59280656 Science Class X

1Quest Tutorials

H-108, New Asiatic Building, Connaught place, New Delhi-1. Ph. 26131625

FORMULABOOKLET

SCIENCE

2Quest Tutorials


ABOUT THE NEW CCE PATTERN

CBSE syllabus for CCE class 10 is divided in two terms. Term-I carries 40% weightage and Term-II carries 60% weightage. There will be two formative and one summative assessment in each term. Summative assessment in term-I will carry 20 marks and summative assessment in Term-II will carry 40 Marks. Formative assessments (total four, two in first term and two in second term) will carry 10 marks each

COURSE STRUCTURECLASS X

FIRST TERM MARKSUNITSI Chemical SubstancesChemical reaction; Acids, bases and salts, Metals and Non Metals

29

II World of livingLife process; control and coordination in animals and plants

19

III Effects of currentElectricity, Magnetic effects of currents

26

IV Natural ResourcesSources of Energy

06

TOTAL 80

The term wise weightage would be as follows:I Term (April – September) FA1 – 10% (pen-paper test)FA2 – 10%SA1 – 30% (pen-paper test)II Term (October - March)FA3 – 10% (pen-paper test)FA4 – 10%SA2 – 30% (pen-paper test)Total Term – 1 and Term – 2

SECOND TERM MARKSUNITSI Chemical SubstancesCarbon compounds; Periodic classification of elements

21

II World of livingReproduction; Heridity and evolution

27

III Natural Phenomena 26IV Natural ResourcesManagement of natural resources; the regional environment;our environment

06

TOTAL 80

3Quest Tutorials


FA1 + FA2 + FA3 + FA4 = 40%SA1 + SA2 = 60%Total = 100%

Grade Marks Range Grade PointA1 93-100 10.0A2 83-92 9.0B1 73-82 8.0B2 63-72 7.0C1 53-62 6.0C2 43-52 5.0D1 33-42 4.0D2 23-32 3.0E 22 and below 2.0CGPA = Total Grade Point / No of Subjects

4Quest Tutorials


CHEMISTRY: Class XChapter 1: CHEMICAL REACTIONS AND EQUATIONS

I. CHEMICAL REACTIONS

DEFINITION: Chemical reactions are the processes in which new substances with new properties are formed

CONSTITUENTS

CHARACTERISTICS:

(a) Evolution of a gas(b) Formation of a precipitate(c) Change in color(d) Change in temperature (e) Change in physical state

II CHEMICAL EQUATION

DEFINITION: Method of representing a chemical reaction with the help of symbols and formulae of the substances involved in it.

E.g.

aqueous-(aq) liquid - lsolid -(s) precipitate-ppt

Chemical Equations are classified as

III STEPS INVOLVING BALANCING OF CHEMICAL EQUATIONS

Consider the reaction: 2 2 2H O H O Step I: Take one element/molecule at a time. Count the no. of atoms on both sides.

Element No. of atoms on reactant side No. of atoms on product side H 2 2No. of Hydrogen atoms are balanced

Reactants[Substances that take part in a chemical reaction]

Products[New substances formed as a result of chemical reaction]

Balanced[Chemical equation having an equal no. of atoms of different elements in the reactants & products]E.g. 2 4 4 2Zn H So ZnSO H

Unbalanced[Chemical equation having unequal no. of atoms of one or more elements in the reactants & products]E.g. 2 2 2H O H O

5Quest Tutorials


Step II: Consider the second elementElement No. of atoms on L.H.S No. of atoms on R.H.S 0 2 1 No. of Oxygen atoms are not balanced.

Step III : Balance the no. of atoms of the required element

2 2 2H O 2H O Now, Hydrogen atoms are not balanced

Step IV : Since there are 4 hydrogen atoms on the R.H.S Therefore to have 4 hydrogen atoms on the L.H.S , we multiply H2 by 2 and write 2H2 .The final equation becomes:

2 2 22H O O 2H O

IV TYPES OF CHEMICAL REACTIONSSome important types of chemical reactions are:

1. COMBINATION REACTIONS: Reactions in which two or more substances combine to form a single substance.Example: Magnesium and Oxygen combine, when heated to form magnesium oxide.

Combination22Mg(s) O (g) 2MgO(s)

Magnesium Oxygen Magnesiumoxide

2. DECOMPOSITION REACTIONS: Reactions in which a compound splits up into two or more simpler substances. They are just the opposite of combination reactions.Example: Calcium carbonate on heating decomposes to give calcium oxide and carbon dioxide.

heat

3 2DecompositionCaCO (s) CaO(s) CO (g)

Calciumcarbonate Calciumoxide Carbondioxide

(limestone) (lime) (dioxide)

3. DISPLACEMENT REACTIONS: Reactions wherein one element takes the place of another element in a compound.Example:

4 4CuSO (aq) Zn(s) ZnSo (aq) Cu(s)

Copper sulphate Zinc Zincsulphate Copper

(Blue solution) (colourless solution)

4. DOUBLE DISPLACEMENT REACTIONS: Reactions in which two compounds reacts by exchange of ions to form new compounds.Example: 2 2 4 4BaCl (aq) Na SO (aq) BaSO (s) 2NaCl

[White ppt] (aq)

5. OXIDATION AND REDUCTION REACTIONS: Oxidation and reduction reactions together are called REDOX REACTIONS.

Example: Removal of hydrogen: oxidation

2 2H S Cl S 2HCl

Addition of hydrogen:Reduction

6Quest Tutorials


(i) Substance oxidized : 2H S

(ii) Substance reduced : 2Cl

(iii) Oxidising agent: 2Cl

(iv) Reducing agent: 2H S

6. EFFECTS OF OXIDATION IN EVERYDAY LIFE:

(i) CORROSION: Process in which metals are eaten up gradually by the action of air, moisture or a chemical (like acid ) on their surface.E.g Rusting of iron.

(ii) RANCIDITY: Condition produced by adding anti-oxidants to foods containing fats & oils. Anti – oxidants are reducing agents & prevents oxidation.

7Quest Tutorials


CHEMISTRY: Class XChapter 2: ACIDS, BASES AND SALTS

I ACIDS

II. BASES[Chemical substances having bitter taste]

III. INDICATORS[Dye that changes colour when put into acid/base]

IV pH Scale: STRENGTH OF ACID & BASE SOLUTIONS

V Some common salts in day to day life

Common Name Chemical Name Formula1. Common salt Sodium chloride NaCl2. Caustic soda Sodium hydroxide NaOH3. Washing soda Sodium carbonate Na2CO3.10H2O4. Baking soda Sodium hydrogen carbonate NaHCO3

5. Bleaching powder Calcium oxy Chloride CaOCl2

6. Hydrogen Salts Iron sulphate hepta hydrate FeSO4.7H2O

pH of SALT SOLUTIONS

Salts of pH Scale pH value(1) Strong acids + Strong Bases Neutral pH=7(2) Strong acid + weak base Acidic pH<7(3) Weak acid + Strong base Basic pH>7

Strong [All mineral acids except carbonic acids]

Weak [Organic acids]

Concentrated[Low amount of water]

Dilute [High amount of water]

Strong[Completely ionized in water]

Weak[Partially ionized in water]

Water soluble[Alkalis]

Water insoluble[Alkaline]

Natural [Litmus, Turmeric]

Synthetic [Methyl orange]

Olfactory[Related to odour changes][Onion, vanilla]

8Quest Tutorials


VI IMPORTANT REACTIONS:

1. formsAcid Metal Salt Hydrogen gas

2 4 4 2Zn(s) H SO (aq) ZnSO (aq) H (g)

Metal Acid(dil) Salt Hydrogen

2. Neutralisation Reaction:

Acid Base Salt Water

2NaOH(aq) HCl(aq) NaCl(aq) H O

Base Acid Salt Water

3. Hydrolysis

All acids produce hydrogen ion (H+) when dissolved in water 2

2 4 4H SO (aq) 2H (aq) SO (aq)

3 3CH COOH(aq) H (aq) CH COOH (aq)

4. Hydrolysis:

All bases produce hydroxide ions (OH-) when dissolved in water 2H ONaOH(s) Na (aq) OH (aq)

2H O 22Mg(OH) (s) Mg (aq) 2OH (aq)

9Quest Tutorials


CHEMISTRY: Class XChapter 3: METALS AND NON–METALS

I. CHEMICAL PROPERTIES OF METALS

1. Reaction of Metals with Oxygen (air):(i) Metal oxygen Metal oxide (basic in nature)

(ii) (a) 2 2 34Al 3O Al O Amphoteric in nature

(b) 2 3 2 2(Base) (Salt) (water)(Amphoteric oxide)

Al O 2NaOH 2NaAlO H O

2. Reaction with water: a) Metal Water Metal oxide + Hydrogen b) Metal oxide + water Metal hydroxide

3. Reaction with Acids(Diluted):Metal Acid(dil) Salt Hydrogen

4. Reaction with Solutions of other metal salts: More reactive metal displaces less

reactive metal from its salt solution. E.g. Metal A being more reactive than Metal B, Then.Metal A + Salt Solution of Metal B Salt solution of Metal A + Metal B

II. CHEMICAL PROPERTIES OF NON – METALS

1. Reaction with Oxygen (air):(i). Non Metal Oxygen Acidic / neutral oxide (ii). Acidic oxide + Water Acid

2. Non – Metals do not react with water and dilute acids to evolve hydrogen gas.

3. Reaction with Chlorine:Non Metal Chlorine Covalent chlorides(non-electrolytes)

4. Reaction with Hydrogen: N.M. Hydrogen Covalent hydrides

III. REACTION OF METALS AND NON – METALS

1. Metal Non metal Ionic / Electrovalent Compounds

Na Cl NaCl 2. Non metal Metal Covalent compounds

4HydrogenCarbon Methane

C 4H CH

IV. EXTRACTION OF METALSMetal Reactivity Series Method of Extraction

Na, K, Ca, Mg & Al High Electrolysis of molten chloride or oxideZn, Fe, Pb & Cu Moderate Reduction of oxide with CarbonCu & Hg Low Heating sulphide in air (Reduction by heat alone)Ag, Au & Pt Non – reactive Found in native state (as metals)

22Cu O 2CuO

10Quest Tutorials


VI. REFINING OF METALS IS DONE BY ELECTROLYTIC REFINING Impure metal is made the anode & pure metal strip is made the cathode.

VII. CORROSION:

Exposed for leads2a longtime to

Metal Air oxygen Moisture H O Corrosion

Metals affected: Fe, Ag, Al, Cu etc.

11Quest Tutorials


CHEMISTRY: Class XChapter 4: CARBON AND ITS COMPOUND

I. ABOUT CARBONSymbol: CAtomic No: 6% age composition: In earth’s crust (mineral form) = 0.02%

In atmosphere (CO2 gas) = 0.03%Carbon Compounds: Called Organic compounds [Found in living beings]

Properties:(i) Valency: tetravalent (4)(ii) Bonding: Forms Covalent Bonds [Sharing of electrons](iii) Catenation: Self combination

Allotropes:(i). Diamond(ii). Graphite(iii).Buckminster fullerene [ C60]

II

III. CARBON COMPOUNDS Or,

HYDROCARBONS

Homologous SeriesA series of carbon compounds having same functional group with a difference of –CH2– unit is called homologous series.

Saturated Unsaturated

Alkenes[Double Bonded]CnH2n

Alkynes[Triple Bonded]CnH2n-2

Alkanes[Single Bonded]CnH2n+2

12Quest Tutorials


CHEMICAL PROPERTIES:

1. Combustion: C + O2 CO2 + heat + light 2. Oxidation: Alcohols converted to carboxylic acids 3. Addition Reaction: GivesUnsaturated hydrocarons Saturated hydrocarbons 4. Substitution Reactions: For saturated hydrocarbons

sunlight4 2 3CH Cl CH Cl HCl

Methane Chlorine Chloromethane Hydrogen Chloride

(saturated

hydrocarbon)

IV IMPORTANT CARBON COMPOUNDS AND THEIR REACTIONS

1. Ethanol: C2H5OH

2 5 2 5 22Na 2C H OH 2C H ONa H

sodium Ethanol sodium ethoxide

2 4

Hot conc2 5 2 2 2H SO

C H OH CH CH H O

Ethanol Ethene

2. Ethanoic Acid: CH3COOH

(i) Esterification: Acid3 2 5 3 2 5CH COOH C H OH CH COOC H

Ethanoic acid ethanol Ester

(ii) Reaction with carbonates & Hydrogen Carbonates

3 2 3 3 2 2CH COOH Na CO 2CH COONa H O CO

(iii) Reaction with a base:

3 3 2CH COOH NaOH CH COONa H O Ethanoic acid Base Sodium acetate Water

13Quest Tutorials


CHEMISTRY: Class XChapter 5: PERIODIC CLASSIFICATION OF ELEMENTS

I CHRONOLOGY OF THE PERIODIC CLASSIFICATION

Modern periodic tableOld to New Mendeleev’s periodic table

Newland’s law of octavesDobereiner’s triads

Out of these Dobereiner’s triads & Newland’s octaves were not of much significance. However Mendeleev’s periodic table was significant in a way that lead to the development of Modern Periodic Table(or long form of Periodic Table)

II MENDELEEV’S CLASSFICATION

MERITS: 1. Predicted the existence of some elements that had not been discovered at that time. 2. It could predict the properties of several elements on the basis of their positions in the periodic

table. 3. It could accommodate noble gases when they were discovred.

LIMITATIONS: 1. Position of isotopes not explained 2. Wrong order of atomic masses of some elements. 3. Correct position of hydrogen not assigned

III. PRESENT BASIS FOR THE CLASSIFICATION OF ELEMENTS: 1. atomic number of the elements 2. electronic configuration of elements.

IV CHARATERISTICS OF THE MODERN PERIODIC TABLE: Also called long form of periodic table.

1. Divided into (i). Periods [Horizontal rows of elements](ii). Groups [ Vertical rows of elements]

2. Number of (i). Periods = 7 [ Contains fixed number of elements](ii). Groups = 18

3. Proper assigning of the Hydrogen atom

4. Periodicity was shown regarding the properties like valency, atomic size, metallic & non-metallic charater.

5. Noble gases, Lanthanoids & Actinoids being given separate space.

14Quest Tutorials


PHYSICS: Class XChapter 1: LIGHT – REFLECTION AND REFRACTION

I LAWS OF REFLECTION OF LIGHT

1. i r 2. The incident ray, the reflected ray & the normal to the mirror, all lie in the same plane.

II THE NEW CARTESIAN SIGN CONVENTION FOR REFLECTION BY SPHERICAL MIRRORS:

1. Object always placed to the left of the mirror inplaying that the light on the mirror falls from the left hand side

2.

3. All distance parallel to the principal axis are measured from the pole of the mirror.

III SOME IMPORTANT FORMULAE

1. Mirror Formula

1 1 1

v u f Where u = Distance of the object from its pole

v=Distance of the image from its pole f = focal length

2. Radius of curvature R = 2f

3. Magnification Formula

h ' v

mh u

Where, h ' = Height of the image

h = Height of the object. m = Magnification produced by a

spherical mirror.

4. The refractive index (constant) = (i angle of incidence)sin i

(r angle of refraction)sin r

15Quest Tutorials


5. 121

2

vn

v Where, 21n refractive index of medium 2 w.r.t 1

1v speed of light in medium 1

212

1

vn

v 2v speed of light in medium 2

6. Lens formula: 1 1 1

f v u

7. Power (P) = 1

f Where, P = Power of the lens

S.I unit of power is Dioptre f = focal length

8. 1 2 3P P P P .......... Where, P is the net power of the lenses

placed in contact

1 2 3P , P , & P are the power of

different lenses.

IV. SOME NUMERICALS BASED ON ABOVE FORMULAE:

1. An object placed 20cm in front of a mirror is found to have an image 15cm(a) in front of it(b) behind it

Find the focal length of the mirror in each case. [Ans: a) 60

cm7

b) 60cm]

2. Describe the nature of the image formed when the object is placed at distance of 20cm from a concave mirror of focal length 10cm.

[Ans: Real image, 20cm in front of the mirror]

3. If the radius of curvature of a concave mirror is 20cm, what is its focal length? [ Ans : 10cm]

4. If the distance between an object and its image in a plane mirror is 5.6 cm. How far is the object from the mirror?

[ Ans : 2.8cm]

5. The speed of light in water is 2.25 × 108 m/s . If the speed of light in vacuum be 3×108 m/s, Calculate the refractive index of water. [ Ans: 1.33]

6. If the magnification of a body of size 1m is 2, what is the size of the image? [ Ans: 2m]

7. If an object of 7 cm height is placed at a distance of 12 cm from a convex lens of focal length 8 cm, Find the magnification, and the height of the image. [ Ans : -2 , -14 cm]

8. What will be the focal length of a lens whose power is given as + 2.0D? [Ans : 0.5 m]

16Quest Tutorials


PHYSICS: Class XChapter 2: THE HUMAN EYE AND THE COLOURFUL WORLD

I DEFECTS OF VISION

1. MYOPIA: Also called near sightedness. Person can see nearby objects but can not see distant objects distinctly.

Correction: Using concave lens of suitable power.

2. HYPERMETROPIA: Or far-sightedness. Person can see distant objects clearly but not the nearby objects.

Correction: Using convex lens of appropriate power

3. PRESBYOPIA: The eye losing its power of accommodation due to old age.

Correction: Use of bi-focal lens

II SOME COMMON TERMS AND FORMULAE: 1. ‘D’: the least distinct vision of human eye for a normal eye, D = 25cm

D = f × m where f = focal length of the lensor m = magnification

Df

m

2. Accommodation of the eye: The ability of the eye to focus both near and distant objects by adjusting its focal length.

3. Tyndall Effect: The phenomenon of scattering of light by the colloidal particles.

III Few Phenomenon Arising Due To REFRACTION OF LIGHT

IV NUMERICALS SECTION

Solved Examples 1. A person with a defective eye-vision is unable to see the objects nearer than 1.5m. He wants to

read books at a distance of 30cm. Find the nature, focal length and power of the lens he needs in his spectacles.

Solution:This person is hypermetropic

Dispersion of white light by a glass prisme.g. Formation of the rainbow colours

Atmospheric Refractione.g. (1) Twinkling of stars (2) Advanced sunrise and delayed sunset

Scattering of lighte.g. (1) Tyndall effect (2) Blue colour of

the sky. (3) Colour of sun at sunrise & sunset.

17Quest Tutorials


his 30cm , 1.5m 150cm

1 1 1 1 1 1 1 4

f 150 30 150 30 150

150

f 37.5cm4

+ve sign shows that he needs a convex lens of focal length 37.5cm.

Now, power of lens(P) = 100

2.67D97.5

2. A myopic power has been using spectacles of power – 1.00D for clear vision. During old age he also needs to use separate reading glasses of power +2.00D. Explain what may have happened to his eye-sight?

Solution: According to the use of lenses,near point (normal) = 25cmfor point = 100cmPower of accommodation = normal

Now, due to old age.

Power of accommodation = abnormalNear point of vision (increase) = 50cmRequired lens of power(P) = 2.00D

1 1f 0.5m 50cm

p 2

25cm (least distance of vision)

1 1 1 1 1 1 1 1

f 50 ( 25) 50 25 50

50cm

The near point is 50cm away from his eyes.

Unsolved Examples

1. For point of a myopic person is 40cm. What type of lens should he wear so as to see the distant objects clearly? Calculate the focal length and the power of the lens he should use.

[Hint : take 1

& 0

] [Ans: f = –0.4 m , P = – 2.5D]

2. Calculate the focal length of the lens, which is used as a simple magnifying & it gives magnification of 6.

[Ans: 4.16cm]

3. The far point of a Myopic person is 80cm in front of the eye. What is the power of the lens required to enable him to see every distant object clearly?

[Ans: 1.25D]

18Quest Tutorials


PHYSICS: Class XChapter 3: ELECTRICITY

I FEW IMPORTANT TERMS

1. What is electricity?A stream of electrons moving through a conductor is known as electric current or electricity.Denoted by ‘I’ and S.I unit is Ampere (A).

QI ,

t

S.I unit of Q is coloumbs [C] 2. Potential Difference (V) between two points = Work done (W)/Charge (Q).

WV

Q Where , 1 volt(v) =

1Joule( j)

1Coulomb(c)

S.I unit ‘V’ is volt (v) 1v = 11 JC

3. Relation between Potential Difference & Current Ohm’s Law:

According to the law: The potential difference across the ends of a resistor is directly Proportional to the current through it, provided its temperature remains the same.

i.e. V I or V = IR (constant) Where, 1 ohm( ) = 1 volt (v)

1 ampere(A)

orV

RI

S.I unit of R is ‘ohm’( )

II FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS

1. Length of the conductor (l). R l

2. Area of the cross section. 1

RA

3. Nature of the material of the conductor.

l

RA

or ,Pl

RA

,

III. RESISTANCE OF A SYSTEM OF RESISTORS

1. Resistors in Series:

By applying ohms Law

1 1 2 2 3 3V IR ,V IR ,V IR

1 2 3IR IR IR IR Or, s 1 2 3R R R R

2. Resistors in parallel:

Where P is the electrical resistivity of the material of the conductor. denoted by ‘rho’ , the constant of proportionality.

S.I. unit is m

Where , ‘Q’ is net charge flowing.‘t’ is the time in seconds (s).

19Quest Tutorials


Total Current,

1 2 3I I I I By applying ohm’s law,

1 2 31 2 3

V V VI , I , I

R R R

p 1 2 3

V V V V

R R R R

p 1 2 3

1 1 1 1

R R R R

IV SOME OTHER FORMULAE 1. For the heating effect of electric current,

QP V P VI

t

Also, for a steady current I, the amount of heat (H) produced in time (t) is , H = VItor, H = I2Rt(By applying Ohm’s law)

2. P = VI

= I2R =2V

RSI unit of electric power is watt (w)1w = 1 VAAlso, 1kwh = 1000 watt × 3600 seconds.

= 3.6 × 106 watt seconds = 3.6 × 106 Joules (J).

Where, P is the power input v = Potential difference

Q = Charges = time in seconds

Where , P is the electric powerw = 1 watt of energyv = 1 volt A = 1 ampere of current

20Quest Tutorials


PHYSICS: Class XChapter 4: MAGNETIC EFFECTS OF ELECTRIC CURRENT

I. MAGNETIC FIELD LINES

The lines drawn in a magnetic field along which north magnetic pole would move.

Properties:

1. Start from the north pole of a magnet & end at the south 2. Closer near the poles. 3. Never intersect each other 4. A magnetic compass, when placed at any of the points on a magnetic line, aligns itself along the

tangent to the line of force at that point.

II MAXWELL’S RIGHT HAND GRIP RULE

It says “ If the current carrying conductor wire is gripped with the right hand in such a way that the thumb gives the direction of the current, the direction of the fingers give the direction of the magnetic field produced around the conductor wire.

III A SOLENOID

A coil of many circular turns of insulted copper wire wrapped closely in the shape of a cylinder is called a solenoid.

21Quest Tutorials


Working of a Solenoid

IV Rules to determine the direction of force

1. When force experienced by a straight conductor carrying current placed in a magnetic field perpendicular to it.

22Quest Tutorials


2. Current induced in a circuit by changing magnetic flux due to the motion of a magnet.

V SOME IMPORTANT TERMS

1. Electric Motor: Converts electric current into mechanical energy

2. Generator: Converts mechanical energy into electrical current.

3. Fuse : Most important safety device, used for protecting circuits due to short-circuiting/ overloading of currents.

4. Earth wire: Also called green-insulator converted to a metallic body deep inside earth.

5. Electromagnetic induction : Phenomenon that produces induced current in a coil placed in a region where the magnetic field changes with time.

23Quest Tutorials


PHYSICS: Class XChapter 5: SOURCES OF ENERGY

I CLASSIFICATION OF SOURCES OF ENERGY

Sources of Energy

Of all these resources , fossil fuels, thermal energy sources and nuclear energy resources are exhaustible resources. While , hydral, wind, solar and sea-energy resources are in exhaustible resources but each has its own limitations. Geothermal energy resources are restricted to very limited places due to less availability of ‘hot-spots’

II. WORKING OF SOME IMPORTANT ENERGY RESOURCES

1. Working of a Bio-gas (Gobar gas) plant:

A very common, low budget and extremely useful energy-resources.

Conventional Non-conventional

Fossil fuels Thermal Hydral Wind Solar Energy Geothermal Sea Energy Nuclear

Coal gas

Burningof fossil fuels

Kinetic energy of flowing water

solar cooker solar cell

Tidal energy

Wave energy

Ocean thermal energy

Petroleum Natural gas

Potential energy of water at height

24Quest Tutorials


2. Working of a Wind-mill:Though this power plant has its own serious limitations but none the less it is highly useful & efficient resource.

25Quest Tutorials


3. Working of a Solar-Water heater:Solar water heaters are an excellent source for heating purposes, they are low-cost,efficient sources.

26Quest Tutorials


SAMPLE PAPER CLASS XPHYSICS – CHEMISTRY (SECTION A)

Time: 2 ½ Hrs Maximum Marks: 60General Instructions:

I) The question paper comprises of two sections A and B. You are to attempt both the sections.ii) The candidates are advised to attempt all the questions of Section A separately and Section B

separately.iii) All questions are compulsory. iv) There is no overall choice. However, internal choice has been provided in two questions of five

marks category in Section A and one question of 2 marks category and one question of 3 marks category in Section B. You are to attempt only one option in such questions.

v) Marks allocated to each question are indicated against it. vi) Questions 1 to 6 in Section A and 17 to19 in Section B are very short answer questions. These are

to be answered in one word or one sentence only. vii) Questions 7 to 10 in Section A and 20 to24 in Section B are short answer questions. These are to

be answered in about 30 - 40 words each.viii) Questions 11 to 14 in Section A and 25 to 26 in Section B are also short answer questions. These

are to be answered in about 40 - 50 words each.ix) Questions 15, 16 in Section A and 27 in Section B are long answer questions. These are to be

answered in about 70 words each.

Section- A1. Write the product formed when Ferrous Sulphate is heated? 2. What is the relationship between angle of Incidence & angle of Reflection? 3. What will happen if the solution Hydrogen Carbonate is heated? Give the equation of the reaction

involved?4. How can Presbyopia be corrected?5. What is the Composition of Aqua regia? 6. Why are copper wire used as connecting wire?7. An element X on burring in air form an oxide Xo2 which when dissolved in water turns blue litmus

red. Identify if X is a metal or non – metal Justify your answer?8. What is the cause of resistance offered by a conductor?9. How is graphite used in lead Pencils10. What are the two methods of producing magnetic field? Which of these is better and why? 11. Classify the element of third period into metal & non – metal of the modern periodic table?12. Briefly explain the mechanism of nuclear fission process? 13. A metal X acquires a green coating on its surface an exposure to air

(a) Identify the metal X and name the process responsible for this change (b) Name the green coating formed on the metal?

14. A potential difference V across a conductor of length L & cross section area A how is the resistance R of conductor affected when only V is halved only L is halved and only A is halved?

15. Describe an experiment in detail to study refraction of light through a prism Draw with a ray diagram too. What do you mean by angle of deviation?

ORDescribe an experiment to study refraction of a light ray through a rectangular glass slab what are the important result obtained from the experiment?

16. The reaction metal X and Fe2 O3 is highly exothermic and is used to join railway (a) Identify metal X and name the reaction (b) Write the chemical equation of its reaction with fe2 O3

27Quest Tutorials


SAMPLE PAPER CLASS XPHYSICS – CHEMISTRY SOLUTIONS (SECTION A)

28Quest Tutorials


59280656 Science Class X

Documents

Transcript of 59280656 Science Class X