CBSE Previous Year Question Papers Class 10 Science SA2 Delhi – 2011

CBSE Previous Year Question Papers Class 10 Science SA2 Delhi – 2011

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 ofSection-A and all questions ofSection-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 How many covalent bonds are there in a molecule of ethane,C₂H₆?
Answer. There are seven covalent bonds in a molecule of ethane:

Question.2 What is Tyndall effect?
Answer. The scattering of a beam of light is called the Tyndall effect. Tyndall effect can be observed when sunlight passes through the canopy of a dense forest. In a dense forest, the mist contains tiny droplets of water, which act as particles of colloid dispersed in air.

Question.3 What will happen if we kill all the organisms in one trophic level?
Answer. It will disturb the food chains and food web, which in turn will decrease the chances of food availability to the succeeding trophic levels and will result in instability of the ecosystem.

Question.4(i) How do you calculate the possible valency of an element from the electronic
configuration of its atoms?
(ii)Calculate the valency of an element X whose atomic number is 9.
Answer.
(i) The valency of an element is determined by the number of valence electrons present in the atom of the element. Electronic configuration gives the number of valence electrons in its atom. The number of valence electrons decides the number of electrons lost or gained by one atom of an element to achieve the nearest inert gas electron configuration and gives the valency of element.

X gains 1 electron to achieve the nearest inert gas configuration , i.e,, 2, 8. Therefore the valency of X is ‘T.

Question.5 Write any two differences between binary fission and multiple fission in a tabular form as observed in cells of organisms.
Answer.

Question.6 How does the electronic configuration of an atom of an element relate to its position in the modern periodic table? Explain with one example.
Answer. The electronic configuration of an atom of an element gives its position in the modem periodic table.

(i) The ‘period number’ of an element is equal to the number of electron shells in its atom.
(ii)

•  The group number of an element having up to two valence electrons is equal to the
  number of valence electrons.
•  The group number of an element having more than 2 valence electrons is equal to the number of valence electrons plus 10.
  Example: If the electronic configuration of an element is 2, 8, 7
  Then its period number is 3 as it has three electron shells.
  Its group number is 17 as it has 7 valence electrons. (.^..  Group no. = 7+10 = 17)

Question.7 (a) What is meant by the dispersion of white light? Draw a diagram to show
dispersion of white light by the glass prism.
(b) Explain why the planets do not twinkle but the stars twinkle.
Answer.

(a) Dispersion of light. The splitting of white light into its component colours on passing through a prism is called dispersion of light.
(b) The planets are much closer to the earth. A planet can be considered as a collection of large number of point-sized sources of light.
So the total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero thereby nullifying the twinkling effect. On the other hand stars twinkle because stars are point-sized sources of energy therefore the continuously changing atmosphere causes atmospheric refraction which causes variation in light.

Question.8 Define homologous and analogous organs.
Classify the following as homologous or analogous organs:
(i) Wings of an insect and wings of bat.
(ii)Forelimbs of a man and forelimbs of a frog.
Answer.
Homologous organs: Those organs which have the same basic structure or same basic design but different functions, are called homologous organs.
Analogous organs: Those organs which have different basic structure or different basic design but have similar appearance and perform similar functions are called analogous organs.

1. Wings of an insect and wings of bat are analogous organs.
2. Forelimbs of a man and forelimbs of frog are homologous organs.

Question.9(a) List any two characteristics of a good fuel.
(b) What are non-renewable resources of energy? Give two examples of such resources.
Answer.(a) Characteristics of a good fuel:

1.  It should have a high calorific value.
2. It should burn without giving out any smoke or harmful gases.
3. Its ignition temperature should neither be too low nor too high.
4.  After burning it should not leave much ash behind. (any.two)

(b) Those sources of energy which have accumulated in nature over a very-very long time and cannot be quickly replaced when exhausted, are called non-renewable sources of energy.
Example: Fossil fuels (coal, petroleum, natural gas) and nuclear fuels (uranium) are non-renewable sources of energy.

Question.10 Explain how equal genetic contribution of male and female parents is ensured in the progeny.
Answer. Both the parents contribute equal DNA material to progeny. Every sexually reproducing organism bears two sets of all genes, one inherited from each parent. Each germ cell must have only one gene set. Thus the male gamete and female gamete carry one gene for each characteristic from the gene pairs of parents. But when a male gamete fuses with a female gamete during fertilisation, they make a new cell called zygote with a full set of genes. Thus zygote grows and develops to form a new organism having equal characteristics from both the parents which it has inherited through genes.

Question.11 Explain the terms:
(i) Speciation (ii) Natural selection
Answer.
(i) Speciation. The process by which new species develop from the existing species is known as speciation.
The important factors which could lead to the formation of new species are:

•  Geographical isolation of a population caused by various types of barriers like mountain ranges, rivers, sea etc. The geographical isolation leads to reproductive isolation due to which there is no flow of genes between separated groups of population.
•  Genetic drift caused by drastic changes in the frequencies of particular genes by chance alone.
•  Variations caused in individuals due to natural selection.

(ii) Natural selection. Natural selection is the process of evolution of a species whereby characteristics which help individual organisms to survive and reproduce are passed on to their offsprings and those characteristics which do not help are not passed on.

Question.12 Explain with examples how the following are evidences in favour of evolution in organisms.

1.  Homologous organs
2. Analogous organs
3.  Fossils

Answer.

1.  Homologous organs. Those organs, which have the same basic structure but different functions, are called homologous organs.
   For example: The forelimbs of a man, a lizard, a frog, a bird and a bat have the same basic design of bones but they perform different functions.
   The presence of homologous organs in different animals provides evidence for evolution that they are derived from the same ancestor who has the basic design of the organ on which all the homologous organs are based.
2.  Analogous organs. Those organs, which have different basic structure but have similar appearance and perform similar functions are called analogous organs.
   For example: The wings of an insect and a bird have different structures but they perform the same function of flying.
   Analogous organs provide evidence for evolution that the organisms from different ancestors perform similar functions to survive, flourish and keep on evolving in prevailing environment.
3.  Fossils. The remains of dead animals or plants that lived in the remote past are known as fossils. The fossils provide evidence for evolution.
   For example: A fossil of a bird called Archaeopteryx looks like a bird but it has many other features found in reptiles. Therefore Archaeopteryx is a connecting link between the reptiles and birds and hence suggests that the birds have evolved from the reptiles.

Question.13 Write one chemical equation to represent each of the following types of reactions of organic substances:
(i) Esterification (ii) Saponification (iii) Substitution
Answer.

Question.14 Two elements X and Y belong to group 1 and 2 respectively in the same period of periodic table. Compare them with respect to:

1. the number of valence electrons in their atoms;
2. their valencies;
3. metallic character;
4. the sizes of their atoms;
5.  the formulae of their oxides;
6. the formulae of their chlorides.

Answer.

•  X and Y belong to same period.
• X belongs to group ‘T.
• Y belongs to group ‘2’.

1. Valence electron in X is 1 whereas valence electrons in Y are 2.
2. The valency of X is 1 whereas valency of Y is 2.
3. X is more metallic than Y because metallic character decreases on moving from left to right in a period.
4.  The size of X is more than Y because size of the atom decreases on moving from left to right in a period.
5. Oxide of X = X₂ 0 Oxide of Y = YO
6.  Chloride of X = XCl Chloride of Y = YCl₂

Question.15 Draw the ray diagram and also state the position, the relative size and the nature of image formed by a concave mirror when the object is placed at the centre of curvature of the mirror.
Answer.

When the object is at the centre of curvature of a concave mirror, i.e., point C:
The image formed is

1. real
2. inverted
3.  same size as the object at C
4.  at C

Question.16 Define, ‘refractive index of a transparent medium.’ What is its unit? Which has a higher refractive index—glass or water?
Answer.

•  The light bending ability of a transparent medium is called the refractive index of that medium.
•  The ratio of speed of light in vacuum to the speed of light in a medium is called the refractive index of that qredium.
  
•  Since refractive index is a ratio of two similar quantities therefore it has no units.
•  The refractive index of glass is more than water.

Question.17 What eye defect is hypermetropia? Describe with a ray diagram how this defect of vision can be corrected by using an appropriate lens.
Answer. Long sightedness is hypermetropia. Due to this defect, a person is not able to see the nearby objects clearly but can see the distant objects clearly.
Causes of long-sightedness. It is caused due to the following reasons:
— Normal increase in the focal length of the eye lens. The lens becomes less convergent.
— Shortening of the eyeball size.
Long sightedness can be corrected by using a convex lens of suitable focal length in the spectacles of such a person.
When a convex lens of suitable power is placed in front of the hypermetropic eye then the diverging rays of light coming from the nearby object are first converged by this convex lens. Due to this, the convex lens forms a virtual image of the nearby object at a point near to the hypermetropic eye. Then the hypermetropic eye can easily focus the image formed ‘ by convex lens on the retina.

Question.18 (a) List two sexually transmitted diseases in each of the following cases:
(i) Bacterial infections
(ii) Viral infections
(b) How may the spread of such diseases be prevented?
Answer.
(a) Sexually transmitted diseases:
(i) Bacterial infections. Gonorrhoea, Syphilis
(ii)Viral infections. Warts, AIDS (Acquired Immuno Deficiency Syndrome)

• Barrier method of contraception provides protection against STDs like AIDS,
  Syphilis etc.
• Observing hygienic conditions.

Question.19 Nalin and his four friends were sitting on his roof on a pleasant day. All of them were enjoying Ludo. Suddenly Ayush saw seven colours in the sky. He jumped with joy and shouted “Look, there is an Indradhanush in the sky”. Then Nalin explained all about the rainbow. After that every one clapped for him.
(a) What information is given by Nalin to his friends about a rainbow?
(b) Is it possible to obtain rainbow phenometwn on the earth?
(c) Which term is used for the seven colours of the rainbow?
(d) Which colour appeals at the top and at the bottom of the rainboiv?
(e) Which moral value is shown by Nalin towards his friends?
Answer.
(a) Rainbow is a natural phenomenon in which an arch of seven colours visible in the sky is produced by the dispersion of white sunlight by raindrops in the atmosphere. Each raindrop acts as a tiny glass prism splitting the sunlight into seven colours.
(b)Yes, in daily life, when white light of the Sun is passed through a glass prism, it splits into seven colours.
(c) The band of seven colours obtained by the splitting of white light is called VIBGYOR.
(d) The red colour appears at the top of the rainbow whereas violet colour appears at its bottom.
(e) Nalin shares his knowledge of Science with his friends. He shares interesting information about the concept of the rainbow with his friends. Sharing infor-mation is the moral value shown by Nalin.

Question.20 (a) What is a soap? Why are soaps not suitable for washing clothes when the water is hard?
(b) Explain the action of soap in removing an oily spot from a piece of cloth.
Answer.
(a) Soaps are sodium salts of fatty acids. Fatty acids are a type of carboxylic acids with
long chain of carbon atoms.
Soaps are not suitable for washing clothes when the water is hard:

• The formation of lather is necessary for removing dirt from clothes during the washing of clothes. Soap does not give lather with hard water as it reacts with the
  calcium and magnesium ions present in hard water to form insoluble precipitates of calcium and magnesium salts of fatty acids.
•  The scum (or precipitate) formed by the action of hard water on soap sticks to the clothes being washed and it interferes with the cleaning ability of soap. This makes the cleaning of clothes difficult.

(b) Action of soap in removing an oily spot from a piece of cloth. Most of the dirt particles on skin or clothes are oily in nature. When a dirty cloth is put in water containing dissolved soap, the oil does not dissolve in

water. The soap molecules are sodium or potassium salts of long chain carboxylic acids. The acid end of soap dissolves in water while its carbon chain dissolves in the oil. In this way soap molecules form rounded special type of structures called micelles in which one end of the soap molecule is towards the oil droplet while the acid end is towards the water. This forms an emulsion in water. In this way soap micelles help in removing the dirt particles from the surface and this dirt dissolves in water and the surface gets cleaned and washed.

Question.21 (a) Draw a diagram of the longitudinal section of a flower and label on it sepal, petal, ovary and stigma.
(b) Write the names of male and female reproductive parts of a flower.
Answer.

Question.22 (a) If the image formed by a lens is diminished in size and erect, for all positions of
the object, what type of lens is it?
(b) Name the point on the lens through which a ray of light passes undeviated.
(c) An object is placed perpendicular to the principal axis of a convex lens of focal
length 20 cm. The distance of the object from the lens is 30 cm. Find

1.  the position
2.  the magnification and
3.  the nature of the image formed.

Answer.
(a) The lens is concave.
(b) Optical centre is the point on the lens through which a ray of light passes undeviated.
(c) Convex lens
Focal length, f = + 20 cm Object distance, u = -30 cm
Image distance, v = ? Magnification, m = ?
Nature of the image = ?

Nature. The +ve sign of v shows that the image is formed on the right side of the convex lens, so the image formed is real.

•  The magnification is two (i.e., more than one) so the image is larger than the object.
•  The -ve sign for m shows that the image is formed below the principal axis. Hence the image is inverted.

.^.. Nature of image: Real, inverted and magnified
Or
(a) One-half of a convex lens is covered with a black paper. Will such a lens produce an image of the complete object? Support your answer with a ray diagram.
(b) An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm.

1.  Draw the ray diagram and
2.  Calculate the position and size of the image formed.
3.  What is the nature of the image?

(a) As we can see in the figure given, when the lower half of the convex lens is covered with a black paper, it still forms lens. However the intensity of the image is reduced when the convex lens is covered with black paper.

   (iii)Nature of the image  

• The +ve sign of v shows that image is real.
•  The -ve sign of h₂ shows that image is inverted.
  .^..  Size of image = 3.3 cm

Question.23(a) In a tabular form, differentiate between ethanol and ethanoic acid under the
following heads:

1.  Physical state
2.  Taste
3. NaHCO₃ test
4.  Ester test

(b) Write chemical reactionto show the dehydration of ethonol
Answer. 

(b) Dehydration of ethanol. When ethanol is heated with excess of cone. H₂SO₄ at 170°C, it
gets dehydrated to fern ethene (an unsaturated hydrocarbon).

Question.24 (a) What is fragmentation in organisms? Name a reproduces by this method.
(b) What is regeneration in organisms? Describe regeneration in Planaria with the help of a suitable diagram.
Answer.
(a) Fragmentation. It occurs in multicellular organisms with relatively simple body organisation, for example, in spirogyra (an algae plant) and planaria (a simple multicellular animal). The body bracks up on maturation into two or more smaller pieces and each piece then develops into a new individual.
(b) The process of getting back a full organism from its body parts is called regeneration. The simple animals like hydra and planaria show regeneration.
Explanation. If the bodu planaria gets cut into a number of pices, then each body pice can regenerate into a complete planaria by growing all the missing parts. The regeneration of an organism from its cut body part occurs by the process of growth and development. The cells of cut body part devide rapidluy to make a ball of cells. The cells then become specilised to from diffrent types of tissues which again from various organs and body parts.

SECTION B
Question.25 Following signs are usually shown on the bottles of commercial acetic acid. The symbols indicate respectively that acetic acid is:

(a) Flammable and corrosive  (b) Radioactive and flammable
(c) Oxidizing and corrosive     (d) Flammable and explosive
Answer.(a) Flammable and corrosive

Question.26 To find the focal length of a concave mirror Rahul focuses a distant object with this mirror. The chosen object should be
(a) a tree (b) a building (c) a window (d) the Sun
Answer.(a) a tree

Question.27 For finding the focal length of a convex lens by obtaining the image of a distant object, one should use as the object
(a) a well lit distant tree
(b) window grill in the class room
(c) any distant tree
(d) a lighted candle kept at the other end of the table
Answer.(a) a well lit distant tree

Question.28 Four students A, B, C and D traced the paths of incident ray and the emergent ray by fixing pins P and Q for incident ray and pins R and S for emergent ray for a ray of light passing through a glass slab.

The correct emergent ray was traced by the student:
(a) A (b) B (c) C (d) D
Answer.(b) B

Question.29 To show that zinc is a more active metal than copper, the correct procedure is to
(a) add dilute nitric acid on strips of both the metals.
(b) observe transmission of heat through strips of zinc and copper.
(c) prepare solution of zinc sulphate and hang strip of copper into it.
(d) prepare solution of copper sulphate and hang strip of zinc into it.
Answer.(d) prepare solution of copper sulphate and hang strip of zinc into it.

Question.30 Acetic acid solution turns
(a) blue litmus red (b) red litmus blue
(c) blue litmus colourless (d) red litmus colourless
Answer.(a) blue litmus red

Question.31 A student was given two permanent slides, one of binary fission in amoeba and other of budding in yeast. He was asked to identify any one difference in the nucleus of the two. One such difference, he identified correctly was
(a) Presence of one nucleus in amoeba, two in yeast cell and one in bud.
(b) Presence of two nucei certrally constricted amoeba, one in yeast cell and one in its bud.
(c) Presence of two distant nuclei in amoeba, one in yeast cell and two in bud.
(d) Presence of a single nucelus each in amoeba, yeast cell and its attached bud.
Answer. (b) Presence of two nucei certrally constricted amoeba, one in yeast cell and one in its bud.

Question.32 Raisins are wiped off gently before final weighing with help of
(a) a filter paper (b) a cotton piece (c) a cloth piece (d) a polythene piece
Answer.(a) a filter paper

Question.33 One of the examples of two analogous organs can be the wings of parrot and
(a) flipper of whale (b) foreleg of horse’ (c) front leg of frog (d) wings of housefly
Answer. (d) wings of housefly

Question.34 How will you test in the laboratory, whether the given sample of water is hard or soft?
Name two salts which make the water hard. 
Answer. When few drops of soap solution are added in the given sample of water, if lather is formed with soap then water is soft water. If lather is not formed then water is hard. Calcium chloride and Magnesium sulphate salts present in water make the water hard.

Question.35 Name two animals which show reproduction by regenration method, which one of these also reproduces by budding method?
Answer. Planaria and Hydra reproduce by regeneration.
Hydra also reproduces by the budding method.

Question.36 Which out of the two spherical mirrors, has positive focal length? Which of the two will form real and inverted image at its ‘¥ point, which can be taken on the screen?
Answer. Convex Mirror has positive focal length. Concave mirror forms a highly diminished sized, ‘ real and inverted image of the object at ‘F point which can be taken on the screen.

SET II

Except for the following questions, all the remaining questions have been asked in Set-I.
SECTION A
Question.1 Which compound are responsible for the depletion of ozone layer?
Answer. Chlorofluorocarbons (CFCs) are the chemicals which are responsible for the depletion of ozone layer.

Question.2 Define, ‘trophic level’.
Answer. The various steps in a food chain at Which the transfer of food (or energy) takes place are called trophic levels. In a food chain, each step representing an organism forms a trophic level.

Question.3 Write the electron dot structure of ethene molecule, C₂H₄.
Answer. Electron dot structure of ethene molecule, C₂H₄:

Question.4 The atomic numbers of three elements, X, Y and Z are 9,11 and 17 respectively. Which two of these elements will show similar chemical properties? Why?
Answer.

Element X and Z show similar chemical properties because both have same number of valence electrons, i.e., 7 as chemical properties of the element depends on the valence electrons.

Question.6 List any four modes of asexual reproduction.
Answer. Four modes of asexual reproduction:

1. Fission: (a) Binary fission, (b) Multiple fission
2.  Budding
3. Spore formation
4. Fragmentation

Question.9 Draw a ray diagram to show the reflection of light through triangular glass prism and mark angle of deviation on it.
Answer. Reflection of light through triangular glass prism:

Question.13 A ray of light travelling in air enters obliquely into water. Does the light ray bend towards or away from the normal? Why? Draw a ray diagram to show the refraction of light in this situation.

Answer. When a ray of light travelling in air enters obliquely into water, the light ray will bend towards the normal. Because when a ray of light travels obliquely from rarer medium to denser medium, it will bend towards the normal.

Question.16 Explain Mendel’s view of a dominant trait. Give an example
Answer. According to Mendel’s view of dominant trait, the traits of an organism are determined by internal factors which occur in pairs. Out of two contrasting traits only one expresses itself in an individual. This trait is called dominant, while the other which has not shown its effect in the presence of dominant trait is called recessive.

Example, When true bred tall pea plants are crossed with true bred dwarf plants, the plants that appeared in Fa generation are tall, although they have received a factor (trait) from dwarf plant. So trait of tallness is dominant one. However this recessive hidden character reappeared, unchanged in F₂ generation.

Question.18 What is an ‘esterification’ reaction? Describe an activity to show esterification.
Answer. Carboxylic acid when reacts with alcohols in the presence of concentrated sulphuric acid

SET III

Except for the following questions, all the remaining questions have been asked in Set-I and
Set-11.
SECTION A
Question.1 What are the various steps in a food chain called?
Answer. The various steps of a food chain are called trophic levels. These steps are:

1. First trophic level consists of producers.
2.  Second trophic level consists of primary consumers or herbivores.
3.  Third trophic level consists of secondary consumers or carnivores.
4.  Fourth trophic level consists of tertiary consumers or top carnivores.

Question.2 What is the important function of presence of ozone in earth’s atmosphere?
Answer. Ozone layer absorbs the harmful UV radiations of the sunlight, so this layer is very important for the survival and existence of life on earth.

Question.3 Write the electron dot structure of ethane molecule, C₂H₆.
Answer. Electron dot structure of ethane, C₂H₆:

Question.4 What makes the earth’s atmosphere a heterogeneous mixture?
Answer. Density of various components of the atmosphere is different. Therefore the earth’s atmosphere is a heterogeneous mixture, for example, the heavier gases like CO₂ and O₂ of the atmosphere occupy the lower level (i.e., near the earth’s surface) of the atmosphere whereas lighter gases like O₃ Occupy the higher level of the atmosphere.
So the components of atmosphere are not equally distributed.

Question.8(a) On the basis of electronic configuration, how will you identify the first and the
last element of a period?
(b) The stars appear higher from horizon than they actually are. Explain why it is so?
Answer. (a) The number of valence electrons increases from 1 to 8 on going from left to right in a period. Therefore the first element in every period has 1 valence electron and the last element in every period has 8 valence electrons (except in the first period where last element helium has only 2 valence electrons).
(b) A ray of light coming from the star travels through earth’s atmosphere. As optical density changes continuously, therefore bending of light takes place continuously. This results in a curved path of light rays due to atmospheric refraction. It appears to the observer as if the ray is coming straight from the source (star). Thus the stars appear higher than their actual positions.

Question.16 Out of HCl and CH₃ COOH, which one is a weak acid and why? Describe ah activity to support your answer.
Answer. Dilute ethanoic acid turns universal indicator paper to orange showing that its pH is about 4. This shows that ethanoic acid is a weak acid. On the other hand dil. hydrochloric acid turns universal indicator paper to red, showing that its pH is about 1. This shows that hydrochloric acid is a strong acid.

Question.18

1. “The refractive index of diamond is 2.42”. What is the meaning of this statement?
2.  Name a liquid whose mass density is less than that of water but it is optically denser than water.

Answer.

1.  The refractive index of diamond is 2.42. It means that the ratio of the speed of light in air and the speed of light in diamond is equal to 2.42.
   Higher is the refractive of a medium, lower is the speed of light in that medium. Because the refractive index of diamond is very high. Therefore the speed of light in diamond is very low.
2. Kerosene has the mass density less than water but it is optically denser than water.