SOLVED C.B.S.E. PAPER 2019 · (ii) The reaction of iron (III) oxide [Fe2O3] with heated aluminium...

SOLVEDPAPER*

(With MarkingScheme)

C.B.S.E.2019

Class–XDelhi & Outside Delhi

Science

*Note : This paper is solely for reference purpose only. The format has now been modified by CBSE for March 2020 examination.

Time : 3 Hours Max. Marks : 80

General Instructions : (i) The question paper comprises five Sections A, B, C, D and E. You are to attempt all the sections. (ii) All questions are compulsory. (iii) Internal choice is given in sections B, C, D and E. (iv) Question numbers 1 and 2 in Section A are one mark questions. They are to be answered in one word or in one sentence. (v) Question numbers 3 to 5 in Section B are two marks questions. These are to be answered in about 30 words each. (vi) Question numbers 6 to 15 in Section C are three marks questions. These are to be answered in about 50 words each. (vii) Question numbers 16 to 21 in Section D are five marks questions. These are to be answered in about 70 words each. (viii) Question numbers 22 to 27 in Section E are based on practical skills. Each question is a two marks question. These are to be

answered in brief.

Delhi Set I Code No. 31/1/1

SECTION – A 1. What is the function of a galvanometer in a circuit ? 1 2. Why is biogas considered an excellent fuel ? 1

SECTION – B

3. How it can be proved that the basic structure of the Modern Periodic Table is based on the electronic configuration of atoms of different elements ? 2

OR The electronic configuration of an element is 2, 8, 4. State its : (a) group and period in the Modern Periodic Table. (b) name and write its one physical property.

4. Write two different ways in which glucose is oxidized to provide energy in human body. Write the products formed in each case. 2

5. Define the term power of accommodation. Write the modification in the curvature of the eye lens which enables us to see the nearby objects clearly ? 2

SECTION – C

6. 2 g of silver chloride is taken in a china dish and the china dish is placed in sunlight for sometime. What will be your observation in this case ? Write the chemical reaction involved in the form of a balanced chemical equation. Identify the type of chemical reaction. 3

OR Identify the type of reactions taking place in each of the following cases and write the balanced chemical equation

for the reactions. (a) Zinc reacts with silver nitrate to produce zinc nitrate and silver. (b) Potassium iodide reacts with lead nitrate to produce potassium nitrate and lead iodide.

7. Identify the acid and the base from which sodium chloride is obtained. Which type of salt is it ? When is it called rock salt ? How is rock salt formed ? 3

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2 Oswaal CBSE Solved Paper - 2019, SCIENCE, Class-X


8. Based on the group valency of elements write the molecular formula of the following compounds giving justification for each : 3

(i) Oxide of first group elements. (ii) Halide of the elements of group thirteen, and (iii) Compound formed when an element, A of group 2 combines with an element B of group seventeen. 9. Write three types of blood vessels. Give one important feature of each. 3 10. Trace the sequence of events which occur when a bright light is focused on your eyes. 3 11. What are plant hormones ? Name the plant hormones responsible for the following : 3 (i) Growth of stem (ii) Promotion of cell division (iii) Inhibition of growth (iv) Elongation of cells 12. Name the plant Mendel used for his experiment. What type of progeny was obtained by Mendel in F1 and F2

generations when he crossed the tall and short plants ? Write the ratio he obtained in F2 generation plants. 3OR

List two differences between acquired traits and inherited traits by giving an example of each. 13. What is a rainbow ? Draw a labelled diagram to show the formation of a rainbow. 3 14. How can we help in reducing the problem of waste disposal ? Suggest any three methods. 3

OR Define an ecosystem. Draw a block diagram to show the flow of energy in an ecosystem. 15. What is water harvesting ? List two main advantages associated with water harvesting at the community level.

Write two causes for the failure of sustained availability of groundwater. 3

SECTION – D

16. (a) List in tabular form three chemical properties on the basis of which we can differentiate between a metal and a non-metal. 5

(b) Give reasons for the following : (i) Most metals conduct electricity well. (ii) The reaction of iron (III) oxide [Fe2O3] with heated aluminium is used to join cracked machine parts. 17. Write the chemical formula and name of the compound which is the active ingredient of all alcoholic drinks. List

its two uses. Write chemical equation and name of the product formed when this compound reacts with : 5 (i) sodium metal (ii) hot concentrated sulphuric acid.

OR What is methane ? Draw its electron dot structure. Name the type of bonds formed in this compound. Why are

such compounds : (i) poor conductors of electricity ? and (ii) have low melting and boiling points ? What happens when this compound burns in oxygen ? 18. Define pollination. Explain the different types of pollination. List two agents of pollination. How does suitable

pollination lead to fertilizaton ? 5OR

(a) Identify the given diagram. Name the parts 1 to 5. (b) What is contraception ? List three advantages of adopting contraceptive measures.

1

2

3

4

5


Oswaal CBSE Solved Paper - 2019, SCIENCE, Class-X 3


19. An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm. 5 (i) Use lens formula to find the distance of the image from the lens. (ii) List four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case. (iii) Draw ray diagram to justify your answer of part (ii).20. (a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of

resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances. 5 (b) In an electric circuit two resistors of 12Ω each are joined in parallel to a 6 V battery. Find the current drawn

from the battery.OR

An electric lamp of resistance 20Ω and a conductor of resistance 4Ω are connected to a 6V battery as shown in the circuit. Calculate :

(a) the total resistance of the circuit. (b) the current through the circuit. (c) the potential difference across the (i) electric lamp and (ii) conductor, and (d) power of the lamp.

+ –––

+–A

6V

K

4

21. What is a solenoid ? Draw the pattern of magnetic field lines of (i) a current carrying solenoid and (ii) a bar magnet. List two distinguishing features between the two fields. 5

SECTION – E

22. Blue litmus solutions is added to two test tubes A and B containing dilute HCl and NaOH solution respectively. In which test tube a colour change will be observed ? State the colour change and give its reason. 2

OR What is observed when 2 mL of dilute hydrochloric acid is added to 1 g of sodium carbonate taken in a clean and

dry test tube ? Write chemical equation for the reaction involved. 23. In three test tubes A, B and C, three different liquids namely, distilled water, underground water and distilled

water in which a pinch of calcium sulphate is dissolved, respectively are taken. Equal amount of soap solution is added to each test tube and the contents are shaken. In which test tube will the length of the foam (lather) be longest ? Justify your answer. 2

24. A student is observing the temporary mount of a leaf peel under a microscope. Draw labelled diagram of the structure of stomata as seen under the microscope. 2

OR Draw a labelled diagram in proper sequence to show budding in Hydra. 25. In the experimental set up to show that “CO2 is given out during respiration”, name the substances taken in the

small test tube kept in the conical flask. State its functions and the consequence of its use. 2 26. While studying the dependence of potential difference (V) across a resistor on the current (I) passing through it,

in order to determine the resistance of the resistor, a student took 5 readings for different values of current and plotted a graph between V and I. He got a straight line graph passing through the origin. What does the straight line signify? Write the method of determining resistance of the resister using this graph. 2

OR What would you suggest to a student if while performing an experiment he finds that the pointer needle of

the ammeter and voltmeter do not coincide with the zero marks on the scales when circuit is open ? No extra ammeter/ voltmeter is available in the laboratory.

27. List four precautions which a student should observe while determining the focal length of a given convex lens by obtaining image of a distant object on a screen. 2




Delhi Set II Code No. 31/1/2

Note : Except these, all other questions are from Set I.

SECTION – A

1. Name and define the SI unit of current. 1 2. Write the name of the main constituent of biogas. Also state its percentage. 1

SECTION – B

3. Write the name, symbol and electronic configuration of an element X whose atomic number is 11. 2OR

Can the following groups of elements be classified as Dobereiner’s triad : (a) Na, Si, Cl (b) Be, Mg, Ca Atomic mass of Be-9; Na-23, Mg-24, Si-28, Cl-35, Ca-40. Justify your answer in each case. 4. How is O2 and CO2 transported in human beings ? 2 5. Write the structure of eye lens and state the role of ciliary muscles in the human eye. 2

SECTION – C

6. Identify the acid and base which form sodium hydrogen carbonate. Write chemical equation in support of your answer. State whether this compound is acidic, basic or neutral. Also write its pH value. 3

9. Define the term transpiration. Design an experiment to demonstrate this process. 3 10. What is feedback mechanism of harmonic regulation. Take the example of insulin to explain this phenomenon. 3

OR List two differences between acquired traits and inherited traits by giving an example of each. 13. Why should there be equitable distribution of resources ? List three forces that would be working against an

equitable distribution of our resources. 3

SECTION – D

17. (a) Write chemical equations for the following reactions : 5 (i) Calcium metal reacts with water. (ii) Cinnabar is heated in the presence of air. (iii) Manganese dioxide is heated with aluminium powder. (b) What are alloys ? List two properties of alloys. 18. An object is placed at a distance of 30 cm from a concave lens of focal length 30 cm. 5 (i) Use lens formula to determine the distance of the image from the lens. (ii) List four characteristics of the image (nature position, size, erect/inverted) in this case. (iii) Draw a labelled diagram to justify your answer of part (ii).

Delhi Set III Code No. 31/1/3


SECTION – A

1. If you could use any source of energy for heating your food, which one would you prefer ? State one reason for your choice. 1

2. Write the function of voltmeter in an electric circuit. 1

SECTION – B

3. What happens to the image distance in the normal human eye when we decrease the distance of an object, say 10 m to 1 m ? Justify your answer. 2

4. List two different functions performed by pancreas in our body. 2

SECTION – C

7. List three resources each of : 3 (i) exploring resources with short term aims, and (ii) using a long term perspective in managing our natural resources.




9. Nervous and hormonal system together perform the function of control and coordination in human beings. Justify this statement with the help of an example. 3

11. What is photosynthesis ? Explain its mechanism. 3 15. Explain the following : 3 (a) Sodium chloride is an ionic compound which does not conduct electricity in solid state, whereas it does

conduct electricity in molten state as well as in its aqueous solution. (b) Reactivity of aluminium decrease if it is dipped in nitric acid. (c) Metals like calcium and magnesium are never found in their free state in nature.

SECTION – D

21. Write the main difference between an acid and a base. With the help of suitable examples explain the term neutralization and the formation of : 5

(i) acidic, (ii) basic and (iii) neutral salts.

SOLUTIONSDelhi Set I Code No. 31/1/1

SECTION – A

1. Detect the presence or direction of current. 1[CBSE Marking Scheme, 2019]

Detailed Answer : The function of a galvanometer in a circuit is to detect and measure the small amount of current flowing in the

circuit. 1

2. It burns completely/ burns without smoke / high calorific value. 1[CBSE Marking Scheme, 2019]

Detailed Answer : Biogas is considered an excellent fuel because (a) It does not cause air pollution as it does not produce smoke. (b) It has high calorific value. 1

SECTION – B

3. Modern periodic table consists of groups and periods. Where number of valence electrons determines the group and number of shells determines the period. 1 + 1

[CBSE Marking Scheme, 2019]

Detailed Answer :

Modern periodic law states that the physical and chemical properties of an element in the periodic table are periodic function of their atomic numbers. Elements with same number of valence electrons are placed in the same group while those having same number of shells are placed in same period.

For example, element with atomic number 11 (2, 8, 1) belongs to group 1 and third period. 2

OR

(a) Group – 14, Period – 3 ½ + ½ (b) Silicon ½ Non – metallic / poor conductor of electricity ½

(or any other property) [CBSE Marking Scheme, 2019]




Detailed Answer :

(a) Electronic configuration = 2, 8, 4

Valence electrons = 4

So, the element belongs to group 14 (10 + 4) and third period. 1

(b) Given element is silicon. Silicon are metalloids thus show properties of both metals and non-metals. 1

4.• Aerobic / Presence of oxygen ½ Product – CO2 and H2O ½ • Anaerobic / Absence of oxygen ½ Product – Lactic acid ½


Detailed Answer :

Glucose can be oxidized aerobically and anaerobically. During aerobic respiration, glucose breaks down into pyruvate followed by formation of carbon dioxide, water and energy (ATP molecules). During anaerobic respiration, glucose breaks down into lactic acid and two molecules of ATP. 1+1

5.• Power of accommodation – Ability of eye lens to adjust its focal length. 1 • Curvature increases/lens becomes thick. 1


Detailed Answer :

The ability of eye lens to adjust its focal length is called as accommodation of eye.

While viewing nearby objects, ciliary muscles contract resulting in the thickening of lens. It gets compressed into a more convex shape decreasing the focal length. The image falls on retina enabling to see the nearby objects clearly. 1 + 1

SECTION – C

6.•Whitesilverchlorideturnsgreyinsunlight 1

•2AgClSunlight → 2 Ag + Cl2 1

•Decompositionreaction/Photolyticdecomposition 1[CBSE Marking Scheme, 2019]

Detailed Answer : When 2 g silver chloride taken in china dish is placed in sunlight, the colour of silver chloride changes to grey. 1 Chemical reaction involved : 1

2AgCl(s) 2Ag(s)+Cl (g)White

Sunlight

(Grey)2 →

This is photochemical decomposition reaction. 1

OR

(a) Displacement reaction ½ Zn + 2AgNO3 → Zn (NO3)2 + 2 Ag 1 (b) Double displacement reaction ½ 2KI + Pb(NO3)2 → PbI2 + 2KNO3 (deduct ½ mark for non balanced equation) 1


Detailed Answer : (a) Displacement reaction

Zn(s) + 2AgNO3(aq) → Zn(NO3)2(aq) + 2Ag(s)

(b) Double displacement reaction

2KI(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbI2(s) 3




7.•Acid–Hydrochloricacid/HCl ½ Base – Sodium hydroxide/NaOH ½ •NeutralSalt ½ •Whenitformsbrowncrystalscombinedwithimpurities ½ •Dryingupofseas 1


Detailed Answer : Sodium chloride is obtained by the reaction between hydrochloric acid (acid) and sodium hydroxide (base).

It is a neutral salt.

It is called as rock salt in crystalline form.

Rock salt is formed from a body of sea water which undergoes an intense evaporation process. This is followed by long process of geologic aging resulting in large masses of salt. 3

8. (i) A2O – Valency of group one is 1 and of oxygen is 2. ½ + ½ (ii) AX3 – Valency of group 13 is 3 and of halogen is 1. ½ + ½ (iii) AB2 – Valency of element A of group 2 is 2 and of element B of group seventeen is 1. ½ + ½


Detailed Answer : (i) Valency of oxygen is 2.

Element

Valency 1 2= M O2

OM

So, the molecular formula of oxide of first group elements is MX2O where 2 is the group 1 element and O is oxygen. 1

(ii) Valency of group 13 elements is 3. Valency of halogens 1.

Element

Valency 3 1= MX3

XM

So, the molecular formula of halide of group thirteen elements is MX3 where M is the group 13 element and X is halogen. 1

(iii) Valency of group 2 elements is + 2. Valency of group 17 elements is 1.

Element

Valency 2 1= AB2

BA

So, the molecular formula is AB2. 1

9. • Arteries – No valves/thick walled/carry oxygenated blood/carry blood away from heart. 1 •Veins–Presenceofvalves/thinwalled/carrydeoxygenatedblood/carrybloodtowardsheart. 1

• Capillaries – Very fine/mixed blood/found in tissues/sites for material exchange. 1[CBSE Marking Scheme, 2019]

Detailed Answer :

Three types of blood vessels are arteries, veins and capillaries. Arteries carry blood away from the heart to different body parts. Veins bring blood from different body parts to the heart. The nutrients, hormones, gases can diffuse into the tissue cells through the walls of capillaries and vice versa. 1+1+1

10. Receptor cells of eyes/retina → Sensory neuron → Brain / CNS

¯ Pupil contracts / Eye lids close/blink ← Eye muscles ← Motor neuron. ½ × 6

(Note : If a child writes spinal cord in place of brain give full credit to him/her)[CBSE Marking Scheme, 2019]




Detailed Answer : When bright light is focused on our eyes, the photoreceptors generate electric impulses and pass it to the sensory

neurons. They carry the stimuli to the spinal cord which transports the message to brain. The brain sends the response to the muscles of the eyelids to close by contracting the pupil.

Receptor → Sensory neuron → Spinal cord → Brain → Motor neuron → Eye → Contraction of eye muscles 3

11. Plant hormones – Chemical substances which help the plant to coordinate growth and development 1 (i) Auxins/ Gibberellins (ii) Cytokinins (iii) Abscisic acid / ABA (iv) Auxins/ Gibberellins ½ × 4


Detailed Answer : Plant hormones or phytohormones are plant growth substances which are secreted in small quantities and are

involved in growth promoting activities such as cell division, cell enlargement, pattern formation, flowering, fruiting, tropic growth and seed formation.

(i) Gibberellins (ii) Cytokinins (iii) Abscisic acid (iv) Auxins 3

12. • Pea Plant / Garden pea / Pisum sativum 1 • F1 – All tall; F2 – Tall and short ½ + ½ • Ratio–Tall:Short 3 : 1 / 1 : 2 : 1 1


Detailed Answer : Mendel used Pisum sativum (Pea plant) for his experiment.

Mendel took a tall pea (TT) plant and a short pea (tt) plant. When he crossed both, the first filial generator (F1) obtained were tall. When F1 progeny was self-pollinated, all plants obtained in F2 generation were not tall. Instead three tall pea (dominant) plants and one short pea (recessive) plant was obtained.

TT(Tall)

Tt × Tt(Tall) (Tall)

T t

TTTall

T

t

TtTall

ttShort

TtTall

Parents :

F1 generation

F2

Selfing of :F1

Phenotypic ratio3 Tall : 1 short

Genotypic ratio1 Pure Tall (TT) : 2 Hybrid (Tt) :

1 Pure short (tt)

tt(Short)

Tt(Tall)

×

Phenotypic ratio and genotypic ratio should be in same horizontal line.

3OR

Acquired Trait Inherited Trait

1. These traits are not transferred from one generation to the next generation.

These traits are transferred from one generation to the next. 1

2. They do not bring about change in DNA. Example : Acquiring any skill.

They bring about changes in DNA. 1Example : Eye colour. 1

(or any other relevant point and example) [CBSE Marking Scheme, 2019]




Detailed Answer : Differences between acquired trait and inherited trait :

Acquired Trait Inherited Trait

1. These traits are the characteristics which are developed during the lifetime of an individual.

Inherited traits are genetically determined characteristic that distinguishes a person.

2. Acquired traits are not passed on to the next generation.

These are the characteristics transmitted from parents to the offspring.

E.g. Learning of dance, music etc, and muscular body of a wrestler.

E.g. Attached or free earlobe and curly hair.

1 + 1 + 1

13. Rainbow – A natural spectrum of sunlight appearing in the sky after a rain shower. 1

Sunlight

Raindrop

Violet

Red

2[CBSE Marking Scheme, 2019]

Detailed Answer :

Rainbow is a natural spectrum appearing in the sky after a rain shower, caused by dispersion of sunlight by tiny water droplets present in the atmosphere.

3

Formation of rainbow :

Sunlight

Raindrop

R

V

14. Segregation of waste; Recycling; Composting: Reducing the use of non – biodegradable material : Reuse. (Any three) 1 × 3


Detailed Answer :

Methods to reduce the problem of waste disposal are :

(i) By minimizing the use of non-biodegradable substances.

(ii) By following the principle of 3Rs-reducing, reusing and recycling.

(iii) By segregating and disposing biodegradable and non-biodegradable substances separately. 3




OR

The system where all the living organisms in an area together interact with the non – living constituents of the environment. 1

Carnivores

Herbivores

Producers

Sunlight 2[CBSE Marking Scheme, 2019]

Detailed Answer : An ecosystem can be defined as a functional unit of nature, where living organisms interact among themselves

and with the surrounding physical environment. Diagram to show the flow of energy in an ecosystem : Assuming 10,000 J of energy is available to the producers,

then 1000 J will be available to the primary consumers, 100 J will be available to secondary consumers and 10 J will be available to tertiary consumers.

TERTIARYCONSUMER

SECONDARY CONSUMERS

PRIMARY CONSUMERS

PRODUCERS

10 J

100 J

1000 J

10000 J

3

15.•Atechniqueusedtocollectandstorewaterforfutureuse 1 •Advantages–Availableresourceintimeofneed ½ + ½ Recharging the ground water level •Causes–Overuseofgroundwater Deforestation ½ + ½


Detailed Answer : Water harvesting is a technique used for collecting, storing and using rainwater for irrigation and other uses. Two main advantages associated with water harvesting at the community level are as follows : (i) It resolves the problem of water scarcity. (ii) It helps to recharge the ground water. Two causes for the failure of sustained availability of ground water are as follows : (i) Pollution of water caused by human activities and industrial effluents. (ii) Massive overuse of ground water resulting in water shortage. 1+1+1

SECTION – D

16. (a)

Metals Non-metals

1. Metals form basic oxides with oxygen. Non – metals form acidic or neutral oxides with oxygen.




2. Metals react with dilute acids to liberate hydrogen. Non metals do not displace hydrogen from dilute acids.

3. Metals form positively charged ions by losing electrons.

Non metals form negatively charged ions by gaining electrons.

1 × 3 (b) (i) Metals have loosely bound electrons / Loose electrons easily / free electrons 1 (ii) Molten iron produced during reaction joins the cracked machine parts. 1 [CBSE Marking Scheme, 2019]Detailed Answer : (a) Differences between metals and non-metals based on chemical properties :

Metals Non-metals

1. Metals react with oxygen to form basic oxides. Non-metals react with oxygen to form acidic oxides or neutral oxides.

2. Metals form ionic chlorides with chlorine, which are electrolytes and non-volatile.

Non-metals form covalent chlorides with chlorine, which are non-electrolytes and volatile.

3. Generally metals do not react with hydrogen, except for sodium, potassium, calcium and magnesium, which combine with hydrogen to form ionic metal hydrides.

Non-metals react with hydrogen to form stable covalent hydrides

3 (b) (i) Because they contain free electrons which can move easily through the metal and conduct electric

current. (ii) In the thermite process, iron (III) oxide is heated with aluminium, which results in evolution of high

amount of heat which melts iron. This molten iron is used to fill the cracked machine parts. Fe2O3 (s) + 2Al(s) → 2Fe(l) + Al2O3 (s) + Heat 2

17.•C2H5OH, Ethanol/Ethyl alcohol ½ + ½ • Good solvent; used in medicines (Any other) ½ + ½ (i) 2C2H5OH + 2 Na ¾® 2C2H5ONa + H2 1 Sodium ethoxide ½ (ii) C2H5OH

Hot Conc. H SO 443 K

2 4 → CH2=CH2+ H2O 1

Ethene ½


Detailed Answer : Ethanol with chemical formula : CH2CH2OH is an active ingredient of all alcoholic drinks. 1 Two uses are : (a) It is used in the manufacture of paints and varnishes. (b) It is used in medical swabs and hand sanitizers. 2 Chemical reactions of ethanol : (i) With sodium metal :

2CH CH OH+ 2Na 2CH CH ONa +2H3 2Ethanol Sodium

3 2Sodium ethoxide

2 →

(ii) With hot concentrated sulphuric acid :

CH CH OH CH =CH +H O3 2

Ethanol

Hot concH SO 2 2

Ethene2

2 4 →

1OR

• CH4/Simplest hydrocarbon ½

C

H

HH

H

× ×

×

×

1




• Covalent bonds ½ (i) No ions or charged particles are formed 1 (ii) Due to weak covalent bonds 1 • Carbon dioxide and water are produced/ CH4 + 2O2 ¾® CO2 + 2H2O 1


Detailed Answer : Methane is a colourless and highly flammable gas produced on decomposition of vegetation naturally in

marshlands. It is the simplest hydrocarbon (CH4). Electron dot structure :

H

×

•

H

×

•

•× HH C •

All the bonds present between four hydrogen atoms and one carbon atom at the center are covalent bonds. (i) Methane is a poor conductor of electricity as all the bonds present are covalent bonds. Hence, no free electrons

are available for conduction of electricity. (ii) As force of attraction between the molecules are not very strong in covalently bonded carbon compounds,

therefore, methane being a covalent compound has very low melting and boiling point. When methane burns in oxygen, carbon dioxide, water and large amount of heat and light is released. CH4 + O2 → CO2 + H2O + heat and light 1+1+1+1+1

18. •Pollination–Transferofpollenfromanther/stamentostigmaoftheflower. 1 • Type of pollination – (a) Self pollination – Transfer of pollen from anther / stamen to stigma occurs in the same flower. ½ + ½ (b) Cross pollination – Pollen is transferred from anther / stamen of one flower to stigma of another flower.

½ + ½ • Agents of pollination – Wind, Water, Insects and Animals. (Any two) ½ + ½ • A tube grows out of the pollen grain and travels through the style, to reach the female germ cell in the ovary

to cause fertilization. 1[CBSE Marking Scheme, 2019]

Detailed Answer : Pollination is the mechanism of transfer of pollen grains (shed from the anther) to the stigma of a pistil. Depending on transfer of pollens, pollination can be of following types : (i) Self pollination : Pollent is transferred to stigma of a flower on the same parent plant. (ii) Cross pollination : Pollen is transferred to stigma of flower of different parent plants. Two agents of pollination are abiotic (i.e. wind) and biotic (i.e. animals). If the pollen of the right type is deposited on to the stigma of the flower of the same species, the pistil accepts the

pollen and promotes post-pollination events. The pollen grains germinates on stigma to produce a pollen tube. The pollen tube carries male gametes to ovule present inside the ovary leading to fertilization. 1+2+1+1

OR

(a) • Female reproductive system ½ • Name of parts – 1 : Fallopian tube/Oviduct 2 : Ovary 3 : Uterus 4 : Cervix 5 : Vagina ½ × 5




(b) • Method to avoid pregnancy ½ • Advantages : - Proper gap between two pregnancies - Avoiding unwanted pregnancy - Keeping population under control ½ × 3


Detailed Answer : (a) 1. Oviduct or Fallopian tube 2. Ovary 3. Uterus 4. Cervix 5. Vagina (b) Contraception is the methods or ways to prevent fertilization and pregnancy in fertile females. Three major advantages of adopting contraceptive measures are as follows : (i) Help in family planning and population control. (ii) Prevention of sexually transmitted diseases like gonorrhoea, HIV-AIDS, etc. (iii) Prevention of unwanted pregnancies. 2½+2½

19. (i) u = – 60 cm, f = –30 cm, v = ?

1f

= 1 1v u

−

½

1v

= 1 1f u

+

= 1

301

603

60( ) ( )−+

−=

−cm cm

\ v = – 20 cm 1

m = vu

= −−

=2060

13

cmcm

½

Distance of the image will be 20 cm in front of lens. ½

(ii) Nature : Virtual ½

Position : 20 cm from lens on the same side as the object ½

Size : Diminished ½

Erect/Inverted : Erect

(iii)

1

A

BF

B' O'

A'

O O30 cm

60 cm

F 2F

[CBSE Marking Scheme, 2019] 5

Detailed Answer : (i) Object distance, u = – 60 cm Focal length, f = –30 cm Using lens formula,

1 1v u

− = 1f




1 160v

−−( )

= 130( )−

1v

= − −130

160

1v

= −360

v = – 20 cm

Distance of the image will be 20 cm in front of lens.

(ii) Nature : Virtual

Position : Between F1 and O or on same side of lens where image is placed.

Size : Diminished and 13

rd of object mvu

hhh

h= = − = = ′ ′ =

2060

13 3

;

Image is erect.

(iii)

O

60 cm

30 cm

F2

F1

2F1

2F2

2 + 2 + 1

20.

(a)

( ( A– +

A– +

R1

R2

R3

M

O

T

L

P

S

–+

1

From figure :

I = I1+I2+I3

I1 =

VR1

, I2 =VR2

, I3 = VR3

\

VRP

= VR1

+

VR2

+

VR2

1

1RP

= 1 1 1

1 2 3R R R+ +

1




(b) R1 = R2 = 12 Ω V= 6 V

1RP

= 1 1 1

121

121 2R R+ = +

½

\ Rp = 6 Ω ½

I = VR

VV

Ap

= =66

1

1


Detailed Answer : (a)

i1

i2

ii

i

iV

i3

in

Rn

R3

R2

R1

1+2+2

Let there be n resistance, each of value R1, R2 .... Rn respectively connected in parallel to a battery of voltage V.

Let current I is sent to the circuit.

If the equivalent resistance is Req, then current drawn I = VReq

According to the above circuit, I = I1 + I2 + I3 + .... + In

VReq

= VR

VR

VR

VRn

n

1

1

2

2

3

3+ + + +....

So, VReq

= 1 1 1 1

1 2 3R R R Rn+ + + +......

Therefore, the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to sum of the reciprocals of individual resistances.

(b) In parallel combination, Rtotal is given as

1RTotal

= 1 1

1 2R R+

= 1

121

12+ =

212

= 6Ω

Hence, current I = V

RTotal =

66

VW

= 1A

OR




(a) R = R1 + R2 1

= 20 Ω + 4 Ω = 24 Ω

(b) I = VR

=

624

VΩ

= 0.25 A 1

(c) (i) For electric lamp : V = IR

=624

20× = 5 V 1

(ii) For Conductor : V = IR

= 624

4× = 1 V 1

(d) P = VI

= 5624

V A× = 1.25 W. 1


Detailed Answer : Given, Voltage of battery, V = 6V Resistance of electric lamp, R1 = 20Ω

Resistance of series conductor, R2 = 4Ω

(a) Total resistance of circuit, RTotal = R1 + R2 = 20Ω + 4Ω = 24Ω

(b) Using Ohm's law I = V

RTotal

= 64

24W = 0.25 A

(c) Potential difference across (i) Electric lamp, V1 = IR1 = 0.25 A × 20Ω = 5V (ii) Conductor, V2 = IR2 = 0.25 A × 4Ω = 1V (d) Power of the lamp = I2R = (0.25)2 × 20Ω = 1.25 W 1+1+2+1 21.•Acoilofmanyturnsofinsulatedcopperwirewrappedcloselyintheshapeofacylinder. 1

(i)

1




(ii)

S N

1

• Distinguishing features :

Solenoid Bar Magnet

1. Field disappear on stopping the current. No effect of current on field.

2. Strength of the field can be changed by changing the Current.

Strength cannot be changed.

3. Direction can be reversed by changing the direction of current through it.

Direction is fixed and cannot be reversed.

(Any two features) 2


Detailed Answer :

Solenoid is a coil of insulated copper wire closely wrapped into a shape of a cylinder. (i) Pattern of magnetic field lines of a current carrying solenoid :

+ –

NS

(ii) Pattern of magnetic field lines of around a bar magnet :

S N

Two distinguishing features between the two fields are :

(i) Magnetic field of solenoid can be changed according to our requirements by just changing current or core of solenoid while the magnetic field of bar magnet is fixed.

(ii) Magnetic field outside the solenoid is negligible in comparison to the bar magnet. 2+2+1




SECTION – E

22. • Test Tube A ½ • It changes the colour from blue to red. ½

Hydrochloric acid turns blue litmus red. 1


Detailed Answer : Dilute HCl (Tube A) being acidic in nature will change the colour of blue litmus solution to red. There will be no

change in the Tube B containing NaOH solution as NaOH is basic in nature. 1+1OR

• Brisk effervescence is produced. 1

• Na2CO3 + 2HCl → 2 NaCl + H2O + CO2 [CBSE Marking Scheme, 2019] 1

Detailed Answer : When 2 ml of dilute hydrochloric acid is added to 1 g of sodium carbonate taken in a clean and dry test tube, a

brisk effervescence is produced due to evolution of carbon dioxide gas. 2HCl + Na2CO3 → 2NaCl + H2O + CO2 2

23. •IntesttubeA 1 •Asdistilledwatercontainsnosalts. [CBSE Marking Scheme, 2019] 1

Detailed Answer : Test tube A is filled with distilled water (which is also considered as soft water) is free from ions like Mg2+, Ca2+,

etc. So, it will give the longest length of foam. Test tube B is filled with underground water which contains ions like Mg2+, Ca2+, etc. On reaction with soap, it form scum which is insoluble in water. So, it will give lesser length of foam than test tube A. Test tube C is filled with distilled water in which CaSO4 is dissolved. Ca2+ ions on reaction with soaps forms scum which is insoluble in water. So, it will also give lesser length of foam than test tube A.

24.

Guard Cells

Stomatal pre

Chloroplast

1(Any one diagram with any two labellings) ½ × 2


Detailed Answer : Structure of stomata :

2

Chloroplast




OR

Bud

1 Drawing in proper sequence Labelling – Bud 1


Detailed Answer : Budding in Hydra :

2

Tentacles

Bud

Parent Hydra Parent Hydra

New Hydra

25. •Substance taken : KOH ½ •Function : It absorbs CO2 produced by the germinating seeds. ½ Consequence : The water level rises in the test tube dipped in the beaker / partial vacuum is created. 1


Detailed Answer : KOH solution is taken in the small test tube kept in the conical flask. The function of KOH is to absorb the carbon dioxide produced during respiration by germinating seeds. Consequence : This creates a vacuum in the conical flask. The air present in the bent glass tube moves into the

conical flask pulling up the water present in the bent tube. 2

26. •Potentialdifference(V)isdirectlyproportionaltocurrent(I)orV∝ I. 1 • Method : Finding slope of the graph. 1


Detailed Answer : The graph between V and I is a straight line and passes the origin, this verifies the Ohm's law.

MP

Q

()

Po

ten

tial

dif

fere

nce

V

( ) CurrentI

The slope gives the resistance of the resistor used in the circuit.

Slope = QMMP

= V VI I2 1

2 1

−−

or R = Value of potential difference at a point

Value of current att the same point 1+1




OR

•Measurethezeroerror. 1 •Valueofzeroerrorshouldbeadjustedtotheobservedvalues. 1


Detailed Answer : This is defined as the zero error of the scale of ammeter or voltmeter. It there is a zero error, then this error is

subtracted from the value that depicts when the circuit is closed, otherwise, the accurate current or potential difference will not be recorded. 2

27. Precautions : (1) Lens should be held in vertical position with its faces parallel to the screen. (2) Clear and sharpest image should be obtained by adjusting the position of lens. (3) Three observations should be taken at least. (4) Base of lens, screen and measuring scale should be in straight line. (or any other) ½ × 4


Detailed Answer : Precautions that should be observed while determining the focal length of a given convex lens by obtaining

image of a distant object on a screen are : (i) While measuring the distance, meter scale should be kept parallel to the ground. (ii) Distance should be measured only when well defined sharp image of the distant object is obtained. 2

Delhi Set II Code No. 31/1 /2


SECTION – A

1.•Ampere ½

•Flow of 1 coulomb of charge per second / 1 ampere =

1 coulomb1 second

[CBSE Marking Scheme, 2019] ½

Detailed Answer :

The SI unit of current is ampere (A). One ampere is defined as the flow of one coulomb of charge per second. 1

2. •Methane ½ •75% [CBSE Marking Scheme, 2019] ½

Detailed Answer :

Methaneisthemainconstituentofbiogas.Itspercentageis75%inthebiogas. ½+½

SECTION – B

3. Name – Sodium ½ Symbol – Na ½ Electronic configuration – 2, 8, 1 1


Detailed Answer : Atomic number : 11 Name : Sodium Symbol : Na Electronic configuration : 1s2 2s2 2p6 3s1 2

OR

(a) Na, Si, Cl – The properties of these three elements are not similar to each other, so no Dobereiner's triads. 1




(b) Be, Mg, Ca – The properties are similar to each other, so it is Dobereiner’s triad. / Atomic mass of Mg

=

Atomic mass of Be+ Atomic mass of Ca2

= 9 402

492

24 5+

= = . [CBSE Marking Scheme, 2019] 1

Detailed Answer :

Dobereiner's triads includes a set of three elements arranged in increasing atomic mass such that the atomic mass of middle element is approximately the average of the atomic masses of other two elements.

(a) Yes, because the atomic mass of silicon is average of atomic masses of sodium (Na) and chlorine (Cl).

Atomic mass of Si = 23 35

2582

29+ = =

(b) Yes, because they have similar properties and the mass of magnesium (Mg) is roughly the average of the atomic mass of Be and Ca.

Mg = 9 40

2492

24 5+ = = . 2

4. O2 is carried by haemoglobin of red blood corpuscles / cells. 1 CO2 is carried by plasma of the blood. [CBSE Marking Scheme, 2019] 1

Detailed Answer :

Transport of oxygen : The oxygen in the alveoli of lungs is taken up by the blood in the alveolar blood vessels to transport to all the cells. The respiratory pigment present in the red blood corpuscles carries oxygen to different tissues of the body.

Transport of carbon dioxide : It is transported in the dissolved form from body tissues to lungs where it diffuses into air in the lungs. 2

5. Structure : Fibrous, jelly like structure 1 Role : To change the curvature of eye lens / to change the focal length of eye lens.


Detailed Answer :

The eye lens is composed of a fibrous, jelly – like material. The key role of ciliary muscles is to modify the curvature of eye lens and thus change its focal length. 2

SECTION – C

6. • Acid — H2CO3 ½ • Base — NaOH ½

• NaOH + H2SO3 → NaHCO3 + H2O 1 [CBSE Marking Scheme, 2019]

Detailed Answer :

NaCl H O CO NHSodiumchloride

BaseAcid

( )( )

+ + +2 2 3 Ammonia → NH Cl NaHCO

Ammoniumchloride

Sodiumhydrogencarbonate

4 3+

The compound is basic (weak). Its pH is more than 7. 2+1

9. Transpiration : Loss of water in vapour form through the surface of leaf / stomata of leaf / aerial parts of the plant. 1

Experiment setup : • Take a potted plant and water it. • Cover the plant / branch with a transparent plastic sheet. • Place it in bright sunlight for half an hour. • Moisture in the form of droplets is observed inside the plastic sheet. [CBSE Marking Scheme, 2019] ½ × 4




Detailed Answer : The loss of water in the form of vapour from the aerial parts of the plant is known as transpiration. Experiment to demonstrate transpiration : Requirements : Two small pots, soil, a green plant, a stick of same height as of green plant and plastic sheets. Method : (i) Take two small pots of approximately equal size having equal amount of soil. (ii) One should have a plant and place a stick in another pot. (iii) Cover the soil in both the pots with a plastic sheet so that moisture cannot escape by evaporation. (iv) Cover both the sets with a separate plastic sheet and place in bright sunlight for half an hour. Observation : Drop of water appear on the inner side of polythene sheet in the pot with a green plant. Whereas

no such drops appear in the pot with a stick. Result : As water drops appear only in the pot with a green plant, it can be concluded that water drops appeared

due to transpiration. While the pot with stick does not have any drop as no plant was present. 1+2

10. Feedback mechanism : Mechanism by which the amount of any chemical increases or decreases resulting in secretion of the related hormone. 1

Example : when sugar level rises, insulin secretion increases. 1 When sugar level falls, insulin secretion reduces. [CBSE Marking Scheme, 2019] 1

Detailed Answer : Feedback mechanism of harmonic regulation is the mechanism to control the timing and amount of hormone

released so that they are secreted in precise quantities. If the sugar levels in blood rise, they are detected by the cells of the pancreas which respond by producing more insulin. As the blood sugar level falls, insulin secretion is reduced. 3

13. • Need for equitable distribution of resources : So that all and not just a handful of rich and powerful people benefit from the development of these

resources / all living beings have a birthright to the available resources. • Forces against equitable distribution of resources : 1 (1) Industrialists who work for their own benefit / profit. (2) When environmental laws / rules are not implemented properly. (3) Mismanagement in the distribution of natural resources or any other relevant answer. (Or any other relevant point) (Any two points) 1 + 1 [CBSE Marking Scheme, 2019]

Detailed Answer : There should be equitable distribution of resources so that natural resources last for future generations. Natural

resources have biological as well as ecological importance in any ecosystem. The three forces that would be working against an equitable distribution of our resources are rich and powerful people, damage to the environment and excessive use of natural resources. 3

SECTION – D

17. (a) (i) Ca + 2H2O → Ca(OH)2 + H2 1 (ii) 2HgS + 3O2 → Heat 2HgO + 2SO2 1 (iii) 3MnO2 + 4Al → 2Al2O3 + 3Mn (b) Alloys are homogeneous mixture of two or more metals or a metal and a non metal 1

Properties : 1 Alloys are stronger / harder / have low melting point / more resistant to corrosion / some are magnetic in

nature. (Any two) ½ + ½ [CBSE Marking Scheme, 2019]

Detailed Answer : (a) (i) Ca s H O l

Calcium Watercold

( ) ( )

( )

+ 2 2 → Ca OH aq H gCalcium hydroxide Hydrogen

( ) ( ) ( )2 2+

(ii) 2 3 2HgS s O gCinnabar

( ) ( )+ → 2 2 2HgO s SO gMercury II oxide

( ) ( )( )

+




2HgO(s) ∆ → 2Hg lMercury

( ) + O2(g)

(iii) 3 42MnO s Al sManganese

dioxide powder

( ) ( )+Aluminium

→ 3 2 2 3Mn l Al O sManganese

oxide

( ) ( )+Aluminium

(b) Alloys are the homogeneous mixture of two or more metals. Example: Duralumin, steel. Two properties of alloys are: (i) They are stronger than metal from which they are made. (ii) They are more resistant to corrosion. 5

18. (i) u = – 30 cm f = – 30 cm v = ? ½

1f

= 1 1v u

−

\ 1v

= 1 1f u

+

1v

=

130

130−( ) +

−( )cm cm

=

−230

cm

v = – 15 cm 1

m =

vu

( )( )--

1530

cmcm =

-

12

½

(ii) Nature — Virtual ½ Position — 15 cm away from the lens, on the same side as the object ½ Size — Diminished ½ Erect / inverted – Erect ½

(iii)

A

30 cm

15 cm 30 cmFB B'

A'

0


Detailed Answer : (i) Given : u = –30 cm, f = 30 cm

Using lens formula, 1f

= 1 1v u

−

⇒ 1v

= 1 1f u

+

= =−( )

+−( )

= − −130

130

1 130

1v

= − = −230

115

So, v = –15 cm

mvu

=

=−−

=+1530

12

cmcm




(ii) Characteristics of the image : (a) Nature : Virtual (b) Position : 15 cm from the lens on the same side as the object. (c) Size of the image : Diminished. (d) Image formed : Erect

(iii)

A

B

B’

A’

O2F1 F

1

15 cm

30 cm 5

Delhi Set III Code No. 31/1/3


SECTION – A

1. Fuel energy / microwave / hot plate / solar energy • Easily available (Or any other source of energy with reason)


Detailed Answer : Natural gas can be used for heating food as it easy to use/ has high calorific value/ produces less amount of smoke

on burning. (Any one) 1

2. To measure potential difference across two points. [CBSE Marking Scheme, 2019] 1

Detailed Answer : The function of voltmeter in an electric circuit is to measure the potential difference. 1

SECTION – B

3. • Image distance remains the same. 1 • It is the distance between the eye lens and retina, which remains the same. 1 [CBSE Marking Scheme, 2019]

4. (i) Pancreas act as a gland by secreting pancreatic juice which contains enzymes. 1 (ii) Secretes hormones like insulin / glucagon 1 [CBSE Marking Scheme, 2019]

SECTION – C

7. Three advantages of exploiting resources with short term aims : (i) Immediate benefit to few people. (ii) Progress in science and technology for development in a country. (iii) Urbanisation and Industrialisation of an area. ½ × 3 Three advantages of using a long time perspective : (i) Resources will be made available for sustainable development. (ii) Provides valuable contribution to the socio-economic development. (iii) Quality of environment will be conserved. ½ × 3 [CBSE Marking Scheme, 2019]

http://bit.ly/2UYX5d7v



Detailed Answer : (i) Advantages of exploiting resources with short-term aims : 1. Beneficial for the present generation. 2. Growth of economy at a faster rate. 3. Comfort will increase. (ii) Advantages of long term perspective in managing our natural resources are : 1. Resources can be used efficiently by present as well as future generations. 2. Sustainable management of resources. 3. Pollution and degradation of environment will be less. 1½ + 1½

9. For nervous and hormonal systems to control and coordinate in human beings, hypothalamus plays an important role in receiving the neural / nerve signals from brain and release hormones. 1

Ex – In situation of iodine deficiency, hypothalamus releases hormones to stimulate pituitary gland, it further sends stimulating hormone to thyroid gland to secrete thyroxine that regulates carbohydrate metabolism. 1 + 1 [CBSE Marking Scheme, 2019]

Detailed Answer : In human beings, nervous and hormonal system work together in controlling and coordinating various life

processes in the body. Nervous system work by generation and transmission of electrical impulse while hormonal system works by secreting chemical messengers called hormones. For e.g. when an emergency stimulus is detected by the nervous system, the stimulus is received and analysed by CNS that send message to effectors to provide proper response. At the same time, nervous system activates adrenal gland to release adrenaline that prepares body by increasing heat rate, blood pressure, respiration etc. Thus, both these systems work together to perform functions of control and coordination. 3

11. A process in which green plants takes carbon dioxide and water and convert them into carbohydrates / food in the presence of sunlight and chlorophyll. 1

Mechanism : (i) Absorption of light energy by chlorophyll.

(ii) Conversion of light energy to chemical energy. (iii) Splitting of water molecules into hydrogen and oxygen. (iv) Reduction of carbon dioxide to carbohydrate. [CBSE Marking Scheme, 2019] ½ × 4

Detailed Answer : Photosynthesis is a process by which green plants prepare their food in the form of carbohydrate with the help

of carbon dioxide, water and sunlight. The mechanism of photosynthesis can be defined by following events : (i) Absorption of light energy by chlorophyll. (ii) Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen. (iii) Reduction of carbon dioxide to carbohydrates. 3

15. (a) In molten state, due to heat the electrostatic forces of attraction between the oppositely charged ions are overcome. So ions move freely and conduct electricity. In aqueous solutions ions are free and conduct electricity. ½ + ½

(b) Due to the formation of a coating of aluminium oxide / Al2O3. (c) Reactive metals like calcium and magnesium react easily with different elements and occur in the form of

ores. [CBSE Marking Scheme, 2019] 1

Detailed Answer : (a) Sodium chloride is an ionic compound formed by ions of sodium (Na+) and chlorine (Cl–). In solid state,

ions are fixed in position so no free electrons are available to conduct electricity. Whereas in molten state and aqueous solution of sodium chloride, free electrons are available to conduct electricity.

(b) On dipping aluminium in nitric acid, a layer of aluminium oxide is formed as nitric acid is a strong oxidizing agent. The layer of aluminium oxide prevents further reaction of aluminium due to which the reactivity of aluminium decreases.

(c) Because these metals are highly reactive and readily react with atmospheric oxygen and other gas. 1+1+1




SECTION – D

21. •

1

Acid Base

1. An acid produces H+ions in aqueous solution.

A base produces OH– ions in aqueous solution.

2. Acids are sour in taste. Bases are bitter in taste.

3. Acids change the colour of blue litmus to red.

Bases change the colour of red litmus to blue.

(Any one) • Neutralization – A reaction of an acid with a base to produce salt and water. 1 (i) Acidic – NH4OH + HCl → NH4Cl + H2O 1 (ii) Basic – NaOH + H2CO3 → Na2CO3 + H2O 1 (iii) Neutral – KOH + HNO3 → KNO3 + H2O (or any other example) 1


Detailed Answer : Acids are the substances that are able to donate their hydrogen ion and accepts the electron whereas bases are the

substances that are able to accept a hydrogen ion and donates electron. Neutralisation is a type of reaction between an acid and a base to give a salt and water with evolution of heat.

Base + Acid → Salt + Water (i) Acidic salts : These are formed by the neutralization of a strong acid with a weak base.

Example : H2SO4 + NaOH → NaHSO4 + H2O (ii) Basic salts : These are formed by the partial neutralization of a strong base by a weak base.

Example : Ca(OH)2 + HCl → Ca(OH)Cl + H2O (iii) Neutral salts : These are formed from complete replacement of hydrogen in acids by other metal cations from

the bases. Example : HCl + NaOH → NaCl + H2O 1+1+1+1+1

Outside Delhi Set I Code No. 31/2/1

SECTION – A

1. Name two industries based on forest produce. 1 2. Why are the heating elements of electric toasters and electric irons made of an alloy rather than a pure metal ? 1

SECTION – B

3. Write the molecular formula of ethene and draw its electron dot structure. 2 4. Give reasons : 2 (a) Platinum, gold and silver are used to make jewellery. (b) Metals like sodium and potassium are stored under oil.

OR Silver articles become black when kept in open for some time, whereas copper vessels lose their shiny brown

surfaces and gain a green coat when kept in open. Name the substances present in air with which these metals react and write the name of the products formed.

5. The absolute refractive index of Ruby is 1.7. Find the speed of light in Ruby. The speed of light in vacuum is 3 × 108m/s. 2

SECTION – C

6. On heating blue coloured powder of copper (II) nitrate in a boiling tube, black copper oxide, O2 and a brown gas X is formed. 3




(a) Identify the type of reaction and the gas X. (b) Write balanced chemical equation of the reaction. (c) Write the pH range of aqueous solution of the gas X. 7. (a) While diluting an acid, why is it recommended that the acid should be added to water and not water to the

acid ? 3 (b) Dry hydrogen chloride gas does not change the colour of dry litmus paper. Why ?

OR How is sodium hydroxide manufactured in industries ? Name the process. In this process a gas X is formed as by-

product. This gas reacts with lime water to give a compound Y, which is used as a bleaching agent in the chemical industry. Identify X and Y and write the chemical equation of the reactions involved.

8. What are amphoteric oxides? Give an example. Write balanced chemical equations to justify your answer. 3 9. What is a homologous series of carbon compounds? Give an example and list its three characteristics. 3 10. List in tabular form three distinguishing features between autotrophic nutrition and heterotrophic nutrition. 3 11. What is transpiration ? List its two functions. 3

OR (a) What is translocation ? Why is it essential for plants ? (b) Where do the substances in plants reach as a result of translocation ? 12. What is carpel ? Write the function of its various parts. 3 13. A student holding a mirror in his hand, directed the reflecting surface of the mirror towards the Sun. He then

directed the reflected light on to a sheet of paper held close to the mirror. 3 (a) What should he do to burn the paper ? (b) Which type of mirror does he have ? (c) Will he be able to determine the approximate value of focal length of this mirror form this activity ? Give

reason and draw ray diagram to justify your answer in this case.OR

A 10 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 12 cm. The distance of the object from the lens is 18 cm. Find the nature, position and size of the image formed.

14. What are solar cells ? Explain the structure of solar panel. List two principal advantages associates with solar cells. 3

15. Write the essential function performed by ozone at the higher levels of the Earth’s atmosphere ? How is it produced ? Name the synthetic chemicals mainly responsible for the drop of amount of ozone in the atmosphere. How can the use of these chemicals be reduced ? 3

SECTION – D

16. (a) List any three observations which posed a challenge to Mendeleev’s Periodic Law. 5 (b) How does the metallic character of elements vary on moving from (i) left to right in a period, (ii) from top to bottom in a period of the Modern Periodic Table ? Given reason for your answer.

OR The electrons in the atoms of four elements A, B, C and D are distributed in three shells having 1, 3, 5 and 7

electrons respectively in their outermost shells. Write the group numbers in which these elements are placed in the Modern Periodic Table. Write the electronic configuration of the atoms of B and D and the molecular formula of the compound formed when B and D combine.

17. (a) Why is the use of iodised salt advisable ? Name the disease caused due to deficiency of iodine in our diet and state its one symptom. 5

(b) How do nerve impulses travel in the body ? Explain.OR

What is hydrotropism ? Design an experiment to demonstrate this phenomenon. 18. (a) What are homologous structures ? Give an example. (b) “The sex of a newborn child is a matter of chance and none of the parents may be considered responsible for

it.” Justify this statement with the help of a flow chart showing sex-determination in human beings. 5 19. When do we consider a person to be myopic or hypermetropic ? List two causes of hypermetropia. Explain using

ray diagrams how the defect associated with hypermetropic eye can be corrected. 5




20. (a) How will you infer with the help of an experiment that the same current flows through every part of a circuit containing three resistors in series connected to a battery ? 5

(b) Consider the given circuit and find the current flowing in the circuit and potential difference across the 15Ω resistor when the circuit is closed.

+ –

5

30V

10 15

OR (a) Three resistors R1, R2 and R3 are connected in parallel and the combination is connected to a battery, ammeter,

voltmeter and key. Draw suitable circuit diagram and obtain an expression for the equivalent resistance of the combination of the resistors.

(b) Calculate the equivalent resistance of the following network:

10

20

20

A B

21. Draw the pattern of magnetic field lines produced around a current carrying straight conductor passing perpendicularly through a horizontal cardboard. State and apply right-hand thumb rule to mark the direction of the field lines. How will the strength of the magnetic field change when the point where magnetic field is to be determined is moved away from the straight conductor ? Give reason to justify your answer. 5

SECTION – E

22. A teacher provided acetic acid, water, lemon juice, aqueous solution of sodium hydrogen carbonate and sodium hydroxide to students in the school laboratory to determine the pH values of these substances using pH papers. One of the students reported the pH values of the given substances as 3, 12, 4, 8 and 14 respectively. Which one of these values is not correct ? Write its correct value stating the reason. 2

OR What would a student report nearly after 30 minutes of placing duly cleaned strips of aluminium, copper, iron

and zinc in freshly prepared iron sulphate taken in four beakers ? 23. What is observed when a pinch of sodium hydrogen carbonate is added to 2mL of acetic acid taken in a test tube?

Write chemical equation for the reaction involved in this case. 2 24. List in proper sequence four steps of obtaining germinating dicot seeds. 2

OR After examining a prepared slide under the high power of a compound microscope, a student concludes that the

given slide shows the various stages of binary fission in a unicellular organism. Write two observations on the basis of which such a conclusion may be drawn.

25. List four precautions which a student should observe while preparing a temporary mount of a leaf peel to show stomata in his school laboratory. 2

26. Draw the path of a ray of light when it enters one of the faces of a glass slab at an angle of nearly 45°. Label on it (i) angle of refraction, (ii) angle of emergence and (iii) lateral displacement. 2

OR A student traces the path of a ray of light through a glass prism as shown in the diagram, but leaves it incomplete

and unlabelled. Redraw and complete the diagram. Also label on it ∠i, ∠e, ∠r and ∠D.




27. The current flowing through a resistor connected in a circuit and the potential difference developed across its ends are as shown in the diagram by milliammeter and voltmeter readings respectively: 2

(a) What are the least counts of these meters ? (b) What is the resistance of the resistor ?

mA

100

200

300

V

1

2

3

Outside Delhi Set II Code No. 31/2/2


SECTION – A

1. Write two advantages associated with water harvesting at the community level. 1 2. Should the resistance of a voltmeter be low or high ? Give reason. 1

SECTION – B

3. Draw electron dot structure of carbon dioxide and write the nature of bonding between carbon and oxygen in its molecule. 2

OR List two properties of carbon which lead to the huge number of carbon compounds we see around us, giving

reason for each. 4. Give reason : (a) Carbonate and sulphide ores are usually converted into oxides during the process of extraction. (b) Aluminium is a highly reactive metal; still it is widely used in making cooking utensils. 2 5. The power of a lens is +5 diopters. What is the nature and focal length of this lens? At what distance from this

lens should an object be placed so as to get its inverted image of the same size ? 2

SECTION – C

6. List two types of the transport system in human beings and write the functions of any one of these. 3 7. Distinguish between pollination and fertilisation. Mention the site and the product of fertilisation in a flower. 3 10. List three environmental consequences of using fossil fuels. Suggest three steps to minimise the pollution caused

by various energy sources. 3 14. Which compounds are called (i) alkanes, (ii) alkenes and (iii) alkynes ? C4H10 belongs to which of these ? Draw

two structural isomers of this compound. 3

SECTION – D

16. (a) What are dominant and recessive traits ? (b) "Is it possible that a trait is inherited but may not be expressed in the next generation ?" Give a suitable

example to justify this statement. 5 20. (a) What is scattering of light ? Explain how the colour of the scattered light depends on the size of the scattering

particles. (b) Explain the reddish appearance of the Sun at sunrise or sunset. Why does it not appear red at noon ? 5




Outside Delhi Set III Code No. 31/2/3


SECTION – A 1. What does the cord of an electric oven not glow while its heating element does ? 1 2. Although coal and petroleum are produced by the degradation of biomass, yet we need to conserve these

resources. Why ? 1

SECTION – B 3. What is atmospheric refraction ? List two phenomena which can be explained on the basis of atmospheric

refraction. 2 4. Name a metal of medium reactivity and write three main steps in the extraction of this metal from its sulphide

ore. 2 5. List two chemical properties on the basis of which ethanol and ethanoic acid may be differentiated and explain

how. 2OR

Unsaturated hydrocarbons contain multiple bonds between two carbon atoms and these compounds show addition reactions. Out of saturated and unsaturated carbon compounds, which compounds are more reactive? Write a test to distinguish ethane from ethene.

SECTION – C 6. What happens to a beam of white light when it gets refracted through a glass prism ? Which colour deviates the

most and the least after refraction through a prism ? What is likely to happen if a second identical prism is placed in an inverted position with respect to the first prism? Justify your answer. 3

OR A student needs spectacles of power – 0.5D for the correction of his vision. (i) Name the defect in vision the student is suffering from. (ii) Find the nature and focal length of the corrective lens. (iii) List two causes of this defect. 7. Define a food chain. Design a terrestrial food chain of four trophic levels. If a pollutant enters at the producer

level, the organisms of which trophic level will have the maximum concentration of the pollutant in their bodies ? What is this phenomenon called ? 3

9. During the reaction of some metals with dilute hydrochloric acid, the following observations were made by a change. 3

(a) Silver does not show any change. (b) Some bubbles of a gas are seen when lead is reacted with the acid. (c) The reaction of sodium is found to be highly explosive. (d) The temperature of the reaction mixture rises when aluminium is added to the acid. Explain these observations giving appropriate reason. 10. Given below are the steps for the extraction of copper from its ore. Write the chemical equation of the reactions

involved in each case. 3 (i) Roasting of copper (I) sulphide (ii) Reduction of copper (I) oxide from copper (I) sulphide (iii) Electrolytic refining 15. (a) Budding, fragmentation and regeneration, all are considered as asexual mode of reproduction. Why ? 3 (b) With the help of neat diagrams, explain the process of regeneration in Planaria.

SECTION – D

16. A 6 cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 30 cm. The distance of the object from the mirror is 45 cm. Use mirror formula to determine the position, nature and size of the image formed. Also, draw labelled ray diagram to show the image formation in this case. 5

OR An object 6 cm in size is placed at 50 cm in front of a convex lens of focal length 30 cm. At what distance from

the lens should a screen be placed in order to obtain a sharp image of the object ? Find the nature and size of the image. Also, draw labelled ray diagram to show the image formation in this case.

20. What is sexual reproduction ? Explain how this mode of reproduction give rise to more viable variations than asexual reproduction ? How does this affect the evolution ? 5




SOLUTIONSOutside Delhi Set I Code No. 31/2/1

SECTION – A

1. Timber / Bidi / Paper / Medicine. (Any two) ½ + ½ [CBSE Marking Scheme, 2019]

Detailed Answer : Paper industry and sports equipment industry. 1

2. Due to high resistivity of alloys rather than its constituting metals. [CBSE Marking Scheme, 2019] 1

Detailed Answer : Because the resistivity of an alloy is higher than a pure metal. Also, alloys do not melt easily at high temperatures.

1

SECTION – B

3. Molecular formula - C2H4. 1

H

HH

H

C C××

××

×

××

×


Detailed Answer : Molecular formula of ethene : C2H4 Electron dot structure :

C× ×

× ×C

H

×

H

×

H

× ×

H1+1

4. (a) Lustre, ductile, malleable, least reactive. (Any two) ½ + ½ (b) Na & K are highly reactive (in air & moisture). [CBSE Marking Scheme, 2019] 1

Detailed Answer : (a) Because of their bright shiny surface and high resistance to corrosion. Also, they have high malleability and

ductility. (b) Because they are very reactive and catches fire when exposed to air. Also, they react with water to form base

therefore, they are kept in oil. 1+1OR

Products Silver Sulphur in air Silver sulphide ½ + ½ Copper Moisture & Carbon dioxide Copper Carbonate [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : Silver articles when exposed to air turn black as it reacts with sulphur in the air to form a coating of silver sulphide. 2Ag + H2S → Ag2

Silver sulphideS H+ 2

Copper vessels lose their shiny brown surface as copper reacts with moist carbon dioxide in the air and forms a green layer of copper carbonate and copper hydroxide. 1+1

2Cu + H2O + CO2 + O2 → Cu OH CuCOCopper

hydroxideCopper

carbonate

( )2 3+




5. m = Speed of light in vacuumSpeed of light in Ruby

=cv

½

v = cµ ½

Where, c = velocity of light m = refractive index

v = 3 10

1 7

8×.

= 1.76 × 108 m/s [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer :

Speed of light in vacuum = 3 × 108 m/s

Refractive index of ruby = Speedof light in vacuum

Speedof light in ruby

1.7 = 3 108×

Speedof light in ruby

Speed of light in ruby = 3 10

1 71 76 10

88×

×.

.= m/s 2

SECTION – C

6. (a) Decomposition / Thermal decomposition, ½ The gas X is NO2 or (nitrogen dioxide) ½ (b) 2Cu (NO3)2

Heat → 2CuO + 4NO2 + O2 1

(c) Range less than 7 (or 0------6.9pH) Note : For (b) ½ mark for equation and ½ mark for balancing the equation 1 [CBSE Marking Scheme, 2019]

Detailed Answer : (a) Type of reaction : Thermal decomposition reaction. The gas X is nitrogen dioxide.

(b) 2 3 2Cu NO sCopper nitrate II

( ) ( )( )

→ 2 4 2 2CuO s NO g O gBlack Brown

( ) ( ) ( )+ +

(c) As oxides of non-metals are acidic in nature, therefore, aqueous solution of nitrogen dioxide will have pH range between 0 and 7. 1+1+1

7. (a) The process of diluting an acid is highly exothermic, and on the addition of acid to the water the excess heat is absorbed by water. 1 + 1

(b) Because HCl does not form H+/H3O+ ions in dry condition. 1 [CBSE Marking Scheme, 2019]

Detailed Answer : (a) Diluting an acid is an exothermic process. When water is added to an acid, the heat generated may cause the

mixture to splash out and may cause serious burns. Therefore, it is recommended that acid should be added to water carefully with constant stirring.

(b) Presence of ions like hydrogen (H+) ions or hydronium (H3O+) ions changes the colour of litmus paper. HCl is able to produce these ions only in the form of aqueous solution. Hence, dry hydrogen chloride gas does not change the colour of dry litmus paper. 1½+1½

OR

• When electricity is passed through an aqueous solution of sodium chloride (brine). ½ • Chlor – alkali process ½ • X - Cl2 ½ • Y = CaOCl2 ½ • 2NaCl(aq) + 2H2O(l) → 2NaOH(aq) + Cl2(g) + H2(g) ½ • Ca(OH)2 + Cl2 → CaOCl2 + H2O [CBSE Marking Scheme, 2019] ½




Detailed Answer : When electricity is passed through an aqueous solution of sodium chloride (brine), it decomposes to form sodium

hydroxide. 2 2 2NaCl aq H O

Sodium chloride( ) ( )+ l → 2 2 2NaOH aq Cl g H g

Sodium hydroxide X( ) ( ) ( )

( )+ +

This process is called as chor-alkali process. In the manufacture of sodium hydroxide, hydrogen gas and chlorine gas (X) are formed as byproducts. When

chlorine gas (X) reacts with lime water, it forms calcium oxychloride (bleaching powder) Y. The reaction is: Ca OH aq Cl g

Lime waterSlaked

( ) ( ) ( )

(

2 2

lime)

+ → CaOCl aq H OCalcium hypochlorite

Y

2 2( ) ( )

( )

+ l

Y is Calcium oxychloride, which is used as bleaching agent in the chemical industry. 1+1+1

8. • Metal oxides showing both acidic and basic nature. ½ • Example : Al2O3 / ZnO (or any other) ½ • Al2O3 + 6HCl → 2AlCl3 + 3H2O 1 Al2O3 + 2NaOH → 2NaAlO2 + H2O [CBSE Marking Scheme, 2019] 1

Detailed Answer : Oxides of metals which have both acidic as well as basic behaviour are known as amphoteric oxides. Examples are

aluminium oxide and zinc oxide. Amphoteric oxides react with acids as well as base to form salt and water. ZnO reacts with hydrochloric acid (acid) to form zinc chloride (salt) and water thus acting as basic oxide.

ZnO S HCl aqZincoxide

Hydrochloric acidAcid

( ) ( )

( )

+ 2 → ZnCl aq H OZinc chloride

SaltWater

2 2( ) ( )

( )

+ l

ZnO reacts with sodium hydroxide (base) to form sodium zincate (salt) and water thus, acting as acidic oxide.

ZnO S NaOH aqZincoxide

Sodium hydroxideBase

( ) ( )

( )

+ 2 → Na ZnO aq H OSodium zincate

SaltWater

2 2 2( ) ( )

( )

+ l

1 + 1 + 1

9. • A series of compounds in which the same functional group substitutes for hydrogen in a carbon chain is called a homologous series. 1

• Example : Alkane / Alkene / Alkyne / Alcohol or any other one correct example. ½ • Characteristics : (i) They have same general formula. (ii) They have same functional group. (iii) The difference in the molecular mass of two successive member in 14μ. (iv) The difference in the molecular formula of two successive member is of CH2 unit. (v) They have similar chemical properties. (Any three points) ½ × 3 [CBSE Marking Scheme, 2019]

Detailed Answer : A homologous series is the family of organic compound having the same functional group, and the successive

(adjacent) members of which differ by CH2 unit or 14 mass unit. Example: A homologous series of alkene includes ethene (C2H4), propene (C3H6), butene (C4H8) and pentene (C5H10).

Characteristics of homologous series are : 1. Succeeding members differ by a –CH2 unit. 2. They show similar chemical properties. 3. They show a steady gradation in physical properties from one member to the next. 1½+1½ 10.

Autotrophic Nutrition Heterotrophic Nutrition

1. They can prepare their own food. They cannot prepare their own food.

2. They require raw materials like CO2, H2O in the presence of sunlight and chlorophyll to prepare their food.

They depend on other plants and animals for their food.

3. They store the food in the form of starch. They store the food in the form of glycogen.

(Any three point) [CBSE Marking Scheme, 2019] 1 × 3




Detailed Answer :

Autotrophic Nutrition Heterotrophic Nutrition

1. It is a mode of nutrition in which organism prepare their own food.

It is a mode of nutrition in which organism cannot prepare their food and obtain it from different sources.

2. The organisms falling in this category are not dependent on any other organism.

The organisms falling in this category are dependent on other organisms for their food.

3. Organisms using this mode are called as producers. Organisms using this mode are called as consumers.

1+1+1

11. The loss of water in the form of vapour from the aerial parts/leaves/stems is known as transpiration. 1 • Functions : (i) It helps in the absorption and upward movement of water. (ii) Movement of dissolved minerals from root to leaves. (iii) It helps in the temperature regulation or cooling of the plant. (Any two points) 1 + 1 [CBSE Marking Scheme, 2019] 2

Detailed Answer : The loss of water in the form of vapour from the aerial parts of the plant is known as transpiration. Functions of transpiration are : 1. Absorption and upward movement of water and minerals dissolved in it from roots to the leaves. 2. It helps in temperature regulation. 1+2

OR

(a) The transport of soluble products of photosynthesis (food or glucose) from one part to the other parts of the plant.

To provide food to all parts of the plant. (b) Root, fruits, seeds and other growing organs/parts of the plant. (any two)


Detailed Answer : (a) Translocation is the process of movement of materials from leaves to all other parts of the plant body. It is essential for the transfer the products of metabolic processes, particularly photosynthesis from leaves to

other parts of the plant. (b) As a result of translocation, the substances in plants reach to the storage organs of roots, fruits and seeds and

to growing organs. 2 + 1

12. Female reproductive part of the plant. 1 Stigma – receive pollen grains Style – passage for the growth of pollen tube Ovary – Site for fertilization 1 + 1 If any two parts with function attempted award 1½ marks only [CBSE Marking Scheme, 2019]

Detailed Answer : Carpel is the female reproductive that produce egg cells. Main parts of carpel are : 1. Stigma being sticky in nature receives pollen grains during pollination. 2. Style connects the stigma and ovary thus, helping with the transfer of pollen through style to the ovary. 3. Ovary is the reproductive organ of carpel which produces the female gamete ovule. 1+1+1

13. (a) Move the mirror/paper to focus the rays at one point. ½ (b) Concave mirror. ½ (c) Yes, distance between mirror and focal point gives approximate focal length. ½ + ½




IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII

C

D

P

F

M

N

i

r

½ + ½[CBSE Marking Scheme, 2019]

Detailed Answer : (a) He should move the sheet of paper back and forth gradually until he finds a bright and sharp spot of lights

i.e. at focus. He should hold the mirror and paper in the same position for a few minutes to burn the paper. (b) He has concave mirror. (c) Yes, he will be able to determine the approximate value of focal length of this mirror from this activity because

the spot of light is the image of the sun which is at the focus of the concave mirror. Therefore, the distance of this image from the position of this mirror gives the approximate value of focal length of this mirror.

M

ir

D

P

F

N

f

C

Light rays

from sun

3OR

1 1 1 1 1 1v u f v f u

− = ∴ = +

½

1 112

118v

= +−( )

1

∴ v = 36 cm ½

m =

vu

hh

='

⇒ ∴ m = 3618 10−

=h '

½

⇒ h' =−20cm(size of the image) ½ Nature of image – Real and inverted [CBSE Marking Scheme, 2019]

Detailed Answer : Given, object size, h = +10 cm; focal length, f = +12 cm (f is positive for a convex lens); object distance, u = –18cm Using lens formula,

1 1v u

− = 1f

1v

= 1 1u f

+

= 118

112−

+ = − +2 336




1v

= 1

36

Therefore, v = 36 cm → Image distance

Magnification, m = hh

vu

' =

Therefore, m = vu

=+−

= −3618

2cmcm

So, image size h' = v hu×

= 36 10

1820

×−( )

= − cm

Positive sign of v signifies that image is 36 cm on right side of lens. So, the image is real. Negative sign of h′ signifies that the image is inverted. The image is two times enlarged in size than the object. 3

14. A device that converts solar energy directly into electrical energy. 1 A large no. of solar cells are combined in an arrangement called Solar Cell Panel. 1 Principal Advantages : (i) They have no moving parts. (ii) Require little maintenance & work quite satisfactorily without the use of any focusing device. (iii) These cells can be set up in remote & inaccessible areas where laying of a power transmission may be

expensive. (Any two) ½ + ½ [CBSE Marking Scheme, 2019]

Detailed Answer : Solar cells are the devices that convert solar energy of sun into electricity. Solar panel consists of a large number of solar cells connected by silver wires. They are made up of special grade

silicon. These are mounted on specially designed inclined roof tops so that more solar energy is incident over it. Advantages of solar cells are : (a) They have no moving parts, require little maintenance and work quite satisfactorily without the use of any

focussing device. (b) They can be set up in remote and inaccessible hamlets or very sparsely inhabited areas. 1 + 2

15. It shields the surface of the earth from the UV radiation from the sun. 1

O2 UV → O + O

O2 + O → O3 or description of this process in words 1 Chloro Fluoro Carbons (CFC’s) ½ Reduce the use of CFC’s by (a) minimizing the leakage through air conditioners and refrigerators / finding

substitute chemicals that are ozone friendly. [CBSE Marking Scheme, 2019] ½

Detailed Answer : Ozone protects the surface of earth from harmful ultraviolet (UV) radiations emitted by the sun. Ozone is a product of UV radiation acting on oxygen (O2) molecule. The higher energy UV radiations split apart

some molecular oxygen (O2) into free oxygen (O) atoms. These atoms combine with molecular oxygen to form ozone.

O O OUV2 → +

O + O2 → OOzone

3( )

Chlorofluorocarbons (CFCs) are mainly responsible for the drop of amount of ozone in the atmosphere. Use of these chemicals can be reduced by using alternate products that do not harm the ozone layer. Also, safe

disposal of old appliances like refrigerators prevents its emission. 3




SECTION – D

16. (a) (i) No fixed position of H in the periodic table. 1 (ii) Position of isotopes not clear. 1 (iii) Atomic mass does not increase in a regular manner. (or any other) 1 (b) (i) Left to right metallic character decreases ½ Reason : Effective nuclear charge increases / tendency to loose electrons decrease / electropositivity

decreases (any one reason) ½ (ii) Top to bottom metallic character increases ½ Reason : Size of atom increase/tendency to loose electron increases (any one reason) ½ [CBSE Marking Scheme, 2019]

Detailed Answer : (a) Three observations which posed a challenge to Mendeleev’s Periodic Law are: (i) Position of hydrogen : Hydrogen resembles both alkali metals and halogens and thus, no fixed position

could be assigned to it. (ii) Position of isotopes : There was no place for isotopes. (iii) Placing of dissimilar elements in same group : Few chemically similar elements fall in different groups

whereas some chemically dissimilar elements are placed together. (b) (i) As we move from left to right in a period, the metallic character decreases as the size of an atom also

decreases and thus, the valence electrons are held tightly by the nucleus. So, it becomes hard to loose electrons resulting in decrease in metallic character.

(ii) As we move down in a group, the metallic character increases as the size of an atom increases and thus, the force of attraction between valence electrons and nucleus is very less. So, the element can easily loose electrons resulting in increase in metallic character. 3+2

OR

A B C D 1 3 5 7 • Group no. 1st 13th 15th 17th ½ × 4 • B = 2, 8, 3 D = 2 , 8 , 7 1 + 1 • BD3 [CBSE Marking Scheme, 2019] 1

Detailed Answer :

Elements Valence electrons Period Group

A 1 3 1

B 3 3 3

C 5 3 5

D 7 3 7

Electronic configuration of B: 1s2 2s2 2p6 3s2 3p1

Electronic configuration of D: 1s2 2s2 2p6 3s2 3p5

Element: B D

13 Valency: So, the molecular formula will be BD3. 5

17. (a) • Iodine is essential for functioning of thyroid / formation of thyroxine hormone 1 • Disease is Goitre 1 • Swollen neck 1 (b) Impulse travels from dendrite to cell body, then along the axon to its end. At the end some chemicals are

released which fill the gap of synapse, and starts a similar electrical impulse to another neuron and the impulse further travel in the body. (Award marks if attempted as a flow chart also) 2





Detailed Answer : (a) Iodine is necessary for thyroid gland to secrete thyroxin hormone which regulates carbohydrate, protein and

fat metabolism in the body. Its deficiency may cause goitre. Therefore, use of iodised salt is advisable. Its key symptom is a swollen neck.

(b) The information received at the end of the dendritic tip, sets off a chemical reaction that creates an electrical impulse. This impulse travels from the dendrite to the cell body and then along axon to its end. Here, the electrical impulse sets off the release of some chemicals which cross synapse inducing similar impulse in a dendrite of the next neuron. 3+2

OR

The movement/response of part of plant (root) towards water 1 Experiment : (i) Soak the seeds in water overnight. ½ (ii) Place moist cotton in a perforated petridish. ½ (iii) Put the soaked seeds in the petridish & place it on a beaker. 1 (iv) Roots pass through pores and grow downwards. 1

(v) After sometime roots will bend towards base of petridish having moisture. 1


Detailed Answer : Hydrotropism is the movement of a plant towards or away from water. Example: Plant roots bend towards the

source of water. Experiment to demonstrate hydrotropism : Requirements : Bean seeds, a deep tray, sand water and a porous flower pot. Procedure : (1) Fill the tray (big enough to accommodate porous pot) with sand and insert some seeds in it. (2) Make a pit in the sand and place the porous pot in it. (3) Fill water in the porous pot. (4) Leave the arrangement for about a week. Observation : After about a week, when seeds are taken out, it s observed that roots grow in the direction of

the porous pot. This shows hydrotropism in roots. 1+4

18. (a) The organs having similar origin / structures but performing different functions. 1

Example : limbs of frog, limbs of lizard, bird, human (Any two) ½ + ½

(b)

Male

X Y X X

XYXX

Girl Boy

,

x

XY XX

FemaleParents

Gametes

Zygote

Sex

Hence, sex determination is purely a matter of chance. [CBSE Marking Scheme, 2019]

Detailed Answer : (a) Homologous structures are structures which have a common basic structure but perform different functions.

Example, forelimbs of reptiles, amphibians and mammals. (b) In human beings, the sex of the individual is determined by the genes inherited from the parents. Women

have a perfect pair of sex chromosomes i.e, XX whereas men have a mismatched pair of normal sized X and short one Y making it XY. All children will inherit an X chromosome from their mother regardless of whether they are boys or girls. Thus, the sex of children is decided by what they inherit from their father. A child who inherits an X chromosome will be a girl and one who inherits a Y chromosome from him will be a boy. So, there is an equal chance of a girl child as well as a boy child.

2 + 3

1

11




Father

XY

XX

Daughter

XX

Daughter

XY

Son

XY

Son

Mother

XX

2 + 3

19. Myopia : Difficult to see the objects placed far away / Hypermetropia : Difficult to see very close or nearby objects. 1

Causes of hypermetropia : (i) The focal length of the eye lens is too long (ii) eye ball has become too small ½ + ½

N'N

Note : Diagram with brief description -03; only correct diagram with labelling -2 or only explanation 01. 3 [CBSE Marking Scheme, 2019]

Detailed Answer : When a person is not able to see distant objects clearly but can see nearby objects clearly then he is considered to

be myopic. If the case is otherwise, then he is hypermetropic. Two causes of hypermetropia are : (i) Focal length of the eye lens becomes too large. (ii) The eyeball becomes too small along its axis. Correction of hypermetropic eye :

(b) Near point ofhypermetropic eye

I

I

I

Near pointof defective eye

Normalnear point

N’

N

N

N

Use of convex lens(c) Correction of

hypermetropic eye

(a) Hypermetropic

25 cm

more than 25 cm

2 + 3

20. (a) (i) Join the three resistors of different values in series. (ii) Connect them with battery, an ammeter and plug key. (iii) Plug the key and note the ammeter reading. (iv) Change the position of ammeter to anywhere in between the resistors and note the ammeter reading

each time. (v) The ammeter reading will remain same everytime. Therefore when resistors are connected in series

same current flows through all resistors, when it is connected to a battery. Note : If explained with the help of diagram give full credit. ½ × 5




(b) Total resistance of the circuit = 1 R = R1 + R2 + R3 = 5 + 10 + 15 = 30 ohm Potential difference across the circuit / By ohm’s law

V = IR or I = VR

VA= =

3030

1ohm

1

Potential difference across 15 ohm Resistor = 1A × 15 ohm = 15 volt [CBSE Marking Scheme, 2019] ½

Detailed Answer : (a) Suppose the experimental set up comprises of three resistors R1, R2 and R3 of three different values which are

connected in series with an ammeter, key and a battery of known voltage is given as below :

+

+

–

–

A

K

I

R1

R2

R3

The key K is closed and the ammeter reading is recorded. Now, the position of ammeter is changed to anywhere in between the resistors. The ammeter reading is recorded each time. It is found that there is an identical reading each time, which shows that same current flows through every part of the circuit containing three resistances in series connected to a battery.

(b) Req = R1 + R2 + R3 = 5Ω + 10Ω + 15Ω = 30Ω

Electric current (I) = VReq

= 3030

1= A

Potential difference across 15Ω resistor = IR = 1 × 15 = 15 V 3+2

OR

(a) (i) Total current I = I1 + I2 + I3

Let Rp be the equivalent resistance of R1, R2, R3. Then the total current I =

VRp

1

(ii) On applying ohm’s law for each R1, R2, R3 Rp = 10 ohms

I1 =

VR1

, I2 =VR2

, I3 =VR3

½

∴ I = VR R R

VR

1 1 1

1 2 3+ +

=p

½

∴ 1

Rp =

1 1 1

1 2 3R R R+ +

½

(b)

( ( A– +

A– +

R1

R2

R3

M

O

T

L

P

S

–+

½




1Rp

= 120

120

220

110

+ = = 1

Rp = 10 ohms Equivalent resistance of the network = Req = R1 + Rp = 10 + 10 = 20 ohm. 1 [CBSE Marking Scheme, 2019]

Detailed Answer :R1

R2

R3

I1

I2

I3

II< <

V

V

R1

R2

R3

A

(•)

(a) Let there be n resistances, each of value R1, R2, …., Rn respectively connected in parallel to a battery of voltage

V. Let current I is sent to the circuit. If the equivalent resistance is Req, then current drawn IVReq

= According to above circuit, I = I1 + I2 + I3

VReq

= VR

VR

VR

1

1

2

2

3

3+ +

So, 1Req

= 1 1 1

1 2 3R R R+ +

(b) Two 20 ohm resistors in parallel are connected to one 10 ohm resistor in series. For resistor in parallel,

1Rp

= 1 1

2 3R R+ = +1

20120

1Rp

= 220

110

= Ω

Rp = 10Ω For resistor in series, RTotal = R1 + Rp

= 10Ω + 10Ω = 20Ω 3+2

21.

+

–

A +–

P

K

Variableresistance

Magnetic compass

X

Y

Diagram 1½ and direction ½




Statement of right hand thumb rule. 1 + 1 The magnetic field strength decreases with increase of distance from the current carrying conductor. 1 Reason : There is inverse relation between field strength and distance from current carrying conductor. 1 Note : Direction of magnetic field should be in accordance with direction of current. 1


Detailed Answer : Pattern of magnetic field lines produced around a current carrying straight conductor :

A

I

+–

K

V

V

V

V

V V

V

Right-hand thumb rule : If we are holding a current carrying straight conductor in right hand such that the thumb points towards the direction of current, then, the fingers will wrap around the conductor in the direction of the field lines of the magnetic field.

Cu

ren

t

Magnetic Field

As the compass is placed farther, deflection in the needle decreases. Thus, the magnetic field produced by given current decreases as the distance from it increases. The concentric circles around the wire become larger as we move away from it. 2+2+1

SECTION – E

22. • The pH value of water given is incorrect. • Its correct value is 7 it is neutral in nature. 1 + 1 [CBSE Marking Scheme, 2019]

Detailed Answer :

The pH of water is incorrect. It should be 7 as it is neutral in nature. 2

OR

• There will be no reaction in the beakers having Fe strip & Cu strip. • The solution having Al & Zn strip will show reaction / the solution of FeSO4 having Al & Zn strip will

become colourless. [CBSE Marking Scheme, 2019] 1 + 1

Detailed Answer :

In the test tube containing Al, the black residue is formed. This is a displacement reaction.

2 3 4Al s FeSO aqLight green

( ) ( )+ → Al SO Fe sColourless Black

2 4 3 3( ) ( )+

In the text tube containing Zn, grayish black precipitate is deposited. This is a displacement reaction.

Zn s FeSO aqLight green

( ) ( )+ 4 → ZnSO aq Fe sColourless Black

4( ) ( )+

In the test tube containing Fe, no reaction takes place. In the test tube containing Cu, no reaction takes place as Cu is less reactive than iron. 2




23. • Brisk effervescense of CO2 evolved. 1 • CH3COOH + NaHCO3 → CH3COONa + CO2 + H2O [CBSE Marking Scheme, 2019] 1

Detailed Answer :

A colourless and odourless gas evolves with a brisk effervescence.

CH COOH aq NaHCOAcetic acid Sodium hydrogen

carbonate

3 3( ) + → CH COONa aq CO g H OSodium acetate Carbon

dioxide

3 2 2( ) ( ) ( )+ + 2

24. (i) Soaking of seeds ½ (ii) Emergence of radicle ½ (iii) Splitting of cotyledons ½ (iv) Emergence of plumule [CBSE Marking Scheme, 2019] ½

Detailed Answer : Seed germination in dicot seeds : Step 1 : Imbibition : The seed imbibes water through micropyle. Step 2 : Rupturing of seed coat. Step 3 : Emergence of radicle and plumule. Step 4 : Development of radicle into roots and plumule into shoot. 2

OR

(i) Elongation of nucleus 1 (ii) Constriction appears due to the division of the cytoplasm [CBSE Marking Scheme, 2019] 1

Detailed Answer : Observations are : (i) An organism having elongated and constricted nucleus. (ii) At the point of binary fission, constriction appears and deepens to divide the cell into two daughter cells. 2

25. (i) Size of the leaf peel should be very small. (ii) Put peel immediately in the drop of water. (iii) Place cover slip carefully to avoid the air bubbles. (iv) It should not be overstained. (v) No fold in the peel. (Any four) ½ × 4


Detailed Answer : Precautions are : (i) Peel should be taken from freshly plucked leaf. (ii) Leaf peel should not be over stained. (iii) Slide should be clean and dry. (iv) Place the coverslip gently, avoiding any air bubbles. 2

26.

i

N

M

L

e

M'

r1

½ Labelling • Angle of refraction (r1) • Angle of emergence (e) • Lateral displacement (ML) ½ × 3 [CBSE Marking Scheme, 2019]




Detailed Answer :

lateral displacement

2

OR

A

G

D

MH

S

R

C

FM'N'

N

Q

B C

P

i r

Labelling of Ði + Ðe + Ðr & ÐD [CBSE Marking Scheme, 2019] ½ × 4

Detailed Answer :

N’ M’

CB

Incident beam Emergent beam

Angle of emergence

Angle of refraction

er

i

MN

AAngle of prism, = 60A

o

Angle of deviationD

i= Angle of incidence

D

E

G

r

2

27. (a) least count of ammeter = 10 mA ½ + ½ least count of Voltmeter = 0.1 V

(b) 2 40 25

..

= 9.6 ohm (250 mA = 0.25 A) [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer :

(a) Least count of milliammeter = 10010

10= mA

Least count of voltmeter =

110

= 0.1 V

(b) Current, I = 250 mA = 250 × 10–3A

Potential difference, V = 2.4 V

Resistance, R = VI

= −

2 4250 10 3

.×

= 9.6Ω 2




Outside Delhi Set II Code No. 31/2/2

SECTION – A

1. (i) Recharged the ground water level ½ (ii) Brought rivers back to life [CBSE Marking Scheme, 2019] ½

2. High, In parallel connection, less current passes through high resistance. [CBSE Marking Scheme, 2019] ½ + ½

SECTION – B

3. Double covalent bond / covalent bond 1

CO O

1 [CBSE Marking Scheme, 2019]

OR

(i) Catenation / ability to form long chains of ½ Reason : very strong carbon – carbon bond ½ (ii) Tetravalency / valency of four ½ Reason : All the four valencies of carbon atom are occupied with other elements. ½


4. (a) It is easier to obtain a metal from its oxide as compared to sulphide and carbonate ore. 1 (b) Aluminium forms a thicker protective oxide layer / anodizing. [CBSE Marking Scheme, 2019] 1

5. P = +5D

f = 1 100

5P= = 20 cm 1

Nature of lens = convex (converging) ½ Distance is 40 cm (at C) [CBSE Marking Scheme, 2019] ½

6. (i) Blood circulatory system ½ (ii) Lymphatic system / lymph or tissue fluid ½ Functions of blood circulatory system : (i) Transport of oxygen (ii) Transport of digested food (iii) Transport of carbon dioxide (iv) Transport of nitrogeneous waste (v) Transport of salts Functions of lymphatic system : (i) Carries digested and absorbed fat (ii) Drains extra fluid from tissue 1 × 2 (extra cellular space) back into the blood Note : Two functions of any one of the transport system to be given. [CBSE Marking Scheme, 2019]

7. Pollination : Transfer of pollen grains from stamen/ anther to stigma. 1 Fertilization : Fusion of male & female gamete (or germ cells) 1 Site of fertilisation : Ovary/ Ovule ½ Product; Zygote. [CBSE Marking Scheme, 2019] ½




10. Consequences : (i) Cause air pollution (ii) The acidic oxides lead to acid rain (iii) High concentration of green house gas (CO2)and its effect (iv) Global Warming (any 3 points) ½ × 3 Steps to minimize the pollution : (i) Use of alternate source of energy (ii) Use of various devices to reduce emission of harmful gases. (iii) By increasing efficiency of combustion process (or any other) ½ × 3


SECTION – C

14. l Alkane Saturated Hydrocarbon with C–C Single Bond lAlkene Unsaturated Hydrocarbon with double bond in C=C lAlkyne Unsaturated Hydrocarbon with triple bond in C ≡ C (or any other) ½ × 3 lAlkane ½ 2 structural isomers

H—C—C—C—C—H

—H —H —H —H

— — — —

H H H H

H—C—C —

—H

—

H

C

C—

H

—H

——H

H

——

—

H

H

—

H

[CBSE Marking Scheme, 2019] ½ + ½

SECTION – D

16. (a) The trait which expresses itself in F1 (first) generation after crossing contrasting (opposite) traits is known as dominant character (trait). 1

Recessive Trait : The trait which is not expressed itself in F1 (first) generation after crossing contrasting (opposite) trait. 1

(b) Yes 1

Tall Dwarf

TT tt×

All tall (Tt)F1

TT × tt

F2 TT ttTt Tt

Tall DwarfTall Tall 2(Or can be explained in words also)


20. (a) Scattering : Direction of ray of light changes when it collides with particles of comparable size. 1 – Fine particle scatter shorter wavelengths like blue light. – Particles of larger size scatter longer wavelengths like red light. – If particle size is large enough scattered light may appear white. ½ × 3




(b) • During sunrise and sunset, sun is near horizon. Sunlight travels longer distance. Most of blue light scatters away by particles. Light of longer wavelength reaches our eye so sun appears reddish.

• Atnoon,sunisdirectlyoverheadsosunlighttravelslesserdistance.Lessamountofbluelightisscatteredgiving white appearance to sun. ½ × 3

Or can be explained with the help of a diagram given below :

Blue scattered awaySun appears reddish

Sun nearhorizon

Less bluescattered

Sun nearlyoverhead

(One Mark with correct labelling) 1 [CBSE Marking Scheme, 2019] 5

Outside Delhi Set III Code No. 31/2/3

SECTION – A

1. Cord is made up of copper wire whereas heating element is made up of alloy. [CBSE Marking Scheme, 2019] 1

Detailed Answer : The cord of an electric oven is usually made of copper or aluminium whose resistance is very low so it does not

glow. Whereas, its heating element is made up of alloy which has very high resistance. So, when current is passed through the heating element it becomes very hot and glows red. 1

2. They are non renewable / their formation takes millions of years / exhausted in the future. (any one point) [CBSE Marking Scheme, 2019] 1

Detailed Answer :

Coal and petroleum take millions of years to form by the degradation of biomass. As these resources are being utilised at a much faster rate than their formation, they will be exhausted in the future very soon. Therefore, they need to be conserved. 1

SECTION – B

3. Refraction of light due to variation in optical density of earth. 1 Two phenomenon : (i) Twinkling of stars (ii) Advance sunrise & delayed sunset. ½ × 2 [CBSE Marking Scheme, 2019]

Detailed Answer :

The refraction of light while passing through earth's atmosphere is called atmospheric refraction.

Twinkling of stars and advance sunrise and delayed sunset are two phenomena which can be explained on the basis of atmospheric refraction. 1+1

4. (i) Iron / Zinc / Lead (any one) ½ (ii) Concentration of ore Enrichment/ of ore ½ ↓ (iii) Roasting / conversion of sulphide into oxide on heating in air ½ ↓ (iv) Reduction of metallic oxide to metal [CBSE Marking Scheme, 2019] ½

Detailed Answer :

Copper is a metal of medium reactivity.




Three main steps of extraction of copper from its sulphide ores are:

(i) Roasting of copper (I) sulphide.

(ii) Reduction of copper (I) oxide with copper (I) sulphide.

(iii) Electrolytic refining. ½+ ½ +½+½

5. (i) Ethanoic acid reacts with NaOH to give sodium salt and water but C2H5OH does not show this reaction. 1 (ii) Ethanoic acid reacts with NaHCO3 (Sodium Bicarbonate) or Na2CO3 (Sodium Carbonate) and given sodium

salt of ethanoic acid, water and carbon dioxide. [CBSE Marking Scheme, 2019] 1

Detailed Answer :

Ethanol Ethanoic acid

1. It does not change the colour of blue litmus paper. It changes the colour of blue litmus paper to red.

2. It does not react with sodium bicarbonate (NaHCO3).

It reacts with sodium bicarbonate (NaHCO3) releasing CO2 gas.

1+1

OR

Unsaturated compounds are more reactive. 1 Test : Bayer’s reagent test / Bromine water test given by ethene not by ethane / Ethane gives clear flame while

ethene gives a yellow flame with lots of black smoke. (any one test) 1 [CBSE Marking Scheme, 2019]

Detailed Answer :

Unsaturated carbon compounds are more reactive than saturated carbon compounds as they contain unstable, weaker pi bonds whereas saturated carbon compounds contain stable, stronger sigma bonds. 1

Test to distinguish ethane from ethene :

Bromine water test : Ethane will not decolorise bromine water and retain reddish brown colour while ethene will decolorise bromine water. 1

SECTION – C

6. The white light splits into seven colours when it gets refracted through the glass prism (VIBGYOR) 1 The colour deviates most – Violet ½ The colour deviates least – Red ½ Colours disappear and again white light obtained. [CBSE Marking Scheme, 2019] 1

Detailed Answer : When a beam of white light gets refracted through a glass prism, it refracts through different angles causing a

splitting of white light into its seven constituent colours (VIBGYOR). This gives rise to the formation of the colour spectrum.

Violet colour deviates the most and red colour deviates the least after refraction through a prism. When a second identical prism is placed in an inverted position with respect to first prism, a beam of white light emerges from the other side of the second prism. The second prism recombines all the seven colours to give a beam of white light.3

OR

(i) Myopia 1 (ii) Concave / diverging lens and focal ½ Length = 200 cm ½ (iii) (a) excessive curvature of the eye lens ½ (b) elongation of eye ball [CBSE Marking Scheme, 2019] ½

Detailed Answer : (i) Negative power shows that lens are concave, so the student is suffering from myopia/near- sightedness. (ii) The nature of lens used to correct the defect is concave lens.




Focal length, f = 1p

= 10 5−( ). =−2m

(iii) Causes of defect are : (a) Excessive curvature of the eye lens. (b) Elongation of the eyeball. 1+1+1

7. Chain of organisms formed as a result of eating or being eaten by organisms is called food chain / A series of organisms feeding on one another, is called food chain. 1

Grass → Insect (grasshopper) → Frog → Snake (Producer) (Herbivore) (Carnivore) (Top Carnivore) 1 (Any other example of food chain) l Tertiary trophic level / snake ½ l Biological magnification / Biomagnification [CBSE Marking Scheme, 2019] ½

Detailed Answer :

Food chain is the series formed of organisms feeding on one another at various biotic levels.

Terrestrial Food Chain :

PRODUCERSPlants( )

→ PRIMARY CONSUMERSRabbit( )

→ SECONDARY CONSUMERS

TERTIARY CONSUMERS

Fox

Lion

( )

( )

↓

The organisms at the highest (top) trophic level have the maximum concentration of the pollutant in ther bodies. This phenomenon is known as biological magnification. 1+1+1

9. (a) Silver is placed below Hydrogen in reactivity series / among least reactive metal / Silver does not react with dil. Hydrochloric acid. ½

(b) Rate of reaction is slow / bubbles of Hydrogen gas are formed / lead lies above hydrogen in reactivity series. 1 (c) Sodium is highly reactive / reaction is highly exothermic, evolving Hydrogen gas, which catches fire. 1 (d) Reaction is exothermic [CBSE Marking Scheme, 2019] ½

Detailed Answer : (a) As silver is less reactive so, does not react with dilute hydrochloric acid. (b) Bubbles are seen due to the evolution of hydrogen gas. Pb(s) + 2HCl (aq) → PbCl2(aq) + H2(g) (c) As sodium is highly reactive metal. It reacts with dilute hydrochloric acid vigorously with evolution of heat. (d) The reaction between aluminium with dilute hydrochloric acid is exothermic thus, the temperature of the

mixture rises on the addition of aluminium. ½+½+1+1

10. (i) 2 Cu2S + 3O2 Heat → 2 Cu2O + 2SO2 1

(ii) 2Cu2O + Cu2S Heat → 6 Cu + SO2 1

(iii) At anode : → Cu → Cu2+ + 2e– ½ At Cathode → Cu2+ 2e → Cu [CBSE Marking Scheme, 2019] ½

Detailed Answer :

(i) Roasting of copper (I) sulphide:

2 3 2 22Cu S s O g Cu O sCopper sulphide

2Heat

2Copper oxide

( ) + ( ) → ( ) + SSO g2 ( )

(ii) Reduction of copper (I) oxide from copper (I) sulphide:




2 2Cu O s Cu S s 6Cu s SO2Copper oxide Copper sulphide

Heat( ) + ( ) → ( ) + 22 g( )

(iii) Electrolytic refining :

At anode : Cu(s) → Cu2+ (aq) + 2e−

At cathode : Cu2+(aq) + 2e− → Cu(s) 1+1+1

15. (a) Because these methods involve only one parent / organisms are formed as a result of mitotic division / progeny (organisms) are similar in their genetic makeup and no variations. (any one) 1

(b)

1 (i) Planaria can be cut into any number of pieces and each piece grows through specialized cells into a

complete organism. [CBSE Marking Scheme, 2019] 1

Detailed Answer : (a) Budding, fragmentation and regeneration, all are considered as asexual mode of reproduction as there is no

gamete formation and a single parent is involved in reproduction. (b) Regeneration in Planaria is carried out by specialized cells which proliferate to make large numbers of cells.

From this mass of cells, different cells undergo changes in sequence to form various cell types and tissues.

1

SECTION – D

16. Given : h = 6 cm f = – 30 cm v = – 45 cm By mirror formula,

1f

=

1 1v u

+

½

1v

= 1 1f u

−




= – 1

30145

−−( )

= – 130

145

+ = –

190

½

f = – 90 cm from the pole of mirror Size of the image

m = −vu

½

= – 9045

= – 2

h1 = – 2 × 6 cm = – 12 cm ½ Image formed will be real, inverted and enlarged. 1

M

N

P

D

FB

A

C

E

B'

2 Well labelled diagram [CBSE Marking Scheme, 2019]

Detailed Answer : Given : h = 6 cm, f =–30 cm, u = –45 cm

Using mirror formula, 1f

= 1 1v u

+

1v

= 1 1f u

−

= 130

145

130

145−( )

−−( )

= − +

1v =

− + = −3 290

190

or v=−90cm So, the image will be formed 90 cm in front from the mirror. The image is real.

Magnification, m = ′

= −hh

vu

Therefore, m = − = −

−( )−( )

= −vu

9045

2

So, the size of the image h' =

−vhu

= −−( )×

−= −

90 645

12cm

So, the image is two times i.e. 12 cm in size. The image formed is real, enlarged and inverted.

. .

v

v

v

v

v

B'

A'

M

AD

E

PF

N

C B

5




OR

Given : f = +30 cm u = – 50 cm h = – 6.0 cm lens formula

1f

=

1 1v u

−

½

1v

=

1 1f u

+

= 130

150

−

5 3150−

=

2150

= 1

75 ½

\ v = + 75 cm ½

m = vu

hh

− 1 ½

= 7550 6

1−

−h

h1 = – 9 cm ½ Image formed is real, inverted and enlarged 1

MA

C1

2F1 F1

2F3F2

C2

BO

N

A'

2 Well labelled diagram [CBSE Marking Scheme, 2019] 5

Detailed Answer : Given : u = –50 cm, f = +30 cm, h = 6 cm

Using lens formula, 1f

= 1 1v u

−

1v

= 1 1f u

+

= 1

30150

5 3150

+−( )

= −

1v

= 2

150

or v = 150

2 = 75 cm

So, the screen should be placed at a distance of 75 cm on other side of optical centre. The image is real and inverted. Also,

Magnification, m = ′

=hh

vu




Height of image, h' = hvu

= ×−

6 7550

= –9 cm (size of image) So, the image is enlarged.

2f2f B

A

B’

A’

f

f

u= –50 cm

h= 6 cm

h= –9 cm

5

20. l When male and female organisms are involved in producing young ones, is known as sexual reproduction / Gametes from two organisms of opposite sex must fuse to produce young ones. 1

l Gametes (germs cells) produced are the products of meiosis / due to combining of DNA from two individuals, this results in mixing of characters and causes variations. 1 + 1

l In asexual reproduction, single parent produces young ones. There is no mixing of characters. 1 l More variations help in the process of evolution. Helpful variations accumulate over time and produce new

species and result in evolution. [CBSE Marking Scheme, 2019] 1

Detailed Answer : Sexual reproduction is a natural mode of reproduction which involves two individuals and gamete formation. During sexual reproduction, at the time of gamete formation, meiotic cell division takes place. During meiosis,

crossing over between fragments of homologous chromosomes occurs which brings about new gene combinations to be transferred to new generation. Crossing over is the fundamental cause of origin of variations in sexually reproducing organisms.

Whereas in asexual reproduction, chance variations can only occur when there is inaccurate copying of DNA as only one individual is involved. The variations caused by crossing over in sexually reproducing organisms are subjected to the selection process. Natural selection selects those variations which have more adaptive value and guide them towards evolution of new species. In this way, sexual reproduction gives rise to more viable variations for evolution. 5


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