) (A)

(5) Nanometre (\(=10^{-9}\) m) (nm)

(6) Micrometre or micron (\(=10^{-6}\) m)

(7) Centimetre (\(=10^{-2}\) m) (cm)

(8) Decimetre ( \(=10^{-1}\) m) (dm)

Wbbse Class 10 Physical Science 

Question 2. What are the bigger units of length?
Answer:

Bigger units of length:

(1) Astronomical unit (AU): The average distance between the earth and the sun.

1 AU = 14.95 X \(=10^8\) km

(2) Light year: The distance that a light wave travels in the year in a vacuum.

1 light year = 9.46 x \(=10^{12}\) km

(3) Parsec: The biggest unit for the measurement of distance. 1 parsec = 3-26 light

years = 30.84 x \(=10^{12}\) km.

Question 3. What are the requisites of a good balance?
Answer:

Requisites of a good balance :

(1) The balance must be true, i-e., the balance beam should be horizontal with no loads or with equal loads on the pans.

(2) The balance must be sensitive.

(3) The balance must be stable, i.e., the balance beam must come quickly to its equilibrium position once it is displaced.

(4) The balance must be rigid, i.e., all the parts of the balance must be strongly built.

Wbbse Class 10 Physical Science 

Question  4. What is a sensitive balance? What are the conditions to be a sensitive balance?
Answer:

Sensitive balance: A balance is said to be sensitive when even for slight differences in weights on the two pans the beam should be tilted from its horizontal position.

Conditions for a balance to be sensitive :

(1) The balance beam should be light with long arms.
(2) The center of gravity of the beam should be near the fulcrum.

Question  5. What is the concept of measurement?
Answer:

Measurements are an integral part of the human race, without them, there would be no trade, no statistics.

Scales are used to measure. One would know a simple ruler or tape could be used to measure small distances. The current system of small units has three standard units meter, kilogram, and second. These three units form the MKS system or the metric system. A meter is a unit of length, currently defined as the distance light travels within 1/299782458th of a second.

A kilogram is a unit of mass. While it was previously defined as a specific volume of water (e.g. 1 liter or a 10 cm? cube), its current definition is based on a prototype platinum-iridium cylinder.

A second is a unit of time. Originally defined as the amount of time the earth needs to make 1/86400 of a rotation, it is now defined as 9192631770 oscillations of a Cesium-133 atom. The bigger units of measurement are astronomical units (AU).

Question  6. Define scalar quantities and vector quantities. Give examples.
Answer:

Scalar quantities: The physical quantities which have only magnitude but no direction are called scalar quantities.

Examples: mass, distance, time, volume, speed, work, energy, power, density, electric current, etc.

Vector quantities: The physical quantities which have magnitude, as well as direction, are called vector quantities.

Examples: displacement, velocity, acceleration, momentum, force, weight, electric field, magnetic field, etc.

Question  7. What do you mean by the unit of a physical quantity? All physical quantities do not have units, explain it.
Answer:

The known standard constant quantity which is used for comparison of a physical quantity is said to be a unit.

For example, the length of an iron rod is 5 meters. It means the unit of length is meter (m) and this unit is contained 5 times the length of that iron rod. Hence,

Physical quantity = numerical value x unit.

All physical quantities do not have units. Those physical quantities which are expressed as the ratio of the same quantities do not have units.

For example, atomic weight of elements, specific gravity, and refractive index.

Question  8. What observations should be taken for the choice of a unit?
Answer :

The selection of a unit to measure a physical quantity should have the following properties :

(1) The unit chosen should be of convenient size.

(2) The unit should be well-defined.

(3) The unit should not change with space, time, and temperature.

(4) It should be reproducible.

Question  9. Define fundamental and derived units. Give examples.
Answer:

Fundamental units: The unit that is independent of any other unit or which can neither be changed nor be related to any other unit is called a fundamental unit.

Examples: Units of length, mass, time, temperature, current, luminous, intensity, etc. are fundamental units.

Derived units: The units which depend on the fundamental units or which can be expressed in terms of the fundamental units are called derived units.

Example: The units of velocity, acceleration, energy, force, momentum, etc. are the derived units.

The velocity of a moving body is obtained by dividing the distance traveled by the body by the time taken. When distance is expressed in the unit of length as centimeters and time in seconds, then

a unit of velocity = unit of length/unit of time = cm/second = cm/sec.

Question  10. Define the C.G.S. system and S.I. system of measurement of units.
Answer:

C.G.S system: In this system, the unit of length is centimeter (cm), a unit of mass is gram (g) and the unit of time is second (s).

S.I. System: In this system units of seven physical quantities have been considered as fundamental units.

In the S.1. system following are considered as the fundamental units :

Question  11. Define the S.I. units of length, mass, and time.
Answer:

The unit of length is meter (m), the mass is kilogram (kg) and that of time is second (s). They are defined as

Metre (m): A meter is defined as 1,553,164.1 times the wavelength of the red line in the spectrum of cadmium.

Kilogram (kg): A kilogram is the mass of a cylindrical piece of platinum-iridium alloy kept at the International Bureau of Weights and Measures at Sevres near Paris.

Second (s): A second is defined as. 1/86400 is the part of a mean solar day.

Question  12. Explain the reason for using different smaller and larger units for measuring a particular quantity.
Answer:

To measure a particular quantity we use both smaller units and larger units as per requirement. For example, to measure the length of a room, we use meter (m), to measure the distance between two cities we use kilometer (km) units.

For the measurement of the distance of a star from the earth, the commonly used unit is light year because a kilometer is considered a very small unit in this case. To measure the length of a very small quantity we consider smaller units of length as centimeters, millimeters, microns, etc.

For example, to measure the thickness (or diameter) of a wire, we use centimeters or millimeters. In this case unit taken as meter is absurd. If the thickness of a paper measured in millimeters (mm) is 1 mm, then expressed in meters is 0.001 m, which looks odd.

Question  13. What do you mean by the dimension of a physical quantity?
Answer:

Generally, the dimension of a physical quantity represents the powers to which the fundamental unit (or units) has to be raised to obtain the derived unit of the physical quantity.

The dimensions of length, mass, and time are expressed as L, M, and T respectively. Dimensions generally represent the nature of the units only. They do not indicate the magnitude of the units.

For example, the volume of a cube L’ units

\begin{equation}
\text { Dimension of density of a cube }=\frac{\text { Dimension of mass of a body }}{\text { Dimension of volume of a body }}
\end{equation}

\begin{equation}
=\frac{M}{L^3}=M L^{-3}
\end{equation}
Dimension of force = dimension of mass x acceleration.

Question  14. State the different measuring devices to measure the different physical quantities.
Answer:

To measure the different physical quantities, the commonly used measuring devices are the following :

(1) Ordinary scale (or meter scale): Used to measure the length of an object.

(2) Common balance:  Used to measure the mass of a body.

(3) Spring balance:  Used to measure the weight of a body.

(4) Measuring cylinder:  Used to measure the volume of a liquid. It is also used to measure the volume of a solid by displacement method.

(5) Clock:  Used for measuring time.

(6) Stopwatch: Used for measuring a time interval between two events such as race, sports, etc.

Question  15. What is ordinary scale? Describe the method of measuring length with the help of an ordinary scale.
Answer:

An ordinary scale is a measuring device used to measure the length of an object. Ordinary scale is generally made up of wood, metal, or plastic. It is graduated in centimeters and millimeters. 1 centimeter is divided into 10 equal parts, each part is 1 millimeter.

To measure the length of a line say AB, we place the scale in such a way that it remains perpendicular to the line AB. Now note the readings of the starting point A and the end point B from the scale. Then,

Length of line AB = Reading of end B – Reading of end A. Let the starting point, A of the line AB at 2 cm, coincide with the scale. If the endpoint, B of the line does not coincide exactly with the graduation of the scale, the reading has to be taken by the eye approximation method. While taking the reading eye should be placed

directly above point B. In the figure given below the correct position of the eye should be at E. If we place our eye sideways, say\(E_1 \text { or } E_2\) then it leads to an error called parallax error. S

Let the end B coincides with the scale at 6.5 cm. Therefore, the length of the line AB = 6.5 cm- 2cm = 4.5 cm.

Question  16. What is density? State its units. How is the density of an irregular body insoluble in water measured?
Answer:

Density: The density of a substance is defined as its mass per unit volume. Density is a scalar quantity and is represented by the letter d. If the mass of a body is M gm, and the volume of the body is V cubic cm, then

d = M/V. Its C.G.S. unit is gm per \(\mathrm{cm}^3\), SI unit is kg/\(m^3\)

To find the density of an irregular body, first find the mass of the body by common balance. Now find the volume of the body insoluble in water by measuring the cylinder. By dividing the mass of the body by volume, we get the density of the body.

Question  17. What is a common balance? How will you measure the mass of a body with the help of a common balance? 
Answer:

Common balance: Common balance is an instrument used to measure the mass of a body. By common balance, we find the mass of a body by comparing its mass with some known standard weights.

Generally, we put the body on the left pan of the balance whose mass is to be determined, and the standard weights are placed on the right pan. When the beam of the balance becomes horizontal (or the pointer rests at zero scale) then the mass of the body is equal to the standard weight.

The different parts of a common balance are shown in the figure given below :

Different parts of common balance:

1- plumb line
2 – levelling screws
3- arrester
4- pans
5- beam
6- adjusting nuts
7-pointer
8- scale

A horizontal rigid beam is placed on a vertical pillar. Two scale pans (marked 4 in figure) of the same mass are suspended on the two ends of the beam supported by two strips placed on knife edges at the two ends.

The centre of the beam is kept on the fulcrum which is also attached with a pointer and a plumb line. The pointer moves over a scale fixed on the pillar when the pillar is raised with the help of a knob. The correct weight is known with the help of the pointer and the scale.

To adjust the balance, it is first leveled by leveling screws (marked 2 in the above figure). The horizontal beam is also adjusted by the two adjusted nuts (marked 6 in the figure).

Question  18. What is the volume of a body? How will you measure the volume of a given liquid by measuring devices?
Answer: The space occupied by a body is called its volume. The volume of a liquid in the laboratory is measured by the following measuring devices:(1) Measuring cylinder
(2) Measuring flasks
(3) Burette
(4) Pipette

A measuring cylinder is commonly used to measure the volume of liquids. It is a glass cylinder of uniform radius. The outer wall is graduated in cubic centimeters and millimeters from the bottom upwards. It is available in different sizes.

Before measuring the volume of a ‘liquid through a measuring cylinder, it is cleaned and kept fixed vertically. The liquid whose volume has to be measured is poured into it and the reading of the upper layer of the liquid is noted. This reading is the volume of the given liquid.

Question  19. What is a spring balance? How will you measure the weight of a body with the help of the spring balance?
Answer:

Spring balance is a type of balance used to measure the weight of a body. It consists of a spring kept in a metal tube. It is partially exposed in front of the metal tube. The Graduate upper end of the spring is rigidly fixed. The lower end of the engram spring is attached to a metal rod.

One hook is attached with a Pointer to the lower end of the metal rod. The body whose weight is to be determined is hanged from this hook. A pointer attached to Spring spring moves on a graduated scale (usually in grams or kilograms). The position of the pointer indicates the weight of the body.

Question  20. Describe a method to find the volume of an irregular solid heavier than and insoluble in water, with the help of a measuring cylinder.
Answer:

To find the volume of an irregularly shaped solid body, we take a graduated cylinder. Fill the cylinder partly with water. Note the reading of the lower level of water, say x, c.c. Now tie the given solid with a thin string and lower it gently into the cylinder so that the solid is completely immersed in water.

The level of water rises again. Note the new reading of the raised water level (say, \(x_2\) C.C.). Since the body displaces water equal to its volume, the volume of the solid = second reading first reading. Thus, the volume of solid \(=\left(x_2-x_1\right)\)cubic centimeters.

Question  21. How will you measure time?
Answer:

Time is measured by a wall clock, wristwatch, and table clock. To measure a fraction of time, the stopwatch is used.

Clock :
In our day-to-day life, time is measured by a pendulum clock, whose functioning is based on the oscillation of a pendulum. In this type of clock, a metal rod is suspended from a rigid support. A heavy metal ball is suspended from the end of the rod. The time taken by the metal ball in one oscillation is regulated as 1 second.

It consists of a circular dial which is divided into 60 equal divisions. Each division represents 1 minute. It is further divided into 12 equal parts, where each part represents 1 hour. Hours are marked from 1 to 12. In this clock, the long arm is the arm of minute and the short arm is the arm of hour.

Spring is wound up with the help of a key in an old type of clock and by a battery in digital clocks or battery-operated clocks. The potential energy stored in the spring is converted to kinetic energy and it rotates the arms of the clock.

Stopwatch:

A stopwatch is used to measure the time interval between two events. it consists of a circular dial which is divided into 60 equal divisions. Each division represents 1 second. A long light needle called the second’s hand, pivoted at the center, rotates over the circular scale.

A big circular dial also contains a small circular dial to read the time in minutes. The smaller dial graduated in 30 or 60 divisions and the small needle called the minute’s hand rotates over the scale. When the second’s hand completes one rotation, the minute’s hand moves through one division or 1 minute.

A knob is provided at the top of the frame for winding the watch. By pressing the knob in succession, the watch can be started and stopped and both the needles can be brought to the starting point for further use.

Question  22. Distinguish between scalar and vector quantities mentioning three points of difference.
Answer:

Question  23. How can you measure (1) the thickness of a page of a book, (2) the diameter of a piece of hair or thread, and (3) the length of a segment of a zigzag line, with the help of a ruler?
Answer:

(1) To determine the thickness of a page of a book :

The thickness of about, say, 25 pages of the book together is measured with a ruler. Dividing this thickness by the number of pages chosen, the thickness of a page is calculated.

The same procedure is followed for different numbers of pages at different sections of the book. The average of these gives the required thickness of a page found out.

(2) To find the diameter of a fine thread or a hair :

For this, a long piece of thread or hair is taken. The piece is then coiled over the millimeter marks on the scale. The coiling should be such that each turn remains in close contact to its adjacent one.

Thus the added diameters of the segments of thread in all the turns cover a few millimetres of the scale which is noted. New dividing the length on the scale by the number of turns, the diameter of the thread or hair is determined.

(3) To measure the length of a curved or a zigzag line:

A piece of thread is taken help of. A required portion of the thread is very carefully laid over the given line. Now, the length of the straightened part of the thread that was used just to cover up the given line is measured with the help of a ruler, which gives the length of the given line.

Question 24. How can you determine the density of a liquid using a common balance and a measuring cylinder?
Answer:

A clean and dry measuring cylinder is placed on the left pan of a good common balance and it is counterpoised with the mass m, gram, say. Now, some quantity of the given liquid is poured cautiously inside the measuring cylinder upto a certain level and the level is noted avoiding parallax error; let it be V cc.

Now mass of the cylinder with the liquid in it is also determined and let it be m, gram. So, the mass of the liquid taken in the measuring cylinder is (m, – m,) gram.

∴ Density of the liquid \(=\frac{\text { mass }}{\text { volume }}\)=\(\frac{\left(m_2-m_1\right) \text { gram }}{V c c}\)

\(=\frac{m_2-m_1}{V} \mathrm{~g} / \mathrm{cc}\)

Question  25. State three differences between common balance and spring balance.
Answer:

The three differences are :

(1) Common balance measures the mass of a body, and spring balance measures the weight of a body

(2) Common balance will show the same value of the mass of a body at different parts of the earth; spring balance will show different weights of a body at different parts of the earth

(3) In a common balance, several standard masses have to be placed in steps on one scale pan to counterpoise with the mass of a given body placed on another scale pan. It is a time-consuming process. But in a spring balance, the weight of the given body can be known instantaneously.

Question  26. How can you measure the exact mass of a solid by a common balance with unequal arm length? 
Answer:

Suppose the length of the left and right arm of the balance are |, and |, respectively. Now, keeping the solid with a mass of m gram at the right-hand pan it is counterpoised with a mass of w, gram at the left-hand pan.

Now,\(w_1 \times I_1\)=\(m \times I_2\)

\(\text { or, } \frac{l_1}{l_2}\)=\(\frac{m}{w_1}\)…………(1)

Again, the solid of mass is kept at the left-hand pan, and let us suppose it is counterpoised with a load of mass w, at the right-hand pan.

So,\(m \times I_1\)=\(w_2 \times I_2\)

\(\text { or, } \frac{I_1}{I_2}\)=\(\frac{w_2}{m}\)………….(2)

Now, from (1) and (2), we get

\(\frac{m}{w_1}\)=\(\frac{w_2}{m}\)

\(\text { or, } m_2\)=\(w_1 w_2\)

\(\text { or, } m\)=\(\sqrt{w_1 w_2}\)

Therefore, the correct mass \(=\sqrt{\text { Product of counterpoising   loads}}\)

Physical Science And Environment Class 9

Question  27. The mass of a piece of gold in the air is 193. g. In a measuring cylinder, there is 50 cm? of water. When the piece of gold is immersed in water in the measuring cylinder, the reading of the volume of water is found to be 60 cm. What is the density of gold?
Answer:

From the problem we get,

Volume of the piece of gold \(=(60-50) \mathrm{cm}^3\) \(=10 \mathrm{~cm}^3(\mathrm{~N})\)

We know, density \(=\frac{\text { mass }}{\text { volume }}\)

So, density of gold \(=\frac{\text { mass of gold }}{\text { volume of gold }}\)
\(=\frac{193 \mathrm{~g}}{10 \mathrm{~cm}^3}=19.3 \mathrm{~g} / \mathrm{cm}^3\)

Question  28. The volume of a can is 100 ml and it weighs 20g. It is then filled fully by a liquid and weighs 180 g. Find the density of the liquid.
Answer:

The volume of the liquid (v) = 100 ml

Mass of the liquid (M) = (180 – 20) g = 160g

So, density of the liquid \(=\frac{\text { mass of the liquid }}{\text { volume of the liquid }}\)
\(=\frac{160 \mathrm{~g}}{100 \mathrm{~cm}^3}\)=\(1.6 \mathrm{~g} / \mathrm{cm}^3\)

Question  29. The density of a liquid is 1°2 g/\(\mathrm{cm}^3\). What will be the mass of 100 \(\mathrm{cm}^3\) of the liquid?
Answer:

Volume of the liquid \(=100 \mathrm{~cm}^3\)

Density of the liquid \(=1^{\prime} 2 \mathrm{~g} / \mathrm{cm}^3\)

We know the mass of liquid  = volume of the liquid x density of the liquid

= (100×1.2)g= 120g.

Question  30. The density of silver in a CGS unit is 10°5 g/cm®. What will be the density of silver in SI unit?
Answer:

From the problem we get,

Density of silver in CGS unit \(=10^{\circ} 5 \mathrm{~g} / \mathrm{cm}^3\)

So, the density of silver in SI unit

\(=10^{\circ} 5 \times 1000 \mathrm{~kg} / \mathrm{m}^3\) \(=10500 \mathrm{~kg} / \mathrm{m}^3\)

Question  31. The radius of a solid sphere is 10 cm. The density of its matter is 8 g/\(\mathrm{cm}^3\). What is the mass of the sphere ?.
Answer: We know,

Volume of a sphere V \(=\frac{4}{3} \pi r^3(r=\text { radius of the sphere) }\)

V\(=\frac{4}{3} \times \frac{22}{7} \times(10)^3 \mathrm{~cm}^3\)          r=10cm

Density of its matter \(=8 \mathrm{~g} / \mathrm{cm}^3\)

So, mass of sphere = volume x density

\(=\frac{4}{3} \times \frac{22}{7} \times(10)^3 \times 8 \mathrm{~g}\)=33.51 kg.

Question  32. The mass of a piece of solid substance is 20-52 g. When the piece is immersed in water its mass becomes 12-48 g. What will be its volume and density?
Answer:

Mass of displaced water by the body = (20.52 – 12.48) g = 8.04 g.

Volume of displaced water \(=\frac{8.04}{1 \mathrm{~g} / \mathrm{cc}}\)

So, the volume of the solid substance =  8.04 c.c.

Question  33. Define liter and gallon and find the relation between them.
Answer:

Litre is the space occupied by the mass of one kilogram of water at 4°C. The volume of 1 liter may also be considered as the space of a cube of length, breadth, and height 10 cm each, occupied by a gas or a liquid.

Thus, 1 litre = 1.0 cm x 10 cm x 10 cm \(=1000 \mathrm{~cm}^3\)

1 litre = 1000 cubic centimeters.

Gallon is the volume of 10 pounds of water at 62°F

1 gallon = 4,536 litres.

Question  34. State the limitation of an ordinary scale for the measurement of length. Why should the zero mark of the scale not be used while measuring length by a meter scale?
Answer:

Limitation of an ordinary scale :

A meter scale can measure length correctly only up to 1 mm (or 0.1 cm), which is the least count of a meter scale. For more accurate measurement, a vernier scale is to be used. Besides, an ordinary meter scale is not useful for the measurement of the diameter of rods, wire, spheres, internal and external diameters of hollow cylinders, etc.

For these, an instrument called a slide caliper is to be used. For measurement of the diameter of very thin objects like hair, or thread, an instrument called a screw gauge is usually used Measurement should be made from the graduation mark other than the ‘0’ mark at one end of the scale to avoid error due to wear and tear of this end.

Question  35. How would you measure the volume of an irregular solid that can not be introduced into a measuring cylinder given to you?
Answer:

If the measuring cylinder is not wide enough for the solid to enter into it, the value of the solid can be measured with the help of an overflow jar which consists of an iron vessel provided with a spout.

The jar is filled with water up to the point of overflowing. The given measuring cylinder is placed under the spout of the overflow jar. The solid is carefully and slowly lowered into the overflow jar.

The solid displaces its volume of water and this water flows out from the spout into the measuring cylinder. The volume of water collected in the measuring cylinder is recorded. This is also the volume of the given solid.

Physical Science And Environment Class 9

Question  36. What will be the mass of 454 liters of water?
Answer:

Mass = volume x density

Density of water = 1 gm/cc.

Now, 1 gm \(=\frac{1}{1000} \mathrm{~kg} .\)=\(10^{-3} \mathrm{~kg}\)

454 litres = 454 x \(10^{3} \)c.c.

A mass of 454 liters of water

= 454 x \(10^{3} \) x1gm

= 454 x \(10^{3} \) x\(10^{-3}

= 454 kg.

Question  37. The length, breadth, and height of a water reservoir are 2 m, 1m, and 1m respectively. How many liters of water it can hold?
Answer:

The volume of the water reservoir = length x breadth x height

=2m x 1m x 1m

=2 x 10dm x 1 x 10dmx 1 x 10dm

= 2000 cubic decimetre

So the volume of water in the reservoir

= 2x [latex]10^{3} \) cubic decimetre

=2x \(10^{3} \) litre

(∴ cubic decimetre = 1000 cubic centimetre = 1 litre)

Question  38. What are the bigger units of length?
Answer:

The bigger units of length are kilometer (km), light year, and Astronomical Unit (A.U.).

1 Kilometre (km) = 1000 m (or\(10^{3} \) m)

A light year is the distance traveled by light in a vacuum in one year.

1 Light year = Speed of light x 1 year

= 3 x \(10^{8} \) m/s x 1 year

=3 x \(10^{8} \) m/s (865 x 24×60 60)s =9.46 x \(10^{12} \) km

Astronomical Unit (AU) is the mean distance of Earth from the sun.

1 A.U. = 1.496 x\(10^{11} \) metre.

Question  39. What are the smaller units of length?
Answer:

The S.I. unit of length is 1 meter. Its smaller units are centimeters (cm), millimeters (mm), micro, and nanometer (nm)

1 Metre =100cm or, 1cm\(=1 / 100 \mathrm{~m}\)

1 Metre = 1,000 mm_ or, 1 mm \(=1 / 1000 \mathrm{~m}\)

1 Metre = 10,00,000 micron or, 1 Micron =\(10^{-5} \)m

1 Nanometre=\(10^{-9} \)m.

Question  40. What is the utility of a unit? Do all physical quantities have units?
Answer:

Utility of a unit: Measurement of a physical quantity is just a process of comparing it with a well-defined standard quantity which is of similar nature to that of the quantity to be measured. This standard quantity, known as a unit, is chosen before measuring a physical quantity.

Now, with a preconception about the magnitude of the chosen unit, a comparison of a certain quantity with it gives a clear idea about the measured quantity. Hence, without

A unit, any measurement is meaningless:
There are some physical quantities, each of which is expressed by a ratio of similar physical quantities, so they have no units.

Examples: Mechanical advantage of a machine, solubility, refractive index, atomic weight, etc.

Question  41. What are the advantages of the metric system?
Answer:

Advantages of the metric system:

(1) In the metric system, multiples and sub-multiples of the units of length and mass are always ten times the preceding ones.

(2) In the metric system, units of length, mass, and volume are related by the same relation.

(3) In the metric system, the same prefixes (such as milli, centi, deci, etc.) are used in the units of length and mass. 2

Question  42. What precautions should be taken while determining the volume and density of a solid body with the help of a common balance?
Answer:

Precautions :

(1) The body must sink completely in the liquid taken in the measuring cylinder.

(2) The body must not react chemically with the liquid and also it must not dissolve in the liquid.

(3) While dipping the body in the liquid taken in the measuring cylinder, it should be done carefully so that no liquid splashes out.

Question  43. State the relationship between
(1) Inch and centimetre
(2) Kilometre and mile
(3) Kilogram and pound
(4) Pound and gram.
Answer:

(1) One Inch = 2.54 Centimetre
(2) One Kilometre = 0.6213 Mile
(3) One Kilogram = 2.2046 Pounds (Ibs.)
(4) One Pound = 453.56 grams.

Question  44. Define the units of the following physical quantities using the fundamental units : area, volume, density, velocity, acceleration, force, work, momentum, power and pressure.
Answer:

Question  45. State the applications of dimensional analysis.
Answer:

Applications of dimensional analysis :

(1) Addition or subtraction can be done with only those physical quantities which have the same dimensions. ‘

Total energy \(=\text { kinetic energy }\left(\frac{1}{2} m v^2\right)\)+Potential energy

Each of energy terms-has the same dimension =\(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)

(2) Identical dimensions can be canceled in the numerator and the denominator.

Acceleration :\(\frac{\text { Velocity }}{\text { Time }}\)=\(\frac{\left.\mid M^0 L T^{-2}\right]}{[T]}\)=\(\left[M 2 L T^{-2}\right]\)

Pressure :\(\frac{\text { Force }}{\text { Area }}\)=\(\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}\)=\(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)

(3) Every correct formula or equation must have the same dimension on both sides of the equation.

Question  46. State the utility of dimensional analysis.
Answer:

Dimensional analysis helps us in deducting certain relations among different physical quantities.

It is of much help in checking the derivation, accuracy, and dimensional consistency of mathematical expressions.

Examples: Deducing a relation

The period of oscillation (T) of a simple pendulum depends on its length (1), mass (m), and g.

Let, T\(=k l^a g^b m^c ; k\)= dimensionless quantity; a, b, c = exponents

\(\left[L^0 M^0 T^1\right]\)=\([L]^a\left[L T^{-2}\right]^b\left[M^c\right]^c\)\(L^{a+b} \cdot T^{-2 b} \cdot M^c\)
On equating the dimensions on both sides, we have
a+b=0,-2b=1,c=0, or b=-½,va=-b=½,c=0

T\(=k l^a g^b m^c ; k\)=\(\left.k\right|^{\frac{1}{2}} g^{\frac{1}{2}} \cdot m^0\)=\(k \sqrt{\frac{1}{g}}\)

Actually, k is observed to be 2π 

T=\(2π \sqrt{\frac{1}{g}}\)

Examples: Dimensional Consistency :

Check whether the equation\(\frac{1}{2} m v^2\)  = mg is dimensionally correct or not.

L.H.S.:  \([M]\left[L T^{-1}\right]^2\)=\(\left[M L^2 T^{-2}\right]\),     R.H.S: \([\mathrm{M}]\left[\mathrm{LT}^{-2}\right][\mathrm{L}]\) =\(   \left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)
L.H.S. = R.H.S., the equation is dimensionally correct.

Physical Science And Environment Class 9

Question  47. Write a note on the different types of errors in measuring physical quantities.
Answer:

Errors :

Least count, Range, Absolute, Relative, and Percentage: Measurement errors, which are, in general, of two types :

(1) Systematic error
(2) Random error.

Systematic error is of three types :

(1) Instrumental error,
(2) Experimental error and
(3) personal error.

They can be minimized or eliminated by improving techniques or experimental procedures, opting for a better instrument, and removing personal error by more careful measurement. unpredictable fluctThe type (b) is irregular and random.

They may arise due to random mutations in experimental conditions. Temperature variation, mechanical vibration, and voltage fluctuation are the possible reasons.

Least count error :

The smallest value that can be measured by an instrument is called its least count. All the readings are. good only upto this value. The least count error is the error associated with the resolution of the instrument. It belongs to the random error category but within a limited size.

A meter scale has the least count of 1 mm or 0.1 cm, and a Vernier caliper has the least count of 0.01 cm. We can improve the least count error by selecting an instrument of higher precision or by repeating the observations several times and then taking the mean or average value.

Absolute and relative error :

Let us consider n number of measurements giving values, \(m_1, m_2, \ldots m_k \ldots m\), and the true value is \(m_{\text {true }}\)

Then mean value \(m_{\text {mean }}\) \(=\left(m_1+m_2 \ldots+m_k+\ldots+m_n\right) / n\)\(=\sum_i m_i / n\)

Absolute error\(\left|\Delta m_k\right|=m_k-m_{\text {mean }}\),k=1,2,………n

Mean absolute error,\(\Delta m_{\text {mean }}\)=\(\sum_k\left|\Delta m_k\right| / n\)

The range of a single measurement \(=m_{\text {mean }} \pm \Delta m_{\text {mean }} \text { i.e. }\)\(=m_{\text {mean }} \pm \Delta m_{\text {mean }}\)
or,\(m_{\text {mean }}-\Delta m_{\text {mean }} \leq m \leq m_{\text {mean }}+\Delta m_{\text {mean }}\)

i.e., any measurement of the physical quantity m is likely to lie between \(\left(m_{\text {mean }}+\Delta m_{\text {mean }}\right)\) and (\(\left(m_{\text {mean }}-\Delta m_{\text {mean }}\right)\)

Relative error =\(=\Delta m_{\text {mean }} / m_{\text {mean }}\)

Percentage error = Relative error 100% \(=\frac{\Delta \mathrm{m}_{\text {mean }}}{\mathrm{m}_{\text {mean }}} \times 100\)

Note that | Δm | is always positive but Δm may be both (+) ve.

WBBSE Class 9 Physical Science Solutions

48. In 5 successive measurements, the values of the period of oscillation of a simple pendulum are obtained as T, = 2.63, T, = 2.56, T, = 2.42, T, = 2.71 and T, = 2.80 (ins). Demonstrate the different errors and range of measurement.
Answer:

\(T_{\text {mean }}=\sum_i T_i / 5\)=(2.63+2.56+2.42+2.71+2.80)/5

\(=\frac{13.12}{5}\) =2.624 s[Mean value of T]

\(\left|\Delta T_i\right|\)=|2.63-2.62|,|2.56-2.62|,|2.42-2.62|,|2.71-2.62|,|2.80-2.62|

=0.01,0.06,0.20,0.09,and,0.18,(s)[absolute errors]

\(\mathrm{T}_{\text {mean }}=\sum_i\left|\Delta \mathrm{T}_{\mathrm{i}}\right| / 5\)=(0.0+0.06+0.20+0.09+0.18)/5=0.54/5

= 0.11 s [mean absolute error]

Range of measurement : \(T_{\text {mean }} \pm \Delta T_{\text {mean }}\)=2.62±0.11s

Relative error \(=\Delta T_{\text {mean }} / T_{\text {mean }}=\frac{0.11}{2.62} \times 100\)=4%

The least count (or the minimum value), the maximum value measurable, and the relative error of a few simple instruments are listed in the following table. Question  49. State the characteristics. of the units of measurement.
Answer:

Characteristics of the units of measurement :

(1) The unit should be well-defined.

(2) The unit should be of considerable size.

(3) The unit should be easily reproducible.

(4) The unit should be imperishable. ‘

(5) The unit should be constant with respect to time or with physical conditions like pressure, temperature, etc.

Question  50. State the different types of variables and constants based on dimensions.
Answer:

Different types of variables and constants :

(1) Dimensional constants :

The quantities which have dimensions but which are of constant value are called dimensional constants.

Example: Gravitational constant, Planck’s constant.

(2) Dimensionless constants :

The constant quantities having no dimensions are called dimensionless constants.

Example: Pure Numbers 1, 2, 3 … 1, π (= 2.718).

(3) Dimensional variables :

The quantities which have dimensions but do not have any fixed value are called dimensional variables.

Example: Volume, Velocity, Force, etc.

(4) Dimensionless variables :

The quantities which have neither dimensions nor any constant value are called dimensionless variables.

Example: Angle, Specific Gravity, Strain, etc.

Question  51. State the limitations of dimensional analysis.
Answer:

Limitations of dimensional analysis :

(1) This method fails to determine the dimensionless constants in the formula.

(2) If a physical quantity depends on more than three factors having dimensions, the formula cannot be determined.

(3) This method cannot be used to derive a relation involving trigonometric or exponential or log functions.

(4) This method fails to derive an exact form of a relation when it consists of more than one part on any one side.

(5) It gives no information regarding whether a physical quantity is scalar or vector.

(6) Even when dimensions are given, the physical quantity may not be unique, as many physical quantities have the same dimensions.

Question  52. Mention some points to be noted while writing SI units.
Answer:

The following points may be noted while writing SI units :

(1) Punctuation marks like full stops, commas, etc. are not written after the symbols of the units. (e.g. we should write cm and not cm.).

(2) The symbols of the units are always used in singular form (i.e., we write 1m, 5 m, etc., and not 5 ms).

(3) The name of a unit is never written in the initial capital letter, even if it is named after some person. (e.g. we write ampere, joule, Newton, etc. and not as Ampere, Joule, Newton, etc.).

(4) If a unit is named after a person, its symbol is written in capital letters but the symbols of other units are not written in capital letters (e.g., we write newton (N), joule (J), metre (m), kilogram (kg), etc.).

(5) When temperature is expresed in kelvin scale, (°) degree sign is not used, i.e., we write 300 K and not 300°K.

(6) Units like metre per-second, etc. are written as m/s or \(\mathrm{ms}^{-1}\)

Question  53. Describe how time is measured by clocks and stopwatch.
Answer:

Clocks and Stop Watch : Time is measured by a clock and small time interval is measured by stopwatch. They are described below

The Pendulum clock : In this clock a simple pendulum is used. At a particular place, a pendulum of a given length takes a fixed time to complete one oscillation, i.e., starting from one extreme position and coming back to that position. This time is called the time period of the pendulum clock.

With the help of a coiled spring a gear (a toothed wheel) is rotated. A second pendulum regulates the speed of this gear being connected with it. Due to this regulated rotation of the gear, the hands of the clock being attached with it turn over a graduated circular dial and indicates the time. The spring requires to be wound after regular interval.

 

The Stop watch : This watch resembles a pocket watch in appearance. On pressing a knob the watch starts and it stops when the knob is pressed for the second time. From the positions of the hands the time interval is obtained. This watch can measure a time interval of one-tenth of a second accurately. The watch is used in sports, scientific work, etc. for measuring time.Question  54. Define the units of area, volume, density and temperature scales in different systems.
Answer:

Units of Area, Volume, Density, and Temperature scales :

(1) The units of area: of a surface in CGS, MKS or SI systems are correspondingly square centimeters (\(\mathrm{cm}^2\), square meter \(\left(m^2\right)\)

(2) Units of volume: of a solid body in CGS, MKS or SI systems are cubic centimeters (cc) and cubic meter (\(\left(m^3\right)\)) respectively.

For measuring volume of a liquid, the units usually used are litres or gallons.

Smaller units of litre are decilitres, millilitres of a cubic centimeter (cc).

10 decilitres = 1 litre; 1000 millitres (ml) = 1 litre or 1000 cubic centimetre (cc) = 1 litre.

(3) Units of density : Density of the material of a body is defined us the quantity of its

Mass per unit volume. Thus ,density \(=\frac{\text { mass }}{\text { volume }}\)

So, the unit of density in CGS is gm/cc or gm \(\left(m^{-3}\right)\), that in MKS or SI is kg/\(\left(m^3\right)\) or Kg \(\left(m^{-3}\right)\). Density of a substance is not dependent on its mass or volume.

If the mass of a substance is increased or decreased, its volume correspondingly increases or decreases proportionately and vice versa, so density, the ratio of mass of volume, remains unchanged, provided temperature remains constant.

For this reason, density of a solid golden bar and that of a solid golden ball is same, although the said bar andthe ball may have different masses, and hence, corresponding different volumes.

(4) Temperature scales : In CGS system, temperature is measured in Celsius scale. A temperature in Celsius scale is written with °C at the end of a number that indicates the magnitude of the temperature. In SI system the temperature scale used is known as Kelvin or Absolute scale. To write any temperature in this scale the degree sign is not written. For example, 300K but not 300°K.

Question  55. Define Ampere, Candela and Mole.
Answer:

Definition of Ampere, Candela, Mole : These three are considered as fundamental physical quantities in SI.

(1) Ampere : Symbol :A. It is the constant current which, flowing in two straight parallel infinite conductors of negligible cross-section placed one metre apart-in vacuum, produces 2 X \(10^{-7} \mathrm{Nm}^{-1}\) force between the conductors. The force is attractive or repulsive accordingly as the currents in the conductors are in same or opposite directions.

(2) Candela: Symbol: Cd. It is the radiant intensity of \(\frac{1}{683}\)watt per steradian (unit of solid angle) in a given direction sent by a source that emits monochromatic radiations of frequency 540 x 1012 Hz.

(3) Mole: Symbol: mole. Mole is a counting unit for chemists and it is the number 6.023

A mole stands for Avogadro’s number of items.
1 mole of atoms                 =  6.023   \(10^{23}\) atoms
1 mole H Molecules           =  6.023   \(10^{23}\) H Molecules
1 mole\(\mathrm{CO}_2\) Molecules       = 6.023  \(10^{23}\)\(\mathrm{CO}_2\) Molecules
1 mole  eletrons                 =  6.023   \(10^{23}\) eletrons
1 mole  rupees                   =  6.023    \(10^{23}\) rupees .

Question  56. How can we measure the area of an irregularly shaped lamina?
Answer:

Measurement of an irregularly shaped lamina : Take a graph paper of 1 m\(m^2\) or \(\frac{1}{100}\) area of each of its small squares. Place the irregular lamina on the graph paper and draw its outline.

Now count the number of complete number of squares inside the outline drawn. In the counting process, the number of squares more than half part of which are within the outline is also added. Let the total number of squares counted as stated above be x. So, the area of the lamina is x m\(\)m^2[/latex

WBBSE Solutions for Class 9 Physical Science And Environment

• Chapter 1 Measurement
• Chapter 2 Force And Motion
• Chapter 3 Matter: Structure And Properties
• Chapter 4 Matter: Atomic Structure; Physical & Chemical Properties of Matter
  • 4.1 Atomic Structure
  • 4.2 Concept Of Mole
  • 4.3 Solution
  • 4.4 Acids, Bases & Salts
  • 4.5 Separation Of Components Of Mixtures
  • 4.6 Water
• Chapter 5. Energy In Action: Work, Power & Energy
• Chapter 6 Heat
• Chapter 7 Sound

Separation Of Components Of Mixture Very Short Answer Type:

Question 1. What is a mixture?
Answer: A mixture refers to the physical combination of two or more elements or compounds at any proportion where the identities of the-all components are retained. Some mixtures can be separated into their components by physical means.

Question 2. What is a homogeneous mixture?
Answer: Most of the materials are neither pure elements nor pure compounds; they are mixtures. A solution of sugar in water is an example.

Question 3. Mention one reason to separate the components of a mixture.
Answer:
To get a pure sample of a substance: Pure substances are required in our daily life for various purposes like, in laboratory for conducting various experiments, in industry, in medicines and for carrying out research works.

West Bengal Board Class 9 Physical Science Solutions

Question 4. Mention the principle of sedimentation and decantation.
Answer: Heavy solid insoluble in a liquid settles down at the bottom forming a sediment. The upper clear liquid is the supernatant liquid.

Read and Learn all WBBSE Solutions for Class 9 Physical Science And Environment

Example: Muddy water (mud forms the sediment).

Question 5. Mention the principle of fillration.
Answer: Light-weighed solid insoluble in a liquid can be separated by filtration.

Example: Sawdust + water (sawdust floats on water).

Question 6. Mention the principle of distillation.
Answer: One of the components is a solute solid. Both components are recovered.

Example: Salt + water.

Question 7. What is distillation?
Answer: The process of converting a liquid into its vapour by heating and the subsequent condensation of the vapours back into the original liquid is called distillation.

Wbbse Madhyamik Physical Science And Environment Class 9 Question 8.

What is fractional distillation?
Answer: The process of separation of two or more miscible liquids by distillation by making use of the difference in their boiling points is called fractional distillation.

Question 9. What is sublimation?
Answer:

Sublimation: A process of phase transition when a solid goes directly into vapour phase without passing through the liquid phase.

Example: lodine or \(\mathrm{NH}_4 \mathrm{Cl}\)

West Bengal Board Class 9 Physical Science Solutions

Question 10. What is alloy?
Answer:
Alloy: Solid-solid mixture of metals or metals with non-metals.

Question 11. What is distillate?
Answer:

Distillate: In distillation process, the condensed vapour is called distillate.

Question 12. What is filtrate?
Answer:

Filtrate: The liquid that passes through the filter.

Question 13. What is residue ?
Answer:

Residue: The insoluble solid left on the filter.

Question 14. What is sedimentation ?
Answer:

Sedimentation: The process of settling down of the insoluble solid when the mixture of insoluble solid and liquid is allowed to stand.

Question 15. What is sediment?
Answer:

Sediment: The solid substance that settles down.

Question 16. What is supernatent liquid?
Answer:

Supernatant Liquid: The clear liquid left in sedimentation.

Question 17. What is decantation?
Answer:

Decantation: The process of pouring out of

Separation Of Components Of Mixture 2 Marks Questions And Answers:

Question 1. What is a mixture? State the types of mixtures.
Answer:

If two or more substances (elements, compounds or both), mixed togther in any proportion, do not undergo any chemical change but retain their characteristics, the resulting mass is called mixture. A mixture may be of two types:

(1) Heterogeneous mixture: A mixture in which various constituents are not mixed uniformly is called a heterogeneous mixture. For example, a mixture of sand, salt, sulphur, etc.

(2)Homogeneous mixture: A mixture in which different constituents are mixed uniformly is called a homogeneous mixture. For example, all solutions are homogeneous mixtures.

Question 2. What are miscible and immiscible liquids? Give examples.
Answer:

When two liquids are dissolved in each other forming a homogeneous solution, it is called a miscible liquids mixture. For example, a solution of-ethy| alcohol and water, a solution of methyl alcohol and acetone. When two liquids do not dissolve in one another, they are called immiscible liquids: For example, mixture of kerosene oil and water, a mixture of carbon disulphide and water.

Question 3. How are the miscible liquid mixture and immiscible liquid mixture separated?
Answer:

The liquid-liquid mixture are separated by use of diffrent physical methods depending on the state and nature of mixture. The miscible liquid mixture is separated by the method of fractional distillation while the mixture of immiscible liquids by separating funnel.

West Bengal Board Class 9 Physical Science Solutions

Question 4. What is the function of a fractionating column?
Answer:

The improved fractionating column is designed in such a way that the liquids of higher boiling point are mostly condensed in it and return in the form of liquid in the distillation flask.Only the liquid of lower boiling point is allowed to pass through it.

Question 5. Mention the principle of mechanical picking with example.
Answer:

Mechanical Picking: The components are to be of different sizes and colours.

Example: Tiny stones from grains and pulses.

Question 6. Mention the principle of magnetic separation with example.
Answer:
Magnetic Separation: One of the components being magnetic, magnetic separation can separate the mixture of of iron and powdered sulphur.

Wbbse Class 9 Physical Science Solutions

Question 7. Mention the principle of gravitational separation with example.
Answer:

Gravitational Separation: One of the component is much heavier than water and the other one is much lighter than water.

Example: A mixture of sand and saw dust.

Question 8. Mention the principle of sublimation with example.
Answer:

Sublimation: One of the components sublimes.

Example: a mixture of sand and iodine, a mixture of common salt (NaCl) and ammonium chloride (\(\mathrm{NH}_4 \mathrm{Cl}\)).

Wbbse Class 9 Physical Science Question Answers

Question 9. Mention the principle of solvent extraction with example.
Answer:

Solvent Extraction : One of the components is soluble in a solvent (say, water) and the other one is not.

Example: Mixture of sodium chloride (soluble in water) and chalk (not soluble in water).

Question 10. Mention the principle of fractional crystallization with example.
Answer:

Fractional Crystallization: The solubility of components in a solvent have to differ widely.

Example: Mixture of potassium nitrate (less soluble in water) and common salt.

Question 11. Mention the principle of sedimentation with example.
Answer:

Sedimentation or Decantation: A heterogeneous mixture containing an insoluble solid in a liquid.

Example: Mixture of sand or mud and water.

Wb Class 9 Physical Science Question Answers

Question 12. Mention the principle of evaporation with example.
Answer:

Evaporation: One component being non-volatile, may be soluble in water or not.

Example: Salt. solution [non-volatile salt recovered].

Wbbse Madhyamik Physical Science And Environment Class 9 Question 13.

Mention the principle of filtration with example.
Answer:

Filtration: A heterogeneous mixture containing an insoluble solid in liquid.

Example: Mixture of sand in water. 

Question 14. Mention the principle of distillation with example. 
Answer:

Distillation: A homogeneous mixture having a dissolved solute in a solvent. (Both components are recovered).

Example: A solution of common salt.

Wb Class 9 Physical Science Question Answers

Question 15. Mention the principle of separation by’ separating funnel with example.
Answer:

By Separating Funnel : A mixture of ifnmiscible liquids, differing widely in density.

Example : A mixture of oil (say, kerosene) and-water. Kerosene being much lighter than water.

Question 16. Mention the principle of fracti6nal -distillation with. example.
Answer:

Fractional Distillation: A homogéneus mixture of two miscible liquids, differing widely in their boiling points.

Example: A solution of alcohol and water. Alcohol boils at much lower temperature than water.

Question 17. What chromatography?
Answer:

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

Wb Class 9 Physical Science Question Answers

Question 18. What is paper chromatography?
Answer:

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

Question 19. Mention two advantages of chromatography.
Answer: Advantages of chromatography :

(1) A small amount of the compound present in the mixture can be separated.
(2) The properties of the individuals present in the mixture do not alter.

Separation Of Components Of Mixture 3 Marks Questions And Answers:

Question 1. Describe the method of separation of a mixture of two immiscible liquids using separating funnel.
Answer: The components of a mixture of two immiscible liquids (such as oil and water) are separated by a separating funnel.Separating funnel is a long glass tube provided with a stop clock. The immiscible liquid is poured in the funnel and allowed to stand for some time. The immiscible liquids separate into two distinct layers. The liquid with lower density (oil) remains in the upper layer and the liquid with higher density (water) will be deposited in the lower layer.

A conical flask is placed under the nozzle of the separating funnel. The stop clock is gently opened so that the heavier liquid trickles in the flask drop by drop. Once the denser liquid is drained out, the stop clock is closed. Another conical flask is placed under the nozzle of the separating funnel. Stop clock is opened to drain the lighter liquid. In this way both the liquids are separated in two separate flasks.

Wbbse Class 9 Physical Science Solutions

Question 2. What is fractional distillation? How is the mixture of two miscible liquids separated by the method of fractional distillation?
Answer:  The process of separation of two miscible liquids by the process distillation, making use of their difference in boiling points, is called fractional distillation. The mixture of two miscible liquids is separated by a fractional distillation apparatus.

The fractional distillation apparatus contains a distillation flask fitted with a fractional columm. One end of a-Leibig’s condenser is connected with fractionating ‘column and other end with a receiving flask to collect the distilate as shown in the given figure. The fractionating column is designed in such a way so that the vapours of the higher boiling liquid are mostly
condensed in it and return into the distilating flask in the form of liquid.

The vapours of the lower boiling liquid passes through the Leibug’s condenser and are collected in a receiver. The thermometer shows a constant reading as long as the vapours of the first liquid are passing to Leibig’s condenser. As soon as the temperature starts rising, the receiver is replaced by another receiver to collect the second liquid.

Question 3. Why do we separate the components of a mixture?
Answer:

Reasons for separating the components of mixture: The components of a mixture are separated because of the following reasons :

(1) To get a pure sample of a substance: Pure substances are required in our daily life for various purposes like, in laboratory for conducting various experiments, in industry, in medicines and for carrying out research works.

(2) To remove any undesirable or harmful components: The undersirable components present in a mixture can sometimes be very harmful, hence they need to be removed in order to get the desirable component. For example, small stones are removed from tice before the rice is cooked, otherwise, they can harm us.

Wbbse Class 9 Physical Science Solutions

Wbbse Class 9 Physical Science Solutions

(3) To obtain the useful components of a mixture: Sometimes a mixture may comprise more than one useful components. These useful components need to be separated for their individual use. The best example of such a mixture is crude petroleum.

Question 4. Explain the method of separation of colouring matters of ink (made of three different coloured substances) by simple paper Chromatography.
Answer:

Take a strip of filter paper about 10 cm long and 5 cm broad and stick its smaller end to a glass plate. On the lower end about 1 cm from the bottom, mark a small point and put one or two drops of ink. Now suspend this filter paper in a tall cylinder and pour some water in the cylinder till the lower end of the filter paper slightly dips in the water. The cylinder is covered with a glass lid to prevent evaporation.

After some time the water rises up the filter paper upto the ink mark and slowly dissalves the various constitutes of ink in it. The various constitutes of ink get absorbed by the filter paper in different amounts. The components of the ink solution rise to different heights on the paper strip depending on their solubility in water. The constitutes that are more absorbed they move upward lesser. Thus different column bands are formed on the paper and colouring matters of the link are separated. The diagram given here shows the method of separation of colouring matters of ink by paper chromatography.

Question 5. Write a short note on distillation.
Answer:

Distillation: This process allows separation and recovery of both components of a solid-liquid mixture by the use of a distillation apparatus. The process of converting a liquid into its vapour by heating and the subsequent condensation of the vapours back into the original liquid is called distillation.

Procedure :

(1) The mixture is taken in a distillating flask fitted with a Leibig’s condenser. At the other end of the condenser a receiver is placed to collect the distance.
(2) When the flask is heated, the vapours of the solvent pass through the condenser. They are cooled, condensed into a liquid and collected as a distillate in the receiver.
(3) The solid component forms the residue in the flask.
Examples: Distillation of solid-liquid mixtures.

Components of the Solid-Liquid Mixture	Soluble solid component	Distillate
1. Iodine + Chloroform	Iodine	Chloroform
2. Salt + Seawater	Common salt and other salts	Pure water
3. Iodine + Alcohol	Iodine	Alcohol

Wbbse Madhyamik Physical Science And Environment Class 9 Question 6.

State the importance of fractionation of petroleum.
Answer:

Importance of fractionation of petroleum: The chief source of liquid fuels is petroleum. However, petroleum cannot be used directly because it burns with a highly sooty flame, producing harmful gases.

The petroleum is subjected to fractional distillation. The crude oil (petroleum) is separately boiled in a ‘retort’so as to form its vapour. The vapours of petroleum are passed into the fractionating column. The vapours of heavier molecules condense in the lower parts of the fractionating column. The vapours of lighter molecules condense in the upper part of fractionating column.

The crude petroleum obtained from mines is a mixture of hydrocarbons containing [laaatex]C_1 \text { to } C_{40}[/latex]   carbon atoms. Its different fractions are collected by fractional distillation. This process of purification of petroleum into different fractions is called refining of petroleum.

Wbbse Class 9 Physical Science Solutions

Question 7. State the procedure of separation by separating funnel.
Answer:

Separation by separating funnel :
(1) Separating Funnel: It is an apparatus used for separating immiscible liquids. The liquids do not mix with each other and due to difference in densities the liquids remain in separate layers in the separatory funnel.

(2) Procedure :
(1) A mixture of the immiscible liquids is taken in a separatory funnel (with its stopcock closed) and allowed to remain on the stand
(2) The immiscible liquids get separated into two x Water distinct layers, the lighter component forming the upper layer and the denser component the lower g Separatory funnel layer.
(3) A conical flask is placed below the stem of the separatory funnel and the stopcock is carefully opened to run out the denser immiscible layer.
(4) The stopcock is then closed and a new conical flask is placed as before and the femaiing lighter upper iayer is carefully taken out.
Example : Separation of Liquid-Liquid pactres using a separatory funnel.

Wbbse Class 9 Physical Science Question 8. State the different types of mixture giving one example each of homogeneous and heterogeneous mixtures.
Answer:

Types of mixture	Homogeneous	Heterogeneous
(1)    Solid in Solid (S in S)	Bronze (Mixture of Cu. Zn, Sn)	Gunpowder (Charcoal. Sulphur, Nitre)
(2)    Solid in Liquid (S in L)	Sugar in water. Iodine in alcohol	Sand in water; Sugar in oil
(3) Liquid in Solid (L is S)	Hg-Au Amalgam	Water m sponge
(4) Liquid in Liquid (L in L)	Ethanol or methanol in water acetone in water	Oil in water
(5) Gas in Liquid (G in L)	Ammonia gas or hydrogen chloride gas in water.	Helium in water.
(6)    Liquid in Gas (L in G)	Moisture in Air	________
(7)    Gas in Gas (G in G)	Air	Industrial release of smoke

Question 9. How does boiling point depend on superincumbent pressure ?
Answer:

Dependence of boiling point on superincumbent pressure : The boiling point of a liquid is directly proportional to the pressure exerted on it. It is known that when the vapour pressure of a liquid becomes equal to the atmospheric pressure, the liquid starts to boil. The vapour pressure of a liquid is also proportional to the pressure exerted on it. At low pressure, the liquid boils and evaporates at a lower temperature.

When any components of a solution decompose at normal boiling point (at normal pressure), then separation of the components is made by distillation at reduced pressure, i.e., the components boil below boiling points.

Wbbse Class 9 Physical Science Question 10. State the common separation techniques and their underlying working principles.
Answer:

Common separation techniques and their underlying working principles :

 

Question 11. Explain the method of separation of the compounds of crude petroleum.
Answer:

Seperation of Components of CrudePetroleum : Crude petroleum contains various hydrocarbon ingredients of close boiling points, the boiling points of these ingredients ranging from <293K to > 573K. These ingredients are collected in 313-443K a range of boiling points using high efficiency

Fractionating column, Example:
(1) Natural gas (293K)
(2)Gasolene (293K-333k)
(3) Naptha (313K-453k)
(4)Kerosene (453K-533k)
(5) Diesel (533K-613K)
(6) Lubricating oil (613K-773K).

WBBSE Solutions for Class 9 Physical Science And Environment

• Chapter 1 Measurement
• Chapter 2 Force And Motion
• Chapter 3 Matter: Structure And Properties
• Chapter 4 Matter: Atomic Structure; Physical & Chemical Properties of Matter
  • 4.1 Atomic Structure
  • 4.2 Concept Of Mole
  • 4.3 Solution
  • 4.4 Acids, Bases & Salts
  • 4.5 Separation Of Components Of Mixtures
  • 4.6 Water
• Chapter 5. Energy In Action: Work, Power & Energy
• Chapter 6 Heat
• Chapter 7 Sound