Category: CBSE

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

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

GENERAL INSTRUCTIONS:

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

SET I

SECTION A
Question.1 How many vertical columns are there in the modem periodic table and what are they called?
Answer. There are 18 vertical columns in the modem periodic table and they are called groups.

Question.2 What is speciation?
Answer. The process by which new species develop from the existing species 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 dust¬bins because biodegradable waste is decomposed by the micro-organisms to form simple harmless substances which can be used as manures for the plants (e.g., in the potted plants in our garden/terrace garden). Non-biodegradable waste cannot be broken down naturally.

Question.4 “The chromosomes number of the sexually reproducing parents and their offspring is the same.” Justify this statement.
Answer. In sexual reproduction, though the genetic material DNA (in the form of chromosomes) from two gametes, male and female gametes, combines together to form a new cell ‘zygote’ but the amount of DNA (the number of chromosomes) in the zygote does not get
doubled because gametes are the special type of cells called reproductive cells which contain only half the amount of DNA (or half the number of chromosomes) as compared to the normal body cells of an organism. So when a male gamete (having half number of chromosomes of the organism) combines with a female gamete (which also contain half the number of chromosomes) during sexual reproduction, then the new cell zygote will have the normal amount of DNA or normal number of chromosomes in it.

Question.5 “A ray of light incident on a rectangular glass slab immersed in any medium emerges parallel to itself.” Draw a labelled ray diagram to justify the statement.
Answer.

Question.6 We often observe domestic waste decomposing in the bylanes of residential colonies. Suggest ways to make people realise that the improper disposal of waste is harmful to the environment.
Answer.

1. When the domestic waste is decomposed by the action of micro-organisms then it becomes the breeding place of flies and mosquitoes. Flies and mosquitoes are the carrier and vector of many dangerous diseases. So some posters related to the spread of such diseases can be pasted on the walls in public places to create awareness.
2.  There should be a strict fine imposed by the sanitary officer from the families which throw their domestic wastes on the roads.

Question.7 Write the name and the structural formula of the compound formed when ethanol is heated at 443 K with excess of cone. H₂SO₄. State the role of cone. H₂SO₄ in this reaction. Write chemical equation for the reaction.
Answer. When ethanol is heated with excess of concentrated sulphuric acid at 443 K, it gets dehydrated to form ethene (an unsaturated hydrocarbon)

In this reaction, concentrated sulphuric acid acts as a dehydrating agent which removes water molecules from the ethanol molecule.

Question.8 Why homologous series of carbon compounds are so called? Write chemical formula of two consecutive members of a homologous series and state the part of these compounds that determines their (i) physical properties, and (ii) chemical properties.
Answer.

1.  All the organic compounds having similar structures show similar properties and
   they are put together in the same groups or series called homologous series.
2. For example, all the alkanes have similar structures with single covalent bonds and show similar chemical properties so they are grouped together in a homologous series having general formula C_nH_2n+2.
3.  The two consecutive members of a homologous series are methane CH₄ and ethane C₂H₆.

Question.9 Given below are some elements of the modem periodic table:

(i) Any two adjacent homologues differ by 1 carbon and 2 hydrogen atoms in their molecular formulae. So the difference in molecular masses of any two adjacent homologues is 14 u. Thus the members of a homologous series show a gradual change in their physical properties with increase in molecular mass.
(ii)All the compounds of a homologous series show similar chemical properties because they have similar structures and similar bonding.
(iii) Select two elements that belong to the same period. Which one of the two has bigger atomic size?
Answer.

Question.10 Write the number of periods the modem periodic table has. How do the valency and metallic character of elements vary on moving from left to right in a period? How do the valency and atomic size of elements vary down a group?
Answer.

•  There are seven periods in the modem periodic table.
•  On moving from left to right in a period, the valency of elements increases from 1 to
  4 and then decreases to 0
•  On moving from left to right in a period, the metallic characters of elements decreases as the electropositive character of elements decreases on moving from left to right in a period.
• All the elements in a group have same valency, because the number of valence electrons in a group is same.
• On going down in a group of the periodic table, the size of atoms increases because a
  : new shell of electrons is added to the atoms at every step on moving down in a
  group.

Question.11 (a) Explain the process of regeneration in Planaria.
(b) How is regeneration different from reproduction?
Answer. (a) Planaria, a flatworm, possesses great power of regeneration. If the body of planaria gets cut into a number of pieces, then each body piece can regenerate into a complete planaria by growing all missing parts.
Diagram. See Q. 24, 2011 (I Delhi).
(b) Regeneration is different from reproduction because most simple animals would not
depend on being cut into pieces to be able to reproduce.

Q.12. Write two examples each of sexually transmitted diseases caused by (i) virus, (ii) bacteria. Explain how the transmission of such diseases be prevented?
Answer. Sexually transmitted diseases caused by
(i) Virus:

•  AIDS (Acquired Immuno Deficiency Syndrome)
•  Genital warts

(ii) Bacteria: • Gonorrhoea – • Syphilis
Transmission of such diseases can be prevented by the following ways:
— Screening tests for blood donors
— Mutually faithful monogamous relationships
— Educating people in high risk groups
— Using condoms etc.

Question.13 Tabulate two distinguishing features between acquired traits and inherited traits with one example of each.
Answer. See Q. 17, 2012 (I Outside Delhi).

Question.14 “The sex of a newborn child is a matter of chance and none of the parents may be considered responsible for it.” Justify this statement with the help of a flow chart showing determination of sex of a newborn.
Answer. Sex determination flow chart. See Q. 14, 2012 (II Outside Delhi).
Thus in human beings, the sex of the baby is determined by the type of sperm that fuses with ovum. As human male produces two types of sperms in equal proportion, so there are 50% chances of a male baby and 50% chances of a female baby.

Question.15 Mention the types of mirrors used as
(i) rear view mirrors,
(ii) shaving mirrors. List two reasons to justify your answers in each case.
Answer.
(i) Convex mirror is used as rear view mirror in vehicles because

• it always produces an erect image of the objects;
•  the image formed in a convex mirror is highly diminished thus it gives a wide field of view.

(ii)Concave mirrors are used as shaving mirrors because

•  when the face is held within the focus of a concave mirror, then an enlarged image of the face is seen in the concave mirror. This helps in making a smooth shave.

Question.16 An object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm.
Answer.


Thus the image is formed at a distance of 3.3 cm from the concave lens. The negative (-) sign for image distance shows that the image is formed on the left side of the concave lens (i.e., virtual). The size of the image is 2 cm and the positive (+) sign for hand image shows that the image is erect.
Thus a virtual, erect, diminished image is formed on the same side of the object (i.e., left side).

Question.17 State the difference in colours of the Sun observed during sunris^sunset and noon.
Give explanation for each.
Answer. The Sim and surrounding sky appear red at sunrise and at sunset because at this time the Sun is near the horizon and sunlight has to travel the greatest distance through the atmosphere to reach us. Thus most of the blue colour present in sunlight has been scattered out and away from our line of sight, leaving behind mainly red colour in the direct sunlight beam that reaches our eyes.
When the Sun is overhead (as at noon) then the light coming from the Sun has to travel a relatively shorter distance through the atmosphere to reach us. Thus only a little of blue colour of the white light is scattered. Since the light coming from the overhead Sun has almost all its component colours in the right proportion, therefore the Sun in the sky overhead appears white.

Question.18 (a) What is an ecosystem? List its two main components.
(b) We do not clean ponds or lakes, but an aquarium needs to be cleaned regularly. Explain.
Answer.
(a) An ecosystem is a self-contained unit of living things (plants, animals and decomposers) and their non-living environment (soil, air and water). An ecosystem needs only the input of sunlight energy for its functioning.
The two main components of an ecosystem are:

• Abiotic component. It includes all non-living components like soil, water, air temperature, light, pressure, etc.
•  Biotic component. It includes all living components like plants, animals, decomposers, etc.

(b) A pond is a self sufficient or independent unit in nature. It contains all the components of the ecosystem. In this ecosystem, producers (hydrophytes) trap the solar energy and then provide the basic food or energy for all other life in the pond. When the producers and consumers die, the decomposers present in the pond act on their dead bodies to return the various elements back to the nutrient pool.
On the other hand, in an aquarium there are not any. producers and nutrient pool to trap solar energy, therefore the fishes living in an aquarium need to be nourished. Moreover due to absence of decomposers the excreta of the fishes cannot be decomposed. Therefore the aquarium needs to be cleaned regularly.

Question.19 (a) Define the term ‘isomers’.
(b) Draw two possible isomers of the compound with molecular formula C₃H₆O and write their names.
(c) Give the electron dot structures of the above two compounds. 
Answer. (a) The organic compounds having same molecular formula but different structures are
known as isomers.

Question.20 (a) List three distinguishing features between sexual and asexual types of reproduction.
(b) Explain why variations are observed in the off springs of sexually reproducing organisms?
Answer.

(b) Reason. There is always a possibility of diversity of characters in the off springs because the offspring is formed as a result of fusion of two gametes produced by two different individuals—the male and the female parents. So there is an opportunity for new combinations of characters.

Question.21 (a) Identify A, 6 and C in the given diagram and write their functions.
(b) Mention the role of gamete and zygote in sexually reproducing organisms

Answer. (a)
A—Stigma. The top part of carpel is called stigma. Stigma is for receiving the pollen grains from the anther of stamen during pollination.
B—Pollen tube. When a pollen grain falls on the stigma, it bursts open and grows a pollen tube downward through the style towards the female gamete in the ovary. A male gamete moves down the pollen tube.
C—Female gamete (ovum). It is a special reproductive female sex cell which combines with male gamete to form zygote.
(b) Sexual reproduction takes place by the combination of special reproductive cells called sex cells. These cells are of two types —male sex cells and female sex cells, which are coming from two different parents—a male and a female. The cells involved in sexual reproduction are called gametes. In sexual reproduction, a male gamete fuses with a female gamete to form a new cell called zygote. This zygote then grows and develops into a new organism in due course of time.

Question.22 (a) State the laws of refraction of light. Give an expression to relate the absolute refractive index of a medium with speed of light in vacuum.
(b) The refractive indices of water and glass with respect to air are 4/3 and 3/2 respectively. If the speed of light in glass is 2×10⁸ ms^-1, find the speed of light in (i) air, (ii) water.
Answer.

Question.23 (a) A person cannot read newspaper placed nearer than 50 cm from his eyes. Name the defect of vision he is suffering from. Draw a ray diagram to illustrate this defect.
List its two possible causes. Draw a ray diagram to show how this defect may be corrected using a lens of appropriate focal length.
(b) We see advertisements for eye donation on television or in newspapers. Write the importance of such advertisements.
Answer.
(a) If a person cannot read newspaper nearer than 50 cm from his eyes then he is suffering from hypermetropia. It is also called long-sightedness.
Ray diagram, causes and correction of this defect. See Q. 17, 2011 (I Delhi).
(b) We have seen advertisements in media persuading people for eye donation. If cornea of the eyes are removed within 6 hours of the death of a person, it can be transplanted to a person suffering from corneal blindness. There are more than 45 lakh cases of corneal blindness and unfortunately out of these a major portion is that of children aged below 12 years. If a person comes forward for eye donation, he can save two corneal blind persons by donating one eye to each.
How excitingjt feels to think that after one’s death, he/she can make two blind persons see this wondeiful world! Therefore it is a must to promote such advertisements to encourage people and make them aware of this noble cause.

Question.24 The elements of the third period of the Periodic Table are given below:

(a) Which atom is bigger, Na or Mg? Why?
(b) Identify the most
(i) metallic and
(ii) non-metallic element in Period 3.
(c) Which is more non-metallic, S or Cl?
(d) Which has higher atomic mass, A1 or Cl?
Answer.
(a) Na is bigger than Mg because on moving from left to right in a period, the atomic
number of elements increases which means that the number of protons and electrons in the atom increases. (The extra electrons being added to the same shell).
(b) (i) Most metallic element is Na. Most non-metallic element is Cl;
Because on moving from left to right in a period the nuclear charge increases thus the valence electrons are pulled in more strongly by the nucleus and it becomes more and more difficult for the atoms to lose electrons so tendency of atoms to lose electrons (i e., metallic character) decreases on moving from left to right in a period. On the other hand, due to increased nuclear charge, it becomes easier for the atoms to gain electrons. So the tendency to gain electrons (i.e. non-metallic character) increases on moving from left to right in a period.
(c) Cl is more non-metallic from S because on moving from left to right in a period,
nuclear charge increases so the tendency to gain electrons increases (i.e., non-metallic character).
(d) Cl has higher atomic mass because on moving from left to right in a period, atomic number increases. Simultaneously, atomic mass increases.

SECTION B
Question.25 A student takes 2 mL acetic acid in a dry test tube and adds a pinch of sodium hydrogen carbonate to it. He makes the following observations:
I. A colourless and odourless gas evolves with a brisk effervescence.
II. The gas turns lime water milky when it is passed through it.
III.The gas bums with an explosion when a burning splinter is brought near it.
IV. The gas extinguishes the burning splinter that is brought near it.
The correct observations are:
(A) I, II and III (B) II, III and IV (C) III, IV and I (D) IV, I and II
Answer.  (D) IV, I and II

Question.26 A student prepared 20% sodium hydroxide solution in a beaker containing water. The observations noted by him are given below.
I. Sodium hydroxide is in the form of pellets.
II. It dissolves in water readily.
III.The beaker appears cold when touched from outside.
IV. The red litmus paper turns blue when dipped into the solution.
The correct observations are:
(A) I, Hand III (B) II, III and IV (C) III, IV and I (D) I, II and IV
Answer.  (D) I, II and IV

Question.27 Hard water required for an experiment is not available in a school laboratory. However, following salts are available in the laboratory. Select the salts which may be dissolved in water to make it hard for the experiment.
1. Calcium Sulphate
2. Sodium Sulphate .
3. Calcium Chloride
4. Potassium Sulphate
5. Sodium Hydrogen Carbonate
6. Magnesium Chloride
(A) 1, 2 and 4 (B) 1, 3 and 6 (C) 3, 5 and 6 (D) 2, 4 and 5
Answer. (B) 1, 3 and 6

Question.28 A student focussed the image of a distant object using a device ‘X’ on a white screen ‘S’ as shown in the figure. If the distance of the screen from the device is 40 cm, select the correct statement about the device.

(A) The device X is a convex lens of focal length 20 cm.
(B) The device X is a concave mirror of focal length 40 cm.
(C) The device X is a concave mirror of radius of curvature 40 cm.
(D) The device X is a convex lens of focal length 40 cm.
Answer.(D) The device X is a convex lens of focal length 40 cm.

Question.29 Select from the following the best set-up for tracing the path of a ray of light through a rectangular glass slab:

(A) I                                                 (B)II                                             (C)III ,                                     (D) IV
Answer. (A) I

Question.30 In an experiment to trace the path of a ray of light through a glass prism for different values of angle of incidence a student would find that the emergent ray:
(A) is parallel to the incident ray
(B) perpendicular to the incident ray
(C) is parallel to the refracted ray
(D) bends at an angle to the direction of incident ray
Answer.(D) bends at an angle to the direction of incident ray

Question.31 Select the correct statements for the process of budding in yeast:
I. A bud arises from a particular region on a parent body.
II. A parent cell divides into two daughter cells, here the parental identity is lost.
III.Before detaching from the parent body a bud may form another bud.
IV. A bud when detaches from the parent body grows into a new individual.
(A) I, II and III (B) II, III and IV (C) III, IV and I (D) IV, I and II
Answer. (C) III, IV and I

Question.32 Study the different conclusions drawn by students of a class on the basis of observations of preserved/available specimens of plants and animals.
I. Potato and swTeet potato are analogous organs in plants.
II. Wings of insects and wings of birds are homologous organs in animals.
III.Wings of insects and wings of bats are analogous organs in animals.
IV. Thoms of citrus and tendrils of cucurbita are analogous organs in plants.
The correct conclusions are:
(A) I and II (B) II and IV (C) I and III (D) III and IV ‘
Answer.  (C) I and III

Question.33 In the figure, the parts marked A, B and C are sequentially:

(A) Plumule, Radicle and Cotyledon
(B) Radicle, Plumule and Cotyledon
(C) Plumule, Cotyledon and Radicle
(D) Radicle, Cotyledon and Plumule
Answer. (A) Plumule, Radicle and Cotyledon

Question.34 Why does a ray of light while passing through a prism, bend towards its base?

Answer. When a ray of light passes through a prism, it under goes refraction twice. First from rarer to denser medium of glass, it bends towards normal which is towards the base of the prism. Second time from denser to rarer medium, i.e., glass to air, it bends away from the normal i.e., towards the base of the prism.

Question.35 Name two types of fissions. Name two living beings of each type which reproduce by these methods of fission.
Answer.
Two types of fissions are:

1.  Binary Fission
2. Multiple Fission

Two examples of binary fission — Amoeba and Paramoecium
Two types of multiple fission are— Plasmodium and Spirogyra

Question.36 Write two tests you would perform to detect, whether the given colourless liquid is Acetic Acid or not.
Answer.Two tests:

1.  If we put a drop of the given colourless liquid on blue litmus paper, if the blue litmus paper changes to red, then the given acid is Acetic acid.
2.  If we smell the given liquid and the liquid gives a smell like that of vinegar, then the given acid is Acetic acid.

SET II

Except for the following questions, all the remaining questions have been asked in Set-I.
SECTION A
Question.l How many horizontal rows are there in the modern periodic table and what are they called?
Answer. There are seven horizontal rows in the modern periodic table. The horizontal rows in a periodic table are called periods.

Question.2 List any two factors that could lead to speciation.
Answer. Two factors that could lead to speciation:

1.  Geographical isolation of a population caused by various types of barriers {such as mountain range, rivers and sea). This leads to reproductive isolation due to which there is no flow of genes between separated groups of population.
2.  Genetic drift caused by drastic changes in the frequencies of particular genes by chance alone.

Question.3 Mention one negative effect of our affluent life, style on the environment.
Answer. Chlorofluorocarbons (CFCs) are the chemicals which are used in refrigerators and air conditioners as coolant when released into the air react with ozone gas present in the ozone layer and destroy it gradually. If the ozone layer in the atmosphere disappears completely, then all the extremely harmful ultraviolet radiations coming from the Sun would reach the earth and cause skin cancer and other ailments in humans, animals and plants.

Question.4 Mention the two functions of human testis.
Answer. Functions of human testis:

1.  Testes make male sex hormones called testosterone.
2.  Testes produce male sex gametes, i.e., sperms.

Question.5 Every one of us can do something to reduce our consumption of various natural resources. List four such activities based on 3-R approach.
Answer.

1. We can reduce the use of LPG by making use of solar cooker for cooking food.
2.  We can reduce the wastage of water by repairing the leakage in taps.
3.  We should collect all the discarded household items like newspapers, broken plastic items, glass bottles etc. and send them to the respective industries for recycling to make fresh paper, plastic and glass objects.
4. We can drink soft cold drinks available in the glass bottles instead of pearipet or metal containers as these glass bottles can be refilled and thus reused again.

Question.8 Name the oxidising agent used for the conversion of ethanol to ethanoic acid. Distinguish between ethanol and ethanoic acid on the basis of (i) litmus test, (ii) reaction with sodium carbonate.
Answer. Acidified K₂Cr₂O₇ can be used as oxidising agent for the conversion of ethanol to ethanoic acid.

(i) Litmus test:

• If an organic compound (to be tested) does not change the colour of either of the litmus solution (blue or red) then the organic compound is neutral in nature. Ethanol is a neutral compound.
•  When some blue litmus solution is added to the organic compound (to be tested), if the blue litmus solution turns red; it shows that the organic compound is acidic and hence it is ethanoic acid.

(ii) Reaction with sodium carbonate:

•  When the organic compound (to be tested) is taken in a test tube and a pinch of sodium carbonate is added to it; if no reaction occurs in the reaction mixture, it is ethanol.

Question.9 (a) Differentiate between alkanes and alkenes. Name and draw the structure of one
member of each.
(b) Alkanes generally burn with clean flame. Why?
Answer.

(b) Alkanes bum in air with a blue and non-sooty flame because the percentage of carbon in the alkane is comparatively low which gets oxidised completely by oxygen . present in air.

Question.11 An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a compound.
(a) Write the position of these elements in the modem periodic table.
(b) Write the formula of the compound formed.
Justify your answer in each case.
Answer. Element ‘X’                     Electronic configuration
Atomic number = 17                        2, 8, 7

Question.16 An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 20 cm.
Answer.

Question.20 (а) Write the function of placenta in human females.
(b) List four ways of preventing pregnancy. State, two advantages of using such preventive methods.
Answer.
(a) The exchange of nutrients, oxygen and waste products between the embryo and the mother takes place through placenta.
(b) Methods to prevent pregnancy. See Q. 15, 2012 (I Outside Delhi). [Page P – 47
Advantages of using preventive methods.

1. Contraceptive methods are used to prevent the spreading of sexually transmitted diseases like syphillis, AIDS etc.
2.  Contraceptive methods are also used to control human population by checking the unwanted pregnancies during sexual activities.

Question.23 State Snell’s law of refraction of light. Write an expression to relate refractive index of a medium with speed of light in vacuum.
The refractive index of a medium ‘a’ with respect to medium ‘b’ is 2/3 and the refractive index of medium ‘b’ with respect to medium ‘c’ is 4/3. Find the refractive index of medium ‘c’ with respect to medium ‘a’.
Answer. Snell’s law states that, “the ratio of sine of angle of incidence to the sine of angle of refraction is constant for a given pair of media”.

SET III

Except for the following questions, all the remaining questions have been asked in Set-I and
Set-11.
SECTION A
Question.1 Write two reasons responsible for late discovery of noble gases?
Answer. The noble gases were discovered very late because they are very unreactive and present in extremely low concentration in the atmosphere.

Question.2 “Cell division is a type of reproduction in unicellular organisms.” Justify.
Answer. Cell division is a type of reproduction in unicellular organisms because many unicellular • organisms split into two identical halves during cell division leading to the creation of new organisms.

Question.3 List any two measures that you suggest for better management of water resources.
Answer. Rain water harvesting and installation of used water recycling plants are the two measures for better management of water resources.

Question.4 (a) Trace the path of sperms from where they are produced in human body to the
exterior.
(b) Write the functions of secretions of prostate gland and seminal vesicles in humans.
Answer.
(a) Testis —> Epididymis —> Vas deferens —> Urethra —> Penis
(b) Along the path of vas-deferens, the glands called seminal vesicles and prostrate gland, add their secretions to sperms to provide nutrition to the sperms and also make their further transport easier.

Question.5 A ray of light falls normally on the surface of a transparent glass slab. Draw a ray diagram to show its path and also mark angle of incidence and angle of emergence.
Answer.

Question.8 What happen when:
(a) ethanol is burnt in air,
(b) ethanol is heated with excess cone.H₂SO₄at 443 K,
(c) a piece of sodium is dropped into ethanol?
Answer.

Question.10 The elements Li, Na and K, each having one valence electron, are in period 2, 3 and 4 respectively of modern periodic table.
(a) In which group of the periodic table should they be?
(b) Which one of them is least reactive?
(c) Which one of them has the largest atomic radius? Give reason to justify your answer in each case.
Answer.

(a) All these elements belong to 1st group because all elements have one electron in their respective outermost shell.
(b) Li is least reactive as reactivity of metals decreases on going down in a group.
(c) K has the largest atomic radius because Li belongs to 2nd period thus has 2 energy shells, Na belongs to the 3rd period thus has 3 energy shells and K belongs to 4th period thus has 4 energy shells. So as the number of shells increases so does the size of the atom.

Question.13 (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 the structure and the function of the structures released from the ‘blobs’ in Rhizopus.
Answer. (a)

1.  Thread like non-reproductive structures present in Rhizopus is hyphae.
2. Blob like structure at the tip of hyphae is sporangia.

(b) The thread like non-reproductive parts develop on the substrate (like bread) called hyphae. At the tip of the hyphae tiny blob like structures develop called sporangia which contain spores. These spores can develop into new Rhizopus individuals. The spores are covered by thick walls that protect them until they come into contact with another moist surface and can begin to grow.

Question.15 What are fossils? State their importance in the study of evolution with the help of a suitable example.
Answer. The remains or impressions of extinct animals or plants that lived in the remote past are known as fossils.
Importance of fossils:

1.  Fossils provide evidence for evolution, e.g., a fossil bird called Archaeopteryx looks like a bird but it has many features which are found in reptiles. Archaeopteryx has feathered wings like birds but teeth and tail like those of reptiles. Therefore Archaeopteryx is a connecting link between the reptiles and birds and hence suggests that the Birds have evolved from the reptiles.
2.  Habits and behaviour of extinct species can be inferred. For example, the estimation of age of dinosaur fossils have told that they first appeared on earth about 250 million years ago and became extinct about 65 million years ago.
3.  Fossils provide direct evidence of past life.
4.  With the help of fossils broad historical sequence of biological evolution can be built up.

Question.16 An object of height 4 cm is kept at a distance of 30 cm from a concave lens. Use lens formula to determine the image distance, nature and size of the image formed if focal length of the lens is 15 cm.
Answer.

Thus the image is formed at a distance of 10 cm from the concave lens.
The negative (-) sign for image distance shows the image is formed on the left side of the concave lens, i.e., it is virtual.

Thus a 1.33 cm high image is formed and positive (+) sign of h₂ shows that image is erect. Thus image distance = 10 cm, image height = 1.33 cm.
Nature of the image is virtual and erect.

Question.20 (a) Draw a sectional view of human female reproductive system and label the following parts:
(i) Where the development of egg occurs.
(ii)Where fertilisation takes place.
(b) Describe the changes the uterus undergoes:
(i) to receive the zygote.
(ii) if zygote is not formed.
Answer.
(a) Human female reproductive system.

Parts where:

1.  Development of egg occurs—Ovary
2.  Fertilisation takes place—Fallopian tube

(b)

1.  When the uterus receives the zygote, zygote divides rapidly by mitosis and forms a hollow ball of hundreds of cells called embryo which gets embedded in the thick lining of the uterus. This process is called implantation. After implantation, a disc like special tissue develops between the uterus wall and the embryo which is called placenta. The exchange of nutrients, oxygen and waste products between the embryo and the mother takes place through placenta.
2. If zygote is not received by the uterus, then the thick and soft uterus lining having lot of blood cappillaries in it is not required. The unfertilised ovum dies within a day and the uterus lining also breaks down. The breaking of the uterus lining produces blood along with tissues in the form of bleeding through the vagina. This vaginal bleeding is called menstrual flow or menstruation.

Question.24 Define the term absolute refractive index. The absolute refractive index of diamond is 2.42. What is the meaning of this statement? Refractive indices of media A, B, C and D are given below:

In which of these four media is the speed of light (i) minimum and (ii) maximum? Find the refractive index of medium C with respect to medium B.
Answer. When light is going from vacuum to another medium, then the value of refractive index is called absolute refractive index to the medium.
The absolute refractive index of diamond is 2.42. It means that the speed of light in diamond is 1/2.42 times the speed of light in vacuum.
As the refractive indices increase, speed of the light decreases in the medium.
(i) The refractive index of medium D is maximum (1.65). So the speed of light in medium D is minimum.
(ii) The refractive index of medium A is minimum (1.33). So the speed of light in medium A is maximum.

CBSE Previous Year Question Papers Class 12 Physics 2012 Delhi

CBSE Previous Year Question Papers Class 12 Physics 2012 Delhi Set-I

Question 1.
When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction ? [1]
Answer :
No, when electric field is applied, the electrons will have net drift from lower to higher field but locally electrons may collide with ions and may change their direction of motion.

Question 2.
The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of earth’s magnetic field at equator ? [1]
Answer:
On the equator, the value of both angle of dip (δ) and vertical component of earth’s magnetic field is zero. So, in this case,
B_v = 0.

Question 3.
Show on a graph, the variation of resistivity with temperature for a typical semiconductor. [1]
Answer :
The following curve shows the variation of resistivity with temperature for a typical semiconductor.

This is because, for a semiconductor, resistivity decreases rapidly with increasing temperature.

Question 4.
Why should electrostatic field be zero inside a conductor ? [1]
Answer :
Charge on conductor resides on its surface. So if we consider a Gaussian surface inside the conductor to find the electrostatic field,
\(\phi=\frac{q}{\varepsilon_{0}}\)
Where, q = charge enclosed in Gaussian surface
q = 0, inside the conductor, Hence the electrostatic field inside the conductor is zero.

Question 5.
Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 A in vacuum. [1]
Answer :
Both microwaves and UV rays are a part of the electromagnetic spectrum. Thus, the physical quantity that remains same for both types of radiation will be their speeds which is equal to c (3 × 10⁸  m/s).

Question 6.
Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid ? [1]
Answer :
A biconvex lens will act like a plane sheet of glass if it is immersed in a liquid having the Same index of refraction as itself.

Question 7.
Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily. [1]

Answer :
Using Lenz’s law we can predict the direction of induced current in both the rings. Induced current oppose the cause of increase of magnetic flux. So,

It will be clockwise in ring 1 and anticlockwise in ring 2.

Question 8.
State de-Broglie hypothesis. [1]
Answer:
de Broglie postulated that the material particles may exhibit wave aspect. Accordingly a moving material particle behaves as wave and the wavelength associated with material particle is
\(\lambda=\frac{h}{m v}\)
where h = Planck’s constant
m = mass of the object
v = velocity of the object

Question 9.
A ray of light incident on an equilateral prism (μ_g = √3) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray. [2]
Answer :
It is given that the prism is equilateral in shape. So, all the angles are equal to 60 . Thus, the angle of prism, A = 60
The angle of refraction in case of a prism,

Question 10.
Distinguish between ‘Analog and Digital signals. [2]
OR
Mention the functions of any two of the following used in communication system:
(i) Transducer
(ii) Repeater
(iii) Transmitter .
(iv) Band pass Filter

Question 11.
A cell of emf E and internal resistance r is connected to two external resistance Ri and R2 and a perfect ammeter. The current in the circuit is measured in four different situations
(i) without any external resistance in the circuit
(ii) with resistance R₁ only
(iii) with R₁ and R₂ in series combination
(iv) with R₁ and R₂ in parallel combination
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in the order. Identify the currents corresponding to the four cases mentioned above.
Answer : .
The current relating to corresponding situations is as follows:
(i) Without any external resistance in the circuit:
\(\mathrm{I}_{1}=\frac{\mathrm{E}}{r}\)
The current in this case would be maximum

Question 12.
The susceptibility of a magnetic material is -2.6 x 10^-5. Identify the type of magnetic material and state its two properties. [2]
Answer : Diamagnetic materials have negative susceptibility. So the given magnetic material is diamagnetic.
Two properties of diamagnetic material:

1. They do not obey Curie’s law.
2. They are feebly repelled by a magnet.

Question 13.
Two identical circular wires P and Q each of radius R and carrying current I are kept in perpendicular planes such that they have a common center as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils. [2]
Answer :
Magnetic field produced by the two coils at their common centre having currents I₁ and I₂, radius a₁ and a₂, number of turns N₁ and N₂ respectively are given by :


Hence, the net magnetic field is directed at an angle of 45° with either of the fields.

Question 14.
When an ideal capacitor is charged by a d.c. battery, no current flows. However, when an a.c. source is used, the current flows continuously. How does one explain this based on the concept of displacement current ? [2]
Answer :
When an ideal capacitor is charged by d.c. battery, charge flows till the capacitor gets fully charged.
When an a.c. source is connected then conduction current  \(i_{c}=\frac{d q}{d t}\) flows in the connecting wire.
Due to changing current, charge deposited on the plates of the capacitor changes with time. Changing charge causes electric field between the plates of capacitor to be varying, giving rise to displacement current \(i_{c}=\frac{d q}{d t}\)
[As displacement current is proportional to the rate of flux variation].
Between the plates, electric field

Displacement current brings continuity in the flow of current between the plates of the capacitor

Question 15.
Draw a plot showing the variation of
(i) electric field (E) and
(ii) electric potential (V) with distance r due to a point charge Q. [2]
Answer:
We know that for a point charge Q.

Question 16.
Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self inductance L is \(\frac{1}{2} \mathbf{L} \mathbf{I}^{2}\) given
Answer :
Self inductance is the inherent inductance of a circuit, given by the ratio of the electromotive force produced in the circuit by self-induction to the rate of change of current producing it. It is also called coefficient of self induction.
Suppose I = Current flowing in the coil at any time.
Φ= Amount of magnetic flux linked.
It is found that Φ ∝I
Φ = LI
where, L is the constant of proportionality and is called coefficient of self induction. SI unit of self-inductance is henry.
Let at t = 0, the current in the inductor is zero. So at any instant t, the current in the inductor is I and rate of growth of I is dl/dt.
Then, the induced emf is E = L x dl / dt
If the source is sending a constant current I
through the inductor for a small time dt, then small amount of work done by the source is given by

The total amount of work done by the source of emf, till the current increases from its initial value I = 0 to its final value I is given by

This work done by the source of emf used in building up current from zero to I is stored in the inductor in energy form. Therefore, energy stored in the inductor is

Question 17.
The current in the forward bias is known to be more (~ mA) than the current in the reverse bias (~ µA). What is the reason, then, to operate the photo diode in reverse bias ? [2]
Answer:
The current in the forward bias is due to majority carriers whereas current in the reverse bias is due to minority carriers. So current in forward bias is more (~mA) than current in reverse bias (~µA).

On illumination of photo diodes with light, the fractional change in the majority carriers would be much less than that in minority carriers. It implies that fractional change due to light on minority carriers dominated reverse bias current is more easily measurable than fractional change in forward bias current. So, photo diodes are operated in reverse bias condition.

Question 18.
A metallic rod of ‘L’ length is rotated with angular frequency of ‘ω’ with one end hinged at the center and the other end at the circumference of a circular metallic ring of radius L about an axis passing through the center and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the center and the metallic ring. [2]
Answer:
The induced emf,


Question 19.
The figure shows a series LCR circuit with L = 5.0 H, C = 80 µF, R = 40 Ω connected to a variable frequency 240 V source, calculate [3]
(i) the angular frequency of the source which drives the circuit at resonance,
(ii) the current at the resonating frequency,
(iii) the rms potential drop across the inductor at resonance.

Answer: Given, L = 5.0 H, C = 80 μF, R = 40 Ω, V = 240 V


Question 20.
A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of 2 A. A straight long wire carrying 5 A current is kept near the loop as shown. If the loop and the wire are coplanar, find [3]
(i) the torque acting on the loop and
(ii) the magnitude and direction of the force on the loop due to the current carrying wire.

Question 21.
(a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the n^th orbital state in hydrogen atom is n times the de Broglie wavelength associated with it.
(b) The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state ? [3]
Answer:
(a) According to Bohr’s second postulate of quantization, the electron can revolve around the nucleus only in those circular orbits in which the angular momentum of the electron is integral

Question 22.
In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs ∈₁ and ∈₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find
(i) ∈₁/∈₂ and
(ii) position of null point for the cell ∈₁. [3]
How is the sensitivity of a potentiometer increased ?
OR
Using Kirchhoff’s rules determine the value of unknown resistance R in the circuit so that no current flows through 4 Ω resistance. Also find the potential between A and D.

Answer:

Question 23.
(i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170 ?
(ii)Show that the density of nucleus over a wide range of nuclei is constant and independent of mass number A. [3]
Answer:
(i) The constancy of BE/A over most of the range is due to saturation property of nuclear force. In heavy nuclei: nuclear size > range of nuclear force. So, nucleons experiences nearly constant interaction.
(ii) To find the density of nucleus of an atom, we have an atom with mass number let say A and let mass of the nucleus of the atom of the mass number A be m A.
Where, m is mass of one nucleon.
Let radius of nucleus be R.


This expression is independent of mass number A and since m, π and R₀ have same values for any atom, therefore, density is constant also.

Question 24.
Write any two factors which justify the need for modulating a signal. Draw a diagram showing an amplitude modulated wave by superposing a modulating signal over a sinusoidal carrier wave. [3]

Question 25.
Write Einstein’s photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation. Write the three salient features observed in photoelectric effect which can be explained using this equation.[3]
Answer:
Einstein’s photoelectric equation,
hv = K_max + Φ
Where, h = Planck’s constant
v = frequency of radiation
Φ = Work function

According to Planck’s quantum theory, light radiations consist of small packets of energy. Einstein postulated that a photon of energy hv is absorbed by the electron of the metal surface, then the energy equal to Φ is used to liberate electron from the surface and rest of the energy hv – Φ becomes the kinetic energy of the electron.
Energy of photon is,
E = hv
The minimum energy required by the electron of a material to escape out of its work function ‘Φ’. The additional energy acquired by the electron appears as the maximum kinetic energy ‘K_max‘ of the electron.

i.e, K_max = hv – Φ
or hv = K_max + Φ
where Φ = eV₀

Sailent features observed in photoelectric effect:

1. The stopping potential and hence the maximum kinetic energy of emitted electrons varies linearly with the frequency of incident radiation.
2. There exists a minimum cut-off frequency V₀, for which the stopping potential is zero.
3. Photoelectric emission is instantaneous

Question 26.
(a) Why are coherent sources necessary to produce a sustained interference pattern ?
(b)In Young’s double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3. [3]
Answer :
(a) Coherent sources have constant phase difference between them i.e., phase difference does not change with time. Hence, the intensity distribution on the screen remains constant and sustained.


Question 27.
Use Huygens’s principle to explain the formation of diffraction pattern due to a single slit illuminated by a monochromatic source of light. When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band ? [3]
Answer :
Consider a parallel beam of monochromatic light is incident normally on a slit of width b as shown in figure. According to Huygens’s principle, every point of slit acts as a ‘ source of secondary wavelets spreading in all directions. Screen is placed at a larger distance.

Consider a particular point P on the screen that ; receives waves from all the secondary sources. All these waves start from different point of the slit and interfere at point P to give resultant intensity.

Point Po is a bisector plane of the slit. At Po, all waves are travelling equal optical path. So all wavelets are in phase thus interfere constructively with each other and maximum intensity is observed. As we move from Po, the wave arrives with different phases and intensity is changed. Intensity at point P is given by
\(\mathrm{I}=\mathrm{I}_{0} \frac{\sin ^{2} \alpha}{\alpha}\)

\(\text { Where } \alpha=\frac{\pi}{\lambda} b \sin \theta\)
For central maxima, α= 0 thus,
I = I_o
When the width of slit is made double the original width intensity will get four times of its original value.
Width of central maximum is given by,
\(\beta=\frac{2 \mathrm{D} \lambda}{b}\)
Where, D = Distance between screen and slit,
λ = Wavelength of the light,
b = size of slit
So with the increase in size of slit the width of central maxima decrease. Hence, double the size of the slit would result in half the width of the central maxima.

Question 28.
Explain the principle of a device that can build up high voltage of the order of a few million volts. Draw a schematic diagram and explain the working of this device. Is there any restriction on the upper limit of the high voltage set up in this machine ? Explain. [5]
OR
(a) Define electric flux. Write its S.I. units.
(b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(c) How is the field directed if
(i) the sheet is positively charged,
(ii) negatively charged ?
Answer :
Van de Graaff generator is the device used for building up high voltages of the order of a few million volts. Such high voltages are used to accelerate charged particles such as electrons, protons, ions, etc.

It is based on the principle that charge given to a hollow conductor is transferred to outer surface and is distributed uniformly over it.
Construction:

It consists of large spherical conducting shell (S) supported over the insulating pillars. A long narrow belt of insulating material is wound – around two pulleys P₁ and P₂, B₁ and B₂ are two sharply pointed metal combs. B₁ is called the spray comb and B₂ is called the collecting comb. Working : The spray comb is given a positive potential by high tension source. The positive charge gets sprayed on the belt.

As the belt moves and reaches the sphere, a negative charge is induced on the sharp ends of collecting comb-B₂ and an equal positive charge is induced on the farther end of B₂.

This positive charge shifts immediately to the outer surface of S. Due to discharging action of sharp points of B₂, the positive charge on the belt is neutralized. The uncharged belt returns down and collects the positive charge from B₁, which in turn is collected by B₂. This is repeated. Thus, the positive charge of S goes on accumulating. In this way, potential differences of as much as 6 or 8 million volts (with respect to the ground) can be built up. The main limiting factor on the value of high potential is the radii of sphere.

If the electric field just outside the sphere is sufficient for dielectric breakdown of air, no more charge can be transferred to it.
For a conducting sphere,
Electric field just outside sphere


Direction of field will be towards the sheet if sheet is negatively charged.

Question 29.
Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length 150 cm and an eye-piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away front the eye-piece. [5]
OR
How is the working of a telescope different from that of a microscope ? The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.
Answer :
Magnifying power of a telescope is defined as the ratio of the angle subtended at the eye by the image formed at the least distance of distinct vision to the angle subtended at the eye by the object lying at infinity, when seen directly.


Negative sign indicates that we get an inverted image.
OR
A microscope is used to look into smaller objects like structure of cells etc. On the other hand, a telescope is used to see larger objects that are very far away like stars, planets etc.

Telescope mainly focuses on collecting the light into the objective lens, which should thus be large, where the microscope already has a focus and the rest is blurred around it. There is a big difference in their magnification factors. For telescope, the angular magnification is given by
\(\mathrm{M}=\frac{f_{o}}{f_{e}}\)


Question 30.
Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, A_v, of the amplifier is given by  \(\mathbf{A}_{\mathbf{V}}=\frac{-\beta_{a c} \mathbf{R} \mathbf{L}}{r_{i}}\)
where β_ac is the current gain,
‘ RL’ is the load resistance and,
‘ r_i ‘ is the input resistance of the transistor.
What is the significance of the negative sign in the expression for the voltage gain ? [5]
OR
(a) Draw the circuit diagram of a full wave rectifier using p-n junction diode. Explain its working and show the output, input wave forms.
(b) Show the output wave forms (Y) for the following inputs A and B of
(i) OR gate
(ii) NAND gate

CBSE Previous Year Question Papers Class 12 Physics 2012 Delhi Set-II

Note : Except for the following questions, all the remaining questions have been asked in previous Set.

Question 1.
Why must electrostatic field be normal to the surface at every point of a charged conductor ? [1]
Answer :
The electrostatic field must be normal to the surface of the conductor at every point because if the electric field is not normal to the surface of the charged conductor, there will be a component of the electric field along the surface of the conductor, which would exert a force on the charges at the surface. Due to this, charge starts flowing which is not possible.

Question 2.
Predict the direction of induced current in a metal ring when the ring is moved towards a straight conductor with constant speed v. The conductor is carrying current I in the direction shown in the figure.

Answer:
Clockwise

Question 3.
Derive the expression for the self inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length. [2]
Answer:
Self-inductance of a long Solenoid :
The magnetic field B at any point inside a solenoid is constant and given by,

Question 4.
Two identical circular loops, P and Q, each of radius r and carrying currents I and 21 respectively are lying in parallel planes such that they have a common axis. The direction of current in both the loops is clockwise as seen from O which is equidistance from the both loops. Find the magnitude of the net magnetic field at point O. [2]

Question 5.
A series LCR circuit with L = 4.0 H, C = 100 μF, R = 60 Ω connected to a variable frequency 240 V source as shown in figure calculate : [3]
(i) the angular frequency of the source which drives the circuit at resonance,
(ii) the current at the resonating frequency,
(iii) the rms potential drop across the inductor at resonance.

Question 6.
A rectangular loop of wire of size 2 cm × 5 cm carries a steady current of 1 A. A straight long wire carrying 4 A current is kept near the loop as shown in the figure. If the loop and the wire are coplanar, find
(i) the torque acting on the loop and
(ii) the magnitude and direction of the force on the loop due to the current carrying wire. [3]


Question 7.
Name the three different modes of propagation of electromagnetic waves. Explain, using a proper diagram the mode of propagation used in the frequency range above 40 MHz. [3]

CBSE Previous Year Question Papers Class 12 Physics 2012 Delhi Set-III

Note : Except for the following questions, all the remaining questions have been asked in previous Set

Question 1.
Why is electrostatic potential constant throughout the volume of the conductor and has the same value (as inside) on its surface ? [1]
Answer :
We know that the electric field inside a conductor is zero. This is the reason why electrostatic potential is constant, as


Question 2.
Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing ? [1]

Answer :
Using Lenz’s law, we can predict the direction of induced current in the ring. Induced current oppose the cause of change of magnetic flux in moving towards the conductor.

In metal Ring 1, induced current will be in is clockwise.

In metal Ring 2, induced current will be in is anticlockwise.

Question 3.
The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its two properties. [2]
Answer :
A magnetic material having relative permeability of 800 would be classified as a ferromagnet. A few examples of such materials include iron and nickel.
Its two properties are :

1. All ferromagnetic materials become para-magnetic when heated to a temperature above the Curie temperature (T_c).
2. These materials show a strong attraction towards magnetic fields and have a tendency to become magnets themselves.

Question 4.
Two identical circular loops, P and Q, each of radius r and carrying equal currents are kept in the parallel planes having a common axis passing through O. The direction of current in P is clockwise and in Q is anti-clockwise as seen from O which is equidistant from the loops P and Q. Find the magnitude of the net magnetic field at O. [2]


Now, as the current flowing in loop P is clockwise, by using right hand thumb’s rule, the direction of the magnetic field will be towards left and as the current in loop Q is clockwise then the direction of magnetic field is towards left. So the net magnetic field at point O will be the sum of the magnetic fields due to loops P and Q.
So, net field

Question 5.
A rectangular loop of wire of size 2.5 cm × 4 cm carries a steady current of 1 A. A straight wire carrying 2 A current is kept near the loop as shown. If the loop and the wire are co-planar, find the
(i) torque acting on the loop and
(ii) the magnitude and direction of the force on the loop due to the current carrying wire. [3]

CBSE Previous Year Question Papers

CBSE Previous Year Question Papers Class 12 Physics 2013 Delhi

CBSE Previous Year Question Papers Class 12 Physics 2013 Delhi Set-I

Question 1.
What are permanent magnets ? Give one example. [1]
Answer :
The magnets which have high retentivity and high coercivity are known as permanent magnets.
For example : Steel, cobalt

Question 2.
What is the geometrical shape of equipotential surface due to a single isolated charge ? [1]
Answer :
The equipotential surfaces of an isolated charge are concentric spherical shells and the distance between the shells increase with the decrease in electric field and vice-versa.

Question 3.
Which of the following waves can be polarized
(i) Heat waves
(ii) Sound waves ? Give reason to support your answer. [1]
Answer:
Heat waves can be polarized as they are transverse waves whereas sound waves cannot be polarized as they are longitudinal waves.

Transverse waves can oscillate in the direction perpendicular to the direction of its transmission but longitudinal waves oscillate only along the direction of its transmission. So, longitudinal
waves cannot be polarized.

Question 4.
A capacitor has been charged by a dc source. What are the magnitude of conduction and displacement current, when it is fully charged ? [1]
Answer: Electric flux through plates of capacitor,

Question 5.
Write the relationship between angle of incidence ‘ i ‘ angle of prism ‘A’ and angle of minimum deviations δ_m for a triangular prism. [1]
Answer :
The relation between the angle of incidence i, angle of prism A, and the angle of minimum deviation δm, for a triangular prism is given by \(i=\frac{\mathrm{A}+\delta_{m}}{2}\)

Question 6.
The given graph shows the variation of photoelectric current (I) versus applied voltage (V) for two different photosensitive materials and for two different intensities of the incident radiations. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation. [2]

Answer :
Curves 1 and 2 correspond to similar materials while curves 3 and 4 represent different materials, since the value of stopping potential for 1, 2 and 3, 4 are the same. For the given frequency of the incident radiation, the stopping potential is independent of its intensity.
So, the pairs of curves (1 and 3) and (2 and 4) correspond to different materials but same intensity of incident radiation.

Question 7.
A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 Ω as shown in the figure. Find the value of the current in circuit. [2]

Answer :
Since, the positive terminal of the batteries are connected together (So they oppose each other), so the equivalent emf of the batteries is given by
E = 200 – 10 = 190 V
Hence, the current in the circuit is given by
\(I=\frac{E}{R}=\frac{190}{38}=5 \mathrm{A}\)

Question 8.
The emf of a cell is always greater than its terminal voltage. Why ? Give reason. [2]
Answer :
The emf of a cell is greater than its terminal voltage because there is some potential drop across the cell due to its small internal resistance.

Question 9.
(a) Write the necessary conditions for the phenomenon of total internal reflection to occur.
(b) Write the relation between the refractive index and critical angle for a given pair of optical media. [2]
Answer :
(a) Necessary conditions for total internal reflection to occur are :

1. The incident ray on the interface should travel in optically denser medium.
2. The angle of incidence should be greater than the critical angle for the given pair of optical media.

(b) \(a_{\mu_{b}}=\frac{1}{\sin C}\)
Where a and b are the rarer and denser media respectively. C is the critical angle for the given pair of optical media.

Question 10.
State Lenz’s law.
A metallic rod held horizontally along east- west direction, is allowed to fall under gravity. Will there be an emf induced at its ends ? Justify your answer. [2]
Answer :
Lenz’s law states that the polarity of induced emf is such that it produces a current which opposes the change in magnetic flux that produces it. Yes, emf will be induced in the rod as there is change in magnetic flux. When a metallic rod held horizontally along east- west direction, is allowed to fall freely under gravity i.e., fall from north to south, the magnetic flux changes due to vertical component of Earth’s magnetic field, which keeps on changing and the emf is induced in it.

Question 11.
A convex lens of focal length 25 m is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature ? [2]
Answer:
We have, focal length of convex lens,

Question 12.
An ammeter of resistance 0.80 Ω can measure current up to 1.0 A. [3]
(i) What must be the value of shunt resistance to enable the ammeter to measure current up to 5.0 A ?
(ii) What is the combined resistance of the ammeter and the shunt ?
Answer :
We have, resistance of ammeter, RA = 0.80 Ω

Question 13.
In the given circuit diagram a voltmeter ‘V’ is connected across a lamp ‘L’. How would
(i) the brightness of the lamp and
(ii) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’ is decreased ? Justify your answer. [3]

Answer :
The given figure is Common Emitter (CE) configuration of an n-p-n transistor. The input circuit is forward biased and collector circuit is reverse biased.

If resistance R decreases, forward biased in the input circuit will increase, thus the base current (I_B) will decrease and the emitter current (I_E) will increase. This will increase the collector current
(I_C) as I_E = I_B + I_C.

When I_C increases which flows through the lamp, the voltage across the bulb will also increase making the lamp brighter and since the voltmeter is connected in parallel with the lamp, . the reading in the voltmeter will also increase.

Question 14.
(a) An EM wave is travelling in a medium with a velocity \(v=v \hat{i}\). Draw a sketch showing the propagation of the EM wave, indicating the direction of the oscillating electric and magnetic fields.
(b) How are the magnitudes of the electric and magnetic fields related to velocity of the EM wave ? [3]
Answer:
(a) Given.
Velocity, \(v=v \hat{i}\) and electric field, E along Y-axis and magnetic field, B along Z-axis.
The propagation of EM wave is shown below :

(b) Speed of EM wave can be given as the ratio of magnitude of electric field (E₀) to the magnitude of magnetic field (B₀),
\(c=\frac{E_{0}}{B_{0}}\)

Question 15.
Block diagram of a receiver is shown in the figure : [3]

(a) Identify ‘X’ and ‘Y’.
(b) Write their functions.

Question 16.
Explain, with the help of a circuit diagram, the working of a photo diode. Write briefly how it is used to detect the optical signals. [3]
OR
Mention the important consideration required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range ?
Answer :
A junction diode made from light sensitive semiconductor is called a photo diode.

An electrical device that is used to detect and convert light into an energy signal with the use of a photo detector is known as a photo diode. The light that falls on it controls the function of pn-junction. Suppose, the wavelength is such that the energy of a photon \(\frac{h c}{\lambda}\) is enough to break a valence bond. There is an increase in number of charge carriers and hence the conductivity of the junction also increases. New hole-electron pairs are created when such light falls on the junction. If the junction is connected in a circuit, the intensity of the incident light controls the current in the circuit.
OR
(i) The reverse breakdown voltage of LED’s are very low, which is around 5V. So enough care is to be taken while fabricating a pn-junction diode such that the high reverse voltages do not occur across them.

(ii) There exist very small resistance to limit the current in LED. So, a resistor must be placed in series with the LED such that no damage is occurred to the LED. The semiconductor used for fabrication of visible LED’s must at least have a band gap of 1.8 eV.

Question 17.
Write three important factors which justify the need of modulating a message signal. Show diagrammatically how an amplitude modulated wave is obtained when a modulating signal is superimposed on a carrier wave. [3]

Question 18.
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 μC. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 μC.
Calculate:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V ? [3]
OR
A hollow cylindrical box of length 1 m and area of cross-section 25 cm² is placed in a three-dimensional coordinate system as shown in the figure. The electric field in the region is given by \(\overrightarrow{\boldsymbol{E}}=50 x \hat{\boldsymbol{i}}\) where \(\mathrm{E} \text { is } \mathrm{NC}^{-1} \mathrm{a}\) and x is in meters. Find:
(i) Net flux through the cylinder.
(ii) Charge enclosed by the cylinder.



Question 19.
(a) In a typical nuclear reaction, e.g.
\(_{1}^{2} \mathrm{H}+_{1}^{2} \mathrm{H} \rightarrow_{2}^{3} \mathrm{He}+\frac{1}{0} n+3.27 \mathrm{MeV}\) although number of nucleons is conserved, yet energy is released. How ? Explain.
(b) Show that nuclear density in a given nucleus is independent of mass number A. [3]
Answer:
(a) In a nuclear reaction, the aggregate of the
masses of the target nucleus \(\overset { 2 }{ 1 } H\) and the bombarding particle may be greater or less than the aggregate of the masses of the product nucleus \(\overset { 3 }{ 2 } H\) and the outgoing particle \(\overset { 1 }{ 0 } H\) So from the law of conservation of mass-energy some energy (3.27 MeV) is evolved or involved in a nuclear reaction. This energy is called Q-value of the nuclear reaction.

Question 20.
(a) Why photoelectric effect cannot be explained on the basis of wave nature of light ? Give reasons.
(b) Write the basic features of photon picture of electro magnetic radiation on which Einstein’s photoelectric equation is based. [3]
Answer :
(a) Wave nature of radiation cannot explain the following : .

1. The immediate ejection of photo electrons.
2. The presence of threshold frequency for a metal surface.
3. The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency. Thus, the photoelectric effect cannot be explained on the basis of wave nature of light.

(b) Photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based on particle nature of light. Its basic features are :

4. By increasing the intensity of light of given wavelength, there is only an increase in the number of photons emitted per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.

5. Photons are electrically neutral and are not deflected by electric and magnetic fields.

6. In a photon-particle collision (such as photon- electron collision), the total energy and total momentum are conserved. However, number of photons may not be Observed.

Question 21.
A metallic rod of length ‘l’ is rotated with a frequency v with one end hinged at the center and the other end at the circumference of a circular metallic ring of radius r, about an axis passing through the center and perpendicular to the plane of the ring. A constant uniform magnetic field B parallel to the axis is present everywhere. Using Lorentz force, explain how emf is induced between the center and the metallic ring and hence obtained the expression for it. [3]
Answer:
Suppose the length of the rod is greater than the radius of the circle and rod rotates anticlockwise and suppose the direction of electrons in the rod at any instant be along +y direction.
Suppose the direction of the magnetic field is along +z direction.
Then, using Lorentz law, we get the following :

Thus, the direction of force on the electrons is along -x axis. So, the electrons will move towards the centre i.e., the fixed end of the rod. This movement of electrons will effect in current and thus it will generate an emf in the rod between the fixed end and the point touching the ring.

Let θ be the angle between the rod and radius of the circle at any time t.
Then, area swept by the rod inside the circle =

Question 22.
Output characteristics of an n-p-n transistor in CE configuration is shown in the figure. [3]
Determine:
(i) dynamic output resistance
(ii) dc current gain and
(iii) ac current gain at an operating point V_CE = 10 V, when I_B = 30 μA.

Question 23.
Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels. [4]
Answer:
In a hydrogen atom, Radius of electron orbit,




When the electron in a hydrogen atom jumps from higher energy level to the lower energy level, the difference of energies of the two energy levels is emitted as a radiation of particular wavelength. It is called a spectral line.
In H-atom, when an electron jumps from the orbit n_i, to orbit n_f, the wavelength of the emitted radiation is given by,


Question 24.
(a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment.
(b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn to study the diffraction taking place at a single slit of aperture 2 x 10-4 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases. [5]
Answer :
(a) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double slit experiment is modified by diffraction from each of the two slits.

Question 25.
In a series LCR circuit connected to an ac source of variable frequency and voltage v = v_m sin ωt, draw a plot showing the variation of current (I) with angular frequency (ω) for two different values of resistance R₁ and R2 (R₁ > R₂). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced ? Define Q-factor of the circuit and give its significance. [5]
Answer :
Figure shows the variation of i_m with ω in a LCR series circuit for two values of Resistance R₁ and R₂ (R₁ > R₂),

The condition for resonance in the LCR circuit is,
\(\omega_{0}=\frac{1}{\sqrt{L C}}\)

We can observe that the current amplitude is maximum at the resonant frequency (ω₀. Since i_m = V_m/R at resonance, the current amplitude for case R₂ is sharper to that for case R₁.
Quality factor or simply the Q-factor of a resonant LCR circuit is defined as the ratio of voltage drop across the capacitor (or inductor) to that of applied voltage.
It is given by \(\mathrm{Q}=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}\)

The Q factor determines the sharpness of the resonance curve and if the resonance is less sharp the maximum current decreases and also the circuit is close to the resonance for a larger range Δω of frequencies and the regulation of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit while higher is the Q, sharper is the resonance curve and lesser will be the loss in energy of the circuit and circuit will be more selective.

Question 26.
While travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm. It became very dark. He stopped driving the car and waited for thunderstorm to stop. Suddenly he noticed a child walking alone on the road. He asked the boy to come inside the car till the thunderstorm stopped. Dr. Pathak dropped the boy at his residence. The boy insisted that Dr. Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak for his concern for safety of the child.
Answer the following questions based on the above information: [5]
(a) Why is it safer to sit inside a car during a thunderstorm ?
(b) Which two values are displayed by Dr. Pathak in his action ?
(c) Which values are reflected in parents’ response to Dr. Pathak ?
(d) Give an example of similar action on your part in the part from everyday life.
Answer :
(a) It is safer to be inside a car during thunderstorm because the car acts like a Faraday cage.
(d) Once I came across to a situation where a puppy was struck in the middle of a busy road during rain and was not able to cross due to heavy flow, So I quickly rushed and helped him.

Question 27.
(a) Draw a ray diagram showing the image formation by a compound microscope. Hence obtain expression for total magnification when the image is formed at infinity.
(b) Distinguish between myopia and hypermetropia. Show diagrammatically how these defects can be corrected. [5]
OR
(a) State Huygens’s principle. Using this principle draw a diagram to show how a plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snell’s law of refraction.
(b) When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons:
(i) Is the frequency of reflected and refracted light same as the frequency of incident light ?
(ii) Does the decrease in speed imply a reduction in the energy carried by light wave ?
Answer :
(a) A compound microscope consists of two convex lenses parallel separated by some distance. The lens nearer to the object is called the objective. The lens through which the final image is viewed is called the eyepiece.

infinity : The magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and objective.
∴ M = M_e x M₀

Where, M_e and M₀ are the magnifying powers of the eyepiece and objective respectively.
If μ₀ is the distance of the object from the objective and v₀ is the distance of the image from the objective, then the magnifying power of the objective is
\(\mathrm{M}_{o}=\frac{h^{\prime}}{h}=\frac{\mathrm{L}}{f_{o}}\)

Where, h, h’ are object and image heights respectively and f₀ is the focal length of the objective and L is the tube length i.e., the distance between the second focal point of the objective and the first focal point of the eyepiece.
When the final image is at infinity,
\(\mathrm{M}_{e}=\frac{\mathrm{D}}{f_{e}}\)
Magnifying power of compound microscope,
\(\mathbf{M}=\mathbf{M}_{o} \times \mathbf{M}_{e}=\frac{\mathbf{L}}{f_{o}} \times \frac{\mathbf{D}}{f_{e}}\)
Magnifying power when final image is at

OR
a) Huygen’s Principle:

1. Each point on the primary wave front acts as a source of secondary wavelets, transferring out disturbance in all directions in the same way as the original source of light does.
2. The new position of the wave front at any instant is the envelope of the secondary wavelets at that instant.

Refraction on the basis of wave theory:

1. Consider any point Q on the incident wave front.
2. Suppose when disturbance from point P on incident wave front reaches point P’ on the refracted wave front, the disturbance from point Q reaches Q’ on the refracting surface XY.

3. Since P’A’ represents the refracted wave front, the time taken by light to travel from a point on incident wave front to the corresponding point on refracted wave front should always be the same.
Now, time taken by light to go from Q to Q’ will be

The rays from different points on the incident wave front will take the same time to reach the corresponding points on the refracted wave front i.e., given equation (iv) is independent of AK. It will happen so, if

This is the Snell’s law for refraction of light.
(b) (i) The frequency of reflected and refracted light remains constant as the frequency of incident light because frequency only depends on the source of light.
(ii) Since the frequency remains same, hence there is no reduction in energy.

Question 28.
(a) State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf’s of two primary cells. Obtain the required expression used for comparing the emfs.
(b) Write two possible causes for one sided deflection in a potentiometer experiment. [5]
OR
(a) State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules, obtain the balance condition in terms of the resistance of four arms of Wheatstone bridge.
(b) In the meter bridge experimental set-up, shown in the figure, the null point ‘D’ is obtained at a distance of 40 cm from end A of the meter bridge wire. If a resistance of 10 Ω is connected in series with R₁, null point is obtained at AD = 60 cm. Calculate the values of R₁ and R₂.

Answer :
(a) Working principle of Potentiometer : When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.
Applications of Potentiometer for comparing emf’s of two cells : The following figure shows an application of the potentiometer to compare the emf of two cells of emf E₁ and E₂.


we can compare the emf’s of any two sources. Generally, one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then calculated from equation (iii).
(b) (i) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared.
(ii) The positive ends of a cells are not connected to the same end of the wire.
OR
(a) Kirchhoff’s first law – Junction rule : The algebraic Siam of the currents meeting at a point in an electrical circuit is always zero.

Let the currents be Ii, I2,13, and I4.
Convention : Current towards the junction positive
Current away from the junction negative
I₃ + (-I₁) + (-I₂) + (-1₄) = 0
This law is in accordance with conservation of charge

Kirchhoff’s second law-Loop rule : In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistance and current flowing through them

Wheatstone bridge : The Wheatstone bridge is an arrangement of four resistances as shown in the following figure


This is the required balanced condition of Wheatstone bridge.
(b) Considering both the situations and writing them in the form of equations Let R’ be the resistance per unit length of the potentiometer wire,

Question 29.
(a) Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field.
(b)A proton and a deuteron having equal momenta enter in a region of a uniform magnetic field at right angle to the direction of the field. Depict their trajectories in the field. [5]
OR
(a) A small compass needle of magnetic moment’m’ is free to turn about an axis perpendicular to the direction of uniform magnetic field ‘B’. The moment of inertia of the needle about the axis is ‘I’. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period.
(b) A compass needle, free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. Find out the values of
(i) horizontal component of earth’s magnetic field and
(ii) angle of dip at the place.
Answer:
(a) Consider a rectangular loop PQRS of length l, breadth b suspended in a uniform magnetic field \(\overrightarrow{\mathrm{B}}\) . The length of loop PQ = RS = l and breadth = QR = SP = b. Let at any instant the normal to the plane of loop make an angle θ with the direction of magnetic field \(\overrightarrow{\mathrm{B}}\) and I be the current in the loop. We know that a force acts on a current carrying wire placed in a magnetic field. Therefore, each side of the loop will experience a force. The net torque acting on the loop will be determined by the forces acting on all sides of the loop. Suppose that the forces on sides

CBSE Previous Year Question Papers Class 12 Physics 2013 Delhi Set-II

Note: Except for the following questions, all the remaining questions have been asked in previous Set.

Question 1.
A cell of emf ‘E’ and internal distance ‘r’ draws a current ‘I’. Write the relation between terminal voltage ‘V’ in terms of E, I, r. [1]
Answer:
When the current I draws from a cell of emf E and internal resistance r, then the terminal voltage is
V = E – Ir

Question 2.
Which of the following substances are diamagnetic ? [1]
Bi, Al, Na, Cu, Ca and Ni
Answer:
Bi and Cu are diamagnetic substances

Question 3.
A heating element is marked 210 V, 630 W. What is the value of current drawn by the element when connected to a 210 V dc source ? [1]
Answer:
In dc source, P = VI
Given that P = 630 W and V = 210 V.
So,
\(I=\frac{P}{V}=\frac{630}{210}=3 \mathrm{A}\)

Question 4.
An ammeter of resistance 1Ω can measure current up to 1.0 A.
(i) What must be the value of the shunt resistance to enable the ammeter to measure upto 5.0 A ?
(ii) What is the combination resistance of the ammeter and the shunt ? [1]
Answer :
We have, Resistance of ammeter,
R_A = 1Ω
Maximum current across ammeter, I_A = 1.0 A
So, Voltage drop across ammeter, V = IR = 1.0 × 1.0 = 1V

Question 5.
A convex lens of focal length 20 cm is placed coaxially in contact with a concave lens of focal length 25 cm. Determine the power of the combination. Will the system be converging or diverging in nature ? [2]
Answer:
We have, Focal length of concave lens f₁ = +20 cm = +0.20 m,
Focal length of concave lens, f₂ = -25 cm = -0.25 m
Power of convex lens,

Question 6.
Using Bohr’s postulates, obtain the expressions for
(i) kinetic energy and
(ii) potential energy of the electron in stationary state of hydrogen atom.
Draw the energy level diagram showing how the transitions between energy levels result in the appearance of Lyman series. [2]
Answer :
According to Bohr’s postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge + e.
And the radius of orbit in which the electron revolves, is given by



Question 7.
Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is moved with a velocity v towards the arm RS Assuming that the arms QR, RS and SP have negligible resistances and the moving arm PQ has the resistance r, obtain the expression for
(i) the current in the loop
(ii) the force and
(iii) the power required to move the arm PQ.


Question 8.
Distinguish between ‘sky waves’ and ‘space waves’ modes of propagation in communication system. [4]
(a) Why is sky wave mode propagation restricted to frequencies upto 40 MHz ?
(b) Give two examples where space wave mode of propagation is used

CBSE Previous Year Question Papers Class 12 Physics 2013 Delhi Set-III

Note : Except for the following questions, all the remaining questions have been asked in previous Sets.

Question 1.
A 5 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 39 Q as shown in the figure. Find the value of the current. [2]

Question 2.
Which of the following substances are paramagnetic ? [2]
Bi, Al, Cu, Ca, Pb, Ni
Answer :
Paramagnetic substances are Aluminium (Al) and Calcium (Ca).

Question 3.
An ammeter of resistance 0.6 Ω can measure current up to 1.0 A. Calculate
(i) The shunt resistance required to enable the ammeter to measure current up to 5.0 A.
(ii) The combined resistance of the ammeter and the shunt. [2]
Answer :
We have, resistance of ammeter, RA = 0.60 ohm.


Question 4.
A convex lens of focal length 30 cm is placed coaxially in contact with a concave lens of focal length 40 cm. Determine the power of the combination. Will the system be converging or diverging in nature ? [3]
Answer:
We have, focal length of convex lens,

Question 5.
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 μC. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 μC. Calculate the potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100 V ? [3]
OR
A hollow cylindrical box of length 0.5 m and area of cross-section 20 cm is placed in a three-dimensional coordinate system as shown in the figure. The electric field in the region is given by




Question 6.
(a) Write two characteristic features distinguishing the diffraction pattern from the interference fringes obtained in Young’s double slit experiment.
(b) Two wavelength of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place due to a single slit of aperture 1 × 10^-4 m. The distance between the slit and the screen is 1.8 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases. [3]
Answer :
(a) Two characteristic features distinguishing the diffraction pattern from the interference fringes obtained in Young’s double slit experiment are:

• The interference fringes may or may not be of the same width whereas the fringes of diffraction pattern are always of varying width.
• In interference the bright fringes are of same intensity whereas in diffraction pattern the intensity falls as we go to successive maxima away from the centre, on either side.

Question 7.
(a) In a nuclear reaction

though the number of nucleons is conserved on both sides of the reaction, yet the energy is released. How ? Explain.
(b) Draw a plot of potential energy between a pair of nucleons as a function of their separation. Mark the regions where potential energy is
(i) positive and
(ii) negative. [3]
Answer:
(a) In a nuclear reaction, the sum of the masses of the target nucleus \(\overset { 2 }{ 3 } He\) may be greater or less the sum of the masses of the product nucleus \(\overset { 4 }{ 2 } He\) and the \(\overset { 1 }{ 1 } He\) . So from the law of conservation of mass energy, some energy, (12.86 MeV) is evolved in nuclear reaction. This energy is called Q-value of the nuclear reaction. The binding energy of the nucleus on the left side is not equal to the right side. The difference in the binding energies on two sides appears as energy released or absorbed in the nuclear reaction.
(b) The potential energy is minimum at r₀. For distance larger than r₀ the negative potential energy goes on decreasing and for the distances less than r₀ the negative potential energy decrease to zero and then becomes positive and increases abruptly. Thus, A to B is the positive potential energy region and B to C is the negative potential energy region.

Question 8.
(b) What is the significance of negative sign in the expression for the energy ?
(c) Draw the energy level diagram showing how the line spectra corresponding to Paschen series occur due to transition between energy levels. [5]
Answer :
(b) Negative sign indicates that revolving electron is bound to the positive nucleus.

CBSE Previous Year Question Papers

CBSE Previous Year Question Papers Class 12 Physics 2017 Delhi

CBSE Previous Year Question Papers Class 12 Physics 2017 Delhi Set-I

Section – A

Question 1.
Does the charge given to a metallic sphere depend on whether it is hollow or solid ? Give reason for your answer. [1]
Answer :
No, because all the charge resides on the surface of the sphere only.

Question 2.
A long straight current carrying wire passes normally through the center of circular loop. If the current through the wire increases, will there be an induced emf in the loop ? Justify. [1]
Answer :
No, the emf will not be induced as the magnetic field lines are parallel to the plane of the circular loop. So magnetic flux will remain zero.

Question 3.
At a place, the horizontal component of earth’s magnetic field is B and angle of dip is 60°. What is the value of horizontal component of the earth’s magnetic field at equator ? [1]


Question 4.
Name the junction diode whose I-V characteristics are drawn below.

Answer:
This is the characteristic of solar cell.

Question 5.
How is the speed of em-waves in vacuum determined by the electric and magnetic fields? [1]
Answer:
The speed of em waves are determined by the ratio of the peak values of electric and magnetic field vectors.
\(c=\frac{E_{0}}{B_{0}}\)

Section-B

Question 6.
How does Ampere-Maxwell law explain the flow of current through a capacitor when it is being charged by a battery ? Write the expression for the displacement current in terms of the rate of change of electric flux. [2]
Answer
During the charging of capacitor, (electric flux between the plates of capacitor keeps on changing, due to which displacement current between the plates is produced. Hence circuit becomes complete and current flow through the circuit.
\(\mathrm{I}_{\mathrm{D}}=\epsilon_{0} \frac{d \phi_{\mathrm{E}}}{d t}\)

Question 7.
Define the distance of closest approach. An a-particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is ‘r’. What will be the distance of closest approach for an a-particle of double the kinetic energy ? [2]
OR
Write two important limitations of Rutherford nuclear model of the atom.
Answer :
It is defined as the minimum distance of the charged particle from the nucleus at which initial kinetic energy of the particle is equal to the potential energy due to the charged nucleus.

Limitations :
(a) It is not in accordance with the Maxwell’s theory and could not explain the stability of an atom.
(b) It did not say anything about the arrangement of electrons in an atom.

Question 8.
Find out the wavelength of the electron orbiting in the ground state of hydrogen atom. [2]
Answer: We know that, according to de Broglie relation,

Question 9.
Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope has short focal lengths ? Explain. [2]
Answer:
Magnifying Power : The magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the object are situated at the least distance of distinct vision from the eye. Magnifying power of the compound microscope when the final image is at infinity,
\(m=-\frac{\mathrm{L}}{f_{0}} \times \frac{\mathrm{D}}{f_{e}}\)
From the above equation, we can see that to achieve a large magnification, the objective and eyepiece should have small focal lengths.

Question 10.
Which basic mode of communication is used in satellite communication ? What type of wave propagation is used in this mode ? Write, giving reason, the frequency range used in this mode of propagation. [2]

Section – C

Question 11.
(a) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown below. Which one leads in phase current or voltage ?
(b) Without making any other change, find the value of the additional capacitor C₁, to be connected in parallel with capacitor C, in order to make the power factor of the circuit unity. [3]



Question 12.
Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction. [3]
Answer :
Two important processes involved during the formation of p-n junction are diffusion and drift. Electron diffusion

Formation of depletion region and barrier potential: At the instant of p-n junction formation, the free electrons near the junction diffuse across the junction into the p region and combines with holes. Thus, on combining with the hole, it makes a negative ion and leaves a positive ion on n-side. These two layers of immobile positive and immobile negative charges form the depletion region.

Further, as electrons diffuse across the junction a point is reached where the negative charge repels any further diffusion of electron. This depletion region now acts as a barrier. Now the external energy is supplied to get the electrons to move across the barrier of electric field. The potential difference required to move the electrons through the electric field is called barrier potential.

Question 13.
(a) Obtain the expression for the cyclotron frequency.
(b) A deuteron and a proton are accelerated by the cyclotron. Can both be accelerated with the same oscillator frequency ? Give reason to justify your answer. [3]
Answer:
(a) Suppose the positive ion with charge q moves in a dee with a velocity v, then
\(\begin{aligned} \frac{m v^{2}}{r} &=\mathrm{qvB} \\ r &=\frac{m v}{q \mathrm{B}} \end{aligned}\)
where m is the mass and r is the radius of the path of ion in the dee and B is the strength of the magnetic field. The time taken by the ion,
\(\begin{array}{l}{\mathrm{T}=\frac{2 \pi r}{v}} \\ {\frac{1}{f}=\frac{2 \pi r}{v}} \\ {f=\frac{\mathrm{v}}{2 \pi r}} \\ {f=\frac{v}{2 \pi \times \mathrm{mv}}[\mathrm{using}(\mathrm{I})]}\end{array}\)

This frequency is called the cyclotron frequency,
(b) Deuteron and proton have different masses and cyclotron on frequency depends inversly on mass. Hence, cannot be accelerated with the same oscillator frequency.

Question 14.
(a) How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?
(b) The work function of the following metals is given: Na = 2.75 eV, K = 23 eV, Mo = 4.17 eV and Ni = 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 A from a laser source placed 1 m away from these metals ? What happens if the laser source is brought nearer and placed 50 cm away ? [3]

The work function of Na and K is less than the energy of incident radiation. ,Therefore, Na and K will cause photoelectric emission while Mo and Ni will not cause photoelectric emission. There will be no effect on photoelectric emission if the source is brought nearer.

Question 15.
A resistance of R draws current from a potentiometer. The potentiometer wire, AB, has a total resistance of R₀. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of potentiometer wire. [3]


Question 16.
Define the term ‘amplitude modulation.’ Explain any two factors which justify the need for modulating a low frequency base-band signal. [3]

Question 17.
(a) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2μF capacitance.

(b) If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network? [3]
Answer :
(a) In the given figure, C₂, C₃ and C₄ are in parallel so, the equivalent capacitance of parallel capacitors is given by C.


Question 18.
(a) Derive the expression for electric field at a point on the equatorial line of an electric dipole.
(b) Depict the orientation of the dipole in
(i) stable,
(ii) unstable equilibrium in a uniform electric field. [3]
Answer :
(a) Consider a point P on broad side position of dipole formed of charges +q and -q at separation 2l. The distance of point P from mid point (O) of electric dipole is r. Let \(\overrightarrow{\mathrm{E}_{1}}\) and \(\overrightarrow{\mathrm{E}_{1}}\) be the electric field strengths due to charges + q and – q of electric dipole.




Question 19.
(a) A radioactive nucleus ‘A’ undergoes a series of decays as given below:

The mass number and atomic number of A₂ are 176 and 71 respectively. Determine the mass and atomic numbers of A₄ and A.
(b) Write the basic nuclear process underlying β⁺ and β^– decays.
Answer:
(a)


Question 20.
(a) A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30°. Calculate the speed of light through the prism.

(b) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC. [3]



Question 21.
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. Given the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ. [3]

Question 22.
Describe the working principle of a moving coil galvanometer. Why is it necessary to use
(i) a radial magnetic field and
(ii) a cylindrical soft iron core in a galvanometer? Write the expression for current sensitivity of the galvanometer. Can a galvanometer as such be used for measuring the current ? Explain. [3]
OR
(a) Define the term ‘self-inductance’ and write its S.I. unit.
(b) Obtain the expression for the mutual inductance of two long co-axial solenoids S₁ and S₂ wound one over the other, each of length L and radii r₁ and r₂ and n₁ and n, number of turns per unit length, when a current I is set up in the outer solenoid S₂.
Answer:
Moving coil galvanometer
Principle : It works on the principle that a current carrying coil placed in a magnetic field experiences a torque.
Radial magnetic field: It is necessary to eliminate the effect of 0 on torque and hence enable us to make the scale of galvanometer linear.
Cylindrical soft iron core: It makes the magnetic lines of force pointing along the radii of the circle and also due to high permeability it intensifies the magnetic field and hence increases the sensitivity of the galvanometer.
Current sensitivity:
\(I_{s}=\frac{N B A}{K} \text { or } \frac{Q}{I}\)

where Q is the deflection of coil No, it can not be used to measure electric current because it is a highly current sensitive device which would show large deflection even with the passage of small current. And the galvanometer coil is likely to damaged by currents in (mA/A) rang.
OR
(a) Self-inductance: The self inductance of a coil may be defined as the induced emf set up in the coil due to a unit rate of change of current through it.
Its S.I. unit is Henry (H).

Section-D

Question 23.
Mrs. Rashmi Singh broke her reading glasses. When she went to the shopkeeper to order new spects, he suggested that she should get spectacles with plastic lenses instead of glass lenses. On getting the new spectacles, she found that the new ones were thicker than the earlier ones, she asked this question to the shopkeeper but he could not offer satisfactory explanation for this. At home, Mrs. Singh raised the same question to her daughter Anuja who explained why plastic lenses were thicker. [4]
(a) Write two qualities displayed each by Anuja and her mother.
(b) How do you explain this fact using lens maker’s formula ?
Answer:
(b) According to lens maker’s formula,
\(\frac{1}{f}=(n-1)\left(\frac{1}{\mathbf{R}_{1}}-\frac{1}{\mathbf{R}_{2}}\right)\)

where f is the focal length, n is the refractive index and R1 and R2 is the radius of curvature of the lens.
Since , n_g > n_p
where, g stands for glass and p stands for plastic. Therefore, we get (n_g -1) > (n_p -1)
Now, using the lens maker’s formula, we see that focal length is inversely proportional to (μ -1).
Hence, f_p > f_g

Thus, in the case of plastic lens the thickness of the lens should be increased to keep the same focal length as that of the glass lens to give the same power.

Section-E

Question 24.
(a) Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.
(b)A circular coil of cross-sectional area 200 cm² and 20 turns is rotated about the vertical diameter with angular speed of 50 rad s^-1 in a uniform magnetic field of magnitude 3.0 x 10^-2 T. Calculate the maximum value of the current in the coil. [5]
OR
(a) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turns and currents in the two coils.
(b) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary winding’s having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V.





Question 25.
(a) Distinguish between unpolarised light and linearly polarised light. How does one get linearly polarised light with the help of a Polaroid ?
(b)A narrow beam of unpolarised light of intensity I₀ is incident on a polaroid P₁ The light transmitted by it is then incident on a second polaroid P₁ with its pass axis making angle of 60° relative to the pass axis of P₁ Find the intenstity of the light transmitted by P₂. [5]
OR
(a) Explain two features to distinguish between the interference pattern in Young’s double slit experiment with the diffraction pattern obtained due to a single slit.
(b) A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen. Estimate the number of fringes obtained in Young’s double slit experiment with fringe width 0.5 mm, which can be accommodated with the region of total angular spread of the central maximum due to single slit. [5]
Answer :
(a) Unpolarised light : If the electric component of light vibrates in all the direction in the plane perpendicular to the direction of propagation, the light is known as unpolarised light.
Linearly polarised light: If the vibration of electric components of light are restricted in one direction only in the plane perpendicular to the propagation of light then light is known as linearly polarised light.

A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors along the direction of the aligned molecules get absorbed. So, when an unpolarised light falls on a polaroid, it lets only those of its electric vectors that are oscillating along a direction perpendicular to its aligned molecules to pass through it. The incident light thus gets linearly polarised.
(b)



Question 26.
(a) Derive an expression for drift velocity of electrons in a conductor. Hence deduce Ohm’s law.
(b) A wire whose cross-sectional area is increasing linearly from its one end to the other, is connected across a battery of V volts. Which of the following quantities remain constant in the wire ?
(i) drift speed
(ii) current density
(iii) electric current
(iv) electric field Justify your answer. [5]
OR
(a) State the two Kirchhoff’s laws. Explain briefly how these rules are justified.
(b) The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for
(i) the current drawn from the cell and
(ii) the power consumed in the network.

Answer :
(a) Expression of Drift velocity : It is defined as the average velocity gained by the free electrons of a conductor in the opposite direction of the externally applied electric field. If an electron have initial velocity u₁ and accelerated for time t₁ then velocity attain by it will be


Hence  Ohm’s law  is derived.
(b) The electric current will remain constant in the wire as it does not depend upon the cross-sectional area whereas the drift speed, current density, electric field depends upon the increasing area of cross-section with the following relations:

OR
(a) Kirchhoff’s laws:
1. Kirchhoff’s first law or junction rule: In an electric circuit, the algebraic sum of currents at any junction is zero.
ΣI= 0
2. Kirchhoff’s second law or loop rule :
Around any closed loop of a network, the algebraic sum of changes in the potential must be zero.
ΣΔV = 0
Justification :
First law is based on the law of conservation of charge and second law is based on the law of conservation of energy.

CBSE Previous Year Question Papers Class 12 Physics 2017 Delhi Set-II

Note : Except for the following questions, all the remaining questions have been asked in previous set.

Question 1.
Find the wavelength of the electron orbiting in the first excited state in hydrogen atom. [2]
Answer: Radius of n^th orbit,

Question 2.
Distinguish between a transducer and a repeater. [2]

Question 3.
Why should the objective of a telescope have large focal length and large aperture ? Justify your answer. [2]

Question 4.
In the study of a photoelectric effect the graph between the stopping potential V and frequency v of the incident radiation on two different metals P and Q is shown below:

Question 5.
(i) Which one of the two metals has higher threshold frequency ?
(ii) Determine the work function of the metal which has greater value.
(iii) Find the maximum kinetic energy of electron emitted by light of frequency 8 × 10^-14 Hz for this metal.
Answer:
(i) Q has higher threshold frequency.
(ii) Work function of Q


Question 6.
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor ? If another capacitor of 6 pF is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor. [3]
Answer :
Energy stored in the capacitor of capacitance 12 pF,

Question 7.
A zener diode is fabricated by heavily doping both p and n- sides of the junction. Explain, why ? Briefly explain the use of zener diode as a dc voltage regulator with the help of a circuit diagram. [3]
Answer :
By heavily doping both p and n sides of the junction, depletion region formed is very thin, i.e. < 10^-6 m. Hence, electric field across the junction is very high (~5 x 10⁶ V/m) even for a small reverse bias voltage. This can lead to a break down during reverse biasing.

If the input voltage increases/decreases then current through resistor R_s and Zener diode also increases/decreases. This increases/decreases the voltage drop across R_s without any change in voltage across the zener diode. This is because in the breakdown region, zener voltage remains constant even though the current through the zener diode changes.

Question 8.
A electron of mass m_e revolves around a nucleuse of charge +Ze. Show that it behaves like a tiny magnetic dipole. Hence prove that the magnetic moment associated with it is expressed as \(\vec{\mu}=-\frac{e}{2 m_{0}} \overrightarrow{\mathrm{L}}\) where \(\overrightarrow{\mathrm{L}}\) is the orbital angular momentum of the electron. Give the significance of negative sign. [3]
Answer :
Electron in circular motion around the nucleus, constitutes a current loop which behaves like a magnetic dipole.
Current associated with the revolving electron,

The negative sign signifies that the angular momentum of the revolving electron is opposite in direction to the magnetic moment associated with it.

Question 9.
(a) Derive the expression for the electric potential due to an electric dipole at a point on its axial line.
(b) Depict the equipotential surfaces due to an electric dipole. [3]
Answer :
(a) Consider an electric dipole having charges -q and +q at separation ‘2a’. The dipole moment of dipole is\(\vec{p}=q(2 \vec{a})\), directed from -q to +q. The electric potential due to dipole is the algebraic sum of potentials due to charges +q and -q.

CBSE Previous Year Question Papers Class 12 Physics 2017 Delhi Set-III

Note : Except for the following questions, all the remaining questions have been asked in previous sets.

Question 1.
When are two objects just resolved ? Explain. How can the resolving power of a compound microscope be increased ? Use relevent formula to support your answer. [2]
Answer:
When the maxima of diffraction pattern from one object coincide with the minima of second object then they are just resolved.

The resolving power of a compound microscope can be increased by increasing µ and by decreasing λ.

Question 2.
(a) What is the line of sight communication ?
(b) Why is it not possible to use sky waves for transmission of T.V. signals ? Up to what distance can a signal be transmitted using an antenna of height ‘h’ ? [2]

Question 3.
An a-particle and a proton are accelerated through the same potential difference. Find the ratio of their de Broglie wavelengths. [2]

Question 4.
(a) State two important features of Einstein’s photoelectric equation.
(b) Radiation of frequency 10¹⁵ Hz is incident on two photosensitive surfaces P and Q. There is no photoemission from surface P. Photoemission occurs from surface Q but photoelectrons have zero kinetic energy. Explain these observations and find the value of work function for surface Q. [3]

Question 5.
(a) Obtain the expression for the torque \(\vec{\tau}\) experienced by an electric dipole of dipole moment \(\overrightarrow{\mathrm{p}}\) in a uniform electric field, \(\overrightarrow{\mathbf{E}}\) .
(b)What will happen if the field were not uniform ? [3]
Answer:
(a)


(b) If the field is non-uniform then there will be a net force acting on the dipole. Also, a net torque acting on the dipole which depends on the location of the dipole.

Question 6.
Explain briefly with the help of necessary diagrams, the forward and the reverse biasing of a p-n junction diode. Also draw their characteristic curves in the two cases. [3]
Answer :
Forward biasing : If the positive terminal of the battery is connected to the p type semiconductor and negative terminal to the n type semiconductor then it is said to be in forward biased.

Reverse Biasing : If the positive terminal of the battery is connected to the n type semiconductor and the negative termal to the p type semiconductor then it is said to be reverse biased.


Question 7.
Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination ? If these were connected in parallel across the same battery, how much energy will be stored in the combination now ? Also find the charge drawn from the battery in each case. [3]



Question 8.
(a) Write the expression for the force \(\overrightarrow{\mathbf{F}}\) acting on a particle of mass m and charge q moving with velocity \(\overrightarrow{\mathbf{v}}\) in a magnetic field \(\overrightarrow{\mathbf{B}}\). Under what conditions will it move in
(i) a circular path and
(ii) a helical path ?
(b) Show that the kinetic energy of the particle moving in magnetic field remains constant. [3]
Answer:
(a) \(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\mathrm{q} \mathrm{vB} \sin \theta\)
(i) If the angle between v and B is 90° then it will move in circular path.
(ii) If the angle is other than, 0°, 90° and 180° the path will be helical.
(b) Since the work done on the charged particle moving in the magnetic field is zero because force experienced is perpendicular cannot bring any change in speed of charged particuler. Hence the change in K.E. is zero.

CBSE Previous Year Question Papers