PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
14. CARBON AND ITS COMPOUNDS 1 Mark Questions I. 1. What are hydro carbons ?
(As–1)
Ans. Hydrocarbons are the cinoiybds if carbon and hydrogen. 2. How many types of Hydrocarbons ? Ans. Hydrocarbons are two types 1) Saturated hydrocarbons (alkanes) and unsaturated Hydrocarbons (alkenes and Alkynces) 3. What is hybridisation ? What are hybrid orbitals ? Ans. the redistribution of orbitals of almost equal energy in individual atoms to give equal number of new orbitals with identical properties like energy and shape is called "Hybridisation" The newly formed orbitals are called as "hybridorbitals." 4. What is allotropy ? What are the allotropy forms of carbon ? Ans. the property of an element to exist in two or more physical forms having more or less similar chemical properties but different physical properties is called allotropy. The allotropy forms of carbon are 1) Amorpho usforms 2) Crystalline forms 5. Write the names of crystalline allotropic forms of carbon ? Ans. Diamond, graphite and buck minister fullerene. 6. Write the chemical equation of preparetian of urea ? Ans.
O
NH4 CNO Heat H H
H
C N
N Urea
H
7. Classify the branched chain and closed chain compounds of the following ? a) CH3 – CH – CH – CH3 CH3 CH2 CH3 b)
CH2 – CH2 H3C – CH – CH2
111
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✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
c) CH2 – CH2 CH – CH2 CH2 CH3 d) HC ≡ C – C – CH3 CH3 Ans. a, b and d are branched chain compounds C is the closed chain compound 8. What are Aldehydes and Ketones ? Ans. The hydrocarbons with functional group of –CHO are called aldehydes.' Ex : Farmaldehyde, Accetal dehyde The hydrocurbons with C C = O functional group are called ketones. C Ex : Acetone, Methyl ketone 9. What is Isomerism ? And what is Iromers ? Ans. The phenomenon of possessing same molecular formula but different properties by the compounds is known as Isomerisom. The compound that eshibit isomerison are called Isomers. 10. What are subtitution reactions ? Ans. A reaction in which an atom or a group of atoms in a given compound is replaced by other atom or groop of atoms is called a substitution reaction. 11. How ethyl alcohol is prepare from ethane ? Ans. Ethanol is prepared on large scale from ethane by the addition of water vapour to it in the presence of catalysts P205, tungesten Oxide at high pressure and temperator CH2 ≡ CH2 + H2O
Catalyst 100 – 300 atm at 300°C
CH3 CH2O H
12. What is Pka? Ans. Pka is the negative value of logarithm of dissociation constant of an acid. Pka = log10 KaOJa is 13. What is saponification reaction ? Ans. The sodium salts of these higher fatty acids being soaps the reaction is the soaps the recton is the soap formation reaction which is generally called as "Saponification reaction". (or) Alkaline hydrolysis of tristers of higher fatty acids producing soaps is called saponification. 112
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
14. What is Micelle ? Ans. A spherical aggregated of soap molecules in water is called micelle. 15. What are hydrophilic and hydrophobie parts in soaps ? Ans. The polar end in soap with carboxy is called hydrophilic end. The non-polar end in soap with hydrocarbon chain in called hydrophobic end. Na+ COO– Hydrophobic end
Hydrophilic end
2 Mark Questions 16. What happens when a small piece of sodium is dropped into ethanol ? Ans. When a small piece of sodium is dropped into ethanol, it shows brisk efferversence and liberats hydrogen gas and forms sodium ethoxide. 2C2H5O5 + 2Na → 2C2H5ONa + H2 Ethanol Sodium ethoxide 17. Define alkanes, alekeneus and alkynes ? Ans. Alkanes : Hydro carbons containing only single bonds between carbon atoms are called alkanes. Alkenes : Hydro carbons containing atleast one double bond between carbon atoms are called Alkenes. Alkynes : Hydro carbons containing atleast one triple bond between carbon atoms are called Alkynes. 18. Suggesta chemical test to distinguish between ethanol and ethanoic acid and explain the procedure ? Ans. 1) Take ethanol and ethonoic acid in two different test tubes. 2) Add nearly 18 ml of sodium bicorbonate to each test tube. 3) Lots and lots of bubbles and form will be observed from the test tube containing ethanoic acid. NaH CO3 + CH3COOH → CH3 COONa + H2O + CO2 ↑ 4) Ethanol will not react with sodium bicarbonate and thus we won't observe any change in the test tube containing ethanol. 19. How do you appreciate the role of esters in everyday life ? Ans. 1) Esters contribute to the flavoures and frogrances of fruits and flowers. 2) They are used as alternative medicine suppliments and vitamins.
(As–6)
20. An orgomic compoundX with molecular formula C2H6O undergoes oxidation with alkaline KMnO4 and forms the compound Y, that has molecular formulae C2H4O2. a) Identify X and Y (b) write your observation regarding the product when the compound X is made to react with compound y which is used as a preservative for pickles. 113
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PHYSCIS - Paper - I
Ans. a) X : C2H6O is Ethanol Y : C2H4O2 is Ethanoic acid. b) Ethyl alcohol undergoes oxidation to form the product accetaldehyde and finally Acetic acid. Alkaline
CH3 CHO → CH3COOH (Ethanol) (Ethnenoic acid) KMnO4 Here CH3COOH is used as preservative for pickles. CH3 CH2 OH
21. Draw the electronic dot structure of ethan molecule (C2H6) Ans. H H C3H6 : H – C – C – H H H Electronic dot structure H
H
H
X X X C X X C X X H
H
X H
22. How do you condern the use of alcohol as a social practice ? Ans. 1) Consumption of alcohol in the form of beverages is harmfol to health. 2) It causes servere damage to blood circulation system. 3) Addition to alcohol drinking leads to heart diseases and damagesthe liver. 4) It also causes ulcers in small interestines due to increased acidity and damages the digestive systerm. 23. Write the names of the following compoundes ? a) CH3 – CH2 – CH2 – CH3 b) CH3 – CH – CH2 – CH3 Cl c) CH3 – CH – CH – CH3 Cl
Cl
d) CH3 – CH2 – CH2 – CH2 – OH e) CH2 – CH2 CH2 CH2
114
PHYSCIS - Paper - I
f)
Br
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
Br
CH – CH CH2 CH2 Ans. a) Butane b) 2 – Chlorobutane c) 2, 3– dichlorobutane d) Butan – 1 – 0 e) eyclobutane f) 1, 2 – dibromo cylobutane
4 Mark Questions 24. Explain with the help of a chemical equation, how an addition reaction is used in vegitable ghee industry ? Ans. Addition reaction which is useful to vegetable ghee industry where the unsaturated oil can be changed into saturaed fat by adding hydrogen R R R R Ni C = C + H2 H – C – C – H H2 R R R R H – C = H + H2
Ni
H H H–C–C–H
H H
H H 25. Write the substitution reactions of alkhanes How CH3Cl, CH2Cl2, CHCl2 and CCl4 are obtained from methane ? Ans. When metane reacts with chlorine in the prusence of sunlight, Hydrogen atoms of CH4 are replaced by chlorine atoms. Sunlight CH4 + Cl2 CH3Cl + HCl (methylchloride) CH3Cl + Cl2
Sunlight
CH2Cl2 + HCl (methyleen chloride)
CH2Cl2 + Cl2 Sunlight CHCl3 + HCl (chloroform) CHCl3 + Cl2 Sunlight CCl4 + HCl (carbon Tetrochlorid) 26. Explain the cleaning action of soap ? Ans. 1) Soaps and detergents make oil and dirt present on the cloth come out into water thereby making the cloth clean. 115
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2) Soap has one polar end to end with carboxyl and one non-polar end. Hydro carbon chain) 3) The polar endis hydrophilic in nature and this end is attraded towards water. 4) The non-polar endis hydrophobic. In nature and it is attracted towards grease or oil on the cloth. But not attracted towards water. 5) When soap is disolved in water, its hydrophotic ends attach themselves to dirt and remove it from cloth. 6) The hydrophobic end of the soap molecules move towards the dirt or greater partice. 7) The hydrophobic ends attraoched to the dirt particle and try to pull out. 8) The molecule of soap surround the dirt particles at the centre of the cluster and form a spherical structure called micelle. 9) These micellies remain suspended in water like partices in a colloidal solution. 10) The various miscelles present in water do not come together to form a precipitate as each micalle repels the other because of the ion-ion repulsion. 11) Thus, the dirt particles remain trapped in micelles and are easily rised away with water. hence, soap micelles remove dirty by dissolving in water. 27. Write the differences of Eterification and Saphoni fication. Ans. Saphssitication Esterification 1. When carboxylic acid is react with in conc. H2SO4 to alcohol esters are formed CH3COOH + C2H5OH
1. When oil is react with carboxylic acid. CH2– OH
Conc H2SO4
CH – OH + Na
CH3COOC2H5 + H2O
CH 2 – OH
2. This is reversible reaction 3. This is use for preparation of different esters.
2. This is irrevarsible reaction. 3. This is use for preparation of soap.
5 Mark Questions 28. Draw the structure of Ethene (C2H4) by using SP2 hybridisation ?
P.No.309
116
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29. Draw the structure of Acetelees (C2H2) ? P.No.308
30. Draw the lattice structure and general structures of Diamond ?
P.No.309
PART – B I. Multiple Choice questions : 1. Bond angle in CH4 is A) 109° 28'
B) 107° 48'
C) 104° 31'
B) Coal
C) Graphite
)
(
)
D) 120°
2. Which of these is not a crstalline form of carbon A) Diamond
(
D) Buckminster fullerence
3. Which of these is a saturated hydrocarbon
(
)
(
)
B) H C ≡ CH = CH2
A) CH – CH2 CH – CH2 C) CH2 – C – CH – CH CH3
D) CH3 – CH2 CH2 – CH3
4. Which of these is a closed chain compound A) CH3 – CH – C – CH3 CH3 CH2 CH2 B)
CH2 – CH2 H3C – CH2
CH2
C) CH – CH2 CH CH2 CH2 117
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PHYSCIS - Paper - I
CH3
D)
H C = C – C – CH3 CH3 5. Which of these represent a ketone
(
)
(
)
(
)
(
)
(
)
A) CH3 – CH2 – C = 0 H O B) CH3 – CH2 – C – CH3 C) H3C – CH2 – OH D) CH3 – C = 0 OH
NH2
6. Find the functional group in CH3 – CH2 – C – CH3 CH3 A) Ester
B) Amine
C) Ether
D) aldehyde
7. Identify the alkene A) C5H12
B) C4H8
C) C6H10
D) C3H8
8. IUPAc Name of carbon compound CH2 – CH – CHO Cl
Cl
A) 1, 2 di chloro Ethanol
B) 2, 3 di chloro propanal
C) 1, 2, 3 di chloro propanol
D) none of the above
9. Identify the compound whose name is pent – 4 – en – 2 – 01 A) CH2 – CH – CH2 – CH – CH3 OH B) CH3 – CH – CH – CH2 – CH3 OH C) CH3 – CH2 – CH – CH2 = CH2 OH D) None of the above
118
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✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
10. The functional group indicates carboxylic acid A) – COOR
B) –COOH
C) –CHO
B) –al
C) –One
B) C2H6
C) C3H8
B) CnH2n
C) CnH2n – 2
)
(
)
(
)
D) C4H10
13. The general formula of homologous series of 'Alkenes' is A) CnH2n + 2
( D) –ene
12. Which one of the following hydrocarbon can show isomerism ? A) C2H4
)
D) –C = 0
11. The suffix used for naming an aldehyde is A) –0l
(
D) C2nHn + 2
14. When acetic acid reacts with ethyl alcohol, we add conc. H2SO4, it acts as ........ and the process is called ( ) A) Oxidizing agent, saponification B) Dehydrating agent, esterification C) Reducing agent, esterification D) Acid & esterification 15. A few drops of ethanoic acid were to solid sodium carbonate. The possible results of the reactions are ( ) A) A hissing sound was evolved B) Brown funes evolved C) Brisk effer vescence occured D) A pungent smelling gas evolved II. Fill in the blanks. 1. Very dilute solution of ethanoic acid .................... 2. Hydrocarbons containing double and triple bonds are called .................... 3. A sweet odour substance formed by the reaction of an alcohol and a carboxylic acid is .................... 4. Hydrocarbons having the general formula CnH2n + 2 are called .................... 5. The reactive part of the organic molecule is called its .................... group. 6. The process of burning of a hydrocarbon in presenue of excess air to give CO2, H2O with evolution of heat and light is known as .................... 7. The organic compounds having the same molecular formular but different structures are known as .................... 8. Type of reactions shown by alkanes is .................... 9. When sodium metal is dropped in ethanol .................... gas will be released. 10. .A compound which is basic constituent of many cough syrups ....................
119
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
II. Matching. I.
A
B
1. Ethene
(
)
A. CH3 – CH2 – CH2 – CH3
2. Butane
(
)
B. CH2 – CH2
3. Propyne
(
)
C. CH3 – C = CH
4. Pentyne
(
)
D. CH3 – CH2 – CH2 – C = CH
5. Propane
(
)
E. CH3 – CH2 – CH3 F. CH3 – CH2 – CH = CH– CH3 G. CH = CH
II.
A
B
1. Aldehyde
(
)
A. – COOH
2. Amine
(
)
B. – C O
3. Ketone
(
)
C. – COOR
4. Acid
(
)
D. – CHO
5. Alcohol
(
)
E. – NH2 F. – OH G. – CONH2
III.
A
B
1. Ethane
(
)
A. C2H4
2. Propene
(
)
B. C2H6
3. Butyne
(
)
C. C3H6
4. Pentene
(
)
D. C2H2
5. Ethyne
(
)
E. C4H6 F. C5H10 G. C2H
IV.
A
B
1. Ethanol
(
)
A. CH3COOH
2. Ethanoic acid
(
)
B. H2C – CH – CH2 OH
3. Ethanal
(
)
C. CH3CH2OH
4. Ciycerol
(
)
D. C17H35COONa
5. Stearic acid
(
)
E. CH3CHO F. C17H35COOH G. CH3 COONa
120
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✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL Answer
I. 1) A
2) B
3) D
4) C
5) B
6) B
7) B
8) B
9) A
10) B
11) B
11) D
12) B
13) B
14) C
II. 1) Vinigar
2) Unsaturated Hydrocarbons
3) Ester
4) Alkanes
5) Functional group
6) Combustion
7) Isomers
8) Substitutional reactions
9) Hydrogen
10) Ethanol (or) Ethyl alcohol
III. A) 1) B
2) A
3) C
4) D
5) E
B) 1) D
2) E
3) B
4) A
5) F
C) 1) B
2) C
3) E
4) F
5) D
D) 1) C
2) A
3) E
4) B
5) F ❖ ❖ ❖ ❖❖
121
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13. PRINCIPLES OF METALLURGY 1 Mark Questions I. 1. What is metallurgy ?
(As–1)
Ans. The process of extraction of metals from their ores is called metallurgy. 2. Do you agree with the statement "All ores are minerals but all minerals need not be ores?" Why. Ans. Yes I agree wish the statement because ore is a mineral from which the metals are extracted wish out economical loss. 3. Which metals exist in the nature in free state ? Ans. Gold (Au), Silver(Ag), Platinum (Pt). 4. What is activity series ? Ans. Arrangement of the metals in decreasing order of their activity is known as activity series. 5. Write the stages involved in the extraction of metals from the ores ? Ans. The extraction of a metal from its ore involves three stages. They are 1) concentration or Dressing (2) Extraction of crude metal 3) Refining (or) purification of the metal. 6. Which method do you suggest for the dressing of a ore containing magnetic and non magnetic and substances ? Ans. I suggest Magnetic separation. Because in magnetic separation method electro magnets are used to separate the magnetic and non-magnetic substances. 7. Write the names of any two ores of Iron ? Ans. Haematite - Fe2O3 Magnetite – Fe3O4 8. Name two metals which corrode easily and two metals which donot corrode readily ? Ans. Iron, and copper corrode easily. Gold and platinum do not corrode. 9. What is Liquation ? Ans. A low melting metal can be made to flow on a slopy surface to separate it from high metting impurity Ex : Tin. 10. Which purification method is used if the impurities have high boiling point ? Ans. Distillation process is used if the impurities have high boiling point. In this method the extracted metal in the molten state is distilled to obtain the pure metal as distillate. 104
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
11. What is gangue ? Ans. The unwanted material in the ore is called gange. 12. What is flux ? Which is used as flux in th extraction of Iron in blast furnace. Ans. Flux is a substance added to the ore to remove the gangue from it by reacting with ore. Calcium carbonate is used as flux in the extraction of Iron in blast furnace. 13. What is slag ? Ans. The substance formed due to reaction of gangue and flux is called slag. Ex : Casio3, Fesio3. 14. Which method is used to extract magnesium from its ore carnalite ? Ans. The formula of carnalite is Kcl. MgCl2.6H2O. Electrolytic reduction method is used to extract magnesium from its ore carnalite. 15. Does the reactivity of a metal and form of its ore has any relatioon with process of extraction ? Ans. Yes, they have relation. Metals like K, Na, Ca, Mg and Al are so reactive and exists in all forms. Moderate reactive metals like Zn, Fe, Pb exists as oxides, sulphires and carbonates. Least reactive metals Au, Ag found in free state
2 Mark Questions II. 1.What is the difference between roasting and calcination ? Give one example for each ? Ans. Roasting Calcination 1. Ore is heated in the presence of air. 2. This method is used for sulphix ores. 3. It dries the ore 4. 2ZnS + 3O2 → 2ZnO + 2SO2
1. Ore is heated in the absence of air. 2. This method is used for carbonates and oxide ores. 3. It makes the ore porus. 4. CaCO3 ∆ CaO + CO2.
2. Which metals are present at the top of the activity series (High reactive metals). how they are extracted. Ans. 1) High reactive metals are – K, Na, Ca, Mg, Al. 2) Their fused components are subjected t electrocysis to get that metal. 3) Ex : Fused Nacl (Sodium chlaix) is electrolysed with steel cathode (–) and graphite anode (+). The metal will be deposited at cathode and chloride liberated at the anode. 4) At Cathode → 2Na+ 2e– → 2Na. At Anose → 2Cl– 2e → Cl2. 3. What is thermite reaction.Where it is used ? Ans. The reacton of Iron (III) Oxide (Fe2O3) with aluminium is used to join vailings of railway tracks or cracked machine parts. This reaction is known as Thermite reaction. (Thermite process) 105
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
2Al + Fe2O3 → Al2O3 + 2Fe + Heat 2Al + Cr2O3 → Al2O3 + 2Cr + Heat 4. Which method is used to purify Blister (copper copper obtained from its sulphide ore) Explain. Ans. Blister copper is parified by poling. In this method the motten metal is stirred with logs (Pores) of green. The impurities are removed either as gases or they ge oxidized and form slag over the motten metal. The reducing gases evolved from the wood prevents the oxidation of copper. 5. Give examples for the metals undago corrosion ? Why do they corrode ? how to prevent corrosion ? Ans. 1) Iron, silver, copper. Generally undergo corrosion. 2) In metallic corrosion a metal is oxidised by loss of electron's generally to oxygen and results in the farmation of oxides. 3) Prevantion : Covering the surface with paint or some chemicals e.g. : Bisphenol. B) Electroplating. 6. What is the main difference in Blast furnace and Reverberatory furnace vegarding fire box and hearth ? Ans. In Blast Furnace both fire box and hearth are combined in big chamber which accomodates both ore and fuel. In Reverbaratory Furnace fire box and hearth are separated, but the vapours obtained due to the burning of the fuel touch the ore in the hearth and heat it.
4 Mark Questions III.1. Which method is used for dressing sulphide ore. Explain with a neat diagram ? Ans. 1) Froth Flotation method is used for dressing the sulphire ore. 2) The ore with impurities is finely powderd and kept in water taken in a flotation cell. 3) Air under pressure is blown to produce forath in water. 4) Froth produced takes the ore particles to the surface. 5) The impurities settle at the bottom. 6) Froth is separated and washed to get ore particles. Froth bubbles carrying sulphide ore particles
Compressed Air Sulphide ore particles Water containing pine oil
Gangue Froth floatiion process for the concentration of sulphide ores 106
PHYSCIS - Paper - I
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2. Ores are given below. Wrie their formula, metal extracted and their reactivity order. ORE Formula Metal Reactivity 1.
Zincite
2.
Magnetite
3.
Galena
4.
Copper Iron pyrites
Ans. 1.
Zincite
2.
Magnetite
3.
Galena
4.
Copper Iron pyrites
Zno
Zn
Moderate
Fe3O4
Fe
Moderate
pbs
pb
Moderate
CuFeS2
Cu
Moderate
Zn < Fe < pb < Cu 3. What is a furnace? Which Furnace is used for smelting. Write the reactions that occur in the furnace ? Ans. 1) The furnace is one which is used to carry out pyrochemical process in Metallurgy. 2) Blast Furnace is used for smetting. Smetting is a pyrochemical process in which the ore is mixed with flux and fuel and strongly heated. 3) All the reactions of Furnace take place in the body. The reactions inside the furnace are 2C + O2 → 2CO Fuel (g) (g) Fe2O3 + 3CO → 2Fe + 3CO2 (S) (l) (g) CaCO3 → CaO + CO2 (Heamatite) (s) (g) Lime stone lime CaO + SiO2 → CaSiO3 (S) (S) (l) lime silica Slag (Calcium silicate) (gangue) 107
PHYSCIS - Paper - I
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4. Describe an experiment to prove that the presence of air and water are essential for corrosion Explain the procedure. (As–3) Ans. 1) Take three test tubes and place clean Iron nails in each of them. 2) Label these test tabes A, B and C. 3) Pour some water in test tub A and cork it. 4) Pour Boiled water in test tube B and add 1 ml of oil and cork it. The oil float on water and prevent the air from dissolving in the water. 5) Put some anhydrous calcium chloride in test tube C and cork it. It absobs any moisture present in the test tube. 6) Leave these test tubes for a few days and then observe. 7) We notice that iron nails rust in test tube A only. Because the nails are exposed to air and water. 8) The nails in B are exposed only to water and Nails in test tube Care exposed to dry air. 9) From this we can say that air and water are essential for corrosion.
5 Mark Questions 1. Write the process involved in Blast furnace in to extraction of metals with a diagram and lable the parts.
Ans. Smetting is carried out in a blast furnace. 108
PHYSCIS - Paper - I
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2. What is the role of a furnace in metallurgy. Draw a neat diagram of Reverberatony furnace ? Ans. Furnace is one which is used to carry out pyrochemical process in meltallurgy.
Ore
Fire box
PART – B I. Multiple Choice questions : 1. The reducing agent used in thermite process is A) Al
B) Mg
C) Fe
B) Reduce
C) Neutralise
B) Calcination
C) Smetting
B) Flux
C) Fuel
B) Zinc
C) Aluninium
A) 16
th
B) 15
th
C) 14
B) Distillation
C) Liquation
B) Fire box
C) Chimney
B) Reduction
C) Both
B) Hearth
C) Chimney
)
(
)
(
)
(
)
(
)
(
)
D) None D) None
10. ........... is the place inside the furnace where the ore is kept for heating purpose A) Fire box
(
D) Electro refining
9. During corrosion ........... process takes place A) Oxidation
)
D) 13
8. The part of Furnace where we keep Fuel for burning is A) Hearth
(
th
7. Low boiling metals are purified by ........... method A) Poling
)
D) Iron
6. ........... group elements are called chalcogens th
(
D) None
5. The most abundent metal in he earth's crust is A) Silver
)
D) None
4. The substance added to remove the impurity is A) Gangue
( D) None
3. The dprocess used to convent sulphise ore into oxide ore is A) Roasting
)
D) Si
2. The purpose of smetting an ore is to .......... it A) 4Oxidse
(
D) None 109
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
II. Matching. I.
A
B
1. Horn silver
(
)
A. Nacl
2. Epsom salt
(
)
B. Pbs
3. Rock salt
(
)
C. MgSO4. 7H2O
4. Cinnabar
(
)
D. Agcl
5. Galena
(
)
E. Hgs F. CaCO3 G. CuFeS2
II.
A
B
1. Oxides
(
)
A. Rocksalt
2. Sulphides
(
)
B. Epsom salt
3. Chlorides
(
)
C. Zincite
4. Carbonates
(
)
D. Zinc Blend
5. Sulphates
(
)
E. Lime stone F. Gold
Answer I. 1) A 8) B I. 1) D II. 1) C
2) B 9) A 2) C 2) D
3) A 10) B 3) A 3) A
4) B
5) C
4) E 4) E
5) B 5) B
6) A
7) B
❖ ❖ ❖ ❖❖
110
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
12. ELECTROMAGNETISM 1 Mark Questions I. 1. Define Magnetic Flux density ? (As–1) Ans. Magnetic flux density (b) is defined as the ratio of flux passing through a plane perpendicular to field and the area of the plane. 2. Are the magnetic linus closed ? Why ? (As–1) Ans. Magnetic lines are imaginarylines or curves forms around the magnet. So the magnaticlines are closed curves. 3. A bar magnet with north pole facing to wards coil moves as shown in figure. What happens to magnetic fluc passing through the coil ? (As–1)
N N
S
Ans. If the magnetic flux passing through a coil then current is generated in the coil. 4. The direction of current flowing in a coil is shown in figure what types magnetic pole is formed at the face that the flow of current as shown in figure ? (As–1)
Ans. North pole is forms at the face that has flow of current as shown in the figure. 5. Why do the picture appear distorted when a bar magnet is brought close to the screen of a television ? (As–1) Ans. The picture appear distorted when a bar magnet is brought close to the screen television due to the fact that magnetci field excerts a force on the moving charge. 6. In the following figure magnetic linus are shown. In what direction does the current through wire flow ? (As–1)
96
PHYSCIS - Paper - I
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Ans. In the diagram the magnetic lines are anticlock wise direction. According to ampere right hand rule the direction of current is vertically up wards. 7. What is the magnetic force on the charge moving parallel to a magnetic field ? (As–1) Ans. When charge moves parallel to the magnetic field the value of θ becomes zero in F = qv B sin θ then F = q vB sin 0 = 0. So the charge experiences no force when it is moving parallel to the magnetic field. 8. State the Faraday's Law ? (As–1) Ans. The induced EMF generated in a closed loop is equal to the rate of change of magnetic flux passing through it. 9. State the Lenz's Law ? (As–1) Ans. The induced current set up in the coil is in such a direction that it opposes the changes in the flux. 10. A coil is kept perpendicular to page. At p current flows in to the page and at Q it comes oot of the page as shown in the following figure. What is the direction of the magnetic field due to coil ? (As–1) Ans. At the top near Q. The direction of magnetic field is anticlock wise direction and at the boton near p, the direction of magnetic field is clock P wise direction Q
P
2 Mark Questions 11.What are the differences between electric motor and a generator ? Ans.
Electric motor
(As–1)
Generator
1. Motor converts electrical energy into mechanical energy.
1. Converts methcanical energy to electrical energy.
2. Works on the principle of Fleming's left hand rule.
2. Wroks on the principle of Flemings right hand rule.l
12. What is a sole noid ? Mention two ways to increase the strength of the field of a solenoid ? (As–1) Ans. A coil of many circular turns of insulated copper wire wrapped closely in the shape of clinder is called a solenoid. 97
PHYSCIS - Paper - I
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Two ways to increase the strength of the field of a solenoid. 1. By increasing the number of turns 2. By increasing current 13. Lateef said to you that the magnetic field lines are open and they start at north pole of bar magnet and end at south pole. What questions do you ask lateef to correct him by saying" field lines are clsoed" ? (As–2) Ans. 1) How can you say the alignonent of lines passing through magnet ? 2) Why the direction of force on a north pole is always same ? 3) If magnetic lines are open why magnetic lines start from north pole ? 4) Why the lines are concentric circle around a current carrying wire ? 14. Collect information about generation of current by using faraday's law ? (As–4) Ans. 1) Dynomo is working on the principle of Faraday's Law. 2) Dynamo consists stator provides constant magnetic field. 3) It rotates armature. 4) Armature turn with in magnetci field induced emf produced which is Faraday's Law. 15. Find the length of the conductor which is moving with a speed of 10m/sec in the perpendicolar to the direction of magnetic field of induction 0.8 T, if it induces an emf of 8v between the ends of the conductor. Ans. Speed v = 10/m sec magnetic field of induction B = 0.8 T E.M.F ε = 8v The electro motive force e.m.f ε = BlV 8 = 0.8 × l × 10' 0.8 × 10 × l = 8 8 ×10 × l 10
l=
8 8
=8
=1
∴ The length of the conductor l = 1 meter 16. How do you appreciate Farwday's Law, which is the consequence of conservation of energy ? (As–6) Ans. 1) Law of conservation of energy says energy neither be created nor be destroyed but can be converted from one form to an other. 2) Faraday's Law says whenever there is a continuous change of magnetic flu linked with a closed coil, a current is generated in the coil. This induced emf is equal to the rate of change of magnetic flux passing through it. 3) We have to do some work to move the magnet through a cell. This work produced energy. 4) This energy is converted into electrical energy in the coil. 5) So conservation of energy takes place in electromagnetic induction. 98
PHYSCIS - Paper - I
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17. How do you appreciate the relation between magnetci field and electricity that changed the life style of mankind ? (As–1) Ans. 1) Changing life style of mankind is a result of many inventions, utilising a lot of scientific principles. 2) Scientiests always going to searching for new principles and new applications to make our life more confortable. 3) If you consider electricity, right from amberstone to nuclear power, so many changes have been incorporated. 4) The idea of ested and Fraday that current carrying wire produces electricity and electromagnetic induction, enable use to electric motors, generators fans, mixers, grinders, induction stoves. All the applications makes our life more confortable. hence Farday and oersted rendered a lot of serves in this field. 18. What are magnetic field lines ? List any two characteristics of field lines ? Ans. A magnetic field line is a patch along which a free north pole tends to move. Characteristics : 1) Outside a magnet the magnetic field lines are directed from N-Pole of magnet towards S-pole however inside a magnetic field lines are directed from spole to N pole. 2) The relative strength of magnetic field lines is given by degree of close ness of the field lines more corocoded field lines means a stronger field.
4 Mark Questions 19. A charged particle q is moving with a speed 'v' perpendicular to the magnetic field of induction B. Find the radius of the path and time period of the particle ? (As–1) Ans. Let us assume that the field is directed into the page as shown infigure. Then the force experienced by the partide is F = mqvB. We know that this force is always directed perpondicular to velocity. Hence the particle moves along a circular path and the magnetic force on acharged particle acts like a centripetal force.
Let r be the radius of the circolar path We know that centripetal force =
mv2 r 99
PHYSCIS - Paper - I
qvB =
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
mv2 r
Solving the equation, we get r = Time period of the particle T =
mv Bq
2πr V
Substituting r in above equation use get T = 20. Collect information about material required and procedure of making a simple elecric motor from internet and make a simple motor on your own. (As–4) Ans. Aim : Preparation of a simple electric motor Materials required : A wire of 15 cm, 1.5V battery, iron mail, strong magnet, paper chip. Procedure : 1) Attach the magnet to the head of the iron nail. 2) Attach a paper clip to the open end of the magnet. 3) Now attach the other end of the nail to the cap of the battery. 4) Now connect the negative terminal of the battery and the head of the iron nail trough a wire. 5) We observe that the paper clip rotates. 21. Explain the working of electric motor with aneat diagram ?
(As–3)
Ans. Electric motor : A motor is a device which converts 2πm electrical energy into mechanical energy. Bq Principle of motor : A motor works on the principle that when a rectangular coil is placed in a magnetic field and current is passed through it. A force acts on the coil which rotates it continuously. anticlockwise rotation
N
B
C
s A
D
C1
Split rng communicator B1
Brushes
B2
Current (reverses at half turn)
Working of electric motor : 1) When electric current is passed into the rectangular coil, this current produces a magnetic field around the coil. 2) The magnetic field of horse shoe type magnet then interacts with the magnetic field of the coil and causes the coil to rotate continuously. 100
PHYSCIS - Paper - I
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3) If ABCD is in horizontal position current from battery enters the coil through brush B1, and commutator half ring C1. The current flows in the direction ABCd and leavces ring and brush B2. 4) The direction of current is from A to B, the direction of current is from C to D. Where as the force on the side C of the coil is in the upward direction. ABCD rotats in anticlock wise direction. 5) While rotating when the coil reaches vertical position then the brushEs B1 and B2 will touch the gap between the commutator rings and current to the coil is cUt of. 6) The coil CD comes on the left side and AB comes to the right side. Again they come in contact with brush B1, current direction is reversed. 7) the revesing of current in the coil is repeated after every half rotation due to which the coil continue to rotate as long as current from the battery is passed through it. 22. Explain the working of DC generator with a neat diagram ?
(As–1)
Ans. 1) If two half slip rings are connected to ends of the coil the AC generator works as DC generator to produce DC current. 2) When the coil is in the vertical position the induced current generated during the first half rotation rises from zero to mascimum and then falls to zero again. 3) As the coil moves further from this position the ends of the coil go to other slip rings. 4) Hence during the second half rotation the current is reversed in the coil itself, the current generated in the second half rotation of the coil is identifical with that during the first half of direct current. Rotation
S N
B
Communication
Brush
101
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
5 Mark Questions 23 . Draw neat diagram of an AC generator ? Ans. Rotation
A N
S
B Slip rings
Brushes 24. Draw the diagrams showing (a) Field lines due to horse shoe magnet between its poles and (b) current into the page Ans. Field lines due to horse shoe magnet between its poles
Current into the page
PART – B I. Fill in the blanks. 1. Electrical energy is converted into mechanical energy in .................... 2. Mechanical energy is converted into electrical energy in .................... 3. the induced current set up in the coil is in such a direction that it opposes the changes in the flux is .................... 4. Current carrying wire produces .................... 5. The S.I. unit of magnetic field induction is .................... 6. Magnetic flux is the product of magnetic field induction and .................... 7. Faraday's Law of induction is the consequences of .................... 8. The device used for producing Electric current is called a .................... 9. The magnetic lines of force of a straight conductor carrying current are .................... 10. The device based on the principle of electromagnetic induction is .................... 11. A current that flows in the some direction is .................... current. 12. A generator with commutator produces .................... current.
102
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
II. Matching. I.
A
B
1. Magnetic field strength
(
)
A. Weber
2. Imaginary linces of force
(
)
B. Tesla
3. Magnetic flux
(
)
C. Oersted
4. Magnetic flux density
(
)
D. Magnetic field
5. current carrying wire
(
)
E. Magnetic lines
II.
A
B
1. Dynamo rule
(
)
A. Gauses
2. Magnetic field
(
)
B. NA–1 m–1
3. Electro magnet
(
)
C. Fleming's right hand rule
(
)
D. BA
(
)
E. Microphones
4. Force on a current carrying conducor 5. Tesla
Answer I. 1) Electric motor 4) Magnetic field
2) Generator
3) Lenzs Law
5) Tesla
6) Area
7) Law of conservation of energy
8) Generator
9) concertric circles
11) Direct
10) Electric generator
12) Direct I. 1) C
2) E
3) A
4) B
5) D
II. 1) C
2) A
3) E
4) D
5) B ❖ ❖ ❖ ❖❖
103
PHYSCIS - Paper - I
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11. ELECTRIC CURRENT 1 Mark Questions 1. Whatis Electric current ?
(As–1)
Ans. Electric current is expressed as the amount of charge flowing through a particular cross section of area in unit time. 2. Why does a bulb glow immediatly when we swich on ?
(As–2)
Ans. When we switch on any elecric circuit, irrespective of length of the connecting wire an elecricfield is set us throughout the conductor instantaneously due to the potential difference of the source connected to the circuit. 3. State the ohm's Law ?
(As–1)
Ans. The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature. 4. What is resistance ?
(As–1)
Ans. The resistance of a conductor is defined as the obstruction to the motion of the electrons in a conductor. 5. What is resistor ?
(As–1)
Ans. The material which offeas resistance to the motion of electrons is called resistor. 6. What are the factors affecting the resistance of a material ?
(As–1)
Ans. The factors affeating the resistance of a material are 1. Temperature 2. Nature of a material 3. Length of a conductor 4. Crossection area 7. What is resistivity ? What is its S.I unit ? Ans. The resistivity of a maerial is the resistance per unit length of a unit cross section of the material the S.I unit of resistivity is ohm meter. 8. When kirchhoff's rules are applicable in Electric current ?
(As–2)
Ans. The kirchhoff's rules are applicable to any DC circuit containing batteries and resistors connected in any way. 9. Name the two kirchhoff's laws ?
(As–1)
Ans. 1. Junction Law 2. Loop Law 84
PHYSCIS - Paper - I
10.
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
Apply the Kirchhoff's Junction Law to the following figure ? I1 I6 I2
I5
I4
I3
Ans. I1 + I4 + I6 = I2 + I3 + I5 The sum of the currents into the junction is equal to the sum of the currents leaving the Junction. 11. What is electric power ? Write the S.I unit of electric power ? Ans. Electric power is the product of potential difference and the current. The S.I unit of electric power is watt.
(As–1)
12. Which type of charge flows through an elecric wire when it is connected in an electric circuit ? Ans. Negative 13. Is there any evidence for the motion of charge in daily life situations ? Ans. Yes, lighting is a live example.
(As–6)
14. What is a value of 1KWH in Joules ? Ans. 1KWH = 1000 × 60 × 60 = 36 ×105 Joules.
(As–1)
15. Find the resistance of a bulb, on which 60 W and 120 V is marked ? Ans. Power P = 60 W, Potential (voltage) V = 120 V 2 Resistance R V = 120 × 120 = 240 Ω P 60 16. Silver is better conductor of electricits than copper why do we use copper wire for conduction electricits. Ans. Silver is better conductor of electricits than copper. But the cost silver is very high when compare with copper. So we use copper wire for conduction of electricity. 17. Why do we consider tungsten as a suitable material for making the filament of a bulb ? (As–2) Ans. Tungsten is consider as a suitable material for making the filament of a bulb because thin wire of tungsten has high resistance and high melting point. When current is passed through it, it becomes hot and emits light. 18. Name the instrument used to measure both electric current and potential difference ? (As–1) Ans. Multimeter is the instrument used to measure both electric current and potential difference. 19. How should we connect the fuse in house wiring circuit ? In series parallel ? Why.(As–6) Ans. The fuse wire in house wiring circuit is connected in series. When the current exceeds the safely limit, the fuse wire melts and breaks the circuit. The electric installations are thus saved from 85
PHYSCIS - Paper - I
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getting damaged. Ω, 4Ω Ω, 6Ω Ω are connected in serias and paralle. Find the equiva20. Three resistors of values 2Ω lent resistance of the combinations ? Ans. R1 = 2 Ω, R2 = 4Ω, R3 = 6Ω Inseries equivalent resistance R= R1 + R2 + R3 = 2 + 4 + 6 = 12 Ω In prallel The eqvivalent resistance
1 1 1 1 = + + R R1 R 2 R 3
=
1 1 1 + + 2 4 6
=
6 + 3 + 2 11 = 12 12
R=
12 11
Ω
2 Mark Questions 21. Write the differences between potential difference and e.m.f ? (As–1) Ans. Potential difference : Work done by the electric force on unit charge is called potential difference. The work has done by the cell to moveunit positive Electromotive force : Charge from negative terminal to positive termenal of the battery. Potential difference =
Workdone Ch arg e
22. How can you veerify that the resistance of a conductor is temperature depends ? (As–1) Ans. 1. When a conductor is connected to a batters, the free elecrons start moving with a drift speed in a special direction. 2. During the motion, the electrons collide with positive ions of the lattice and come to halt. 3. This means that they loss mechanical energy in the form of heat. 4. Thisproves that the resistance of a conducter is temperature dependent. 23. How do you verify that resistance of a conductor is proportional to the length of the conductor for constant cross section area and temperature ? (As–1) Ans. 1. Make a circuit as shown in the figure.
A B
(.) K
86
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PHYSCIS - Paper - I
2. 3. 4. 5.
Connect one of the iron spoke say 10 cm length between A and B. Measure the value of correct using the current not down in your note book. Repeate this for other lengths of the iron spokes. We note that the current decreases with increasing the length of the spoke. So the resistance of a conductor is proportional to its length.
24. By using the following circuit, apply the Kirchhott's second Law or the Loop Law to the I1 EFDCE and Loop ACDBA, LOOP loop EF BAE ? (As–5) V1 the R 1 B A C
I2
V2
I3 + I3
R2 R3
E
D
F
Ans. By using Kirchhoff's Loop Law or second Law. For the Loop ACDBA V1 – I1R1 + I2 R2 – V2 = 0 For the Loop EFDCE I2 R2 + (I + I2) R3 – V2 = 0 For the Loop EFBAE I1 R1 + (I1 + I2) R3 – V1 = 0 25. Explain overloading of house hold current ? (As–1) Ans. 1. The lien wires that are etering into the meter have a potential difference about 240V and limit of current from the mains is 5-20 A. 2. If we consume above 20 A, then circuit resistor in overheating that mars cause a firce. Twisis called overloading. 3. To preventthe damagedue to overloading we connec electric fuse to the household circuit. 26. Two bulbs have rating 100 W, 220 V and 60 W, 220 V. which one has the gratest resistance? V2 Ans. Power P = R 2 For the first bulb P = V = 220 × 220 = 484 Ω R 100 For the second bulb R 2 =
V2 P
=
220 × 220 = 806 . 6 Ω 60
So the second bulb having 60W, 220V has greater resistance. 27. We use parallel arrangement of electrical appiances like Fan, bulb and T.V.** why ? Ans. In a series combination, if any electrical appliance is switched off, all the electrical appliamces in house will be off. Therefore we use parallel arrangement of electrical appliances like Fan, 87
PHYSCIS - Paper - I
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bulb and T.V indmestic circuits. 28. Suppose that you have three resistores each of value 20Ω Ω. How many resitors can you obtain by various combinations of these three resistors ? Draw diagramsin support to your predictions ? (As–1) 20Ω 20Ω 20Ω Ans. 1. Connecting them inseries. 20Ω 2. Connecting them in parallel
= 20 + 20 + 20 = 60Ω
20Ω 20Ω
1 1 1 1 1 +1 +1 3 = + + = = R 20 20 20 20 20
R=
20 = 6 .67 Ω 3
20Ω 20Ωand the remaining third one in parallel 3. Connecting any two of them in series
20Ω
5Ω = 0potential at 'A' is ........ when the potential at B is zero ?(As–6) 29. In 1A the following figureVthe B Ans. 2V The potential at A is V = iR = 1 × 5 = 5 Volt. The potential V = V1 + V2 = 5 + 2 = 7V. 30. A uniform wire of resistance 100 Ω is melted and recasts into wire of length double that of the original. What would be resistance of the new wire formed ? (As–1) Ans. R1 = 100 Ω l1 = t R2 = ? l2 = 2 l R 1 W 11 100 l = 2 = = = R2 t 2 l2 R2 2 × 2l
or
100 1 = R2 4
88
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
R2 = 400Ω
4 Mark Questions 31. Explain how electron flow causes electric current with lorentz – Drude theory of electrons. Ans. 1. Drude and Lorentz are proposed that conductors like metals contain large number of free electrons as while the positive ions are fixed in their locations. The arrangement of the positive ions is called lattice. 2. When the conductor is in an open circuit. The elctrons move randomly in latice space. 3. If we imagine any cross section, the number of electrons, crossing the cross section from left to right in one second is equal to that of electrons passing the cross section from right to left in one second. Hence the net charge moving along a conductor through any cross section is zero. 4. When the ends of the conduct are connected to the battery through a bulb, the bulb flows because energy flow takes place from battery to the bulb. This is because the orderly motion of electrons. 5. When the electrons are in ordered motion, there will be a net charge crossing through any cross section of the conductor. This orchered motion of electrons is called electric current.
Random motion of electrons (in open circuit) 32. Explain Kirchhol's Laws with examples. Ans. 1. Junction Law : At any Junction point in a circuit where the current can divide the sum of the currents into junction must equal the sum I1 of the currents leaving the Junction. I6 I2
I5
I4
I3
I1 + I4 + I6 = I2 + I3 + I5 2. Loop Law : The algetoric sum of the increases and decreases in potential difference across various components of the circuit in aclosed circuit loop must be zero. This Law is based on the conservation of energy. Ex : Let us imagine a circuit loop the poential difference between the two points at the beginning of the loop has a certain value. As we move around the circuit looop and measure the 89
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
potential difference across each component in the loop, the potential difference may decrease on increase depending upon the nature of the element. But when we have completely traversed he circuit loop and arrive I1 back to our R1 the net change in the poential difference must V1 starting point, A B be zero.
C
I2
R2
V2
I1 + I2
R3
E
D
F
Apply loop law to the circuit For the LooP ACDBA – V2 + I2 R2 – I1 R1 + V1 = 0 For the LooP EFDCE – (I1 + I2) R3 – I2 R2 + V2 = 0 For the LooP EFBAE – (I1 + I2) R3 – I1 R1 + V1 = 0 33. State Ohm's Law. Suggest an experiment to verify it and explain the procedure ? (As–1) Ans. Ohm's Law : Ohm's Law states that the potential difference between the ends of a conductor is directly proportional to the electrical current passing through it. Verification of Ohm's Law Aim : To verify Ohm's Law Material required : Batters, rheostat, Resistance, Ammeter, Voltmeter and wire. Procedure : 1. Connections are made as shown in the circuit diagram. 2. By changing the positions of Iron the rheostat, spoke change the flow of current in the circuit. Battery A
Key
3. Note the reading in the voltmeter and ammeter and tabulated below S.No Voltmeter reading Ammeter Reading (V) (I)
R=t
90
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4. From the table we observe that
V i
= Constant. Masis equal to resistance of the wire. So Ohm's
+ Law is verified. V1 34. Observe the circuit and answer the questions given below. – VB – V3 +
–14V
+ V2 –
+ –8V V4 –
1. Are resistors 3 and 4 in series ? Ans. No, they are not in series. They are in parallel. 2. Ans. 3. Ans.
Are resistors 1 and 2 in series ? Yes, they are in series. Is the battars in series with any resistors ? No
4. What is the total emf in the circuit if the potential drop across the resistor 1 is 6 V ? Ans. The total emf in the circuit. V1 + V2 + V3 + V4 = 6 14 + 8 + 8 = 36 V. 35. A house has 3 tube lights, two fans and Television. Each tube light draws 40 W. The fan draws 80 W and the television draw 60 W on he average, it alll the tube lights are kept on fortive hours. Two fans for 12 hours and the television for five hours every day. Find the cost of electric energy used in 30 days at the rate of Rs. 3.00 per KWH. (As–1) Ans. Total consumption of current in 30 days =
(3 × 40 × 5) + (2 × 80 × 12) + (5 × 60 × 30) ×100 1000
=
(600 + 1920 + 300 ) 30 1000
=
(600 + 1920 + 300 ) 30 1000
=
2820 × 30 282 × 3 = = 84 .6 Walts 1000 10
Cost of 1 unit charge = Rs. 3.00 Cost of 84.6 Watts + 846 × 3 = Rs. 253.80 Ps. 36. Deduce the expression for the equivalent resistance of the three resistors connected in series. (As–1) Ans. Let R1, R2 and R3 are the three resistances and V1, V2, V3 are the potential difference across the 91
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PHYSCIS - Paper - I
resistances R1, R2 and R3respectively when they are connected in series. Let I be the current V flowing through each of the resistance and V is the total potential difference in the circuit.
A
V1
V2
R1
R2
V3
R3
B
From Ohm's Law at V1 = IR1, V2 = IR2 and V3 - IR3 Since the resistors are connected in services V = V1 + V2 + V3 ...... (1) Substituting the values of V1, V2 and V3 in (1) V = IR1 + IR2 + IR3 ...... (2) But V = IR ......... (3) From (2) and (3) IR = IR1, + IR2 + IR3 IR = I (R1, + R2 + R3) ∴ R = R1 + R2 + R3 Result : When two or more resistores are connected in series combination then the equivalent resistance is equal to the sum of the individual resistors. 37. Deduce the expression for the equivalent resistance of the three resistors connected in Parallel. (As–1) Ans. Consider R1, R2 and R3 are three resistors connected in parallel. Suppose a current. I flows through the circuit where R1 'V' is connected across the combination. The current I1 a cell of voltage I is divided as I1, I2 and I3 which are flow through R1, R2 and R3 respectively. I = I1 + I2 + I3 ........ (1) I I
2
R2
A
B R3 I3
Ohm's Law as
I1 =
V V V and I 3 = , I2 = R3 R2 R1
92
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PHYSCIS - Paper - I
Substituting the values of curents in (1) I1 =
V V V + + R1 R 2 R 3
But I =
V R
From (2) and (3) V V V V = + + R R1 R 2 R 3
1 1 1 1 + + V = V R R R R 2 3 1
1 1 1 1 = + + R R1 R2 R3
Result : The resiprocal of equivalent resistance is equal to the sum of the resiprocal of the individual resistance
5 Mark Questions → I(In ampere)
38. Draw theYshape of V – I graph for a conductor and the shape of V – I graph for a semi
Ans.
→ I(In ampere)
conductor.
o
→ V(in volt) For conducor
X
o
→ V(in volt) For Semi conductor
39. Explain over loading of house hold circuit with diagram. Ans. 1) Electricity enters our house through two wires called lines these lines are low resistance and the potential difference between the wires as usually about 240 V. 2) All electrical devies are connected in parallel in our home. The P.D drop across each devide is 240 V. 3) Based on the resistance of each electric device. It draws some current from the supply. Total current drawn from the mains is equal to the sum of the currents passing through each de93
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL Heater
vice.
4) If we add more devices to the hosuehold circuit the current from the mains also increases. 4A Fan Bulb 8A Fridge 6A TV Fuse 20A 20V
5) This leads to overheating and may cause a fire. This is called overloading. V V V circuit V2 2 V1 1diagrams
V
V 40. Draw the of three V3 3resistance sare connected in (a) series and (b) parallel I to find the equvivalent resistance. Req I I R1 B R2 R3
I
I1
R1
I2
R2
I3
R3
V
A A
I
I
I Req
A
PART – B I. 1. 2. 3. 4. 5. 6. 7.
Fill in the blanks. .................... is an electric discharge between two clouds or between cloud and earth. .................... conduct electricity. The ordered motion of electron is called .................... The S.I unit of electric current is .................... The obstraction of the flow of current is called .................... The metals which obey Ohm's Law are called .................... conductors. The melting point of tungsten is ....................
94
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PHYSCIS - Paper - I
8. The filament of an electric bulb is made of .................... 9. The S.I unit of power is .................... 10. The Kilowatt hour is the unit of ....................
11. Three resistors of 2Ω, 3Ω, 5Ω are connected to serives. The equivalent resistance of combination of resistors is .................... 12. A thick wire has a .................... resistance than a thin wire. 13. An unknown circuit draws a current of 2A from a 12 V battery its equivalent resistance is .................... 14. If two or more resistors are connected to series, these .................... flows through them is same. 15. 1KW = .................... Watt. 16. Electricity enters our homes through two wires called .................... II. Matching. I.
A
B
1. Electric energy
(
)
A. Volt
2. Potential difference
(
)
B. Ampere
3. Current
(
)
C. Ohm
4. Resistance
(
)
D. Watt
5. Electric power
(
)
E. KWH
II.
A
B
1. Ohm's Law
(
)
A. R = R1 + R2
2. Series connection
(
)
B. V = iR
3. Parallel connection of resistors
(
)
C. P =
4. Power
(
)
D. V = q
5. Potential difference (pd)
(
)
E. R = R + R 1 2
III.
A
1.
W t W
1
1
A
B (
)
A. Volmeter
2.
(
)
B. Battery
3.
(
)
C. Ammeter
4.
(
)
D. Rheostat
5.
(
)
E. Resistance
V
1
Answer I. 1) Lightning 5) Resistance 9) Watt 13) 6Ω I. 1) E 2) A II. 1) B 2) A III. 1) B 2) C
2) Conductor 6) ohms 10) electrical energy 14) current 3) B 4) C 3) E 4) C 3) A 4) E
3) Electric current 7) 3422°c 11) 10Ω 15) 1000 5) D 5) D 5) D ❖ ❖ ❖ ❖❖
4) 8) 12) 16)
Ampere Jungsten Less lines 95
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PHYSCIS - Paper - I
6. REFRACTION OF LIGHT OF CURVED SURFACES 1 Mark Questions 1. Write the names of the lenses shown in figure. a)
(As–1)
b)
Ans. a) Plano – convex lens b) Biconvex lens 2. A convex lens is made up of three different materials as shown in the figure. How many images does it form ? (As–2) Ans. 1) Given convex lens is made up to three different materials have different refractive indices. 2) So the given lens has three different focal lengths. Hence it forms three images. 3. How does alight ray behave when it is passing through the focus of a lens ? (As–1) Ans. When a light ray passing through the focus will take a path parallel to principal axis after refraction. 4. Suppose you are inside the water in a swimming pool near an edge. A friend is standing on 1 1 height ? Why ? (As–7) the edge. Do you find your friend taller or 1shorter than = (n − 1) − usual his f to be taller 2 actual height. R1thanRhis Ans. When I saw my friend through water, he seems This is due to light rays bend away from the normal as they enters rarer medium (air) from denser medium (water)
2 Mark Questions II.5. Write the lens formula and explain the terms in it ? Ans. Lens formula :
(As–1)
1 1 1 = − ..... (1) f v u Where f = the focal length of the lens u = object distance v = image distance
6. Write the lens maker's formula and explain the terms in it ? Ans. Lens marker's formula : Where f = focal length of the lens n = absolute refractive index of the lens R1 = Radius of curvature of first surface R1 = Radius of curvature of second surface
(As–1)
..... (1)
45
PHYSCIS - Paper - I
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7. Two converging lenses are to be placed in the path of parallel rays so that the rays remain parallel after passing through both lenses. How should the lenses be arranged ? Explain with a neat ray diagram. (As–1) Ans. i) A parallel beam of light rays will converge on focal point of the lens after refraction. ii) Light rays passes through the focal point will parallel to principal axis after refraction. iii) So the two lenses are arranged on a common principal axis such that their focal points coincide with each other, then the rays remain parallel after passing through both lenses.
f1
f2
8. Write the names of different types of lenses. Draw their diagrams ? Ans. Different types of lenses : 1) Biconvex lens
–
2) Biconcave lens
– ⇒
3) Plano – Convex lens
–
4) Plano – Concave lens
–
5) Concavo – Convex lens
–
(As–1)
1 1 1 1 1 = − = − v f u 20 60
9. The focal length of a converging lens is 20 cm. An object is 60 cm from the lens. Where will the image beformed and what kind of image is it ? (As–1) Ans. f = 20 cm; u = 60 cm; image distance (v) = ? 1 1 1 Lens formula : = − f v u
1 3 −1 2 1 = = = v 60 60 30 ∴ v = 30 cm; Here a real diminished and inverted image is formed.
46
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
10. What is the focal length of double concave lens kept in air which two spherical surfaces of radii R1 = 30 cm and R2 = 60 cm. Take refractive index of lens as n = 1.5 ? (As–1) Ans. n = 1.5; R1 = 1.5 ? R2 = 60 cm (Q double concave lens)
−20 −200 = 0.5 5 ∴ f = –40 cm Here minus sign indicates that the lens is divergent. ⇒f =
11. A double convex lens has two surfaces of equal radii 'R' and refractive index n = 1.5. Find the focal length 'f'. (As–1) Ans. R1 = R2 = R −0.5 3 − 22−1111 1.01 11− 11 = (0.5) 0.5 (1.5 − 1) =−=+− 0.5 (1.5 (n20−− 1)1) For a double convex lens, R1 = R, R2 = –R, nf== 1.5 − 60 30R 2 60 f 1 1 RR RR R 260 1 1 1 = (n −1) − f R1 R 2
∴f=R ∴ The focal length of a double convex lens is 'R'. 12. Find the refractive index of the glass which is a symmetrical convergent lens if its focal length is equal to the radius of curvature of its surface ? (As–7) Ans. Given lens is symmetrical convergent lens R1 = R; R2 = –R Focal length of the lens f = R Lens maker's formula
1 1 1 = (n −1) − f R1 R 2 47
PHYSCIS - Paper - I
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⇒ a = 2(n – 1) ⇒ n – 1 =
⇒n=1+
= 1.5
∴ n = 1.5 13. Find the radii of curvature of a convexo – concave convergent lens made of glass with refractive index n = 1.5 having focal length of 24 cm. One of the radii of curvature is double the other ? (As–7) Ans. n = 1.5; f = 24 cm; R2 = 2R1 Lens maker's formula :
=
=
=
151 11 1 22 1f d+ 1f 1 −11−11+1−− 12 1 = +− = ×=(n ×−1) 24 = (1.5 1) (n (0.5) − 1) 10 ff224 f12 f 2 2R ff fR 2 2 RfR 11 1122R2R
R1 = ∴ R1 = 6 cm R2 = 2R1 = 2 × 6 = 12 cm 14. Let us assume a system that consists of two lenses with focal length f1 and f2 respectively. How do you find the focal length of the system experimentally, when (As–3) i) two lenses are touching each other ii) they are separated by a distance d' with common principal axis. Ans. i) Two lenses are touching each other : The focal, lengths of two lenses having f1 and f2 respectively if they touch each other then the focal length of the system is
F=
f1 f 2 ..... (1) f1 + f 2
ii) If the lenses are separated by a distance 'd' with common principal axis. Then the focal length of the system ..... (2) 48
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
15. Draw a ray diagram for an object placed between F1C1 of convex lens. Explain the nature and position of image ? (As–5) Ans. Object is placed between F1C1 of convex lens : J C1
C2
F2
I
P
O F1
G
(Note : OJ = object; IG = Image) Charactenstics : 1) Image is formed beyond C2. 2) Real image is formed. 3) Inttered image. 4) Image is magnified 16. Draw a ray diagram for the image formed at C2. Explain the nature of image and position of the object ? (As–5) L Ans. F2 C1
C2
F1
If the image formed at C2. 1) The object should be placed at C1. 2) The image is real. 3) The image is inverted. 4) The image is same size as that of object.
4 Mark Questions III.17. How do you find the focal length of a lens experimentally ? (As–1) Ans. Aim : To determine the focal length of a convex lens. Apparatus : Convex lens, meter scale, V-stand, screen, candle. procedure : 1) Take a v-stand and place it on a long (nearly 2 m) table at the middle. 2) Place the convex lens on a 'V' stand. 3) Now place the candle at a distance of 60 cm from the lens, such that the flame of the candle lies along the principal axis of the lens. 4) Place the screen on the other side of the lens and adjust to get a clear image on it. 5) Now measure the image distance (V) between the lens and the screen. 49
PHYSCIS - Paper - I
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6) Now repeat the experiment for various object distances like 50 cm, 40 cm, 30 cm etc. 7) Measure the image distances in allcases and note them in table. 8) Calculate the focal length in each case by using the formula, f =
uv and noteit in the table. u+v
9) We find that the focal length remains constant in each case. Table : S.No. Object distance Image distance Focal length (u) (v) f = uv/u + v 1. 60 cm 2. 50 cm 3. 40 cm 4. 30 cm 18. Draw ray diagrams (for a convex lens) for the following positions and explain the nature and position of image. (As–5) i) Object is placed at infinity. ii) Object is placed beyond C2. iii) Object is placed at C2. iv) Object is placed between F2 and C2. v) Object is placed at F2. vi) Object is placed between P and F2. Ans. i) Object at infinity :
C2
L
F2
F1
C1
Characteristics : 1) Image is formed at the focal point. 2) The image is real, inverted and point size. ii) Object is placed beyond C2: L F1 C2
F2
C1
P
Characteristics : 1) The image is formed between F1 and C1. 2) The image is real, inverted and diminished. 50
PHYSCIS - Paper - I
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iii) Object is placed at C2 :
L F1
C2
F2
C1
P
Characteristics : 1) The image is formed at C1. 2) The image is real, inverted and same size as that of object. iv) Object is placed between C2 and F2.
L F1
C1 P
F2
C2
Characteristics : 1) The image is formed beyond C1. 2) The image is real, inverted and magnified. v) Object is placed at F2 : L F1 P
F2
C2
C1
Characteristics : 1) The image is formed at infinity. 2) We cannot discuss the nature of image. vi)Object is placed between F2 and P : L F1
C2
F2
C1
P
Characteristics : 1) The image is formed on the same side of the object. 2) The image is virtual, erect and magnified. 51
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL BITS
I. Multiple Choice questions : 1. Wich one of the following materials cannot be used ot make a lens A) Water
B) Glass
C) Plastic
(
)
(
)
D) Clay
2. Which of the following is true ?
A) The distance of virtual image is always greater than the object distance for convex lens B) The distance of virtual image is not greater than the object distance for convex lens C) Convex lens always forms a real image
D) Convex lens alwyas forms a virtual image
3. Focal length of the plano - convex lens is ........ when its radius of curvature of the surface is R and n is the refractive index of the lens ( ) A) f = R
B) f =
C) f =
D) f =
4. The value of the focal length of the lens is equal to the value of the image distance when the rays are ............. ( ) A) passing through the optic centre
B) Parallel to the principal axis R D) R− 1) 1 1(n 1 C) Passing through the focus (nIn+ −all 1) the cases + = − (n f2 R− 1) R R2 5. Which of the following is the lens maker's formula ? 1 A)
B)
C)
D)
6. The distance between focus and 'Optic centre' is .............. A) Focal length
B) Radius of curvatureC) Principal axis
(
)
(
)
D) None
II. Fill in the blanks. 7. The rays from the distance object, falling on the convex lens pass through .................... 8. The rays passing through the .................... of the lens is not deviated. 9. Lens formula is given by .................... 10. Lens makers formula is given by .................... 11. The focal length of the plano-convex lens is '2R' where R is the radius of curvature of the surface. Then the refractive index of the material of the lens is .................... 52
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
12. The lens which can form real and virtual images is .................... 13. The lens which always form virtual imagesis .................... 14. Converging lens is .................... 15. For drawing ray diagrams, we represent canvex lens with a symbol .................... III. Matching. 16. Object is beyond C2
(
)
A. Image is formed beyond C1
17. Object is at C2
(
)
B. Image is formed between F1 and C1
18. Object is between C2 and F2
(
)
C. Image is formed at infinity
19. Object is at F2
(
)
D. Image is formedis formed at F1
20. Object is between F2 and P
(
)
E. Image is formed on same side of object F. Image is formed at C1
Answer I. 1) D
2) A
3) C
4) B
II. 7) FOCAL point (focus)
5) C
6) A
10)
8) Optic centre 9) 1 1 1 1 b 1 1 = = (n −− 1) − ff v u R 1 R2 11) 1.5 12) Convex lens
13) Concave lens
14) convex lens
III. 16) B
17) F
18) A
19) C
15)
20) E
❖ ❖ ❖ ❖❖
53
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
7. HUMAN EYE AND COLOURFUL WORLD 1 Mark Questions 1. What is least distance of distinet vision ? What is its value for a human being ? Ans. The minimum distance at which an object is to be placed so that it can be viewed distinetly and confortably is known as least distance of distinct vision. The value of least distance of distinet vision is 25 cm. (This value change from person to person and with age of person) 2. What is angle of vision ? What is its value for healthy human being ? Ans. The maximum angle at which we are able to see the whole object is called angle of vision. 3. What is accommodation of eye lens ? Ans. The ability of eye lens to change its focal length is called accommodation of lens. 4. How many types of eye defects ? What are they ? Ans. There are mainly three common defects of vision. They are i) Myopia ii) Hypermetropia iii) Presbyopia 5. What is Presbyopia ? Ans. Presbyopia is a vision defect when the ability of accommodation of the eye usually decreases with agening. 6. What is the difference between for point and near point Ans. The point of maximum distance at which the eye lens can form an image on the retina is called 'far point'. The point of minimum distance at which the eye lens can form an image on the retina is called near point. 7. What is Dispersion of light. Give an example. Ans. The splitting of white light into its constituent colour (VIBGYOR) is known as dispersion. Ex : The phenomenon 'Dispersion' is resparsible for producing rainbow. 8. What is scattering of light. Give an example. Ans. The process of re-emission of absorbed light in all directions with different intensities by atoms or molecules is called scattering of light. 9. What is 'Minimum deviation angle' ? Ans. The angle between incident ray and emergent ray is known as angle of deviation. When the angle of incidence is equal to angle of emergence the angle of deviation attains least value. This is known as 'angle of minimum deviation'. 10. What is the role of rods and cones in human eye ? Ans. Rods – identify the colour Cones – Identify the intensity of light 11. What are the maximum and minimum focal lengths of eye lens ? Ans. Maximum focal length of eye lens = 2.5 cm Minimum focal length of eye lens = 2.27 cm 54
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
12. To corrent Myopia and Hypermetropia. Mention the focal lengths of bi-concave and biconvex lens ? Ans. i) To correct the Myopia The focal length of bi-concave lens is f = – D Here D = Distance to far point from eye ii) To correct the Hypermetropia The foal length of bi–convex lens is f = Here d = Distance to near point from eye. 13. Define power of lens ? What are its units ? Ans. The reciprocal of focal length is called power of lens. Let f be the focal length of lens, Power of lens P =
1 f (in m)
(or) P = The unit of power of lens is 'Dioptre' (D) 14. Write a formula to find the refractive index of the material of the prism. And explain the 25 100 terms. dA + D sin cm)2 Ans. Refractive index of the material of a prism df (in − 25 A sin 2 n=
Here A = Angle of prism D = Angle of minimum deviation
2 Mark Questions II.1. Explain briefly the reason for the blue of the sky. Ans. 1) Sky appears blue due to a phenomenon known as scattering of light. 2) Our atmosphere contains different types of molecules and atoms. The reason for blue sky is due to the molecules N2 and O2. 3) The sizes of these molecules are comparable to the wave length of blue light. These molecules act as scattering centres for scattering of blue light. 4) That is why sky appears in blue colour. 2. How do you appreciate the role of molecules in the atmosphere for the blue colour of the sky ? Ans. 1) The sky appear blue due to atmospheric refraction and scattering of light through molecules. 55
PHYSCIS - Paper - I
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2) 3) 4) 5)
Molecules behaves as scattering entres. The reason to blue sky is due to the molecules N2 and O2. The sizes of these molecules are comparable to the wavelength of blue light. In the absence of molecules there will be no scattering of sunlight and the sky will appear dark. 6) We should appreciate the molecules which are scattering centres. 3. Why does the sky sometimes appear write when you view in certain direction on hot days ? Ans. 1) Our atmosphere contain atoms and molecules of different sizes. 2) According to their sizes, they are able to scatter different wavelengths of light. 3) For example, the size of the water molecule is greater than the size of the N2 or O2. 4) It acts as a scattering centre for other frequencies which are lower than the frequency of blue light. 5) On a hot day, due to rise in the temperature, water vapour enters into atmosphere which leads to abundant presence of water molecules in the atmosphere. 6) These water molecules scatter the colours of other frequencies other than blue. All such colours of other frequencies reach your eye and the sky appears white. 4. Why the reasons for appearance the red colour of sun during sunrise and at sunset ? And does not appear red during noon hours ? Ans. 1) The atmosphere contains free molecules and atoms with different sizes. These molecules and atoms scatter light of different wavelengths which are comparable to their size. 2) Molecules having a size that is comparable to the wavelength of red light are less in the atmosphere. 3) Hence scattering of red light is less when compared to the other colours of light. 4) The light from the sun needs to travel more distance in atmosphere during sunrise and sunset to reach our eye. 5) In morning and evening times, during sunrise and sunset except red light all colours scatter more and vanish before they reach us. 6) Since scattering of red light is very small, it reaches 4s. As a result sun appears red in colour during sunrise and sun set. 7) During noon hours, the distance to be travelled by the sun rays in the atmosphere is less than that compared to morning and evening hours. 8) There fore all colours reach our eye without much scattering. Hence the sun appears white during noon hours. 5. How do you appreciate the working of cilliary muscles in the eye ? Ans. 1) The ciliary muscles to which eye lens is attached helps the eye lens to hange its focal length by changing the radii of curvature of the eye lens. 2) When the eye is focussed on a distant object, the ciliary muscles are released so that focal length has its maximum value and when the eye is focussed on a closer object, the ciliary muscles are strained so that focal length of eye lens decreases. 56
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3) Thus ciliary muscles adjust the focal length. This accommodation is a wonderful phenomenon through which we are able to see the distant and near objects. 4) If this ciliary muscles are not present, we cannot see the objects beyond a certain distance. If we imagine this we cannot guess our normal life. Hence the role of ciliary musceles is highly appreciable. 6. Glass is known to be a transparent material. But ground glass is opaque and white in colour. Why ? Ans. 1) Glass is generally a transparent material because it transmits most of the light incident on it. 2) when glass is ground it's surface becomes rough due to microscopic unevenness. 3) When light is incident on such a rough surface it is reflected in many directions. 4) This type of reflections is known as diffuse reflection. Due to this ground glass is opaque and white in colour. 7. If a white sheet of paper is stained with oil the paper turns transparent. Why ? Ans. The refractive index of oil and refractive index of paper is same then light passes from oil to paper with out scattering hence the paper become transperent. 8. Geetha can read a book but she does not able to read the letters on backboard clearly. i) What defect has Geetha ? ii) Which type of lens she has to use to correct her eye defect ? 100100 = 20 cm Ans. i) Geetha has eye defect 'Myopia'. f p2(in cm) ii) She has to use 'Biconcave lens'' to correct her eye defect. 9. Ramu can see the name boards of buses cearly from long distance. But he can not read paper clearly. i) What type of eye defect has Ramu ? ii) What kind of lens does Ramu use to correct his eye defect ? Ans. i) Ramu has eye defect – 'Hyper metropia' ii) 'Hypermetropia' can be corrected by using 'biconvex lens'. 10. Doctor advised to use 2D lens. What is its focal length ? Ans. Given power of lens P = 2D We know that power lens P = Where f = focal length of the lens ∴ Focal length f = = 57
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
11. A prism with an angle A = 600 produces an angle of minimum deviation of 300. Find the refractive index of material of the prism. Ans. Given angle of prism A = 600 Angle of minimum deviation D = 300 ∴ Refractive index of the material of the prism A+D sin 2 n = A sin 2
n =
1 sin 45 2 = 2 = = 2 = 1.414 = o 1 sin 30 2 2 o
60o + 30o 12. Explain an activity for the formation of artificial sin rain bow. 2 Ans. 1) Take a metal tray and fill it with water. o 2) Place a mirror in the water such that it makes an angle 60 to sin the water surface. 2 3) Now focus white light on the mirror through the water. 4) Try to obtain colours on a white cardboard sheet kept above the water surface. 5) We can observe the seven colours of VIBGYOR on the card board sheet.
Dispersed light
white light
Mirror Metal tray
4 Mark Questions III.1. What is the eye defect Myopia ? How do you correct the Myopia ? Ans. 1) Some people cannot see object at long distances but can see nearby objects clearly. This type of defect in vision is called myopia (near sightedness) 2) For the people of myopia, maximum focal length is less than 2.5 cm so that the rays coming nfrom distant object after refraction through the eye lens form an image before the retina as shown in the figure.
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PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3) The point of maximum distance at which the eye lens can form an image on the retina is called far point (M). 4) The eye lens can form clear image on the retina when an object is placed between far point (M) and point of least distance of distinct vision (L). 5) If we are able to bring the image of the object kept beyond far point, between far point and the point of least distance of distinct vision using a lens. 6) This image acts as an object for the eye lens. This can be possible only when a bi-concave lens is used.
7) To correct one's myopia, we need to select a lens which forms an image at the far point for an object at infinity. 8) We need to select bi-concave lens to achieve this. 2. What is the eye defect hypermetropia ? Explain the correction of the eye defect Hypermetropia. Ans. 1) Hypermetropia is also known as far sightedness. A person with hypermetropia can see distant objects clearly but cannot see objects at near distances, because the minimum focal length of eye for the person of hypermetropia is greater than 2.27 cm. 2) In such cases the rays coming from a near by object, after refraction at eye lens, forms an image beyond the retina as shown in figure.
H
L
3) The point of minimum distance at which the eye lens can form an image on the retina is called near point. 4) The people with defect of hypermetropia cannot see objects placed between near point (H) and point of least distance of distinct vision (L). 5) Eye lens can form a clear image on the retina when any object is placed beyond near point. 6) To correct the defect of hypermetropia, we need to use a lens which forms an image of an object beyond near point (H), when the object is between near point (H) and least distance of distrinct vision (L). This is possible only when a double convex lens is used.
H
L
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PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3. How do you find experimentally the refractive index of material of a prism ? Ans. Aim : To find the refractive index of a prism. Material required : Prism, Piece of white chart, pencil, pens, scale and protractor. Procedure : 1) Take a prism and place it on the white chart and draw line around triangular base and measure the angle of prism (A) between the sides PQ, PR. P
Normal
angle of the prism angle of deviation
A O
incident ray
d
i1 B
A
Normal
M
r N
i2 angle of emergency emergent ray
C Q
R
D
angle of deviation d
2) Draw a normal to PQ at M and draw a line with 300 to the normal. This is incident ray AB . Fix two ball pins on this ray at A and B. 3) Place the prism in its exact position and Fix another two pins at C and D such that all four pins appear to lie along the same line by seeing the images of pins through the prism from the other side PR. 4) Draw line joining C and D and extend it to meet PR at N this is emerging ray. Draw normal at A + D emergent ray CD . PR at N and measure the angle between normal sin at N and 2 5) If we extent the incident ray AB and emergent ray CD, they meet at O. Measure angle beA tween these two ray and note as angle of deviation. sin 2 6) Similarly repeat the process for different angles of incidence and measure corresponding angle of deviation. 7) We draw a graph by taking angles of incident on X-axis and angles of deviation (d) on y-axis, we will get a curve as D shown in the figure. Find angle of minimum deviation (D). 8) Now we can calculate the refractive index of the material of i1 = i2 O the prism by using the formula. i1
Angle of incidence
n=
.
4. Explain the formation of rainbow. Ans. 1) The beautiful colours of the rainbow are due to dispersion of the sunlight by millions of tiny sun ligh water droplets. t
water drop
Violet 400 450 Red
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✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
2) The rays of sunlight enter the drop near its top surface. At this first refraction, the white light is dispersed into its spectrum of colours, violet being deviated the most and red the least. 3) Reaching the opposite side of the drop, each colour is reflected back into the drop because of total internal reflection. Arriving at the surface of the drop, each colour is again refracted into air. 4) At the second refraction the angle between red and violet rays further increases when compared to the angle between those at first refraction. 5) The angle between the incomins and outgoing rays can be anything between 00 and about 420. We observe rain bow when the angle between incoming and outgoing rays is near the maximum angle of 420. 6) Although each drop disperses a full spectrum of colour an observer is in a position to see only a single colour from any one drop depending upon its position. 7) The colour red will be seen when the angle between a beam of sunlight and light sent back by a drop is 420. The colour violet is seen when the angle between a sun beam and light sent back by a drop is 400. 8) If we look at an angle between 400 and 420 we can observe the remaining colours of VIBGYOR. sun
ligh
t
water drops viole
t
100 d25t d 3d d443− 25
viole Red
Red
5. Find focal length of a lens suggested to a person with hypermetropia is 100 cm. Then find i) the distance of near point ii) power of the lens Ans. Given focal length of bi convex lens f = 100 cm i) Let d be the distance of near point. Then Focal length of lens f = 25 d 100 = d − 25 100 (d – 25) = 25 d d d – 25 = ; d – = 25 4
= 25 f=
= 33.33 cm
∴ Distance of near point d = 33.33 cm 61
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
PHYSCIS - Paper - I
ii) We know that power of lens P = p=
( f = 100 cm)
p = 1D 6. A light ray fals on one of the focas of prism at an angle 400. So that it suffer angle of minimum deviation of 300. Find the angle of prism and angle of refraction at the given surface. Ans. Given that angle of incidence i1 = 400 Angle of minimum deviation D = 300 i) at angle of minimum deviation i1 – i2 (Here i2 is angle of emergency) so A + D = 2i, ∴ A = 2i1 – D = 2(400) – 300 = 500 ii) at angle of minimum deviation Angle of refraction r = =
= 250
Q o×d50 100 A25 25 100 7. A person is unable to see the objects nearer50 than 50 cm. He wants to read a book placed at 100 fd250 −2525 cm) 2(in distance of 25 cm.
i) Name the defect of the vision he is suffering from ii) How can it be corrected iii) Ans. i) ii) iii)
What is the power of such lens ? Since the person is not able to see the objects nearer than 50 cm. This defect of vision 'hypermetropia' can be corrected by using 'bi–convex lens'. Given that distance of near point d = 50 cm If f be the focal length of the bi convex lens we have f= 25 × 50 f = 50 − 25 =
= 50 cm
Now power of lens p = ∴p=
=2D
∴ By using convex lens of power 2D, we can correct given defect of vision. 62
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
8. A person cannot see objects beyond a distance of 2 m. Then find i) What type of eye defect he has ? ii) What kind of lens he has to use to overcome his eye defect ? iii) What is focal length of the lens ? iv) What is power of the lens he has to use ? Ans. i) Since the person is not able to see the objects beyond 2m, he is suffering from 'Myopia' or short sightedness. ii) Thiss defect of vision – myopia can be corrected by using 'bi–concave lens'. iii) Given that distance of far point D = 2m We know that focal length of lens using to correct myopia is f = –D Where D is distance of far point. ∴ Focal length of f = – 2m = – 200 cm iv) Now power of lens p = =
= –0.5 dioptre
Here – indicates that it is concanve lens.
5 Mark Questions 100 100 f−(in 200cm)
1. Draw the figure showing eye parts.
(T.B. Page 138)
Ciliary muscles
Iris
Retina
Pupil Aqueous humar Cornea Optic nurve
2. Draw the figures showing how the eye defects occured and how will they corrected. Myopia : (T.B. Page 141)
fig : 5(a)
M
fig : 5(e) M
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Hypermetropia : (T.B. Page 142) fig : 6(a) H fig : 6(e) H
BITS I. Multiple Choice questions : 1. The size of an object as perceived by an eye depends primarily on
(
A) actual size of the object
B) distance of the object form the eye
C) aperture of the pupil
D) size of the image formed on the retina
)
2. When objects at different distances are seen by the eye which of the following remain constant ? A) focal length of eye - lens
B) object distance from eye lens
C) the radii of curvature of eye - lens
D) image distance from eye lens
(
)
3. During refraction ................... will not change. ( ) 100 1 A) wave length B) frequency C) speed of light D) all the above (cmm) s) f (mm) (in 4. Far point of a person is 5 m. In order that he has normal vision what kind of spectcles should he use ( ) A) concave lense with focal length 5 cm
B) concave lense with focal length 10 cm
C) convex lense with focal length 5 cm
D) convex lense with focal length 2.5 cm
5. The process of re-emission of absorbed light in all directions with different intensities by the atom or molecule is called ( ) A) scattering of light
B) dispersion of light C) reflection of light D) refraction of light
6. Power of lens = 1 A) f (cm 5)
( B)
C)
)
D)
7. The vision defect when the ability of accommodation of the eye usually decreases with agening A) Hypermetropia
B) Myopia
C) Presbyopia
D) Blindness
(
)
8. Some time a person may suffer from both myopia and hypermetropia with agening. Then he need to use ( ) A) bi-concave lens
B) bi-convex lens
C) bi-focal lens
D) cooling lens 64
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
9. The splilting of white light into seven colours is due to A) dispersion
B) scattering
C) reflection
B) Refraction
C) Dispersion
B) 2.5 cm / 2.27 cm B) cones
)
(
)
(
)
C) 25 mm/22.7 mm D) B and C
12. Which part of retena identify the colour A) rods
( D) Scattering
11. The maximum and minimum focal lengths of eye lens are A) 24 cm / 22.7 cm
)
D) refraction
10. The reason for the blue colour of the sky is A) Reflection
(
C) A and B
D) none of the above
13. The angle of minimum deviation for an equalateral triangle prism is found to be 300. Its refractive index is ( ) B)
A)
C)
D)
14. An equilateral triangle prism is arranged in minimum deviation position for an angle of incidence of 450. The angle of minimum deviation is ( ) A) 450
B) 600
C) 300
D) 00
15. The colour which has least wavelength in visible spectrum VIBGYOR A) Violet
B) Red
16. Refractive index of material of prism
A)
B)
Green sin 12C)(A ++B) 1A D +A sin sin− −D 22 2sin 2A sin sin (A sin +AB) 2 C)
B) 50 cm
C) 200 cm
)
(
)
(
)
(
)
D) Blue
D)
17. If power of lens of 2D then focal length of that lens A) 100 cm
(
D) 25 cm
18. If power of lens is + 0.5 D then that lens is A) Concave lens of focal length 50 cm
B) Convex lens of focal length 50 cm
C) Convex lens of focal length 200 cm
D) Concave lens of focal lengeth 200 cm
19. m
L
The adjacent figure shows which eye defect ? And how it will correct ? A) Hypermetropia, bi-concave lens
B) Myopia, bi-convex lens
C) Hypermetropia, bi-convex lens
D) Myopia, bi-concave lens
(
)
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PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
20. Eye lens adjust its focal length according to distance of object. Which helps for this ? ( A) Carnia
B) Retina
)
C) Ciliary muscles D) Rods and Cones
II. Fill in the blanks. 1. The value of least distance of distinct vision is about .......................... 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
Myopia can be corrected by using .......................... lens. In minimum deviation position of prism, the angle of incidence is equal to angle of .......................... The maximum focal length of the eye lens is about .......................... The distance between the eye lens and retina is about .......................... The eye lens can change its focal length due to working of .......................... muscles. Angle of vision in healthy human is .......................... The splitting of white light into different colours (VIBGYOR) is called .......................... If a person can see only near object clearly. Then he is suffering with eye defect .......................... The ability of eye lens to change its focal length is called .......................... The shape of rainbow when observed during travel in an aeroplane is .......................... The reason for formation of rainbow is .......................... During refraction of light the character of light which does not change is .......................... The reciprocal of focal length is called .......................... The angle between incident ray and emergent ray is ..........................
Answer I. 1) B
2) B
3) B
4) A
5) A
6) C
7) C
8) C
9) A
10) D
11) D
12) A
13) B
14) A
15) A
16) B
17) B
18) D
19) D
20) C
II. 1) 25 cm
2) bi-concave lens
3) angle of emergent
5) 2.5 cm
6) ciliary muscles
7) 600
9) Myopia
10) Accommodation of lens
11) circular
12) dispersion
13) frequency
15) Angle of Deviation
14) power of lens
4) 2.5 cm 8) dispersion
❖ ❖ ❖ ❖❖
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PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
8. ATOMIC STRUCTURE 1 Mark Questions 1. What is a spectrum ? How many types of spectrums are there ? Ans. A group of wave lengths is called spectrum. Spectra are of 2 types. A) Emmisson spectrum B) Absorption spectrum. 2. Which type of spectrum is rain bow ? Ans. Continuous spectrum 3. What is electron configuration ? Ans. The arrangement of electrons in shells, sub-shells and orbitals of an atom is called electron configuration. 4. The electron configuration of Helium is 1s2. Write the information conveyed by it ? Ans. He = 1s2 '1' denotes principal quantum number 'S' denotes angular momentum quantum number '2' denotes the number of electrons present in that orbital 5. Raju represented the configuration of Helium atom as
. Which rule is violated in this ? 1s2
Ans. Pauli's exclusion principle is violated here. Electrons with paired spoins are donotated by Opposite spins.
.
6. Write the values of magnetic quantum number for the sub shell d. Ans. The angular momentum quantum number (l) for sub shell d is '2'. So magnetic quantum number m = 2l + 1 i.e., 2 × 2 + 1 = 5. The values are –1 – 10 + 1 + 2 7. What is the maximum value of 'l' for n = 4 ? Ans. n = 4 so l = n – 1 = 4 – 1 = 3 i.e., 'f' orbital 8. Haw many spin orientations are possible for an electron in an orbital ? Ans. Two spin orientations are possible for an electron in an orbital i.e., clockwise and anti clock wise. If both are positive values than the spins are parellel otherwise anti parellel. 9. State Pauli's exclusion principle ? Ans. Pauli's exclusion principle states that no two electrons of the same atom have four quantum numbers same. 10. State Aufbau principle ? Ans. Aufbau principle states that the lowest energy orbitals are filled first. 67
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
11. State Hund's rule ? Ans. Hund's rule states that the orbitals of equal energy are occupied with one electron each before pairing start's. 12. Write the electronic configuration of calcium atomic number 20 ? Ans. Ca Atomic number (z) = 20 = 1s2 2s22p63s23p64s2 13. Among Red and blue colours which is having high energy – Give reason ? Ans. Blue colour is having high energy. This is because the wavelength of blue colour is less than red colour. The colour with lower wave length has higher frequency. 14. Following orbital diagram shows the electron configuration of oxygen atom. Which rule does not support this ? (As–1) O=8
1s2 2s2 2px 2py 2px
Ans. Hund 's rule because degenerate orbitals are filled first then pairing starts. 15. An electron in an atom has the following set of four quantum numbers to which orbital it belong to and name that element ? n
l
m
s
2
0
0
–
1 2
1 2
Ans. It belong to 2s2. Name of the element is Beryllium. 16. Write the four quantum numbers for 1s1 electron ? Ans. n l m s 1
0
0
+
17. Why ar Bohr's orbits called stationary orbits ? Ans. As long as the electron is revolving in an orbit it's energy is same. Hence these orbits are called stationary orbits. 18. Write Planck's equation and write what each letter represents in that equantion. Ans. E = h λ Here E = Energy of radiation h = Planck's constant λ = frequency of the radiation absovbed / emitted 19. When do we get Atomic line spectra ? Ans. Atomic line spectra arise because of absorption or emmission of certain frequencies of light energy. 68
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
2 Mark Questions II.1. Write the differences between Orbit and Orbital ? Ans. Orbit Orbital 1. The path of the electrons which re1. The space around the nucleus volves around the the nucleus is where the probability of finding the called orbit. electron is maximum is called orbital 2. Orbitals have definite shape. Ex2. Orbits are circular and non-direccept 's' orbital other orbitals are ditional. rectional. 2. The wave length of a radio wave is 1 m. Find its frequency ? Ans. Wave length of a wave = 1 m Speed of light = 3 × 108 m/sec. Frequency = ? C= λ 3 × 108 = ×1 = 3 × 108 Hez.
=
8 3 × 10 3. Krishna said that, "The velocity of an electron and its exact position can not be determined at a time". Do you agree with this statement -1 explain. Ans. Yes, I agree with the statement of Krishna. Because electrons are very small. To know its position light of very short wave length is required. This short wave length light interacts with the electron and disturbs the motion of the electron. Hence it is not possible to determine its position and velocity accurately.
4. Write the four quantum numbers for the differenting electron of potassium atomic number 19 ? Ans. K = 19 = 1s2 2s2 2p6 3s2 3p6 4s1. The last electron enters into 4s orbital. So the four quantum numbers are as follows n l m s 4
0
0
+
1 2
5. The Electronic configuration of an element 'X' is given as belon, observe it and Answer the Questions ? X = 1s2 2s2 2p6 3s2 3p6 4s2 3d1 A) Name the element 'X' B) Which is the outer most shell ? Ans. A) Name of the element 'X' is Scandium. B) Outer most shell is '4'. 69
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✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
6. Write the electronic configuration of copper atomic number 29. Which rule is deviated ? Then why is it so ? Ans. Cu = 29 = 1s2 2s2 2p6 3s2 3p6 4s1 3d10 Aufbau principle is deviated here. Lower energy 4s is not complete filled. But this is because half filled and completely filled orbitals are stable. 7. What is ground state and excited staten? Does the electron remains in the excited state forever ? Ans. Lowest energy stte of the electron is known as ground state. By gaining energy it moves to a high energy level called excited state. The electron does not remain in the excited state for ever. By loosing energy the electron come back to its ground state.
4 Mark Questions III.1. Write the values of Angular momentum Quantum number, Magnetic quantum number, number of electrons present in the orbitals of principal quantum number '2'. Ans. Principal Angular Magnetic Orbital Number of quantum momentum qunatum notation electrons present number qunatum number in the orbitals number (l) 2l + 1 2
0
0
2s
2
1
–10 + 1
2p
6
2. What is (n + l) rule ? On its basis explain the order of filling of 3d, 4s and 4p orbitals. Ans. 1) Electrons enter into the orbitals in the increasing order of (n + l) values. 2) If two orbitals have same (n + l) value, the electrons enter into the orbital which has lower 'n' value. 3)
Orbitals
(n + 1) value
3d
3 + 2= 5
4s
4+0=4
4p
4+1=5
4) So, the electrons first enter into 4s then 3d and then 4p. 3. By considering carbon as an example explain Hunds rule ? Ans. 1) Hunds rule states that electron pairing takes place only after filling all the available degenerate orbitals with one electron each. 2) Ex : Carbon. Its atomic number is '6'. So it is having six electrons. 3) The first 4 electrons goes into 1s and 2s orbitals. 4) The next two electrons enter into 2p orbitals with both electrons having same spin. 1s2 2s2 zpz C=6= 70
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
4. In an atom the number of electrons in M shell is equal to the electrons present in the L shell. Consideing K shell as the first shell. Answer the following questions. A) Which is the outer most shell. B) How many electrons are there in its outer most shell ? C) What is the atomic number of that element ? D) Write the electronic configuration of that element ? Ans. Number of electrons in K shell = 2 Number of electrons in L shell = 8 Number of electrons in M shell = 8 A) Outer most shell is M. B) 8 electrons are there in its outer most shell. C) The atomic number of that element is 18. D) The electronic configuration is 1s22s22p63s23p6
5 Mark Questions 1. Draw the diagrams of the orbitals having. Spherical, Dumbell, double dumbell. Ans. Y
Y
Z X
s-orbital
Y Z
X pX-orbital
X pY-orbital
pZ-orbital
All orbital
2. Draw the chart showing the filling order of atomic orbitals (Moeller chart).
71
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I. Multiple Choice questions : 1. Planck's constant A) 6.626 × 10–32 JS
B) 6.626 × 10–34 JS B) Band spectrum B) Sommer feld
C) m
B) 2
C) 3
C) 2l + 1
B) 5
8. In the equation c = A) C
C) 10
λ the letter that represents1 frequency is 1 + − 2 λ2 B) C) B) Phosphorus
C) Sulphur
B) 2
C) 3
(
)
(
)
(
)
(
)
(
)
D) h D) Argon
10. 'P' orbital starts from .......... orbit. A) 1 II. Matching.
)
D) 14
9. 1s2 2s2 2p6 3s2 3p4 is the electronic configuration of A) Potassium
(
D)
7. Maximum number of electrons that can be accomorated in 'd' sub shell is A) 7
)
D) 4
6. Magnetic quantum number (ml) values can be known from B) n – 1
( D) s
5. The 'l' value of 'f' orbil is
A) n
)
C) Ludwig planck D) Erwin schordinger
B) l
A) 1
)
(
4. Size and energy of the main shell was given by A) n
( C) Green spectrum D) Violet spectrum
3. Quantum mechanical model of an atom was proposed by A) Neil Bohr
)
C) 6.626 × 10–27 JS D) 6.626 × 1039 JS
2. Hydrogen spectrum is A) Line spectrum
(
D) 4
1. Scandium
(
)
A. 1s2 2s2 2p6 3s2 3p6 3d10 4s1
2. Aluminium
(
)
B. 1s2 2s2 2p6 3s2 3p6 4s2 3d1
3. Copper
(
)
C. 1s2 2s2 2p6 3s2 3p6 3d5 4s1
4. Neon
(
)
D. 1s2 2s2 2p6 3s2 3p1
5. Chromium
(
)
E. 1s2 2s2 2p6 F. 1s2 2s2 2p6 3s2 3p6 3d2
Answer I. 1) B
2) A
3) D
8) B
9) C
10) B
II. 1) B
2) D
3) A
4) A
5) C
4) E
5) C
6) C
7) C
❖ ❖ ❖ ❖❖ 72
PHYSCIS - Paper - I
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9. CLASSIFICATION OF ELEMENTS - PERIODIC TABLE 1 Mark Questions 1. Which group of elements are called chalcogens. Ans. The elements in a group 16 (VIA) which form ores with elements are called chacogens. 2. Which group elements has electronic configuration of ns2 or ns2np6. Ans. VIII A group elements have the configuration of ns2 or ns2np6. 3. How does the metallic character changes in periods and groups. Ans. 1) The metallic character increases as we move along a group. 2) The metallic character decreases as we move along a period (from left to right). 4. How does the valency changes in periods from left to right. Ans. In periods the valence has increased from 1 to 4 and after decreased from 3 to 1. In group valence has not changed. 5. The Element X Belongs to 4th period and 5th group. Write the no of valence electrons, Valency and state whether it is metal or non-metal. Ans. X element belongs to 4th period and 5th group : Arsenic Valence : 03 Metallic character : metalloid
2 Mark Questions II.1. Write the limitations of law of triads. Ans. Limitations : i) All the then known elements could not be arranged in the form of triads. ii) The law failed for very low mass or for very high mass elements. In case of F, Cl, Br, the atomic mass of Cl is not an arithmetic mean of atomic masses of F and Br. iii) As the techniques improved for measuring atomic masses accurately, the law was unable to remain strictly valid. 2. Why the elements are classified. Ans. They approximately more than 115 elements. We cannot easily understand chemical and physical properties of it. So there is a necessicity to classify the elements. 3. What is Dobereiner triad ? Give Two examples to it. Ans. Dobereiner discovered that ''The relative atomic mass of the middle element in each triad was close to the average of the relative atomic masses of the other two elements". This statement is called the Dobereiner's law of Triads. Li : 07
Ca : 40
Cl : 35.5
Na : 23
Sr : 87.5
Br : 80
K : 39
Ba : 137
I : 127 73
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
4. Why lanthanides and actinides are placed separately at the bottom of the periodic table. Ans. The properties of these elements does not coincide with other elements because the valance electron enters in to 4f and 5f orbitals respectively. So they are placed separately at the bottom of the periodic table. 5. Define modern periodic law ? How many periods and groups are present. Ans. Modern Periodic Law : "The physical and chemical properties of elements are the periodic function of the electronic configurations of their atoms". The modern periodic table has eighteen vertical columns known as groups and seven horizontal rows known as periods. 6. The Atomic Numbers of elements A, B, C, D and E are 7, 10, 12, 41, 88 respectively. a) Which have same chemical properties. b) Which are inert in nature. c) Which belongs to 3rd period. d) Which belongs to non metallic group. Ans. A) C & E B) B C) C D) A 7. The atomic number of element is 16. a) What is its valency b) belongs to which group c) Name of the element d) Metal or non-metal Ans. a) 2 b) VI A Group c) Sulphur
d) Non metal
8. X, Y and Z elements has electronic configuration as mentioned below. X = 2, Y = 2, Z = 2, 8, 2 a) Which belongs to Group II b) Which belongs to II period c) Which is inert in nature Ans. a) The element ''Y'' belongs to II Group b) The Valence of the element Z : 2 so it belongs to II period c) The X element has completely filled outer most orbital. So it belongs to Inert Group. 9. Identify the element that has the lower Ionization energy in each pair of the following i) Mg or Na ii) Li or O iii) Br or F iv) K or Br Ans. Ionization energy decreases as we go down in a group and generally increases from left to right in a period. a) Mg Or Na : Mg. Na are belongs to same period. Na has lower I.E Value b) Li or O : Li has lower I.E Value. c) Br or F : Br and F are belongs to same group. Br has less I.E Value. d) K or Br : K and Br are belong to same period. K has less I.E Value. 10. The electronic Configuration of the A, B, C, D elements are A) 1s22s2 B) 1s22s22p63s2 C) 1s22s22p23s23p6 D) 1s22s22p6 74
PHYSCIS - Paper - I
1. 2. 3. 4. Ans. 1) 2) 3) 4)
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
Which belongs to same period Which belongs to same group Which are inert C element belongs to which Group and Period B and C are belongs to same period. A and B are belongs to same group. D belongs to inert gas. C element belongs to 3 period and V A group.
4 Mark Questions III. 1.Write the newlands law of octaves and drawbacks of this rule ? Ans. Newlands law of octaves states that when elements are arranged in the ascending order of their atomic masses they fall into a pattern in which their properties repeat at regular intervals. Every eighth element starting from a given elements resembles in its properties to that of the starting element. a) Certain elements, totally dissimilar in their properties, were fitted into the same group. b) The law was not valid for elements that had atomic masses higher than Calcium. c) Newland's periodic table was restricted to only 56 elements. 2. a) Why the elements are classificed : b) What is the properity has to taken for classifying the elements by mendeleeff. c) In mendeleefs periodic table some gaps are not filled by the elements ? Why. Ans. a) They approximately more than 115 elements. We cannot easily understand chemical and physical properties of it. So there is a necessicity to classify the elements. b) Mendeleeff's states that the physical and chemical properties of the elements are a periodic function of their atomic weights". c) Based on the arrangement of the elements in the table he predicted that some elements were missing and left blank spaces at the appropriate places in the table. Mendeleff believed that some new elements wouldbe discovered definitely. 3. What are the main properties of mendeleef periodic table ? What are the limitations. Ans. The Periodic Law : Based on Mendeleeff's observations regarding the properties of elements in the periodic table, a law known as the periodic lw of the properties of elements was proposed. "The law states that the physical and chemical properties of the elements are a periodic function of their atomic weights". Salient features and achievements of the Mendeleeff's periodic table : 1) Groups and sub-groups : There are eight vertical columns in Mendeleff's periodic table called as groups. They are represented by Roman numerals I to VIII. Elements present in a given vertical column (group) have similar properties. Each group is divided into two sub-group 'A' and 'B'. The elements within any sub-group resemble each other to great extent. For example, sub-group IA elements called 'alkali metals' (Li, Na, K, Rh, Cs, Fr) resemble each other very much. 2) Periods : Thehorizontal rows in Mendeleeff's periodic table are called periods. There are sevenperiods in the table, which are denoted by Arabic numerals 1 to 7. A period comprises the entire range of elements after which properties repeat themselves. 75
PHYSCIS - Paper - I
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3) Predicting the properties of missing elements : Based on the arrangement of the elements in the table he predicted that some elements wre missing and left blank spaces at the appropriate places in the table. Mendeleef believed that some new elements would be discovered definitely. He predicted the properties of these new additional elements in advance purely depending onhis table. His predicted properties wre almost the same as the observed Properties of those elements after their discovery. He named those elements tentatively by adding the prefix 'eka' (eka is a Sanskrit numeral means one) to the name of the element immediately above each empty space. The predicted the properties of elements namely eka-aluminium, ekaboron, eka-aluminium and eka-silicon were close to the observed properties of Scandium, Gallium and Germanium respectively whichwere discovered later. Limitations of Mendeleeff's periodic table : 1. Anomalous pair of elements : Certain elements of highest atomic mass precide those with lower atomic mass. For example, tellurium (atomic mass 127.6) precedes iodine (atomic mass 126.9) 2. Dissimilar elements placed togehter : Elements with dissimilar properties were placed in same group as sub-group A and sub-group B. For example, alkali metal like Li, Na, K etc., of IA group have little resemblance with coinage metals like Cu, Ag, Au of IB group. 3. Some similar elements separated : Some similar elements like 'copper and mercury' and 'silicon and thalium' etc are placed in different groups of the periodic table. 4. Write the name of elements of imagine by mendeleef ? Write their names given by mendeleef ? Ans. Predicting the properties of missing elements : Based on the arrangement of the elements in the table he predicted that some elements were missing and left blank spaces at the appropriate places in the table. He named those elements tentatively by adding the prefix 'eka' (eka is a Sanskrit numeral means one) to the name of the element immediately above each empty space. The predicted the properties of elements namely eka-aluminium, eka-boron, eka-aluminium and eka-silicon were close to the observed properties of Scandium, Gallium and Germanium respectively which wre discovered later. 5. Define the modern periodic law ? Discuss the contructionof the long form of periodic table. Ans. Modern Periodic table : Based on the modern periodic law the physical and chemical properties of atoms of the elements depend not on the number of protons but on the number of electrons and their arrangements (electronic configurations) in atoms. Therefore, the modern periodic law may be stated as "The physical and chemical properties of elements are the periodic function of the electronic configurations of their atoms". 1. The modernperiodic table has eighteen vertical columns known as groups and seven horizontal rows known as periods and also arranged the elements in a period according to Atomic number. 2. The elements with similar outer shell (valence shell) electronic configurations in their atoms are inthe same column called group. 3. Depending upon to which sub-shell the diferentiating electron. i.e., the last coming electron enters in the atom of the given element, the elements are classified as 's' 'p', 'd' and 'f' block elements. 76
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
a) The elements with valence shell electronic configuration from ns2np2 to ns2np2 are called pblock elements. b) The s-block and p-block elements together known as Representative elements. The elements with valence electronic configuration ns2np6. c) nd to ns2np6 nd10 are called d-block elements. In d-block elements as we move from left to right in periodic table, we observe a transition of elements from metals, to non-metals. Hence we call d-block elements as "Transition elements". d) The elements in which f-orbitals are being filled in their atoms called f-block elements. These elements are also called "inner transition elements". Periods : The horizontal rows in the periodic table are called periods. They are seven periods in form of periodic table. These periods are represented by Arabic numerals 1 through 7. 1. They are only two elements in first period e.g., hydrogen (H) and helium (He). 2. They are only Eight elements in Second and Third periods. 3. The fourth and 5th period contains 18 elements. 4. The 6th period contains 32 elements. 5. The seventh period is incomplete period. '4f' elements are called Lanthanoids or Lanthanides. Elements from 58Ce to 71Lu possess almost the same properties as 57La. So they are called lanthanoids. The 5f elements are called Actinoids also known as Actinides they are from 90Th to 103Lr. The f-block elements known as lanthanoids and actinoids are shown separately at the bottom of the periodic table. 6. Explain how the elements are classified into S, P, D and Fblock elements in the periodic table and give the advantage of t his kind of classification. Ans. Depending upon the which sub-shell the differentiating electron i.e., the last coming electron enters in the atom of the given element, the elements are classified as 's' 'p', 'd' and 'f' block elements. S Block Elements : The elements with valence shell electronic configuration ns1 np6 are called P-block elements : The s-block and p-block elements together known as Representative elements. D Block Elements : The elements with valence electronic configuration ns2np6nd1 to ns2np6nd10 are called d-block elements. In d-block elements as we move fromleft to right in periodic table, we observe a transition of elements from metals, to non-metals. Hence we call d-block elements as "Transition elements". F Block Elements : The elements in which f-orbitals arebeing filled in their atoms called fblock elements. These elements are also called ''inner transition elements". Advantages : 1) The systematic grouping of elements into groups made the study simple. 2) Each period begins with the electron entering a new shell and ends with the complete filling of S and P sub shells of that shell. 7. Elements in a group generally posses similar properties but elements along a period have different properties. How do you explain this statement. Ans. 1) Physical and chemical properties of elements are related to their electronic configuration, particularly the outer shell configuration. 77
PHYSCIS - Paper - I
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2) Therefore all the elements in a group should have similar chemical properties. 3) Similarly, across the table from left to right in any period, elements get an increase in the atomic number by one unit between any two successive elements. 4) Therefore, the electronic configuration of valence shell of any two elements in a period is not same. Due to this reason, elements along a period possess diferent hemical properties. 8. Identify the element that has the alrger atomic radius in each pair of the following. i) Mg or Ca ii) Li or Cs iii) N or P iv) B or Al Ans. Atomic radii of elements decrease across a period from left to right and increase in groups from top to bottom. i) Mg or Ca : The atomic radius of Ca is greater than Mg because both element are belongs to same group but the atomic number of Calcium is more than Mg. Therefore to accommodate more number of electrons, more shells are required. ii) Li or Cs : The atomic radius of Ca is greater than Li because both element are belongs to same group but the atomic number of Cs is more than Li. Therefore to accommodate more number of electrons, more shells are required. iii) N or P : The atomic radius of P is greater than N because both element are belongs to same group but the atomic number of P is more than N. Therefore to accommodate more number of electrons, more shells are required. ii) B or Al : The atomic radius of Al is greater than B because both element are belongs to same group but the atomic number of Al is more than B. Therefore to accommodate more number of electrons, more shells are required. 9. What is ionization energy ? What are the factors influence it. Ans. Ionization Energy : The energy required to remove electron from the outer most orbit or shell of a neutral gaseous atom is called ionization energy. The following factors are influenced on Ionization energy. a) Nuclear Charge b) Secreening Effect c) Penetrating power of the orbitals d) Stable configuration e) Atomic Size 10. Define atomic radius ? How does the property has change in Groups and periods. Ans. The half of the distance between two nuclei outer most orbital is called atomic radius. Atomic radii increase from top to bottom in a group (column) of the periodic table. As we go down in a group, the atomic number of the element increases. Atomic radii of elements decrease across a period from left to right. As we go to right, electrons enter into the same main shell. Hence, the nuclear attraction on the outer shell increases. As a result the size of the atom decrease. 11. How do the following properties change in Groups and periods a) Ionisation Energy b) Electronic affinity c) metallic and nonmetallic nature Ans. a) Ionisation Energy : Period : When we move from left to right it does follow a regular trend but generally increases due to increase in atomic number. Group : In groups from top to bottom, the ionization energy decrease due to increase in atomic size. 78
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
b) Electronic affinity : Period : Electron affinity values increases from left to right in a period. Group : Electron affinity values decrease from top to bottom in a group. c) metallic and nonmetallic nature : Period : Metallic nature decrease from left to right in a period and increase nonmetallic nature. Group : Metallic nature increase from top to bottom in a group. 12. How do you appreciate the role of electronic configuration of the atoms of elements in periodic classification ? Ans. 1) Modern periodic table is based on electronic configuration. So elements are arranged in ascending order of their atomic number. 2) The chemical properties of elements depends on valence electrons the elements in same group has same number of valence electrons. So the elements belongs to same group have similar properties. 3) So the construction of modern periodic table mainly depends on electronic configuration. 4) Certain elements of highest atomic mass precede those with lower atomic mass. This type of Anomalous pair of elements are also rectified in Modern periodic table. 5) Hence electronic configuration play a major role in the preparation of Modern periodic table. So it is appreciated.
BITS I. Fill in the blanks. 1. The scienctist who classified elements first is ............ 2. Li, Na, K are examples for ............ 3. According Mendeleeff physical and chemical properties of elements are periodic functions of ............ 4. Ek-Born proposed by mendelee H after named as ............ 5. Modern periodic table was prepared based on the elements ............ 6. In mendaleeff's periodic table ............ group are prensent. 7. Modern periodic table was prepared by ............ 8. The in complete period in modern period table is ............ period. 9. In the modern periodic table ............ groups are present. 10. In the modern periodic table ............ periods are present. 11. The electronic configuration of as element is 1s22s22p63s2 the it belongs ............ period. 12. The electronic configuration of as element is 1s22s22p4 then it belongs to ............ block. 13. The valency of noblegases is ............ 14. The same group elements has same ............ 15. In a group from top to bottom metlic nature ............ 16. It atomic size of 'A' is less than 'B' than ionisation energy of is ............ than B. 17. Transition elements are called ............ block elements. 18. f-block elements are known as ............ elements. 79
PHYSCIS - Paper - I
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19. 20. 21. 22.
Among Na, Na+ ............ has least atomic radiis The elements from atomic number 58 to 71 are called as ............ The elements from atomic number 90 to 103 are called as ............ If 'A' belongs to 3rd period 'B' belongs 4th periods then ............ has more atomic radius.
23. 24. 25. 26. 27. 28. 29. 30. II.
Most electronegative element is ............ Least electronegative element is ............ Due to increase in screening effect value ............ decreases. The electronic configuration of Cr is ............ The electronic configuration of Cu is ............ If an element beongs to 3rd period then its valency is ............ Electromegetivty scale was invented by ............ The noble gas which does's has octet configuration in outer most shall is ............ Matching.
a) 1. Alkali earth metals
(
)
A. IA groups
2. Helogens
(
)
B. Gallions
3. Noblegases
(
)
C. IIA groups
4. Alkali meatals
(
)
D. VIIA groups
5. Ek-Aluminium
(
)
E. O' group
(
)
A. pM
2. Atomic radius
(
)
B. KJ mol–1
3. Ionisction energy
(
)
C. Dobereiser
4. S, P block element
(
)
D. configuration
5. ns2, np6
(
)
E. representative elements
b) 1. Cl, Br, I
Answer I. 1) Doberiner
2) Dobreiner triod
3) Atomir weight
4)
5) electronic configuration
6) 8
7) H–J Moseley
8) 7th period
9) 18
10) 7
11) 3
12) p
13) 0
14) valency eletronic configuration
15) increase
16) more
17) 'd' block
18) inner transtion
19) Na+
20) Lanthanides
21) Actinides
22) 'B' element
23) Fluorine (F)
24) Cesium (Cs)
25) Ionisation energy
26) [Ar] 4s1 3d5
27) [Ar] 4s13d10
28) 3
29) pauling
30) Helium (He)
II. a) 1) C
2) D
3) E
4) A
5) B
b) 1) C
2) A
3) B
4) E
5) D
Seandium
❖ ❖ ❖ ❖❖ 80
PHYSCIS - Paper - I
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10. CHEMICAL BONDING 1 Mark Questions 1. Give examples of elements which are stable in their atomic state ? Ans. Helium, Neon, Argon, Krypton, Xenon, Radon are stable in their Atomic state. 2. Why the Noble gases are least reactive ? Ans. Except Helium other Noble gases have eight electrons in their outer most shell. i.e., why Noble gases are least reactive. 3. What is octet rule ? Ans. Presence of eight electrons in the outermost orbit is called octet rule. 4. Why is the chemical formula of water is H2O why not HO2 ? Ans. The valency of hydrogen is one (1) Valency of oxygen is two (2) So two hydrogen atoms shares 'Their electron with one oxygen atom. i.e., why the chemical formula of water is H2O. 5. Name the bond formed between Alkalimetals and Halogens ? Ans. Inoic bond is formed between Alkalimetals and Halogens. 6. What type of bond is formed in 17th group / VII A group elements ? Ans. Covalent bond 7. What is coordination number ? Write the coordination number of sodium chloride ? Ans. The number of ions of opposite charge that surround a given ion of given charge is known as coordination number. Co-ordination number of solid sodium chloride is 6. 8. Write the conditions that are favourable for the formation of cations ? Ans. The atoms of elements with low conisation energy, low electron affinity high atomic size and low electro negativity form cations. 9. Write the conditions that are favourable for the formation of anions ? Ans. The atoms of elements with high ionisation potential, high electron affinity, small atomic size and high electro negitivity form anion. 10. Give example for ionic compounds ? Ans. Sodium chloride (NaCl), Magnesium chloride (MgCl2), Aluminium chloride (AlCl3). 11. Give examples for covalent compounds ? Ans. Oxygen molecule (O2), Nitrogen molecule (N2), Methane molecule (CH4), Ammonia molecule (NH3), water molecule (H2O). 12. Explain the difference between the valence electrons and the covalency of an element ? Ans. Valence electrons : The electrons present in the outer most orbit of an atom are called valence electrons. Covalency : The total number of covalent bonds that an atom of an element forms is called its covalency. 81
PHYSCIS - Paper - I
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13. Bond length of F2 is 1.44 Aº. What does it mean ? Ans. The equilibrium distance between the nuclei of two fluorine atoms is 1.44 A0. 14. The Bond dissociation energy of H – F molecule (Hydrogen fluorid) molecule is 570 KJ mol–1. What does it mean ? Ans. 570 KJ mol–1 energy is needed to break the covalent bond of hydrogen fluoride molecule.
2 Mark Questions II.1. What is VSEPRT (theory) ? Who proposed it ? Ans. 1) To explain the bond angles inthe molecules through covalent bonds the valence-shell-electron pair repulsion theory was proposed. 2) This theory was proposed by sidgwick and powell and improved by Gillespie and Nyhlan. 2. A chemical compound has the following Lewis notation : A) Write the valence electrons of A. B × B) Write the valence electrons of B. C) How many covalent bonds are there in the molecule. B× A ×B D) Suggest a name for the elements A and B. × B Ans. A) 4
B) 1
C) 4
D) CH4
3. The electronic configuration of X, Y are given belon. X = 1s22s22p63s23p1 Y = 1s2 2s22p4 Answer the following questions. 1) What is the valency of X ? 2) Which will form Anion ? 3) Which will form cation ? 4) What is the valency of X ? Ans. 1) Valecny of X = 3 2) 'Y' will form Anion 3) 'X' will form Cation 4) Valency of Y = 2 4. Why do only valence electrons involve in bond formation why not electrons of inner shells ? Ans. The nucleus and the electrons in the inner shell remain unaffected as they arenot exposed directly when atoms come close together. But the electrons in the outer most shell of an atom get affected. Thus the valence electrons are responsible for the formation of bond between atoms. 5. What are Lewis symbols ? What is the significance of Lewis symbols ? Ans. The symbol of the element, surrounded by the valence electrons of its atom, represented in the form of dots around it is known as Lewis symbols or electron dot symbols. 82
PHYSCIS - Paper - I
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Significance : 1) The number of dots present around the symbol gives the number of electrons present in the outer most shell. i.e., number of valence electrons. 2) The common valency of the element is equal to the number of dots around the symbol. 6. What are the Draw backs of electronic theory of valence ? Ans. 1) When covalent bond is formed between any two atoms, irrespective of the nature of the atoms, the bond lengths and bond energies are expected to be the same. But practically it was observed that bond lengths and bond energies are not same when the atoms that form the bond are different. 2) If fails to explain the shapes of the molecules. 7. Why VESPER THEORY is proposed ? What is the draw back of this theory ? Ans. To explain the bond angles in the molecules with 3 or morethan 3 atom with all atoms attached to a central atom through covalent bonds. VESPER Theory was proposed. Drawback : VESPER theory does not explain the strongths of the bonds.
4 Mark Questions III.1. An element X has electronic configuration 1s2 2s2 2p6 3s2 An element Y has electronic configuration 1s2 2s2 2p6 3s2 3p5 What type of bond is formed between then ? Explain ? Ans. 1) The electronic configuration of X = 1s2 2s2 2p6 3s2 − ze − X → X 2+ 2) It contains 2 electrons in its outer most orbit. 3) It looses its 2 electrons to attain the stable structure of Ne. 4) The electronic configuration of Y = 1s2 2s2 2p6 3s2 3p5 5) It has 7 electrons in its outer most orbit. To attain the stable structure it requires are electron. −
+e 6) Y → Y−
7) The two electrons of X is transformed to 2y atoms. 8) The bond formed is Ionic bond X2+ + 2y– → XY2 2. Two chemical reactions are described below. 1) Nitrogen and hydrogen react to form ammonia. 2) Carbon and hydrogen react to form a molecule of methane. For each reaction give a) The valency of each of the atoms involves in this reaction. b) The Lewis structure of the product that is formed. Ans. A) The valency of each of the atoms involved in this reaction. N = 7 = 1s2 2s2 2p3 Valency = 3 1 H = 1 = 1s Valency = 1 C = 6 = 1s2 2s2 2p3 Valency = 4 83
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
Note : 1) Valency of VA group and above is 8 – group number. 2) Group number becomes valency up to IV A. B) N + 3H → NH3 (Ammonia) N + 3 H x →
N × × × H H H
C + 4 H → CH4 (Methane) H × C
+
4 HX → H×
H
C × H
×H
or
C H
H
H
3. Usha said the bond angle of water 1090281. But priya told the bond angle of water is decreased to 1040311. How priya told this – Explain ? Ans. 1) Priya told the bond angle of water is 1040311 on the basis of Hybridisation. 2) Hybridisation is a phenomenon of inter mixing of atomic orbitals of almost equal energies of an atom and their redistribution into an equal number of identical orbitals. 3) Oxygen = 8 = 1s22s22p4. The valence orbital of oxygen atom has one 2s orbital and three 2p orbitals (2px, 2py, 2pz) inter mix and redistribute into for identical sp3 orbitals. 4) Among the six outer most electrons two sp3 orbitals get pairs and the sp3 orbitals get one electron each. 5) The two sp3 orbitals of oxygen atom overlap with 's' orbitals of two hydrogen atoms giving σ sp3 – s bonds. 6) Due to the lone pair - lone pair repulsions and lone pair bond pair repulsions the bond angle of water decreases from 1090 281 to 1040311.
H H 4. Explain the formation of Nitrogen molecule on the basis of valence bond theory ? Ans. 1) A covalent bond between two atoms is formed when the two atoms approach each other closely and one atom overlaps its valence orbital containing unpaired electron with the valence orbital of the other atom that contains un paired electron of opposite spin. The paired electrons in the over lapping orbitals are attracted to the nuclei of both the atoms. 2) Ex : Nitrogen = Z = 7 = 1s2 22 2p3 px py pz 84
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3) px orbital of one Nitrogen atom overlaps with the px orbital of the other Nitrogen atom giving σ px – px bond along the inter nuclear axis. 4) py and pz orbitals of one 'N' atom overlap the py and pz orbital of other 'N' atom laterally, perpendicular to inter nuclear axis giving πpy – py and πpz – pz bonds. 5) Thus N2 molecule has a triple bond between two Nitrogen atoms. πpy – py πpz – pz σ px – px +
5 Mark Questions 1. Draw the Lewis structure of following. A) Oxygen molecule B) Methane molecule C) Ammonia molecule D) Water mole A) Oxygen molecule
B) Methane molecule
C) Ammonia molecule D) Water mole
2. Hybridisation structure of A) Beryllium chloride
B) Boran trifluoride
BITS I. Multiple Choice questions : 1. Which one of the following is an ionic compound A) HCl B) CO2 C) H2O 2. In N2 molecule the atoms are bonded by A) 1 σ 3 π B) 1 σ 2 π C) 3 σ 2 π
(
)
(
)
D) CaO D) 2 σ 1 π 85
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3. Covalent compounds are soluble in ( ) A) polar solvants B) Non–polar solvents C) Concentrated ions D) All solvents ( ) 4. The bond angle in NH3 molecule is 0 1 0 0 1 0 A) 109 28 B) 90 C) 107 48 D) 105 5. Polar covalent bond is present in ( ) A) HCl B) NaCl C) MgO D) Na2O 6. Valency of Noble gases is ( ) A) 0 B) 1 C) 3 D) 5 7. The number of unpaired electrons in oxygen atom is ( ) A) 0 B) 1 C) 2 D) 3 8. The number of valency electrons in carbon atom is ( ) A) 0 B) 2 C) 4 D) 6 9. Among the following electron deficinet molecule is ( ) A) NaCl B) MgO C) CH4 D) BeCl2 10. Ionic bond is formed between atoms of elements with electro negitivity difference equal to or greator than ( ) A) 0.1 B) 1.0 C) 1.9 D) 9.1 II. Fill in the blanks.
11. Transfer of electrons between atoms of two dissimilar elements leads to the formation of .............. bond. 12. Elements with more electro negative character forms .............. 13. The equilibrium distance between the nuclei of tw atoms which form a covalent bond is called .............. 14. The bond angle in BeCl2 is .............. 15. Ionic compounds have .............. melting point. 16. In shared electron pair is called .............. 17. .............. proposed the Ionic bond. 18. Covalent bond was proposed by .............. 19. VSEPRT was proposed by .............. 20. Valence bond theory was proposed by .............. III. Matching. Group – A Group – B 1. BeCl2 ( ) A. Pyramid 2. H2O ( ) B. linear 3. CH4 ( ) C. face centred cubic structure 4. NH3 ( ) D. V shape 5. NaCl ( ) E. Tetrahedral F. Trigonal bipyramid
Answer I. 1) D 2) B 3) B 4) C 5) A 6) A 7) C 8) C 9) D 10) C II. 11) Ionic 12) Anions 13) Bond length 14) 1800 15) High 16) Lone pair 17) Kossel 18) G.N. Lewis 19) Sidgwick and Powell 20) Linus pauling III. 1) B 2) D 3) E 4) A 5) C ❖ ❖ ❖ ❖❖ 86
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
1. HEAT 1 Mark Questions I. 1. Define 'Heat' ?
(As–1)
Ans. Heat : Heat is a form of energy that flows from a hotter body to a colder body. 2. Define 'Temeperature' ?
(As–1)
Ans. Temperature : The degree of hotness or coldness of a body is called 'Temperature'. 3. Convert 200C into kelvin scale ?
(As–1)
Ans. 200C = (20 + 273) K0 = 293 K0 4. What is 'Evaporation' ?
(As–1)
Ans. Evaporation : "The process of escaping of molecules from the surface of a liquid at any temperature is called evaporation". 5. What is humidity ?
(As–1)
Ans. Humidity : The amount of water vapour present in air is called 'humidity'. 6. Why does ice floats on water ?
(As–1)
Ans. The density of ice is less than the density of water. So ice floats on water. 7. If we put the glass bottle into the fridge for a few hours which is completely filled with water and fix the lid tightly will break. Why ? (As–1) Ans. If we put the glass bottle into the fridge for a few hours, the water in the bottle freezes to ice. The volume increases. So the bottle is broken. 8. A samosa appears to be cool outside but it is hot when we eat it. Why ?
(As–7)
Ans. A samosa appears to be cool outside but it is hot when we eat it because the curry inside the samosa contains ingredients with higher specific heats. 9. How much energy is released or obsorbed when 1 gm of water at 00C freezes to ice at 00C ? (As–1) Ans. 80 cal. of heat is released when 1 gm of water at 00 C freezes to ice at 00C. (Note : Q = mL, m = 1 gm; L = 80 Cal/gm) 10. How much energy is transferred when 1 gm of boiling water at 1000C condenses to water at 1000C ? (As–1) Ans. 540 cal of energy is transferred, when 1 gm if boiling water at 1000C condenses to water at 1000C. (Note : Q = mL, m = 1 gm; L = 540 Cal/gm) 1
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
2 Mark Questions II.11. Write the differences between Heat and Temperature ? Ans.
Heat
(As–1)
Temperature
1. Heat is a form of energy that flows from a hotter body to a colder body.
1. The degree of hotness or coldness of a body is called 'Temperature'.
2. The SI unit of heat is Joule.
2. The SI unit of temperature is Kelvin.
3. Heat is measured by calorimeter.
3. Temperature is measured by thermometer. 12. Write the differences between evaporation and boiling ? Ans.
Evaporation
Boiling
1. The process of escaping of molecules from the surface of a liquid at any temperature is called evaporation.
1. A process in which the liquid phase changes to gaseous phase at a constant temperature and costant pressure is called boiling.
2. Evaporation takes place at any temperature.
2. Boiling takes place at boiling pint 3000only. 100 3. Boiling is a bulk phenomenon.
3. Evaporation is a surface phenomenon.
(As–1)
13. What would be the final temperature of a mixture of 50 gm of water at 200C temperature and 50 gm of water at 400C temperature ? (As–1) Ans. m1 = 50 gm; T1 = 200C; m2 = 50 gm; T2 = 400C Final temperature (T) = =
m1T1 + m 2 T2 m1 + m 2 (50 × 20) + (50 × 40) 50 + 50
2000 + 1000 = 100 ∴ Final temperature = 300C
=
14. Write the formula for the amount of heat absorbed (released) by a substance and explain the terms in it ? (As–1) Ans. Q = mS ∆T ..... (1) Where Q = Amount of heat absorbed by a substance m = Mass of the substance s = Specific heat of the substance and ∆T = Change in temperature. 2
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
15. The temperatures of two substances A and B in different cases are given in the table. Case Substance
1
2
3
4
5
A
300C
500C
–2730C
100K
400K
B
300C
300C
00 K
100C
400C
Write the answer for the following questions using the table :
(As–1)
i) In which cases A and B are in thermal equilibrium ? Give reason ? ii) In which cases heat transfers from A to B ? Ans. i) In 1st and 3rd cases A and B are in thermal equilibrium because they have equal temperatures. ii) In 2nd case heat transfers from A to B. 16. Explain why dogs pant during hot summer days using the concept of evaporation ?(As–1) Ans. 1) Dogs do not have sweat poress on their body. 2) Dogs pant during hot summer days to reduce their internal temperature. 3) When the dogs pant, the water molecules present on the tongue gets evaporated by absorbing some heat from their body. 4) As a result the interior of dog's body gets cooled. 17. Equal amounts of water are kept in a cup and in a dish. Which will evaporate faster ? Why ? (As–3) Ans. 1) Rate of evaporation of a liquid increases with increase in surface area. 2) The surface area of water in a dish is more than the surface area of water in a cup. 18. Why do we prefer to sip hot tea with a saucer rather than a cup ?
(As–3)
Ans. 1) The rate of evoporation of a liquid increases with increase in surface area. 2) The surface area of a saucer is more than the surface area of a cup. 3) So tea in a saucer evaporates faster. 4) As a result tea in a saucer becomes cool fastly than in a cup. Hence we prefer to sip hot tea with a saucer rather than a cup. 19. Why do we sweat while doing work ? Why do we feel cooler after sweating ?
(As–7)
Ans. 1) When we do work, we spend our energy mostly in the form of heat from the body. 2) As a result the temperature of the skin becomes higher. 3) Then the water in the sweat glands comes out so we get sweat. 4) The sweat evaporates by absorbing the heat from our body. This make us to feel cool. 20. What role does specific heat play in keeping a watermelon cool for a long time after removing it from a fridge on a hot day ? (As–7) Ans. 1) Watermelon contains large percentage of water. 2) Water has higher specific heat value. Hence water melon takes a lot of time for cooling after removing it from a fridge on a hot day. 3
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
21. If you are chilly outside the shower stall, why do you feel warm after the bath if you stay in the bathroom ? (As–7) Ans. 1) In the bathroom the number of vapour molecules per unit volume is greater than the number of vapuor molecules per unit volume outside the bathroom. 2) When we try to dry ourselves with a towel, the vapour molecules surrounding us condense on our skin. 3) Condensation is a warming process. 4) Hence we feel warm.
4 Mark Questions III.22. Your friend is asked to differentiate between evaporation and boiling. What questions could you ask him to know the differences between evaporation and boiling ? (As–2) Ans. I would ask the following questions : i) If we heat water, at which temperature it convert into vapour ? ii) What we call this process ? iii) By which process wet clothes become dry ? iv) Does evaporation take place at all temperatures or not ? v) Is the temperature if liquid increases or decreases in boiling ? vi) Is the temperature of liquid increases or decreases in evaporation ? vii) Is boiling a surface phenomenon/bulk phenomenon ? viii) Is evaporation a surface phenomenon / bulk phenomenon ? (Note : Ask any question which is relevant to the above processes) 23. Explain the procedure of finding specific heat of solid experimentally ? (As–1) Ans. Aim : To find the specific heat of given solid. Material required : Calorimeter, thermometer, stirrer, water, lead shots and woodenbox. Procedure : 1) Measure the mass of the calorimeter along with stirrer. Mass of the calorimeter = m1 = ...... 2) Now fill one third of the volume of calorimeter with water. Measure its mass and its temperature. Mass of the calorimeter + Water = m2 = ......... Mass of the water = m2 – m1 = ............. Temperature of water in calorimeter = T1 = ........... C0. 3) Take a few lead shots and place them in water. Heat them upto a temperature 1000C. Let this temperature be T2. Transfer the hot lead shots quickly into the calorimeter. The mixture settles to a certain temperature after some time. Let this temperature be T3. Mass of calorimeter + Water + Lead shots = m3 = ............ Mass of lead shots = m3 – m2 = ................. 4
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
4) Let the specific heats of the calorimeter, Lead shots and water be SC, SL and SW respectively. 5) According to the method of mixtures, Heat lost by the solid = Heat gain by the calorimeter + Heat gain by the water. 6) Hence (m3 – m2) SL (T2 – T3) = m1SC (T3 – T1) + (m2 – m1) SW (T3 – T1) = [m1SC + (m2 – m1) SW] (T3 – T1) SL =
..... (1)
We can calculate the specific heat of a solid (lead shots) by using the above formula. 24. F D
1000C
00 C
E
B
C
–50C A
Heat (Q)
[m1SC + (m 2 − m1 ) SW ] (T3 − T1 ) (m 3 − m 2 ) (T2 − T3 )
By observing the graph, answer the following
(As–1)
i) Which part indicates conversion of ice into water ? ii) What does the part DE represent ? iii) At which point ice starts melting ? iv) What does the point 'E' represent ? Ans. i) BC ii) DE represents latent heat of vaporization i.e., the conversion of water into vapour at 1000C. iii) At B iv) Completely vapour at 1000C.
BITS I. Multiple Choice questions : 1. Which of the following is a warming process. A) Evaporation
B) Condensation
( C) Boiling
)
D) All the above
2. Three bodies A, B and C are in thermal equilibrium. The temperature of B is 450C. Then the temperature of C is .......... ( ) A) 450C
B) 500C
C) 400C
D) any tempterature 5
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3. The temperature of a steel rod is 330 K0. Its temperature in C0. is ......... A) 550C
B) 570C
C) 590C
B) Q∆T
C) Q/m∆T
B) increases
C) decreases
12. 13. 14. 15. 16. 17. 18. III. 19. 20. 21. 22. 23.
)
(
)
(
)
D) cannot say
6. Which of the following is a cooling process II. 7. 8. 9. 10. 11.
( D) m∆T/Q
5. When ice melts, its temperature A) remains constant
)
D) 530C
4. Specific heat S = .............. A) Q/∆T
(
A) Evaporation B) Condensation C) Boiling D) All the above Fill in the blanks. The SI unit of specific heat is .................... The latent heat of fusion of ice is .................... The latent heat of vaporization of water is .................... Temperature of a body is directly proportional to .................... According to the principle of method of mixtures, the net heat lost by the hot bodies is equal to .................... by the cold bodies. The sultryness in summer days is due to .................... .................... is used as a coolant. Ice floats on water because .................... 00C = .................... 0K. 1 cal = .................... Joules. The amount of water vapour present in air is called .................... The water droplets keep floating in the air and form a thick mist is called .................... Matching. Melting ( ) A. Vapour to liquid Boiling ( ) B. Liquid to vapour Freezing ( ) C. Surface phenomenon Condensation ( ) D. Solid to liquid Evaporation ( ) E. Liquid to solid F. Solid to vapour
Answer I. 1) B 2) A 3) B 4) C 5) A 6) A II. 7) J/Kg–K 8) 80 Cal/gm 9) 540 cal/gm 10) average kinetic energy of the molecules 11) heat gained 12) Humidity 13) water 14) the density of ice is less than the density of water 15) 273 16) 4.186 17) Humidity 18) fog III. 19) D 20) B 21) E 22) A 23) C ❖ ❖ ❖ ❖❖ 6
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
2. CHEMICAL REACTIONS AND EQUATIONS 1 Mark Questions I. 1. Write two reactions which proves that the substance act as base.
(As–1)
Ans. 1) Bases are soapy to touch and turn red litmus to blue. 2) The methyl orange indicator convert Base solution into yellow colour. 2. Write balanced chemical reaction between Calcium Oxide and water.
(As–1)
Ans. CaO + H2O → Ca(OH)2 3. Write world equation of Zn + dil. HCl2 → ZnCl2 + H2.
(As–1)
Ans. Zinc + dilute Hydro chloric acid → Zinc Chloride + Hydrogen 4. Write the substances which undergoes chemical reaction.
(As–1)
Ans. The subtances which undergo chemical change in the reaction are called reactants. 5. Write the substances which are present left side of the arrow mark in the chemical equation ? And also right of the arrow mark in the chemical equation. (As–1) Ans. The reactants are written on the left side of arrow and the final substances, or products are written on the right side of the arrow. 6. A) C(g) + O2 (g) → CO2 + Q B) N2(g) + O2 (g) → 2NO(g) – Q State the type of chemical reaction.
(As–1)
Ans. A) Exothermic Reaction B) Endothermic Reaction 7. What is meant by Chemical reaction.
(As–1)
Ans. The making and breaking of chemical bonds are called Chemical reactions. 8. X react with Y and forms Ca(OH)2 and Heat. Name X and Y the substances paricipated in the reaction. (As–1) Ans. X : Cao
Y : H2O
9. Write the reaction involved in the whitening of walls.
(As–1)
Ans. A solution of slaked lime produced in the reaction is used to white wash walls. Calcium hydroxide reacts slowly with the carbon dioxide in air to form a thin layer of calcium carbonate on the walls. It gives a shiny finish to the walls. Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l) 10. How to determine the realeased gas is CO2 in the chemical reaciton.
(As–1)
Ans. In the chemical reaction the gas is put off burning match stick with "TUP'' sound then is confirmed the released gas is CO2. 7
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
11. Which gas evolved with brown colour in the Heat reaction of Lead Nitrate.
(As–1)
Ans. On heating lead nitrate decomposes to lead oxide, oxygen and Nitrogen dioxide. You observe the brown fumes liberating in the boiling tube. These brown fumes are of Nitrogen dioxide (NO2). 2Pb (NO3)2 → 2PbO(s) + 4NO2(g) + O2 12. Why they add Dil HCl in electrolysis of water.
(As–1)
Ans. The Dil HCl is added to water in the electrolysis process for improve the Conductivity of Electricity. 13. Write the chemical reaction of Lead Nitrate decomposition.
(As–1)
Ans. On heating lead nitrate decomposes to Head oxide, oxygen and Nitrogen dioxide. The brown fumes are liberated by the boiling tube. These brown fumes are due to liberation of Nitrogendioxide (NO2). 2Pb (NO3)2 → 2PbO(s) + 4NO2(g) + O2 14. When the reactions are said to be endothermic ?
(As–1)
Ans. The reactions require energy in the form of heat, light or electricity for converting the reactants to products. The reactions are Called endothermic Reactions. 15. Which substance has undergone oxidation reaction in the following.
(As–1)
Ans. H2S + Br → 2 HBr + S In the above reactions H2S lost its Hydrogen. So H2S is oxidized by Bromine. 16. Write the role of Vitamin C & E in preservation of food.
(As–1)
Ans. Usually substances which prevent oxidation (Antioxidants) are added to food. The spoilage of food can be prevented by adding preservatives like Vitamin C and Vitamin E. 17. What is the use of keeping food in air tight containers.
(As–1)
Ans. Chips manufacturers usually flush bags of chips with gas such as nitrogen to prevent the chips from getting oxidized. 18. Write the chemical reaction, which the silver metal exposed to moisture.
(As–1)
Ans. The black coatings on silver 4Ag + 2H2S + O2 → 2Ag2S + 2H2O 19. Define rancidity
(As–1)
Ans. When fats and oils in food material that were left for a long period are responsible for spoiling of food. The process is known as Rancidity. Rancidity is an oxidation reaction. 20. Write the name of metal which is not undergone oxidation process.
(As–1)
Ans. Gold 21. Write the substances are used for manufacturing of stainless steel.
(As–1)
Ans. Iron is mixed with carbon, nickel and chromium to get an alloy stainless steel. 8
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
2 Mark Questions II.22.Give examples for exothermic reactions.
(As–1)
Ans. i) Burning of Coal : When coal is burnt in oxygen, carbon dioxide is produced. C(s) + O2(g) → CO2(g) + Q (heat energy) ..... (1) ii) Slaked lime is prepared by adding water to quick lime. Ca O(s) + H2O(l) → Ca(OH)2 (aq) + Q (heat energy) ..... (2) Large amount of heat energy is released on reaction of water with CaO(s). If you touch the walls of the container you will feel the hotness. Such reactions are called exothermic reactions. 23. Write a chemical change between Barium chloride and Sodium Sulphate ? Determine the colour of the end product for the above reaction.
(As–1)
Ans. Sodium sulphate reacts with barium chloride to give white precipitate, barium sulphate and sodium chloride. Na2SO4 + BaCl2 → BaSO4 + NaCl 24. What are the changes do you observed between the chemical reaction of Zinc and Dilute Hidro chloric acid in conical plask.
(As–1)
Ans. Zinc metal reacts with dilute HCl to yield ZnCl2 and liberates Hydrogen gas. Zn + HCl → ZnCl2 + H2 25. What is meant by chemical equation ? Write chemical equation of the reaction between Barium chloride and Sodium Sulphate.
(As–1)
Ans. Describing a chemical reaction using least possible words or symbols is called a chemical equation. Sodium sulphate reacts with barium chloride to give white precipitate, barium sulphate and sodium chloride. Na2SO4 + BaCl2 → BaSO4 + NaCl 26. Balance the following chemical equation a) C3H8 + O2 → CO2 + H2O b) Fe2O3 + Al → Fe + Al2O3 c) CO2 + H2O → C6H12O6 + O2 d) Pb(NO3)2 → PbO + NO2 + O2
(As–1)
Ans. a) 2C3H8 + O2 → 6CO2 + 8H2O b) Fe2O3 + 2Al → 2Fe + Al2O3 c) 6CO2 + 6H2O → C6H12O6 + 6O2 d) 2Pb(NO3)2 → 2PbO + 4NO2 + O2 9
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
27. Write the photolytic reaction by taking example as Silver bromide decomposition reaction. (As–1) Ans. Silver bromide decomposes to silver and bromine in sunlight. Light yellow coloured silver bromide turns to gray due to sunlight. 2AgBr(s) → 2 Ag(s) + Br2(g) This decomposition reaction occurs in presence of sunlight and such reactioins are called photochemical reactions. 28. When you dippled iron nails in Copper sulphate solution, becoming brown. Write causes for loss of colour of copper sulphate solution. (As–1) Ans. The Iron nail dipped in copper sulphate solutions compared before and after the experiment. Fe(S) + CuSO4(s) → FeSO4(aq) + Cu(g) Iron is more reactive than copper, so it displaces copper from copper sulphate. 29. What do you observe in the chemical displacement reaction. How do you satisy the above exaplanation with suitable example. (As–1) Ans. In the displacement reaction the element in the one compound is displaced by another. Zinc pieces react with dilute hydrochloric acid and liberate hydrogen gas as shonw below. Zn(S) + 2HCl(aq) → ZnCl2(aq) + H2(g) In reaction the element zinc has displaced hydrogen from Hydrochloric acid. 30. What are precipitative reaction ? Give examples.
(As–1)
Ans. Lead nitrate solution with potassium iodide solution. A yellow colour substance which is insoluble in water, is formed. This insoluble substance in known as precipitate, and the reactions are called precpitative reactions. Pb(NO3)2(aq) + 3Kl(aq) → PbI2(s) + 2KNO3(aq) 31. In the following chemical reaction name the compound which is oxidized and which is reduced. (As–1) Ans. MnO2 + HCl → MnCl2 + 2H2O + Cl2 In the above reaction MnO2 loss it Oxygen. So MnO2 is reduced. Hcl accept Oxygen, So it is Oxidized. 32. What do you mean by corrosion ? Give precausationary measure for prevention of corrossion. (As–1) Ans. When some metals are exposed to moisture, acids etc., they tarnish due to the formation of respective metal oxide on their surface. This process is called corrosion. Corrosion can be prevented or at least minimized by shielding the metal surface from oxygen and moisture. It can be prevented by painting, oiling, greasing, galvanizing, chrome plating or making alloys. Galvanizing is a method of protecting iron from rusting by coating them a thin layer of Zinc. 10
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
33. Why the apple change their colour in to brown, when you cut with knife. (As–1) Ans. Apples produce an enzyme called polyphenol oxidase or tyrosinase, that reacts with oxygen. When the fruit is cut, it damages the cells in the fruit and allows the oxygen in the air to react with the enzyme and other chemicals of fruit. This reaction leads to browning of cut surface of fruit. 34. What is galvonisation ? Write their uses. (As–1) Ans. Galvanizing is a method of coating a metal with a thin layer of Zinc. It is essential for protection of metals from rusting. 35. Write the bleaching reaction of chlorine. Ans. Bleaching of coloured objects using moist chlorine. Cl2 + H2O → HOCl + HCL HOCL → HCl + (O) Coloured object + (O) Colourless object.
(As–1)
36. Give two examples of reactions in w hich oxidation - reduction process has observe in the same reaction. (As–1) Ans. Oxidation and reduction occur in the samireaction. If one reactant gets oxidized, the other gets reduced. Such reactions are called oxidation-reduction reactions or redox reactions. CuO + H2 → Cu + H2O In the CuO, H2 reaction CuO is reduced and H2 is oxidized. 37. Why the formers add Cao or CaSO4.2H2O or CaCO3 to the soil. Give me reasons. (As–1) Ans. The formers are used the Cao or CaSO4.2H2O or CaCO3 to the soil to increase the PH of soil and to reach the ideal soil pH for the growth of plants.
4 Mark Questions III.38. How to determine the substances undergone a chemical reaction ? (As–1) Ans. We can determine when the substance undergone a chemical reaction the following changes are observed. 1) A change that changes state and colour of substance. 2) A change that release heat energy. 3) A change which forms an insoluble substance as precipitate. 4) A change that liberate a gas. 39. Write the steps involved in the balancing of chemical equation with suitable examples ? (As–1) Ans. The chemical equation is balanced by using a systematic method. Step 1 : Write the unbalanced equation using the correct chemical formula for each reactant and products. In the reaction of hydrogen with oxygen to yield water, you can write unbalanced chemical equations as follows : H2 + O2 → H2O 11
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
Step 2 : Compare number atoms of each element on both sides. Find the suitable coefficients – the numbers placed before formula to indicate how many formula units of each substance are required to balance the equation. Only these coefficients can be changed when balancing an equation, the formulas themselves can't be changed. We take the reaction of hydrogen with oxygen as an example; we can balance the equation by adding a coefficient of 2 to both H2 and H2O. By so doing there are 4 hydrogen and 2 oxygen atoms on each side of the equation : 2H2 + O2 → 2H2O Step 3 : Reduce the coefficients to their smallest whole number values, if necessary by dividing them all by a common devisor. Step 4 : Check the answer by making sure that the numbers and kinds of atoms are the same on both sides of the equation. Example 2 : Propane, C3H8 is a colorless, odorless gas often used as a heating and cooking fuel. Write the chemical equation for the combustion reaction of propane with oxygen to yield carbon dioxide and water and balance it. Follow the four steps described in previous discussion. Step 1 : Write the unbalanced equation using correct chemical formulas for all substances. C3H8 + O2 → CO2 + H2O Step 2 : Compare number of atoms of each element on both sides. Find the coefficients to balance the equation. It is usually best to start with the most complex substance – in this case C3H8 – and to deal with one element at a time. Look at the unbalanced equation, and note that there are 3 carbon atoms on the left side of the equation but only 1 on the right side. If we add a coefficient of 3 to CO2 on the right side the carbon atoms balance. C3H8 + O2 → 3CO2 + H2O Element
No of atoms LHS
RHS
C
3
1
H
8
2
O
2
3
Now, look at the number of hydrogen atoms. There are 8 hydrogens on the left but only 2 on the right side. By taking a coefficient of 4 to the H2O on the right, the hydrogens balance. C3H8 + O2 → 3CO2 + H2O Finally, look at the number of oxygen atoms. There are 2 on the leftside but 10 on the right side, by taking a coefficient of 5 to the O2 on the left, the oxygen balance.' C3H8 + 5O2 → 3CO2 + H2O Step 3 : Make sure the coefficients are reduced to their smallest wholenumber values. In fact, the equation – 11 we got is already coefficients in smallest whole number. There is no need to reduce its coefficients, but this might not be achieved in each chemical reaction. lLet us assume that you have got chemical equation as shown below : 2C3H8 + 10O2 → 6CO2 + 8H2O 12
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
Though the equation – 12 is balanced, the coefficients are not the smallest whole numbers. If would be necessary to divide all coefficients by 2 to reach the final equation. Step 4 : Check the answer. Count the numbers and kinds of atoms on both sides of the equationto make sure that are the same. 40. What is meant by balanced equation ? Why the chemical equation must be balanced. (As–1) Ans. A chemical equation in which the numbers of atoms of different elements on the reactants side (left side) are same as those on product side (right side) is called a balanced reaction. According to the law of conservation of mass, the total mass of the substances thatare taking part in chemical reaction must be the same before and after the reaction. You know an atom is the smallest particle of an element that take part in a chemical reaction as it is the atom which account for the mass of any substance. The number of atoms of each element before and after reaction must be the same. All the chemical equations must balance, because atoms are neither created nor destroyed in chemical reactions. 41. What type of information acquired from Chemical equation.
(As–1)
Ans. i) Expressing physical state : To make the chemical equation more informative, the physical states have to be mentioned along with their chemical formulas. The different states i.e., gaseous, liquid, and solid states are represented by the notations (g), (l) and (s) respectively. If the substance is present as a solution in water the word aqueous is written as (aq). The balanced equation is written along with the physical states as : Fe2O3(s) + 2Al(s) → 2 Fe(l) + Al2O3(s) In the above equation Fe(l) indicates that iron is produced in liquid state, remaining all the substances are in solid state. ii) Expressing the heat changes : You know heat is liberated in an exothermic reactions and heat is absorbed in endothermic reactions. See the following examples. 1) C(s) + O2(g) → CO2 + Q (exothermic reaction) 2) N2(g) + O2(g) → 2NO(g) – Q (endothermic reaction) iii) Expressing the gas eolved : If a gas is evolved in a reaction, it is denoted by an upward arrow. Example : Zn + H2SO4 → ZnSO4 + H2 iv) Expressing precipitate formed : If a precipitate is formed in the reactions it is denoted by a downward arrow. Example : AgNO3 + NaCl → AgCl + NaNO3 42. Define Chemical combination ? Explain with Examples.
(As–1)
Ans. A reaction in which single product is formed from two or more reactants is known as chemical combination reaction. 13
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
1) Take a small piece (about 3 cm long) of Magnesium ribbon. 2) Rub the Magnesium ribbon with sand paper. 3) Hold it with a pair of tongs. 4) Burn it with a spirit lamp or burner. Magnesium burns in oxygen by producing dazzling white flame and changes into white powder. The white powder is Magnesium oxide. 2 Mg(s) + O2(g) → 2MgO(g) In this reaction Magnesium and oxygen combine to form a new substance magnesium oxide, such a reaction in which single product is formed from two or more reactants is known as chemical combination reaction. 43. Define Chemical decomposition ? Explain with examples.
(As–1)
Ans. Chemical decomposition : In a decomposition reaction a single substance decomposes to give two or more Substances. 1) Take a pinch of Calcium Carbonate (lime stone) in a boiling tube. 2) Heat the boiling tube over the flame of spirit lamp or burner. 3) Now take a burning match stick near the mouth of boiling tube. In the above activity, on heating calcium carbonate, it decomposes to calcium oxide and carbon dioxide. CaCO3 (s) → CaO(S) + CO2(g) It is a thermal decomposition reaction. When a decomposition reaction is carried out by heating, it is called thermal decomposition reaction. 44. Explain the Electrolysis experiment to release of H2 and O2.
(As–1)
Ans. 1) Take a plastic mug. Drill two holes at its base. 2) Fit two rubber stoppers in these holes. 3) Insert two carbon electrodes in these rubber stoppers. 4) Connect the electrodes to 6V battery as shown in fig. 5) Fill the mug with water, so that the electrodes are immersed. 6) Add few drops of dilute sulphuric acid to water. 7) Take two test tubes filled with water and invert them over the two carbon electrodes. 8) Switch on the current and leave the apparatus undisturbed for sometime. The liberation of gas bubbles at both the electrodes. These bubbles displace the water in the test tubes. Once the test tubes are filled with gases take them out carefully. Test both the gases separately by bringing a burning candle near the mouth of each test tube. In the above activity on passing the electricity, water dissociates to Hydrogen and oxygen. 2H2O(l) → 2 H2(g) + O2(g) 14
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
45. Define displacement reaction ? Explain with Examples.
(As–1)
Ans. The element displaces another element from its compund, the reactions called displacement reaction. Take a small quantity of zinc dust in a conical flask with nozzle. Add dilute Hydrochloric acid slowly. Now take a balloon and tie it to the mouth of the conical flask. Closely observe the changes in the conical flask and ballon. The gas bubbles coming out from the solution and the balloom bulges out. Zinc pieces react with dilute hydrochloric acid and liberate hydrogen gas as shown below. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) In reaction the element zinc has displaced hydrogen from Hydrochloric acid. This is displacement reaction. 46. Define double displacement reaction ? Explain with Examples.
(As–1)
Ans. Double displacement reaction : Two different atoms or ions are Exchanged in double displacement reactions. 1) Take a pinch of lead nitrate and dissolve in 5.0 mL of distilled water in a test tube. 2) Take a pinch of potassium iodide in a test tube and dissolve in distilled water. 3) Mix lead nitrate solution with potassium iodide solution. A yellow colour substance which is insoluble in water, is formed. This insoluble substance in known as precipitate. The precipitate is Lead Iodide. Pb(NO3)2(aq) + 3Kl(aq) → PbI2(s) + 2KNO3(aq) In the above reaction, lead ion and Potassium ion exchange their places each other. Lead ion combines with iodide ion and forms PbI2 as precipitate and KNO3. 47. Define oxidation and reduction ? Give the example for oxidation and reduction reactions occus in the same reaction. (As–1) Ans. 'OXIDATION' is a reaction that involves the addition of oxygen or removal of hydrogen. 'REDUCTION' is a reaction that involves the addition of hydrogen or removal of oxygen. On heating coper it reacts with oxygen present in the atmosphere to form coper oxide. The reaction is shown below. Cu(2) + O2(s) → 2CuO(s) Here copper combines with oxygen to form coper oxide. Here oxygen is gained and the process is called oxidation. Generally oxidation and reduction occur in the same reaction. If one reactant gets oxidized, the other gets reduced. Such reactions are called oxidation-reduction reactions or redox reactions. CuO(2) + H2(s) → Cu(s) + H2O(g) In the CuO, H2 reaction CuO is reduced and H2 in oxidized. 15
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
48. Write the balanced chemical equation for the following and identify the type of reaction in each case. (As–1) A) Ca(OH)2 + HNO3 → H2O + Ca(NO3)2 B) Ma + I2 → MgI2 C) Mg + HCl → MgCl2 + H2 D) Zn + CaCl2 → ZnCl2 + Ca Ans. A) Ca(OH)2 + 2HNO3 → 2H2O + Ca(NO3)2 This is double displacement reaction. B) Ma + I2 → MgI2 This is chemical combination reaction. C) Mg + 2HCl → MgCl2 + H2 This is chemical displacement reaction D) Zn + CaCl2 → ZnCl2 + Ca This is Chemical Displacement reaction.
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16
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3. REFLECTION OF LIGHT BY DIFFERENT SURFACES 1 Mark Questions 1. State Fermat's principle. Ans. Fermat's principle states that the light selects the path which takes the least time to travel. 2. State the laws of reflection of light. Ans. i) When light gets reflected from a surface, the angle of reflection is equal to the angle of incidence. ii) The incident ray, the normal at the point of incidence and the reflected ray all lie in the same plane. 3. Why does an image of plane mirror suffer lateral (right-left) inversion ? Ans. The light rays which come from our right ear get reflected from the plane mirror and reach our eye. Our brain feels that the ray (reflected ray) is coming from the inside of the mirror. That is why our right ear looks like left ear in the image. 4. Define magnification. Ans. The ratio of the height of the image to the height of the object is called magnification. Magnification (m) =
Height of the image (h i ) Height of the object (h 0 )
1 1 is +1. 1 What does this mean ? 5. The magnification produced by a plane mirror = + u vand (ii) the size of the image is equal to Ans. Magnification +1 indicates that (i) the imagef is erect
size of the object. 6. Write the mirror formula and explain the terms. Ans. Mirror formula Here f = focal length u = object distance v = image distance
2 Mark Questions II.1. State the differences between convex and convace mirror. Ans.
Convex mirror
Convave mirror
1. A spherical mirror whose reflecting surface is curved outward is called convex mirror.
1. A speherical mirror whose reflecting surface is curved inward is called concave mirror.
2. After reflection from the mirror the light rays diverge.
2. After reflection from the mirror the light rays converge. 17
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3. The image formed due to convex mirror is always virtual.
3. The image formed due to concave mirror is generally real. But when an object is placed between pole and focus, it is virtual.
4. The image formed due to convex mirror is always diminished.
4. The image formed due to concave mirror is either diminished, or magnified orin same size.
2. Distinguish between real and virtual image. Ans.
Real image
Virtual image
1. Real image formed due to converging of light rays.
1. Virtual image formed due to diverging of light rays.
2. Real image can be formed on screen.
2. This image cannot be formed on screen.
3. Real image is always inverted.
3. Virtual image is always erected.
4. Here the rays actually meet at the image point.
4. Here the rays appears to diverge from the image point.
3. List out any four properties of the image formed by a plane mirror. Ans. 1) The image by a plane mirror is virtual and erect. 2) Size of the image is equal to the size of the object. 3) Image distance is equal to object distance. (i.e., the distance of the image in a plane mirror seems to be equal to the distance between the mirror and object). 4) The image suffers lateral inversion. 4. Why does the size of the image seem to be decreased when you move the object towards your eye ? Ans. When we move the object from the mirror to our eye the image in the mirror seems to move back in the mirror. Then the distance from the image to our eye increases. The angle made by image at our eye is smaller than the angle made by the object. That is why the image looks smaller than the object. 5. Define the following terms in connection with spherical mirrors. i) Pole
ii) Centre of curvature
iii) Principal axis
iv) Radius of curvature
v) Focal pint (focus)
vi) Focal length 18
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
Ans.
i) Pole : The geometrical centre of the spherical mirror is called 'pole' (P). ii) Centre of curvature : It is the centre of the sphere of which the mirror forms a part. iii) Principal Axis : The line passing through the centre of curvature and pole is called principal axis. iv) Radius of curvature : The distance between pole and centre of curvature is called radius of curvature. v) Focus : It is a point on the prinipal axis where a beam of light parallel to the principal axis either actually converges to or appears to diverge from, after reflection from a mirror. vi) Focal length : The distance between the focus and the pole of the mirror is called focal length. 6. Write the rules for sign convention. Ans. i) All distance should be measured from the pole. ii) The distances measured in the direction of incident light to be taken positive and those measured in the direction opposite to incident light to be taken negative. iii) Height of object and height of image are positive if measured upwards from the axis and negative if measured downwards. 7. Imagine that spherical mirrors were not known to human beings. Guess the consequences. Ans. i) Without rear view mirrors safe driving in automobiles will not be possible. ii) Dentists may not have proper diagnosis of teeth and also by ENT specialists not possible to see the inner parts of ear and nose without concave mirrors. iii) Automobile head lights, torch light, search lights can not give bright lighting. iv) Solar cookers are also made from concave mirrors. Construction of reflecting telescopes would not be possible. Thus every where we are using spherical mirrors and without these life of human begings is miserable. 8. How do you appreciate the role of spherical mirrors in daily life ? Ans. Spherical mirros are very useful to our life. i) Concave mirrors are used by dentists to see the large images of the teeth of patient. ii) Spherical mirrors are used in telescopes. iii) Concave mirrors are used as reflectors in torches and vehicle head lights. 19
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
iv) Convese mirrors are used as rear view mirrors in vehicles. v) Concave mirrors are used in solar furnaces. vi) Concave mirrors are used by ENT specialists to see the affected part more visible. So i can appreciate the role of spherical mirrors. 9. Why do we prefer a convese mirror as a rear-view mirror in the vehicles ? Ans. i) Convex mirror always forms virtual, erect and diminished images irrespective of distance of the object. ii) A convex mirror enables a driver to view large area of the traffic behind him. iii) Convex mirror forms very small image than the object. Due to these reasons convex mirrors are used as rear view mirrors in vehicles. 10. By observing steel vessels and different images in them, Surya, a third class student asked some questions his elder sister Vidya. What may be those questions ? Ans. i) How these steel vessels form images like mirrors ? ii) One side they form large image and other side they formed small images. Why ? iii) Why is the image size changing when the vessel is moved away or towards the face ? iv) Why the image is blurred not as clear as in mirror ?
4 Mark Questions III.1. What do you infer from the experiment which you did with concave mirrors and measure the distance of object and distance of image ? Ans. Position of the object Between mirror
Position of
Enlarged/
Inverted/
Real/
the image
diminished
erected
virtual
beind mirror
Enlarged
Erect
Virtual
At focal point F
at infinity
–
–
–
Between F and C
beyond C
Enlarged
Inverted
Real
At centre of
at centre of
Same size
Inverted
Real
curvature C
curvature C
Beyond C
between F and C
Diminished
Inverted
Real
at infinity
at focus
Very diminished
Inverted
Real
and F
(Point sized) 2. How do you find the focal length of a concave mirror ? Ans. i) Hold a concave mirror perpendicular to the direction of sunlight. ii) Take a small paper and slowly move it in front of the mirror and findout the point where you get smallest and brightest spot, which will be the image of the sun. iii) The rays coming from sun parallel to concave mirror are converging at a point. 20
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
iv) This point is called focus or focal point of the concave mirror. v) Measure the distance of this spot from the pole of the mirror. vi) This distance is the focal length of the given concave mirror.
5 Mark Questions 1. Draw suitable rays by which we can guess the position of the image formed by a concave mirror. Ans. 1) A ray parallel to the principal axis passing through principal focus (F) after reflection from a concave mirror.
2) A ray passing through Focus (F) becomes parallel to principal axis after reflection from a concave mirror.
3) A ray passing through C is reflected back along the same path after reflection from a concave mirror.
21
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
2. Explain the nature of imges formed with the help of ray diagrams when an object is placed at different points between a concave mirror. Ans. 1) When object is at infinity : When object is at infinity, the image is formed at focal point (F) of the mirror. The image is real and point sized.
2) When the object is (between infinity and centre of curvature) beyond C : When the object is beyond C (centre of curvature), the image is formed between Focus (F) and C. This is real, inverted and diminished image. 3) When the object is at C : When the object is at C, the image also forms at C. This is a real, inverted and same sized image.
4) When the object is between C and F : When the object is between C and F, the image is formed beyond C. This is real, inverted and enlarged image.
5) When the object is at F : When the object is at F, the image is formed at infinity. This is real, inverted and highly enlarged. 6) When the object is between F and Pole : When the object is between F and pole, the image forms behind the mirror. This is virtual, enlarged and erect image. 22
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
Related questions : 1. Where will the image form when we place an object, on the principal axis of a concave mirror at a point between focus and centre of curvature ? 2. How do you get a virtual image using a concave mirror ? 3. Show the formation of image with a ray diagram when an object is placed on the principal axis of a concave mirror beyond the centre of curvature and explain the nature of the image. 4. Top form the image on the object itself, how should we place the object in front of a concave mirror ? Explain with a ray diagram. 5. Draw the formation of image with the help of suitable rays. And explain the nature of the image.
6. Indicate the pole, Focus, centre of curvature, position of the image in the following figure.
O
3. With the help of ray diagrams, explain the formation images by a convese mirror when i) Object between pole and infinity ii) Object at infinity Ans.
Object between pole and infinity
Object at infinity 23
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
Position of the
Position of the
Nature of the
Size of the
object
image
image
image
Between pole and
Between P and F
Virtual and
Diminished
object
image
image
image
At infinity
At the focus,
Virtual and
Highly diminished
behind the mirror
erect
(point-sized)
Problems : 1. If the radius of curvature of a spherical mirror is 20 cm what is its focal length ? Ans. Focal length f = Given radius of curvature R = 20 cm ∴f=
= 10 cm
2. Find the distance of the image when an object is placed on the principal axis at a distance of 10 cm in front of a concave mirror whose radius of curvature is 8 cm. Ans. Object distance u = 10 cm Radius curvature R = 8 cm 1520 R− 21183 1 = − −= + f4v2220 10 uf20 2 uv = 4 cm ∴ Focal length f = Image distance V = ? From mirror formula (or) =
=
20 = 6.67 cm 3 ∴ Image distance v = 6.67 cm.
∴v=
3. An object is placed at a distance of 10 cm from a convese mirror of focal length 15 cm. Find the position and nature of the image. Ans. Object distance u = – 10 cm Focal length f = 15 cm Image distance v = ? 24
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
From mirror formula (or) = =
1 1 2+3 + = 15 10 30
∴v=
=6
∴ Image is formed behind 6 cm from the mirror. Image is virtual, erect and diminished. 4. A convese mirror with a radius of curvature of 3 m is used as rear view in an automobile. If a bus is located at 5 m from this mirror, find the position and size of the image. Ans. Radius of curvaturer = 3 Focal length =
= 1.5 m
Object distance u = –5 m
15 10 130 Radius 2 1 (R) 3 111 +13 111of113vurvature =−−+ + −= = + = 1.5 13 3fv5/15 2 uf(−−5510) uv1532 5 2 15
Image distance v = ? From mirror formula (or) = = = ∴v=
= 1.15 m
∴ Image is formed behind 1.15 m of mirror. Image is virtual, erect and diminished. 25
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
TRY THESE : 1. Comment on the following ray diagrams.
2. Explain the nature of the image in the following figure.
3. Raju saw his face in a mirror and observed that 1 his 1 face 1 became smaller in size. − + = i) Which type of mirror is that ? uv uv f ii) What is the nature of the image ? 4) i) Which is type of mirror is used in head lights of vehicles like car ? ii) What is the position of the bulb with respect to the mirror ? iii) In a head light of the car, light rays coming from the bulb incident on mirror and reflects. Show this with a ray diagram. 5. What questions do you ask to confirm the factors shoud know to draw ray diagrams of a spherical mirror. 6. In olden days kings burned the ships and tents of enemies by using mirrors. i) What type of mirrors they used ? ii) How they burned the ships ?
BITS I. Multiple Choice questions : 1. The ratio of the focal length of spherical mirror to its radius of curvature is A) 1 : 2
B) 2 : 1
C) 1 : 3
(
)
D) 1 : 4
2. The object distance u, image distance v and focal length f for a spherical mirror are related as A)
B)
C) v + u = f
D) f =
(
) 26
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3. The image formed by a concave mirror A) is always real
B) is always virtual
C) can be both real and virtual
D) none of these
4. The image formed by a convese mirror is always A) real and magnified
B) real and diminished
C) virtual and diminished
D) virtual and magnified
5. The mirror which has a wide field of view must be A) concave
B) convex
C) plane
(
)
(
)
(
)
D) none of these
6. A real and inverted image of the same size is forme dby a concave miror when the o bject is placed ( ) A) between the mirror and its focus
B) between the focus and the centre of curvature
C) at the centre of curvature
D) beyond the centre of the curvature
7. A concave mirror always forms real and inverted image except when the object is placed( ) A) at infinity
B) between F and C
C) at F
D) between F and pole of the mirror
uhv 0i a concave mirror of focal length 15 cm. The 8. An object is placed at a distance of 30 cm from uhv 0i image will be ( )
A) real and same size
B) real and magnified
C) real and diminished
D) virtual and magnified
9. If an object is placed at C on the principal axis in from of a concave mirror, the position of the image is ( ) A) at infinity
B) between F and C
C) at C
D) beyond C
10. A ray which seems to be traelling through the focus of a convese mirror passes ......... after reflection. ( ) A) paralled to the principal axis
B) along the same path in
C) through F
D) through C
11. Magnification m = A)
B)
C)
B) concave
C) plane
)
(
)
D)
12. The driver's mirror used in automobiles is A) convex
(
D) none of these 27
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
II. Fill in the blanks. 1. Light selects the least time path to travel between two points. This principle was states by .................... 2. The relation between focal length and radius of curvature is given by .................... i
r
3.
In the adjacent figure, Lr indicates ....................
4. The rays which are parallel to the principal axis of a concave mirror on reflection, meet at the .................... 5. The distance between pole and centre of curvature is .................... 6. The distance between pole and focus is .................... 7. The equation of mirror formula is .................... 8. The relation between the angle of incidence and angle of reflection is given by .................... 9. If m > 1, the size of the image is .................... than the size of the object. 10. The dentists use .................... mirrors to see large image of the teeth of patients. 11. The geometric centre of the mirror is .................... 12. The line which passes through the centre of curvature and pole is .................... III. Matching. i =1Behind ∠r 1 the mirror 1RA. 1. Beyond C ( ) ∠ = + v 2. On C ( ) f2B. uAt infinity 3. Between C and F ( ) C. Between F and C 4. On F ( ) D. On F 5. Between F and P ( ) E. Beyond C 6. At infinity F. At C
Answer I. 1) A 7) D
2) B 8) A
II. 1) Fermat
3) C 9) C 2) t f =
4) C 10) A (or) R = 2f
5) B 11) D
6) C 12) A
3) angle of reflection 4) Focus
5) Radius of curvature
6) Focal length
7)
8)
10) concave
11) pole 12) principal axis
III. 1) C
9) more (greater) 2) F
3) E
4) B
5) A
6) D
❖ ❖ ❖ ❖❖ 28
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
4. ACIDS AND BASES 1 Mark Questions I. 1. Which substance involved in the chemical reaction for formation of Sodium Zinkate. Ans. Zinc metal react in a bottle and add sodium hydroxide (NaOH) solution and warm the contents. The reaction is written as follows. 2 NaOH + Zn → Na2ZnO2
+ H2
(Sodium zincate) 2. What type of reaction in stomach when an antacid tablet is consumed. Ans. These antacids neutralize the excess acid in the stomach. Magnesium hydroxide (milk of magnesia), a mild base, is often used for this purpose. Being alkaline, it neutralizes excess acid in the stomach and provides relief. 3. Write the reaction between bases with non-metal oxides with suitable examples. Ans. Calcium hydroxide, which is a base, reacts with carbon dioxide to produce a salt and water. This reason is similar to the reaction between a base and an acid. Thus we can conclude that carbon dioxide which is a non metallic oxide is acidic in nature. Ca(OH)2(aq) + CO2(g) → CaCO3 + H2O(l) 4. A calcium compound react with Dil.HCl and forms a gas with hiss sound. The released gas convert the lime water into milkly white. In this reaction the product formed is Calcium Chloride. Write balanced equation for the above reaction. Ans. Ca(OH)2(aq) + CO2(g) → CaCO3 + H2O(l) 5. How to form hydronium ions. Ans. Hydrogen ions cannot exist as bare ions. They associate with water molecules and exist as hydronium ions (H3O+). H+ + H2O → H3O+ 6. What is meant by dilution of acids. Ans. Mixing an acid or base with water result in decrease in the concentration of ions (H3O/OH) per unit volume. Such a process is called dilution of acids. 7. How to reduce the pain when stung by the honey-bee Ans. Honey-Bee sting leaves an acid which causes pain and irritation. Use of a mild base like baking soda on the stung area gives relief. 8. Define salts Ans. Salts are the ionic compounds which are produced by the neutralization of acid with base. 9. Write the salts produced from common salt. Ans. The common salt is an important raw material for various materials of daily use, such as sodium hydroxide, baking soda, washing soda, bleaching powder and many more. 29
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
10. How to prepared brine solution. Ans. An aqueous solution of sodium chloride is called brine, it is prepared by dissolving of NaCl in Distilled water. 11. Which substance are you added for making of the cake soft and spongy. Ans. Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid. When baking powder is heated or mixed in water, the following reaction takes place. NaHCO3 + H → CO2 + H2O + Sodium Salt of acid. Carbon dioxide produced during the reaction causes bread or cake to rise making them soft and spongy. 12. Why the plaster of pairs is called Calcium Sulphatre Hemi Hydrate. Ans. The two formula units of CaSO4 share one molecule of water. So 1/2 molecule of water present in each Plaster of paris unit. So Plaster of paris is called Calcium Sulphatre Hemi Hydrate.
2 Mark Questions 1. Write the tests to identify the nature of substances which acts as acid and bases. Ans. 1) Blue litmus, red litmus, methyl orange, phenolphthalein indicators are used to identy the acidic and basic nature substances in solution. 2) A blue litmus paper turn the colour to red indicates the presence of acidic nature solution in a substance. 3) A red litmus paper turn the colour to blue indicates the presence of basic nature solution in a substance. 4) The acidic solution change the colour methyl orange into red. 5) The basic solution change the colour of methyl orange in to yellow. 6) The Acidic solution is not change the colour of Phenolphthalein. 7) The basic solution change the colour of Phenolphthalein in to pink. 2. Write the reaction between the acids and metallic carbonatates and bi carbonates. Ans. Take two test tubes; label them as A and B. Take about 0.5 gm of sodium carbonate (Na2CO3) in test tube A and about 0.5 gm of sodium hydrogen carbonate (NaHCO3) in test tube B. Add about 2 ml of dilute HCl to both test tubes. Pass the gass produced in each cash through lime water (calcium hydroxide solution). The reactions occurring in the above activity are as follows : Na2CO3(s) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g) NaHCO 3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g) Pass the gas evolved through lime water. Ca(OH)2(aq) + CO2(g) → CaCO3 + H2O(l) 30
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
3. Which substance are formed due to reaction between acids with metal oxides ? Give an example. Ans. The general reaction between a metal oxide and an acid can be written as : Metal oxide + acid → salt + water The copper oxide present in the beaker dissolves in dilute HCl and the colour of the solution becomes blueish-green. The reason for this change is the formation of Copper (II) Chloride in the reaction. CuO + HCl → CuCl2 + H2O 4. Explain the activity the acids conduct the electricity. Repeat this activity with glucose and alcohol and write the causes for non conductivity property. Ans. Prepare solutions of glucose, alcohol, hydrochloric acid, sulphuric acid etc. Drill two holes on a rubber cork and introduce two nails into it. Connect two different coloured electrical wires and keep it in a 100 ml beaker. Connect free ends of the wire to 6 volts DC battery and complete the circuit. Now pour some dilute HCl in the beaker and switch on the current. Repeat activity with dilute sulphuric acid and glucose and alcohol solutions separately. It confirms that the bulb glows only in acid solutions but not in glucose and alcohol solutions. Glowing ofbulb indicates that there is flow of electric current through the solution. Acid solutions have ions and the moment of these ions in solution helps for flow of electric current through the solution. The positive ion (cation) present in HCl solution is H. This suggests that acids produce hydrogen ions H+ in the solutiion, which are responsible for their acidic properties. Inglucose and alcohol solution the bulb did not glow. This indicates the absence of ions in these solutions. 5. Explain the activity the HCl released Hydrogen ion in the presence of water only. Ans. The HCl gas evolved at delivery tube dissociates in presence of water to produce hydrogen ions. In the absence of water dissociation of HCl molecules do not occur. 6. What do you observe when the water is mixed with acid or bases. Ans. The process of dissolving an acid or a base in water is an exothermic process. Care must be taken while mixing concentrated nitric acid or sulphuric acid with water. The acid must always be added slowly to water with constant stirring. If water is added to a concentrated acid, the heat generated may cause the mixture of splash out and cause burns. The glass container may also break due to excessive local heating. 7. What are the uses of acid – base universal indicators. Ans. The universal indicator can be used to know the strength of acid or base. Universal indicator is a mixture of several indicators. The universal indicator shows different colours at different concentrations of hydrogen ions in a solution. 8. What is PH ? Note the points of Acids, Bases and Neutral solutions in the PH Scale. Ans. A Scale for measuring hydrogen ion concentration in a solution is called PH scale. The pH of neutral solutions is 7, values less than 7 on the pH scale represent an acidic solution. As the pH value increases from 7 to 14, it represents a decrease in H3O+ ion concentration or an increases in OH– ion concentration in the solution. i.e., if pH value of a solution above is '7' then it represents a basic solution. 31
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
9. Write the PH value and nature of the each solution on the basis of the following table.
Ans. 1) 2) 3) 4) 5) 6)
Battery acid Vinegar Milk Blood Ammonia solution Sodium Hydroxide solution
– 0.2 –3 – 6.4 – 7.4 – 11.4 – 13.8
– Strong acid – 11 – Weak acid – Weak Base – Strong base – Strong base
10. What effect of lowers the PH values of river water on aquatic life. Ans. Living organisms can survive only in a narrow range of ph change between 7.0 to 7.8. When pH of rain water is less than 5.6, it is called acid rain. When acid rain flows in to the rivers, it lowers the pH of the river water, the survival of aquatic life in such riversbecomes difficult. 11. Write the chemical the following a) Sodium Sulphate b) Potassium Chloride c) Magnesium Sulphate d) Sodium Carbonate c) MgSO4.7H2O d) Na2SO4 Ans. a) Na2SO4 b) KCl 12. What are the factors influence on PH of salts. Ans. Salt of a strong acid and a strong base are neutral and the pH value is 7. The salts of a strong acid and weak base are acidic and the pH value is less than 7. The salts of a strong base and weak acid are basic in nature and the pH value is more than 7. 13. How is bleaching powder produced ? Write some uses of it. Ans. The chlorine gas is used for the manufacture of bleaching power. Bleaching power is produced by the action of chlorine on dry slaked lime [Ca(OH)2]. Bleaching powder is represented by formula CaOCl2, through the actual composition is quite complex. Ca(OH)2 + Cl2 → CaOCl2 + H2O User of Bleaching Power : 1) It is used for bleaching cotton and linen in the textile industry for bleaching wood pulp in paper industry and for bleaching washed clothes in laundry. 2) Used as an oxidizing agent in many chemical industries. 3) Used for disinfecting drinking water to make it free of germs. 4) Used as a reagent in the preparation of chloroform. 14. A milkman adds a very small amount of banking soda to fresh milk. Why does this milk take a long time to set a curd ? Ans. Fresh milk has a pH of 6. A milkman adds a very small amount of baking soda to fresh milk. They shift the pH of the fresh milk from 6 to slightly alkaline. In this condition the milk stored for long time but it cannot set a curd because inconvenient environment for the curd formation causing bacteria. 32
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
15. How is baking soda produced ? Write some uses of it. Ans. This is commonly used in the kitchen for making tasty crispy pakodas is baking soda. Sometimes it is added for faster cooking. The chemical name of the compound is sodium hydrogen carbonate (NaHCO3). It is produced using sodium chloride as one of the raw materials. NaCl + H2O + CO2 + NH3 → NH4Cl + NaHCO3 Uses of sodium hydrogen carbonate : i) Backing powder is a mixture of baking soda and a mild edible acid such as tartaric acid. When baking powder is heated or mixed in water, the following reaction takes place. NaHCO3 + H+ → CO2 + H2O + Sodium Salt of acid. Carbon dioxide produced during the reaction causes bread or cake to rise making them soft and spongy. ii) Sodium hydrogen carbonate is also an igredient in antacids. Being alkaline, it neutralizes excess acid in the stomach and provides relief. iii) It is also used in soda-acid fire extinguishers. iv) It acts as mild antiseptic. 16. Write the general name of the Sodium Carbonate ? Give the importance of Sodium Carbonate. Ans. Another Name sodium carbonate is washing soda. It is o btained by heating baking soda. Na2CO3 + 10 H2O → Na2CO3. 10 H2O Sodium carbonate and sodium hydrogen carbonate are useful chemicals for many industrial processes. Uses of Washing Soda : 1) Sodium carbonate (washing soda) is used in glass, soap and paper industries. 2) It is used in the manufacture of sodium compounds such as borax. 3) Sodium carbonate can be used as a cleaning agent for domestic purposes. 4) It is used for removing permanent hardness of water. 17. Write the procedure for manufacturing of Plaster of Paris ? Why they the material in air locked plastic bags. Ans. On careful heating of gypsum (CaSO4 2 H2O) at 373 K it loses water molecule partially to become calcium sulphate hemihydrate (CaSO4. 1/2 H2O). This is called plaster of paris, the substance which doctors use as plaster for supporting fractured bones in the right position. Plaster of paris is a white powder and on mixing with water, it sets into hard solid mass due to the formation of gypsum. CaSO4. 1/2 H2O + 11/2 H2O → CaSO4 2 H2O It is stored in airlock plastic bags for prevention of formation of hard solid mass. 18. What is meant by water Crystallization ? Describe an activity to show the water crystallization. The presence of water molecules in the salts is called Crystallisation. Ans. Copper sulphate crystals which seem to be dry contain the water of crystallization, when these crystals are heated, water present in crystals is evaporated and the salt turns white. When the crystals are moistened with water, the blue colour reapears. Water of crystallization is the fixed number of water molecules present in one formula unit of a salt. Five water molecules are present in one formula unit of copper sulphate. Chemical formula for hydrated copper sulphate is CuSO4.5H2O. 33
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
19. A white powder has used for supporting of fractured bones a) State the name of white powder b) Write the chemical name of it c) Write the reaction between white powder and water Ans. A) Plaster of paris B) CaSO4.1/2H2O C) CaSO4.1/2 H2O + 11/2 H2O → CaSO4.2H2O
4 Mark Questions 1. Note down the responses of various test soluntions with the following indicators. Ans. S.No. Test Blue Red Methyl Phynolphthalin solution litmus litmus orange 1.
Acetic acid
Red
No change
Red
No change
2.
Nitric acid
Red
No change
Red
No change
3.
Sodium hydroxide
No change
Blue
Yellow
Pink
4.
Ammonium hydroxide
No change
Blue
Yellow
Pink
5.
Sulfuric acid
Red
No change
Red
No change
6.
Potasium hydroxide
No change
Blue
Yellow
Pink
2. Write the activity in the chemical reaction between metals with acids to release H2 gas. Ans. Take 10 ml of dilute HCl in one of the bottles. Add few pieces of Zinc granules to it. Insert one end of the plastic tube in to the bottle through its rubber cork. Fill the other bottle with water and invert it. The gas coming out from the mouth of the bottle burns with a pop sound indicating H2. The chemical reaction of the above activity is Acid + metal → Salt + Hydrogen 2 HCl(aq) + Zn(s) → Zn Cl2 + H2(g) 3. Define Neutralisation reaction ? Write the activity for this reaction. Ans. The reaction of an acid with a base to give a salt and water is known as a neutralization reaction. In general, a neutralization reaction can be written as : Base + Acid → Salt + Water Take about 2 ml of dilute NaOH solution in a test tube and add two drops of phenolphthalein indicator. Observe the colour of the solution after adding phenolphthalein. Now add one or two drops of NaOH to the above mixture. In the above activity, when HCl is added to the solution you observe that the pink colour of the Phenolphthalein indicator disappears because of NaOH present in the solution reacts with HCl. i.e., the effect of base is nullified by an acid and vice-versa. The reaction that is taking place between acid and base in above activity can be written as : NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) 34
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
4. How to find whether the acid is strong or weak. Ans. A test to know whether the Acid is trong or weak. Take two beakers A and B. Fill the beaker A with dil. CH3COOH (Acetic acid) and beaker B with dil.HCl (Hydrochloric Acid). Arrange the apparatus and pass electric current. When we will notice that the bulb glows brightly in HCl solution while the glowing intensity of the bulb is low in acetic acid solution. This indicates that there are more ions in HCl solution and fewer ions in acetic acid solution. More ions in HCl solution mean more H3O+ ions. Therefore it is a strong acid. Whereas acetic acid has fewer H3O+ ions and hence it is weak acid. 5. Explain the effect of PH Changes on dental and digestive systems. Ans. PH Change cause of teeth decay : Tooth decay starts when the pH of the mouth is lower than 5.5. Tooth enamel, made of calcium phosphate is the hardest substance in the body. It does not dissolve in water, but is corroded when the pH in the mouth is below 5.5. Bacteria present in the mouth produce acids by degradation of sugar and food particles remaining in the mouth. The best way to prevent this is to clean the mouth after eating food. Using tooth pastes, which are generally basic neutralize the excess acid and prevent tooth decay. pH in our digestive system : It is very interesting to note that our stomach produces hydrochloric acid. It helps in the digestion of food without harming the stomach. During indigestion the stomach produces too much acid and this causes pain and irritation. To get rid of this pain, people use bases called antacids; These antacids neutralize the excess acid in the stomach. Magnesium hydroxide (milk of magnesia), a mild base, is often used for this purpose.
BITS I. Multiple Choice questions : 1. When red litmus paper is dippled in X solution then the litmus paper changes to blue colour. The nature solution X is ( ) A) Salt B) Acid C) Base D) All of the above 2. The reaction between sodium sulphate and barium chloride gives barium sulphate. The colour of the product is ( ) A) Yellow B) White C) Brown D) Black 3. N2 + O2 2Na–Q in this reaction the negation indicates ( ) A) Exothermic reaction B) Endothermic reaction C) Electro chemical reaction D) All of the above 4. 2 Grams of H2 Occupies how much volume at STP condition ( ) A) 112 lit B) 22.4 lit C) 24.2 lit D) 211 lit 5. What is reason for whitening of wall when wet lime is applied ( ) A) Ca(OH)2 B) CaO C) CO2 D) CaCO3 6. Take X substance and heat it with a Bunsen burner. In this reaction they give brown colour gas. State the gas realeased from the following ( ) A) Nitrous oxide B) Nitric Oxide C) Nitrogen dioxide D) Nitrogen 7. Take yellow coloured Silver bromide in a watch glass and keep it in sunlight. Then the colour changes to ash colour. State the type of reaction occurred ( ) A) Chemical combination B) Decompositon C) Displacement
D) Double displacement 35
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
8. The colour of Precipitate formed when potassium lodide is reacted with Lead Nitrate.( A) Yellow
B) White
C) Brown
)
D) Black
9. In a chemical reaction the reactants changes their positive and negative radicals mutually. The reaction is called ( ) A) Chemical combination
B) Decompositon
C) Displacement
D) Double displacement
10. Which type of reaction occurred in crakers A) Neutralization
B) Oxidation
( C) Reduction
)
D) Mixed reaction
11. Sravanthi anklets changes its colour to black after some days. What is reason for the formation of the ......... A) AgO
B) Ag(OH)2
C) Ag2S
D) AgCl
(
)
12. An Apple slice changes it colour to brown immediately after cutting. What is reason for this. A) Oxidation
B) Reduction
C) Double displacement
D) Decomposition
(
)
13. Wet yello wflowers are placed in chlorine gas. Then they losses their colour. State the reason A) Cl2
B) O
C) H2O
D) HCl
14. Stainless steel is mixture off
(
)
(
)
(
)
A) Iron mixed with Carbon, Nickel and Chromium B) Silver mixed with Carbon, Iron, NIckel C) Copper mixed with Carbon, Chromium and Iron D) Iron mixed with Copper, Carbon and Chromium 15. Which gas is filled in the pocket for long time storage of chips. II. 16. 17. 18. 19. 20. 6.
A) Oxygen B) Nitrogen Matching. Group – A Chemical combination ( Decomposition ( Chemical displacement ( Double displacement ( Down arrow mark ( At infinity
C) Carbon Dioxide D) Air
) ) ) ) )
A. B. C. D. E. F.
Group – B 2 AgCl → 2Ag + Cl Pb + CuCl2 → PbCl2 + Cu C + O2 → CO2 Precipitation NaCl + AgNO3 → AgCl + NaNO3 Air
Answer I. 1) C 7) B 13) B II. 1) C
2) B 8) D 14) A 2) A
3) B 9) D 15) B 3) B
4) B 10) B
5) D 11) C
6) C 12) A
4) E 5) D ❖ ❖ ❖ ❖❖ 36
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
5. REFRACTION OF LIGHT AT PLANE SURFACES 1 Mark Questions I. 1. Write the formula for refractive index ?
(As–1)
Ans. Refractive index n = Here c = speed of light in vacuum v = speed of light in medium 2. The speed of light in a diamond is 1,24,000 km/s. Find the refreactive index of diamond if the speed of light in air is 3,00,000 km/s. (As–1) Ans. c = 3,00,000 km/s; v = 1,24,000 km/s Refractive index of diamond, n = Refractive index of diamond, n = Refractive index of diamond, n = 2.42 3. Write the formula for relative refractive index 9ncv1gw200,?9000 3, 8 == 8nv 221g24, 000 1, 89 Ans. Relative refractive index (n21) =
.... (1)
Where n2 = Refractive index of second medium n1 = Refractive index of first medium Relative refractive index (n21) =
.... (2)
Where v1 = Speed of light in first medium v2 = Speed of light in second medium 4. Refractive index of glass relative to water is
. What is the refractive index of water rela-
tive to glass ? Ans. Refractive index of glass relative to water ngw = ∴ ngw = Refraftive index of water relative to glass ngw = 37
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
5. Media (Material) Refractive index
Ice
Water
Benzene
Carbon disulphide
1.31
1.33
1.5
1.63
By observing the above table, answer the following ?
(As–1)
i) In which material, speed of light is high ? ii) In which material, speed of light is low ? Ans. i) In Ice speed of light is high. ii) In carbon-di-sulphide speed of light is low. 6. What is snell's law ?
(As–1)
Ans. Snell's law : (or) n1 sin i = n2 sin r N
7.
} a (Glass) O
M
M
sin i n 2 = sin r n1
} b (Water)
N
In the above figure, MM is the plane separating two media 'a' and 'b'. NN is the normal drawn at '0' to the plane MM. 1) Then, In a and b which is the denser medium ? 2) Which is the rarer medium ?
(As–1)
Ans. 1) 'a' is the denser medium and b is the rarer medium. 8. Observe the following table. Medium Refreactive Index
(As–1)
Water
Crown glass
1.33
1.52
i) Which is the deuser medium in water and crown glass ? Ans. Crown glass is the denser medium. (Note : d & n) 38
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
9.
By observing the above figure, write the answer for the following questions.
(As–1)
i) What is the relation between i and r ? ii) In first and second media, which is douser ? Ans. i) i > r (or r < i ii) Second medium is denser. 10. In what cases does a light ray not deviate at the interface of two media ?
(As–7)
Ans. 1) When the incident ray strikes normally at the interface of two media, it does not deviate from H 4 its path. (Q ∠i = ∠r = 0) 33 2) If the refractive indices of two media are equal, the light ray passes without any deviation at the boundary. 11. Define critical angle ? (As–1) Ans. Critical angle : The angle of incidence at which the light ray travelling from denser to rarer medium grazes the interface is called 'Critical angle' for denser medium. (Or) The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 900 is called 'Critical angle'. 12. Define total internal reflection ?
(As–1)
Ans. Total internal reflection : When the angle of incidence is greater than the critical angle, the light ray is reflected into denser medium at interface. This phenomenon is called 'Total internal reflection'. 13. The absolute refractive index of water is
. What is the critical angle ?
(As–1)
Ans. c = ? Absolute refractive index of water n12 = 39
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
PHYSCIS - Paper - I
sin c =
sin c =
c = sin–1
= sin–1 (0.75)
∴ c = 48.5 14. What is the reason behind the shining of diamonds ?
(As–1)
Ans. Total internal reflection is the main reason for brilliance of diamonds.
2 Mark Questions II. 15.Why is it difficult to shoot a fish swimming in watern?
(As–1)
Ans. 1) Due to refraction of light, it is difficult to shoot a fish swimming in water. 2) The fish and observer are in two different media. 3) When the fish is in water (denser medium) and observer is in air (rarer medium) due to refraction at water-air interface, the fish appears to be raised and seems to be close to surface Q which is called 'Apparent position'. 1131=1 3 sin c 4o n14412 30 4) The shooter aims the gum to apparent position of the fish instead of real position. 32 Hence it is difficult to shoot a fish swimming in water. 16. Look at the picture.
N
a) What is the
n1
b) Find the refractive index of denser medium with respect rarer medium ? Ans. a) Critical angle c = 30°
(As–1)
30°
n2
b) Refractive index of denser medium w.r.t rarer medium = n21 = =
(
i = c, r = 90°) (
c = 30°)
= ∴ n21 = 2 40
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
17. What is the reason behind the spining of diamonds and how do you appreciate it ?(As–6) Ans. 1) The reason behind the shining of diamonds is total internal reflection. 2) The critical angle of a diamond is very low (24.40). 3) So if a high ray enters a diamond it undergo total internal reflection which makes the diamond shine. 18. Why does a diamond shine more than a glass piece cut to the same shape ? Ans. 1) The critical angle of a diamond is very low (24.40). 2) Hence all the light rays enter the diamond, undergo total internal reflection and make the diamond shines more. 3) The critical angle of a glass (420) is more than diamond. 4) Hence most of the incident rays reflects and less number of rays get total internal reflection. So glass shines less than diamond. 19. Why do stars appears twinkling ? Ans. 1) Due to refraction of light stars appears twinkling. 2) The light coming from stars when entering our atmosphere undergo multiple refractions continuously. Hence stars appears twinkling.
4 Mark Questions III. 20. Explain the refraction of light through a glass slab with a neat ray diagram ? Identify lateral shift ? Ans. Aim : Determination of position and nature of image formed by a glass slab. Material required : Plank, chart paper, Clamps, Scale, Pencil, Thin glass slab and pins. Procedure : 1) Place a piece of chart (paper) on a plank clamp it. 2) Place of glass slab in the middle of the paper. Draw D border line along the edges of the slab by using a pencil. Remove the glass slab. We will get a figure of a rectangle. Name the vertices of the rectangle as A, B, C and D. 3) Draw a perpendicular at a point 'L' on the longer side 'AB' of the rectangle.
A
C
B
4) Now draw a line from 'L' in such a way that it makes 300 angle with normal. Mark two points P, Q on this line. 41
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
5) The line PQ represents the incident ray. The angle it makes with normal represents the angle of incidence (i = 300) 6) Now place the glass slab in the rectangle ABCD. Fix two identical pins at 'P' and 'Q' such that theystand vertically with equal height. 7) By looking at the two pins from other side (CD) of the slab, fix two more pins at R, S in such a way that all pins appear to be along a straight line. 8) Remove glass slab and the pins. Draw a straight line by joining R, S upto the edge CD of the rectangle. This line (RS) represents emergent ray of the light. 9) Draw a perdicular to the line CD at 'M' where the line 'RS' meets the line 'CD'. 10) Measure the angle between emergent ray (RS) and normal. This is called 'angle of emergence (x)'. 11) The angles of incidence and emergence are equal (∠i = ∠e) 12) The incident and emergent rays (PQ, RS) are parallel. 13) Measure the distance between the parallel rays (PQ, RS). This distance is called 'Lateral shift'. 21. Explain the woring of optical fibres. Prepare a report about various uses of optical fibres in our daily life ? (AS–4) Ans. Optical fibre : Principle : Total internal reflection is the basis principle behind working of optical fibre. Description : An optical fibre is very thin fibre made of glass (or) plastic having radius about a micrometer (10–6m). A bunch of such thin fibres form a light pipe. Working : The light going into pipe makes a nearly glancing incidence on the wall. The angle of incidence is greater than the critical angle and hence total internal reflection takes place. In this way light is transmitted along the fibre.
Fig (1) : Show the principle of light transmission by an optical fibre.
Uses of optical fibres in our daily life : 1) Communication : Different telephone signals by superposing on the optical beam can be transmitted through fibres without any interference. 2) Medical investigation : Optical fibres are used in endoscope, Laproscope for visual examination of inaccessible regions in the human body. 3) Photometric sensors : Measuring the blood flow in the heart. 4) Sensors : To measure temperature and Pressure. 5) Refractometers : To determine the refractive indices of liquids. 42
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL BITS
I. Multiple Choice questions : 1. Which of the following is snell's law
(
A) n1sin i = sin r /n2
B) n1/n2 = sin r / sin i
C) n2/n1 = sin r / sin i
D) n2sin i = constant
)
2. The refractive index of glass with respect to air is 2. Then the critical angle of glass-air interface is ............. ( ) A) 00
B) 450
(Note : sin C =
; C = ∠300)
C) 300
D) 600
3. Total internal reflection takes place when the high ray travels from ........ A) rarer to denser medium
B) rarer to rarer medium
C) denser to rarer medium
D) denser to denser medium
4. The angle of deviation produced by the glass slab is .......... A) 00
B) 200
(
)
(
)
C) 900
D) depends on the angle formed by the light ray and normal to the slab (Note : ∠i = ∠e) 5. 1 micrometer = .............. metres A) 10–8
B) 10–9
31 1 = 4n12C) 10 2 –4
B) 900
C) 1200
B) m/s2
C) kg/m3
B) Water
C) Light pipe
)
(
)
(
)
D) No units
8. ............ is used for visual examination of in accessible regions in the human body A) Vacuum
(
D) 450
7. The unit of refractive index is A) m/s
)
D) 10–6
6. At critical angle of incidence, the angle of refraction is ................... A) 600
(
D) None
II. Fill in the blanks 9. Speed of light in vacuum is ............... 10. The refractive index of a transparent material is 11. 12. 13. 14. 15.
. The speed of the light in that medium is
............... Mirage is an example of ............... The reason behind the shining of diamond is ............... ............... is the basic principle of optical fibre. The reason for a coin placed in the water appears to be raised is ............... Twinking of starts is due to ............... 43
PHYSCIS - Paper - I
✤ ✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻✻ ✤ S.S.C. STUDY MATERIAL
III. Matching. Group – A
Group – B
16. Water
(
)
A. 1.50
17. Kerosene
(
)
B. 2.42
18. Flint glass
(
)
C. 1.52
19. Benzene
(
)
D. 1.65
20. Diamond
(
)
E. 1.33 F. 1.71 G. 1.44
Answer I. 1) B
2) C
7) D
8) C
II. 9) 3 × 108 m/s
3) C
4) A
10) 2 × 108 m/s
5) D
6) B
11) Total internal reflection
12) Total internal reflection
13) Total internal reflection
14) Refraction of light
15) Refraction of light
II. 16) E
17) G
18) D
19) A
20) B
❖ ❖ ❖ ❖❖
44
