Set 1

Section A

1. Define resistance. Give its SI unit.

2. Name any two elements that are used infabricating solar cells.

Section B

3. State laws of reflection of light.

Or

Define absolute refractive index and expressit mathematically.

4. Draw magnetic field lines around a barmagnet.

5. What happens when 5% alkaline potassiumpermanganate solution is added drop by dropto warm propyl alcohol (propanol) taken in atest tube? Explain with the help of a chemicalequation.

Section C

6. What are fossils? Describe briefly

two methods of determining the age of

fossils.

7. What is the cause of dispersion of white

light through a glass prism? Draw a ray

diagram to show the path of light when

two identical glass prisms are arranged

together in inverted position with

respect to each other and a narrow beam

of white light is allowed to fall obliquely

on one of the faces of the prisms.

Or

What is scattering of light? Use this

phenomenon to explain why (i) the Sun

appears reddish at sunrise and (ii) the

clear sky appears blue?

Fully Solved

SCIENCE

1. The question paper comprises of five Sections A, B, C, D and E. You are to attempt

all the sections.

2. All questions are compulsory.

3. Internal choice is given in Sections B, C, D and E.

4. Question number 1 and 2 in Section A are one mark questions. These are to be

answered in one word or in one sentence.

5. Question numbers 3 to 5 in Section B are two marks questions. These are to be

answered in about 30 words each.

6. Question numbers 6 to 15 in Section C are three marks questions. These are to be

answered in about 50 words each.

7. Question numbers 16 to 21 in Section D are five marks questions. These are to be

answered in about 70 words each.

8. Question numbers 22 to 27 in Section E are questions based on practical skills. Each

question is of two marks. These are to be answered in brief.

CBSE EXAMINATION PAPER 2019

Max. Marks : 80

GENERAL INSTRUCTIONS

(All India)

8. (a) Classify the following reactions intodifferent types :

(i) AgNO NaCl AgCl3 ( ) ( ) ( )aq aq s+ →+ NaNO3 ( )aq

(ii) CaO H O Ca(OH)2 2( ) ( ) ( )s l aq+ →(iii) 2KClO 2KCl 3O3 2( ) ( ) ( )s aq g→ +∆

(iv) Zn CuSO ZnSO + Cu4 4+ →(b) Translate the following statement into a

balanced chemical equation :

‘’Barium chloride reacts withaluminium sulphate to give aluminiumchloride and barium sulphate.’’

Or

When potassium iodide solution is addedto a solution of lead (II) nitrate in a testtube, a precipitate is formed.

(a) What is the colour of this precipitate?Name the compound precipitated.

(b) Write the balanced chemical equationfor this reaction.

(c) List two types of reactions in which thisreaction can be placed.

9. (a) Natural water bodies are not regularlycleaned whereas an aquarium needsregular cleaning. Why?

(b) What are decomposers? What will be theconsequence if the decomposers arecompletely eradicated from anecosystem? Give justification in supportof your answer.

Or

How is ozone formed in the upperatmosphere? State its importance. What isresponsible for its depletion? Write oneharmful effect of ozone depletion.

10. A white powder is used by doctors tosupport fractured bones.

(a) Write the name and chemical formula ofthe powder.

(b) How is this powder prepared?

(c) When this white powder is mixed withwater, a hard solid mass is obtained.Write a balanced chemical equation forthe change.

(d) Give one more use of this white powder.

11. A coil of insulated copper wire is connectedto a galvanometer. What would happen, if astrong bar magnet is

(a) pushed into the coil?

(b) withdrawn from inside the coil?

(c) held stationary inside the coil?

Give justification for each observation.

12. (a) Write the function of the following in thehuman alimentary canal

(i) Saliva (ii) HCl in stomach

(iii) Bile juice (iv) Villi

(b) Write one function each of the followingenzymes

(i) Pepsin

(ii) Lipase

13. (a) Plants do not have any nervous systembut yet, if we touch a sensitive plant,some observable changes take place inits leaves. Explain how could this plantrespond to the external stimuli and howit is communicated.

(b) Name the hormone that needs to beadministered to

(i) increase the height of a dwarf plant.

(ii) causes rapid cell division in fruitsand seeds.

14. What is biodiversity? Why are forestsconsidered as ‘’biodiversity hotspots’’?List two factors responsible for causingdeforestation.

15. How is the method of extraction of metalshigh up in the reactivity series differentfrom that for metals in the middle? Whycan the same process not be applied forthem? Name the process used for theextraction of these metals.

Section D

16. (a) Distinguish between esterification andsaponification reactions with the help ofchemical equations for each.

(b) With a labelled diagram describe in briefan activity to show the formation of anester.

Or

What is the difference between soaps anddetergents? State in brief the cleansingaction of soaps in removing an oily spotfrom a fabric. Why are soaps not veryeffective when a fabric is washed in hardwater? How is this problem resolved?

17. A person is unable to see objects distinctlyplaced within 50 cm from his eyes.

(a) Name the defect of vision the person issuffering from and list its two possiblecauses.

(b) Draw a ray diagram to show the defect inthe above case.

(c) Mention the type of lens used by himfor the correction of the defect andcalculate its power. Assume that the nearpoint for the normal eye is 25 cm.

(d) Draw a labelled diagram for thecorrection of the defect in the above case.

18. (a) How is the valency of an elementdetermine if its electronic configurationis known? Determine the valency of anelement of atomic no. 9.

(b) Given below are some elements ofthe modern periodic table atomicnumbers of the elements are given inparentheses :

A(4), B(9), C(14), D(19), E(20)

(i) With the help of the electronicconfiguration, find out which one ofthe above elements will have oneelectron in its outermost shell.

(ii) Which two elements belong to thesame group? Give reasons for youranswer.

(iii) Which one of the above elementsbelonging to the fourth periodhas bigger atomic radius and why?

19. (a) Define electric power. An electricaldevice of resistance R is connectedacross a source of voltage V and draws acurrent I. Derive an expression forpower in terms of current andresistance.

(b) Two electric bulbs rated 100 W, 220 Vand 60 W, 220 V are connected inparallel to an electric mains of 220 V.Find the current drawn by the bulbs fromthe mains.

Or

(a) How will you infer with the help of anexperiment that the same current flowsthrough every part of the circuitcontaining three resistors R R1 2, and R3

in series connected to a battery of Vvolts?

(b) Study the following circuit and findout the

(i) current in12 Ω resistor.

(ii) difference in the reading of A1 and A2, ifany.

20. (a) Define vegetative propagation. List itstwo methods.

(b) Why is this mode practised for growingsome types of plants?

(c) Explain the process of budding in Hydrawith the help of labelled diagrams.

Or

What is contraception? List its four

different methods. State four reasons for

adopting contraceptive methods.

21. (a) List two visible traits of garden pea that

Mendel considered in his experiments.

How do Mendel’s experiments show that

traits can be dominant or recessive?

(b) With the help of a flow diagram, how

would you establish that in human

beings the sex of a newborn is purely a

matter of chance and none of the parents

may be considered responsible for any

particular sex of a newborn child?

Section E

22. A student mixes sodium sulphate powderin barium chloride. What change wouldthe student observe on mixing the twopowder. Justify your answer and explainhow he can obtain the desired change?

A1

A2

+

–

6V 12Ω

24Ω

24Ω

Or

(a) Arrange the following metals in theincreasing order reactivities :

Copper, zinc, aluminium and iron

(b) List two observations you would recordin your 30 minutes after adding ironfilings to copper sulphate solution.

23. A solution ‘X’ gives orange colour when adrop of it falls on pH paper, while anothersolution ‘Y’ gives bluish colour when adrop of its falls on pH paper. What is thenature of both the solutions? Determine thepH of solutions ‘X’ and ‘Y’.

24. Study the following ray diagram and listtwo mistakes committed by the studentwhile tracing it. Rectify these mistakes bydrawing the correct ray diagram to showthe real position and size of the imagecorresponding to the position of the objectAB.

Or

A student has to trace the path of a ray oflight through a glass prism. List fourprecautions he should observe for betterresults.

25. Which one of the following is the correctset-up for studying the dependence of thecurrent on the potential difference acrossa resistor and why?

26. Write four sequential steps of the

procedure of the experiment ‘’Preparing a

temporary mount of a leaf peel to show

stomata.’’

Or

In the experimental set-up to show that

‘’the germinating seeds give out

carbon dioxide’’, answer the following

questions :

(i) Why do we keep the conical flaskairtight?

(ii) Name the substance kept in the small testtube inside the conical flask. Write itsrole.

(iii) Why does water rise in the delivery tube?

27. List two observations on the basis of which

it may be concluded that the given slide

shows binary fission in Amoeba.

+ –

A+

–

V+–

R

+ –

+ –

R

V+ –

A

C D

2F1 F1B O

F2 2F2

A

+ –

A+

–V

+ –

R

+ –

V+

–A

+ –

R

A B

Set 2

Section A

2. Name the component of sunlight whichfacilitates drying of wheat after harvesting.

Section B

3. List four characteristics of the imageformed by a concave mirror of focal length40 cm when the object is placed in front ofit at a distance of 20 cm from its pole.

4. Draw magnetic field lines in and around acurrent carrying straight solenoid.

5. Write the name and molecular formula of acarbon compound having its name suffixedwith ‘’-ol’’ and having two carbon atoms inits molecule. With the help of a chemicalequation indicate what hapens when thiscompound is heated with excess conc.H SO2 4?

Section C

6. Define the term evolution. ‘Evolutioncannot be equated with progress.’ Giveexamples to justify this statement.

9. Salt ‘P’, commonly used in bakeryproducts, on heating gets converted intoanother salt ‘Q’ which itself is used for theremoval of hardness of water and a gas ‘R’is evolved. The gas ‘R’ when passedthrough freshly prepared lime water turnsmilky. Identify ‘P’, ‘Q’ and ‘R’, givingchemical equation for the justification ofyour answer.

12. The following diagram shows two parallelstraight conductors carrying samecurrent. Copy the diagram and draw thepattern of the magnetic field lines aroundthem showing their directions.

What is the magnitude of magnetic fieldat a point X. which is equidistant from theconductors? Give justification for youranswer.

15. What is exploitation of resources withshort term aims? List its four advantages.

Section D

16. (a) If we cross pure-bred tall (dominant)pea plants with pure-bred dwarf(recessive) pea plants, we get pea plantsof F

1-generation. If we now self-cross the

pea plant of F1-generation, then we

obtain pea plants of F2-generation.

(i) What do the plants of F1-generation

look like?

(ii) What is the ratio of tall plants todwarf plants in F

2-generation?

(iii) State the type of plants notfound in F

1-generation but

appeared inF

2-generation, mentioning the

reason for the same.

(b) What are homologous structures? Givean example. Is it necessary thathomologous structures always havecommon ancestors?

18. (a) List two limitations of Newland’s law ofoctaves.

(b) Write the electronic configuration of twoelements A and B whose atomic numbersare 20 and 17 respectively. Write themolecular formula of the compoundformed when element A reacts withelement B. State whether this compoundis acidic, basic or neutral. Give reason tojustify your answer.

X

Set 3

Section A

1. Mention the condition under which a

current can flow in a conductor.

2. List two merits of solar cells.

Section B

3. List four characteristics of the image

formed by a convex lens of focal length

20 cm when the object is placed in front of

it at a distance of 10 cm from its optical

centre.

Or Define refractive index of transparentmedium. The speed of light in a medium ofabsolute refractive index 1.5 is 2 108×ms −1. What is the speed of light in vacuum?

4. It is established that an electric currentthrough a conductor produces a magneticfield around it. Is there a similar magneticfield produced around a thin beam ofmoving (i) alpha particles, (ii) neutrons?Justify your answer in each case.

5. ‘Conversion of ethanol to ethanoic acid isan oxidation reaction.” Justify thisstatement giving the relevant equation forthe chemical reaction involved.

Section C

6. List three roles of forests in conserving theenvironment. How do the forests getdepleted? State two consequences ofdeforestation on the environment.

8. State right hand thumb rule to determinethe direction of magnetic field around acurent carrying conductor. Apply this ruleto find the direction of magnetic fieldinside and outside a circular loop of wirelying in the plane of a table and current isflowing through it clockwise.

10. A metal X, which is used in thermiteprocess, when heated with oxygen gives anoxide Y which is amphoteric in nature.Identify X and Y. Write balanced chemicalequations of the reactions of oxide Y withhydrochloric acid and sodium hydroxide.

13. Draw a diagram of human excretorysystem and label the following

(i) Urinary bladder (ii) Left kidney

(iii) Left ureter

Section D

17. (a) List in tabular form two differencesbetween acquired traits and inheritedtraits.

(b) Give an example of body characteristicsused to determine how close two speciesare in terms of evolution and explain it.

19. (a) List two criteria Mendeleev’s used in hisperiodic table to classify the elements.State Mendeleev’s periodic law andexplain why no fixed position wasassigned to hydrogen in Mendeleev’speriodic table?

(b) How and why does the atomic size ofelements vary as we move

(i) from left to right in a period, and

(ii) down a group in the modern periodictable?

Answers

Set 1

1. Resistance is the property of a conductor by virtue

of which it opposes/resists the flow of charges

through it. Its SI unit is ohm and is represented by

the Greek letter Ω.

Resistance of a conductor is given by RV

I= .

2. Two primary elements that are used to fabricate and

manufacture a solar cell are silver and silicon.

3. There are two laws of reflection:(i) Angle of incidence is always equal to the angle

of reflection

i.e. ∠ = ∠i r

(ii) The incident ray, the reflected ray and thenormal at the point of incidence, all lie in thesame plane.

Or

The refractive index of a medium with respect to

vacuum is called absolute refractive index of the

medium. The absolute refractive index of a medium

is simply called its refractive index.

Mathematically it can also be expressed as,

1 2µ = Speed of light in medium

Speed of light in vaccum

For glass/water pair,

w ga g

a w

µµµ

=

4.

5. When 5% alkaline potassium parmanganatesolution is added drop by drop to warm propanol,then oxidation reaction occurs and propanoic acidis formed.

CH CH CH — OH3 2 2Propanol

→Heat

Alk.KMnO 4

CH CH COOH3 2Propanoic acid

6. Fossils are the preserved traces of living organismsthat lived millions of years ago. These help in tracingevolutionary relationships. The two methods ofdetermining the age of fossils are as follows:

• Digging the earth to find fossils, the deeperlayers generally contain older fossils ascompared to the surface layers.

• Carbon dating method can be used todetermine the age of fossils by detecting theratio of different isotopes of the carbon element inthe fossil material.

N S

Magnetic field due to a bar magnet

7. Light rays of different colours, travel with the samespeed in vacuum and air. But in any other medium,they travel with different speeds, because they havedifferent wavelengths and bend through differentangles, which is the cause of dispersion of lightthrough a glass prism.

The ray diagram following the path of ray throughtwo inverted glass prisms are shown below.

Or

The reflection of light from an object in all directionsis called scattering of light. The colour of scatteredlight depends on the size of scattering particles andwavelength of light.(i) At sunrise and sunset, the Sun and the sky

appears red. Light from the Sun near thehorizon passes through thicker layers of air andcovers larger distance in the atmospherebefore reaching our eyes.

Near the horizon, most of the blue light andshorter wavelength light rays are scatteredaway by the particles. Therefore, the light thatreaches our eyes are of longer wavelengths, asred has the longer wavelength. This gives riseto the reddish appearance of the Sun andthe sky.

(ii) During the day time, sky appears blue. This isbecause the size of the particles in theatmosphere is smaller than the wavelength ofvisible light, so they scatter the light of shorterwavelengths.

8. (a) (i) Double displacement reaction

(ii) Combination reaction

(iii) Thermal decomposition reaction

(iv) Single displacement reaction

(b) Balanced chemical equation is given as follows :

3BaCl + Al (SO )2Bariumchloride

2 4 3Aluminiumsulphate

3Aluminiumchloride

4Bariumsulphat

2AlCl 3BaSO→ +

e

Or

(a) A yellow precipitate of lead iodide (PbI )2 isformed.

(b) Pb(NO ) 2KI3 2Leadnitrate

Potassiumiodide

( ) ( )aq aq+ →

PbI ( ) 2KNO2Lead iodide(yellow ppt.)

3Potassiumn

s +

itrate

( )aq

(c) Precipitation reaction or double displacementreaction.

9. (a) Natural water bodies such as a pond or a lakeare self-sustaining and complete. In them, allthe organisms of a food chain are available.

If any organism dies, microbes like bacteria andfungi decompose they dead bodies intosimpler substances. An aquarium, on thecontrary is an artificial and incompleteecosystem. Abiotic components are notsupplied naturally to it. It may also not have allthe biotic components in it. Thus, if a fish dies inan aquarium, due to the absence ofdecomposers, the dead body will rot andpollute the water of aquarium. Hence anaquarium needs to be cleaned regularly.

(b) Organisms which breakdown the complexorganic compounds present in dead anddecaying matter are called decomposers, e.g.certain bacteria and fungi. Decomposers act ascleaning agents of environment bydecomposing dead bodies of plants andanimals. They also help in recycling ofmaterials, replenishment of soil’s nutrients, etc.

The consequence of their absence in anecosystem can be disastrous. The dead bodieswould persist for long durations, leading to theiraccumulation and thus, pollute theenvironment.

Nutrients associated with the dead remains willnot be returned back to the environment.

Or

Ozone at the higher levels of the atmosphere is aproduct that is obtained by the action of UVradiations on oxygen (O2) molecule. The highenergy UV radiations split apart some molecularoxygen (O2) into free oxygen (O) atoms. Theseatoms are very reactive and combine with othermolecular oxygen present to form ozone.

O O O2molecularoxygen

UV

Oxygenatoms

→ +

O O O2 3Ozone

+ →

It shields the surface of the earth from harmfulUltraviolet (UV) radiations of the sun.Chlorofluorocarbons (CFCs) are mainly responsiblefor the depletion of ozone layer.

The ozone depletion allows more amount of UVradiations to enter the earth. These radiations arehighly damaging to organisms as these can causeskin cancer in humans, decrease crop yield, etc.

10. (a) Plaster of Paris[Calcium Sulphate Hemihydrate

(CaSO . 1 / 2H O)4 2 ]

(b) It is obtained by heating gypsum

(CaSO 2H O4 2⋅ ) at 373 K.

V

White light

Dispersion

Recombination

First prism Second prism

White light

ScreenA

P1 A

P2

Slit

R

VV

R R

Recombination of the spectrum of

white light

At this temperature, gypsum loses water

molecules and forms plaster of Paris.

CaSO . 2H O4 2Gypsum Heat

373 K→

CaSO . 1

2H O 1

1

2H O4 2

Plaster of Paris

2+

(c) Plaster of Paris is a white powder and on mixingwith water, it changes to gypsum giving a hardsolid mass.

CaSO . ½ H O + 11

2H O CaSO4 2

Plaster of Paris2 4→ ⋅ 2H O2

Gypsum

(d) It is also used for making decorative pieces andfor making designs on ceilings.

11. (a) When bar magnet is pushed into the coil, then amomentary deflection is observed in thegalvanometer. This deflection indicates that amomentary current is produced in the coil.

(b) While withdrawing the magnet out of the coil, adeflection in opposite direction is observed. Itindicates that the current produced in the coil isin opposite direction.

(c) When the magnet is held stationary or at rest,then there is no deflection in the galvanometer.It indicates that no current is produced in thecoil in this case.

12. (i) Saliva contains an enzyme salivary anylase(ptyalin) that converts starch into sugar at anoptimum pH of about 7.

(ii) Hydrochloric acid (HCl) in the stomach kill thebacteria that have been ingested with food. Itcreates an acidic medium of pH about 2facilitating the action of pepsin enzyme.

(iii) Bile juice is secreted by the liver and acts onlarge fat molecules converting them to smallglobules for increasing the efficiency of enzymeaction.

(iv) Villi are numerous finger-like projectionspresent on the inner lining of the small intestinewhich increase the surface area for absorption.(a) Pepsin is a digestive enzyme secreted from the

gastric glands present in the walls of stomachwhich converts proteins into peptones.

(b) Lipase breaks down emulsified fats into fattyacids and glycerol.

13. (a) Plants do not have nervous system or muscletissue like animals. However, they still showmovement and responsiveness. Sensitiveplants give an immediate response to astimulus. Mimosa pudica (touch sensitive plant)shows folding up and drooping of leaves if itsleaves are touched. In such movements, plantcells change shape by changing the amount ofwater in them.

(b) (i) Gibberellin is the hormone needed to beadministrated to increase the height of a dwarfplant.

(ii) Cytokinin is the hormone needed to beadministrated to cause rapid cell division and itshighest concentration occurs in seeds and fruits.

14. Biodiversity refers to the variety and variability of lifefound in a particular area. Forests are considered as“biodiversity hotspots” because these are extremelyrich in species which are found only in nativehabitats and are under constant threat of extinction.Thus, the conservation of forests is important inorder to preserve the forest biodiversity.

Factors responsible for deforestation (cutting downtrees) are as follows :

• Forests are cleaned to meet the requirement oflarge quantities of wood, timber, medicines,food, etc.

• Due to the increasing population, forests arebeing cleaned to make homes for humans.

15. The metals in the middle of the reactivity series(such as iron, zinc, lead, copper etc.) is moderatelyreactive. Thus, to obtain such metals from theircompounds, their sulphides and carbonates arefirst converted into their oxides by the process ofroasting and calcination respectively and then themetal oxides are reduced to corresponding metalby using suitable reducing agents such as carbon.On the other hand, metals which is high up in thereactivity series (such as sodium, magnesium,calcium, aluminium, etc.) is very reactive andcannot be obtained from its compound by heatingwith carbon. Therefore such metals are obtained byelectrolytic reduction of their molten salt.

16. (a) Esterification When ethanol (an alcohol) reactswith acetic acid (a carboxylic acid) in thepresence of an acid as catalyst, a fruity (sweet)smelling liquid called ester is obtained. Thisreaction is called esterification.

CH COOH CH CH OH3Ethanoic

acid

3 2Ethanol

–H O

Con

2

+ →c.H SO2 4

CH —C

O

O CH CH3 2 3

Ester

Soaps are made from animal fats or vegetableoils by heating it with sodium hydroxide. Thisprocess of preparation of soap is calledsaponification.

Fat or Oil Alkali(Ester) (Sodium

hydroxide)

He+ →at

(Sodium saltof fatty acid)

Soap + Glycerol(An alcohol)

(b) Activity1. Take 1 mL ethanol and 1 mL glacial acetic acid

alongwith a few drops of concentrated sulphuricacid in a test tube.

2. Warm the contents in a water bath for atleast 5min.

3. Pour into a beaker containing 20-50 mL of waterand smell the resulting mixture.

4. Sweet smell would be observed.Reaction

CH COOH CH CH OH3Ethanoic acid

3 2Alcohol

Conc. H2SO+ 4 →

CH C

O

OC H H O3

Ester

2 5 2

+

Or

Soaps are sodium or potassium salts of long chaincarboxylic acids and have general formula

R COO Na– + .

where, R =C H , C H15 31 17 35 etc.

Detergents are usually ammonium or sulphonatesalts of long chain carboxylic acids. They are alsocalled as soapless soap (soaps and detergents).

Soaps when added to hard water, reacts to form acalcium or magnesium salt of the organic acidwhich being insoluble gets precipitated and agreyish soap scum is formed with no lather.

Detergents on the other hand, form lather in hardwater as well as in soft water. So detergents can beused to resolve the problem.

17. (a) As the person can not see within 50 cm from hiseyes so, he is suffering from hypermetropia orlong-sightedness.

This defect arises due to following reasons :

• Focal length of eye lens becomes large.

• Eyeball becomes too short, so that the image isformed behind retina.

(b) Actually the image is formed behind the retinaas the figure (i) shown below. Near point shiftsfrom normal figure (ii).

(c) This defect can be corrected by using a convex lensof suitable power. This will bring the image back onretina.

Here, v = −50 cm, u = −25 cm

Using lens formula,1 1 1

f v u= −

⇒ 1 1

50

1

25

1 2

50f=

−−

−= − +

= 1

50⇒ f = 50cm

Power of lens, Pf

= = = = +1 1

50

100

502

( )inmD

(d)

18. (a) It depends upon the number of valenceelectrons present in the outermost shell of its atom.

For the elements of group 1, 2, 13 and 14, valency =number of valence electron(s), whereas for theelements of group 15 onwards, valency = 8 −valence electrons.

Atomic number = 9 (i.e. F)

∴ Electronic configuration = 2 7K L

,

Valency = 1 ( )Q 8 7 1− =(b) (i) D (19) has one electron is its outermost shell.

Electronic configuration of

DK L M N

( ) , , ,19 2 8 8 1=

Ethanol + Acetic acid (1:1)+ Conc. H SO2 4

Burner

Water

Sweet smelling liquid

Tripod stand

Wire gauze

Beaker

Water

Test tube containing

reaction mixture

Near point

of the eye

Eye lensRetina

Image is formed behind the retina

N ′N

Normal near

point

Figure (i)

N′N

Image isformedon theretina

Eye lens

Near pointof the eye

Normalnear point

Retina

Near point

of the eye

Eye lens

Retina

Image is formed on the retina

N ′N

Normal near

point

Figure (ii)

(ii) A(4) and E( )20 belongs to same group as theyhave same number of valence electrons.

AK L

( ) ,4 2 2=

EK L M N

( ) , , ,20

2 8 8 2=

(iii) D( )19 and E( )20 belong to fourth period and D( )19

has bigger radius than E(20) as atomic size

decreases on moving left to right in a period.

19. (a) Electric power is defined as the amount of electricenergy consumed in a circuit per unit time. If anelectrical device of resistance R is connectedacross a source of voltageV, drawing current I,then

According to Ohm’s law, V IR=Electron power, P VI=

= × =IR I I R2

(b) For first bulb, P1 100= W, V1 220= V

∴ 112

1

220 220

100= = ×V

P= 484 Ω

For second bulb, P2 60= W, V2 220= V

∴ RV

P2

22

2

= = ×220 220

60= 8066. Ω

Since, both bulbs are connected in parallelcombination.

Hence, equivalent resistance,

RR R

R R=

+1 2

1 2

R = ×+

484 8066

484 8066

.

.= 302 5. Ω

Current drawn by the bulbs,

IV

R= = 220

302 5.= 073. A

Or

(a) The experimental set up comprises three

resistors R R1 2, and R3 of three different values

such as 1Ω, 2Ω and 3Ω which are connected in

series. Connect them with a battery of 6V, an

ammeter and plug key, as shown in figure,

The key K is closed and the ammeter reading isrecorded. Now, the position of ammeter ischanged to anywhere in between the resistorsagain, the ammeter reading is recorded eachtime. It’s observed that there were identicalreadings each time, which proved that samecurrent flows through every part of the circuitcontaining three resistances in seriesconnected to a battery.

(b) (i) Equivalent resistance of given circuit is R, thenR = +( || )24 24 12

= ×+

+24 24

24 2412 = +12 12 = 24 Ω

∴ Current through 12 Ω resistor,

IV

R= = 6

24= 025. A

(ii) Difference in reading ofA1 and A2

( . . )025 025− A = 0 A

20. (a) Vegetative Propagation It is a type of asexualreproduction in plants in which, new plants areobtained from a part of the parent plant like root,stem and leaves. Under favourable conditions,various structures take part in this type ofreproduction. This is mode of reproduction is of twotypes; natural vegetative propagation, where plantparts like root, stem and leaves develop into newplants and artificial vegetative propagation.

Which involves cutting, layering and grafting. Theyare used for many plants like sugarcane, roses,grapes, etc., by farmers.

(b) Vegetative propagation is practiced for growing sometypes of plants because of the following reasons(i) Plants that have lost their capability to produce

seeds can be propagated by this method.(ii) It helps to grow plants bearing superior traits, as

young plants obtained through asexual

reproduction are genetically identical to the

parent plant.

(iii) It is used for growing plants which require a

longer time to grow and become mature.

(c) Budding is a type of asexual reproduction

where a daughter organism is formed from a

small projection known as bud. It develops as

an outgrowth due to repeated cell division on

the parent body.

When fully grown, the bud detaches to grow into

a new independent individual, e.g. Hydra.

.

Or

The prevention of pregnancy is called

contraception or birth control.(i) Barrier Methods The physical devices such as

condoms and diaphragms are used. These are

physical barriers, which prevent sperms from

meeting the egg.

K+ –

A

II+ –

+

R1 R2 R3

V

–

Tentacles

Growing

budNew Hydra

Parent

Hydra

Budding in Hydra

(ii) Hormonal Methods These contain hormonalpreparations in the form of ‘pills’ which preventthe release of ovum.

(iii) Chemical Methods The vaginal pills containchemicals called spermicides which kill thesperms.

(iv) Surgical Methods In males, a small portion ofsperm duct is cut by surgical method. The cutend is tied properly to prevent the sperms fromcoming out. In females, a small part of theFallopian tube/oviduct is cut and tied to preventthe egg from entering the oviduct.

Four reasons for adopting contraceptivemethods are as follows :

• To avoid unwanted pregnancies.

• To prevent the transmission of sexuallytransmitted diseases (STDs).

• To control population size.

• To maintain a gap of few years between2 consecutive children (family planning).

The prevention of pregnancy is calledcontraception or birth control.

21. (a) Mendel used a number of visible contrastingtraits of pea plants and two such characters areround/wrinkled seeds and tall/short plants.

Mendel crossed a pure tall pea plant (TT) withpure dwarf pea plant (tt) and observed that allthe progeny were hybrid tall (Tt), i.e. only one ofthe traits was able to express itself in theF1-generation, which is the dominant trait.

The other trait is called the recessive trait whichremains suppressed. However, when heself-crossed plants of F1-generation, heobserved that one fourth of the plants weredwarf and three fourth were tall. The expressedtrait T for the tallness is dominant trait, while thetrait ‘t’ of dwarfness is recessive. Thus,Mendel’s experiments show that traits may bedominant or recessive.

(b) In humans, there are 23 pairs of chromosomes,out of which 22 pairs are autosomes and onepair is the sex chromosome. Females have aperfect pair of sex chromosome, both called Xand thus, contribute X-chromosome to both thesexes of progeny. But males have amismatched pair in which one is X(normal-sized) and the other is Y-chromosome(short in size).

An egg fertilised by X-chromosome carrying spermresults in a zygote with XX, which becomes a femaleand if an egg is fertilised by Y-chromosome carryingsperm, it results in a XY zygote that becomes male.

There are equal chances of an egg being fertilisedby either X-chromosome carrying sperm orY-chromosome carrying sperm. Thus, it is right tosay that sex of a newborn is purely a matter ofchance and none of the parents may be consideredresponsible for any particular sex of a new bornchild.

22. When the student mixes sodium sulphate powder inbarium chloride in a dry state, no change would beobserved. But when he dissolved them in water,barium sulphate precipitates out and sodiumchloride remains in solution. This is called a doubledisplacement reaction.

Refer to text on pg 5 (Double displacement reaction)

Or

(a) Copper < iron < zinc < aluminium

(b) (i) The iron fillings become brownish in colour.

Male Female

Gametes

Zygote

X Y X

XX XY

Offspring

Female Male

XYXX

Sex-determination in human beings

(ii) The blue colour of copper sulphate solution

fades. This is because iron being more reactive

than copper, displaces copper from copper

sulphate solution.

Fe( ) CuSO ( )Iron fillings

4Copper sulphate(blue)

s aq+

→ +FeSO ( ) Cu( )4Ferrous sulphate(Green)

Copper met

aq sal

23. Solution X has pH value around 4 and acidic innature.

Solution Y has pH value around 10 and basic innature.

Explanation According to pH colour chart orangecolour corresponds to the pH value 4 and bluishcolour represent pH value 10. Also, compoundshaving pH value less than 7 are acidic in nature andwhich have pH value greater then 7 are basic innature.

24. As the object is placed between F1 and 2 1F , so the

mistakes commited by the student are(i) The image should be formed beyond 2 2F .

(ii) The size of the image should be magnified innature. The correct ray diagram is

Or

Following precautions should be taken:(i) Prism should be within the boundary made all

through the experimentation.

(ii) Pins should be fixed vertically to the plane of thepaper.

(iii) The distance between the pins should be at least5 cm.

(iv) Angle of incidence should be taken between30° and 60° to observe the refraction clearly.

25. C represents correct set-up for studing the

dependence of the current on the potential

difference across a resistor because ammeter A is

connected in series while voltmeter V is connected

across the resistor R, parallely.

26. The sequential steps of the procedure of the

experiment “preparing a temporary mount of a leaf

peel to show stomata” are as follows :

(i) Take a freshly plucked leaf of Balsam. Removethe peel of the leaf from its lower surface by

tearing the leaf. Put the leaf peels in a watchglass containing water.

(ii) Add 1-2 drops of safrannin to the watch glasscontaining leaf peels to stain them. Now, selecta thin leaf peel and place it on a clean slide withthe help of a brush.

(iii) Put a drop of glycerine on the slide over thepeel. Now, with the help of forcaps gently placea coverslip over the peel.

(iv) Remove the excess stain and glycerine with ablotting paper. Then initially observe the slideunder low magnification of a compoundmicroscope and further under highermagnification.

Or

(i) We keep the conical flask airtight in theexperiment of germination of seeds becausepartial vacuum condition is needed to show therelease of CO2 from germinating seeds.

(ii) Freshly prepared 20% KOH solution is kept inthe small test tube inside the conical flask. KOHabsorbs CO2 gas, which creates a partialvacuum in the conical flask.

(iii) The partial vacuum created in the conical flaskby the KOH solution causes a rise in water levelin the U-shaped delivery tube.

27. Two observations on the basis of which it may beconcluded that the given slide shows binary fissionin Amoeba are as follows :

• Amoeba reproduces by the method of binaryfission where division of parent cell and itscomponents occur by stretching.

• Nucleus breaks into two daughter nuclei(karyokinesis). It slowly followed by division ofcytoplasm (cytokinesis).

Set 2

2. Infrared rays facilitate the drying of wheat afterharvesting.

3. Given, f = − 40 cm, u = − 20 cm

By mirror formula,1 1 1

v u f+ =

1 1 1

v f u= − =

−−

−1

40

1

20

= −1

20

1

40=

−2 1

40= 1

40

∴ v = 40 cm

Hence, virtual, erect, enlarged image is formed at adistance of 40 cm from pole on the back side ofmirror.

2F1 F12F2

F2

A

OB

B′

A′

4.

5. Name of the compound : Ethanol

Molecular formula : CH CH OH3 2

Ethanol is treated at 443 K with excess conc.H SO2 4,the water molecules gets removed from it andethene is obtained.

CH CH OH3Ethanol

2Hot

Con. H SO , 443 K2 4

→ CH CH H O2 2 2== +

Ethene Water

6. Evolution is the sequence of gradual changes thattake place in the heritable traits of a population oversuccessive generations.

Evolution should not be equated with progress.There is no real progress in the concept of evolution.It is simply the generation of diversity by naturalselection and its shaping by environmental selection.

Take for example, the case of human beings andchimpanzees. It is not true that human beings haveevolved from chimpanzees. Rather, humans andchimpanzees share more similarities than the othertypes of apes, which means that they probablyshare a more recent ancestor or a commonancestor than others.

9. Since, the gas ‘R’ evolved turned lime water milky.Thefore, it must be carbon dioxide. Moreover, salt‘Q’ produced is used for removal of hardeness ofwater, it must be washing soda.

Chemical equations as follow :

2NaHCO Na CO H O3 2 3Sodium carbonate

2( ) ( )s s l∆ → + ( )

‘P’ ‘Q’ ‘R’

+CO 2( )g ↑Na CO H O NO CO H O2 3 2 2 3 210 10( ) ( ) ( )s R s+ → ⋅

‘Q’

P : Sodium hydrogen carbonate (Baking soda)

Q : Sodium carbonate (washing soda)

R : Carbon dioxide (CO )2

12. Magnetic field lines due to parallel current carryingconductors is shown in figure.

Magnetic field at X = 0, because magnetic field at Xdue to both conductors are equal in magnitude butopposite in direction.

15. The exploitation of resources with a short spanobjectives (short terms aims) suggests theconsumption of resources for urgent necessitieswithout their maintenance for future use.

The following are the four benefits of utilisingresources with short terms aims :(i) It provides the immediate advantage of

meeting current basic human needs.

(ii) It provides the growth of industries.

(iii) It can lead to rapid development of people andnation.

(iv) It provides large number of products for useand comfort.

16. (a) (i) If pure-bred tall (dominant) plants were crossedwith pure-bred dwarf (recessive) plants, then theentire progeny of F-generation would to tall.

(ii) The phenotypic ratio of tall plants to dwarf plantsin F2-generation is 3 : 1. The genetypic ratio ofpure tall plants to hybrid tall plants to dwarfplants is 1 : 2 : 1.

(iii) In theF2-generation, Mendel found that all plantswere not tall, three quarter were tall and onequarter of them were short. This observationindicated that both the traits of shortness andtallness were inherited in F1-generation. But onlythe tallness trait was expressed inF1-generation.TT and Tt are phenotypically tall plants, whereastt is a short plant. For a plant to be tall, a singlecopy of ‘T’ is enough. Therefore, in trait Tt, ‘T’ isdominant while ‘t’ is recessive. Thus, inF2-generation, both the characters arerecovered, though one of these is not seen in F1

stage as tallness expressed by T is dominantover dwarfness expressed by t.

(b) Organs, which have the same basic structure(or same basic design), but perform differentfunctions are called homologous structures,e.g. forelimbs of reptiles, amphibians, birdsand mammals. Yes, homologous structures arealways inherited from a common ancestor, e.g.the mammals, birds, reptiles and amphibiansall have four limbs. The basic structure of limbsis similar though it has been modified toperform different functions in variousvertebrates. This shows that they have evolvedfrom a common ancestor.

+ –

K

S N

Magnetic field lines of

force due to a current

X

18. (a) This law was applicable only upto calcium. Aftercalcium, every eighth element did not possessthe same properties similar to that of the first.

Newland assumed that there were only 56elements existed in nature and no moreelements would be discovered in the future.But, later on, several new elements werediscovered, whose properties did not fit into theLaw of Octaves.

(b) EC of a ( )Z = 20 = 2 8 8 2K L M N

, , ,

EC of b ( )Z = 17 = 2 8 7K L M

, ,

Molecular formula of the compound formed :AB2. Since A is calcium and B is chlorine so themolecular formula of the compound formed isCaCl2. It is a salt of strong acid and strong base.Thus, it is neutral.

Also, valency of Ca = 2

and of Cl = 1

∴ Formula = CaCl2

Set 3

1. A current can flow in a conductor when potentialdifference across the ends of conductor is applied.

2. (i) Solar cells have no moving parts and requires littlemaintenance.

(ii) They work without the use of any focussingdevice.

3. Given, f = 20 cm, u = −10 cm

By lens formula,1 1 1

v u f− =

⇒ 1 1 1

v f u= +

1 1

20

1

10

1 2

20v= − = − = − 1

20

v = −20cm

Hence, virtual, erect and magnifield image isformed on the same side where the object is kept.

Or

For any two transparent media, the refractive indexof the second medium with respect to first mediumis equal to the ratio of the velocities of light in themedia.

Given, speed of light in the medium = ×2 108 m/s

Refractive index, n = 15.

Hence, n = Speed of light in vacuum

Speed of light in medium

⇒ 1 52 108

. =×vvac

where, vvac = speed of light in vauum.

⇒ vvac = × ×1 5 2 108. = ×3 108 ms −1

4. (i) Alpha particles are positively charged, hencethey can also produce magnetic field aroundthe moving beam.

(ii) Neutrons do not have any charge, hence theycannot produce magnetic field.

5. The chemical reaction for the convertion of ethanolto ethanoic acid are given below:

(i) CH CH OH + [O] CH CHO + H O3 2Ethanol

3Ethanal

2 →

(ii) CH CHO + [O] CH COOH3 3Ethanoic acid

→

As, in the first reaction, H2 is given out from ethanol.

We know that removal of hydrogen is oxidation. So,

this reaction is an oxidation reaction.

Similarly, is second equation an oxygen atom is

added to ethanal to form ethanoic acid. As we know,

addition of oxygen is called oxidation. So, the

second reaction is also an oxidation reaction.

6. Any three roles of forests in conserving theenvironment are as follows :

• They provide habitat to numerous species ofplants and animals. All inhabitants get food andprotection from the forests.

• They help in protecting the soil from erosion.

• They play a vital role in atmospheric circulationand thus, control the global climate.

The forests get depleted due to forest fires,urbanisation, industrialisation, overgrazing byanimals, etc.

The consequences of deforestation on theenvironment are as follows:(i) Changes in the climate and global warming.

(ii) Soil erosion and floods.

(iii) Extinction of wildlife.

8. Right hand thumb rule states that, if you hold the

current carrying straight wire in the grip of your right

hand in such a way that the stretched thumb points

in the direction of current, then the direction of the

curl of the fingers will give the direction of the

magnetic field. This rule is also called Maxwell’s

corkscrew rule.

Applying the right hand thumb rule, the magnetic

field inside the loop is in vertically downward

direction and outside the loop, it is in vertically

upward direction.

10. Since, the metal ‘X’ is used in thermite process,

therefore

‘X’ is aluminum and ‘Y’ is aluminium oxide, Al O2 3

(amphoteric in nature).

Balanced chemical equation is as follow :

(i) Al O ( ) 6HCl( ) 2AlCl ( )2 3Aluminiumoxide

3Al

s aq aq+ →uminium

chloride

23H O( )+ l

(ii) Al O ( ) 2NaOH( )2 3Aluminiumoxide

Sodiumhydroxid

s aq+

e

→ +2NaAlO ( ) H O( )2Sodium aluminate

2aq l

13. The labelled diagram of human excretory system is as follows :

17. (a) The differences between acquired and inherited raits are as follows:

Acquired traits Inherited traits

These characters develop in the

organism during their lifetime.

These characters are inherited by the

organisms from their parents.

These do not bring about any change

in the genes of organisms.

These bring about changes in the

genes of organisms.

These are lost with the death of the

individual, e.g., good physique of an

athlete, intelligence.

These are transferred to the next

generation, e.g., fused and free

earlobes.

(b) Forelimbs of humans and wings of birds show the common ancestry between humans and birds since, theforelimbs of a human and wings of birds have similar structure but perform different functions (homologousorgans).

Thus, the presence of homologous organs in different animals provide evidence for evolution by indicatingthat they have evolved from the same ancestor who had a similar basic design of the organ on which all thehomologous organs are based.

19. (a) To classify these elements:

He examined the relationship between the atomic masses of the elements and their physical and chemicalproperties.

Among chemical properties, he concentrated on the compounds formed by the elements with oxygen andhydrogen as they are very reactive and form compounds with most elements.

On this basis, Mendeleev gave a Periodic Law, which states that ‘the properties of elements are the periodicfunction of their atomic masses’.

The electronic configuration of hydrogen resembles with alkali metals. Moreover, it combines with halogen, oxygenand sulphur to give similar type of compounds as given by alkali metals, e.g. (HCl, NaCl), (H O, Na O)2 2 and(H S, Na S)2 2 . But just like halogen it exists in diatomic form and combines with metals and non-metals to formcovalent compounds. Thus, its position was not fixed in the Mendeleev’s periodic table but it was kept withalkali metals.

(b) The atomic size increases down the group. This is because new shells are being added as we go down thegroup. This increases the distance between the outermost electrons and the nucleus. As a result, the atomicsize increases inspite of the increase in nuclear charge.

The atomic radius decreases on moving from left to right along a period. This is due to an increase in nuclearcharge which tends to pull the valence electrons closer to the nucleus and reduces the size of the atom.

(i) Urinary

bladder

(iii) Left ureter

(ii) Left kidney