Nirmala

CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2011

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. There are 26 questions in all.
2.  This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
4.  There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
5. You may use the following values of physical constants wherever necessary:

SET I

Question.1.Define electric dipole moment. Write its S.I. unit.
Answer : Electric dipole moment : Dipole moment is a measure of strength of electric dipole. It is vector quantity whose magnitude is equal to product of magnitude of charge and the distance between them.
p = q x d
SI unit of dipole moment is coulomb-meter (C-m)

Question.2.Where on the surface of Earth is the angle of dip 90°?
Answer : Magnetic dip is the angle made by a compass needle with the horizontal point on earths surface. Positive value of inclination indicates that the field is pointing downwards, into the earth, at the point of measurement. The angle of dip is 90° at the poles.

Question.3. A hollow metal sphere of radius 3 cm is charged such that the potential on its surface is 10 V. What is potential at the centre of the sphere?
Answer : Potential inside the charge sphere is constant and equal to potential on the surface of the conductor so, therefore, potential at the centre of sphere is 10V.

Question.4. How are radiowaves produced?
Answer : Production of Radiowaves : These are the electromagnetic waves of frequeficy ranging from 500 kHz to about 1000 MHz. Radiowaves are made by various types of transmitter, depending on the wavelength. They are also given off by starts, sparks and lighting. These waves are used in the field of radio communication.

Question.5. Write any two characteristic properties of nuclear force.
Answer : Characteristic properties of nuclear forces are :

1.  Nuclear forces are strongest forces in nature : Magnitude of nuclear forces is 100 times that of electrostatic force and 1038 times the gravitational force.
2. Nuclear forces are charge independent : Nuclear forces between a pair of protons, a pair of neutrons or a pair of neutron and proton act with same strength.

Question.6.Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in .the figure. Predict the polarity of the capacitor.

Question.7. What happens to the width of depletion layer of a p-n junction when it is

1. forward biased,
2. reverse biased?

Answer :

1. Reverse biased : Potential drop across the junction increases and diffusion of holes and electrons across the junction ‘decreases. It makes the width of the depletion layer larger.
2. Forward biased : Potential drop across the junction decreases and diffusion of holes and electrons across the junction increases. It makes the width of the depletion layer smaller. .

Question.8. Define the term ‘stopping potential’ in relation to photoelectric effect.
Answer : Stopping potential is the minimum negative (retarding) potential of anode for which photocurrent stops or becomes zero. It is denoted by Vs. The value of stopping potential is different for different metals but it is independent of the intensity of incident light.

Question.9. A thin straight infinitely long conducting wire having charge density X. is enclosed by a cylindrical surface of radius r and length 4 its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.

Question.10

Question.11. Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity v in a magnetic field B. Show that no work is done by this force on the charged * particle.

Question.12. What are eddy currents? Write any two applications of eddy currents.
Answer : When a bulk piece of conductor is subjected to
changing magnetic flux, the induced current, developed in it is called eddy current.
Applications of eddy currents :

1. Magnetic brakes in trains.
2.  Electromagnetic damping.
3.  Induction furnaces.
4. Electric power meter.

Question.13. What is sky wave communication? Why is this mode of propagation restricted to the frequencies only upto few MHz?
Answer : Ionosphere reflection of radiowaves, back towards the earth, is known as sky wave communication. It is the mode of Wave propagation in.which the radiowaves emitted from the transmitter antenna reach the receiving antenna after reflection by the ionosphere.
The electromagnetic waves of frequencies greater than  40 MHz penetrate the ionosphere and escape. The ionosphere
layers act as a reflector for a certain range of frequencies (3-30 MHz).

Question.14. In the given circuit, assuming point A to be at zero potential, use KirchhofFs rules to determine the potential at point B.

Question.16. Net capacitance of three identical capacitors in series is 1 μF. What will be their net capacitance if connected in parallel?
Find the ratio of energy stored in the two configurations if they are both connected to the same source.

Question.17. Using the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release in .energy in the processes of nuclear fission and nuclear fusion can be explained.

Answer : The above curve shows that:

1.  When a heavy nucleus breaks into two medium sized
   nuclei (in nuclear fission) the BE/nucleon increases resulting, in the release of energy.
2. When two small nuclei combine to form a relatively bigger nucleus in nuclear fusion BE/nucleon increases, resulting in the release of energy.

Question.18. In the meter bridge experiment, balance point was observed atJ with AJ = l.

1.  The values of R and X were doubled and then interchanged. What would be the new position of balance point?
2. If the galvanometer and battery are interchanged at
   the balance position, how will the balance point get affected?
   

Answer :

1.  By doubling and interchanging R and X, there is no change in the position of balance point because
   
2.  By interchanging galvanometer and battery, there will be no change in the balance point position.

Question.19. A convex less made up of glass of refractive index 1.5 is dipped, in turn, in (i) a medium of refractive index 1.65, (ii) a medium of refractive index 1.33.
(a) Will it behave as a converging or a diverging lens in the two cases?
(b) How will its focal length change in the two media?
Answer : (i) When convex lens is dipped in medium of refractive index 1.65, its focal length (f₁) is given by

Question.20. Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, v₁ > v₂ of incident radiation having the same intensity. In which case will be stopping potential be higher? Justify your answer.
Answer:

Stopping potential is more for the curve corresponding to the frequency V₂ (.^..v₁ > v₂)
This is due to the fact that with increase in the frequency, the kinetic energy of emitted photoelectrons also increases. Therefore we need more negative potential to stop these electrons.

Question.22. Use the mirror equation to show that
(a) an object placed between f and 2 f of a concave mirror produces a real image beyond If.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) an object placed between the pole and focus of a concave
mirror produces a virtual and enlarged image.
Answer: As


Therefore, v is always +ve and always less than d, So we can say that convex mirror always produces a virtual image between pole and focus.
(c) Object between pole and F we have 0 < 0 < 1 In this case : v is +ve (virtual image) and /v/>c Therefore, we get a virtual and enlarge image.’

Question.23. Draw a labelled diagram of a full wave rectifier circuit state its working principle. Show the inp ut-output waveforms.
Answer:

Rectification : Rectification means conversion of ac into dc. A p-n diode acts as a rectifier because an ac changes polarity periodically and a p-n diode conducts only when it is forward biased; it does not conduct when reverse biased.
Working : The ac input voltage across secondary S₁ and S₂ changes polarity after each half cycle. Suppose during the first cycle of input ac signal, the terrginal S₁ is positive relative to centre tap and S₂ is negative relative to it Then diode Di is forward biased and D₂ is reverse biased. Therefore, diode D₁ conducts while D₂ does not. Thus, die-current in load resistance R_L is in the same direction for both half cycles of input ac signal the output current is a continuous series of unidirectional pulses.
In a full wave rectifier, if input frequency is Hertz, then output frequency will be 2f Hertz because for each cycle of input, two positive half cycles of output are obtained.

Question.24.(a) Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
(b) The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state? ‘
Answer.

Question.25. You are given, a circuit below. Write its truth table. Hence, identify the 1 qgic operation carried out by this circuit. Draw the logic symbol of the gate it corresponds to.

Question.26. A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope.
OR
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification „ of the telescope.
If this telescope is used to ‘view the moon,’ what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 x 106 m and the radius of the lunar orbit is 3.8 x 10⁸m.
Answer:

Question.27. Two heating elements of resistances Ri and R2 when operated at a constant supply of voltage, V, consume powers P₁ and P₂ respectively. Deduce the expressions for the pqwer of their combination when they are, in turn, connected in (i) series and (i) parallel across the same voltage supply.
Answer: Two resistance are Ri and R2 Both consumed the power Pi and P2 respectively.

Question.28. (a) Using Amperes circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside . the solenoid on its axis.
(b) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.
(c) How is the magnetic field inside a given solenoid made
strong?
OR
(a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram.
(b) “Increasing the current sensitivity of a galvonometer may not necessarily increase its voltage sensitivity.” Justify this statement.
(c) Outline the necessary steps to convert a galvanometer of resistance RQ into an ammeter of a given range.
Answer: Magnetic field inside a long solenoid is uniform energy where and approximately zero outside it. Fig shows a sectional view of long solenoid, current comming out of the plane of the papers at points marked and current entering the plane of the paper at point marked at any point inside consider a rectangular loop abed as Amperian loop.
(b) Difference: In a toroid, magnetic lines do not exist outside the body. Toroid is closed whereas the solenoid is open on both sides.
Magnetic field is uniform inside a toroid whereas for solenoid, it is different at the two ends and centre
to Ampere circuital law.



(c) Strength of magnetic field :
(1) By inserting the ferromagnetic substance inside the solenoid.
(2) By increasing the current through the solenoid.
Answer: (a)

Principle : Its working based on the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque.
Working: When current (I) is passed in the coil, torque x acts on the coil, given by

Question.29. State the working of a.c. generator with”the help of a labelled diagram.
The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity to. Deduce the expression for the alternating e.m.f. generated in the coil. What is the source of energy generation in this device.
OR
(a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
(b) A horizontal straight wire of length L extending from east to west is hilling with speed v at right angles to the horizontal component of Earth’s magnetic field B.

1.  Write the expression for the instantaneous value of the e.m.f. induced in the wire.
2. What is the direction of the e.m.f.?
3. Which end of the wire is at the higher potential?

Answer : Working : When a coil (armature) rotates inside a uniform magnetic field, magnetic flux linked with the coil changes w.r.t. time. This produces an e.m.f according to Faradays law.
For first half of the rotation the current will be from one end (first ring) to the other end (second ring). For second half of the rotation it is in opposite sense.
To calculate the magnitude of e.m.f. induced,

Question.30. State the importance of coherent sources in the phenomenon of interference.
In Youngs double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence deduce the expression for the fringe width. How does the fringe width get affected, if the entire experimental apparatus ofYoung is immersed in water?
Answer : Two sources of light which continuously emit light waves of same frequency with a zero or constant phase difference between them are called coherent sources. They are necessary to produce sustained interference pattern.
A thin film of oil spread over water shows beautiful colors due to interference of light.
If coherent sources are not taken, the phase difference, between the two interfering waves, will change continuously and a sustained interference pattern will not be obtained.

SET II

Question.1. A hollow metal sphere of radius 10 cm is charged such that
the potential on its surface is 5 V. What is the potential at the centre of the sphere?
Answer : Potential inside the charged sphere is constant and equal to potential on the surface of conductor.
Therefore, potential at the centre of sphere is 5V.

Question.2. How are X-rays produced?
Answer : X-rays are produced when electron strike a metal target. The electrons are liberated from the heated filament and accelerated from the high voltage towards the metal target. X-rays are produced’when electrons collide with the atom and nuclei of metal target.

Question.4. Where on the surface of Earth is the angle of dip zero?
Answer : At magnetic equator angle of dip is zero.

Question.12. State the principle of working of a transformer. Can a transformer be used to step up or step down a d c. voltage? Justify your answer.
Answer : Transformer—Principle : It is a device which converts high voltage A.C. into low voltage A.C. and vice -versa. It is based upon the principle of mutual induction. When alternating current passed through a coil, an induced e.m.f. is set up in the neighbouring coil.

Working : When an alternating current is passed through the primary, the magnetic flux through the iron core changes which does two things. It produces e.m.f. in the primary and an induced e.m.f. is also set up in the secondary. If we assume that the resistance of primary is negligible, the back e.m.f. will be equal to the voltage applied to the primary.


A transformer can not be used to step up or step down a d.c. voltage because d.c. can not produce a changing magnetic flux in the core of the transformer.

Question.14. In the given circuit, assuming point A to be at zero potential, use KirchhofFs rules to determine the potential at point B.

Question.17. What is ground wave communication? On what factors does the maximum range of propagation in this mode depend?
Answer : A radiowave that can travel directly from on point to another following the surface of earth is called a ground wave.
The maximum range of ground wave propagation depends upon the following factors :

1.  The frequency of the transmitted wave
2.  Power of transmitter.

Question.22. A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (i) a medium of refractive index 1.6, (ii) a medium of refractive index 1.3
(a) Will it behave as a converging or a diverging lens in two cases?
(b) How will its focal length change in the two media?

SET III

Note : Except for the following questions, all the remaining question have been asked in Previous Set.
Question.1. A hollow metal sphere of radius 6 cm is charged such that the potential on its surface is 12 V. What is the potential at the centre of sphere?
Answer : Potential inside the charged sphere is constant and equal to potential on the surface of conductor. So therefore, potential at the centre of sphere is 12V.

Question.3. How are microwaves produced?
Answer : Microwaves are electromagnetic waves with wavelength ranging from as long as meter to as short as one millimeter, or equivalently with frequencies between 300 . MHz and 300 GHz. Microwaves are produced by vacuum
tubes devices that operate on the ballistic motion of electron controlled by magnetic or electric fields. Some different kinds of microwaves emitters are the cavity magnetron, the klystron, the traveling wave tube (TWT), the gyrotron and all starts.

Question.12. Mention various energy losses in transformer?
Answer : Transformers energy losses tend to worsen with increasing frequency. Magnetic core losses are exaggerated with higher frequencies, eddy currents in the iron core, resistance of windings or copper loss, hysteresis lpss and flux leakage are energy losses in transformers.

Question.14. In the given circuit, assuming point A to be at zero potential, , use Kirchhoff’s rules to determine the potential at point B.

Question.18. A thin straight infinitely long conducting wire having charge density X is enclosed by a cylindrical surface of . radius r and length 4 its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.
Answer: Electric flux through the cylinderical surface

Question.20. A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. It is immersed in a liquid of refractive index 1.3. Calculate its new focal length.
Answer :

A Thing of Beauty Important Questions CBSE Class 12 English

Extract Based Questions (4 Marks)

Question.1. Read the extracts given below and answer the questions that follow.
And such too is the grandeur of the dooms
We have imagined for the mighty dead;
All lovely tales that we have heard or read;
An endless fountain of immortal drink,
Pouring unto us from heaven’s brink
(a) Name the poem and the poet.
(b) What is the thing of beauty mentioned in these lines?
(c) What image does the poet use in these lines?
(d) What is the effect of this ‘immortal drink’ on us?
or
(a) Who are the ‘mighty dead’? How do we know about them?
(b) What images does the poet use to convey that beauty is everlasting?
(Foreign 2009, Delhi 2014; Modified)
(c) What is the effect of the immortal drink?
(d) Write the words from the extract which mean
(i) stories
(ii) magnificence
Answer.(a) The poem is ‘A thing of beauty’ and the poet is John Keats.
(b) The thing of beauty mentioned in these lines is the group of stories celebrating the glory of our powerful ancestors, which we have read or heard.
(c) The poet uses the image of an ‘endless fountain’ coming down to us from heaven. According to him, nature and other beautiful things around us are like a gift from God, which give us infinite delight and Joy.
(d) The immortahtirrnk gives us immense joy, a joy that is everlasting, one that will never come to an end.
or
(a) Our ancestors, who were great in their own ways and the dead emperors have been referred to as the mighty dead. We come to know about them by reading or hearing valorous tales, which speak of their innumerable sacrifices. They are a source of motivation for all who go through them.
(b) ‘Endless fountain’ of joy is the image that has been used to convey that beauty is everlasting. Things of beauty are an eternal source of motivation, a precious gift from heaven which give us infinite pleasure and defight.
(c) The immortal drink that nature’s endless fountain pours on us gives us immense joy and pleasure.
(d) (i) tales (ii) grandeur

Question.2. Read the extract given below and answer the questions that follow.
Spite of despondence, of the inhuman dearth
Of noble natures, of the gloomy days,
Of all the unhealthy and o’er-darkened ways
Made for our searching: yes in spite of all,
Some shape of beauty moves away the pall
From our dark spirits.
(a) Name the poem and the poet.
(b) Why are we despondent?
(c) What removes “the pall from our dark spirits”?
(d) Explain, “the inhuman dearth of noble natures.” ’
or
(a) What are we doing every day?
(b) Which evil things do we possess and suffer from?
(c) What are the circumstances that contribute towards making humans unhappy and disillusioned with life? (Delhi 2013; Modified, All India 2014; Modified)
(d) What removes the pall from our dark spirits?
Answer. (a) The poem is’A thing of beauty’and the poet is John Keats.
(b) Human iife is full of trials and tribulations. We add to our troubles and make our lives more difficult by following evil ways. We hurt others as well as ourselves. This makes us despondent.
(c) The various objects of beauty around us remove the pall from our dark spirits. Beautiful things have the power to make us happy.
(d) The poet here means that the world lacks in men of noble nature or qualities. There are few people who are noble in character; most of the people are self-centred and pursue evil ways.
or
(a) The memory of beautiful things remains in our minds and strengthens our bond with the Earth. Every day, these bonds get strengthened by the flowery bands that we prepare.
(b) Humans suffer from hopelessness and sadness. Our poor style of living make us grim.
(C) Hopelessness, sadness and lack of noble ways are a part of human life. We harm ourselves by following unhealthy and wicked paths. These circumstances contribute towards making humans unhappy and disillusioned with life.
(d) Anything beautiful in our surroundings or even the memories of our beautiful experiences removes the pall from our dark spirits.

Question.3. Read the extract given below and answer the questions that follow.
“Some shape of beauty moves away the pall
From our dark spirits,”
(a) How does beauty help us when we are burdened with grief?
(b) Explain; “Some shape of beauty.”
(c) Identify the figure of speech in the above lines.
(d) Why are our spirits referred to as ‘dark’?
Answer. (a) Whenever we look at a beautiful object, we are filled with comfort and joy. When we are burdened with grief, a thing of beauty comes as a ray of hope, and makes us forget our sorrow and suffering at least for some time.
(b) Beauty is an abstract idea and has no specific shape. The poet here means beauty in some form or some beautiful object which pleases us.
(c) The above lines present an example of metaphor. Our souls are steeped in sadness and disappointment which are compared to a pall or shroud covering the dead.
(d) Our spirits are dejected due to extreme sadness and disappointment which may be the result of our own evil actions. So, the poet refers to them as dark, because of the hopelessness we feel.

Question.4. Read the extract given below and answer the questions that follow.
Therefore, on every morrow, are we wreathing
A flowery band to bind us to the Earth,
Spite of despondence, of the inhuman dearth
Of noble natures, of the gloomy days,
Of all the unhealthy and o’er-darkened ways
Made for our searching
(a) What are the flowery bands that bind us to the Earth?
(b) What message do the above lines convey? (Delhi 2009)
(c) Why is there an “inhuman dearth of noble natures”?
(d) What do you understand by “unhealthy and o’er-darkened ways”?
Answer. (a) Here beautiful and pleasant things have been referred to as flowery bands. It is only those things which have beauty of some or the other kind that bind us to this Earth, or in a way, keep us going. The memory of our beautiful experiences helps us strengthen our bond with the Earth.
(b) The message that these lines convey is that the world is full of sorrow, sadness and depression. People have become selfish,-and there is a scarcity of noble people.
Despite this, life is possible because there are some beautiful things around, which help us to look towards the positive side of life. It is the beauty of these things which brings a smile on our face and gives us happiness in our lives.
(c) Today, there are few people who can rise above petty differences and be generous towards others. These days people are mostly selfish and do not think about others.
(d) The unhealthy and o’er-darkened ways refer to the trials and tribulations in our life, and the dishonest and unfair means people adopt to achieve their goals.

Question.5. Read the extract given below and answer the questions that follow.
A thing of beauty is a joy forever
Its loveliness increases, it will never
Pass into nothingness; but will keep
A bower quiet for us and a sleep
Full of sweet dreams and health and quiet breathing
(a) How does a thing of beauty remain a joy forever?
(b) Mention any two sources of joy which a thing, of beauty provides to us.
(All India 2008; Modified)
(c) Explain, “never pass into nothingness.”
(d) What is meant by‘bower’?
Answer. (a) A thing of beauty provides eternal and everlasting joy to us. Whenever we are in a pensive mood we can relive the joyful experience and happiness it gave us.
(b) A thing of beauty gives us a quiet bower to rest peacefully, and a blissful sleep full of
pleasant dreams.
(c) The joy that a thing of beauty give us is eternal. It never loses its importance, but it only increases with time.
(d) A ‘bower’ refers to a peaceful, quiet place under the shade of trees.

Short Answer Type Questions p Marks, 30-40 words)

Question.1. How do beautiful things help us to live a happy life? (Foreign 2014)
Answer. We derive comfort and happiness when we look at objects of beauty, and the lasting impression they leave on our minds is a source of permanent joy. Beautiful things bind us to the Earth and help us strengthen our relationship with nature. They provide us relief from our everyday trials and tribulations and help us live a happy life.

Question.2. What does Keats consider an endless fountain of immortal drink and why does he
call its drink immortal? (All India 2013)
Answer. Keats considers nature’s beauty as an endless fountain of immortal drink. He calls this drink immortal as the pleasure and delight given by it never passes into nothingness. It only increases with time.

Question.3. According to Keats, what spreads the pall of despondence over our dark spirits? How
is it removed? (All India 2013)
or
What spreads the pall of despondence over our dark spirits? How is it removed?
Answer. John Keats says that pain and suffering are a part of life. These bring us sadness and despondence. But still, we can find hope. The hope and motivation come from nature and remove the pall of despondence from our spirits. The beautiful sights give us inspiration to continue the journey of life.

Question.4. How does a thing of beauty provide shelter and comfort? (Modified India 2013)
Answer. A thing of beauty provides shelter and comfort in many ways. It is like a bower, a place under the shade of a tree where we can rest. It gives us a sleep filled with sweet dreams. The streams and forests provide a cooling covert for us in the hot weather.

Question.5. How is a thing of beauty a joy forever? (Delhi 2012)
Answer. A thing of beauty is a joy forever because it is a permanent source of joy and inspiration and leaves a lasting impression on us. The pleasure and happiness given by it never fades away but keeps on increasing.

Question.6. What makes human beings love life in spite of troubles and sufferings? .
(Ml India 2012,2010; Delhi 2008)
Answer. Human beings suffer from hopelessness and sadness. Our style of living often makes us hopeless and grim. We at times also, suffer due to the self-centred and wicked nature of others. In spite of all our troubles and sufferings, we are able to derive pleasure from our surroundings. The beautiful objects from nature help us shed our gloom, give us a glimmer of hope, and make us love life amidst all troubles.

Question.7. Why is ‘grandeur’ associated with the mighty dead? (Delhi 2011)
Answer. The mighty dead are those who have laid down their lives for a great or noble cause. Their death is replete with intellectual and spiritual beauty. Birth, growth and decay, all are inseparable. Each has its own. beauty. Their death is grand or beautiful because it makes them immortal through the grand tombs and memorials erected in their honour.

Question.8. What is the message of the poem, ‘A thing of beauty’? (Modified India 2011)
Answer. The message of the poem ‘A thing of beauty’ is that beautiful things are a source of constant joy and inspiration. They give us eternal pleasure and leave an indelible image on our minds and, therefore, every beautiful thing and aspect is worth being treasured.

Question.9. What is the ‘endless fountain’ and what is its effect? (Foreign 2011)
Answer. Here beauty has been referred to as an ‘endless fountain’, which is an eternal source of motivation, an elixir of life and a precious gift from heaven. It gives us infinite pleasure and delight, and motivates us to continue our journey of life.

Question.10.What images does the poet use to describe the beautiful bounty of the Earth? (Delhi 2010)
Answer. The poet uses the image of an ‘endless fountain’, which showers bounties on Earth. He calls this an ‘immortal drink’ from heaven. The sun, the moon, the trees, the daffodils, the lush green forests and streams, all reflect the beautiful bounties which God has given us.

Important Questions for Class 12 English

CBSE previous Year Solved  Papers  Class 12 Biology Delhi 2011

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. Name the type of cell division that takes place in the zygote of an organism exhibiting haplontic life cycle.
Answer : Meiosis occurring in the zygote of an organism exhibiting haplontic life Cycle.

Question.2. Write the scientific name of the microbe used for fermenting malted cereals and fruit juices.
Answer: Saccharomyces cerevisiae.

Question.3. Write the unit used for measuring ozone thickness.
Answer : Dobson units (DU).

Question.4. Name the event during cell division cycle that results in the gain or loss of chromosome.
Answer: Aneuploidy results in the loss or gain of chromosome during cell division cycle.

Question.5. How can bacterial DNA be released from the bacterial cell for biotechnology experiments?
Answer : To release the bacterial DNA from the bacterial cell for biotechnology experiments the cell wall need to be open using enzyme lysozyme and DNA is isolated along with RNA and protein.

Question.6. Write the importance of cryopreservation in conservation of biodiversity.
Answer : Cryopreservation is used in storing the gametes of endangered species that can be fertilized in-vitro followed by propagation through tissue culture methods, seeds can be preserved in seed banks, preserving vegetatively propagated crops.

Question.7. Mention the role of the codons AUG and UGA during protein synthesis.
Answer : AUG is the start codon that initiates the protein synthesis. UGA is the stop codon that signals for the termination of protein synthesis.

Question. 8. Normally one embryo develops in one seed but when an orange seed is squeezed many embryos of different shapes and sizes are seen. Mention how it has happened.
Answer : The nucellus cells start dividing and protrude into the embryo sac to develop into embryos. Hence, each ovule may contain several embryos. The presence of many embryos of different shapes and sizes in a seed is called Polyembryony.

SECTION-B

Question.9. How do histones acquire positive charge ?
Answer : Histones acquire a positive charge due to presence of abundance of basic amino acid residues such as lysine and arginine carry positive charges in their side chains.

Question.10. Why is CuT considered a good contraceptive device to space children ?
Answer : CuT is a method of Intrauterine device (IUD) considered to be a good contraceptive device because it releases copper ions which increase phagocytosis of sperms and decrease sperm motility and fertilizing capacity of the sperm. Hence, the sperms lose their motility before reaching the ova.

Question.11. Differentiate between albuminous and non-albuminous seeds, giving one example of each.
Answer : The differences between albuminous and non-albuminous seeds are as under :

Question.12. Explain the process of RNA interference.
Answer: RNA interference is the cellular defense mechanism in all eukaryotic organisms. This method involves silencing of a specific mRNA of the parasite due to complementary dsRNA molecule that binds and prevents translation of the mRNA(silencing). RNA interference is a method adopted to prevent infestation of roots of tobacco plants by a nematode Meloidogyne incognita.

Question.13. List the key tools used in recombinant DNA technology.
Answer: The key tools used in recombinant DNA technology are restriction enzymes, Polymerase Chain Reaction, competent host and cloning vectors, polymerase enzymes, ligases and host organisms.

Question.14. Name the two types of immune systems in a human body. Why are cell mediated and humoral immunities so called ?
Answer : The two types of immunity found in humans are Humoral and Cell-mediated immunity. Humoral immunity is called so because, B lymphocytes produce antibodies against pathogens into our blood stream to fight with them. Cell-mediated immunity is called so because it is performed by specialised cells in the body known as T-lymphocytes. T-cell also help B cells to produce antibodies.
OR
Write the scientific names of the causal organisms of elephantiasis and ringworm in humans. Mention the body parts affected by them.
Answer: Elephantiasis
The Scientific name of the causal organism of Elephantiasis in human is Wuchereria bancrofti. The body part affected by them is Lymphatic vessel of lower Limbs.
Ringworm in human is, Microsporum, Trichophyton and Epidermophyton. The body part affected is skin.

Question.15. Justify with the help of an example where a deliberate attempt by humans has led to the extinction of a particular species.
Answer: The deliberate attempt by human has led to the extinction of a particular species. For example, the introduction of Nile perch in Lake Victoria that led to the extinction of more than 200 species of cichlid fish in the lake. Abingdon tortoise in Galapagos islands became extinct, after goats were introduced.

Question.16. Identify A, D, E and F in the diagram of an antibody molecule given below:
Answer : A- Antigen binding site.

D- Light chain
E – Heavy chain (constant region)
F – disulphide bond

Question.17. Study the graph given below. Explain how is oxygen ‘ concentration affected jn the river when sewage is discharged into it.

Answer : The introduction of sewage caused a sudden and sharp decline of dissolved oxygen at the point of sewage discharge in the river. This is because oxygen is used up by microorganisms involved in biodegradation resulting in an increase in the biological oxygen demand (BOD). This in turn leads to the death of fishes and other aquatic life. As the sewage gets diluted due the flow of the water, the river slowly regains some of its oxygen concentration.

Question.18. Explain how a hereditary disease can be corrected. Give an example of first successful attempt made towards correction of such diseases.
Answer : Gene therapy is the correction of a genetic defect by introduction of normal gene into an individual or embryo to take over or compensate the function for a non-functional gene. The first disease to have a gene therapy is ADA (Adenosine deaminase) deficiency. In this, the gene coding for enzyme ADA gets deleted leading to deficiency of ADA and problems in immune system.
Gene therapy for ADA deficiency includes :

1.  In in-vitro culturing of isolated lymphocytes from the patients blood.
2. Then introduced fuctional ADA cDNA into the cluttered lymphocytes.
3. These lymphocytes are returned back to the patients body.
4.  Permanent care of this problem is the introduction of isolated gene from bone marrow cells producing ADA into cells at early embryonic stages.

SECTION – C

Question.19. Draw a diagram of a, male gametophyte of an angiosperm. Label any four parts. Why is sporopollenin considered the most resistant organic material ?
Answer: Following is the labelled diagram of a male gametophyte of an angiosperm.

Sporopollenin is chemically very stable and is usually well preserved in soils and sediments. It can withstand high temperatures, acidic and alkaline conditions, and enzymes.

Question.20. How are dominance, codominance and incomplete dominance patterns of inheritance different from each other ?
Answer: Dominance: An allele expresses itself in the presence of the hybrid heterozygous condition. For example, tall plant, round seed, violet flower, etc. are dominant characters in a pea plant.
Co-dominance : Co-dominance is the phenomenon in which both the alleles of a trait are expressed in heterozygous condition. Both the alleles of a gqne are equally dominant. Example: ABO blood group system. .
Incomplete dominance : Neither of the two alleles of a gene is completely dominant over the other in heterozygous condition, the hybrid is intermediate. For example, a monohybrid cross between the plants having red flowers and white flowers in Antirrhinum species will result in all pink flower plants in Fj generation.

Question.21. The base sequence in one of the strands of DNA is TAGCATGAT.
(i) Give the base sequence of its complementary strand.
(ii) How are these base pairs held together in a DNA molecule?
(iii) Explain the base complementarity rules. Name the scientist who framed this rule.
Answer : (i) The base sequence of the complementary strand will be ATCGTACTA.
(ii) The base pairs in the DNA molecules are held with hydrogen bonds, between A and T and C and G on the two strands.
(iii) Base Complementarity Rule: A purine will always pair with a pyrimidine in a DNA molecule i.e. A will pair with T and G will pair with C. The ratio of A and T or C and G will always he 1. The base complementarity rule was framed by Erwin Chargaff.

Question.22. (a) Sickle cell anaemia in humans is a result of point mutation. Explain.
(b) Write the genotypes of both the parents who have produced a sickle celled anaemic offspring.
Answer : (a) Sickle celled anaemia in humans is a result of point mutation beacuse mutation of a single base change (substitution) in the gene leads to the replacement of GAG
by GUG. This leads to the substitution of the amino acid Glutamine (Glu) by Valine (Val) at sixth position of beta globin chain of haemoglobin.
(b) Genotype of parents who have produced a sickle cell anaemic offspring. Father – HbA HbS, Mother – HbA HbS.

Question.23. What is inbreeding depression and how is it caused in organisms ? Write any two advantages of inbreeding.
Answer : Inbreeding depression is the reduction in the fertility and productivity of an organism due to continuous inbreeding.
Advantages : (i) Production of pure lines
(ii) Elimination inferior genes,
(iii) Accumulation of superior genes.

Question.24. (a) Identify (A) and (B) illustrations in the following:

(b) Write the term given to (A) and (C) and why ?
(c) Expand PCR. Mention its importance in biotechnology.
Answer : (a) The part labelled A-AATTC is the sticky end. The part labelled B is the foreign DNA insert.
(b) The term used For A and C are called the palindrofmc nucleotide sequence, because the sequence of base pairs reads same on the two strands when the orientation of the reading is kept the same.
(c) PCR stands for Polymerase Chain Reaction.
PCR is a technique in molecular biology, used to amplify a gene or a piece of DNA to .obtain its several copies. Biotechnological Importance – amplification of gene of interest in vitro.

Question.25.

The diagram above is that of a typical biogas plant. Explain the sequence of events occurring in a biogas plant. Identify a, b and c.
Answer: The’sequences of events occurring in a biogas plant are:

1.  The complex molecules are broken down into simple sugars, amino acids & fatty acids.
2. Further breakdown of remaining , components by fermentative bacteria.
3. Simple molecules obtained by acidogenesis phase further digested by acetogens to produce large amount of acetic acid, CO2 and Ify.
4.  Methanogens utilise the intermediate products of the previous steps and convert them into methane, carbondioxide and water.
   (a) Sludge loader.
   (b) Gas holder CH4 and GCK
   (c) Inlet to feed slurry.

Question.26. How can crop varieties be made disease resistant to overcome food crisis in India ? Explain.
Name one disease resistant variety in India of:
(a) Wheat to leaf and stripe rust
(b) Brassica to white rust
Answer: A wide range of fungal, bacterial and viral pathogens affect the yield of cultivated crops. Disease resistant can be provided by conventional breeding, mutational breeding or genetic engineering.

1. Conventional breeding: It includes the basic steps of screening, germplasm, hybridisation, selection, testing and release.
2. Mutational breeding: In this method, genetic variations
   are created, which then result in the creation of traits not found in the parental type.
3.  Genetic engineering: Certain wild varieties have disease- resistant characteristics, but they are low yielding, Disease- resistant genes from such varieties are introduce in high- yielding varieties through recombinant DNA technology. One disease resistant variey in India of:
   (a) Himgiri (b) Pusa swarnim

OR
Write the source and the effect on the human body of the following drugs:
(i) Morphine (ii) Cocaine (iii) Marijuana The source and effect of the following drugs are :
Answer: (i) Morphine :
Source—Latex of poppy plants (Papaver somniferum)
Effect—It is a depressant; slows down body functions
(ii) Cocaine:
Source—Coca plant Erytjfroxylum coca
Effect—Stimulates the CNS, producing a sense of euphoria
and increased energy and hallucination.
(iii) Marijuana:
Source—Inflorescences of the plant Cannabis sativa
Effect—Effects the cardiovascular system.

Question.27. Name the type of interaction seen in each of the following examples:
(1) Ascaris worms living in the intestine of human
(2) Wasp pollinating fig inflorescence
(3) Clown fish living among the tentacles of sea-anemone
(4) Myeorrhizae living on the roots of higher plants
(5) Orchid growing on ai branch of a mango tree
(6) Disappearance of smaller barnacles when Balanus dominated in the Coast of Scotland.
Answer :

1.  Ascaris worms living in the intestine of human- Parasitism
2.  Wasp pollinating fig inflorescence – Mutualism
3. Clown fish living among the tentacles of sea-anemone Commensalism
4. Myeorrhizae living on the roots of higher plants – Mutualism
5. Orchid growing on a branch of a mango tree – Commensalism
6.  Disappearance of Smaller barnacles when Balanus dominated in the Coast of Scodand – Competition

SECTION-D

Question.28. (a) Draw a labelled diagram of the human female repro-ductive system.
(b) Enumerate the events in the ovary of a human female during:
(i) Follicular phase (ii) Luteal phase of menstrual cycle
Answer: (a) Female reproductive system :

(b) (i) Follicular phase : The growth of the primary ovarian follicle and also causes maturation of the primary oocyte in this follicle. The follicular cells of the Graafian follicle secrete estrogen. Due to which the uterine endometrium becomes thick, more vascular and more glandular.
(ii) Luteal phase : Development of corpus luteum continues to release the hormone, progesterone. It lasts for about 12-14 days and extends from the 16th – 28th day of the menstrual cycle.
OR
(a) Write the specific location and the functions of the following cells in human males:
(i) Leydig cells (ii) Sertoli cells (iii) Primary spermatocyte
(b) Explain the role of any two accessory glands in human male reproductive system.
Answer : (a) (i) Leydig cell are found in interstitial spaces between seminiferous tubules in the testes. They synthesise and secrete testicular hormones called androgens (testosterone).
(ii) Sertoli cells are found in between spermatocyte inside seminiferous tubules. They act as nurse cell and provide nutrition to the male germ cells.
(iii) Primary spermatocytes are found in between spermatocyte inside seminiferous tubules. They are diploid” cells that are derived from the spermatogonia. They undergo meiotic division to give rise to secondary spermatocyte and thereby male gamete-sperm.
(b) The male accessory glands include the seminal vesicles, prostrate gland, and the bulbourethral gland.
Prostate Gland: It stores and secretes an alkaline, milky fluid known as seminal plasma, which provide alkalinity of the semen is to neutralize the acidity of the vaginal tract. Bulbourethral glands: This gland secretes a viscous secretion during sexual arousal which helps to lubricate the urethra for easy passage of spermatozoa to pass through, and to help ‘ flush out any residual urine or foreign matter and also help in lubrication of penis.

Question.29. Explain the salient features of Hugo de Vries theory of mutation. How is Darwin’s theory of natural selection different from it? Explain.
Answer : Salient features of theory of Hugo de Vries :

1. Mutations cause evolution.
2. New species originate due to large mutations
3. Evolution is a discontinuous process and not gradual
4. Mutations are directionless
5.  Mutations appear suddenly
6.  Mutations exhibit their effect immediately
   

OR
(a) Name the primates that lived about 15 million years ago. List their characteristic features.
(b) (i). Where was the first man-like animal found?
(ii) Write the order in which Neanderthals, Homo habilis and Homo erectus appeared on earth. State the brain capacity of each one of them.
(iii) When did modern Homo sapiens appear on this planet?
Answer: (a) Dryopithecus (ape-like) and Ramapithecus (man¬like). These primates were hairy and their walk was similar to that of chimpanzees.
(b) (i) The first man-like animal was found in Africa.
(ii) Order = Homo habilis , Homo erectus , Neanderthals Cranial capacity =Homo habilis = 650 – 800 cc,
Homo erectus = 900 cc, Neanderthals = 1400 cc
(iii) During ice age / 75000 – 10000 years ago

Question.30. (a) Explain primary productivity and the factors that influence it.
(b) Describe how do oxygen and chemical composition of detritus control decomposition ?
Answer : (a) The amount of biomass produced per unit area over a time period by plants during photosynthesis is defined as the primary productivity. It is expressed as weight (g-2) or energy (Kcal m-2).
Factors : Availability of nutrients, quality of light available, availability of water, temperature of the given place, type of plant species of the area, photosynthetic capacity of the plants; (b) Decomposition is an oxygen consuming process thus it is directly proportional to the concentration of the oxygen in the environment. It is controlled by the chemical composition of detritus and climatic conditions. The decomposition rate decreases when the detritus is rich in lignin and chitin.
OR
(a) What is El Nino effect? Explain how it accounts for biodiversity loss.
(b) Explain any three measures that you as an individual would take, to reduce environmental pollution.
Answer : (a) Rise in temperature leading to deleterious changes in the environment and resulting in odd climatic changes is El Nino effect. An El Nino event is a temporary change in the temperature, surface air pressure and currents of the Pacific Ocean in the region surrounding the equator. El Nino causes global warming and the melting of polar ice cap. This leads to the submerging of coastal regions and loss of species endemic to that area.
(b) The measures that we as an individual can take in order to reduce environmental pollution can be :

1.  Reducing our garbage generation and dumping of organic waste in places far from residential areas.
2.  Less use of fossil fuel and using car pool so as to reduce pollution and fuel consumption.
3.  Turning off air conditioners when not in use.
4. Planting more and more trees and deducing the deforestation.

SET-II

SECTION-A

Question.1. How does Penicillium reproduce asexually ?
Answer : Penicillium reproduces asexually by means of spores called conidia which are non-motile and developed on conidiospores.

Question.5. Mention the uses of cloning vector in biotechnology.
Answer : The uses of cloning vector in biotechnology:

1.  Help in linking the foreign DNA with the host.
2.  Help in the selection of recombinants from the non-recombinants.

Question.8. Mention the function of trophoblast in human embryo.
Answer: Functions of Trophoblast in human embryo :

1.  During pre embryonic development stage (6 or 7 Day), the trophoblast cells contact with the endometrial lining and secrete enzyme that digest endometrial cells.
2.  The trophoblast releases the hormone human chorionic gonadotrophin, it keeps corpus luteum alive and is useful for detection of pregnancy.

SECTION-B

Question.10. Name the oral pills used as a contraceptive by human females. Explain how does it prevent pregnancy.
Answer: The oral pill used as a contraceptive by human females is ‘Saheli’. Saheli inhibits ovulation and implantation. It alters the quality of cervical mucus and prevents the entry of sperms into the cervix.

Question.14. Explain the advantage of crossbreeding of the two species of sugarcane in India.
Answer : Saccharum barberi, grown in North India, had poor sugar content and yield, whereas saccharum officinarum, grown in South India, had thicker stem and higher sugar content, but did not grow well in North India. The hybrid produces by cross breeding of these two species has the following desirable traits :

1.  High Yield
2.  Thick stem
3. High sugar content
4.  Ability to grow in North Indian sugarcane fields.

OR
How do cellular barriers and cytokine barriers provide innate immunity in humans ?
Answer : Cellular Barriers : Leukocytes and monocytes are natural killers in the blood and they destroy microbes. Macrophages are present in the tissues. They phagocytose and destroy microbes.
Cytokine Barriers : Virus infected cells secrete proteins called interferons and protect non-infected cells from the viral infection.

SECTION-C

Question.19. Differentiate between geitonogamy and xenogamy in plants. Which one between the two will lead to inbreeding depression and why ?
Answer:

Question.22. A pea plant with purple flowers was crossed with white flowers producing 50 plants with only purple flowers. On selfing, these plants produced 482 plants with purple flowers and 162 with white flowers. What genetic mechanism accounts for these results ?
Answer:

Reason:
(i) Factors segregate from each other that remained together in a hybrid during gamete formation.
(ii) A homozygous parent produces all gamete that are similar while heterozygous parent produces two kinds of gametes in equal ratio.

Question.23. Mention the property of plant cells that has helped them to grow into a new plant in in-vitro conditions. Explain the advantages of micropropagation.
Answer : The property of plant cells that has helped them to grow into a new plant in in-vitro condition is totipotency. Advantages of micropropagation are :

1.  Many important food plants like tomato, banana, apple etc. can be produced on a commercial scale.
2.  It helps in recovery of healthy plants from diseased plants because of the presence of virus free meristem in tb,e plant.
   One can remove the meristem and grow it in vitro to obtain virus free plant. Scientists have succeeded in culturing meristem of banana, sugarcane, potato etc.

SECTION-D

Question.30.

(a) (i) Name the biogeochemical (nutrient) cycle shown above.
(ii) Name an activity of the living organisms not depicted in the cycle by which this nutrient is returned to the atmosphere.
(b) How would the flow of the nutrient in the cycle be affected due to large scale deforestation ? Explain giving reasons.
(c) Describe the effect of an increased level of this nutrient in the atmosphere on our environment.
Answer : (a) (i) The biogeochemical (nutrient cycle) shown is carbon cycle.
(ii) Decomposition of organic wastesby microbes is an activity of the living organisms is not depicted in the cycle.
(b) Deforestation leads to increase in carbon dioxide levels in the air. Because the CO2 present is not being utilized for photosynthesis in the absence of plants.
(c) The effect of an increased level of this nutrient in the atmosphere on our environment is due to human activity. Decomposers also contribute substantially to CO2 pool by their processing of waste materials and dead organic matter of land or oceans. The increased level of this gas produced higher global temperature and changes in precipitation patterns.
OR
(a) Healthy ecosystems are the base of wide range of (ecosystem) services. Justify.
(b) Explain the differences and the similarities between hydrarch and xerarch succession of plants.
Answer : (a)Healthy ecosystem provides following ecological services:

1.  Purification of air and maintenance of gaseous composition.
2. Mitigation of drought and floods.
3. Cycling of nutrients.
4.  Storehouse of carbon.
5.  Maintainance of biodiversity.
6.  Habitat for a number of wildlife.
7. Influence of hydrological cycle.

(b) Differences between hydrarch and xerarch succession of plants :

Similarities: Both hydrarch and xerarch succession lead towards mesic conditions, e.g., with moderate water conditions.

SET-III

SECTION-A

Question.3. Why green plants are not found beyond a certain depth in the ocean ?
Answer : The green plants are not found beyond a certain . depth in the ocean because sunlight cannot penetrate beyond a certain depth.

Question.4. Write the scientific name of the microbe used for fermenting malted cereals and fruit juices.
Answer : Saccharomyces cervisiae is the scientific name of the microbe used for fermenting malted cereals ahd fruit juices.

SECTION-B

Question.9. State a dual role of deoxyribonucleoside triphosphate during DNA replication.
Answer: Deoxyribonucleoside triphosphate serves dual purpose

1.  It acts as substrate.
2.  It also provides energy for polymerization reaction.

Question.13. Explain the role of Ti plasmids in biotechnology.
Answer : Ti Plasmid (tumor-inducing plasmid) refers to the plasmid of Agrobacterium tumefacims. A tumefaceiens is a plant pathogen and is known to produce tumours in the plants it infects.
Role of Ti plasmids in biotechnology; Ti plasmid can be modified into a cloning vector by removing the genes responsible for pathogenicity.

Question.16. State the functions of primary and secondary lymphoid organs in humans.
Answer : Role of primary and secondary lymphoid organs in humans are:
Primary Lymphoid Organs: Here immature lymphocytes are differentiated into antigen sensitive lymphocytes.
Secondary Lymphoid Organs: Where lymphocytes interact with antigen accessory cells and proliferate to become effector cells, resulting in the initiation of the antigen specific response.

Question.21. Why do you see two different types of replicating strands in the given DNA replication fork ? Explain Name these strands.

Answer :

1. The DNA dependent DNA polymerase catalyse polymerization of the nucleotides only in 5’—>3’ direction. Since the two strands of DNA run in opposite direction, one is in 5’—>3’ and the other is in 3’—>5’ direction.
2. Replication proceeds in two opposite directions. DNA synthesis is continuous on the 3’—>5’ template strand while on the other template strand with polarity 5’—>3’ direction, the synthesis of DNA is discontinuous i.e short stretches of DNA are formed.
   5’—>y Continuous sysnthesis 3’—>5’ Discontinuous sysnthesis

Question.25. (a) Why do farmer prefer biofertilizers to chemical fertilizer these days ? Explain.
(b) How do Anabaena and mycorrhiza acts as biofertilisers ?
Answer : (a) Farmer prefer biofertilizers to chemical fertilizer because:

1.  Biofertilisers would be better in the long run because they contain organic materials while fertilizers contain chemical substanes that when used in excess become harmful to the consumers.
2.  Continuous application of fertilizers to the soil, decreases its fertility and also leads to increase in the acidity of the soil.
3.  Fertilizer run off from the fields also causes water pollution,

(b) Anabaena and mycorrhiza acts as biofertiliser, fungi sucii as Glomus form symbiotic associations with plants. This relationship is known as Mycorrhiza. They absorb phosphorus from the soil and pass it to plants.
Cyanobacteria such as Anabaena and Nostocalsoftx atmospheric nitrogen and act as bio – fertilisers especially in paddy fields.

Question.26. (a) Name the stage of plasmodium that gains entry into the human body.
(b) Trace the stages of plasmodium in the body of female Anopheles after its entry.
(c) Explain the cause of periodic recurrence of chill and high fever during malarial attack in humans.
Answer: (a) The stage of the plasmodium is Sporozoites.
(b) Plasmodium requires two hosts to complete its life cycle. When the Anopheles mosquito bites a diseased person, gametocytes of Plasmodium are introduced into the mosquito.
Gametocytes fertilises and developed inside the intestine of mosquito to form sporozoites. Sporozoites are stored in the salivary glands of mosquito and are released into a healthy person who is bitten by the mosquito.
(c) Parasites initially multiply within the liver and then attack the red blood cells resulting in their rupture. The rupture of RBCs is associated with the release of a toxic substance called haemozoin. Haemozoin is responsible for the chill and high fever recurring every three to four days.
OR
Trace the events that occur in human body to cause immunodeficiency when HIV gains entry into the body.
Answer: Events occurring in the human host after the entry of HIV:

1. After entering in the human body, viruses enter into the macrophages.
2.  Macrophages becomes a virtual HIV factory.
3.  Thereafter HIV enters helper T-Lymphocytes, replicates and produces progenies.
4. As the progenies are released they attack other T-Lymphocytes.
5.  Therefore, T-lymphocytes start decreasing in number and immune response of the person become weak.
6. Even infection that Could be overcome easily starts aggravating.

SECTION-D

Question.28. (a) Describe the stages of oogenesis in human females,
(b) Draw a labelled diagram of human ovum released after ovulation?
Answer: (a) Oogenesis is the formation of the egg (ovum) in the female.

1. During fetal development females have oogonia which . are diploid sex cells.
   (i) While still in the womb the oogonia. divide by mitosis to form 14-1 million primary oocytes.
   (ii) These primary oocytes will begin the first meiotic division but stall during prophase I.
   (iii) The female is born with these primary oocytes.
2.  By the time the female reaches puberty approximately 40,000 of the primary oocytes will remain.
3. Beginning during puberty, each month hormones from the anterior pituitary stimulate a primary oocyte to complete the first meiotic division generating two secondary oocytes of unequal size.
   (i) The smaller secondary oocyte is called a polar body, containing one set of chromosomes.
   (ii) The larger secondary oocyte is the ovum (egg) that will be released from the ovary for fertilization by the spermatozoa.
   (a) Only if the ovum is fertilized will it continue the second meiotic division.
   (b) If fertilized the ovum divides again to produce a second polar body, with the fertilized ovum forming the diploid zygote.
   (c) If the ovum is not fertilized within 24 hours after release it will be broken down.
   (b)
   
   OR
   (a) When and where does spermatogenesis occur in human male ?
   (b) Draw a diagram of a mature human male gamete. Label the following parts: acrosome, nucleus, middle piece and tail.
   (c) Mention the function of acrosome and middle piece.
   Answer: (a) Spermatogenesis occur in testis at the time of puberty.
   (b) Diagram
   
   (c) Functions of acrosome : Acrosome contains hydrolytic enzymes that help in dissolving membranes of the ovum fo fertilization.
   Middle piece: It contains a number of mitochondria that provide energy for the movement of the tail and provides motility to sperm.

CBSE Sample Papers for Class 8 Science Practice Paper 2

Test Paper II
Class VIII

Time : 2 1/2 Hrs                                                                                                        Maximum Marks : 100 

General Instructions:

1. Read the content of the question carefully and attempt them there after.
2. Follow the specified word limit, wherever mentioned.
3. Try to be brief and concise and have clarity in expression.
4. Make diagrams and figures wherever necessary.

1. Fill in the blanks:                                                                                              1 x 10=10

1. …………is a chemical reaction in which the surface of a metal changes from metal atom to metal ion.
2. Shooting stars are actually not…………
3. The body cover that protects brain is called …………
4. The female sex harmones are oestrogen and …………
5. The ductless glands are also called …………
6. …………is the periods of rapid physical growth and mental development taking place between childhood and adulthood.
7. Brown coal is commonly called…………
8. Reducing friction by…………an aircraft or car give it a higher speed.
9. The…………is the black circle present in the eye.
10. The process of fertilisation takes place in human being (female) in…………

2. Write the full form of these abbreviations:                                       1 x 10 = 10

1. OIL-RIG
2. ONGC
3. STD
4. OPEC
5. CNG
6. HIV
7. TSH
8. GH
9. AIDS
10. AMS

3.Give one word answer:                                                                             2 x 10 = 20

1. The coating of an electrically conductive item with a layer of metal using electric current.
2. A branch of chemistry that studies the reaction which takes place at the interface of an electronic conductor.
3. A reaction in which both oxidation and reduction are occurring.
4. Part of the female reproduction system, the embryo development takes place.
5. The pituitary hormone which facilitates child birth.
6. The product of fertilization.
7. The poisonous gas produced when coal is burnt in a closed room.
8. The mining of oil under sea.
9. A code, which enables blind persons to read and write.
10. The instrument that measures air pressure.

4. Define the following:                                                                                  2 x 5 = 10

1. Hermaphrodite
2. Destructive Distillation
3. Carbonisation
4. Dispersion
5. Electrolysis

5.Difference between the following pair (any three)                          4 x 3 = 12

1. Menarche and Menopause
2. External fertilization and Internal fertilization
3. Regular reflection and irregular reflection
4. Real Image and Virtual image

6.(A) Write answer in short:                                                                          4 x 3 = 12

1. State Faraday’s first law of electrolysis.
2. What is natural gas? What are its component?
3. What are petro chemicals? Why are they so important?
4. What is an electrochemical reaction?

(B) Give reasons for the following:                                                             4 x 3 = 12

1. Sportsmen use shoes with spikes.
2. Petroleum is also called ‘Black Gold’.
3. Water must not be used on burning oil.
4. It is dangerous to touch a switch with wet hands.

7.Give answer in detail (any two)                                                                5 x 2 = 10

1. Write in three points on the important life style that the adolescents should take care of.
2. Explain with diagram how coal formed in the earth.
3. What is friction? Explain the types of friction.

8.Draw neat and labelled diagram of anyone:                                                        4

1. Destructive distillation of coal
2. Electric bell

CBSE previous Year Solved  Papers  Class 12 Biology Delhi 2012

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. Mention the unique flowering phenomenon exhibited by Strobilanthus kunthiana (neelakuranji).
Answer : The unique flowering phenomenon exhibited by Strobilanthus kunthiana (neelakuranji) is that it flowers once in 12 years.

Question.2. How does smoking tobacco in human lead to oxygen deficiency in their body ?
Answer: Smoking tobacco lead to oxygen deficiency in human body because smoking increases carbon monoxide (CO) content in blood. The greater affinity of CO to haemoglobin/ CO forms a stable bond with haemoglobin. Presence of CO does not allow pxygen to bind with haemoglobin and reduces the concentration of haem-bound oxygen. This causes oxygen deficiency in the body.

Question.3. A garden pea plant (A) produced inflated yellow pod, and another plant (B) of the same species produced constricted
green pods. Identify the dominant traits.
Answer : The trait to produce inflated yellow seeds is dominant over the trait producing constructed green seeds.

Question.4. Why is Eichhomia crassipes nicknamed as “Terror of Bengal” ?
Answer : It grows at an alarming rate and spreads on . the surface of the water body, causes oxygen depletion leading to death of aquatic life of fishes and other aquatic organisms.

Question.5. Write the location and function of the sertoli cells in humans.
Answer : Location of Sertoli cells – In the testis Function of Sertoli cells – Provide nutrition to the developing sperm cells.

Question.6. Name the following:
(a) The semi-dwarf variety of wheat is high-yielding and disease-resistant.
(b) Any one inter-specific hybrid mammal.
Answer : (a) Sonalika and Kalyan Sona are the semi-dwarf.
(b) Mule/Hinny/Liger/Tigon.

Question.7. Write the similarity between the wing of a butterfly and
the wing of a bat. What do you infer from the above with reference to evolution ?
Answer: Analogous organs as their origin is not the same but they perform similar functions. From the above reference we can infer the evolutionary relationship between organisms and show convergent evolution.

Question.8. Write what do phytophagous insects feed on.
Answer : As the name suggest phytophagous insects feed on plants/ Plant sap.

SECTION – B

Question.9. Draw a neat labelled sketch of a replicating fork of DNA.
Answer:

Question.10. Where is sporopollenin present in plants ? State its significance with reference to its chemical nature.
Answer : Present in exine, of pollen or pollen grain Sporopollenin is the most resistant organic polymer hence protects the pollen/gamete and provides protection to pollen from unfavorable conditions.

Question.11. (a) Highlight the role of thymus as a lymphoid organ.
(b) Name the cells that are released from the above mentioned gland. Mention how they help in immunity. .
Answer: (a) Immature lymphocyte differentiate into mature lymphocyte in thymus. T-cells produced in the bone marrow get matured in the thymus and are released from here and migrate to secondary lymphoid organs.
(b) The cells are released from the thymus gland are known as T-lymphocyte. These T-cells help B-cells to produce antibodies and takes part in immunity.

Question.12. Explain the work carried out by Cohen and Boyer that contributed immensely in biotechnology.
Answer : Stanley Cohen and Herbert Boyer constructed the first artificial recombinant DNA. They isolated the antibiotic resistant gene, from the plasmid of a bacterium that was resistant to the antibiotic drug , and then linked this gene with the plasmid of Salmonella typhimurium, construction of artificial recombinant DNA molecule.

Question.13. Why do clownfish and sea anemone pair up ? What is this relationship called ?
Answer : Clown fish lives in tentacles of sea anemone and gets protection from its predators by moving around the stinging tentacles of the sea anemone.
The sea anemone is neither helped or harmed by the inte-raction with the fish, the relationship is called commensalism.

Question.14. (a) State the difference between meiocyte and gamete with
respect to chromosome number.
(b) Why is a whiptail lizard is referred to as partheno-genetic ?
Answer : (a) Meiocyte – diploid (2n)
Gamete – haploid (n)
(b) Whiptail lizard referred as parthenogenetic because it is a female and gives rise to new female reptile without fertilization and female gamete undergoes development to form new organism from unfertilised egg.

Question.15. Name the plant source of the drug popularly called “smack”. How does it affect the body of the abuser ?
Answer: Plant source of smack: Papaver somniferum (Poppy) which is a stronger analgesic than morphine.
Effect of smack on the body of abuser: It reduces heart rate and increase blood sugar and it is a depressant which slow down body function.
OR
Why is Rhizobium categorized as a ‘symbiotic bacterium’ ? How does it act as a biofertiliser ?
Answer : Rhizobium, a symbiotic bacteria, which live in the . root nodule of leguminous plants, fixes atmospheric nitrogen into organic forms to be used by plants. It is a biofertilizer as it is a living organism that enriches nutrient content of the plant. Bacterium gets food and shelter from the plant.

Question.16. (a) State the role of DNA ligase in biotechnology.
(b) What happens when Meloidegyne incognitia consumes cells with RNAi gene ?
Answer : Role of DNA ligase in biotechnology, (a) Joining of DNA fragment is done by DNA ligase, linking of Okazaki fragments or discontinous synthesis fragments and linking of desired gene with plasmid to form recombinant DNA.
(b) If Meloidegyne incognitia consumes cells with RNAi gene so specific mRNA of the nematode is silenced and prevents the translation of mRNA. Thus causes death of parasites.

Question.17. Some organisms suspend their metabolic activities to survive in unfavourable conditions. Explain with the help of any four examples.
Answer : Some examples with their unfavourable conditions under which they suspend their metabolic activities:

1.  Polar bear – hibernation during winter
2.  Snails / fishes – Aestivation during summer
3.  Species of zooplankton – diapauses
4.  Higher plants/ spores of bacteria/ fungi — become dormant , and cyst formation as in case of amoeba .

Question.18. (a) Name the Protozoan parasite that causes amoebic dysentery in humans.
(b) Mention two diagnostic symptoms of the disease.
(c) How is this disease transmitted to others?
Answer : (a) The Protozoan parasite that causes amoebic dysentery in humans is Entamoeba histolytica
(b) Symptoms : Constipation, abdominal pain , stools with mucus and blood clot.
(c) This disease is transmitted to others by fecally contaminated food and water.

SECTION-C

Question.19. It is established that RNA is the first genetic material. Explain giving three reasons.
Answer : Reason for RNA is the first genetic material:

1.  Processes like metabolism, translation, splicing evolved around RNA.
2.  RNA is reactive and catalyses reaction and in some virus it is the hereditary material.
3.  It is unstable and hence would have mutated to lead to , evolution.

OR
(a) Name the enzyme responsible for the transcription of tRNA and the amino acid the initiator tRNA gets linked with.
(b) Explain the role of initiator tRNA in initiation of protein synthesis.
Answer : (a) Enzyme responsible for the transcription of tRNA: RNA polymerase in prokaryotes and RNA polymerase III in eukaryotes.
The amino acid the initiator tRNA get linked with Formyl methionine in prokaryotes and methionine in eukaryotes.
(b) The role of initiator tRNA in initiation of protein synthesis.

1.  This tRNA combine with amino acid methionine in presence of amino acyl-tRNA synthetic enzymes resulting in the formation of charged tRNA.
2.  The mRNA attaches to smaller subunit of ribosome and charged initiator tRNA. After that tRNA joins the initiator codon and signals the start of translation.
3. The initiator tRNA function on the ability of anticodon sequence mutants of initiator tRNA to initiate, protein synthesis. The anticodon UAC recognizes the mRNA AUG codon and binds by forming complementary base pairs, leaves the amino acid, initiating protein synthesis.

Question.20. State the theory of Biogenesis. How does Miller s experiment support this theory ?
Answer : Theory of Biogenesis : The first forih of life could have come , from pre-existing, non-living organic molecules (e.g, RNA, protein etc.) proposed by Oparin and Haldane. This theory suggest that origin of life is first abiogenesis and biogenesis later.

Question.21. Name the two different categories of microbes naturally occurring in sewage water. Explain their role in cleaning sewage water into usable water.
Answer: Different categories of microbes naturally occurring in sewage water, all aerobic and anaerobic bacteria, fungi, filamentous fungi.
Role in cleaning sewage water into Potable water :

1.  The primary effluent is passed into large aeration tanks where if is constantly agitated.
2.  This allows abundant growth of aerobic microbes like bacteria and filamentous fungi.
3. The growth of these microbes reduces BOD of effluents. Once the BOD is reduced significantly, then the effluent is passed into settling tanks where the bacterial floes are allowed to sediment.
4.  This sediment is called activated sludge. A small part of activated sludge is again introduced into large tanks called anaerpbic sludge digesters.
5. flere anaerobic bacteria digest the bacterial and fungi in the sludge.
6.  This digestion produces methane, H2S and COz gas. These gases form biogas.
7.  The effluent from secondary treatment is them released
   into natural water bodies.

Question.22. Write the function of each one of the following :
(a) (Oviducal) Fimbriae (b) Coleoptile (c) Oxytocin
Answer : (a) (Oviducal) Fimbriae : Collection of ovum released by ovary.
(b) Coleoptile : Protects the plumule of the monocot embryo.
(c) Oxytocin : Causes uterine contraction for partu- rition(child birth) and also promotes milk ejection.

Question.23. Name the genes responsible for making Bt cotton plants resistant to bollworm attack How do such plants attain resistance against bollworm attacks ? Explain.
Answer: The Bt toxin is encoded by the cry gene, crylhC and crylLAb, which produces protein crystals during a particular phase of their growth. This toxin provides resistance to plants against lepidoptern, coleoptern and dipterans pests. Bt cotton, is the variety of cotton in which the gene from the bacterium that encodes for the toxin is incorporated. When boll worm bites the cotton fruits, it consumes the toxic insecticidal protein. The protoxin gets activated by the alkaline pH of the gut of the insect and binds to the surface of the midgut epithelium of the insects and causes swelling and cell lysis, which eventually leads to the death of the insects.

Question.24. Study a part of the life cycle of malarial parasite given below. Answer the questions that follow:

(a) Mention the roles of ‘A’ in the life cycle of the malarial parasite.
(b) Name the event ‘C’ and the organ where this event occurs.
(c) Identify the organ ‘B’ and name the cells being released from it.
Answer: (a) A represents the female mosquito. Gametocytes of Plasmodium enter the mosquito when it bites an infected person and takes the malaria parasite along with the blood meal.
(b) C is the fertilization stage and it takes place in the intestine of the mosquito.
(c) B is the salivary gland of the female anopheles mosquito and the sporozoites escape out of the mosquitos salivary gland.

Question.25. Given below is the representation of amino acid composition not the relevant translated portion of P-chain of haemoglobin, related to the shape of human red blood cells.

(a) Is this representation indicating a normal human or a sufferer from certain related genetic disease ? Give reason in support of your answer.
(b) What difference would be noticed in the phenotype of the normal and the sufferer related to this gene ?
Who are likely to suffer more from the defect related to the gene represented the males, the females or both males and females equally ? And why ?
Answer : (a) Yes, this representation is indicating a normal human. Normal person mRNA contains GAG codon which codes for glutamic acid, at the 6th position.
(b) In a sufferer who exhibit sickle cell trait, the codon GAG is replaced by GUG in the mRNA. Hence, during translation of the defective mRNA, Glutamic acid is replaced by Valine at the 6th position of Beta globin chain of the haemoglobin.
(c) Both the males and females suffer equally because sickle cell anaemia is not a sex linked disease. It is an autosomal recessive disease and sickle shaped RBC will cause equal deficiency of oxygen in both males and females.

Question.26. By the end of 2002 the public transport of Delhi switched over to a new fuel. Name the fuel. Why is this fuel considered better? Explain.
Answer : Delhi Government switched to CNG (Compressed Natural Gas). Delhi had been categorized as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi. CNG is considered better due to following reasons :

1. CNG is a clean fuel that produces very little unburnt particles and thus it is considered as eco-friendly.
2. CNG burns most efficiently, unlike petrol or diesel, in the automobiles and very little of it is left unburnt.
3. Moreover, CNG is cheaper than petrol or diesel.

Question.27. Draw a schematic sketch of pBR 322 plasmid and label the following in it:
(a) Any two restriction sites. (b) Ori and rop genes.
(c) An antibiotic resistant gene.
Answer : Sketch of pBR 322 plasmid :

(b) The Restriction sites mentioned in the figure are Hind III, EcoR I, Bam HI, Sal I, Pvu II, Pst I, Cla I and Antibiotic resistant.
(c) genes: \({ amp }^{ R }\) and \({ tet }^{ R }\)

SECTION – D

Question.28. Explain the carbon cycle with the help of a simplified model.
Answer: Key process involved in carbon cycle is photosynthesis and respiration.

1.  In photosynthesis, carbon-dioxide and water produce carbohydrate and oxygen while respiration oxidizes food to give carbon-dioxide and water.
2.  71% carbon is found dissolved in oceans. In rock it gets stored as calcium carbonate.
3. Carbon can either be liberated to atmosphere through respiration or can be passed to. animal when being eaten or remain in plant even after death.
4.  A considerable amount of COz is fixed annually by process of photosynthesis.
5.  Excess of carbon-dioxide can cause global warming.

OR
Explain how does:
(a) A primary succession start on a bare rock and reach a climax community ?
(b) The algal bloom eventually choke the water body in an industrial area ?
Answer : (a)

1.  The species of organisms that first invade a bare area are called pioneer species (lichens).
2.  Lichens secrete acids which dissolve rocks, thereby leading to weathering and soil formation.
3. Next serai stage will be bryophytes which can hold in the small amount of soil.
4.  Bryophytes are then succeeded by grasses.
5.  They are succeeded by bigger plants, and ultimately, an entire forest gets established. This remains stable as long as the environment remains unchanged.

(b) Effluent from industries contains large amount of nutrients. This causes excessive growth of free-floating algae causing algal bloom. Later, the decomposition of these algae depletes the supply of oxygen, leading to the death of other aquatic animal life thus choking the water body. This ageing process of a lake caused due to nutrient enrichment is called Eutrophication.

Question.29. The following is the illustration of the sequence of ovarian events (a – i) in a human female.

(1) Identify the figure that illustrates ovulation and mention the stage of oogenesis it represents.
(2) Name the ovarian hormone and the pituitary hormone that have caused the above mentioned event.
(3) Explain the changes that occur in the uterus simultaneously in anticipation.
(4) Write the difference between ‘c’ and ‘h’.
(5) Draw a labelled sketch of the structure of a human ovum prior to fertilization.
Answer:

1. Figure ‘f’ illustrates stage of ovulation.
2.  Ovarian hormone = estrogen
   Pituitary hormone = Luetenizing hormone
3.  Endometrium lining gets thickened and highly vascularised, high regeneration anticipating implantation of the fertilized ovum.
4.  ‘c’ is developing Graafian follicle while ‘h’ is regressing corpus luteum Human ovum prior to fertilization.
   

OR
How does the megaspore mother cell develop into 7-celled,-» 8 nucleate embryo sac in an angiosperm ? Draw a labelled diagram of a mature embryo sac.
Answer : The megaspore mother cell undergoes mitosis to form one functional and viable megaspore. The functional megaspore divides mitotically to produce two nuclei which migrate to opposite poles, forming a 2-nucleate embryo sac.

1. Further mitotic divisions lead to the formation of 4-nucleate followed by 8-nucleate stages of the embryo sac.
2.  Among the eight nuclei, six are enclosed by cell walls and organised into cells, while the remaining two nuclei (called polar nuclei) are situated above the egg apparatus in a large central cell.
   Out of the six cells, three are grouped at the micropylar end, and constitute the egg apparatus made up of two synergids and one egg cell.
3. The other three cells are located at the chalazal end, and are called antipodals. Thus, a typical angiosperm embryo sac after maturity is 8-nucleated and 7-celled.
   

Question.30.What is the inheritance pattern observed in the size of starch grains and seed shape of Pisum sativunP. Workout the monohybrid cross showing the above traits. How does this pattern of inheritance deviate from that of Mendelian law of dominance?
Answer : In pea plants (Pisum sativum), a single gene controls the expression of a number of traits, namely starch synheiss & size of starch grains. This phenomenon is called pleiotropy & genes are called pleiotropic & genes are called pleiotropic genes. It has two alleles B and b. BB homozygotes produced large starch grains as compared to that produced by bb homozygotes. The cross involved is

Deviation from Mendels law of dominance : If starch grain size is considered as the phenotype, the trait of size of starch grain shows incomplete dominance. Hence in heterozygous condition the starch grain are of intermediate size. The trait of seed shape follows Law of Dominance and the hybrid will show only dominant trait.
OR
State the aim and describe Messelson and Stahl’s experiment. Answer: Messelson and Stahl in 1958 aimed at proving that the DNA replicates in a semi-conservative fashion. The semi-conservative DNA replication suggests that after the completion of replication, each DNA molecule will have one parental and one newly-synthesised strand.
Experiment done by Messlson & stahl:

1. They grew colonies of E.Coli for several generations on a culture medium haring NH4C1 with heay isotope of Nitrogen, 15N.
2. This heavy isotope got incorporated in Nitrogen containing compounds, like DNA.
3. When the whole DNA was found to contain heavy isotope of N2, the bacteria were shifted to culture medium having normal nitrogen, 14N.
4.  After 20 minutes, 1st generation of E. Coli was obtained. Its DNA was isolated & tested for presence of heavy DNA & normal DNA. This was done for several generations.
5.  To test the presence of ype of isotope, the isolated DNA was added to tubes having CsCl. The tube were provided with centrifugal force for many hours, till the DNA come to lie at a fixed position in the tube.
6. DNA of I generation, having completely heavy isotope was heaviest.
7. When E. Coli Was shifted to normal N2 Containing culture, the DNA of its I generation was slighdy less dense than its parental material.
8. In the 2nd generation, DNA formed two sedimens, one like the FI generation & second lighter than it.
9. I generation DNA had intermediate density having both 15N & 14N. 2nd generation was one intermediate with (15N & 14N) & other lighter with only 14N. Their ratio was 50-50 in 2nd generation.
10.  In 3rd generation, same two types were produced, but ratio of intermediate reduced to 25% & intermediate percentae reduced to 12.5% & lighter one at 8.75%.
11.  ‘This is possible only when in each generation out of the two stands of DNA, one was obtained from parent double stran & other is formed new. This proves semi-conservative nature of DNA replication.
    

SET-II

SECTION-A

Question.1. Cucurbits and papaya plants bear staminate and pistillate flowers. Mention the categories they are put under separately on the basis of the type of flowers they bear.
Answer : Papaya is dioecious because the staminate and pistillate flowers are borne in two different plants while cucumber is monoecious because it bears both staminate and pistillate flowers in the same plant.

Question.4. What is the interaction called between Cuscuta and shoe flower bush ?
Answer : The interaction between Cuscuta and shoe fl&wer bush is called parasitism.

Question.5. When do the oogenesis and the spermatogenesis initiate in human females and males respectively ?
Answer: Oogenesis starts in females in their foetal stage while spermatogenesis in males starts at puberty.

Question.7. State the significance of the study of fossils in evolution.
Answer : The significance of the study of fossils in evolution :
(i) Study of fossils indicates the geological period in which various life forms were arisen.
(ii) The calculation of geological period can be done via radioactive dating.
(iii) We can know the morphological details of the organisms in the past and can relate them to the organisms in the present to understand the process of evolution.

SECTION-B

Question.13. Draw a schematic diagram of a part of double stranded dinucleotide DNA chain having all the four nitrogenous bases and showing the correct polarity.
Answer: Schematic diagram of a double stranded dinucleotide DNA chain having all the four nitrogenous bases with polarity.

Question.14. Name of parasite the causes filariasis in humans. “Mention its two diagnostic symptoms. How is this transmitted to others?
Answer : Wuchereria (W bancrofti and W. malayi), are the filarial worms that cause filariasis in humans.

Diagnostic symptoms :
1. Chronic inflammation of the organs in which they live for many years. Lymphatic organs are most affected.
2. The genital organs are also often affected resulting in gross deformities.
Transmission: The pathogens are transmitted to a healthy person through the bite by the female mosquito vectors.

Question.15. Name the source of streptokinase. How does this bio molecule function in our body ?
Answer : Streptokinase enzyme is produced by the bacterium Streptococcus. It is modified by genetic engineering and is used as a clot buster for removing clots from the blood vessels of patients who have suffered from myocardial infraction leading to hqart attack.
OR
How do mycorrhizae act as biofertilizers? Explain. Name a genus of fungi that forms a mycorrhizal association with plants.
Answer : Mycorrhizae act as biofertilizers : The fungal symbiont in these associations absorbs phosphorus from soil and passes it to the plant. The fungi help the plant in the absorption of essential nutrients from the soil while the plant in turn provides the fungi with energy yielding carbohydrates.
The fungi belonging to the genus Glomus form mycorrhizal associations with plants.

SECTION-C

Question.19. Write the function of each of the following :
(a) Middle piece in human sperm.
(b) Tapetum in anthers.
(c) Luteinizing hormone in human males.
Answer:
(a) Middle piece in human sperm : It possesses numerous mitochondria, which produces energy for the movement of tail that facilitate sperm motility essential for fertilization.
(b) Tapetum in anthers : Tapetum is the inner most layer in the main function of tapetum is to provide nourishment to the developing pollen grains.
(c) Luteinizing hormones in human males : It stimulates the leydig cells to produce androgen like testosterone.

Question.26. How does an algal bloom cause eutrophication of a water body? Name the weed that can grow in such a eutrophic lake.
Answer : Algaes are the major producers of any aquatic ecosystem. Presence of large amount of nutrients in water. also causes excessive growth of planktonic algae called an algal bloom which imparts a distinct colour to the waste bodies. Later, the decomposition of these algae depletes the supply of oxygen, leading to the death of other aquatic animal life. This phenomenon is called eutrophication.
Water hyacinth (Eichhornia crassipes) is a weed that can grow in such an eutrophic lake.

SECTION-D

Question.28. (a) Draw a ‘pyramid of numbers’ of a situation where a large population of insects feed upon a very big tree. The insects in turn, are eaten by small birds which in turn are fed upon by big birds.
(b) Differentiate giving reason, between the pyramid of biomass of the above situation and the pyramid of numbers that you have drawn.
Answer : (a) Pyramid of numbers showing interaction between trees, insects, birds and big birds.

(b) (i) ‘Pyramid of number’ is spindle shaped as the number of insects is maximum. The number of trees and birds are less than the insects. The numbers is gradually decreasing at each trophic level.

(ii) The Pyramid of Biomass in this ecosystem is erect because the biomass decreases at each tropic level.
OR
(a) What are the two types of desirable approaches to conserve biodiversity? Explain with examples bringing out the difference between the two types.
(b) What is the association between the bumblebee and its favourite orchid Ophrys ? How would extinction or change of one affect the other ?
Answer : (a) Two approaches to conserve biodiversity are :
(i) In situ conservation
(ii) Ex situ conservation

(b) Mutualism is the type of association seen between the bumblebee and the orchid Ophrys. In this type of association, Orchids show a bewildering diversity of floral patterns many of which have evolved to attract the right pollinator insect (bees and bumblebees) and ensure guaranteed pollination by it. The petals of ophrys recemble the female bundle. If the female bee’s colour patterns change even slightly for any reason during evolution, pollination success will be reduced unless the orchid flower co-evolves to maintain the resemblance of its petal to the female bee.
-Extinction of bumble bee will definitely affect the orchid flower because these bees are the means of pollination for the flower and if they get extinct then the pollination percentage will be reduced. But the extinction of the orchid will not affect the bumble bee population.

SET -III

SECTION-A

Question.1. Mention the difference between spermiogenesis and spermiation.
Answer: Difference between spermiogenesis and spermiation:- Spermiogenesis : It is the process of transforming spermatids into matured spermatozoa or sperms. Spermiation : It is the process when mature spermatozoa are released from the Sertoli cells into the cavity of seminiferous tubules.

Question.3. What is an interaction called when an orchid grows on a mango plant ?
Answer : The relationship between a mango tree and an orchid is an example of commensalism.

Question.4. Write the names of the semi-dwarf and high yielding rice varieties developed in India after 1966.
Answer: Jaya and Ratna are two semi-dwarf and high yielding rice varieties developed in India after 1966,

Question.6. Mention the unique feature with respect to flowering and fruiting in bamboo species.
Answer : Bamboo species flowers only once in their life time, generally after 50-100 years, bamboo species produce large number of fruits and then die.

Question.8. State the significance of biochemical similarities among diverse organisms in evolution ?
Answer : The significance of biochemicals similarities in bio-chemicals such as DNA, help in deriving the line of evolution.

SECTION-B

Question.15. Mention the importance of Lactic acid bacteria to humans other than setting milk into curd.
Answer : (i) Lactic acid bacteria play a very beneficial role in checking disease causing microbes.
(ii) It also used to produce acid called lactic acid which is an important industrial product. It is used in bakery products, beverages, meat products, confectionery, dairy products, etc.
OR
How do methanogens help in producing biogas ?
Answer : Methanogens such as Methanobacterium act on excreta of cattle and anaerobic sludge and grow anaerobically, producing large amount of methane along with C02 and H2. Methanogens do not use oxygen to respire. In fact, oxygen inhibits the growth of methanogens.

Question.19. (a) Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical template strand given below:

(b) Write the RNA strand transcribed from the above transcription unit along with its polarity.
Answer:

OR
How are the structural genes inactivated in lac operon in E. coli ? Explain.
Answer : The structural gene in the lac operon consists of three genes, lac z, y and a. The structural gene is inactivated in the presence of repressor and absence of inducer (lactose). The repressor of the operon produced by the i gene. The repressor protein when produced binds to the operator region of the operon and prevents RNA polymerase firom transcribing the operon.

SECTION-C

Question.20. Write the function of each of the following:
(a) Seminal vesicle (b) Scutellum (c) Acrosome of human sperm.
Answer: (a) Seminal vesicle : It secretes an alkaline fluid that helps in neutralizing the acidity of the vaginal tract and thereby increasing the life-span of sperms.
(b) Scutellum : Scutellum is the tissue present in seed. It is the papery cotyledon of the monocot seed and acts as a passage for movement of nutrients from the endosperm to the developing embryo.
(c) Acrosome of human sperm : It is present at the tip of sperm. It is a cap-like structure which contains hydrolytic enzymes that help in penetration of egg during fertilization.

Question.25. (a) Why are the colourful polystyrene and plastic packagings used for protecting the food, considered an environmental menace ?
(b) Write about the remedy found for. the efficient use of plastic waste by Ahmed khan of Bangalore. [3]
Answer : (a) Polysterene and plastic packaging used for protecting food cause environmental pollution as these are non-biodegradable substances and its recycling process is very costly and includes manual participation thus exposing workers to toxic substances produced during recycling process.These polybags cause harm to aquatic life, choke the water pipes, etc.
(b) The efficient use of plastic waste by Ahmed Khan in Bangalore. He developed polyblend, a fine powder of recycled modified plastic. This mixture is mixed with the bitumen that is used to lay roads. Ahmed Khan proved that blends of – Polyblend and bitumen, when used to lay roads, enhanced the bitumens water repellant properties, and helped to increase road life by a factor of three.

SECTION-D

Question.30. Name the scientists who proved experimentally that DNA is the genetic material ? Describe their experiment.
Answer : Hershey and Chase worked with bacteriophage, E.coli and they proved that DNA is the genetic material. They worked on different radioactive isotopes to label DNA and protein coat of the bacteriophage.
Both of them grew some bacteriophages on a medium containing radioactive phosphorus (P32) to identify DNA and some on a medium containing sulphur (S35) to identify protein. These radioactive labelled phages were allowed’ to infect E.coli bacteria. After infecting, the protein coat of bacteriophage was separated from the bacterial cell by blending and then subjected to the process of centrifugation. Since, the protein coat was lighter, it was found in the
whereas the infected bacteria got settled at the bottom of the centrifuge tube. Hence, it was proved that DNA is the genetic material as it was transferred from virus to bacteria.
OR
(a) List the three different allelic forms of gene T in humans. Explain the different phenotypic expressions, controlled by these three forms,
(b) A woman with blood group A’ marries a man with blood group ‘O’. Discuss the possibilities of the inheritance of the blood groups in the folllowing starting with yes’ or ‘no’ for each :
(i) They produce children with blood group ‘A’ only.
(ii) They produce children some with ‘O’ blood group and some with ‘A’ blood group.
Answer : (a) In humans, the ABO blood groups are controlled by a gene called gene ‘I’. It has three alleles, namely IA, IB and i.
Table showing the Genetic Basis of Blood Groups in Human population.

CBSE Previous Year Solved  Papers  Class 12 Maths Delhi 2010

Time allowed : 3 hours                                                                                           Maximum Marks: 100

General Instructions:

1.  All questions are compulsory.
2.  Please check that this question paper contains 26 questions.
3.  Questions 1-6 in Section A are very short-answer type questions carrying 1 mark each.
4.  Questions 7-19 in Section B are long-answer I type questions carrying 4 marks each.
5. Questions 20-26 in Section C are long-answer II type questions carrying 6 marks each.
6.  Please write down the serial number of the question before attempting it.

SET I

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

SECTION – A

Question.1. What is the range of the function

Question.2.

Solution:

Question.3.

Solution:A = I

Question.4.What is the value of the determinant

Question.5.

Question.6.What is the degree of the following differential equation?

Solution:Degree of given differential equation is 1 as the index of highest derivative is one.

Question.7.Write a vector of magnitude 15 units in the direction of

Question.8.Write the vector equation of the following line:

Question.9.

Solution:

Question.10.What is the cosine of the angle which the vector

Solution: The cosine of angle which the given vector

SECTION – B

Question.11.On a multiple choice examination with three possible answer (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Question.12.
Find the position vector of point R which divides the line
joining two points P and Q whose position vectors

Question.13 Find the cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, -1,2) and parallel to the.line

Question.14.Using elementary row operations, find the inverse of the following matrix:

Question.15.Let Z be the set of all integers and R be the relation on Z defined as R = {(a, b):a,b ∈ Z and (a – b) is divisible by 5}. Prove that R is an equivalence relation.

Question.16.Prove the following:


Solution:

Question.17.Show that the function/defined as follows, is continuous at x – 2, but not differential there at:

Solution:Test of continuity at x = 2

Question.18.

Solution:

Question.19.

Question.20.Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinate of the point.
Solution: Given curve is y = x³

Question.21.Find the general solution of the differential equation

Solution:Given differential equation is

Question.22. Find the particular solution of the differential equation satisfying the given conditions:
x²dy + (xy + y²) dx = 0; y = 1 when x = 1.
Solution: Given differential equation is
x²dy + (xy + y²) dx= 0

SECTION – C

Question.23. A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is atmost 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on the ring is ? 300 and that on a chain is ? 190, find the number of rings and chain that should be manufactured per day, so as to earn the maximum profit. Make it as an L.P.P. and solve it graphically.
Solution : Let x be number of gold rings and y be the number of chain we make the following table from the given data:

Question.24.A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to both clubs. Find the probability of the lost card being of clubs.

Question.25.The points A(4,5,10), B(2,3,4) and C(1, 2, -1) are three vertices of a parallelogram ABCD. Find the vector equations of the sides AB and BC and also find the coordinates of point D.

Question.26. Using integration, find the area of the region bounded by the curve x²= 4y and the line x = 4y – 2.
Solution: Given equations of curve and line are
x² = 4y                                                                                                …(1)

Question.27.Show that the right circular cylinder, open at the top, and of given surface area and maximum volume is such that its height is euqal to the radius of the base.

Question.28. Find the values of x of which f(x) = [x(x – 2)]² is an increasing function. Also, find the points on the curve, where the tangent is parallel to x-axis.

Question.29.Using properties of determinants show,the following:

SET II

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

SECTION – A

Question.1. Find the minor of the element of second row and third column (a₂₃) in the following determinant:

SECTION – B

Question.11. Find all points of discontinuity of f, wheref is defined as follows:

Question.14. Let * be a binary operation on Q defined by a*b = 3ab/5. Show that * is commutative as well as associative. Also find its identity element, if it exists.

Question.18.

Solution:

Question.20. Find the equations of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x+ 14y + 4 = 0

SECTION – C

Question.23.

Question.29. Write the vector equations of the following lines and hence determine the distance between them:

SET III

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

SECTION – A

Question.1.

Question.2.What is the degree of the following differential equation?

Solution:The degree of the given differential equation is

Question.9.If A is a square matrix of order 3and |3A| =K | A |, then write the value of K.

SECTION – B

Question.11.There are two Bags, Bag I and Bag II. Bag I contains 4 white and 3 red balls while another Bag II contains 3 white and 7 red balls. One ball is drawn at random from one of the bags and it is found to be white. Find the probability that it was drawn from Bag I,

Question.14.

Question.17.Show that the relation S in the set S of real numbers, defined as S = {(a, b): a, b∈ R and a≤ b³ } is neither reflexive, nor symmetric nor transitive.
Solution: We have S = {(a, b): a, b ∈ R and a ≤ b³} is

Question.19. Find the equation of tangent to the curve

SECTION – C

Question.23. Find the intervals in which the function f given by f(x) = sin x – cos x, 0 ≤ x ≤2π is strictly increasing or strictly decreasing.
Solution:
f(x) = sin x – cos x
f(x) = cosx + sinx

Question.24.

CBSE Previous Year Solved  Papers  Class 12 Chemistry Outside Delhi 2012

Time allowed: 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. 
2.  Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
3.  Questions number 6 to 10 are short-answer questions and carry 2 marks each.
4.  Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
5.  Questions number 23 is a value based question and carry 4 marks.
6.  Questions number 24 to 26 are long-answer questions and carry 5 marks each. 
7.  Use log tables, if necessary. Use of calculators is not allowed.

SET I

Question.1. How may the conductivity of an intrinsic semiconductor be increased?
Answer : By doping it with elements like arsenic and phosphorous.

Question.2. Define ‘peptization.
Answer : The process of conversion of freshly prepared precipitate into a colloidal solution by adding suitable electrolyte is called peptization.

Question.3. How is copper extracted from a low grade ore of it?
Answer : Copper from its-low grade ore is leached out using acid or bacteria and then Cu²⁺ ions in the solution are reduced to copper by treating with Hydrogen or Iron. This method is called hydrometallurgy.

Question.4. Which is a stronger reducing agent, SbH₃ or BiH₃, and why?
Answer: BiH₃ is stronger reducing agent because its tendency to liberate H is maximum.

Question.5. What happens when bromine attacks

Question.6. Write the IUPAC name of the following:

Question.7.Write the structure of the product obtained when glucose is oxidised with nitric acid.
Answer:

Question.8. Differentiate between disinfectants and antiseptics.
Answer : Antiseptics are used on living tissues to kill or prevent the growth of microorganisms. It does not cause any harm to the living tissues e.g. 0.2% solution of phenol. Disinfectants are used on non-living objects to kill the microorganisms. They are harmful to the living tissues and hence, cannot be applied to the skin e.g. 1% solution of phenol.

Question.9.Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity
OR
The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm² mol^-1. Calculate the conductivity of this solution.

Question.10. A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

Question.11. Which methods are usually employed for purifying the following metals:

1.  Nickel
2. Germanium.

Mention the principle behind each one of them.
Answer :

1. Mond’s process : In this method Nickel form volatile complex with Co which decomposes on heating to give pure nickel.
2. Zone refining : It is based on the principle that “the impurities are more soluble in the molten state than in the solid state of metal.

Question.12. Explain the following facts giving appropriate reason in each case:
(i) NF₃ is an exothermic compound whereas NCl₃ is not.
(ii)All the bond in SF₄ are not equivalent.
Answer :
(i) The bond energy of F-F bond is lower that of N-F bond So NF₃ is an exothermic compound whereas bond energy of Cl-Cl bond is higher than N-Cl bond so NCl₃ is an endothermic compound. .
(ii) SF₄has see-saw structure with bond pairs at two equatorial and two axial positions. Hence, all the bonds in SF₄ are not equivalent.

Question.13. Complete the following chemical reaction equations:

Question.14. Explain the mechanism of acid catalysed hydration of an alkene to form corresponding alcohol.

Question.15. Explain the following behaviours:

1.  Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.
2. Qrtho-nitrophenol is more acidic than ortho- methoxyphenol.

Answer:

1.  Due to the formation of hydrogen bonds by alcohols with water molecules.
2. The phenoxide ion is more stable for o-nitrophenol as nitro group is electron withdrawing due to resonance while methoxy group is electron donating via resonance.

Question.16. Describe the following giving the relevant chemical equation in each case:

1. Carbylamine reaction
2. Hofmann’s bromamide reaction.

Answer:

1.  Carbylamine reaction : When aliphatic and aromatic primary amines are heated with chloroform and ethanolic KOH solution, they form isocyanides or carbylamines which are foul smelling substances.
   
2. Hofmanns bromamide reaction : In this reaction, primary amines are prepared by treating an amide 3vith Br₂ in an aqueous or. alcoholic soln of NaOH .
   

Question.17. Complete the following reaction equations:

Question.18. What are food preservatives? Name two such substances.
Answer : Chemicals used to prevent spoilage of food by preventing growth of microorganisms like bacteria, fungus etc are called food preservatives. Sodium benzoate, nitrogen gas are two such substances of food preservatives.

Question.19.Copper crystallizes with face centred cubic unit celLIf the radius of copper atom is 127.8 pm, calculate the density of copper metal. (Atomic mass of Cu = 63.55 u and Avogadro’s number N_A = 6.02 x 10²³ mol^-1)
OR
Iron has a body centred cubic unit cell with the cell dimension of 286.65 pm. Density of iron is 7.87 g cm-3. Use this information to calculate Avogadro’s number. (Atomic mass of Fe = 56.0 u)

Question.20. The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 x10³ohm. Calculate its resistivity, conductivity and
molar conductivity.
Answer:

Question.21. The reaction contributes to air pollution whenever a fuel is burnt in air at a high temperature.At 1500 K, equilibrium constant K for it is 1.0 x 10^-5. Suppose in a case [N₂] = 0.80 mol L^-1 and [O₂] = 0.20 mol IT1 before any reaction occurs.Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K.

Question.22. Explain the following terms giving a suitable example for each:

1. Aerosol
2. Emulsion
3. Micelle.

Answer :

1.  An aerosol is a colloid in which dispersed phase is a solid and dispersion medium is a gas eg. Dust, smoke.
2. Emulsion is a colloid solution in which both the dispersed phase and dispersion medium are in liquid state, eg. Milk and
   cod liver oil.
3. They are associated colloids showing colloidal behavior at high concentration and strong electrolytes at low concentration, eg. Soap.

Question.23. How would you account for the following:

1.  Among lanthanoids, Ln (III) compounds are predominant. However, occasionally in’ solution or in solid compounds, +2 and +4 ions are also obtained.
2. The E⁰_M²⁺_/M  for copper is positive (0.34 V).Copper is the only metal in the first series of transition elements showing this behavior.
3. The metallic radii of the third (5d) series of transition. metals are nearly the same as those of the corresponding members of the second series.

Answer :

1.  Some lanthanoid show +2 and +4 oxidation states in ionic solutions or solid components due to the extra stability that arises because of empty half-filled or fully filled 4ƒ-subshell.
2. This is because copper has high enthalpy of atomization and low enthalpy of hydration. Hence, E⁰_Cu²⁺_/Cu is positive.
3. This is because of lanthanoid contraction the metallic radii of the third (5d) series of transition metals are nearly the same as those of the corresponding members of the second series.

Question.24. Name the following coordination entities and draw the structures of their steroisomers :

Question.25. Answer the following questions :
(i) What is meant by chirality of a compound? Give an example.
(ii)Which one of the following compounds is more easily hydrolyzed by KOH and why?
CH₃CHClCH₂CH₃ or CH₃CH₂CH₂Cl

Answer : (i) Chirality is the property of a molecule to have non-superimposable mirror image. These molecules contain one asymmetric carbon atom.
e.g., Butan – 2- ol
(ii) CH₃CHClCH₂CH₃ is more easily hydrolyzed due to the formation of more stable secondary carbocation.
(iii)CH₃CH₂CH₂Cl undergoes SN₂ substitution reaction faster because it is a better leaving group due to its large size and less electronegativity.

Question.26. What is essentially the difference between CX-glucose and (3 -glucose? What is meant by pyranose structure of glucose?
Answer : CX-glucose and (3-glucose are two cyclic hemiacetal forms of glucose which differ only in the configuration of hydroxyl group (-OH) at anomeric carbon. Such isomers are called anomers. The six-membered cyclic structure of glucose is called pyranose structure.

Question.27. Differentiate between thermoplastic and thermosetting polymers. Give one example of each.
Answer: Thermoplastic Polymers : These polymers donot have cross-links between their chains and hence can be reshaped upon heating eg. Polyethylene, Polypropene etc. Thermosetting Polymers : These polymers have cross-links between their chains and hence cannot be reshaped upon heating, eg. Bakelite, Melamine etc.

Question.28. (a) Define the following terms :
(i) Mole fraction
(ii) Ideal solution.
(b) 15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at -0.34 °C. What is the molar mass of this material? (K_ƒ for water = 1.86 K kg mol^-1 )
OR
(a) Explain the following:
(i) Henry’s law about dissolution of a gas in a liquid.
(ii) Boiling point elevation constant for a solvent.
(b) A solution of glycerol (C₃H₈O₃) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass of glycerol was dissolved to make this solution? (KA for water = 0.512 K kg mol-1).

(a) (i) Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. i.e. p = K_Hx
(ii) The boiling point elevation constant for a solvent is defined as the elevation in boiling point when the molality of the solution is unity.

Question.29. (a) Draw the molecular structures of the following compounds:
(i) N₂OS
(ii)XeOF₄.
(b) Explain the following observations:
(i) Sulphur has a greater tendency for catenation than oxygen.
(ii)I-Cl is more reactive than I₂.
(iii)Despite lower value of its electron gain enthalpy with negative sign, fluorine (F₂) is a stronger oxidizing agent than Cl₂.
OR
(a) Complete the followingchemical equations:

(b)Explain the following observations :
(i)Phosphorus has greater tendency for catenation than nitrogen.
(ii)Oxygen is a gas but sulphur a solid.
(iii)The halogens are coloured. Why?

(b) (i) Due to strong S-S bond and less interelectronic repulsion, sulphur has greater tendency for catenation.
(ii) I-Cl bond is polar and hence more reactive compound to I₂ in which I-I bond is non-polar.
(iii) Due to high electronegativity and small size of fluorine, it act as stronger oxidizing agent

(b) (i) Catenation (i.e. linking of atoms of the same kind with one another) is related to the atom-atom bond energy. Greater the atom-atom bond energy, greater is the catenation. Because of low N-N bond energy (163.8 Kj mol^-1) nitrogen shows little tendency for catenation. P-P bond energy (201 x 10^-16 kj/mol) is quite high, hence, it shows more tendency for catenation than nitrogen.
(ii) Oxygen forms pπ-pπ multiple bonds. Due to small size and high electronegativity oxygen exists as diatomic (O₂) molecule. These molecules are held together by weak van der Waals’ forces. Hence O₂ is a gas at room temperature. Sulphur because of its bigger size and lower electronegativity, prefer to form S-S single bonds. Further because of stronger S-S than 0-0 single bonds, sulphur has a much greater tendency for catenation than oxygen. Thus, sulphur because of its higher tendency for catenation and lower tendency for pπ — pπ multiple bonds, forms octa atomic (S₈) molecules, having eight-memb’ered puckered ring structure. Because of bigger size, the force of attraction holding the S₈ molecule together are much stronger. Hence, sulphur is a solid at room temperature.
(iii) All halogens are coloured. It is due to the reason that their molecules absorb light in the visible region as a result of which their electrons get excited to higher energy levels while the remaining light is transmitted. The colour of the halogens is actually the colour of this transmitted light.

Question.30. (a) Write a suitable chemical equation to complete each of the following transformations :
(i) Butan-l-ol to butanoic acid
(ii)4-Methylacetophenone to benzene-1,4-dicarboxylic acid
(b) An organic compound with molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid. Identify the compound.
OR
(a) Give chemical tests to distinguish between:
(i) Propanol and propanone
(ii)Benzaldehyde and acetophenone
(b) Arrange the following compounds in an increasing order of their property as indicated :
(i) Acetaldehyde, Acetone,’ Methyl tert-butyl ketone (reactivity towards HCN)
(ii) Benzoic add, 3, 4-dinitrobenzoic add, 4-Methoxy- benzoic add (add strength)

(b) (i) The given compound with molecular formula C₉H₁₀O forms a 2,4,-DNP derivatitve and reduces Tollen’s reagent, It must be an aldehyde.

(ii)As the undergoes cannizzaro reaction, therefore CHO group is directly attached to the benzene ring.
(iii)On vigrous oxidation, it gives 1, 2-benzene dicarboxylic acid, therefore, it must be an ortho-substituted benzaldehyde.
Answer : (a)
(i) Iodoform Test : This test is given by propanone and not by propanol. Propanone on reacting with hot NaOH/I₂ gives a yellow precipitate of CHI₃while propanol does not.

(ii) Silver Mirror Test : Benzaldehyde being an aldehyde reduces Tollens’ reagent to give silver mirror test but acetophenone being a ketone does not give this test.

SET II

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

Question.1. Which stoichiometric defect increases the density of a solid?
Answer: Interstitial defect.

Question.2. What is meant by ‘shape selective catalysis’?
Answer : Zeolites are shape selective catalysts and shape selective catalysis depends upon the structure of the pores present in the catalyst and size ofjhe reactant and product molecules. ZSM-5 is an example of shape selective catalyst, which is used in converting alcohol directly into gasoline.

Question.3. What is the role of collectors in Froth Floatation process?
Answer : Collectors help in attachment of ore particle to an air bubble in froth eg. Sodium xanthates.

Question.6. Write the IUPAC name of Ph – CH = CH – CHO.
Answer: 3-phenyl prop-2-enal

Question.17. Explain the cleaning action of soap. Why do soaps not work in hard water?
Answer : When soap is rubbed on dirty cloth in water, concentration of soap become greater than (CMC) critical micelle concentration, micelle formation takes place. These micelles get adsorb at the dirt or grease

In such a manner that a hydrophobic end get adsorbs and anionic head remains at the surface. Therefore dirt particle become negatively charged and when rinsing is done, this particle moves along with water and clothes become free from dirt or grease cloth.
In soap sodium or potassium salts of fatty acids are present e.g.C₁₇H₃₅COO^–Na⁺ (sodium stearate) when it is added in hard water, presence of calcium or magnesium salts makes insoluble salts of Ca or Mg with carboxylate ion called scum which stick to cloth as gummy mass.

Question.20. A voltaic cell is set up at 25°C with the following half cells: Al/Al³⁺ (0.001 M) and Ni/Ni²⁺ (0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.

Question.23. Explain the following observations :

1. Many of the transition elements are known to form interstitial compounds.
2. There is a general increase in density from titanium t (Z = 22) to copper (Z = 29).
3. The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series.

Answer :

1. Because these elements are capable of entrapping smaller atoms of other elements such as H,C and N in the interstitial sites in their crystal lattice.
2. It is due to increase in atomic mass whereas atomic size volume decreases from Ti to Cu .Therfore density goes on increasing.
3. Because of the comparable energies of 5f 6d and 7s orbitals in actinoids, they exhibit larger number of oxidation states than the corresponding members of lanthanoid series.

Question.27. Explain the following terms giving a suitable example for each: ‘

1. Elastomers
2.  Condensation polymers
3. Addition polymers.

Answer :

1. Elastomers : Their polymer chain are held together by weakest intermolecular forces so that they can be stretched. That is why they exhibit elastic properties and are rubber-like solids, eg: Buna-S,Buna-N etc.
2.  Condensation Polymers : They are formrd by repeated condensation reaction between two different bi or tri-functional monomeric units with the elimination of smaller molecules such as water, alcohol, HCl etc. eg : Nylon 6, 6.
3. Addition Polymers : They are formed by the repeated addition of monomer molecules having double or triple bonds. Eg: Polyethylene.

Question.30. (a) Draw the structures of the following molecules:
(i) H₃PO₂ (ii) ClF₃ (b) Explain the following observations:
(i) Nitrogen is much less reactive than phosphorus.
(ii)Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
(iii)Sulphur has a greater tendency for catenation than oxygen in the same group.
OR
(a) Draw the structures of the following molecules:
(i) N₂OS (ii) HClO₄
(b) Explain the following observations :
(i) H₂S is more acidic than H₂O.
(ii)Fluorine does not exhibit any positive oxidation state.
(iii) Helium forms no real chemical compound.

b(i) Due to N = N triple bond in N₂ molecule, it is inert and hence is less reactive as compared to phosphorous which has P-P single bond,which makes -it more reactive.
(ii)In case of HF, average two intermolecular hydrogen bond is present. As a result, Vander Waals forces of attraction increases in water moleculer and hence, boiling point increases.
(iii)Because S-S bond is stronger than O — O bonds as there is more interelectronic repulsion in O — O than in S – S.

(b) (i) Because bond dissociation enthalpy of H-S bond in H₂S is less than that of H-O bond in H₂O .
(ii)Flourine is the most electronegative element in nature and it has no d-orbitals and therefore,there is no scope for electron promotion.Hence,it can show only -1 oxidation state in its compounds.
(iii)Because it is an inert gas and has very high ionization
enthalpy, therefore no real chemical compound of helium is known.

SET III

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

Question.1. What are w-type semiconductors?
Answer : Semiconductors which are doped with Group 15 elements like phosphorous are n-type semiconductors.

Question.4. What is the basicity of H₃PO₂ acid and why?
Answer : H₃PO₂ is monobasic as it has one replaceable hydrogen atom.

Question.5. Write the IUPAC name of the following :

Answer : 3 Bromo-2-methyl propene.

Question.7. Write a reaction which shows that all the carbon atoms in glucose are linked in a straight chain.
Answer:

Question.8. What is the cause of a feeling of depression in human beings? Name a drug which can be useful in treating this depression.
Answer :The inability to achieve one’s goal and extra work may cause the level of noradrenaline low, as a result the signal-sending activity becomes low and the person suffers from depression. Tranquilizers such as Equanil can help in treating depression.

Question.11. Explain the role of each of the following:
(i) NaCN in the extraction of silver.
(ii)SiO₂ in the extraction of copper.

Question.22. Write three distinct features of chemisorptions which are not found in physisorptions.
Answer : The three distinct features of chemisorptions are :
(i) It is irreversible.
(ii)It is used by chemical bond formation and hence requires activation energy.
(iii)It is highly specific in nature.

23.Explain each of the following observations :

1. With the same d-orbital configuration(d⁴), Cr²⁺ is a reducing agent while Mn3+ is an oxidising agent.
2. Actinoids exhibit a much larger number of oxidation states than the lanthanoids.
3. There is hardly any increase in atomic size with increasing atomic numbers in a series of transition metals.

Answer :

1. Cr²⁺ is reducing because when it loses one electron to form Cr³⁺: [Ar]3d³, it has three unpaired electrons in lower energy d-orbitals which are more stable whereas M³⁺ is oxidizing because after gaining one electron it becomes Mn²⁺ which has more stable electronic configuration due to half-filled d-orbitals Mn²⁺:{Ar]3d5
2. Due to comparable energies of 5f, 6d and 7s orbitals and unpaired electrons in these orbitals, actinoids exhibit much larger number of oxidation states than the lanthanoids.
3. Because along transition series, nuclear charge increases which tends to decrease the size but the addition of electrons in the penultimate d-subshell increases the screening effect which counter balances the effect of increased nuclear charge. Thus, atomic radii does not change.

24.Name the following coordination entities and describe their structures:

Question.26. Define the following as related to proteins :

1.  Peptide linkage
2. Primary structure
3. Denaturation.

Answer :

1. Peptide linkage is the amide linkage between two amino acids to form proteins and polypeptides. Co-NH is peptide linkage.
2. The simple linear structure of a protein molecule in a specific sequence in which various amino acids are present.
3. When a protein in fts native form is subjected to any physical change like change in temperature or chemical change like change in pH, the H-bonds gets disturbed. Due to this
   globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.

CBSE Previous Year Solved  Papers  Class 12 Chemistry Outside Delhi 2010

Time allowed: 3 hours                                                                                          Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. 
2.  Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
3.  Questions number 6 to 10 are short-answer questions and carry 2 marks each.
4.  Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
5.  Questions number 23 is a value based question and carry 4 marks.
6.  Questions number 24 to 26 are long-answer questions and carry 5 marks each. 
7.  Use log tables, if necessary. Use of calculators is not allowed.

SET I

Question.1.What type of interactions hold the molecules together in a polar molecular solids?
Answer : The molecules of polar molecular solid are held together by strong dipole-dipole interactions.

Question.2.What is meant by ‘limiting molar conductivity7?
Answer: Molar conductivity at infinite dilution is known as limiting molar conductivity,

Question.3.Fluorine does not exhibit any positive oxidation state. Why ?
Answer : Fluorine is the most electronegative element therefore, does not exhibit any positive oxidation state.

Question.4.Given the IUPAC name of the following

Answer: 4-bromo-3-methyl-pent-2-ene

Question.5.Write the . structure of the molecule of a compound whose IUPAC name is l-phenylpropan-2-ol.
Answer:

Question.6.What is Tollen’s reagent ? Write one use of this reagent.
Answer : Tollen’s reagent is an ammonical solution of silver
nitrate. It is used for the silver mirror test of aldehydes (Both aromatic and aliphatic aldehydes)

Question.7.What is meant by ‘reducing sugars’?
Answer : Sugars that can reduce Tollen’s or Fehlings solution are called reducing sugars. Reducing sugars contain tree aldehyde group and gets reduced in a chemical reaction to -OH group.

Question.8.What does the designation, 6, 6, mean in the name nylon-6,6?
Answer : The monomeric unit of Nylon-6, 6 polymer is derived from the two monomers hexamethylene diamine and adipic acid [NH-(CH₂)₆-NH-CO(CH₂)₄-CO] in which 6-6 represents, the number of carbon atoms in the two.

Question.9.Define the terms, ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions?
Answer : The flow of solvent from solution of lower concentration to higher concentration of solution through a semi-permeable membrane is called osmosis. Minimum external pressure that prevents osmosis is called osmotic pressure. Osmotic pressure can be measured at room temperature and for very dilute solutions. Its magnitude is large even for very dilute solution.

Question.10. Express the relation among the cell constant, the resistance of the solution in the cell and the conductivity of the solution. How in the conductivity of a solution related to its molar conductivity.

Question.11. Given that the standard electrode potentials (E°) of metals are:

The half reactions of an electrochemical cell are given below :

Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured.

Question.12. Describe the following:

Tyndall effect
Shape-selective catalysis
Answer :

1. When strong beam of light is passed through a colloidal solution it gets scattered and path of light beam gets illuminated by a bluish light. This phenomenon is known as Tyndall effect.
2.  The catalytic action that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is known as shape-selective catalyst, e.g. zeolites.

Question.13. What is meant by coagulation of a colloidal solution? Name any method by which coagulation of lyophobic sols can be carried out.
Answer : The bringing together of colloidal particles such that they either float on the surface or precipitate by adding suitable electrolyte is called coagulation of colloidal solutions. Methods : addition of electrolyte, electrophoresis.

Question.14. Complete the following reactions equation :

Question.15. Draw the structural formulae of the following compounds :

Question.16. Give the chemical tests to distinguish between the following pairs of compounds :
(i) Ethylamine and Aniline
(ii)Aniline and Benzylamine
Answer :

Question.17. Identify A and B in each of the following processes :

Question.18. Draw the molecular structure of the monomers of

Question.19. The density of copper metal is 8.95 g cm^-3. The radius of copper is 127.8 pm. Is the copper unit cell simple cubic, body-centred cubic or face-centred cubic?
(Given : atomic mass of Cu = 63.54 g mol-1 and NA = 6.02 x 10²³mol^-1)

Question.20. What mass of NaCl (molar mass = 58.5 g mol^-1) be dissolved in 65 g of water to lower the freezing point by 7.5°C ? The freezing point depression constant,k_f for water is 1.86  kg mol^-1. Assume van’t Hoff factor for NaCl is 1.87.

Question.21. Describe the role of the following:
(i) NaCN in the extraction of silver from a silver ore
(ii)Iodine in the refining of titanium.
(iii)Cryolite in the metallurgy of aluminium.
OR
Describe the principle involved in each of the following processes of metallurgy :
(i) Froth floatation method
(ii)Electrolytic refining of metals
(iii)Zone refining of metals
Answer :
(i) NaCN is Used to leach the silver ore in the presence of air. Pure silver is obtained by replacement in the process of extraction of silver.

(i) In this method, the ore particles are wetted with oil and come on the surface as froth and gangue particles get wetted by water and settle down.
(ii)In this method, impure metal is made to act as anode and a stripe of pure metal is made as cathode. Electrolyte is used in the cell to conduct electricity. When current is passed, metal ions from the electrolyte are deposited at cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form of ions.
(iii)Zone refining is used to obtain high purity metal using the principle that impurities are more soluble in the molten state than in the solid state of metal,

Question.22. Explain the following cases giving appropriate reasons:

1. Nickel does not form low spin octahedral complexes.
2. The πcomplexes are known for the transition metals only.
3. CO²⁺ is easily oxidized to CO³⁺ in the presence of a strong ligand.

Answer :

1. The low spin octahedral complexes involve d²sp³ hybridisation which is not possible for nickel due its configuration [Ar] 3d⁸4s² therefore, Ni does not form low spin octahedral complexes.
2.  Transition elements have vacant and partially filled d-orbitals leading to variable valency and hence π-complex formation.
3. CO²⁺ ions are easily oxidized to CO³⁺ in presence of strong ligand because crystal field stabilization energy of CO³⁺ion is higher than CO²⁺ ion as CO³⁺has d 6 configuration and CO²⁺ has d⁷ configuration.

Question.23. How would you differentiate betweenS_N1 and S_N2 mech-anisms and give examples of each.
Answer: S_N1generally involves a weak nucleophile in aqueous solution to form carbocation intermediate. The rate of reaction depends upon the concentration of only one reactant.

Question.24. How would you convert the following:
(i) Phenol to benzoquinone
(ii)Propanone to 2-Methypropan-2-ol
(iii)Propene to propan-2-ol Answer:
Answer:

Question.25. How would you account for the following :

1.  NCl₃ is an endothermic compound while NF₃ is an exothermic one.
2. XeF₂ is linear molecule without a bend.
3. The electron gain enthalpy with negative sign for fluorine is less than that for chlorine; still fluorine is a stronger oxidizing agent than chlorine.

Answer :

1. This is because bond dissociation energy of F₂ is lower than Cl₂. Morever, fluorine forms stronger bond with nitrogen due to comparable size.
2. XeF₂has sp3d hybridization with 3 lone pairs in equatorial position and (F) attached at axial positions.
3. Fluorine has smaller size compared to chloride and has low enthalpy of dissociation of F-F bond. Therefore it is a stronger .oxidizing agent than chlorine.

Question.26. Amino acids may be acidic, alkaline or neutral. How does this happen? ‘What are essential and non-essential amino acids? Name one of each type.
Answer : Amino acids can be acidic, basic on neutral depending up on the relative number of amine and carboxyl group in their molecule. Equal number of amine and carboxyl group make it neutral; more no. of amine (—NH₂) than carboxyl groups make it basic and more carboxyl group as compared to amine group make it acidic.
The amino acids which the body is not able to synthesise are taken through diet are called essential amino acids. E.g. valine, while the amino acids which the body can synthesise are called non-essential amino acids, e.g Glycine.

Question.27. Explain the following terms with one example in each case:

1. Food preservatives
2. Enzymes
3. Detergents

Answer :

1. Food preservatives are chemical substances used to prevent microbial growth enhancing their appeal in food without changing the nutritive vklue. e.g. Sodium benzoate, sodium metasuphite.
2. Enzymes are biocatalysts of body and are made up of amino acids. The enzyme is so built that it binds to the substrate in specific manner. pH as well as temperature specific, e.g. E. Coli
3. Detergents are sodium salts of long chain alkylbenzene sulphonic acids. They do not form scum with hard water. e.g. Sodium laurylsulphonate, sodium n-dodecyl benzene sulphonote, etc.

Question.28. (a) Explain the following term:

1. Rate of a reaction
2. Activation energy of a reaction

(b) The decomposition of phosphine, PH3, proceeds according to the following equation :
It is found that the reaction follows the following rate equations:
4PH₃(g)→P₄(g)+ 6H₂(g)
It is found that the reaction follows the following rate equations:
Rate =k[PH₃]
The half-life of PH3 is 37.9 s at 120°C
(i) How much time is required for 3/4th of PH₃ to decompose ?
(ii)What fraction of the original sample of PH₃remains behind after 1 minute ?
OR
Explain the following term :
(i) Order of a reaction
(ii) Molecularity of a reaction
(b) The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. (R = 8.314 J K^-1mol^-1)
Answer :
(a) (i) The rate of decrease (or increase) in the concentration of reactants (or product) per unit time is called rate of reaction.
(ii) The minimum energy required to be given to the reactant to start a reaction is called activation energy.
(b) (i) Hence t_1/2 = 37.9 s


(a) (i) The sum of powers of the concentrations of the reactants in the rate law expression is called the order of that chemical reaction.
(ii) The number of components i.e. atoms, ions or molecules taking part in elementary reaction and collide with one another to start the reaction is. called molecularity of a reaction.

Question.29. (a) Complete the following chemical equations :

(b)Explain the following observations :
(i)La³⁺ (Z = 57) and Lu³⁺ (Z = 71) do not show any colour in solutions.
(ii)Among the bivalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism.
(iii)Cu+ ions is not known in aqueous solutions.
Answer: (a)

Question.30.(a) Illustrate the following name reactions giving a chemical equation in each case :

1. Clemmensen reaction
2. Cannizzaro’s reaction

(b) Describe how the following conversion can be brought about:
(i) Cyclohexanol to cyclohexan 1-one
(ii)Ethylbenzene to benzoic acid
(iii)Bromobenzene to benzoic acid
OR
(a) Illustrate the following name reactions :
(i) Hell-Volhard-Zelinsky Reaction (ii) Wolfif-Kishner reduction reaction
(b) How arc the following conversions carried out:
(i) Ethylcyanide to ethanoic acid .
(ii)Butan-l-ol to butanoic acid
(iii)Methylbenzene to benzoic acid

SET II

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

Question.1. Which type of semi-conductor is obtained when silicon is doped with arsenic ?
Answer:

Question.2. Nitrogen is relatively inert as compared to phosphorus. Why?
Answer: Since P-P single bond is much weaker than N = N bond. Therefore, P is more reactive and N is inert.

Question.6. What are monosaccharides?
Answer : Monosaccharides are the carbohydrates that can’t be further hydrolysed to simpler smaller units, e. g. Glucose, fructose.

Question.8. What is meant by‘copolymerisation’?
Answer : The process of polymerisation from a mixture of two or more unsaturated monomer units is called copolymerization e. g. Buna-S, Nylon-6,6.

Question.13. Define the following:

1.  Peptization
2.  Reversible sols

Answer :

1.  The process of converting 4 precipitate into a colloidal solution by shaking in dispersion medium and adding suitable electrolyte is called peptization.
2. Reversible sols are lyophil’ic sols that can be remade by simple mixing dispersed phase with the dispersing medium.

Question.15. Complete the following chemical reaction equation:

Question.17. Give the chemical tests to distinguish between the following pairs of compounds :
(i) Methylamine and Dimethylamine
(ii)Aniline and N-Methylamine
Answer: (i) Methylamine gives carbylamine test when heated with alcoholic solution of KOH and chloroform and gives foul smell of methyl isocyanide. Dimethylamine does not give carbylamines test.

Question.18. Draw the structure of the monomers of the following

Question.19. Silver crystallizes in face centred cubic unit cell. Each side of the unit cell is of length 409 pm. What is the radius of t silver atoms?

Question.20. What mass of ethylene glycol (molar mass = 62.0 g mol1) must be added to 5.50 kg of water to lower the freezing point of water from 0° C to – 10.0°C? ( K_f  for water = 1.86 K kg mol^-1)

Question.27. What are analgesic drugs? How are they classified and when are they usually recommended for use?
Answer:  Analgesics drugs : The class of drugs used to relieve pain and inflammation are called analgesic drugs.
They are classified as :

1. Narcotic analgesics
2. Non-narcotic analgesics

Non-narcotic drugs are used in prevention of heart attacks like aspirin and Narcotic drugs are required for relief of cardiac pain, post operative pain. e. g. Morphine, heroin, etc.

SET III

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

Question.1.Write a distinguishing feature of metallic solids.
Answer : Metallic solids have metallic bonds. They have positive metal centres and mobile electrons due to which they have high electrical and thermal conductivity.

Question.3.Differentiate between molarity and molality of a solution.
Answer: Molarity is the number of moles of solute dissolved per litre of the solution

Question.8.What are the products of hydrolysis of sucrose?
Answer : Dextrorotatory glucose and laevorotatory fructose

Question.20. 15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at -0.34°C. What is the molar mass of the substance? (K_f for water = 1.6 Kkg mol^-1 )
Answer: Given,

Question.22. How would you account for the following:

1.  The electron gain enthalpy with negative sign is less for oxygen than- that for sulphur.
2. Phosphorus shows greater tendency for catenation than nitrogen.
3. Fluorine never acts as the central atom in polyatomic interhalogen compounds.

Answer:

1. Due to high electronegativity and small size of oxygen compared to sulphur.
2. Due to low electronegativity, large size and presence of d-orbitals in phosphorus.
3. Fluorine has small size therefore, it cannot accommodate larger sized halogen atoms around it, d-orbitals are absent therefore, it does not show positive oxidation state of +3, +5 and +7 hence does not act as a central atom in polyatomic inter-halogen compounds.

Question.25. Write the name, the state of hybridization, the shape and the magnetic behaviour of the following complexes :

Question.26. Differentiate between fibrous proteins and globular proteins. What is meant by the denaturation of a protein?
Answer:


The breaking down of proteins into amino acids due to high temperature is known as denaturation.

Question.27. Explain the following terms with an example for each :

1. Antibiotics
2. Antiseptics
3. Analgesics

Answer :

1. Antibiotics are substances obtained from micro¬organisms to inhibit the growth or kill microorganisms, e. g. penicillin, ampicillin, etc.
2. Antiseptics are antimicrobial substances that are applied
   to living tissues present microbial growth and also kill them. e. g. dettol, iodine, etc.
3. Analgesics are substances that are used to get relief from pain. e.g. paracetamol aspirin, etc.

CBSE previous Year Solved  Papers  Class 12 English Outside Delhi 2011

Time allowed : 3 hours                                                                                           Maximum Marks: 100
General Instructions :

1. This paper is divided into three sections : A, B and C. All the sections are compulsory.
2. Separate instructions are given with each section and question, wherever necessary. Read these instructions very carefully and follow them faithfully.
3. Do not exceed the prescribed word limit while answering the questions.

SET I

SECTION – A
(Reading)

Question.1. Read the passage given below and answer the questions
that follow:

1.  For many years now the governments has been promising the eradication of child-labour in hazardous industries in India. But the truth is that despite all the rhetoric, neither the government so far has succeeded in eradicating this evil, nor has been able to ensure compulsory primary education for every Indian child. Between 60 and 100 million children are still at work instead of going to school, and around 10 million are working in hazardous industries. India has the largest child population of 380 million in the world, plus the largest number of children who are forced to earn a living.
2.  We have many laws that ban child-labour in hazardous
   industries. According to the Child-Labour (Prohibition
   and Regulation) Act 1986, the employment of children below the age of 14 in hazardous occupations has been strictly banned. But each state has different rules regarding the minimum age of employment. This makes implementation of these laws difficult.
3.  Also, there is no ban on child-labour in non-hazardous occupations. The Act applies to the organised or factory sector and not the unorganised or informal sector where most children find employment as cleaners, servants, porters, waiters etc. among other forms of unskilled
   work. Thus, child-labour continues . because the
   implementation of the existing laws is lax.
4.  There are industries, which have a special demand for child labour because of their nimble fingers, high level of concentration and capacity to work hard at abysmally low wages. The carpet industry in U.P. and Kashmir employs children to make hand-knotted carpets. There are 80,000 child workers in Jammu and Kashmir alone. In Kashmir because of the political unrest, children are forced to work while many schools are shut. Industries like gem cutting and polishing, pottery and glass want to remain competitive by employing children.
5.  The truth is that it is poverty which is pushing children into the brutish labour market. We have 260 million people below the poverty line in India, a large number of them are women. Poor and especially woman-headed families, have no’option but to push their little ones in this hard life, in to hostile conditions, with no human or labour rights.
6.  There is a lobby which argues that there is nothing wrong
   with children working as long as the environment for i work is conducive to learning new skills, but studies have
   shown that the children are made to do boring, repetitive and tedious jobs and are not taught new skills as they grow older. In these, hell-holes like the sweet shops for the old there is no hope.
7.  Children working irr hazardous industries are prone to debilitating diseases which can cripple them for life. By sitting in cramped, damp and unhygienic spaces, their limbs become deformed for life, inside matehstick, fire¬works and glass industries they are victims of bronchial diseases andT.B. Their mental and physical development is permanently impaired by long hours of work. Once trapped, they can’t get out of this vicious circle of poverty. They remain uneducated and powerless. Finally, in later years, they too are compelled to send their own children to work. Child-labour perpetuates its own nightmare.
8.  If at all the government was serious about granting children their rights, an intensive effort ought to have been made to implement the Supreme Court’s Directive of 1997, which laid down punitive action against employers of child-labour. Only compulsory primary education can eliminate child-labour,
9. Surely, if 380 million children are given a better life and elementary education, Indians human capital would be greatly enhanced. But that needs, as former President Abdul Kalam says, “a second vision”.

(a) (i) On which two accounts has the government not succeeded so far in respect of children ?
Answer : The government has not succeeded so far in respect of children on account of eradication of child-labour in hazardous industries in India and to ensure compulsory primary education for every Indian child.
(ii) What makes the implementation of child-labour law
difficult ? 
Answer : Each state has different rules regarding the minimum age of employment. Also, there is no ban on child-labour in non-hazardous occupation, The Act does not apply to the unorganized or informal sector. This makes the implementation of child-labour law difficult.
(iii) Why do the industries prefer child-labour ?
Answer : There are industries which have a special demand for child-labour, it is because of their nimble fingers, high level of concentration and capacity to work hard at low wages.
(iv) What are the adverse effects of hazardous industries
on children ? Give any two.
Answer: Children working in hazardous industries are prone to debilitating diseases which can cripple them for life. By . sitting in cramped, damp and unhygienic spaces, their limbs become deformed.
Secondly, they become victim of bronchial diseases by working in matehstick, fire-works and glass industries.
(v) What does the Supreme Court’s Directive of 1997 provide ?
Answer : The Supreme Court’s Directive of 1997 laid down punitive action against employers of child-labour. Only compulsory primary education can eliminate child-labour.
(b) Find words from the passage which mean the same as the following ?
(i) risky/dangerous (para 1)
Answer: hazardous .
(ii) very unfriendly (para 5)
Answer: hostile
(iii) intended as punishment (para 8)
Answer: punitive

Question.2. Read the passage given below and answer the questions that follow: 
There is nothing more frustrating than, when you sit down at your table to study with the most sincere of intentions and instead of being able to finish the task at hand you find your thoughts wandering. However, there are certain techniques that you can use to enhance your concentration. “Your concentration level depends on a number of factors,” says Samuel Ghosh, a social counsellor. “In order to develop your concentration span, it is necessary to examine various facets of your physical and internal environment,” she adds.
To begin’ with one should attempt to create the physical environment that is conducive to focussed thought. Whether it is the radio, TV or your noisy neighbours identify the factors that make it difficult for you to focus. For instance, if you live in a very noisy neighbourhood, you could try to plan your study hours in a nearby library.
She disagrees with the notion that people can concentrate or study in an environment with distractions like a loud television, blaring music etc. “If you are distracted when you are attempting to focus, your attention and retention powers do not work at optimum levels,” cautions Ghosh. “Not more than two of your senses should be activated at the same time,” she adds. What that means is that music that sets your feet tapping is not the ideal accompaniment to your books’?
Also do not place your study table or desk in front of a window. “While there is no cure for a mind that wants to wander, one should try and provide as little stimulus as possible. Looking out of a window when you are trying to concentrate will invariably send your mind on a tangent,” says Ghosh.
The second important thing, she says, is to establish goals for oneself instead of setting a general target and then trying to accomplish, what you can in a haphazard fashion. It is very important to decide what you have to finish in a given span of * time. The human mind recognises fixed goals and targets and appreciates schedules more than random thoughts”. Once your thoughts and goals are in line, a focused system will follow.
She recommends that you divide your schedule into study and recreation hours. When you study, choose a mix of subjects that you enjoy and dislike and save the former for the last so that you have something to look forward to. For instance, if you enjoy verbal skill tests more than mathematical problems, then finish Maths first. Not only will you find yourself working harder, you will have a sense of achievement when you wind up.
Try not to sit for more than 40 minutes at a stretch. Take a very short break to make a cup of tea or listen to a song and sit down again. Under no circumstances, should one sit for more than one and a half hours. Short breaks build your concentration and refresh your mind. However, be careful not to overdo the relaxation. It may have undesired effects.
More than anything else, do not get disheartened. Concentration is merely a matter of disciplining the mind. It comes with practice and patience and does not take very long to become a habit for life.
(a) On the basis of your reading of the above passage, make notes on it in points only using abbreviations wherever necessary. Supply a suitable title.
1. Notes
(i) Tech, to enhance concentration
(a) create phy envirmnt.
(b) identify the factors resp
(c) music not the ideal accomp
(ii) Recommendations
(a) dvd schedule
(b) choose sub that you like or dislike
(c) don’t sit for more than 40 min
(iii) More suggestions
(a) don’t get dishrtnd
(b) concentrn is merely a discipline of mind
(c) rqurs practice and patience

Abbreviations used
Tech                technique
Phy                  physical
Envirmnt       environment
Resp                responsible
Accomp          accompaniment
Dvd                 divide
Dishrtnd        disheartened
Concentrn     concentration
Rqurs              requires
Title : Concentration – Key to Success (or) Techniques to Enhance Concentration
(b) Write a summary of the above passage in about 80
words.

SUMMARY

According to Samuel Ghosh, a social counsellor, study needs concentration which depends on various factors. These include physical environment and avoiding distractions.
Music should not be listened as it lessens the ‘attention and retention power. One should -not keep study table or desk in front of a window to provide little stimulus to mind. Another important thing is to set goals. Once thoughts and goals are in line, a focused system will flow. Ghosh recommended that the schedule to study and recreation hours should be divided. Choose the subject first which is disliked. Do not sit for more than 40 minutes, take a break in between and do not get disheartened.

SECTION-B
(Advanced Writing Skills)

Question.3.You are Secretary of Gymkhana Club, Madurai. Write a
notice in not more than 50 words informing the members to attend an extraordinary meeting of governing body. Include details like date, time, venue etc. Sign as Prabhu/ Pratibha.
OR
Due to sudden landslide and inclement weather, St. Francis School, Vasco has to be closed for a week. As a Principal of that school, draft a notice in not more than 50 words to be displayed at the school’s main gate notice board.
Answer:

Answer:

Question.4.You are Poorva/Partha, Cultural Secretary of your school,
D. B. Senior Secondary School, Ambur. A week long Music and Dance festival was organized by your school. Write a report in 100-125 words for your school magazine. Invent details.
OR
The Debating Society of your school has recently held a workshop on Continuous and Comprehensive Evaluation (CCE) introduced for the students of class X in all the schools. The students discussed the assessment made by the school on the basis of their participation in various activities and the system of grading. Write a report in 1 GO- 125 words for your school magazine. You are Parveen/ Payal, Secretary of Society.
Answer.
Music And Dance Festival
Report: By Poorva (Cultural Secretary)
D.B. Sen. Sec. School
25th November, 2014 ,
D. B. Senior Secondary School, Ambur, organized a week long music and dance festival in the school auditorium last week. The chief guest of the function was none other than very popular Ad Guru, Mr. Piyush Pandey. The function commenced with the garlanding of Lord Ganesha and lightening up of the lamp. Students gave cultural performance on this occasion. A group of 50 girls presented Rajasthani folk dance accompanied with a band of 25 girls playing the music at the background, which received a huge applaud from the audience. Our Principal, honourable Mr. M. K. Singh read the annual report of the school. The festive cheer echoed the building of the school. One could see the exuberant enthusiasm of the artists which made the festival special. The function concluded with the speech by the Chief Guest.
OR
Workshop on Continuous and ”
Comprehensive Evaluation
Organized by : Debating Society
Report: By Parveen (Secretary of Society)
26 November 20XX
Continuous and Comprehensive Evaluation, refers to a system of school based evaluation of students that covers all aspects of students’ development. Keeping this in mind, the debating society of our school has recently held a workshop on CCE.
The students discussed about the assessment made by the school based on their performance in scholastic as well as co¬scholastic areas. They found that this system has lessened the burden of children to a great extent and it will go a long way. It includes the Formative and Summative Assessment. With this new system it has become possible to learn through do-it- yourself and fun-filled way. With the introduction of grading system, it encourages specific abilities of the students who do not excel in academics. Learning has become a pleasure as it has taken away the fear of examination. The students told that they are enjoying it and it is a positive approach towards learning.

Question.5.You are Raman / Roma, a member of Parents Teacher Association of Little Valley Senior Secondary School, Hyderabad. Write a letter to the Principal of the school asking him to introduce vocational stream in the school providing facility of teaching such.subjects as computer, insurance etc. so that the students may not needlessly continue academic studies. You are residing at 15, Anand Colony, Hyderabad.
OR
Write a letter to the Manager (Publication) of Little Flower Company, Hyderabad, placing an order for four books on management and administration recently published by them. You are Ronit/Rohini, Librarian, H. P. Engineering College, Tirupathi.
Answer:
15, Anand Colony
Hyderabad
10th May, 2014
The Principal
Little Valley Senior Secondary School Hyderabad
Subject: Introduction of Vocational Stream in School Dear Sir,
I, would like to draw your attention towards the changing need of time. The world is growing at a faster speed and we have to keep our pace with it. Students of today are the visionaries of tomorrow. So. we have to develop their personality and skills in such a way that they develop into responsible human beings. Keeping this in mind, vocational stream should be started in the school so as to facilitate students in such a way that they do not have to study academics without purpose. This will provide educational and vocational guidance in the field in which they are interested etc. The introduction of the subjects as computer, insurance would not only help them in ; choosing their career, but interest them also. It would assist
them in exploring various career options. Thus, I request you to introduce vocational stream in the school.
Hope you will give your due attention to my request. Thanking you,
Yours sincerely,
Roma (Member of Parents Teacher Association)
OR
H.P. Engineering College
Tirupathi
12th April, 2014
The Manager (Publication)
Little Flower Company,
Hyderabad
Subject: Order for the books of Dear Sir.
On behalf of the college, I want to place an order for four books
on Management and four books on Administration which you
have recently published. As our college has urgent need of these books, I would be grateful if you supply them at the earliest, as the session is soon’going to start. You could send these books
by V.EP. and attach the bill along with. I will dispatch a cheque of the required amount in favour of Little Flower Company, Hyderabad within 10 days after receiving the books. I hope that a discount of 25% on the printed price will be applicable to us as per rules for Libraries and Education Institutions.
I would feel obliged if the delivery of the books will be done within 10 days.
Yours truly,
Rohini (Librarian)
H.P. Engineering College Tirupathi

Question.6.The invention of mobile phone has brought about a revolution in the lives of the people in the country. If used properly it can be a blessing but if misused it can prove to be a curse. Write an article in 150-200 words on ‘Mobile phone – a boon or bane.’ You are Kartik/Krishna. [10] OR
With the rising number of people in almost all the big cities of the country, the rate of crime has also increased proportionately. The police needs to be trained in new methodology of combating the crime besides changing its mindset. Write an article in 150-200 words on “The Role of Police in maintaining law and order in the Metropolitan Cities”. You are Ravi/Ravina.
Answer:
‘Mobile phone — a boon or bane’
By: Krishna
Modern Technology and Science has brought a revolution on earth. The best of it, is the invention of gadgets. We have become so prone to these gadgets that we feel handicapped without them. Among them one is mobile phones which are most important. Earlier it was supposed to be a luxury but now it has become necessity. It has reached in the hands of every common man. Even children carry it with them. Mobile phones have brought a revolution in the world of communication. We can say that they are blessings if used in a proper way. They can be useful at the time of any emergency, conveying important message or saving a precious life. But on the other hand, if they are not used wisely, they can prove to be a curse. Using mobiles while driving can prove fatal to -» any life, excessive use by the children make them addicted to it or sometimes the family life gets disturbed. There are certain laws which restrict the use of mobile phones while driving because many accidents have taken place due to it but how many of us abide by the laws. So mobile phones should be used only when required and not as a source of entertainment.
OR
“The Role of Police in maintaining law and order in the Metropolitan Cities”
By: Ravina
The law and order situation of the country is deteriorating day by day. The crime graph of the Metropolitan Cities is heating up. There is not a single day when we do not read the cases of burglary, murder, accident, kidnapping or stabbing in newspapers. It is the symbol of total disrespect towards the laws and regulations of the Constitution. Chain snatching has become the most common problem. Girls don’t feel safe while moving out alone especially after evening. Now personal security has become the major issue and police will have to take strict action against it adopting some different methods to tackle these criminals. They should be given training in such a way that they are able to read the minds of the criminals and know their planning before they commit any barbarous crime. People living in Metropolitan Cities are suffering from a strong feeling of insecurity, lawlessness and ignorance. They are losing their faith in the police. In order to gain the faith of the public; police should study the psychology of the criminals and then initiate adequate steps towards stopping the crime.

SECTION -C
(Text Books)

Question.7. Read the extract given below and answer the questions that follow:
Break O break open till they break the down And show the children to green fields, and make their world Run azure on gold sands, and let their tongues Run naked into books the white and the green leaves open
History theirs whose language is the sun.
(a) To whom does ‘they’ refer ?
Answer : ‘They’ refers to the children of the slum sitting in the classroom of an elementary school.
(b) What would they break ? .
Answer : They would break free from the chains of the slum. They would break all the windows which have sealed their fate and enjoy a new lease of life and freedom.
(c) What other freedom should they enjoy ?
Answer : They should enjoy equal rights like other citizens and given proper education and should have a bright future like others.

Question.8. Answer any three of the following in 30-40 words each :
(a) What is the sadness that the poet, Pablo Neruda refers to in the poem, “Keeping Quiet” ?
Answer : The poet talks about the sadness that is within the people. In order to fulfill their desires and completing their duty, they become so selfish that they forget others and live without happiness and universal brotherhood.
(b) What is the message of the poem, ‘A Thing of Beauty’ ?
Answer : The poet conveys the message that the world is full of unwavening beauty that gives eternal pleasure. Such beautiful things are valuable and never pass into nothingness. So they should not be destroyed.
(c) What were Kamala Das’ fears as a child ? Why do they surface when she is going to the Airport ?
Answer : Kamala Das, fears that her mother would leave her alone one day. They surface when she takes an intense look at her mother who is seated beside her while going to the Airport.
(d) Why do you think Aunt Jennifer created animals that are so different from her own character ?
Answer: Aunt Jennifer had experienced constraints through-out in her married life. The ring around her finger symbolizes the weight of her marriage so she created tigers which are representative of the freedom that she aspires for.

Question.9. Answer the following in 30-40 words each :
(a) Franz thinks, “Will they make them sing in German, even the pigeons ?” What could this mean ?
Answer: This means that the Germans had made it compulsory to teach German in the schools but they cannot take away their love for their mother tongue, French which was mixed in their blood.
(b) Which factors led Douglas to decide in favour of Y.M.C.A. pool ?
Answer : Douglas wanted to learn swimming so he decided in favour of Y.M.C.A. pool because it was not too deep at the shallow end and he found it safe as the drop at the other end was gradual. ,
(c) Why was peddler surprised when he knocked at the door of the cottage ?
Answer : The peddler was surprised to see that the owner of the cottage did not get irritated when he knocked at the door at night, but gave him a warm welcome and provided him with food and shelter. He was treated like a guest.
(d) What thoughts came to Sophie’s mind as she sat by the canal ?
Answer : While sitting by the canal, Sophie was thinking about Danny Casey only. She started fantasizing, his actions and the dialogues as he was a hero in her eyes. She did not even care about other’s view about her.

Question.10. Answer the following in 125-150 words each :
Why do you think Gandhiji considered the Champaran episode to be a turning point in his life ?
OR
‘Lost Spring’ explains the grinding poverty and traditions that condemned thousands of people to live a life of abject poverty. Do you agree ? Why/why not ?
Answer : Champaran episode was the turning point in Gandhiji’s life because it released the farmers from the British Landlords and made them realize their rights. It was also Gandhiji’s first non-cooperative movement. Gandhiji went to Bihar and there he heard the problems of peasants and came to know about their pathetic condition. He resolved to fight for the peasants and release the poor sharecroppers from the British landlords. From Patna, he went to Muzaffarpur where he met the lawyers and convinced them that they were there to serve the peasants. He also rebuked them for charging a hefty fee from the poor sharecroppers. His main aim was to make the peasants fearless, as the freedom from fear was more important than legal justice for them. The British planters were forced to leave the land of the owners.
Therefore,Champaran episode proved to be a turning point in his life as he understood that people of India could win freedom by truth and non-violence.
OR
Yes, ‘Lost Spring’ explains the grinding poverty and traditions, that condemned thousands of people to live a life of abject poverty. The slum of Seemapuri, in Trans-Yamuna area of Delhi is full of people living in pathetic conditions.
The family of Saheb-E-Alam, also came from Bangladesh and settled there. They have no work to do, no shelter, no sufficient food, no proper clothes and even no chappals. So they began to do the work of rag picking. The chapter deals with the lives of children like Mukesh and Saheb-E-Alam who worked in bangle making factory and did the work of rag picking respectively. These children were deprived of basic necessities like going to school and having fun. They are forced to do work in an unhealthy environment at a very early stage. There, they are exploited and become the victims of the cruel masters. Thus, the title is justified as the spring which is a ‘season of bloom’ is lost from their lives.

Question.11. Answer the following in 125-150 words each :
Dr. Sadao was compelled by his duty as a doctor to help the enemy soldier. What made Hana, his wife, sympathize with him in the face of open defiance from the domestic staff ?
Answer : Dr. Sadao was a maft of principles. Once when he was standing out of his house, situated on the beach, he saw a man flung out of the ocean. He noticed that the man was an enemy soldier and was wounded badly. He thought of throwing him back into the ocean but could not do so. Although, he had no love for Americans and he was a true patriot still he saved him realizing his sacred duty to save a, dying man. This brought an open defiance of his domestic staff and they all left him. His wife had to do all the household work. But when she saw her husband doing his duty, she sympathized with him and helped her husband in giving anesthesia to the wounded man. She even, washed the wound’s of the enemy solider when the maid refused to do so. She set an example that feeling of humanity was way above the feeling of nationalism.

Question.12. Answer the following in 30-40 words each :
(a) Why does Mr. Lamb leave his gates always open ? Answer : Mr. Lamb lived alone in his house. He had no company so he left the gates open so as to let the children enter. He had an apple orchard from where he pulled down the crab apples and save them to the children.
(b) How did the governor react to the two phone calls he
received in quick succession ? :
Answer : The Governor accepted the calls in a strict manner but replied positively. The call was from the secretary who informed that they had forgotten to put the correction slip in the examination package. The Governor allowed them to explain the correction to Evans without any doubt.
(c) What advice did Annan offer Bama ?
Answer : Annan advised Bama to study with care and learn all she could. If one is ahead in lessons, people would attach themselves to you. He told her that education was the only key to open the gates of respect and dignity. The community would also be benefitted by this.

SET II

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

SECTION B
(ADVANCED WRITING SKILLS)

Question.3. You are Secretary of J. P. Narain Housing Society, R. W. A. Meerut. Draft a notice in not more than 50 words stating that the second installment of maintenance
charges falls due on 31st March, 2011 and requesting the members to pay before the due date. Sign as Anil/Anita.
OR
You are Scout Master/Guide Captain of K. R. Public School, Mysore. You have decided to send a troop of scouts and guides of your school to the jamboree to be held at Lucknow for a week. Draft a notice in not more than 50 words to be placed on the school notice board inviting the names of those scouts and guides who are interested to participate in the jamboree. Invent the necessary details. .
Answer:

Question.4. You visited a Job fair organized by Ability Foundation at Chennai recently. You were impressed to see that nearly 55 companies from various sectors such as information technology, telecommunication, electronics etc. offered jobs to the final year students of various colleges. As a reporter of ‘The Deccan Times’, Chennai, prepare a report in 100-125 words. You are Peeyush/Priya.
OR
You are Anand/Anandi, a ‘Times Of India’ correspondent. You attended the inaugural function of Tamilnadu Hospital, Chennai. Mentioning the specialities of the hospital, the number of beds, all available facilities etc. and details about the ceremony, write a report in 100-125 words for your newspaper.
Answer:
Ability Foundation Organised
Job Fair, At Chennai
Report by : Priya (‘The Deccan Times’),
Chennai
16th May, 20XX
Recendy a Job Fair was organized by Ability Foundation at Chennai. It was a two day job fair and was held at Madras University sports complex. It was a unique event which saw nearly, 55 companies from various IT sectors, offering jobs to the final year students. The Chief Minister of Tamil Nadu inaugurated the job fair and called it a constructive step as, large number of students will be benefitted by this. They were able to gain employment adeast with one of the organization. It was able to satisfy the students completely because they are now looking for offbeat jobs like Fashion designing, Publishing, Aviation, Film making, Geologists, Journalists, and many more. Seeing the queue at the registration desk, one could guess at first sight that it would be a hopeful event. Companies at job fair included Wipro, Max Bupa, Convergys, etc. The Minimum package offered to the students at the fair was 2 Lakh per annum and the Maximum was 8 Lakh per annum.
OR
Tamilnadu Hospital Inaugurated
Report By : Anandi (Toi Correspondent)
17th June, 20XX
A Swanky 18 th floor structure of Tamilnadu Hospital in the heart of Chennai city was inaugurated by the Chairman, Apollo Group of Hospitals. It is furnished with world class facilities. Every type of quality healthcare services is promised to be provided by the hospital. It has a capacity of 140 beds, housed with ultra-modern facilities. The womens health care, High Tech Laboratory, ICU facilities, Digital Imaging, On Call Doctor Service, Emergency are some of the features offered by the hospital. Other facilities would include round the clock frequent visits by doctors and Pharmacy. No consultation fees will be charged for the first week of the inauguration. It would provide a unique opportunity to the patients as well as the Medical Students. Beside treatment, the hospital will work on different research areas as told by senior doctor, Dr. Swami Iyer. Overall it can be said that, we all have high expectations ftom this hospital.

SECTION – C

Question.9. Answer the following in 30-40 words each : 
(a) “What a thunder clap these words were to me!” Which were the words that shocked and surprised little Franz ?
Answer : The words that shocked little Franz were that it was their last French lesson and these were spoken by their French Teacher, M. Hamel. There was an order from Berlin that German would be taught in the schools of Alsace and Lorraine and the new teacher would be arriving from the next day.
(c) What made the peddler change his ways ?
Answer : The behavior and hospitality shown by Edla, the ironmasters daughter, changed the ways of peddler. He was, living a monotonous life of a vagabond, but love and kindness of Edla changed him completely. ‘
(e) Which was the only occasion when Sophie got to see Danny Casey in person ?
Answer : The only occasion when Sophie met Danny Casey in person, was when she went to see football match with her family. Although she saw him from a distance but she fancied . her actions and dialogues.

Question.12. Answer the following in 30-40 words each :
(c) “I felt like sinking to the floor”, says Zitkala-Sa.
When did she feel so and why ?
Answer : The tradition of the community to which Zitkala-Sa belonged was to keep long hair. When her long hair was cut, it was like sinking to the floor for her. Short hair in her community %as the sign of slavery and worn by mourners and cowards.

SET III

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

SECTION – B
(ADVANCED WRITING SKILLS)

Question.3.You are the Sports Secretary of Lalwani Public School, Udaipur. Draft a notice in not more 50 words for your school notice board asking the students to give their names for participation in various events to be held on the Annual Sports Day of your school. Invent the details of the events. Sign as Lalit/Lalita.
OR
You are the Secretary of Sri Parthasarthy Sahha, Salem. Draft a notice in not more than 50 words to be sent to the members of the Sabha, requesting them to attend an extraordinary meeting. Invent necessary details such as items on the agenda, date, time, etc. Sign as Kamal/ Komal.
Answer:

Question.4. You witnessed a programme by differendy abled persons on Zee TV You were very much impressed by their performance and were emotionally touched. Highlighting their talent, reaction of the judges to their performance etc. prepare a report in 100-125 words for your school magazine. You are Gopal/Gopi, student editor of the magazine.
OR
You are Latha/Lalith of Gurgaon. You had attended the inaugural ceremony of the newly opened vocational stream and hobby classes at the secondary school level in Paras Public School, Gurgaon. Draft a report in 100-125 words to be published in ‘Gurgaon Times’. Invent other details.
Answer:
Wonderful Performance by
Differently Abled
Report By : Gopi (Student Editor) .
27th January, 20XX
Yesterday I witnessed a programme on Zee TV presented by the differendy abled people, which moved me a lot. The roles, characters, performances done by them were marvelous. All the actors managed an enthralling performance. They even portrayed the emotional and personal side of the characters perfecdy. No one could tell that this was a performance by special people of our society. Hats off to them who managed to do so in spite of their disabilities. Even the judges were overwhelmed. The performances kept the judges as well as public gripped throughout the programme. I felt glad to have watched this, as it opened the gates of inspiration for many. Although it was hard to do but it included the resdess efforts of the directors and organizers.
OR
Answer:
Inaugural Ceremony of Voctional
Stream And Hobby Classes
A Report: Latha
Gurgaon Times
15th November, 20XX
The opening ceremony of the vocational stream and the hobby classes witnessed a huge gathering on the grounds of secondary school level in Paras Public School, Gurgaon. Honorable Principal extended a warm welcome to the HRD Minister who was the Chief Guest for the ceremony. The faculty for the newly opened streams were also welcomed whole heartedly. The Principal revealed that these new streams and hobby classes would meet the need of the students which would help them in their future prospects. They will be able’ to actualize their potential to the fullest instead of competing for the top three positions. It would also provide a platform for the budding artists. To enhance the learning skills of t children, certain measures were counted. Keeping in mind the requirements and interest of the children, these streams would assist them in vocational guidance and choosing their profession. In the end the ceremony was concluded with vote of thanks by the Head of the Department, Humanities.

SECTION – C

Question.9. Answer the following in 30-40 words each:
(a) What announcement did Mr. M. Hamel make ? What was the impact of this on Franz ?
Answer: Mr. M. Hamel made an announcement that French would no longer be taught hence forth in the schools of Alsace and Lorraine and it was their last lesson in French. This announcement came as a shock to not only Franz but all the other people. Franz realized the importance of his mother tongue and thought that he would not be able to read and speak in French again.
(c) Why did the peddler keep to the woods after leaving crofter’s cottage ? How did he feel ?
Answer : After leaving crofter’s cottage, the peddler kept to the woods in order to avoid the police. He felt that he was caught in the worlds rattrap because after stealing crofters money he realized that it was bait for him and if he took to the road, police would catch him.
(e) Why did Jansie discourage Sophie from entertaining thoughts about the sports star, Danny Casey ?
Answer: Jansie very well knew that Sophie was a day dreamer and she fancied wild stories whereas Jansie herself was a realist and knew that they belonged to a middle class family and Sophies dreams would never come true. So she discouraged her to avoid any type of disappointment.

Question.12. Answer the following in 30-40 words each :
(c) What did Judewin tell Zitkal-Sa? How did she react to it ?
Answer : Judewin told Zitkal-Sa to get ready for hair loss as the school authorities were strong enough. But Zitkal-Sa reacted very furiously over it because keeping long hair was a symbol of bravery and courage in her community.

CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2011

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. There are 26 questions in all.
2.  This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
4.  There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
5. You may use the following values of physical constants wherever necessary:

SET I

Note : Except for the following questions, all the remaining question have been asked in Previous Set.
Question.1.A point charge is placed at point O as shown in the figure. Is the potential difference V_A-V_B positive, negative or zero, if Qis (i) positive (ii) negative

Answer : If Q is positively charged, V_A-V_B = positive If Q is negatively charged, V_A-V_B = negative.

Question.2.A plane electromagnetic wave travels in vacuum along Z-direction. What can you say about the direction of electric and magnetic field vector?
Answer : In electromagnetic wave, the electric field vector E and magnetic field vector g show their variations perpendicular to’ the direction of propagation of wave as well as perpendicular to each other.. As the electromagnetic wave is travelling along z-direction, hence E and B show their variation in xy-plane..

Question.3.A resistance R is connected across a cell of emf e and internal resistance r. A potentiometer now measures the potential
difference between the terminals of the cell as V. Write the expression for ‘r’ in terms of ε, V and R.
Answer:

Question.4.The permeability of magnetic material is 0.9983. Name the type of magnetic materials it represents.
Answer: Paramagnetic material.

Question.5.Show graphically, the variation of the de-Broglie wavelength (λ) with the potential (V) through which an electron is accelerated from rest.
Answer:

Question.6. In a transistor, dopping level .in base is increased slightly. How will it affect

1.  collector current and
2. base current?

Answer :

1. Collector current decreases slightly
2.  Base current increases slightly

Question.7. Define the term ‘wattless current’.
Answer : Current flowing in a circuit without any net dissipation of power is called wattless current.

Question.8. When monochromatic light travels from one medium to another its wavelength changes but frequency remains the same, Explain.
Answer : Atoms (of the second medium) oscillate with the same (incident light) frequency and in turn, emit light of the same frequency.

Question.9. Two uniformly large parallel thin plates having charge densities +σ and – σ are kept in the X-Z plane at a distance ‘d apart. Sketch an equipotential surface due to electric field between the plates. If a particle of mass m and charge ‘-q’ remains stationary between the plates, what is the magnitude and direction of this field?
OR
Two small identical electrical dipoles AB and CD, each of dipole moment ‘p’ are kept at an angle of 120° as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to electric field () directed along +X direction, what will be the magnitude and direction of the torque acting on this?

Question.10. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place.

Question.11. Figure shows two identical capacitors, C₁ and C₂, each of lpF capacitance connected to a battery of 6 V. Initially switch ‘S’ is closed. After sometime ‘s’ is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?

Question.12. Two convex lenses of same focal length but of aperture A₁ and A₂ (A₂ < A₁), are used as the objective lenses in two astronomical telescopes having identical eyepieces. What is the ratio of their resolving power? Which telescope will you prefer and why? Give reason.

Question.13. Draw the output waveform at X, using the given inputs A and B for the logic circuit shown below. Also, identify the logic operation performed by this circuit.

Question.14. Name the semiconductor device that can be used to regulate an unregulated dc power supply. With the help of I-V characteristics of this device, explain its working principle.
Answer : A Zener diode is a specially designed diode which is operated in reverse breakdown region continuously with any damage. When Zener diode is operated in the reverse break down region, the voltage across it remains practically constant (V_z) for a large change in reverse current. Therefore, for any increase/decrease of the input voltage there is a increase/deer ease of the voltage drop across series resistance (Rs) without any change in the voltage across Zener diode,

Question.15. How are infrared waves produced? Why are these referred to as ‘heat wave’? Write their one important use.
Answer: Infrared rays are produced by hot bodies and molecules. Infrared waves are Called heat waves as they cause heating effect. Infrared waves are used to maintain earths warmth, in physical therapy, remote switches etc.

Question.16. Draw the transfer characteristic curve of a base biased transistor in CE configuration. Explain clearly how the active region of the V₀ versus V_i– curve in a transistor is used as an amplifier.
Answer:

Question.17.

1.  Define modulation index.
2. Why is the amplitude of modulating signal kept less than the amplitude of carrier wave?

Answer :

1.  Modulation index is defined as the ratio of amplitude of modulating signal to amplitude of carrier wave. It is given by
   
2.  The amplitude of modulating signal is kept less than the amplitude of carrier wave to avoid noise.

Question.18. A current is induced in coil C₁due to the motion of current carrying coil C₂. (a) Write any two ways by which a large deflection can be obtained in the galvanometer G. (b) Suggest an alternative device to demonstrate the induced current in place of a galvanometer.

Answer : (a) Any two ways to obtain large deflection in G :
(i) Moving C₂ faster towards C₁
(ii)Insertion of soft iron core in C₁
(b) Alternative device that can be used in place of galvanometer is LED.

Question.19. Define the terms (i) drift velocity, (ii) relaxation time. A conductor of length L is connected to a dc”source of emf 8. If this conductor is replaced by another conductor of same material and same area of cross-section but of length 3L, how will the drift velocity change?
Answer : (i) Drift velocity : The average velocity with which the free electrons drift towards positive terminal under the influence of an external field is called drift velocity.
(ii) Relaxation time : Average time interval between two successive collisions of an electron with the ions / atoms of the conductor.

Question.20.Using Gauss’s law obtain the expression for the electric field due to a uniformly charged thin spherical shell of radius R at a point outside the shell. Draw a graph showing the variation of electric field with r, for r > R and r < R.
Answer : Electric field at point P at a distance V’outside the spherical shell.

Question.21.An electron and a photon each have a wavelength 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon and
(c) the kinetic energy

Question.22. Draw a schematic diagram showing the (i) ground wave (ii) sky wave and (iii) space wave propagation modes for em waves. ,
Write the frequency range for each of the following :
(i) Standard AM broadcast
(ii) Television
(iii) Satellite communication

Question.23. Describe Youngs double slit experiment to produce interference pattern due to a monochromatic source of light. Deduce the expression for the fringe width.
OR
Use Huygens principle to verify the laws of refraction.
Answer : S is a narrow slit (of width about 1 mm) illuminated by a monochromatic source of light. At a suitable distance (= 10 mm) from S, two slits S₁ and S₂ are placed parallel to S. When a screen is placed at a large distance (about 2m) from the slit S₁ and S₂, alternate dark and bright bands appear on the screen. These are the interference bands or fringes. The band disappear when either slit is covered.

Explanation : According to Huygens principle, the monochromatic source of light illuminating the slit S tends out spherical wavefront.
The two waves of same amplitude and same frequency super impose on each other, Dark fringes appear on the screen when the crest of one wall falls on the trough of other and they neutralize the effect of each other. Bright fringes appear on the screen when the crest of one wave coincides with the crest of other and they reinforce each other.
Expression for the fringe width :
Let d = distance between slits S₁ and S₂
D = distance of screen from two slits and
x = distance between the central maxima
‘O’ and observation point P.


Now the intensity at point P is maximum :or minimum according as the path difference is an integral multiple of wavelength or an odd integral multiple of half wavelength.
OR
Wavefront : The continuous locus of all the particles of a medium, which are vibrating in the same phase is called wavefront.
Laws of refraction : Let PP’ represent the surface separating medium 1 and medium 2 as shown in fig.


Which is Snell’s Law of refraction of light (first law)
Second Law : Incident wavefront, refracted wave front, normals all lie in the same plane.

Question.24. (a) Describe briefly, with the help of suitable diagram, how the transverse nature of light can be demonstrated by the phenomenon of polarization.
(b) When unpojarized light passes from air to a transparent medium, under what condition does the reflected light get polarized?
Answer:
(a) When a polaroid P₁ is rotated in the path of an unpolarised light, there is no change in transmitted intensity. v The light transmitted through polaroid P₁ is made to pass through polaroid P₂. On rotating polaroid P₂, in path of light transmitted from P₁ we notice a change in intensity of transmitted light. This shows the light transmitted from P₁ is polarized. Since light can be polarized, it has transverse
nature.

(b) Whenever unpolarised light is incident from air to a transparent medium at an angle of incidence equal to polarizing angle, the reflected light gets fully polarized.

Question.25. The energy levels of a hypothetical atom are shown below. Which of the shown transitions will result in the emission of a photon of wavelength 275 nm? 
Which of these transitions correspond to emission of radiation of (i) maximum and (ii) minimum wavelength?

Hence transition B would result in the emission of a photon of wavelength 275 nm.
(i) Transition A corresponds to maximum wavelength .
(ii)Transition D corresponds to minimum wavelength

Question.26. State the law of radioactive decay. Plot a graph showing the number (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half lifeT_1/2.
Depict in the plot the number of undecayed nuclei at (i) t = 3 T_1/2. and (ii) t = 5 T_1/2..
Answer : The number of nuclei undergoing decay per unit time, at any instant, is proportional to the total number of nuclei in the sample at that instant.

Question.27. In the circuit shown, R₁ = 4 Ω, R₂ = R₃ = 15 Ω, R₄ = 30Ω and E = 10 V. Calculate the equivalent resistance of the circuit and the current in each resistor.

Question.28.State Biot-Savart law, giving the mathematical expression for it. Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its
How does a circular loop carrying current behave as a magnet?
OR
With the help of a labelled diagram, state the underlying principle of a cyclotron. Example clearly how it works to accelerate the charged particles.
Show that cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason.




Working : High frequency oscillator maintains moderate alternating potential difference between the dees. This potential difference establishes an electric field that reverse its direction periodically. Suppose a positive ion of moderate mass produced at the centre of the dees, finds D₂ at negative potential. It gets accelerated towards it. A uniform magnetic field, normal to the plane of the dees, makes it move in a circular track. Particle t races a semicircular track and returns back to the region between the dees. The moment it arrives in the region electric field reverses its direction and accelerate the charge towards. D₁. This way charge keeps on getting accelerated until it is removed out of the dees.
Centripetal force, neede d by the charged particle to move in circular track, is provided by the magnetic field.

Thus, frequency of revolut/ion is independent of the energy of the particle. Yes, there is an upper limit on the energy acquired by the charged particle. The charged part icle gains maximum speed when it moves in a path of radius equal to the radius of the dees.

Question.29. (a) Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism.
Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation.
(b) Explain briefly how the phenomenon of total internal reflection is used in fibre: optics.

Here the ray of light propagation from a rarer medium of refractive index ( n₁) to a denser medium of refractive index ( n₂) is incident on the convex side of spherical refracting surface of radius of curvature R.
(b) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the image found.
Answer: (a)


Each optical fibre consists of a core and cladding, refractive index of the material of the core is higher than that of cladding. When a signal, in the form of light, is directed into the optical fibre, at an angle greater than the critical angle, it undergoes repeated total internal reflections along the length of the fibre and comes out of it at the other end with almost negligible loss of intensity.

Question.30.(i) With the help of a labelled diagram, describe briefly the underlying principle and working of a step up transformer.
(ii) Write any two sources of energy loss in a transformer.
(iii) A step up transformer converts a low input voltage into a
high output voltage. Does it violate law of conservation of energy? Explain.
OR
Derive an expression for the impedance of a series LCR circuit connected to an AC supply of variable frequency. Plot a graph showing variation of current with the frequency of the applied voltage.
Explain briefly how the phenomenon of resonance in the circuit can be used in the tuning mechanism of a radio or a TV set.

Principle : It works on the principle of mutual induction. When alternating current is passed through a coil an induced emf is set up in the neighbouring coil.
Working : When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links with both primary and secondary windings. Let (ø) be the flux in each turn in the core at any time t due to current in rlie primary when a voltage Vp is applied to it.
Then the induced emf or voltage Es, in the secondary with


(ii) Sources of energy loss in transformer are : Joule’s loss in the resistan ce of windings, loss due to eddy currents.
(iii) No. A step up transformer steps up the voltage while it steps down the current. So the input and output power remain srime (provided there is no loss). Hence there is no violation of the principle of energy conservation.
OR
impedance (Z) in LCR circuit. The effective resistance offered by a series LCR circuit to the flow of current is called its impedance. It is denoted by Z.



The given graph showing’variation of curr ent with the frequency of the applied voltage.
The radio and TV receiver sets are the practical applications of series resonant circuits. Signals of several different frequencies
are available in air. By turning the tuning knob of the radio set, we vary the frequency of the LC circuit till it mat ches the frequency of the desired signal.

A series resonant circuit allows maximum current through it. So, it is called acceptor circuit. A series resonance circuit is that circuit in which inductance L, capacitance C and resistance R are connected in series. The impedance of this circuit has a minimum value and the current through the circuit is maximum.

SET II

Note : Except for the following questions, all the rem aining question have been asked in Previous Set.

Question.2. The susceptibility of magnetic material is 1.9 x 10 ^-5. Name the type of magnetic materials it represents.
Answer : Paramagnetic substance because susceptibility of magnetic material is positive.

Question.4. A plane electromagnetic wave travels in vacuum along x-direction. What can you say about the direction of elect tic and magnetic field vectors?
Answer: The electric and magnetic field vectors are in YZ-plane.

Question.10. A magnet is quickly moved in the direction indicated by am arrow between coils C₁and C₂ as shown in figure. What will be the direction of induced current in each coil as seen, from the magnet? Justify your answer.

Answer: For coil C₁: As the N-pole of the magnet is moving away from the coil C₁, the end of the coil will behave as S-pole so as to oppose the motion of the magnet. Therefore, looking from the end, the current in the coil C₁ will be in clockwise direction.
For coil C₂ : The end of the coil should behave as S-pole so as to repel the approaching magnet. Looking from the end, the direction of current in the coil of C₂ will be clockwise direction.

Question.11. Figure shows two identical capacitors C₁ and C₂ each of 2μF capacitance, connected to a battery of 5V. Initially switch ‘S’
is dosed. After sometime ‘S’ is left open and dielectric slabs of dielectric constant k = 5 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?

Question.16. How is forward biasing different from reverse biasing in a p-n junction diode?
Answer: Forward Biasing : The positive terminal of the external battery is connected to p-side and negative terminal of battery to n-side of p-n junction. The forward bias voltage oppose the potential barrier. Due to this, the potential barrier is reduced and hence the depletion becomes thin.
Reverse biasing : The negative terminal of the external battery is connected to p-side and positive terminal of battery to n-side of p-n junction. The reverse bias voltage supports the potential barrier. Due to this, the potential barrier is increased. The resistance of p-n junction becomes high.

Question.20. In the circuit shown, R₁ = 4 Ω, R₂ = R₃= 5 Ω, R₄ = 10 Ω, and E = 6V Work out the equivalent resistance of the circuit and the current in each resistor.

Question.23. An electron and a photon each have a wavelength of 1.50 nm. Find.
(i) their momenta
(ii) the energy of the photon
(iii) the kinetic energy of the electron
Answer:

SET III

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

Question.2.A plane electromagnetic wave travels in vacuum along y-direction. What can you say about the direction of electric and, magnetic field vectors?
Answer : For an electromagnetic wave propagating along T-direction, the electric and magnetic field vectors vary sinusoidally along X-direction and Z-direction respectively.

Question.7.The susceptibility of a magnetic material is -4.2 x 10 ^-6 Name the type of magnetic materials it represents.
Answer : Diamagnetic substance.

Question.9.Explain how a depletion region is formed in a junction diode.
Answer : Depletion region is formed in a junction diode ‘ because depletion region is created around the p-n junction w ho is devoid of free charge carriers and has immobile ions. It is created due to diffusion of majority carriers across the junction when p-n junction is formed.

Question.14. Figu re shows two identical capacitors C₁and C₂ each of 1.5 μF capacitance, connected to a battery of 2V. Initially switch
S’ is closed. After sometime ‘S’ is left open and dielectric slabs of dielectric constant K=2 are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge ahd (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?

Question.16. Write the truth table for the logic circuit shown below and identify the logic operation by this circuit.

Question.17. Predict the polarity of the capacitor when the two magnets are quickly moved in the directions marked by arrows.

Answer: In this situation shown, A will become positive with respect to B, as current induced is in clockwise direction.

Question.23. In the circuit shown, R₁ = 2 Ω, R₂ = R₃ = 10 Ω, R₄ = 20 Ω, and E = 6V. Work out the equivalent resistance of the circuit and the current in each resistor.

Answer.

Question.25. An electron and a photon each have a wavelength of 2 nm. Find
(i) their momenta
(ii)the energy of photon
(iii)the kinetic energy of the electron
Answer:

CBSE Previous Year Solved  Papers  Class 12 Chemistry Delhi 2012

Time allowed: 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1.  All questions are compulsory. 
2.  Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
3.  Questions number 6 to 10 are short-answer questions and carry 2 marks each.
4.  Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
5.  Questions number 23 is a value based question and carry 4 marks.
6.  Questions number 24 to 26 are long-answer questions and carry 5 marks each. 
7.  Use log tables, if necessary. Use of calculators is not allowed.

SET I

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

Question.1. What is meant by ‘doping’ in a semiconductor ?
Answer : Addition of impurities (like phosphorous and arsenic) to a semiconductor (like silicon) to improve the
conductivity is called doping.

Question.2. What is the role of graphite in the electrometallurgy of aluminium?
Answer : Graphite is used as anode and useful for the reduction of Al₂O₃ into Al.

Question.3. Which one of PCl⁺₄and PCl^–₄ is not likely to exist and why?
Answer: The oxidation state of P in PCl⁺¹₄ is +5 while that PCl^-1₄ is +3. As we move down the group, the stability of +5 oxidation decreases and +3 oxidation state increases. Chlorine belongs to third period. Hence, PCl⁺¹₄ compound is more likely to exist.

Question.4. Give the IUPAC name of the following compound.

Question.5. Draw the structural formula of 2-methylpropan-2ol molecule.

Question.6.Arrange the following compounds in ait increasing order of their reactivity in nucleophilic addition reactions : ethanol, propanal, propanone, butanone.
Answer : Butanone < Propanone < Proponal < Ethanol

Question.7. Arrange the following in the decreasing order of their basic strength in aqueous solutions:

Question.8. Define the term, ‘homopolymerisation’ giving an example
Answer : Polymerisation involving the presence of one monomer is called homopolymerisation, e.g. polyethene is a homopolymer

Question.9. A 1.00 molal aqueous solution of trichloroacetic acid (CCl₃  COOH) is heated to its boiling point. The solution has the boiling point of 100.18 °C. Determine the Van’t Hoff factor for trichloroacetic acid. (K_b for water = 0.512 K kg mol^-1)
OR
Define the following terms :

1.  Mole fraction
2. Isotonic solutions
3. Van’t Hoff factor
4. Ideal solution

1.  Ratio of the number of moles of a component in a mixture to the total number of moles in the mixture is called the mole fraction of that component. It is denoted by ‘x
2. Two solutions having the same molar concentration are said to be isotonic solutions, eg : All intravenous injections must be isotonic with body’fluids.
3. The ratio of observed colligative property to the calculated colligative property is called the Van’t Hoff factor. It is denoted by Y.
4. Solutions that follow Raoult s law at all temperatures and concentrations are called ideal solutions.

Question.10. What do you understand by the ‘order of a reaction’ ?
Identify the reaction order from each of the following units of reaction rate constant:

1. L^-1 mol s^-1
2. L mol^-1 s^-1

Answer : The sum of the powers to which the concentration of reactants are raised in the rate law expression is called the order of a reaction.

1. Zero order reaction
2. Second order reaction.

Question.11. Name the two groups into which phenomenon of catalysis can be divided. Give an example of each group with the chemical equation involved.
Answer: Catalysis can be positive, that is, it increases the rate of the reaction or negative i.e. decreases the rate of reaction. Depending on the phase of the reactants and the catalyst, catalysis can be :
(i) Homogenous catalysis : The reactants and catalyst are in the same phase.

Question.12. What is meant by coagulation of a colloidal solution?
Describe briefly any three methods by which coagulation of lyophobic sols can be carried out.
Answer : The process of setting of colloidal particles is called coagulation of sol. Methods of coagulation are :

1.  Electrophoresis : In this process, the colloidal particles move towards opposite charged electrodes and get discharged and precipitated.
2. Mixing two oppositely charged sols : Equal proportions of oppositely charged sols are mixed, they get neutralized and get precipitated.
3. Dialysis : Electrolytes are- removed from the sol and colloid becomes unstable and gets coagulated.

Question.13. Describe the principle involved in each of the following processes.

1.  Mond process for refining of Nickel.
2. Column chromatography for purification of rare elements.

Answer :

1.  Nickel combined with carbon monoxide to form volatile complex within further be decomposed to get back pure nickel.
2. The basic principle involved in column chromatography is that different elements present in a mixture are adsorbed on adsorbent at different extents.

Question.14. Explain the following giving an appropriate reason in each case.

1.  O₂ and F₂both stabilize higher oxidation states of metals but O₂ exceeds F₂ in doing so.
2. Structures of xenon fluorides cannot be explained by Valence Bond approach.

Answer :

1. Due to the difference in atomic size of oxygen and fluorine and the property of oxygen to form multiple bonds with metals,O₂ exceeds F₂, stabilize higher oxidation states,
2. For explaining the structures of xenon fluorides, we need to use VSEPR and hybridization theories because in VBT, covalent bonds are formed by overlapping of half filled atomic orbital. But xenon has fully filled electronic configuration.

Question.15. Complete the following chemical equations :

Question.16. What is meant by

1.  peptide linkage
2. biocatalysts?

Answer:

1. Peptides linkage is the amide bond that helps to connect amino acids to form proteins. It is formed between -COOH and -NH₂ group of two amino acids with the loss of water molecule.
2. Biocatalysts are enzymes that catalyses the biochemical reactions in the bodies of living organisms, e.g. Amylase.

Question.17. Write any two reactions of glucose which cannot be explained by the open chain structure of glucose molecule.
Answer : Two reactions which can’t be explained by open chain structure of glucose are :

1.  Despite having the aldehyde group, glucose does not give 2,4 —DNP test.
2. The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free -CHO group.

Question.18. Draw the structure of the monomer for each of the following polymers:
(i) Nylon 6 (ii) Polypropene.
Answer:

Question.19. Tungsten crystallizes in body centered cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom!
OR
Iron has a body centered cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm^-3. Use this information to calculate Avogadro’s number. (At mass of Fe = 55.845 u)

Question.20. Calculate the amount of KCl which must be added to 1kg of water so that the freezing point is depressed by 2K.
(K_f for water = 1.86 K kg mol^-1)

Question.21.For the reaction
2NO(g)+Cl₂(g)→2NdCl(g)
the following data were collected. All the measurements were taken at 263 K:


(a) Write the expression for rate law.
(b) Calculate the value of rate constant and specify its units.
(c) What is the inidal rate of disappearance of Cl₂ in exp. 4?

Question.22. How would you account for the following?

1.  Many of the transition- elements are known to form interstitial compounds.
2. The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the second (4d) series.
3. Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical.

Answer :

1.  Transition metal lattice have voids and hence these voids can trap small atoms like H, C and N to form interstitial compounds.
2. It is because electrons first fill the 4forbitals and than the 5d orbitals causing decline in radii of third (5d) series.
3. This is because of the comparable energies of 5f 6d and Is orbitals so all of them can participate

Question.23. Give the formula of each of the following coordination
entities:
(i) CO³⁺ ion is bound to one Cl^–one NH₃ molecule and two bidentate ethylene diamine (en) molecules.
(ii)Ni²⁺ ion is bound to two water molecules and two oxalate ions.
Write the name and magnetic behavior of each of the above coordination entities.

Question.24. Although chlorine is an electron ‘withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Explain why it is so?
Answer : Due to resonance, electron density is increased on ortho and para positions for electron donating (via resonance) groups like chlorine.

Question.25. Draw the structure and name the product formed if the following alcohols are oxidized. Assume that an excess of oxidizing agent is used.
(i) CH₃CH₂CH₂CH₂OH (ii) 2-butenol
(iii)2-methyl-l-propanol
Answer: (i)

Question.26. Write chemical equations for the following conversions :
(i) Nitrobenzene to benzoic acid.
(ii)Benzyl chloride to 2-phenylethanamine.
(iii)Aniline to benzyl alcohol.
Answer: (i)

Question.27. What are the following substances? Give one example of each one of them.

1. Tranquilizers
2. Food preservatives
3. Synthetic detergents.

Answer :

1.  Tranquilizers are drugs that act on the central nervous system to get relief from anxiety, stress etc. They are used in treatment of stress related mental disorders.
2. These are chemicals used to preserve food by protecting it against microbial growth, eg. Sodium benzoate.
3. Synthetic detergents have all the properties of soap but they do not- precipitate in hard water, eg. Sodium p-dodecyl benzenesulphonate.

Question.28. (a) What type of a battery is the lead storage battery? Write the anode and the cathode reactions and the overall reactions occurring in a lead storage battery when current is drawn from it.
(b) In the button cell, widely used in watches, the following reaction takes place.

(a) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte.
(b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell
constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 x 10^-3S cm^-1?

Question.29. (a) Complete the following chemical reactions. equations:

Question.30. (a) Illustrate the following name reactions giving suitable example in each case :
(i) Clemmensen reduction
(ii)Hell-Volhard-Zelinsky reaction
(b) How are the following conversions carried out?
(i) Ethyicyanide to ethanoic acid
(ii)Butan-l-ol to butanoic acid
(iii)Benzoic acid to m-bromobenzoic acid
OR
(a) Illustrate the following reactions giving a suitable example for each:
(i) Cross aldol condensation
(ii) Decarboxylation
(b) Give simple tests to distinguish between the following pairs of compounds :
(i) Pentan-2-one and Pentan-3-one
(ii)Benzaldehyde and Acetophenone
(iii)Phenol and Benzoic acid
Answer : (a)
(i) Aldehydes and ketones are reduced to CH₂group on treatment with zinc-amalgam and cone. HCl.


(ii) Carboxylic acids containing a-hydrogen atom gives halo carboxylic acids on treatment with halogens in the presence of red phosphorous.


(ii) Carboxylic acid lose carbon dioxide to form hydrocarbon, when their salts are heated with sodium. The reaction is known as decarboxylation reaction

SET II

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

Question.1.Write a point of distinction between a metallic solid and an
ionic solid other than metallic luster.
Answer : Metallic solids do not ionize in aqueous state but ionic solids ionize immediately

Question.11.Describe a conspicuous change observed when :

1.  a solution of NaCl is added to a sol of hydrated ferric oxide.
2. a beam of light is passed through a solution of NaCl and then through a sol.

Answer :

1.  Coagulation of ferric hydroxide sol. would take place.
2. NaCl solution is transparent so when beam of light is passed, no tyndall effect is produced. But on passing through soln the path of light becomes visible due to Tyndall effect.

Question.13. Describe the following:

1.  The role of cryolite in electro-metallurgy of aluminium.
2. The role of carbon monoxide in the refining of crude nickel.

Answer :

1.  Cryolite lowers the melting point of the mixture and brings conductivity. Therefore, it is mixed with alumina during metallurgy of aluminium.
2.  Carbon monoxide forms a volatile complex with nickel.
   The volatile complex is then subjected to high temperature to get pure metal through decomposition.
   

Question.14. What is meant by

1.  peptide linkage
2.  biocatalysts?

Answer:

1.  Peptide linkage: It is the linkage between amino acids, formed due to loss of a water molecule and
   amide
   
2. Biocatalysts : Biocatalysts are enzymes which are used to perform chemical reactions on organic compounds.

Question.18. Write the main structural difference between DNA and RNA. Of the two bases, thymine and uracil, which one is present in DNA?
Answer:

Question.23. How would you account for the following?

1.  With the same d-orbital configuration (d⁴) Cr²⁺ is a reducing agent while Mn³⁺ is an oxidizing agent.
2. The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series.
3. Most of the transition metal ions exhibit characteristic in colours in aqueous solutions.

Answer:

1. Cr²⁺ has d⁴ configuration. It gets oxidized , to Cr³⁺ with electronic configuration d³ which is more stable. Therefore, Cr²⁺ is a reducing agent. Mn3+ has d⁴ configuration. It gets reduced to Mn²⁺with d⁵ configuration. This is half-filled d-orbital and is stable. Therefore Mn³⁺is an oxidizing agent.
2. Because 5f, 6d and 7s energy levels has small enery gap in the actinoid series. Due to these orbitals actinoids exhibit large number of oxidation states.
3. Due to partial absorption of visible light the electron from one orbital gets promoted to another orbital of the d subshell. Due to presence of unpaired electrons transition metals are coloured.

Question.30. (a) Give a possible explanation for each one of the following:
(i) There are two -NH₂ groups in semicarbazide. However, only one such group is involved in the formation of semicarbazones.
(ii) Cyclohexanone forms cyanohydrin in good yield but 2, 4, 6-trimethylcydohexanone does not.
(b) An organic compound with molecular formula C₉H₁₀O forms 2, 4, -DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1,2-benzene-di-carboxylic acid. Identify the compound.
OR
(a) Give chemical tests to distinguish between
(i) Phenol and Benzoic acid
(ii)Acetophenone and Benzpphenone
(b) Write the structures of the main products of following
reactions:

Answer : (a)
(i) Semicarbazide show resonance involving one of the two -NH₂ groups, which is attached to the carboxyl carbon atom. Due to which electron density on —NH₂ group involved in resonance decreases. So it cannot act as nucleophile. Other -NH₂ group can act as nucleophile to produce semicarbazones with aldehydes and ketones.
(ii) In cyclohexanone CN^– can easily attack without any steric hindrance. But in 2, 4, 6 — Trimethylcyclohexanone due to presence of methyl groups steric hindrance is produced and CN^– cannot attack effectively.
(iii) C₉H₁₀Ois aldehyde because it reduces Tollens reagent. It undergoes Cannizaro’s reaction therefore it is substituted benzaldehyde. It gives 1, 2, Benezene-di carboxylic acid. The compound is 2-Ethylbenzaldehyde.

SET III

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

Question.3. Of PH₃ and H₂ S which is more acidic and why?
Answer : H₂S is more acidic than PH₃ due to smaller size and higher electronegativity of sulphur. Therefore S-H bond is polar than P-H bond and easy to remove.

Question.5. Draw the structure of hex-l-en-3-ol compound.
Answer :

Question.12. Explain the following terms-giving one example for each :

1. Miscelles
2. Aerosol.

Answer :

1.  Micelles are aggregates which exhibit colloidal behavior at higher concentration, with the hydrophilic part outside and the hydrophobic part towards the oil and dirt particle, eg. soap.
2. An aerosol is a colloid of fine solid particle or liquid drops in air or another gas. It can be natural or artificial. Example dust particle and smoke.

Question.20. 15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at -0.34 °C. What is the molar mass of this material? (K_f for water = 1.86 K kg mol^-1)

Question.22. Explain the following observations giving an appropriate reason for each :

1.  The enthalpies of atomization of transition elements are quite high.
2. There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e. 3rd series).
3. Mn²⁺is much more resistant than Fe²⁺ towards oxidation.

Answer:

1.  Due to the presence of metallic bonds as a result of large number of valence electrons.
2. The presence of valence electrons and unpaired d-orbital electrons help heavy transition metals to form metallic bonds.
3. Due to stability of Mn²⁺ because of half filled-subshell (3d⁵) it does not gets oxidized. But Fe²⁺ has 3d⁶configuration and it can lose one electron to become 3d⁵ which is stable. Therefore it is easily oxidized.

Question.23. Write the name, the structure and the magnetic behavior of each of the following complexes:

Answer :
(i) Amminechloridonitrito—N-platinum (II),
trigonal planar, diamagnetic.
(ii)Tetra-ammine dichlorido cobalt (III) chloride, octahedral, diamagnetic
(iii)Tetracarbonyl nickel (0), Tetrahedral, diamagnetic

Question.27. Explain the following terms giving one example of each , type:

1. Antacids, .
2. Disinfectants,
3. Enzymes.

Answer :

1. Substances consumed to reduce acidity in the stomach by neutralizing excess HCl produced by the stomach, eg. Milk of magnesia.
2. Disinfectants are chemicals used to kill microorganism, applied only to non-living objects like floors and drains, e.g. 1% phenol solution.
3. Enzymes are structurally globular proteins that catalyse biochemical reactions in living organisms eg. Trypsin.

Question.30. Draw the molecular structures of following compounds:
(i) XeF₆ (ii) H₂S₂O₈
(b) Explain the following observations :
(i) The molecules NH₃ and NF₃ have dipole moments which are of opposite direction.
(ii)All the bonds of PCl₅ molecule are not equivalent.
(iii)Sulphur in vapour state exhibits paramagnetism.


(b) (i) Fluorine is more electronegative than nitrogen while hydrogen is less electronegative than nitrogen resulting in opposite c(ipole moments of NH₃ and NF₃. Dipole points towards N in NH₃ and towards F in NF₃.
(ii) PCl₅ has trigonal bipyramidal shape. The two axial bonds feel more repulsion from the three equatorial bond pairs and are hence longer. Therefore all bonds in PCl₅ are not equivalent.
(iii)Sulphur in vapour state exists asS₂ and hence likeO₂ has two unpaired electrons in antibonding n orbitals leading to paramagnetism.

(b)(i) Due to the small size, absence of d-orbitals and presence of triple bond in nitrogen, it is less reactive than phosphorous.
(ii)Due to the increase in size and inert pair effect, +5 oxidation state decreases down the group.
(iii)Bond angles in NO^–₂ and NO⁺₂ are not of same value because NO⁺₂ is linear and NO^–₂ has bent shape.