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CBSE Previous Year Question Papers Class 10 Science SA2 Delhi 2015

CBSE Previous Year Question Papers Class 10 Science SA2 Delhi 2015

Time allowed: 3 hours                                                                                        Maximum marks: 90
General Instructions:

    1. The Question Paper comprises two Sections, A and B. You are to attempt both the Sections.
    2. All questions are compulsory.
    3. There is no choice in any of the questions.
    4. All questions of Section-A and all questions of Section B are to be attempted separately.
    5. Question numbers 1 to 3 in Section A are one mark questions. These are to be answered in one word or in one sentence.
  1. Question numbers 4 to 6 in Section A are two marks questions. These are to be answered in about 30 words each.
  2. Question numbers 7 to 18 in Section A are three marks questions. These are to be answered in about 56 words each.
  3. Question numbers 19 to 24 in Section A are five marks questions, these are to be answered in about 70 words each.
  4.  Question numbers 25 to 33 in Section B are multiple choice questions based on practical skills. Each question is a one mark question. You are to select one most appropriate response out of the four provided to you.
  5. Question numbers 34 to 36 in Section B are two marks questions based on practical skills. These are to be answered in brief.


Question.1. Write the name and formula of the 2nd member of homologous series having general formula CnH2n.
Answer. CnH2n = Alkene
2nd member = C3H6 (Propene)

Question.2. List two functions performed by the testis in human beings.
Answer. Two functions of testes are:

  1. To produce the male sex cells (gametes) called sperms.
  2. To secrete the male sex hormone called testosterone which is responsible for secondary sexual changes in males.

Question.3. What is the function of ozone in the upper atmosphere?
Answer. Ozone layer is very important for the existence of life on earth. The function of the ozone layer in the upper atmosphere is to absorb most of the harmful ultraviolet radiations coming from the sun and prevent them from reaching the earth’s surface.

Question.4. List four characteristics of the images formed by plane mirrors.
Answer. The characteristics of the images formed by plane mirrors are:

  1. The image formed by a plane mirror is virtual and erect. It cannot be received on a screen.
  2. The image formed by a plane mirror is of the same size as the object.
  3. The image formed by a plane mirror is at the same distance behind the mirror as the object is in front of the mirror.
  4.  The image formed in a plane mirror is laterally inverted.

Question.5. Why are forests considered “biodiversity hot spots”? List two ways in which an individual can contribute effectively to the management of forests and wildlife.
Answer. Due to the presence of a large number of species (of plants and animals) threatened with extinction, forests are said to be ‘biodiversity hotspots’.

The two ways in which an individual can contribute effectively to the management of forests and wildlife are:

  1. As individuals we can try to cut down the use of products directly obtained from forests such as gum, rubber, wood, paper, etc., thereby protecting against the excessive cutting of trees.
  2.  As individuals we must discourage and protect against the poaching of wild animals for commercial use.

Question.6. What is meant by “sustainable management”? Why is reuse considered better than recycling?
Answer. The development and management of resources in such a way which meets the current basic human needs and also preserves the resources for the needs of future generations, is called sustainable management.
The process of ‘reuse’ is considered better than the process of ‘recycling’ because recycling requires the use of a large amount of energy and money whereas no energy is required for reusing materials.

Question.7. With the help of an example, explain the process of hydrogenation. Mention the essential conditions for the reaction and state the change in physical property with the formation of the product.
Answer. Process of hydrogenation:
The addition of hydrogen to an unsaturated hydrocarbon to obtain a saturated hydrocarbon is called hydrogenation.
Essential conditions for the reaction are:

  1. Presence of an unsaturated hydrocarbon.
  2. Presence of a catalyst such as nickel
  3. or palladium.

Changes observed:

  • Change observed in the physical property is the change of unsaturated compound from the liquid state to saturated compound in solid state.
  • The boiling or melting points of a product is increased.

Question.8. What is the difference between the molecules of soaps and detergents, chemically? Explain the cleansing action of soaps.

Question.9. How many groups and periods are there in the modern periodic table? How do the atomic size and metallic character of elements vary as we move:
(i) down a group and (ii) from left to right in a period?
Answer. There are 18 groups and 7 periods in the modem periodic table.
(i) Atomic size and metallic character of the elements increases down a group.

  • Atomic size and metallic character of elements decreases from left to right in a period.
  • Metallic character of the element is decreased.

Question.10. From the following elements:
(i) Select the element having one electron in the outermost shell.
(ii) two elements of the same group.
Write the formula of and mention the nature of the compound formed by the union of 19K and element X (2,8,7).

Question.11. What is DNA copying? State its importance.
Answer. A process in which a DNA molecule produces two similar copies of itself in a reproducing cell through chemical reaction is a called DNA copying.

  1. It transmits the characteristics from parents to the next generation (offspring).
  2. It causes increased the variations in the population.

Question.12. Explain budding in. Hydra with the help of labelled diagrams only.
Answer. See Q. 20, 2011 (I Outside Delhi).

Question.13. List any four contraceptions used by humans. How does their use have a direct effect on ’ the health and prosperity of a family.’
Answer. See Q. 12,2014 (I Delhi).

Question.14. “We cannot pass on to our progeny the experiences and qualifications earned during our life time”. Justify the statement giving reason and examples.
Answer. All traits of the parent cannot be transmitted to the offspring because all traits are not inherited traits, some of the characteristics are acquired. The experiences and qualifications earned during our lifetime are examples of acquired traits.
A trait of an organism which is ‘not inherited’ but develops in response to the environment is called an acquired trait. The acquired traits of organisms cannot be passed on to their future generations.
Only those traits can be transmitted to future generations in which changes have occurred in the genes present in the reproductive cells of parent organisms.
For example, If we breed some mice, all the progeny of mice will have tails, just like parents. Now, if we cut the tails of these first generation mice surgically and breed them, we will get new mice, all with full tails. Despite the cutting of tails of mice for a number of generations, a tailless mouse is never born.
This is because the cut tail of mice is an acquired trait which is never passed on to their progeny. This is because cutting the tails of mice does not Change the genes of their reproductive cells.

Question.15. (a) Planaria, insects, octopus arid vertebrates all have eyes. Can we group eyes of these animals together to establish a common evolutionary origin? Justify your answer, (b) “Birds have evolved from reptiles”. State evidence to prove the statement.
Answer. (a) The eyes seen in planaria, insects, octopus and vertebrates vary greatly in their structure. These Organisms can be used for studying evolution of eyes as the eyes of planaria are simple without lens, insects have compound eyes and vertebrates have highly specialised eyes. However, all of them perform the same function, that is, vision. Thus, a common evolutionary origin can be established.
(b) Birds have evolved from reptiles as the connecting link between reptiles and birds is Archaeopteryx (flying dinosaur). Also, there are some similarities in birds and reptiles. Birds have a four-chambered heart/ which is also a feature of some reptiles. Both birds and reptiles have separate sexes and internal fertilization occurs in both.

Question.16. To construct a ray diagram we use two rays of light which are so chosen that it is easy to determine their directions after reflection from the mirror. Choose these two rays and state the path of these rays after reflection from a concave mirror. Use these two rays to find the nature and position of the image of an object placed at a distance of 15 cm from a concave mirror of focal length 10 cm.
Answer. Ray 1. When an incident ray of light is parallel to the principal axis of a concave mirror, its reflected ray must pass through the principal focus of the concave mirror.
Ray 2. A ray passing through the principal focus of a concave mirror after reflection will emerge parallel to the principal axis.
• Focal length = 10 cm; Then centre of curvature, C = 20 cm
. Object, is placed at 15 cm, i.e., between F & C
when the object is between F and C (centre of curvature):
The image formed is real, inverted and magnified.It is formed beyond C.

Question.17. With the help of a labelled diagram, explain why the sun appears reddish at the sunrise and the sunset.
Answer. At the time of sunrise and sunset when the sun is near the horizon, the sunlight has to travel the greatest distance through the atmosphere to reach us. During the long journey of sunlight, most of the shorter wavelength blue-colour present in it is scattered out and away from our line of sight so, the light reaching us directly from the rising sun or setting sun consists mainly of longer wavelength red colour due to which the sun appears red. Due to the same reason, the sky surrounding the rising sun and setting sun also appears red. Thus, at sunrise and sunrise, the sun itself as well as the surrounding sky appear red.

Question.18. After the examinations Rakesh with his friends went on a picnic to a nearby park. All friends carried cooked food packed in plastic bags or plastic cans. After eating the food some friends collected the leftover food and plastic bags etc. and planned to dispose them off by burning. Rakesh immediately checked them and suggested to segregate the leftover food and peels of fruits from the plastic materials and respectively dispose them off separately in the green and red dustbins placed in the comer of the park.
(a) In your opinion, is burning plastic an eco-friendly method of waste disposal? Why? State the advantage of the method suggested by Rakesh.
(b) How can we contribute in maintaining the parks and roads neat and clean?
Answer. (i) Burning of plastic is not an eco-friendly method of waste disposal because it
produces toxic gases which cause too much air pollution. It has an adverse effect on the health of all types of living organisms including human beings. –
The method of waste disposal Used by Rakesh is advantageous as the leftover food and peels of fruits are biodegradable and can be used as manure.
The plastic bags and cans should be disposed off in red dustbins from where they can be sent for recycling, thereby keeping the environment clean.
(ii) • We should not litter on roads and parks.
• We should not spit on roads and parks.
• We should make it a point to throw the garbage in dustbins and learn to segregate garbage into biodegradable and non-biodegradable.

Question.19. Explain why carbon forms compounds mainly by covalent bond. Explain in brief two main reasons for carbon forming a large number of compounds. Why does carbon form strong bonds with most other elements?
Answer. The atomic number of carbon is 6 which means that a neutral atom of carbon retains 6
electrons. So, the electronic configuration of carbon is K, L. Since, carbon has 4 electrons in its outermost shell so, it should either lose or gain 4 electrons to achieve the inert gas configuration and become stable.

  1. It could gain four electrons forming C4-anion. But it would be difficult for the nucleus with six protons to hold on to ten electrons due to inter electronic repulsion.
  2. It could lose 4 electrons forming C4+ cation. But it would require a large amount of energy to remove four electrons from its outermost shell.
    Thus, it forms compounds mainly by covalent bonds.

Two properties of carbon which lead to huge number of carbon compounds are:

  1. Catenation. Catenation is the unique property of carbon atoms to form bonds with other atoms of carbon giving rise to large molecules.
  2. Tetravalency. Since carbon has a valency of 4, it is capable of bonding with 4 other atoms of carbon or atoms of some other monovalent element.

The reason for the formation of strong bonds by the carbon atoms is their small atomic size. Due to the small size of carbon atoms their nuclei hold the shared pairs of electrons between atoms strongly, leading to the formation of strong covalent bonds. The carbon atoms also form strong covalent bonds with the atoms of other elements such as hydrogen, oxygen, nitrogen, sulphur, chlorine and other elements.

Question.20. Write the functions of the following in human female reproductive system:
Ovary, oviduct, uterus
How does the embryo get nourishment inside the mother’s body? Explain in brief.
Answer. Functions of:

  • Ovary—Ovaries are the primary reproductive organs in a woman (or female). The function of ovaries is to produce mature female sex cells called ‘ova’ or ‘eggs’ and also to secrete the female sex hormones.
  • Oviduct—The ovum (or egg cell) released by an ovary goes into the oviduct through its funnel shaped opening. The fertilisation of egg by a sperm takes place in the oviduct.
  • Uterus—The growth and development of a fertilised ovum (or fertilised egg) into a baby takes place in the uterus.

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 finger-like projections called 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, the embryo gets its nutrition.

Question.21. How many pairs of chromosomes are present in human beings? Out of these how many are sex chromosomes? How many types of sex chromosomes are found in human beings? “The sex of a newborn child is a matter of chance and none of the parents may be considered responsible for it”. Draw a flow chart showing determination of sex of a newborn to justify this statement.

  • There are 23 pairs of chromosomes present in human beings.
  • There is 1 pair of sex chromosomes present in human beings.
  •  The chromosomes which determine the sex of a person are called sex chromosomes. There are two types of sex chromosomes, one is called X chromosome and the other is called Y chromosome. Males contain one X chromosome and one Y chromosome (XY), while females contain two X chromosomes (XX).
  • Flow chart showing determination of sex of a child. See Q. 14, 2014 (II Outside Delhi).

Question.22. (a) State the laws of refraction of light. Explain the term absolute refractive index of a medium and write an expression to relate it with the speed of light in vacuum. (b) The absolute refractive indices of two media ‘A’ and ‘B’ are 2.0 and 1.5 respectively. If the speed of light in medium ‘B’ is 2 x 108 m/s, calculate the speed of light in: (i) vacuum, (ii) medium ‘A’.
Answer. (a) Laws of Refraction:
(i) The first law of refraction of light states that the incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.
(ii) The second law of refraction of light is the Snell’s Law of Refraction. It states that the ratio of sine of the angle of incidence to the sine of angle of refraction is a constant for a given pair of medium.

Question.23. “A convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it.” Draw ray diagram to justify this statement stating the position of the object with respect to the lens in each case.
An object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed.
Answer. • A convex lens can form a magnified erect image when the object is placed between the optical centre and principal focus of the convex lens (i.e., between O and F).
• A convex lens can form a magnified inverted image when the object is placed between focus and the centre of curvature (i.e., between F’ and 2F’).

Question.24. A student is unable to see clearly the words written on the blackboard placed at a distance of approximately 4 m from him.
Name the defect of vision the boy is suffering from. Explain the method of correcting this defect. Draw ray diagram for the:
(i) defect of vision and also (ii) for its correction.

  • As the student is unable to see clearly the Words written on the blackboard at a distance of 4 m, the student is suffering from myopia or near-sightedness:
  • This defect is corrected by using spectacles fitted with concave lens of suitable focal length is used. The image is allowed to form at the retina by using a concave lens of suitable power.
    Ray diagrams for the defect and for its corrections. See Q. 10, 2012 (1 Delhi).

Question.25. A student adds 2 mL of acetic acid to a test tube containing 2 mL of distilled water. He then shakes the test tube well and leaves it to settle for some time. After about 5 minutes he observes that in the test tube there is:
(A) A clear transparent colourless solution
(B) A clear transparent pink solution
(C) A precipitate settling at the bottom of the test tube
(D) A layer of water over the layer of acetic acid
Answer. (A)

Question.26. A student prepared 20% sodium hydroxide solution in a beaker to study saponification reaction. Some observations related to this are given below:
(I) Sodium hydroxide solution turns red litmus blue
(II) Sodium hydroxide readily dissolves in water ,
(III) The beaker containing solution appears cold when touched from .outside
(IV) The blue litmus paper turns red when dipped, into the solution
The correct observations are:
(A) I, II and IV (B) I, II and III
(C) only-III and IV (D) only I and II
Answer. (D)

Question.27. Hard water is not available for an experiment. Some salts are: given below:
(I) Sodium chloride (II) Sodium sulphate
(III) Calcium chloride (IV) Calcium sulphate
(V) Potassium chloride (VI) Magnesium sulphate
Select from the following a group of these salts, each member of which may be dissolved in water to make it hard.
(A) I, 11 and V (B) I, III, V
(C) III, IV, VI . (D) II, IV, VI
Answer. (C)

Question.28. A student identified the various parts of an embryo of a gram seed and listed them as given below:
(I) Testa (II) Plumule
(III) Radicle (IV) Cotyledon
(V) Tegmen
Out of these the actual parts of the embryo are:
(A) I, II, III (B) II, III, IV
(C) Ill, IV, V (D) II, IV, V
Answer. (B)

Question.29. Four students A, B, C and D reported the following set of organs to be homologous. Who is correct?
(A) Wings of a bat and a butterfly
(B) Wings of a pigeon and a bat
(C) Wings of a pigeon and a butterfly
(D) Forelimbs of cow, a duck and a lizard
Answer. (D)

Question.30. Study the given diagram and select the correct statement about the device ‘X’:
(A) Device ‘X’ is a concave mirror of radius of curvature 12 cm
(B) Device ‘X’ is a concave mirror of focal length 6. cm
(C) Device ‘X’ is a concave mirror of radius of focal length 12 cm
(D) Device ‘X’ is a convex mirror of focal length 12 cm
Answer. (C)

Question.31. A student has obtained a point image of a distant object using the given convex lens. To find the focal length of the lens he should measure the distance between the :
(A) lens and the object only
(B) lens,and the screen only
(C) object and the image only
(D) lens and the object and also between the object and the image
Answer. (B)

Question.32. Four students P, Q, R and S traced the path of a ray of light passing through a glass slab for an angle of incidence 40° and measured the angle of refraction. The values, as measured by them were 18°; 22°; 25° arid 30° respectively. The student who has performed the experiment methodically is
(A) P (B) Q (C) R (D) S
Answer. (C)

Question.33. After tracing the path of a ray of light through a glass prism a student marked the angle incidence (∠i), angle of refraction (∠r), angle of emergence (∠e) and the angle of deviation (∠D) as shown in the diagram. The correctly marked angles are:
(A) ∠i and ∠r (B) ∠i and ∠e
(C) ∠i, ∠e and ∠D (D) ∠i, ∠r and ∠e
Answer. (B)

Question.34. List two observations which you make when you add a pinch of sodium hydrogen carbonate to acetic acid in a test tube. Write the chemical equation for the reaction that occurs.
Answer. When a pinch of sodium hydrogen carbonate is added to acetic acid in a test tube, there are two observations:
(i) Brisk effervescence .
(ii) Evolution of a colourless and odourless gas which is C02.

Question.35. Name the type of asexual reproduction in which two individuals are formed from a single parent and the parental identity is lost.
Draw the initial and the final stages of this type of reproduction. State the event with which this reproduction starts.

  • Binary fission is a type of asexual reproduction in which two individuals are formed from a single parent and the parental identity is lost.
  • This reproduction starts with the elongation of the nucleus.
  • Stages of Binary fission. See Q. 20(Or), 2011 (I Outside Delhi).

Question.36. To find the image-distance for varying object-distances in case of a convex lens, a student obtains on a screen a sharp image of a bright object placed very far from the lens. After that he gradually moves the object towards the lens and each time focuses its image on the screen.
(a) In which direction-towards or away from the lens, does he move the screen to focus the object?
(b) What happens to the size of image-does it increase or decrease?
(c) What happen when he moves the object very close to the lens?
Answer. (a) He moves the screen away from lens to focus the object.
(b) The size of the image increases.
(c) When the object is placed very close to the lens, then no image will be formed on the screen.


Except for the following questions, all the remaining questions have been asked in Set-I.
Question.1. Write the name and formula of the 2nd member of homologous series having general formula CnH2n+2.
Answer. CnH2n+2: Alkane series Name: Ethane (2nd member) Formula: C2H6

Question.2. What is the magnification of the images formed by plane mirrors and why?
Answer. The magnification of the image formed by a plane mirror is 1 because size of the image is equal to the size of the object.

Question.3. What is meant by power of a lens?
Answer. The power of a lens is a measure of the degree of convergence or divergence of light rays falling on it. The power of a lens is defined as the reciprocal of its focal length in metres.
The unit of power is dioptre.

Question.4. Write two differences between binary fission and multiple fission in a tabular form.
Answer. See Q. 4, 2011 (I Delhi).

Question.5. (a) Why do we need to manage our resources carefully?
(b) Why management of natural resources requires a long term perspective?
Answer. (a) We need to manage our natural resources because of the following reasons:
The resources of the earth are limited. As the human population is increasing rapidly, the demand for resources increases day by day. The proper management can ensure that these resources last for the generations to come.
(b) The proper management of natural resources takes into consideration a long-term perspective so that these natural resources are used judiciously at present and their exploitation to the hilt for short term gains is prevented. This is essential to meet their future needs and demands.

Question.6. List four measures that can be taken to conserve forests.
Answer. Forests can be conserved in the following ways:

  1. By silviculture—It means reforesting those forest lands from where large number of trees have been cut in a planned way.
  2. By taking the help of the local people in conserving the forests, e.g., Sal forests of Arabari forest range of West Bengal were conserved with the help of the local people by the Forest Department of that state.
  3. Encouraging the people not to overuse the materials obtained from the forests, e.g., timber.
  4. Indiscriminate felling of trees for commercial gains should be stopped.
  5. Overgrazing in forests should be discouraged and forest fires should be prevented.

Question.7. Na, Mg and Al are the elements of the same period of Modem Periodic Table having one, two and three valence electrons respectively. Which of these elements (i) has the largest atomic radius, (ii) is least reactive? Justify your answer stating reason for each case.
Answer. Valence Electrons
(i) Na has the largest atomic radius because on moving from left to right in the periodic table, lb . ‘omic radius decreases due to increase in positive charge on the nucleus which pulls the outermost electrons in more close to the nucleus and the size of atom decreases.
(ii) Al is least reactive because on moving from left to right in the periodic table the nuclear charge increases and the valence electrons are pulled in more close to the nucleus.

Question.8. From the following elements:
(i) Select the element having one electron in the outermost shell.
(ii) two elements of the same group.
Write the formula of and mention the nature of the compound formed by the union of 19K and element X(2,8,7).

Question.9. What is meant by isomers? Draw the structures of two isomers of butane, C4H10. Explain why we cannot have isomers of first three members of alkane series.

  • The organic compounds having the same molecular formula but different structures are known as isomers.
  •  Isomers of C4H10:
  •  Isomers of first three members, i.e., methane, ethane and propane are not possible because they are too short and cannot be branched.

Question.11. What are sexually transmitted diseases? List two examples of each diseases caused due to (i) bacterial infection and (ii) viral infection. Which device may be used to prevent the spread of such diseases?

  • Diseases which are transmitted from an infected person to a healthy person due to ‘ ‘ sexual contact are called sexually transmitted diseases or STD.
  • Examples of STD caused by:
    Bacterial infection—Gonorrhoea and Syphilis Viral infection—AIDS and Herpes
  • Use of condoms can prevent the spread of such diseases.

Question.19. The image of a candle flame placed at a distance of 30 cm from a spherical lens is formed on a screen placed on the other side of the lens at a distance of 60 cm from the optical centre of the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 3 cm, find the height of its imaged
Answer. Since the image is formed on the screen, the image is real. A concave lens cannot form a real image. Therefore, the lens is convex.


Except for the following questions, all the remaining questions have been asked in Set-1 and Set-11.
Question.1. Write the name and formula of the 2nd member of homologous series having general formula CnH2n-2.
Answer. CnH2n-2: Alkynes Name: Propane (2nd member) Formula: C3H4

Question.2. What is speciation?
Answer. The process by which new species develop from the existing species by evolution or genetic modification is known as speciation.

Question.3. Why should biodegradable and non-biodegradable wastes be discarded in two separate dustbins?
Answer. Biodegradable and non-biodegradable wastes should be discarded in two separate dustbins so that the time and energy required in segregation later may be saved and the waste may be recycled accordingly. Non-biodegradable waste is recycled into new products.

Question.4. List four specific characteristics of the images of the objects formed by convex mirrors.
Answer. The images of the objects formed by convex mirrors are always—(i) virtual, (ii) erect, (iii) diminished and (iv) formed behind the mirror between focus and pole of the mirror.

Question.5. List two advantages associated with water harvesting at the community level.
Answer. Advantages of water harvesting at the community level:

  1. Water harvesting improves the quality of ground water thereby improving vegetation in and around the area. It also prevents soil erosion.
  2. Water harvesting in rural areas not only increases the agricultural production and income of the farmers but also makes less severe the effect of droughts and floods, and increases the life of downstream dams and reservoirs.

Question.6. Everyone of us can do something to reduce our personal consumption of various natural resources. List four such activities based on 3-R approach.
Answer. See Q. 5, 2013 (II Delhi).

Question.7. Write the name and structural formula of the compound obtained when ethanol is heated at 443 K with excess of cone. H2S04. Also write chemical equation for the reaction stating the role of cone. H2S04 in it.

Question.10. Write the number of periods the Modern Periodic Table has. State the changes in valency and metallic character of elements as we move from left to right in a period. Also state the changes, if any, in the valency and atomic size of elements as we move down a group.
Answer. No. of periods in the modern periodic table is 7.

  • On moving from left to right in the periodic table the valency of the elements first increases from 1 to 4 and then decreases to zero.
  • On moving from left- to right in a period, the metallic character of elements decreases because on moving from left to right in a period, the electro positive character of elements decreases.
  •  All the elements in a group have the same valency.
  • On going down in a group, the atomic size increases because a new shell of electrons is added to the atoms at every step.

Question.12. (a) Name the following:
(i) Thread like non-reproductive structures present in Rhizopus.
(ii) ‘Blobs’ that develop at the tips of the non-reproductive threads in Rhizopus.
(b) Explain how these structures protect themselves and what is the function of the structures released from the ‘blobs’ in Rhizopus.
Answer. (a) (i) Thread like non-reproductive structures present in Rhizopus are Hyphae.
(ii) ‘Blobs’ at the tips of these threads are called Sporangia.
(b) Sporangia contain hundreds of minute spores enclosed in a spore case. When the spore case hurts, the tiny spores are dispersed in air. These spores are the asexual reproductive units which can produce more Rhizopus under suitable conditions. During infavourable conditions, these spores are protected by their thick walls.

Question.13. Name the parts A, B and C shown in the diagram and write their functions.

  1.  A—Stigma. Stigma receives the pollen grains from the anther of stamen (during pollination).
  2. B—Pollen tube. When male gamete falls on the stigma, it moves down to the ovary through the pollen tube for fertilisation.
  3. C — Ovule/germ cell. Ovule contains the female gamete of the plant which fuses with male gamete by the process of fertilisation and forms zygote.

Question.15. List in tabular form, two distinguishing features between the acquired traits and the inherited traits with one example of each.
Answer. See Q. 17, 2012 (I Outside Delhi).

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