CBSE Previous Year Question Papers Class 10 Science SA2 Delhi 2016
- The Question Paper comprises two Sections, A and B. You are to attempt both the Sections.
- All questions are compulsory.
- There is no choice in any of the questions.
- All questions of Section-A and all questions of Section B are to be attempted separately.
- Question numbers 1 to 3 in Section A are one mark questions. These are to be answered in one word or in one sentence.
- Question numbers 4 to 6 in Section A are two marks questions. These are to be answered in about 30 words each.
- Question numbers 7 to 18 in Section A are three marks questions. These are to be answered in about 56 words each.
- Question numbers 19 to 24 in Section A are five marks questions, these are to be answered in about 70 words each.
- 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.
- 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 next homologue of each of the following:
(i)C2H4 (ii) C4H6
Answer. (i) Next homologue of C2H4 is C3H6.
(ii) Next homologue of C4H6 is C5H8.
Question.2. Name the part of Bryophyllum where the buds are produced for vegetative propagation.
Answer. The leaves produce buds in Bryophyllum which are used for vegetative propagation.
Question.3. List two natural ecosystems.
Answer. Two natural ecosystems are: Lake and Pond
Question.4. State two positions in which a concave mirror produces a magnified image of a given object. List two differences between the two images.
Answer. The two positions are:
- When an object is placed between the pole (p) and focus (f) of a concave mirror, the image formed is larger than the object (or magnified).
- When an object is placed between the focus (f) and centre of curvature (c) of a concave mirror, the image formed is larger than the object.
Question.5. List four advantages of properly managed watershed management.
Answer. Four advantages of properly managed watershed management are:
- Watershed management emphasizes scientific soil and water conservation in order to increase the biomass production. It strictly maintains the water quality.
- Watershed management develops primary resources of land and water, to produce secondary resources of plants and animals for use in a manner which will not cause ecological imbalance.
- Watershed management increases the production and income of the watershed community.
- It mitigates droughts and floods and increases the life of the downstream dam and reservoirs.
Question.6. Explain giving example where active involvement of local people lead to efficient management of forest.
- People’s participation in the management of forests can help in increasing forest produce as well as in their conservation. ‘
- In 1972, the West Bengal Forest Department formulated a novel scheme to revive the degraded Sal forests by involving the local people.
- A far sighted forest officer A.K. Banerjee involved the villagers of the area around the forest in the protection of 1272 hectares of badly degraded Sal forest.
- In return for help in protecting the forest, the villagers were given employment in both silviculture and harvesting operations of the forest, 25% of the final harvest produce, and were allowed to collect firewood and fodder from the forest area on nominal payment.
- With the active and willing participation of local people living around the forest, the degraded Sal forest of Arabari became thick and green within ten years.
Question.7. What are covalent compounds? Why are they different from ionic compounds? List their three characteristic properties.
- The chemical bond formed by the sharing of electrons between two atoms is known as a covalent bond. The molecules formed by sharing of electrons between two or more same atoms or between two or more non-metals are called covalent compounds.
- Covalent compounds are different from ionic compounds as ionic compounds are formed by transference of electrons.
Characteristics of covalent compounds:
- Covalent compounds usually have low melting and boiling points as they are formed by electrically neutral molecules. So, the force of attraction between the molecules of covalent compounds is very weak. Only a small amount of heat energy is required to break these forces.
- Covalent compounds are usually insoluble in water but they are soluble in organic solvents.
- Covalent compounds do not conduct electricity as they do not contain ions.
Question.8. When ethanol reacts with ethanoic acid in the presence of cone. H2SO4 a substance with fruity smell is produced. Answer the following:
- State the class of compounds to which the fruity smelling compounds belong. Write the chemical equation for the reaction and write the chemical name of the product formed.
- State the role of cone. H2SO4 in this reaction.
- Esters are the fruity smelling compounds.
- Esterification takes place in the presence of catalyst concentrated H2SO4. It acts as a dehydrating agent, i.e., helps in the removal of water formed in the reaction between alcohol and carboxylic acid.
Question.9. Calcium is an element with atomic number 20. Stating reason answer each of the following questions:
- Is calcium a metal or non-metal?
- Will its atomic radius be larger or smaller than that of potassium with atomic number 19?
- Write the formula of its oxide.
Answer. Calcium (Ca)
Atomic number = 20 Electronic configuration = 2, 8, 8, 2
- It is a metal as it has two electrons in its outermost shell and it loses these two electrons to acquire inert gas configuration and form +2 ion.
- Potassium (atomic number 19) is placed before Calcium (atomic number 20) in the same period (IV period). Therefore, size of Ca is smaller than K since on moving from left to right in a period of the periodic table, the size of the atoms decreases. As on moving from left to right in a period, the atomic number of the elements increases which means that the number of protons and electrons in the atoms increases (the extra electrons being added to the same shell). Due to large positive charge on the nucleus, the outermost shell is pulled with more force by the nucleus, it moves closer to the nucleus and the size of the atom decreases.
- The oxide of calcium is calcium oxide (CaO).
Question.10. An element ‘M’ with electronic configuration (2, 8, 2) combines separately with (N03)–, (S04)2- and (P04)3- radicals. Write the formula of the three compounds so formed. To which group and period of the Modem Periodic Table does the elements ‘M’ belong? Will ‘M’ form covalent or ionic compounds? Give reason to justify your answer.
- The electronic configuration (2, 8, 2) of the element ‘M’ suggests that it belongs to group 2 and period 3 of the Modern Periodic Table and its valency is 2.
- The chemical formulae of the compounds formed are:
(i) M(N03)2 (ii) MS04 (iii) M3(P04)2
- ‘M’ will form ionic compounds by losing two valence electrons to achieve a noble gas configuration, that is, a stable octet in the valence shell.
Question.11. How do organisms, whether reproduced asexually or sexually maintain a constant chromosome number through several generations? Explain with the help of suitable example.
- When organisms reproduce asexually, a basic event in reproduction is the creation of a DNA copy. Cells use chemical reactions to build copies of their DNA. This creates two copies of the DNA in a reproducing cell, and they will need to be separated from each other. However, keeping one copy of DNA in the original cell and simply pushing the other one out would not work, because the copy pushed out would not have any organised cellular structure for maintaining life processes. Therefore, DNA copying is accompanied by the creation of an additional cellular apparatus, and then DNA copies separate, each with its own cellular apparatus. Effectively, a cell divides to give rise to two cells. Thus, chromosome number remains unchanged. For example, reproduction of amoeba by binary fission.
- In sexual reproduction, organisms produce gametes through a special type of division, meiosis —reductional division, in which the original number of chromosomes becomes half. These two gametes then combine to form the zygote and the original number of chromosomes is restored as in the case of human beings.
Question.12. Name the parts A, B and C shown in the following diagram and state one function of each.
Answer. A—> Anther. It produces pollen grains which are the male gametes.
B —> Style. It provides the path through which the pollen tube grows and reaches the ovary.
C—> Ovary. It contains ovules and each ovule has an egg cell/female gamete. It develops into fruit after fertilization.
Question.13. Suggest three contraceptive methods to control the size of human population which is essential for the health and prosperity of a country. State the basic principle involved in each.
Answer. There are three different methods of contraception:
- Barrier methods. In these methods, physical devices such as condoms, diaphragms and cervical caps are used. These devices prevent the entry of sperm in the female genital tract, thus acting as a barrier between them.
- Surgical methods. There are surgeries that can be carried out in males and females. In males, a small portion of the sperm duct (vas deferens) is blocked by a surgical operation. This prevents the sperms from coming out.
In females, a small portion of the fallopian tubes (oviducts) is blocked by a surgical operation. It prevents the egg from reaching the uterus. In both the cases, fertilisation will not take place.
- Chemical methods. This category of contraceptives acts by changing the hormonal balance of the body so that eggs are not released and fertilisation cannot occur. Females use two types of pills for preventing pregnancies, i.e., oral pills and vaginal pills. The oral pills contain hormones which stop the ovaries from releasing ovum into the fallopian tube. This is also called oral contraceptives (OC). Other contraceptive devices such as loop or the copper-T are placed in the uterus to prevent pregnancy.
Question.14. In one of his experiments with pea plants Mendel observed that when a pure tall pea plant is crossed with a pure dwarf pea plant, in the first generation, Fa only tall plants appear.
- What happens to the traits of the dwarf plants in this case?
- When the F1 generation plants were self-fertilised, he observed that in the plants
of second generation, F2 both tall plants and dwarf plants were present. Why it happened? Explain briefly.
- Mendel first crossed pure-bred tall pea plants with pure-bred dwarf pea plants and found that only tall pea plants were produced in the first generation or F1 generation. No dwarf pea plants were obtained in the first generation of progeny. From this Mendel concluded that the F1 generation showed the traits of only one of the parent plants: tallness being the dominant trait. The trait of other parent plant, dwarfness, being recessive did not show up in the progeny of first generation.
- In the F2 generation, both the tall and dwarf traits are present in the ratio of 3 :1. This showed that the traits for tallness and dwarfness are present in the F1 generation, being the recessive trait does not express itself in the presence of tallness, the dominant trait.
Question.15. List three distinguishing features, in tabular form, between acquired traits and the inherited traits.
Question.16. Draw the following diagram, in which a ray of light is incident on a concave/convex mirror, on your answer sheet. Show the path of this ray, after reflection, in each case.
Question.17. Why does the sun appear reddish early in the morning? Will this phenomenon be observed by an observer on the moon? Justify your answer with a reason.
Answer. The sun at sunrise (early in the morning) is located near the horizon of the earth. Light from the sun near the horizon has to pass through thick layers of air and a large distance through the earth’s atmosphere before reaching our eyes. Near the horizon, most of the blue light rays with shorter wavelength are scattered away by the particles in the atmosphere. Therefore, the light that reaches our eyes is the red light of longer wavelengths. This gives rise to the reddish appearance of the sun.
This phenomenon will not be observed by an observer on the moon because there is no atmosphere on the moon to. scatter light.
Question.18. Give reason to justify the following:
- The existence of decomposers is essential in a biosphere.
- Flow of energy in a food chain is unidirectional.
- Decomposers break down complex organic substances (dead remains and waste products of organisms) into simpler inorganic substances that can be absorbed by the plants. They are essential for the proper functioning of an ecosystem.
• Decomposers play an important role in the cycling of materials in the biosphere.
• By decomposing dead bodies of plants and animals they help irt cleaning the environment.
• They replenish the soil naturally.
- In the ecosystem energy flows from one trophic level to the next trophic level of the food chain. Energy flows from producers, i.e., green plants to the consumers. It does not flow from the last consumers to its previous consumers and so on. The energy captured by the autotrophs does not go back to the solar input. Thus the energy does not flow back from consumers to the producers. Hence the flow of energy in a food chain is unidirectional.
Question.19. (a) Give a chemical test to distinguish between saturated and unsaturated hydrocarbon.
(b) Name the products formed when ethane burns in air. Write the balanced chemical equation for the reaction showing the types of energies liberated.
(c) Why is reaction between methane and chlorine in the presence of sunlight considered a substitution reaction?
Answer. (a) Bromine water test. The addition of bromine (Br2 ) gives addition reactions with unsaturated compounds (like alkenes and alkynes). The addition of bromine is used as a test for unsaturated compounds. For this purpose, bromine is used in the form of bromine water. A solution of bromine in water is called bromine water. Bromine water has a red-brown colour due to the presence of bromine in it. When bromine water is added to an unsaturated compound, then bromine gets added to the unsaturated compound and the red-brown colour of bromine water is discharged. So, if an organic compound decolourises bromine water, then it will be an unsaturated hydrocarbon (containing a double bond or a triple bond), but saturated hydrocarbon (alkanes) do not decolourise bromine water.
(b) When ethane bums in air, carbon dioxide and water vapours are formed along with heat and light.
(c) Methane reacts with chlorine in the presence of sunlight to form chloromethane and hydrogen chloride.
In this reaction, one ‘H’ atom of methane has been substituted (replaced) by a ‘Cl’ atom, converting CH4 to CH3Cl. Hence, it is considered a substitution reaction.
Question.20. (a) Write the functions of the following parts in human female reproductive system: (i) Ovary (ii) Oviduct (iii) Uterus
(b) Describe the structure and function of placenta.
Answer. (a) (i) Ovary. The ovaries produce thousands of eggs in the female body.
• It produces female hormones called estrogen.
• It produces female gamete ova.
• Transfer of female gamete (egg) from the ovary to the uterus takes place through the oviduct (or fallopian tube).
• The fertilisation of egg (or ovum) by a sperm takes place in the oviduct.
(iii) Uterus. The growth and development of a fertilised ovum into a baby takes place in the uterus.
(b) Structure of placenta. Placenta is the link between the mother’s body and the baby (embryo). It is a disc like structure embedded in the uterine wall connected to the embryo. It has villi on the embryo’s side of the tissue and on the mother’s side, it has blood spaces which surround the villi.
Function of placenta. It provides a large surface area for nutrients (glucose) and oxygen to pass from the mother’s side to the embryo and waste substances from the embryo’s side to the mother’s blood.
Question.21. What is meant by speciation? List four factors that could lead to speciation. Which of these cannot be a major factor in the speciation of a self-pollinating plant species. Give reason to justify your answer.
Answer. Speciation. The process by which new species develop from the existing species by evolution or any genetic modification of previous 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.
- Drastic change in the genes or DNA called mutation is also a cause of speciation. Reproductive isolation can not be a major factor for the speciation of a self-pollinating plant species as it does not depend on any other plant for its reproduction process.
Question.22. (a) Define the following terms in the context of spherical mirrors:
(i) Pole (ii) Centre of curvature
(iii) Principal axis (iv) Principal focus
(b) Draw ray diagrams to show the principal focus of a:
(i) Concave mirror (ii) Convex mirror
(c) Consider the following diagram in which M is a mirror and P is an object and Q is its magnified image formed by the mirror.
State the type of the mirror M and one characteristic property of the image Q.
(i) Pole. The middle point of the reflecting surface of a spherical mirror is called letter P represents pole, MP = M’P.
(ii) Centre of curvature. It is the centre of the sphere of glass of which the mirror is a part,The letter C represents the centre of curvature.
(iii) Principal axis of a spherical mirror is the straight line joining the centre of and pole of the mirror.
(iv) Principal focus. The mid-point of CP is called focus (F). It is the point on the axis of a spherical mirror where all incident rays parallel to the Principal axis meet or appear to diverge after reflection.
(b) Ray diagrams of principal focus:
(i) Concave mirror. In a concave mirror the reflected rays of incident rays parallel to principal axis actually pass through the focus concave mirror has a real principal focus.
(ii) Convex mirror. In a convex mirror the reflected rays do not actually pass through the focus (F). Thus, a convex mirror has a virtual principal focus situated behind the mirror.
(c) The mirror used in the given diagram is a concave spherical mirror. Image formed (Q) is virtual and magnified.
Question.23. (a) Draw a ray diagram to show the formation of image by a convex lens when an object is placed in front of the lens between its optical centre and principal focus.
(b) In the above ray diagram mark the object-distance (u) and the image-distance (ν) with their proper signs (+ve or -ve as per the new Cartesian sign convention) and state how these distances are related to the focal length if) of the convex lens in this case.
(c) Find the power of a convex lens which forms a real, and inverted image of magnification -1 of an object placed at a distance of 20 cm from its optical centre.
Answer. (a) When the object is placed in front of a convex lens between its optical centre and principal focus (i.e., between O and F’): –
The image formed is (i) behind the object (on the same date)
(b) —u (OB)
as shown in the above diagram.
Question.24. (a) Write the function of each of the following parts of human eye:
cornea; iris; crystalline lens; ciliary muscles (b) Millions of people of the developing countries of the world are suffering from corneal blindness. These persons can be cured by replacing the defective cornea with the cornea of a donated eye. A charitable society of your city has organised a campaign in your neighbourhood in order to create awareness about this fact. If you are asked to participate in this mission how would you contribute in this noble cause?
(i) State the objective of organising such campaigns.
(ii) List two arguments which you would give to motivate the people to donate their eyes after death.
(iii) List two values which are developed in the persons who actively participate and contribute in such programmes.
Answer. (a) Functions of the following parts of human eye:
- Cornea. The front part of the eye is called cornea. It is made up of a transparent substance. The light coming from objects enters the eye through cornea.
- Iris. This is a flat, coloured, ring-shaped membrane behind the cornea. Pupil is a hole in the middle of the iris. Iris controls the size of the pupil.
- Crystalline lens. Eye lens is a convex lens which focuses the image of the object on the retina.
- Ciliary muscles. Ciliary muscles hold the eye lens and changes the thickness of eye-lens while focussing.
(b) (i) Objective of such campaigns. To make people aware of corneal blindness and make them realise their duties towards the society by taking pledge for eye donation.
(ii) • One pair of eyes can give eyesight to two corneal blind persons (each getting one eye), and make them see this beautiful world.
• Our eyes can live even after our death. People belonging to all age groups, even people with medical conditions like cataract, diabetes, hypertension can donate their eyes.
(iii) Values developed in persons who actively participate in such programmes:
• Concern for others and social welfare.
• Responsible behaviour and awareness.
Question.25. Which of the following sets of materials can be used for conducting a saponification reaction for the preparation of soap?
(a) Ca(OH)2 and neem oil (b) NaOH and neem oil
(c) NaOH and mineral oil (d) Ca(OH)2 and mineral oil
Question.26. A student takes four test tubes marked P, Q, R and S of 25 mL capacity and fills 10 mL of distilled water in each. He dissolves one spoon full of four different salts in each as — KCl in P, NaCl in Q, CaCl2 in R and MgCl2 in S. He then adds about 2 mL of a sample of soap solution to each of the above test tubes. On shaking the contents of each of the test tubes, he is likely to observe a good amount of lather (foam) in the test tubes marked:
(a) P and Q (b) R and S (c) P, Q and R (d) P, Q and S
Question.27. Consider the following comments about saponification reactions:
(I) Heat is evolved in these reactions. ‘
(II) For quick precipitation of soap sodium chloride is added to the reaction mixture.
(III) Saponification reactions are special kind of neutralisation reactions.
(IV) Soaps are basic salts of long chain fatty acids.
The correct comments are:
(a) I, II and III (b) II, III and IV
(c) I, II and IV (d) Only 1 and IV
Question.28. A student has to perform the experiment “To identify the different parts of an embryo of a dicot seed.” Select from the following an appropriate group of seeds:
(a) pea, gram, wheat (b) red kidney bean, maize, gram
(c) maize, wheat, red kidney bean (c) red kidney bean, pea, gram
Question.29. Which of the following is a correct set of homologous organs?
(a) Forelimbs of frog, bird and lizard (b) Spine of cactus and thorn of bougainvillea
(c) Wings of bat and wings of butterfly (d) Wings of a bird and wings of a bat
Question.30. A student obtained a sharp image of a candle flame placed at the distant end of the laboratory table on a screen using a concave mirror to determine its focal length. The teacher suggested him to focus a distant building about 1 km far from the laboratory, for getting more correct value of the focal length. In order to focus the distant building on the same screen the student should slightly move the
(a) mirror away from the screen (b) screen away from the mirror
(c) screen towards the mirror (d) screen towards the building
Question.31. To determine the approximate focal length of the given convex lens by focussing a distant object (say, a sign board), you try to focus the image of the object on a screen. The image you obtain on the screen is always:
(a) erect and laterally inverted (b) erect and diminished
(c) inverted and diminished (d) virtual, inverted and diminished
Question.32. Select from the following the best experimental set-up for tracing the path of a ray of light passing through a rectangular glass slab:
(a) P (b) Q (c) R (d) S
Question.33. Study the following figure in which a student has marked the angle of incidence (∠i), angle of refraction (∠r), angle of emergence (∠e), angle of prism (∠A) and the angle of deviation (∠D).The correctly marked angles are:
Question.34. What do you observe when you drop a few drops of acetic acid Pi a test tube containing:
- distilled water
- universal indicator
- sodium hydrogen carbonate powder
Answer. Action of acetic acid on
- phenolphthalein —> no change/ remains colourless
- distilled water —> no change, acetic acid dissolves in distilled water.
- universal indicator —> Turns orange
- sodium hydrogen carbonate powder —> evolution of colourless, odourless gas with brisk effervescence.
Question.35. Draw a labelled diagram to show that particular stage of binary fission in amoeba in which its nucleus elongates and divides into two and a constriction appears in its cell membrane.
Question.36. A student focuses the image of a well illuminated distant object on a screen using a convex lens. After that he gradually moves the object towards the lens and each time focuses its image on the screen by adjusting the lens.
- In which direction—towards the screen or away from the screen, does he move the lens?
- What happens to the size of the image —does it decrease or increase?
- What happens to the image on the screen when he moves the object very close to the lens?
- He should move the lens towards the screen.
- Size of the image increases.
- 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. Name the following compounds:
Question.2. What is DNA?
Answer. DNA. Deoxyribonucleic acid is a genetic material found in all the living organisms.
It is the carrier of hereditary information from parents to the next generation.
Question.3. List two biotic components of a biosphere.
Answer. Biotic components:
- Micro organisms
- Decomposers, etc.
Question.4. A ray of light is incident on a convex mirror as shown. Redraw the diagram and complete the path of this ray after reflection from the mirror. Mark angle of incidence and angle of reflection on it.
Question.7.What is an oxidising agent? What happens when an oxidising agent is added to propanol? Explain with the help of a chemical equation.
- The substance that supplies oxygen in a reaction for oxidation is called an oxidising agent. Examples of oxidising agents are potassium permanganate, potassium dichromate etc.
- When propanol is heated with alkaline potassium permanganate solution (or acidified potassium dichromate solution), it gets oxidised to propanoic acid.
Question.10. Name any two elements of group one and write their electronic configurations. What similarity do you observe in their electronic configurations? Write the formula of oxide of any of the aforesaid element.
Answer. Two elements of group 1
Na = Sodium K = Potassium
Electronic configuration of:
Na = 2, 8,1 K = 2, 8, 8,1
Similarity between Na and K. Both have one electron in their outermost shell. Thus both have one valence electron-and valency one.
Oxide of Sodium => Na20 Oxide of Potassium => K20
Question.11. What are the functions of testes in the human male reproductive system? Why are these located outside the abdominal cavity? Who is responsible for bringing about changes in appearance seen in boys at the time of puberty?
Answer. Functions of testes are:
- to make male sex cells (or male gametes) called sperms.
- to make the male sex hormone called testosterone.
- The testes of a man lie in a small muscular pouch called scrotum outside the abdominal cavity of the body. This is because the sperm formation requires a lower temperature than the normal body temperature. Being outside the abdominal cavity, the temperature of scrotum is about 3° C lower than the temperature inside the body and thus the testes are provided an optimal temperature for the formation of sperms. The male sex hormone testosterone is responsible for bringing about changes in ‘ appearance seen in boys at the time of puberty.
Question.13. What is multiple fission? How does it occur in an organism? Explain briefly. Name one organism which exhibits this type of reproduction.
Answer. Multiple fission. The process of asexual reproduction in which many individuals are produced from the parent cell Process. Sometimes during unfavourable conditions a protective wall or cyst is formed around the cell of a single celled organism (like in plasmodium). Inside the cyst, the nucleus of the cell splits several times to form many smaller nuclei called daughter nuclei. Each nucleus gathers a bit of cytoplasm around itself, develops a membrane around each structure. Thus many daughter cells develop which on liberation grow into adult organisms. Plasmodium exhibits multiple fission.
Question.14. How did Mendel interpret his result to show that traits may be dominant or recessive? Describe briefly.
- In Mendel’s experiment with pea plants, when he cross bred a pure tall pea plant with a pure dwarf pea plant, he found that the first generation F1 was of only tall plants. Tallness is the dominant trait.
- Then he produced F2 generation by selling of hybrids F1 .
- In the T1 progeny, no dwarf plants were obtained. However in F2 generation, both tall and dwarf plants were obtained in the ratio 3 : 2 respectively.
- The trait which remains hidden in F, generation, i.e., dwarfness is the recessive trait.
- He observed that even when riot expressed in the first generation, alternate forms of a trait could retain their identify in the hybrid and could re-emerge in the next generation.
Question.16. What is meant by scattering of light? The sky appears blue and the Sun appears reddish at sunrise and sunset. Explain these phenomena with reason.
Answer. Scattering of light. The phenomenon of spreading of light when it falls on various types of suspended particles in its path is called scattering of light.
- The sky appears blue because of scattering of white light that takes place in the atmosphere. The molecules of air and other fine particles suspended in the atmosphere have a size smaller than the wavelength of visible light. So these particles scatter more effectively the light rays of shorter wavelength at the blue end than light rays of longer wavelengths at the red end. Thus, when sunlight passes through the atmosphere, the fine particles in air scatter blue light. When the scattered blue light enters our eyes, it gives us the feeling of a blue sky.
- The sun at sunrise and at sunset time is located very near to the horizon of the earth. Hence light has to travel a long distance through the earth’s atmosphere. Near the horizon most of the blue light of shorter wavelength is scattered away by the particles in the atmosphere. Therefore, the red light of longer wavelengths is least scattered by the air particles and is able to reach our eyes. This gives rise to the reddish appearance of the Sun at sunrise and sunset.
Question.22. (a) Draw a ray diagram to show the formation of image by a concave lens when an
object is placed in front of it.
(b) In the above diagram mark the object-distance (u) and the image-distance (ν) with their proper signs (+ve or -ve as per the new Cartesian sign convention) and state how these distances are related to the focal length (f) of the concave lens in this case.
(c) Find the nature and power of a lens which forms a real and inverted image of
magnification -1 at a distance of 40 cm from its optical centre.
Answer. (a) Concave lens. See Q. 24(a), 2011 (I Outside Delhi).
(b) CB = —u CF = -f CB=-ν
Except for the following questions, all the remaining questions have been asked in Set-I and Set-II.
Question.1. Select saturated hydrocarbons from the following:
C3H6 ; C5H10 ; C4H10 ; C6H14 ; C2H4
Answer. Saturated hydrocarbons:
General formula = CnH2n+2
Question.2. What happens when a Planaria gets cut into two pieces?
Answer. When a Planaria gets cut into two pieces, then each piece regenerates into a new planaria.
Question.3. Why are green plants called producers?
Answer. Green plants are called producers because green plants synthesize their own food during photosynthesis by taking raw materials from the earth and energy from the Sun.
Question.4. What is meant by power of a lens? What does its sign (+ve or -ve) indicate? State its S.I. unit. How is this unit related to focal length of a lens?
- The power of a lens is a measure of the degree of convergence or divergence of light rays falling
- +ve sign —> converging lens/’convex lens -ve sign —> diverging lens/concave lens
- The SI unit of lens is dioptre. One dioptre is the power of a lens whose focal length is 1 metre.
Question.6. “Reuse is better than recycling of materials”. Give reason to justify this statement.
Answer. The ‘reuse’ strategy is better than ‘recycling’ because even the process of recycling uses large amount of energy and money. In the reuse strategy we can preserve energy. For instance, plastic bottles in which we buy various food items like jam and pickle can be used for storing things in the kitchen. Whereas, if we send the discarded plastic bottles for recycling to the industry, we will be using energy sources for the recycling process.
Question.7. Name the compound formed when ethanol is heated is excess of cone, sulphuric acid at 443 K. Also write the chemical equation of the reaction stating the role of cone, sulphuric acid in it. What would happen if hydrogen is added to the product of this reaction in the presence of catalysts such as palladium or nickel?
- When ethanol is heated with excess of concentrated sulphuric acid at 170° C (443K) it gets dehydrated to form ethene (an unsaturated hydrocarbon).
- In this reaction, concentrated sulphuric acid acts as a dehydrating agent which removes water molecule from the ethanol molecule.
- If hydrogen is added to the ethene (product of above reaction) in the presence of catalysts like palladium or nickel then one atom of H adds to each carbon atom of ethene due to which the double bond opens up to form a single bond in ethane.
Question.9. Two elements ‘A’ and ‘B’ belong to the 3rd period of Modern periodic table and are in group 2 and 13 respectively. Compare their following characteristics in tabular form:
(a) Number of electrons in their atoms
(b) Size of their atoms
(c) Their tendencies to lose electrons
(d) The formula of their oxides
(e) Their metallic character
(f) The formula of their chlorides 3
Answer. Electronic configuration of A = 2, 8, 2
Electronic configuration of B = 2, 8, 3
Question.11. What is meant by pollination? Name and differentiate between the two modes pollination in flowering plants.
Answer. See Q. 21, 2013 (I Outside Delhi).
Question.14. In a mono hybrid cross between tall pea plants (TT) and short pea plants (tt) a scientist obtained only tall pea plants (Tt) in the F1 generation. However, on selfing the F2 generation pea plants, he obtained both tall and short plants in F2 generation. On the basis of above observations with other angiosperms also, can the scientist arrive at a law? If yes, explain the law. If not, give justification for your answer.
Answer. Law of Segregation:
When plants of two different traits of character are cross bred to get a progeny (F1 generation), only the dominant trait is visible in this generation. But when plants of F1 generation are self bred then the two traits of character get separated and the recessive traits also appears in the plant of F2 generation. This is known as Law of Segregation (separation) of traits.
Law of Dominance:
According to this law, the characteristics (or traits) of an organism are determined by internal ‘factors’ which occur in pairs. Only one of a pair of such factors can be presented in a single gamete. This law explains expression of only of the parental character in F1 generation and expression of both in F2 generation.
Question.16. (a) Draw a ray diagram to show the refraction of light through a glass slab and mark angle of refraction and the lateral shift suffered by the ray of light while passing through the slab.
(b) If the refractive index of glass for light going from air to glass is 3/2, find the refractive index of air for light going from glass to air.
Answer. (a) Ray diagram of refraction through a glass slab.
Question.17. State the cause of dispersion of white light passing through a glass prism. How did Newton show that white light of Sun contains seven colours using two identical glass prisms. Draw a ray diagram to show the path of light when two identical glass prisms are arranged together in inverted position with respect to each other and a narrow beam of white light is allowed to fall obliquely on one of the focus of the first prism.
- White light is a mixture of lights of seven colours, red, orange, yellow, green, blue, indigo and violet. The dispersion of white light occurs because colours of white light travel at different speeds through the glass prism. The amount of refraction depends on the speed of coloured light in glass.
- When white light consisting of seven colours falls on a glass prism, each colour in it is refracted by a different angle, with the result that seven colours are spread out to form a spectrum. The red light bends the least, while violet bends the most.
Newton s experiment with two identical prisms:
- When a beam of white light is passed through a glass prism, a band of seven colours is formed on a white screen. This band of seven colours is called spectrum of white light.
- Newton showed that the seven coloured lights of the spectrum can be recombined to give back white light.
- First he tried to split the colours of the spectrum of white light using a prism.
- He then placed a second identical prism in an inverted position with respect to the first prism. This allowed all the colours of the spectrum to pass through the second prism. He found a beam of white light emerging from the other side of the second prism.