CBSE Previous Year Question Papers Class 10 Science SA2 Outside Delhi – 2013

CBSE Previous Year Question Papers Class 10 Science SA2 Outside Delhi – 2013

ScienceMathsSanskritEnglishComputer ScienceHindiSocial Science
Time allowed: 3 hours                                                                                           Maximum marks: 90

GENERAL INSTRUCTIONS:

1.  The Question Paper comprises of two Sections, A and B. You are to attempt both the Sections.
2. All questions are compulsory.
3. All questions of Section-A and all questions of Section-B are to be attempted separately.
4. Question numbers 1 to 3 in Section-A are one mark questions. These are to be answered in one word or in one sentence.
5. Question numbers 4 to 6 in Section-A are two marks questions. These are to be answered in about 30 words each.
6. Question numbers 7 to 18 in Section-A are three marks questions. These are to be answered in
   about 50 words each.
7. Question numbers 19 to 24 in Section-A are five marks questions. These are to be answered in about 70 words each.
8.  Question numbers 25 to 36 in Section-B are questions based on practical skills. Question nos. 25 to 33 are MCQs. Each question is a one mark question. You are to select one most appropriate response out of the four provided to you. Question nos. 34 to 36 are short answer questions carrying two marks each.

SET I

SECTION A
Question.1 State modern periodic law of classification of elements.
Answer. Modem periodic law states “that the properties of elements are a periodic function of their atomic numbers”.

Question.2 No two individuals are absolutely alike in a population. Why?
Answer. No two individuals are absolutely alike in a population because DNA copying is an essential part of the process of reproduction. The process of DNA copying brings some variations which brings variations among individuals of the same species.

Question.3 Give one example each from your daily life where the household waste can be effectively reused and recycled respectively.
Answer. Example of Reuse. Plastic bottles that we get with jams and pickles can be reused for storing things in the kitchen.
Example of Recycling. We should collect used and discarded items of paper, plastic, glass and metals and send them to the respective industries for making fresh paper, plastic, glass or metal objects.

Question.4 State the importance of chromosomal difference between sperms and eggs of humans.
Answer. A male has one X chromosome and One Y chromosome. Thus half the sperms will have X chromosomes and the other half will have Y chromosome.
A female has two X chromosomes. So all the female gametes will have only X chromosomes. If a sperm carrying X chromosome fertilizes an ovum then the child bom will be a girl. If a sperm carrying Y chromosome fertilizes an ovum then the child bom will be a boy.
Thus the chromosomal difference between sperms and eggs of humans determines the sex of the child.

Question.5 “A concave mirror of focal length ‘f’ can form a magnified erect as well as an inverted image of an object placed in front of it.” Justify this statement stating the position of the object with respect to the mirror in each case for obtaining these images.
Answer.
When the object is placed at a distance less than ‘f’ (focal length) from the mirror, the image formed is virtual, magnified and erect.
When the object is placed at a distance of ‘f’ (focal length) from the mirror, the image is real, inverted and highly magnified.
When the object is placed at a distance more than (focal length) but less than ‘2f’, the image formed is real, inverted and magnified.
When the object is placed at the distance of “If from the mirror, the image is real, inverted and same size of the object.
When the object is placed at a distance more than ‘If from the mirror, the image formed is real, inverted and diminished.

Question.6 Write the harmful effects of using plastic bags on the environment. Suggest alternatives to plastic bags.
Answer. Harmful effects of using plastic bags:

1. Plastic is non-biodegradable so it will remain as such and pollute the environment.
2. Burning of plastic bags produces toxic gases.
3. Plastic bags can block the drainage system.
4. Discarded plastic bags when eaten by cows and other stray animals can block their allimentary canal and cause harm to them.
5.  Plastic bags when thrown in the water bodies, can cause water pollution as these do not decompose.
   Alternatives to plastic bags. Use paper or home made cloth bags to carry goods.

Question.7 (a) Differentiate, between alkanes and alkenes. Name and draw the structure of one member of each.
(b) Alkanes generally give clean flame. Why?
Answer. See Q. 9, 2013 (II Delhi).

Question.8 A carboxylic acid C₂H₄O₂ reacts with an alcohol in the presence of H₂SO₄ to form a compound ‘X’. The alcohol on oxidation with alkaline KMnO₄ followed by acidification gives the same carboxylic acid, C₂H₄O₂. Write the name and structure of

1.  Carboxylic acid,
2.  alcohol and
3.  the compound ‘X’.

Answer. Carboxylic acid, C₂H₄O₂:

Question.9 The electronic configuration of an element ‘X’ is 2, 8, 8, 2. To which (a) period and (b) group of the modern periodic table does ‘X’ belong? State its valency. Justify your answer in each case.
Answer. Electronic configuration of ‘X’ is 2, 8, 8, 2.
(a) It belongs to the 4th period because it has four energy shells and period number of an element is equal to the number of energy shells.
(b) It belongs to 2nd group because it has 2 valence electrons and the group number of an element having up to two valence electrons is equal to the number of valence electrons.
Its valency is ‘2’. To acquire inert gas configuration ‘X’ either loses 2 electrons or gains 6 electrons. Since the shorter route is to lose 2 electrons for obtaining inert gas configuration, hence its valency is 2.

Question.10 Four elements P, Q, R and S have atomic numbers 12,13,14 and 15 respectively.
Answer the following questions giving reasons:

1. What is the valency of Q?
2.  Classify these elements as metals and non-metals.
3.  Which of these elements will form the most basic oxide?

Answer.

1.  Valency of Q is 3 as Q loses 3 electrons to obtain inert gas configuration.
2.  P and Q are metals as these elements lose 2 and 3 electrons respectively to obtain inert gas configuration and those elements which lose electrons are metals.
   R and S are non-metals as these elements either share or gain electrons for attaining inert gas configuration and those metals which accept electrons either by sharing or gaining are non-metals.
3. Element P will form the most basic oxide because P is the most metallic element among the 4 given elements and thus forms the most basic oxide.

Question.11 Explain budding in Hydra with the help of diagrams only.
Answer. Budding in Hydra. Hydra reproduces by budding which is an asexual type of reproduction.
See Q. 20, 2011 (I Outside Delhi).

Question.12 List any four methods of contraception used by humans. How does their use have a direct effect on the health and prosperity of the family? 
Answer. Methods of Contraception. See Q. 15, 2012 (I Outside Delhi).
Effect of these methods on the health. Contraceptive methods are used to prevent the spreading of STDs like syphillis, AIDS, etc.
Effect of these methods on the prosperity of the family. If a couple has less number of children, they can provide good clothes, good food and a good education to each child. This will make the parents as well as children happy.

Question.13 Explain with the help of suitable examples why certain traits cannot be passed on to the next generation? What are such traits called?
Answer. A trait (or characteristic) of an organism which is ‘not inherited’ but develops in response to the environment is called an acquired trait.
The acquired traits of an organism cannot be passed on to its future generations.
For example, ‘low weight’ of beetle, ‘cut tail’ of a mouse.
Changes brought in the non-reproductive tissues cannot be passed on to the DNA of the germ cells. So the experiences of an individual during its lifetime cannot be passed on to its progeny. Thus acquired characters of an individual cannot be inherited.

Question.14 A cross was carried out between a pure bred tall pea plant and a pure bred dwarf pea plant and F₁ progeny was obtained. Later, the F₁ progeny was selfed to obtain F₂ progeny. Answer the following questions:
(a) What is the phenotype of the F₁ progeny and why?
(b) Give the phenotypic ratio of the F₂ progeny.
(c) Why is the F₂ progeny different from the F2 progeny?
Answer. When plants of two different traits of character are crossbred to get a progeny (F₁ generation), only the dominant trait is visible in this generation. But when plants of F₁ generation are selfbred then the two traits of character get separated and the recessive traint also appears in the plant of F₂ generation. This is known as Law of Segregation (separation) of traits.
flow chart

Question.15 Name the type of mirror used

1. by dentists and
2.  in solar furnaces. Give two reasons why such mirrors are used in each case.

Answer.

1.  Concave mirrors are used by dentists to see the large images of the teeth of patients
   because when a tooth is within the focus of a concave mirror, then an enlarged image of the tooth is seen in the concave mirror. Thus it becomes easier to locate the defect in the tooth.
2.  Large concave mirrors are used in solar furnaces as reflectors. Solar furnace is placed at the focus of the concave reflector which focusses the Sun’s heat rays on the furnace due to which the solar furnace gets very hot. Even steel can be melted in this solar furnace.

Question.16 An object placed on a metre scale at 8 cm mark was focussed on a white screen placed at 92 cm mark, using a converging lens placed on the scale at 50 cm mark.
(i) Find the focal length of the converging lens.
(ii) Find the position of the image formed if the object is shifted towards the lens at a position of 29.0 cm.
(iii) State the nature of the image formed if the object is further shifted towards the lens.
Answer.

Question.17 When and where do we see a rainbow? How is a rainbow formed? Draw a labelled diagram to illustrate the formation of a rainbow.

Answer. Rainbow is caused by dispersion of sunlight by tiny water droplets present in the atmosphere. The water droplets act like small prisms. They reflect the incident sunlight and then reflect it internally and finally refract it again when it comes out of the rain-drop. A rainbow is always formed in the direction opposite to that of the Sin

Question.18 State in brief two ways in which non-biodegradable substances would affect the environment. List two methods of safe disposal of the non-biodegradable waste.
Answer. Non-biodegradable wastes cannot be decomposed into simpler harmless substances by the action of micro-organisms. These substances persist in the enviromnent and cause adverse effects on the environment.

1. The use of pesticides, insecticides has increased agricultural output but the chemicals present in them enter water and food chains. This affects the fertility of the soil and causes water and soil pollution.
2. Plastic bags haVe changed the modem lifestyle but cause severe problems. They cause blockages in the drainage systems and these plastic bags when eaten by stray animals cause harm to them.

Methods of safe disposal of the non-biodegradable waste:

1. Molten plastic waste mixed with asphalt can be used for making roads.
2. Solid wastes should be buried in the urban areas as land fills.

Question.19 An organic compound ‘X’ on heating with cone. H₂SO₄forms a compound ‘Y’ which on addition of one molecule of hydrogen in the presence of nickel forms a compound ‘Z’. One molecule of compound ‘Z’ on combustion forms two molecules of CO₂ and three molecules of H₂O. Identify giving reasons the compounds ‘X’, ‘Y’ and ‘Z’. Write the chemical equations for all the chemical reactions involved.
Answer.

Question.20 Write the two causes of human population explosion. Explain with the help of suitable examples how this explosion can be checked.
Answer. There are three main causes of human population explosion:

1.  Decline in death rate from 13.8 per 1,000 in 1985 to 8 per 1,000 in 2001.
2. Increase in longevity.
3. Increase in birth rate.
   The uncontrolled growth of human population causes many socio-economic problems. To mitigate the various ill-effects of increasing population, there is a need of population control by reducing the birth rate through family planning programmes.

1. By adopting various methods of contraception:

•  Barrier methods—Physical devices such as condoms, diaphragm and cervical caps are used.
•  Surgical methods
•  Chemical methods—Oral and vaginal pills used by females.

2. People in reproductive age group should be educated about the advantages of a small family with the help of mass media like radio, T.V., newspapers, etc. and educational institutions.
3. Marriage age should be increased.
4. Social organisations should be involved in family planning programmes.
5. Literacy rate, especially that of females should be increased.

Question.21 What is pollination? How does it occur in plants? How does pollination lead to fertilization? Explain.
Answer. The transfer of pollen grains from anthers (male sexual parts) of a flower to the stigma part of the pistil (female sexual part) is known as pollination. Pollination is done by insects, birds, wind and water.
Pollination can occur in two ways:

1.  Self pollination. When the pollen grains from the anther of a flower are transferred to the stigma of the same flower or another flower on the same plant, it is called self pollination.
2. Cross pollination. When the pollen grains from the anther of a flower on one plant are transferred to the stigma of a flower on another similar plan it is called cross pollination.
   When a pollen grain falls on the stigma of the carpel, it grows a pollen tube downwards through the style towards the female gamete in the ovary. A male gamete moves down the tube. When the pollen tube enters the ovule, its tip bursts open and male gamete comes out of the pollen tube and combings with the nucleus of the female gamete and forms zygote. This process is known as fertilization.

Question.22(a) To construct a ray diagram we use two light rays which are so chosen that it is easy
to know their directions after reflection from the mirror. List these two rays and state the path of these rays after reflection. Use these two rays to locate the image of an object placed between infinity and the centre of curvature of a concave mirror.
(b) Draw a ray diagram to show the formation of image of an object placed between the pole and principal focus of a concave mirror. How will the nature and size of the image formed change, if the mirror is replaced by a converging lens of same focal length?
Answer.

Question.23(a) A student cannot see clearly a chart hanging on a wall placed at a distance of 3 m
from his eyes. Name the defect of vision he is suffering from. Draw a ray diagram to illustrate this defect. List its two possible causes.
(b) Draw a ray diagram to show how this defect may be corrected using a lens of apropriate focal length.
(c) An eye donation camp is being organised by social workers in your locality. How
and why would you help in this cause?
Answer.
(a) & (b) See Q. 17, 2011 (I Outside Delhi).
(c) There are millions of blind people in our country who cannot see at all. The eyesight of most of these blind people can be restored if they are given the eyes donated by other persons after their death. Our two eyes can give eyesight to two blind persons.
In the eye donation camps, we can even today, make a pledge in writing, i.e., whenever we die, our eyes should be removed and given to the blind persons to light up their dark world.
We should be greatful to God that he has given us the gift of vision to see this wonderful world. We must try and pass on this priceless gift of vision to our less fortunate blind brothers and sisters by registering our name for eye donation.

Question.24 Explain the trends in the Modern Periodic Table of various properties like valency, atomic size, metallic and non-mettalic properties of the atoms of elements.
Answer.

1.  Valency. Elements belonging to same group have same number of valence electrons and thus same valency.
   Valency in a particular period from left to right first increases as positive valency and ‘ then decreases as negative valency.
   Example, In elements of 2nd period:
   Li ha 1⁺valency, then Be^2+, Be³⁺, C⁴⁺ covalency, N^3-valency, then  O^2- and F^(-) valency.
2. The atomic size or atomic radius increases as we move down in a group and it decreases as we move from left to right in a period. Atomic size increases down a group due to increase in the number of shells. Atomic size decreases along a period
   due to increase in the nuclear charge which tends to pull the electrons closer to the nucleus and reduces the size of the atom.
3.  Metallic and non-metallic properties. In the modern periodic table metals are on the left side and non-metals are on the right side. A zigzag line of metalloids separates metals from non-metals. Metallic character decreases from left to right in a period and increases while going down in a group. Non-metallic character increases from left to right in a period due to increase in the electro-negativity and this character decreases from top to bottom in a group due to decrease in the electro-negativity of atoms while going down in a group.

SECTION B
Question.25 Select the correct observation about dilute solution of acetic acid.
(A) It smells like rotten egg and turns blue litmus red
(B) It smells like vinegar and turns red litmus blue
(C) It smells like rotten egg and turns red litmus blue
(D) It smells like vinegar and turns blue litmus red
Answer.(D) It smells like vinegar and turns blue litmus red

Question.26 In the preparation of soap, a small amount of sodium chloride (common salt) is added to the mixture of fat and sodium hydroxide. The role of common salt is to
(A) favour the precipitation of soap
(B) enhance the cleansing capacity of soap
(C) increase the weight of the soap to earn money
(D) decrease the acidity of the soap
Answer.(A) favour the precipitation of soap

Question.27 A student takes about 6 mL distilled water in four test tubes marked P, Q, R and S. He dissolves sodium sulphate in P, potassium sulphate in Q, calcium sulphate in R and magnesium sulphate in S. After that he adds equal amount of soap solution in each test tube. On shaking these test tubes, he would observe a good amount of lather in the test tubes marked
(A) P and Q (BfQandR (C) R and S (D) P and S
Answer.(A) P and Q

Question.28 A student focussed the image of a distant object using a device X on a white screen S as shown in the figure. If the distance of the screen from the device is 30 cm, select the correct statement about the device X.

(A) The device X is a concave mirror of focal length 15 cm
(B) The device X is a concave mirror of focal length 30 cm.
(C) The device X is a concave mirror of radius of curvature 30 cm.
(D) The device X is a convex mirror of focal length 30 cm.
Answer.(B) The device X is a concave mirror of focal length 30 cm.

Question.29 A student traces the path of a ray of light through a rectangular glass slab for four different angles of incidence. He very cautiously measures the angle f, angle r and the angle e. On analysing his measurements, he is likely to draw the following conclusion:
(A) ∠i = ∠e > ∠r       (B) ∠i > ∠r > ∠e         (C) ∠i = Zr < Ze         (D) ∠i = ∠e< ∠r
Answer.(A)∠i = ∠e > ∠r

Question.30 For the refraction of a ray of light through a glass prism, the path of a ray of light is shown below:

The angle of incidence, the angle of emergence and the angle of deviation respectively have been represented by
(A) Y, N, Z (B) X, Z, M (C) X, N, Z (D) X, M, Z
Answer.(D) X, M, Z

Question.31 Slides of binary fission in amoeba and budding in yeast were given for observations to a group of students. Some of the observations reported by the group are given below:
I. Cytokinesis was observed in the yeast cell.
II. A chain of buds were observed due to reproduction in amoeba.
III. Single cell of amoeba and single cell of yeast were undergoing binary fission and budding respectively.
IV. Elongated nucleus was dividing to form, two daughter nuclei in amoeba.
The correctly reported observations are:
(A) I and II (B) II and III (C) III and IV (D) I and IV
Answer.(C) III and IV

Question.32 Study the different conclusions drawn by students on the basis of their observations of fresh available specimens of plants and animals:
I. Potato and sweet potato are homologous organs.
II. Wings of insects and wings of bird are analogous organs.
III. Wings of insects and wings of bats are homologous organs.
IV. Thoms of citrus arid tendrils of cucurbita are homologous organs.
The correct conclusions are:
(A) I and II (B) II and IV (C) I and III (D) III and IV
Answer.(B) II and IV

Question.33 You have a basket of vegetables which contains carrot, potato, tomato, ginger, radish, sweet potato. Select two vegetables to represent the correct homologous structures.
(A) Potato and sweet potato (B) Carrot and radish.
(C) Potato and carrot (D) Carrot and tomato
Answer.(B) Carrot and radish.

Question.34 Name two salts each of calcium and magnesium which make the water hard?
Answer. These salts are:
Calcium chloride—CaCl₂ Magnesium chloride—MgCl₂
Calcium sulphate—CaSO₄ Magnesium sulphate—MgSO₄

Question.35 Name the process of asexual reproduction shown by yeast. What type of living being is yeast? What is its commercial importance?
Answer. Yeast reproduces by method of budding. Yeast is a unicellular fungus.
Yeast forms lots of buds in the sugar solution. These buds become young yeast cells which change sugar of the solution to produce ethyl alcohol by fermentation in the distillaries.

Question.36 List the factors on which the angle of deviation through a prism depend?
Answer. Angle of deviation through a prism depends

•  on the ∠A called ∠ of prism;
•  on the ∠ of incidence;
• on the optical density of the material of the prism.

SET II

Except for the following questions, all the remaining questions have been asked in Set-I.
SECTION A
Question.1 The automic numbers of three elements X, Y and Z are 3, 11 and 17 respectively. State living reason which two elements will show similar chemical properties.

X and Y will show similar chemical properties as both elements have same number of valence electrons, i.e., 1,
Answer.

Question.2 Why is DNA copying necessary during reproduction?
Answer. DNA copying is essential during reproduction for the inheritance of features from parents to the next generation.

Question.3 Name any two items which can be easily recycled but are generally thrown in the dustbins by us.
Answer. Cold drink cans and empty milk polypacks can be easily recycled but are generally thrown in the dustbins.

Question.4 List two advantages of growing grapes or banana plants through vegetative propagation.
Answer. Two advantages of growing grapes or banana plant through vegetative propagation:

1.  The fruit trees grown through vegetative propagation bear fruits much earlier.
2.  Banana and grapes produce either very few seeds or do not produce viable seeds. Therefore their plants are grown by vegetative propagation.

Question.5 “A convex lens of focal length ‘F can form a magnified erect as well as inverted image.” Justify this statement stating the position of the object with respect to the lens in each case for obtaining these images.
Answer. The type of image formed by a convex lens depends on the position of the object in front of the lens.

•  When the object is placed between the optical centre and the focus (i.e., between O and F’), the image formed is behind the object (on the same side). It is virtual, magnified and erect.
•  When the object is placed at the focus of a convex lens, the image formed is at infinity. It is real, inverted and highly magnified.
•  When the object is placed between focus and the centre of curvature (i.e., between F’ and 2F’) the image formed is beyond 2F. It is real, inverted and magnified.
•  When the object is placed at the centre of curvature (i.e., at 2F’), the image formed is at 2F. It is real, inverted and of same size of the object.
•  When the object is beyond 2F’ (or beyond centre of curvature), the image formed is between F and 2F. It is real, inverted and diminished.
•  When the object is at infinity, the image formed is at focus. It is real, inverted and much smaller than the object (highly diminished).

Question.8 What is meant by isomers? “We cannot have isomers of first three members of alkane series.” Give reason to justify this statement. Draw the structures of two isomers of pentane, C₅H₁₂.
Answer. The organic compounds having the same molecular formula but different structures are known as isomers.
Isomers of first three members of alkane series are not possible because only one arrangement of carbon atoms is possible in their molecules.

Question.10 The atomic number of an element is 17. Predict (a) its valency, (b) whether it is a metal or non-metal, (c) its relative size with respect to other members of its group. Justify your answer in each case.
Answer. Atomic number of given element = 17
.^.. Electronic configuration of given element = 2, 8, 7
(a) Valency = 1
Since this element requires one electron to complete its octet (outermost shell).
(b) This element is a non-metal as it gains one electron to complete its outermost shell and elements which gain electrons are non-metals.
(c) This element is the 2nd member of this group. Its size is greater than the 1st member (2, 7) and smaller than the rest of the other members of that group because size of the atom increases on going down in a group as one energy shell is added at every step.

Question.11 The electronic configuration of an element ‘X’ is 2, 8, 8, 2. To which (a) period and (b) group of the modern periodic table does ‘X’ belong? State its valency. Justify your answer in each case.
Answer.

Element: ‘X’,                                    Electronic configuration: 2, 8, 8, 2
(a) ‘X’ belongs to 4^th period because X’ has four energy shells and energy shell number
corresponds to the period number.
(b) ‘X’ belongs to 2^nd group because ‘X’ has 2 electrons in its outermost shell and valence electrons (1, 2) corresponds to the group number of the element.
Valency: 2

Question.12 Explain the process of regeneration in planaria.
Answer. The process of getting back a full organism from its body parts is called regeneration. Planaria is a flat worm which is found in fresh water ponds. It possesses great power of regeneration. If the body of planaria somehow gets cut into a number of pieces, then each body piece can regenerate into a complete planaria by growing all the missing parts. Diagram. See Q. 24, 2011 (I Delhi).

Question.14 A blue colour flower plant denoted by BB is crossbred with a white colour flower plant denoted by ww.
(a) State the colour of flower we would expect in their Fj progeny.
(b) Write the percentage of plants bearing white flower in F2 generation when the flowers of Fj plants were selfed.
(c) State the expected ratio of the genotype BB and Bw in the F2 progeny.
Answer. See Q. 16, 2012 (I Delhi).

Question.16 A student focussed the image of an object on a white screen using a converging lens. He noted down the positions of the object, screen and the lens on a scale as given below: Position of object = 10.0 cm Postion of lens = 50.0 cm Position of screen = 90.0 cm
(a) Find the focal length of the converging lens.
(b) Find the position of the image if the object is shifted towards the lens at a position of 30.0 cm.
(c) State the nature of the image formed if the object is further shifted towards the lens.
Answer.

Question.20 List the sign conventions that are followed in case of refraction of light through spherical lenses. Draw a diagram and apply these conventions in determining the nature and focal length of a spherical lens which forms a four times magnified real image of an object placed 20 cm from the lens.
Answer. Sign conventions in case of refraction of light through spherical lenses.

1.  All the distances are measured from the optical centre of the lens.
2.  The distances measured in the same direction as that of incident light are taken as positive (+ve).
3.  The distances measured in the same direction as that of incident light are taken as negative (-ve).
4. The perpendicular distances to the principal axis in the upward direction are taken as positive (+ve).
5.  The perpendicular distances to the principal axis in the downward direction are taken as negative (-ve).

u = object distance from O (optical centre)
v = image distance from O, f = focal length

These rules conclude that:
(1) ‘u’ (object distance) is always -ve
(2) ‘f’ of convex lens is always +ve
(3) ‘f’ of concave lens is always-ve
(4) ‘V’ in case of virtual image is always -ve
(5) ‘V’ in case of real image is always +ve
(6) Real image is always inverted thus h2 (size of the image) in case of real image is always -ve.

SET III

Except for the following questions, all the remaining questions have been asked in Set-I and
Set-II.
SECTION A
Question.1 The atomic numbers of three elements A, B and C are 11,17 and 19 respectively. State jiving reason which two elements will show similar chemical properties.
Answer.

A and C will show similar chemical properties as both elements have the same number of valence electrons, i.e., 1

Question.2 Name the body part where fertilisation occurs in human female.
Answer.
Fallopian tube.

Question.3 Write the main objective of conservation of biodiversity.
Answer.
The main objective of conservation of biodivesity is that the loss of diversity may lead to loss of ecological stability.

Question.4 (a) How do Leishmania and Plasmodium reproduce?
(b) State one difference in their mode of reproduction.
Answer.
(a) Leishmania and Plasmodium reproduce by fission (asexual mode of reproduction).
(b)

1.  Leishmania reproduce by binary fission in which the parent organism splits to
   form two new organisms.
2.  Plasmodium reproduce by multiple fission in which the parent organism splits to form many new organisms at the same time.

Question.5 What is meant by power of a lens? Define its SI unit.
Answer.
The power of a lens is a measure of the degree of convergence or divergence.
The power of a lens is defined as the reciprocal of its focal length in metres.
The SI unit of power is dioptre denoted by the letter D. One dioptre is the power of a lens whose focal length is 1 metre.

Question.8 (a) Distinguish between ethanol and ethanoic acid on the basis (i) litmus test, (ii) reaction with sodium hydrogen carbonate.
(b) Name the oxidising agents used in the conversion of ethanol to ethanoic acid.
Answer.See Q. 10, 2012 (III Outside Delhi).

Question.10 The atomic number of an element ‘X’ is 20. Write
(a) its valency,
(b) whether it is a metal or non-metal,
(c) the formula of compound formed when the element ‘X’ reacts with an element ‘Y’ of atomic number 8.
ustify your answer in each case.
Answer.

(a) Valency of X is 2 as it loses two electrons to acquire nearest inert gas configuration.
(b) X is a metal as it loses electrons to acquire inert gas configuration and an element which loses electrons for acquiring inert gas configuration is a metal.

Question.13 What is placenta? Explain its function in human female.
Answer. After fertilisation, zygote is formed which develops into an embryo. The embryo gets nutrition from the mother’s blood with the help of a special tissue called placenta. This is a disc which is embedded in the uterine wall. It contains villi on the embryo’s side of the tissue. On the mother’s side are blood spaces, which surround the villi. This provides a large surface area for glucose and oxygen to pass from the mother to the embryo. In this way, embryo gets its nutrition.

Question.15 How are fossils formed? State two methods of determining the age of fossils.
Answer.Formation of a fossil. Usually when organisms die, their bodies get decomposed by the action of micro-organisms in the presence of oxygen, moisture, etc. Sometimes, the conditions in the environment like oxygen, moisture, etc. are absent which do not let the body of the organism decompose completely. So fossils are such remains or impressions of dead animals or plants that lived in the remote past but which were not decomposed completely. For example, if a dead leaf gets caught in mud, it will not decompose quickly. The mud around the leaf will set around it as a mould and gradually harden to form a rock and retain the impression of the whole leaf. This forms a leaf fossil which can be dug out from the earth a long time after.
Determination of age of the fossil.

1.  Fossils closer to the surface are more recent compared to those found in the deeper layers of the earth’s surface.
2. Another way is by detecting the ratio of different isotopes of the same element in the fossil material.

Question.16 A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the positions of the candle flame, screen and the lens as given below: Position of the candle flame = 12.0 cm Position of the lens = 50.0 cm Position of the screen = 88.0 cm
(i) Find the focal length of the convex lens.
(ii) Find the position of the image of the candle flame if it is shifted towards the lens at a position of 31.0 cm.
(iii) State the nature of the image formed if the candle flame is further shifted towards the lens,
Answer.

Question.20 A spherical mirror ‘A’ always forms an erect image of an object and another spherical mirror ‘B’ forms erect as well as inverted image of an object. State with reasons the type of spherical mirrors ‘A’ and ‘B’ and draw ray diagrams showing formation of images to justify your answer.

Answer. Mirror ‘A’ is a convex mirror as this mirror always forms an erect and virtual image.
When an incident ray is parallel to the principal axis, the reflected ray appears to be coming from the focus and when incident ray moves towards the centre of curvature, the reflected ray retraces the path.
These two rules show that the reflected rays are diverging and when produced back-wards, the rays appear to meet. Thus, this mirror always produces a virtual image which is always erect.
Mirror ‘B’ is a concave mirror. The nature, position and size of the image formed by a concave mirror depends on the position of the object in relation to points P(Pole), F(focus) and C(centre of curvature) of the mirror.

Case 1. When the object is placed between P and
Case 2. As the object moves away from the mirror beyond F the two reflected rays actually meet below the principal axis which produces a real and inverted image.
F—the two reflected rays (one passes through the focus and other retraces through centre of curvature) are diverging and appear to meet in backward direction. This produces a virtual and erect image.

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