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CBSE Sample Papers for Class 12 Physics Paper 5 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 5.

CBSE Sample Papers for Class 12 Physics Paper 5

Time Allowed : 3 Hours

Max. Marks : 70

General Instructions 

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

Questions
SECTION : A

Question 1.
The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of earth’s magnetic field at equator?

Question 2.
A heating element is marked 210 V, 630 W. What is the value of current drawn by the element when connected to a 210 V dc source?

Question 3.
Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily.

Question 4.
Show on a graph, the variation of resistivity with temperature for a typical semiconductor.

Question 5.
Why should electrostatic field be zero inside a conductor?

SECTION : B

Question 6.
Draw a plot showing the variation of
(i) electric field (E) and
(ii) distance r due to a point charge Q.

Question 7.
Distinguish between ‘Analog and Digital signals’

OR

Mention the functions of any two of the following used in communication system :
(a) Transducer
(b) Repeater
(c) Transmitter
(d) Bandpass Filter

Question 8.
A ray of light incident on an equilateral prism (μ_g = \(\sqrt { 3 } \)) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.

Question 9.
The susceptibility of a magnetic material is -2.6 x 10^-5. Identify the type of magnetic material and state its two properties.

Question 10.
Two identical circular wires P and Q each of radius R and carrying current ‘I’  are kept in perpendicular planes such that they have a common centre. Find the magnitude and direction of the net magnetic field at the common centre of the two coils.

SECTiON : C

Question 11.
A cell of emf E and internal resistance r is connected to two external resistance R₁ and R₂ and a perfect ammeter. The current in the circuit is measured in four different situations :
(i) without any external resistance in the circuit
(ii) with resistance R₁only
(iii) with R₁ and R₂ in series combination
(iv) with R₁ and R₂ in parallel combination.
The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in the order.Identify the currents corresponding to the four cases mentioned above.

Question 12.
Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self inductance L is given by 1/2 LI²

Question 13.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
(b) In Young’s double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3.

Question 14.
When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source is used the current flows continuously. How does one explain this based on the concept of displacement current?

Question 15.

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

Question 16.
In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs ε₁ and ε₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A.
Find
(i) ε₁/ε₂ and
(ii) position of null point for the cell ε₁

How is the sensitivity of a potentiometer increased?

OR

Using Kirchhoffs rules determine the value of unknown resistance R in the circuit so that no current flows through 4 Q resistance. Also find the potential between A and D.

Question 17.
Write any two factors which justify the need for modulating a signal. Draw a diagram showing an amplitude modulated wave by superimposing a modulating signal over a sinusoidal carrier wave.

Question 18.
The current in the forward bias is known to be more (~ mA) than the current in the reverse bias (~ μA). What is the reason, then, to operate the photodiode in reverse bias?

Question 19.
A metallic rod of length ‘L’ is rotated with angular frequency of ‘ω’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the induced em/between the centre and the metallic ring.

Question 20.
Write Einstein’s photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation. Write the three salient features observed in photoelectric effect which can be explained using this equation.

Question 21.

The figure shows a series LCR circuit with L = 5 H, C = 80 μF, R = 40 Ω connected to a variable frequency 240 V source, calculate
(i) the angular frequency of the source which drives the circuit at resonance,
(ii) the current at the resonating frequency,
(iii) the rmspotential drop across the inductor at resonance.

Question 22.
(i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?
(ii) Show that the density of nucleus over a wide range of nuclei is constant independent of mass number A.

SECTION : D

Questoin 23.
Renu and her friend went to see an exhibition. There, security guard standing on the entrance gate, asked them to come through a metal detector gate. Her friend was scared of it. But Renu convinced her and explained the purpose and working of a metal detector.
(i) What values Renu possess?
(ii) What is a metal detector and how it works?

SECTION : E

Question 24.
A parallel plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following will change;
(i) electric field between the plates
(ii) capacitance, and
(iii) energy stored in the capacitor

OR

(a) Define electric flux. Write its S.I. units.
(b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(c) How is the field directed if
(i) the sheet is positively charged,
(ii) negatively charged?

Question 25.
Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens of focal length 150 cm and an eye-piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye-piece.

OR

How is the working of a telescope different from that of a microscope? The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.

Question 26.
Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain A_v of the amplifier is given by

is the current again ; R_L is the load resistance and r. is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain?

OR

(a) Draw the circuit diagram of a full wave rectifier using p-n junction diode. Explain its working and show the output, input waveforms.
(b) Show the output waveforms (Y) for the following inputs A and B of
(i) OR gate
(ii) NAND gate.

 Answers :
SECTION : A

Answer 1.
On the equator, the value of both angle of dip (5) and vertical component of earth’s magnetic field is zero. So, in this case,
B_v = 0.

Answer 2.
In dc source, P = VI
Given that P = 630 W and V = 210 V
So, I = P/V = 630/210 = 3 A.

Answer 3.
Using Lenz’s law we can predict the direction of induced current in both the rings. Induced current oppose the cause of increase of magnetic flux. So, it will be clockwise in ring 1 and anticlockwise in ring 2.

Answer 4.
The following curve shows the variation of resistivity with temperature for a typical semiconductor.

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

Answer 5.
Charge on conductor resides on its surface. So if we consider a Gaussian surface inside the conductor to find the electrostatic field,

Where, q = charge enclosed in Gaussian surface. q = 0, inside the conductor, hence the electrostatic field inside the conductor is zero.

SECTION : B

These Sample papers are part of CBSE Sample Papers for Class 12 Biology. Here we have given CBSE Sample Papers for Class 12 Biology Paper 4.

CBSE Sample Papers for Class 12 Biology Paper 4

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 4 of Solved CBSE Sample Paper for Class 12 Biology is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. There are total 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 Questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer Type Questions of three marks each.
5. Section D contains question number 23, Value Based Question of four mark.
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 examiner is to attempt any one of the question out of the two given in the question paper with the same question number.
8. No. of printed pages are three.

SECTION-A

Question 1.
What are cleistogamous flowers?

Question 2.
What is coleorrhiza?

Question 3.
What are alleles?

Question 4.
What are introns?

Question 5.
What is apiculture?

SECTION-B

Question 6.
Explain MOET.

Question 7.
Write a note on discovery of penicillin.

Question 8.
What does the diagram below signify?

Question 9.
What are the basic steps involved in genetically modifying an organism?

OR

What is humus?

Question 10.
Represent the age pyramids for human population.

SECTION-C

Question 11.
Explain the parts of an ovule with a diagram

Question 12.
How is RNA synthesized in bacteria? Illustrate

Question 13.
Discuss the barrier methods for contraception

Question 14.
Explain convergent evolution with examples.

Question 15.
What are the major causes of cancer?

Question 16.
Is it possible to obtain large quantities of DNA from a single cell?

Question 17.
What are the advantages of GM plants?

Question 18.
Explain T_i plasmid of Agrobacterium tumefaciens.

Question 19.
Give the equations of both exponential and logistic growth curves. Also represent them graphically.

OR

Present a case study for remedy of plastic waste

Question 20.
Complete the diagram below:

Question 21.
Haploid content of human DNA is 3.3 x 10⁹ bp and the distance between two consecutive
bp is 0.34 x 10 ⁹. What is the length of the DNA molecule?

Question 22.
Label the diagram given below:

SECTION-D

Question 23.
Rita and her parents were watching a TV serial in the evening. During a commercial break
an advertisement flashed on the screen which was promoting use of sanitary napkins. Rita
was still watching the TV when the parents got embarrassed and changed the channel. Rita
objected to her parent’s behaviour and explained the need for these advertisements.
(a) What values did the parents show?
(b) Briefly describe the phases of a menstrual cycle.

SECTION-E

Question 24.
What are the post pollination events?

OR

Explain endosperm development.

Question 25.
What are chromosomal disorders?

OR

List the observations of Human Genome Project.

Question 26.
Explain some interspecific relationship where no species is harmed.

OR

Explain ecological succession.

Answers

SECTION-A

Answer 1.
Self pollinating flowers in which stamens and pistil are in close proximity.

Answer 2.
In embryos of monocots the root cap and radicle are enclosed in an undifferentiated sheath j called coleorhiza.

Answer 3.
Genes which code for a pair of contrasting traits.

Answer 4.
Intervening sequences in DNA which are not expressed in mature or processed RNA.

Answer 5.
Apiculture is the maintenance of hives of honeybees for the production of honey.

SECTION-B

Answer 6.
Multiple Ovulation Embryo Transfer Technology (MOET) is a programme for herd improvement. In this method, a cow is administered hormones, with FSH-like activity, to induce follicular maturation and super ovulation – instead of one egg, which they normally yield per cycle, they produce 6-8 eggs. The animal is either mated with an elite bull or artificially inseminated. The fertilised eggs at 8-32 cells stages, are recovered non-surgically and transferred to surrogate mothers. The genetic mother is available for another round of super ovulation.

Answer 7.
Alexander Fleming while working on Staphylococci bacteria, once observed a mould growing in one of his unwashed culture plates around which Staphylococci could not grow. He found out that it was due to a chemical produced by the mould and he named it Penicillin after the mould Penicillium notatum.

Answer 8.
The first diagram is that of the normal cells which show controlled growth due to property of contact inhibition.
The second diagram shows loss of contact inhibition by cancer cells. This uncontrolled cell growth leads to tumor

Answer 9.
Three basic steps in genetically modifying an organism are:

• Identification of DNA with desirable genes;
• Introduction of the identified DNA into the host;
• Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.

OR

Humus is a dark coloured amorphous substance which is highly resistant to microbial action and undergoes decomposition at an extremely slow rate. Being colloidal in nature it serves as a reservoir of nutrients. The process of humus formation is called humification.

Answer 10.

SECTION-C

Answer 11.

The ovule is a small structure attached to the placenta by means of a stalk called funicle.The body of the ovule fuses with funicle in the region called hilum. Each ovule has one or two protective envelopes called integuments. Integuments encircle the ovule except at the tip where a small opening called the micropyle is organised. Opposite the micropylar end, is the chalaza, representing the basal part of the ovule. Enclosed within the integuments is a mass of cells called the nucellus. Cells of the nucellus have abundant reserve food materials. Located in the embryo sac or female gametophyte.

Answer 12.

In a bacteria, there are three major types of RNAs: mRNA (messenger RNA), tRNA (transfer RNA), and rRNA (ribosomal RNA). The mRNA provides the template, tRNA brings aminoacids and reads the genetic code, and rRNAs play structural and catalytic role during translation. There is single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria. RNA polymerase binds to promoter and initiates transcription (Initiation). It uses nucleoside triphosphates as substrate and polymerises in a template depended fashion following the rule of complementarity. It also facilitates opening of the helix and continues elongation. Only a short stretch of RNA remains bound to the enzyme. Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination of transcription.

Answer 13.
In barrier methods, ovum and sperms are prevented from physically meeting. Such methods are available for both males and females.
(1) Condoms are made of thin rubber/latex sheath that are used to cover the penis in the male or vagina and cervix in the female, just before coitus so that the ejaculated semen would not enter into the female reproductive tract. This can prevent conception.

(2) Diaphragms, cervical caps and vaults are also barriers made of rubber that are inserted into the female reproductive tract to cover the cervix during coitus. They prevent conception by blocking the entry of sperms through the cervix. They are reusable.

(3) Spermicidal creams, jellies and foams are usually used along with these barriers to increase their contraceptive efficiency.

(4) Intra Uterine Devices (IUDs): These devices are inserted by doctors or expert nurses in the uterus through vagina. These Intra Uterine Devices are presently available as the non- medicated IUDs (e.g., Lippes loop), copper releasing IUDs (CuT, Cu7, Multiload 375) and the hormone releasing IUDs (Progestasert, LNG-20). IUDs increase phagocytosis of sperms within the uterus and the Cu ions released suppress sperm motility and the fertilising capacity of sperms. The hormone releasing IUDs, in addition, make the uterus unsuitable for implantation and the cervix hostile to the sperms.

Answer 14.
Convergent evolution is a evolution of different structures for the same function and hence having similarity. It is the similar habitat that has resulted in selection of similar adaptive features in different groups of organisms but toward the same function.
For example the eye of the octopus and of mammals, the flippers of Penguins and Dolphins, sweet potato (root modification) and potato (stem modification) similarities in proteins and genes performing a given function among diverse organisms give clues to common ancestory.

Answer 15.
The major causes of cancer are:

1. Transformation of normal cells into cancerous neoplastic cells which may be induced by physical, chemical or biological agents called carcinogens.
2. Ionising radiations like X-rays and gamma rays and non-ionizing radiations like UV cause DNA damage leading to neoplastic transformation.
3. The chemical carcinogens present in tobacco smoke have been identified as a major cause of lung cancer.
4. Cancer causing viruses called oncogenic viruses have genes called viral oncogenes. Cellular oncogenes (cone) or proto oncogenes have been identified in normal cells which,when activated under certain conditions, could lead to oncogenic transformation of the cells.
   

Answer 16.
Yes, it is possible to make large quantities of DNA from a single cell by

PCR stands for Polymerase Chain Reaction. In this reaction, multiple copies of the gene (or DNA) of interest is synthesised in vitro using two sets of primers (small chemically synthesised oligonucleotides that are complementary to the regions of DNA) and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the genomic DNA as template. If the process of replication of DNA is repeated many times, the segment of DNA can be amplified to approximately billion times, i.e., 1 billion copies are made. Such repeated amplification is achieved by the use of thermostable DNA polymerase (isolated form a bacterium, Thermus aquaticus), which remain active during the high temperature induced denaturation of double stranded DNA. The amplified fragment if desired can now be used to ligate with a vector for further cloning.

Answer 17.
Advantages of GM plants are:

• GM crops are more tolerant to abiotic stresses (cold, drought, salt, heat).
• These have reduced reliance on chemical pesticides (pest-resistant crops).
• These crops helped to reduce post harvest losses.
• GM crops have increased efficiency of mineral usage which prevents early exhaustion of fertility of soil.
• These crops have enhanced nutritional value of food, e.g., Vitamin ‘A’ enriched rice.
• These crops have also helped in creation of tailor-made plants to supply alternative resources to industries, in the form of starches, fuels and pharmaceuticals.

Answer 18.
Agrobacterium tumifaciens is a pathogen of several dicot plants which has an ability to deliver ‘T-DNA’ that further transforms normal plant cells into a tumor and direct these tumor cells to produce the chemicals required by the pathogen. The tumor inducing (Ti) plasmid of Agrobacterium tumifaciens has now been modified into a cloning vector which is no more pathogenic to the plants but is still able to use the mechanisms to deliver genes of our interest into a variety of plants.

Answer 19.

Answer 20.

Answer 21.
Haploid content = 3.3 x 10⁹.
Therefore, diploid content = 6.6 x 10⁹
Distance between bp = 0.34 x 10 ⁹
Therefore length = diploid content x distance between bp
⇒ 6.6 x 10⁹ x 0.34 x 10~⁹ = 2.24 m

Answer 22.

SECTION-D

Answer 23.
(a) The parents were traditional but understood the need for such advertisements.
(b)

SECTION-E

Answer 24.

Following compatible pollination, the pollen grain germinates on the stigma to produce a pollen tube through one of the germ pores. The contents of the pollen grain move into the pollen tube. Pollen tube grows through the tissues of the stigma and style and reaches the ovary. In plants, where the pollen grains are shed at the two celled stage, the generative cell divides and forms the two male gametes during the growth of pollen tube in the stigma. In plants which shed pollen in the three-celled condition, pollen tubes carry the two male gametes from the beginning. Pollen tube after reaching the ovary, enters the ovule through the micropyle and then enters one of the synergids through the filiform apparatus. Filiform apparatus present at the micropylar part of the synergids guides the entry of pollen tube. All these events are together referred to as pollen-pistil interaction.

OR

Endosperm development precedes embryo development. The primary endosperm cell divides repeatedly and forms a triploid endosperm tissue. The cells of this tissue are filled with reverse food materials and are used for the nutrition of the developing embryo. In the most common type of endosperm development, the PEN undergoes successive nuclear divisions to give rise to free nuclei. This stage of endosperm development is called free-nuclear endosperm. Subsequently cell wall formation occurs and the endosperm becomes cellular. The number of free nuclei formed before cellularisation varies greatly

Answer 25.
The chromosomal disorders are caused due to absence or excess or abnormal arrangement of one or more chromosomes. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome), called aneuploidy.

Failure of cytokinesis after telophase stage of cell division results in an increase in a whole set of chromosomes in an organism and is called polyploidy.

Sometimes, either an additional copy of a chromosome may be included in an individual or an individual may lack one of any one pair of chromosomes: These situations are known as trisomy or monosomy of a chromosome, respectively.

Common examples of chromosomal disorders are Down’s syndrome, Turner’s syndrome, Klinefelter’s syndrome.

Down’s Syndrome: Caused by trisomy of 21 chromosome number.

Klinefelter’s Syndrome: Caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 447, XXY.

Turner’s Syndrome: Caused due to the absence of one of the X chromosome, i.e. 45 with XO.

OR

The salient observations drawn from human genome project are:

1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.
3. The total number of genes is estimated at 30,000-much lower than previous estimates of 80,000 to 1,40,000 genes. Almost all (99.9 per cent) nucleotide bases are exactly the same in all people.
4. The functions are unknown for over 50 per cent of the discovered genes.
5. Less than 2 per cent of the genome codes for proteins.
6. Repeated sequences make up very large portion of the human genome.
7. Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.
8. Chromosome 1 has most genes (2968), and the Y has the fewest (231).
9. Scientists have identified about 1.4 million location where single base DNA differences (SNPs- single nucleotide polymorphism, pronounced as ‘snips’) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease-associated sequences and tracing human history.

Answer 26.
Some interspecific relationship where no species is harmed are:
(1) Commonalism: This is the interaction in which one species benefits and the other is neither harmed nor benefited. For example: an orchid growing as an epiphyte on a mango branch, barnacles growing on the back of a whale, the cattle egret and grazing cattle, and Sea anemones stinging tentacles protect the clown fish from predators that lives among them.

(2) Mutualism: This interaction confers benefits on both interacting species. For example, Lichens represent an intimate mutualjstic relationship between a fungus and photosynthesising algae or cyanobacteria.

(3) Plant-Animal Relationships: Plants need the help of animals for pollinating their flowers and dispersing their seeds; in return plants offer rewards or fees in the form of pollen and nectar for pollinators and juicy and nutritious fruits for seed dispersers.

The gradual and fairly predictable change in the species composition of a given area is called Ecological succession. During succession some species colonise an area and their populations become more numerous, whereas populations of other species decline and even disappear. The entire sequence of communities that successively change in a given area are called sere(s). The individual transitional communities are termed serai stages or serai communities. In the successive serai stages there is a change in the diversity of species of organism, increase in the number of species and organisms as well as an increase in the total biomass. Succession is a process that starts where no living organisms are there-these could be areas where no living organisms ever existed, i.e., bare rock; or in areas that somehow, lost all the living organisms that existed there. The former is called primary succession, while the latter is termed secondary succession.

We hope the CBSE Sample Papers for Class 12 Biology Paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 12 Biology Paper 4, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 12 Physical Education Paper 1 are part of CBSE Sample Papers for Class 12 Physical Education. Here we have given CBSE Sample Papers for Class 12 Physical Education Paper 1.

CBSE Sample Papers for Class 12 Physical Education Paper 1

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 1 of Solved CBSE Sample Paper for Class 12 Physical Education is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 70

General Instructions:

• All questions are compulsory.
• Answers to questions carrying 1 mark should approximately 10-20 words.
• Answers to questions carrying 3 marks should approximately 30-50 words.
• Answers to questions carrying 5 marks should approximately 75-100 words.

Questions.

Question 1.
Define Planning? 1

Question 2.
List down the components of Barrow General Motor Ability Test? 1

Question 3.
What do you understand by Balance Diet? 1

Question 4.
Define disability and disorder among children? 1

Question 5.
Define Posture? 1

Question 6.
What do you mean by Motor Development? 1

Question 7.
List down the physiological factors determining Physical Fitness? 1

Question 8.
Define Vital Capacity? 1

Question 9.
What do you understand by Sports Medicine? 1

Question 10.
Define Soft Tissue Injury? 1

Question 11.
List Newton’s Law of motion? 1

Question 12.
Describe the Axis and Plane used in movements? 3

Question 13.
Describe the importance of intramural in physical education? 3

Question 14.
Describe various types of biomechanical movements? 3

Question 15.
Differentiate between Micro and Macro Nutrient? 3

Question 16.
Describe the benefits of yoga for healthy life? 1

Question 17.
Explain the factors effecting motor development? 3

Question 18.
Children of a particular residential school were very indisciplined, fighting, and creating problems around, resulting in poor academic achievement. The new school principal asked the teachers to start inter-section sports program for children along with extra PE classes at weekends. After the year end, students attended for assessment test related to psychomotor skills. The analysis of report highlighted increase in academics performance and various other 1 parameters.
(i) What would those various parameters which developed among the participating students?
(ii) Which type of sports competition was organized by the school?
(iii) What psychomotor test could be planned for testing fitness of school children? 3

Question 19.
List down the test components of Rikli and Jones for elderly people? 3

Question 20.
Appraise coping strategies for stress management among athletes? 5

Question 21.
Analyze the muscles involved in Throwing a cricket ball? 5

Question 22.
Differentiate Physical and Physiological differences between male and female athletes? 5

Question 23.
Categorize various types of sports injuries and explain the preventive measures? 5

Question 24.
Explain corrective measures through physical activity and exercise for improving posture? 5

Question 25.
Draw a fixture of 13 Football teams participating in a Tournament on the basis of knock out? 5

Question 26.
Describe advantages of physical activities for children with special needs? 5

Answers.

Answer 1.
Planning is basic management function involving formulation of detailed plans to achieve optimum balance of need or demands with available resources.

Answer 2.
The Barrow Motor Ability Test consisting of three items designed to test the motor ability of high school boys.
Test 1 – Standing Jump Test – It is used as a measure of explosive leg power.
Test 2 – Zigzag Run Test – Its objective is to measure agility.
Test 3 – Medicine Ball Put Test – Its objective is to measure arm and shoulder girdle strength.

Answer 3.
A diet which consists of all the essential food constituents which are necessary for growth and maintenance of the body. Protein, carbohydrates, fat, minerals, vitamins are essential micro and macro nutrients which form part of balance diet.

Answer 4.
Disability: Any degree of physical disability, malformation or disfigurement that is caused by bodily injury, birth defect or illness. Disorder to is the disturbance which affects normal functions performed by an individual.

Answer 5.
Posture is the alignment of the body part for producing various task. Posture involves of muscles, ligaments and tendons along with joints to work together for performing any action, it also effects the functioning of the organic systems.

Answer 6.
Motor development refers to the development of a child’s bones, muscles and ability to move around and manipulate his or her environment. Motor development can be divided into two sections: gross motor development and fine motor development.

• Gross motor development involves the development of the large muscles in the child’s body. These muscles allow us to sit, stand, walk and run, among other activities.
• Fine motor development involves the small muscles of the body, especially in the hand.

Answer 7.
Physiological factors determining physical fitness are:

• Cardiovascular Endurance
• Lung Capacity
• Muscle Composition
• Muscle Fibre size
• Somatotype
• Muscular Strength
• Muscular Speed

Answer 8.
Vital Capacity (VC) is the maximum amount of air a person can expel from lungs after a maximum inhalation.

Answer 9.
A field of medicine concerned with the functioning of the human body during physical activity and with the prevention and treatment of athletic injuries.

Answer 10.
Soft tissue injuries are the most common injury to tissues that connect, support, or surround other structures and organs of the body. Soft tissue includes muscles, tendons, ligaments, fascia, nerves, fibrous tissues, fat, blood vessels, and synovial membranes.

Answer 11.
There are three Newton’s Law of motion.

1. The Law of Inertia: A body at rest tends to remain at rest. A body in motion tends to continue in motion with consistent speed and in the same direction unless acted upon by an outside force.
2. The Law of Acceleration: The velocity of a body is changed only when acted upon by an additional force.
3. The Law of Counterforce: The production of any force will create another force opposite and equal to the first force.

Answer 12.
Human movements are described in three dimensions based on a series of planes and axis,

(i) A plane is the surface on which movement take place. There are three planes of motion that pass through the human body.

• The sagital plane lies vertically and divides the body into right and left parts.
• The frontal plane also lies vertically and divides the body into anterior and posterior parts.
• The transverse plane lies horizontally and divides the body into superior and inferior parts.

(ii) An axis is a straight line around which an object rotates. Movement at a joint takes place in a plane about an axis. There are three axis of rotation.

• The sagital axis passes horizontally from posterior to anterior and is formed by the intersection of the sagital and transverse planes.
• The frontal axis passes horizontally from left to right and is formed by the intersection of the frontal and transverse planes.
• The vertical axis passes vertically from inferior to superior and is formed by the intersection of the sagital and frontal planes.

Answer 13.
Intramural activities are performed within the institution, where outside students can’t participate and the main focus is on maximal participation. Following are the importance of intramurals:

• Opportunities for physical, mental, emotional and social development of students
• Inculcation of moral and ethical values through sports
• Awareness of health and wellness among children
• Source of enjoyment, fun and recreation among children.
• Opportunities for maximization of sports participation among students. Develop Leadership, group cohesion among students.

Answer 14.
There are various types of biomechanical movements:

• Flexion
• Adducation
• Extension
• Ulnar deviation
• Abduction

Extension – It is a straightening movement that increases the angle between body parts. When a joint can move forward and backward, such as the neck and trunk, extension refers to movement in the posterior direction. For example, when standing up, the knees are extended. Extension of the hip or shoulder moves the arm or leg backward. When the chin is against the chest, the head is flexed, and the trunk is flexed when a person leans forward.

Abduction – It refers to a motion that pulls a structure or part away from the midline of the body. In the case of fingers and toes, it refers to spreading the digits apart, away from the centerline of the hand or foot. Abduction of the wrist is also called radial deviation For example, raising the arms up, such as when tightrope-walking, is an example of abduction at the shoulder. When the legs are splayed at the hip, such as when doing a star jump or doing a split, the legs are abducted at the hip.

Adduction – It refers to a motion that pulls a structure or part toward the midline of the body, or towards the midline of a limb. In the case of fingers and toes, it refers to bringing the digits together, towards the centerline of the hand or foot. Adduction of the wrist is also called ulnar deviation. Dropping the arms to the sides, and bringing the knees together, are examples of adduction.

Answer 15.

Answer 16.
More importantly, yoga is extremely effective in:

1. Increasing Flexibility – Yoga has positions that act upon the various joints of the body including those joints that are never really on the ‘radar screen’ let alone exercised.

2. Increasing lubrication of the joints, ligaments and tendons – Yoga positions exercise the different tendons and ligaments of the body. Surprisingly it has been found that the body which may have been quite rigid starts experiencing a remarkable flexibility in even those > parts which have not been consciously work upon. Why? It is here that the remarkable research behind yoga positions proves its mettle. Seemingly unrelated “non strenuous” yoga positions act upon certain parts of the body in an interrelated manner. When done together, they work in harmony to create a situation where flexibility is attained relatively easily.

3. Massaging of ALL Organs of the Body – Yoga is perhaps the only form of activity which massages all the internal glands and organs of the body in a thorough manner, including those – such as the prostate – that hardly get externally stimulated during our entire lifetime. Yoga acts in a wholesome manner on the various body parts. This stimulation and massage of the organs in turn benefits us by keeping away disease and providing a forewarning the first possible instance of a likely onset of disease or disorder.

4. Complete Detoxification – By gently stretching muscles and joints as well as massaging the various organs, yoga ensures the optimum blood supply to various parts of the body.This helps in the flushing out of toxins from every nook and cranny as well as providing nourishment up to the last point. This leads to benefits such as delayed ageing, energy and a remarkable zest for life.

5. Excellent toning of the muscles – Muscles that have become flaccid, weak or slothy are
stimulated repeatedly to shed excess flab and flaccidity.

Answer 17.
There are many factors that effect motor development:

1. Nutrition: Nutritious food promotes good motor development. Sensory motor development is dependent upon nutrition that the child gets to a great extent. Children get stronger and development is good if they get nutritious food.
2. Immunisation: If mother and child both are immunized at a proper time it leads to good sensory motor development.
3. Environment: Encouragement, love and security help the child to take risk to explore fearlessly and to know more about environment which leads to a better sensory development.

Answer 18.
(i) Parameters developed are:

• Leadership
• Motivation
• Personality
• Posture
• Fitness

(ii) Intramural competitions
(iii) Test for School Children:

• AAPHER physical fitness test

Answer 19.
Rikli and Jones is the senior citizen test. It was developed by Rikli and Jones in 2001. This test is beneficial for various senior citizens. It helps the early identification of at-risk participants. It is significant to plan safe and effective physical exercise programmes for senior citizens because individual’s health and fitness level can be known better with the help of this test.

1. Chair Stand Test – This test measures the lower body strength, particularly legs.
2. Arm Curl Test – The arm curl test is a test for upper body strength and endurance which is required for performing household activities.
3. Chair sit and reach test: It assess the lower body flexibility, which is important for good posture.

Answer 20.
Coping refers to the thoughts and actions which we usually use to deal with a threatening situation. Lazarus and Folkman said “Coping is a process of constantly changing cognitive and behavioral efforts to manage specific external and/or internal demands or conflicts appraised as taxing or exceeding one’s resources”. These are the following types of strategies:
(i) Problem focused coping strategies – They aim at changing or eliminating the authentic source of stress by:

• analyze the stressful situation
• Seek professional help
• reframing
• slow down pace
• stay focused
•  plan properly

(ii) Emotion focused coping strategies – tried to reduce the negative emotional responses linked with stress by:

• denial of reality
• blaming
• mentally disengaging from stressful situation or people
• humorous attitude
• seek support
• be positive

Answer 21.
Throwing comprises of two phases, the preparatory phase and the throwing phase.
Most actions are rotational in the transverse plane and longitudinal axis and the two joints primarily involved are the elbow and shoulder.
The elbow is a hinge joint formed by the humerus and ulna.
The shoulder is a ball and socket joint formed between the humerus and the scapula.

Answer 22.
The male and female athletes have various physiological differences. These differences generally result in difference in performance ouputs interms of strength, speed, endurance.

1. Cardiovascular Fitness: Athletes’ cardiovascular fitness is measured by their maximum oxygen consumption, also known as VO2 max, which measures their capacity to transport and use oxygen during exercise. This is measured by calculating the point at which an athlete’s oxygen consumption remains steady despite an increase in an exercise intensity. Elite male athletes have a higher oxygen carrying capacity than women, which allows them to reach their maximum training peak earlier. According ACSM’s Primary Care Sports Medicine reference book, this is probably due to women’s lower hemoglobin levels and men’s larger body size. Maximum oxygen consumption is directly related to body size,

2. Bones and Ligaments: Male athletes have longer and larger bones, which provide a clear mechanical advantage over female athletes. The increased articular surface and larger structure of male bones provide them with a greater leverage and a wider frame on which to support muscle. Similarly, the ligaments of female athletes are generally more lax and fragile than those of their male counterparts. This gives male athletes an advantage in sports that involve throwing, kicking and hitting, and explains the higher incidence of musculoskeletal injuries among female athletes. On the other hand, female athletes have a wider pelvis and a lower center of gravity, which provides excellent balance.

3. Strength: Male athletes have a higher ratio of muscle mass to body weight, which allows for greater speed and acceleration. This explains why female speed records in running and swimming are consistently 10 percent slower than men’s, and why, on average, they have two thirds of the strength of men. However, when you factor out the larger muscle mass in men and compare muscular strength relative to cross-section area of muscle, the strength of male and female athletes is nearly equal.

4. Endurance: Endurance is largely determined by a body’s efficiency when converting calories into energy. Female athletes are more efficient than male athletes at converting glycogen to energy. Glycogen is a secondary source of fuel you use when glucose levels drop. This is why female athletes excel in ultra-long-distance sports and rarely hit the wall during long races. It also explains why ultra-running, which includes races longer than a marathon, is one of the few sports where elite female and male athletes regularly compete together, and in which female athletes sometimes win.

Answer 23.
Classification of injury:
(i) Soft Tissue
(ii) Bone Injury
(iii) Joint Injury

(i) Soft Tissue Injury:

• Contusion: Direct impact with blunt object which causes bleeding deep with muscles due to damage in capillaries.
• Strain: Injury to musculo-tendon injury
• Sprain: Ligament injury.
• Abrasion: Loss of epidermis (Outer layer of skin) Superficial injury with loss of skin
• Incision: Cut on arteries, tendon, veins, nerves due to sharp objects
• Laceration: Irregular tear in skin, cut in epidermis and dermis with blunt edge objects

(ii) Bone Injury-Fracture:
A.Close Fracture

• Transverse Fracture
• Oblique Fracture
• Spiral Fracture
• Comminute Fracture
• Impact Fracture

B.Open Fracture

• Compound Fracture

(iii) Joint Injury: Joint injuries often occur as a result of bicycle wrecks, falling is contact sports, and car accidents. They can range from sprains to fractures and dislocations. Common joints injuries include:

• Runners knee
• Plica syndrome of the knee
• Rotator cut injury (shoulder)
• Sprained Ankle

Preventive Measures:

• Wear protective gear such as helmets, protective pads, etc.
• Warm up and cool down
• Keep in mind the rules of the game.

Answer 24.
Exercise for Posture Correction:

1. Back exercises – to put shoulder at original place, some workouts of back strengthening are required.
2. Trunk exercises – twisting of trunk often strengthens the back and abdomen.
3. Leg press, hamstring, curls and leg extensions.
4. Aerobics
5. Walking, running, jumping, etc.

Answer 25.
Total number of teams =13
Upper half = n + 1/2 = 13 + 1/2 = 7
Lower half = n – 1/2 = 13 – 1/2 = 6
Power of two’s = 24 n. of teams
= 16 -13 = 3 No. of Byes = 3
No. of Byes in Lower Half =nb + 1/2 = 3 + 1/2 = 2
No. of Byes in Upper Half = nb – 1/2 = 3 – 1/2 = 1

Answer 26.
Scientific research has demonstrated repeatedly that physical education can enhance academic performance and cognitive function. However, for children with special needs, it’s valuable for so many reasons, from providing an opportunity to build collaborative and social skills, to teaching individuals how to focus on specific goals and overcome obstacles.

When students with special needs participate in physical activity and sports, they see improvements in everything from their hand-eye coordination and flexibility, to their muscle strength, endurance, and even cardiovascular efficiency. These are all simply the natural benefits of exercise — a development of better motor skills and enhanced physical health that helps individuals to fight back against problems such as obesity, and the health complications that follow.

1. Mental Improvements in Confidence and Well-Being: Regular exposure to sports through physical education classes isn’t only good for a child’s body — it’s beneficial to their mind, too. Physical activity improves general mood and wellness in psychiatric patients suffering from anxiety and depressive disorders. What’s more, regular fitness links to improvements in self-esteem, social awareness, and self-confidence — all essential for empowering the lives of young people with special needs.

2. Reduces stress and anxiety: Providing a physical outlet may help students reduce or cope with anxiety, stress and depression — while interaction and involvement with other students will help to give children a sense of accomplishment and confidence. Their physical education teachers to involve them in environments where they can feel as though they’re successfully contributing to a group and their abilities in other areas will improve according to their positive self-image and confidence.

3. Behavioral Improvements in Attention, Relationships, and Academics: Physical education is about a lot more than simply learning how to engage in a particular sport — it teaches children a range of skills, from how to work as a team, to how to solve problems, increase attention span, and focus on task-based behavior. Eventually, those skills can transfer into other classroom settings too, so that students with special needs have a greater ability to leam and engage with their peers outside of physical education.

4. Self-Esteem: Developing a sense of self-esteem and confidence is an extremely important part of special education. These children need to be involved in environments where they feel that they are contributing successfully to a group. Their abilities in all other skill areas will improve as a result of a positive self-image and confidence.

5. Cognitive Benefits: The hands-on nature of physical education leads to cognitive improvements in children with special needs, allowing them to access skills that they couldn’t challenge within a traditional classroom setting. The structure of sport – which comes with a set of rules and organization, can be a learning tool that helps children to practice self-regulation and enhance their decision making skills. On top of that, children with special needs can leam to focus on specific goals, and work on their verbal communication by interacting with peers through sport.

6. Improves appetite

7. Improves quality of sleep
As these children often feel isolated, they

We hope the CBSE Sample Papers for Class 12 Physical Education Paper 1 help you. If you have any query regarding CBSE Sample Papers for Class 12 Physical Education Paper 1, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 12 Biology. Here we have given CBSE Sample Papers for Class 12 Biology Paper 6.

CBSE Sample Papers for Class 12 Biology Paper 6

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 6 of Solved CBSE Sample Paper for Class 12 Biology is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. There are total 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 Questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer Type Questions of three marks each.
5. Section D contains question number 23, Value Based Question of four mark.
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 examiner is to attempt any one of the question out of the two given in the question paper with the same question number.
8. No. of printed pages are three.

SECTION-A

Question 1.
What is coleorhiza?

Question 2.
What is the advantage of use of biotechnology in molecular biology over traditional pathological tests?

Question 3.
What are cleistogamous flowers?

Question 4.
How is Agrobacterium tumefaciens considered useful?

Question 5.
Which attribute of human population do the following figures represent?

SECTION-B

Question 6.
What are the basic steps involved in genetically modifying an organism?

Question 7.
Explain the four types of barriers of Innate immunity.

Question 8.
What do you mean by inbreeding depression? How this problem should be solved during animal breeding?

OR

Write with examples, how use of microbes helps us to make different types of cheese with specific texture & flavors?

Question 9.
S strain → Inject into mice→ ……
….. → Inject into mice → Mice live
Complete the diagram above. What was this experiment about and who performed it?

Question 10.
What may be the reasons for low productivity of ocean?

SECTION-C

Question 11.
Discuss the barrier methods for contraception.

Question 12.
An individual has genotype with an extra chromosome 21.
(a) What is this disorder called?
(b) What will be the physical appearance?

Question 13.
Discuss the role of microbes in sewage treatment

Question 14.
What is DNA finger printing? On what principle does it work? Mention its two applications.

OR

Explain Miller’s experiment to prove the ‘theory of chemical origin of life’ as proposed by Oparin and Haldane.

Question 15.
Differentiate between spermatogenesis and oogenesis with a diagram.

Question 16.
What are the advantages of GM plants?

Question 17.
What do you understand by the term bio-pesticide? Name and explain the mode of action of a popular bio-pesticide

Question 18.
Represent schematically the life cycle of malarial parasite.

Question 19.
Compare and contrast: isogamy and anisogamy. With examples.

Question 20.
Answer the following:  
(a) Expand IUT.
(b) In which part of the female reproductive system the 8-celled embryo will be transferred during test tube baby programme.

Question 21.
Halpoid content of human DNA is 3.3 x 10⁹ bp and the distance between 2 consecutive bp is 0.34x 10^-9. What is the length of the DNA molecule?

Question 22.
Explain convergent evolution with examples.

SECTION-D

Question 23.
Rakhi and her parents were watching a TV serial in the evening. During a commercial break, an advertisement flashed on the screen which was promoting use of sanitary napkins. Rakhi was still watching the TV. The parents got embarrassed and changed the channel. Rakhi objected to her parent’s behaviour and explained the need for these advertisements.
(a) What values did the parents show?
(b) Briefly describe the phases of a menstrual cycle.

SECTION-E

Question 24.
Plant breeding programmes are carried out in a systematic way worldwide. Explain the five main steps in breeding a new genetic variety.

OR

Why are CO₂, CH₄, N₂Oetc known as greenhouse gases?
Why is CNG better than Diesel?

Question 25.
What are chromosomal disorders?

OR

(a) Explain primary productivity and the factors that influence it. 5
(b) Describe how oxygen and chemical composition of detritus control decomposition.

Question 26.
Draw schematically a single polynucleotide strand (with at least three nucleotides). Provide labels and directions.

OR

What are the post pollination events? Explain it.

Answers

SECTION-A

Answer 1.
In embryos of monocots the root cap and radicle are enclosed in an undifferentiated sheath called coleorhiza.

Answer 2.
Biotechnological methods make early diagnosis possible to detect a disease and start treatment at an early stage rather conventional methods of diagnosis where it is not possible.

Answer 3.
Self pollinating flowers in which stamens and pistil are in close proximity.

Answer 4.
Agrobacterium tumefaciens help a tumor causing gene in the bacteria to be substituted with a gene of interest and its introduction into plants.

Answer 5.
The figures represent the age pyramids of human population as Expanding Stable and Declining

SECTION-B

Answer 6.
Three basic steps involved in genetically modifying an organism are:

1. Identification of DNA with desirable genes.
2. Introduction of the identified DNA into the host.
3. Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.

Answer 7.
Innate immunity consists of four types of barriers. These are:

1. Physical barriers: Skin on our body is the main barrier which prevents entry of the micro­organisms.
2. Physiological barriers: Acid in the stomach, saliva in the mouth, tears from eyes-all prevent microbial growth.
3. Cellular barriers: Certain types of leukocytes (WBC) of our body like polymorpho-nuclear leukocytes (PMNL-neutrophils) and monocyte can phagocytose and destroy microbes.
4. Cytokine barriers: Virus-infected cells secrete proteins called interferons which protect non-infected cells from further viral infection.

Answer 8.
Continued inbreeding, especially close inbreeding, usually reduces fertility and even productivity. This is called inbreeding depression. This problem can be solved by selecting animals of the breeding population to be mated with unrelated superior animals of the same breed which usually helps restore fertility and yield.

OR

The specificity of texture, flavour and taste in different variety of cheese comes from microbes used. For example,

1. The large holes in ‘Swiss Cheese’ are due to production of a large amount of CO, by Propionibacterium sharmanii.
2. The ‘Roquefort cheese’ are ripended by growing a specific fungi on them, which gives them a particular flavor

Answer 9.
S strain → Inject into mice → Mice die
R strain → Inject into mice →Mice live

Answer 10.
Low productivity of ocean is due to:

1. Lack of light.
2. High salinity.
3. High pressure and
4. Waves and tides

Answer 11.
Various barrier methods available for both males and females are:

(1) Condoms: Barriers made of thin rubber/latex sheath that are used to cover the penis in the male or vagina and cervix in the female, just before coitus so that the ejaculated semen would not enter into the female reproductive tract.

(2) Diaphragms, Cervical Caps and Vaults: Barriers made of rubber that are inserted into the female reproductive tract to cover the cervix during coitus. They prevent conception by blocking the entry of sperms through the cervix. They are reusable. Spermicidal creams, jellies and foams are usually used along with these barriers to increase their contraceptive efficiency.

(3) Intra Uterine Devices (IUDs): These devices are inserted by doctor or expert nurses in the uterus through vagina. These are presently available as the non-medicated IUDs (g., Lippes loop), copper releasing IUDs (CuT, Cu7, Multiload 375) and the hormone releasing IUDs (Progestasert, LNG-20). IUDs increase phagocytosis of sperms within the uterus and the Cu ions released suppress sperm motility and the fertilising capacity of sperms. The hormone releasing IUDs, in addition, make the uterus unsuitable for implantation and the cervix hostile to the sperms.

Answer 12.
An individual has genotype with an extra chromosome 21.
(a) The disorder is called Down’s syndrome or trisomy of 21.
(b) The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Palm is broad with characteristic palm crease. Physical, psychomotor and mental development is retarded.

Answer 13.
Heterotrophic microbes naturally present in the sewage plays a major role in the treatment of sewage which is carried out in two stages:

Primary Treatment: This initially involves removal of floating debris by sequential filtration followed by grit (soil and small pebbles) removal by sedimentation. All solids that settle down form the primary sludge, and the supernatant forms the effluent. The effluent from the primary settling tank is taken for secondary treatment.

Secondary Treatment or Biological Treatment: The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it. This allows vigorous growth of useful aerobic microbes into floes. The significantly reduced BOD (biochemical oxygen demand) of the effluent is then passed into a setting tank where the bacterial ‘floes’ are allowed to form activiated sludge. A small part of the activated sludge is pumped back into the aeration tank to serve as the inoculum. The remaining major part of the sludge is pumped into anaerobic sludge digesters, where, other kinds of bacteria digest the bacteria and the fungi in the sludge. During this digestion, bacteria produce a mixture of gases such as methane, hydrogen sulphide and carbon dioxide. These gases form biogas and can be used as source of energy as it is inflammable.

Answer 14.
DNA finger printing is a genetic molecular method that identifies and evaluates an individual from another individual on the basis of unique patterns (polymorphisms) in their DNA. DNA finger printing works on a principle of repetitive DNA sequences that gives a unique identity to an individual. These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation and categories, into micro-satellites, mini-satellites. These sequence normally do not code for any proteins, but they form a large portion of human genome. These sequence show high degree of polymorphism and form the basis of DNA finger printing. Its two applications are:

1. As an identification tool in forensics.
2. Paternity testing, in case of disputes.

OR

Oparin and Haldane proposed that the first form of life could have come from pre-existing non-living organic molecules (e.g. RNA, protein, etc.) and that formation of life was preceded by chemical evolution, i.e. formation of diverse organic molecules from inorganic constitutents. The conditions on earth were—high temperature, volcanic storms, reducing atmosphere containing CH₄, NH_r Miller created similar conditions in a laboratory scale. He created electric discharge in a closed flask containing CH₄, H₂, NH₃ and water vapour at 8000° C. He observed formation of amino acids. In similar experiments others observed, formation of sugars, nitrogen bases, pigment and fats. Analysis of meteorite content also revealed similar compounds indicating that similar processes are occurring else where in space.

Answer 15.

Answer 16.
Genetically Modified crops/plants have the following advantages:

1. Crops are more tolerant to abiotic stresses (cold, drought, salt, heat).
2. Have reduced reliance on chemical pesticides (pest-resistant crops).
3. Helpful in reduction of post harvest losses.
4. Plants have increased efficiency of mineral usage to prevent early exhaustion of fertility of soil.
5. Crops yield enhanced nutritional value of food, e.g., Vitamin ‘A’ enriched rice.
6. Tailor-made plants have created alternative resources for industries, in the form of starches, fuels and pharmaceuticals.

Answer 17.
Bio-pesticides are derived from natural materials such as animals, plants, bacteria and certain minerals. This microbial bio-control agents that is used to control butterfly caterpillars is the bacteria Bacillus thuringiensis (often written as Bt). These are mixed with water and sprayed onto vulnerable plants such as brasssicas and fruit trees, where these are eaten by the insect larvae. In the gut of the larvae, the toxin is released and the larvae get killed. The bacterial disease will kill the caterpillars, but leave other insects unharmed.method of controlling pests relies on natural predation rather than introduced chemicals. An example of B. thuringiensis toxin genes has also been introduced with the help of genetic engineering into plants which offers resistance to insect pests. Bt-cotton is one such example

Answer 18.

Answer 19.

Answer 20.
(a) IUT stands for Intra Uterine transfer. The 8-celled embryo developed by In-vitro fertilization or ICSI is transferred in this technique.
(b) The 8-celled embryo is transferred to the uterus.

Answer 21.
Haploid content is 3.3 x 10⁹
Therefore, diploid content is 6.6 x 10⁹ Distance between bp is 0.34 x 10^-9
Therefore length is diploid content x distance between bp = 6.6 x 10⁹ x 0.34 x 10 ⁹ = 2.24 m

Answer 22.
Convergent evolution occurs when more than one adaptive radiation appear to have occurred in an isolated geographical area of different habitats. The analogous organs have almost similar appearance and perform the same function but develop in different groups and are totally different in their basic structure and development origin. For example, Darwins’ Finches and Australian Marsupials where each marsupial differ from each other.

SECTION-D

Answer 23.
(a) The parents were traditional but understood the need for such advertisements. They showed maturity and openness later.
(b)

1. Menstrual phase
2. Proliferative phase
3. Ovulatory phase
4. Secretory phase.

Answer 24.
The main steps in breeding a new genetic variety of a crop are as follows:

(1) Collection of Variability: The entire collection of plants/seeds having all the diverse alleles for all genes including all the different wild varieties, species and relative of the cultivated species in a given crop called as germplasm collection, is collected.

(2) Evaluation and Selection of Parents: The germplasm is evaluated so as to identity plants with desirable combination of characters. The selected plants are multiplied and used in the process of hybridization.

(3) Cross Hybridization among the Selected Parents: The desired characters are combined from two different plants (parents), to produce hybrids that genetically combine the desired characters in one plant. This is a very time-consuming and tedious process since the pollen grains from the desirable plant chosen as male parent have to be collected and placed on the stigma of the flowers selected as female parent. Usually only one in few hundred to a thousand crosses shows the desirable combination, for example high protein quality of one parent may need to be combined with disease resistance from another parent.

(4) Selection and Testing of Superior Recombinants: Progency plants that are superior to both of the parents are selected. The selection process is crucial to the success of breeding objective and requires careful scientific evaluation of the progency.

(5) Testing, Release and Commercialization of new Cultivars: The newly selected lines are evaluated for their yield and other agronomic traits of quality, disease resistance etc. The material is then evaluated in comparison to the best available local crop cultivar usually by a check or reference cultivar.

Answer 25.

OR 

(a) Primary productivity is defined as the amount of biomass or organic matter produced per unit area
over a time period by plants during photosynthesis. It is expressed in terms of weight (g ²) or energy
(kcal nr²). It is also expressed in terms of g ² yr¹ or (kcal nr²) yr¹ to compare the productivity of different ecosystems.

It is divided into gross primary productivity (GPP), net primary productivity (NPP) and secondary productivity.

Gross primary productivity of an ecosystem is the rate of production of organic matter ditfing photosynthesis.

Net primary productivity is the available biomass for the consumption to heterotrophs (herbivores and decomposers), i.e., NPP = GPP – R.

Secondary productivity is defined as the rate of-formation of new organic matter by consumers.

The factors on which primary productivity depends are: (z) Plant species inhabiting a particular area, (ii) On a variety of environmental factors such as availability of nutrients. (iii) Photosynthetic capacity of plants.

(b) Decomposition is an oxygen-requiring process. The rate of decomposition is controlled by both chemical composition of detritus and climatic factors. In a particular climatic condition, decomposition rate is slower if detritus is rich in lignin and chitin, and quicker, if detritus is rich in nitrogen and water-soluble substances like sugars. Temperature and soil moisture are the most important climatic factors that regulate decomposition through their effects on the activities of soil microbes. Warm and moist environment favour decomposition whereas low temperature and anaerobiosis inhibit decomposition resulting in build up of organic matter.

Answer 26.

The post pollination events are:
1. The pollen grain germinates on the stigma to produce a pollen tube through one of the germ pores.
2. The contents of the pollen grain move into the pollen tube. Pollen tube grows through the tissues of the stigma and style and reaches the ovary.
3. The generative cell divides and forms the two male gametes during the growth of pollen tube in the stigma.
4. Pollen tube, after reaching the ovary, enters the ovule through the micropyle and then enters one of the synergids through the filiform apparatus.
All these events are together referred to as pollen-pistil interaction also.

We hope the CBSE Sample Papers for Class 12 Biology Paper 6 help you. If you have any query regarding CBSE Sample Papers for Class 12 Biology Paper 6, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 12 Biology. Here we have given CBSE Sample Papers for Class 12 Biology Paper 5.

CBSE Sample Papers for Class 12 Biology Paper 5

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 5 of Solved CBSE Sample Paper for Class 12 Biology is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. There are total 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 Questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer Type Questions of three marks each.
5. Section D contains question number 23, Value Based Question of four mark.
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 examiner is to attempt any one of the question out of the two given in the question paper with the same question number.
8. No. of printed pages are three.

SECTION-A

Question 1.
What is the number of chromosome in human zygote?

Question 2.
What is totipotency?

Question 3.
What are palindromic sequences?

Question 4.
What is Allen’s rule?

Question 5.
How do you define NPP?

SECTION-B

Question 6.
Write the transcription product sequence for
(a) 5 ‘ATGCACTGATCCAA 3’
(b) 3 ‘ GTACGTACGTAC 5’

Question 7.
Complete the table:

 Question 8.
What are the types of acquired immunity?

Question 9.
Which microbe converts milk to curd?

Question 10.
Give some examples of diseases and their insect vector.

OR

What are the different methods of breeding?

SECTION-C

Question 11.
List the salient features of DNA double helix model.

Question 12.
What is the fate of the product of fertilization in humans?

Question 13.
How was the genetic code elucidated?

Question 14.
p² + 2pq + q² = 1.Explain this equation.

Question 15.
What are the different levels at which gene regulation can be achieved?

Question 16.
What are the primary lymphoid organs?

Question 17.
Explain gene therapy with an example?

Question 18.
Diagrammatically represent the replication of retrovirus.

Question 19.
How have cry proteins been utilized?

OR

Explain carbon cycle with diagram.

Question 20.
Explain two reasons for loss of biodiversity.

Question 21.
Give some adaptations of desert plants to survive the heat.

Question 22.
What does the picture represent?

SECTION-D

Question 23.
In art class the teacher asked Sunita to mix green and red paint and report on the combined colour formed. Sunita could not find red colour in her box and was scolded by the teacher who found it lying right in front. Suddenly Vijay realized that Sunita was not able to identify red colour and reported the matter to the teacher who was of the opinion that she lacked colour concept. After school was over, Vijay reported this matter to Sunita’s parents.                                                                                                (a) What values did Vijay possess?
(b) Did Sunita lack knowledge of colours? If not, give the biological reason for the same.
(c) Give the technical term for this type of inheritance. Explain with a typical example

SECTION-E

Question 24.
Explain with diagram the experiment that proved that DNA is a genetic material.

OR

Explain pollination by wind and water.

Question 25.
Give the journey of sperm formation with diagram. What are the hormones involved?

OR

Explain the technique of fingerprinting with diagram.

Question 26.
What is parasitism? What are the types?

OR

What are ecosystem services?

Answers

SECTION-A

Answer 1.
46

Answer 2.
Capacity of generating a whole plant from cell/explant is called totipotency.

Answer 3.
The palindrome in DNA is a sequence of base pairs on the two strands that reads the same when orientation of reading is kept same

Answer 4.
Mammals from colder climates generally have shorter ears and limbs to minimise heat loss. This is known as Allen’s rule.

Answer 5.
NPP stands for Net Primary Productivity. It is the available biomass for the consumption by heterotrophs (herbivores and decomposers) NPP = GPP – R. Where ‘GPP’ is Gross Primary Productivity, ‘R’ is Respiration losses.

SECTION-B

Answer 6.
(a) 3 ‘TACGTGACTAGGTT 5’                                      ‘
(b) 5 ‘AUGCACUGAUCCAA 3 ’

Answer 7.

Answer 8.
1. Antibody mediated immunity or humoral immune response by the production of antibodies against antigens.
2. Cell mediated immunity initiated by T lymphocytes.

Answer 9.
Micro-organisms such as Lactobacillus and others commonly called lactic acid bacteria (LAB) grow in milk and convert it to curd. During growth, the LAB produce acids that coagulate and partially digest the milk proteins. A small amount of curd added to the fresh milk as inoculum or starter contain millions of LAB, which at suitable temperatures multiply, thus converting milk to curd.

Answer 10.
(1) Malaria – female Anopheles mosquito
(2) Dengue or Chikungunya – Aedes mosquito.

OR

Different methods of breeding are:
(a) Inbreeding;
(b) Out breeding;
(c) Out crossing;
(d) Cross breeding; and
(e) Interspecific hybridisation

SECTION-C

Answer 11.
The salient features of the Double-helix structure of DNA are as follows:

1. It is made of two polynucleotide chains, where the backbone is constituted by sugar phosphate, and the bases project inside.
2. The two chains have anti-parallel polarity. It means, if one chain has the polarity 5’→3′, the other has
   3′ → 5′.
3. The base in two strands are paired through hydrogen bond (H-bonds) forming base pairs (bp). Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa. Similarly, Guanine is bonded with Cytosine with three H-bonds. As a result, always a purine comes opposite to a pyrimidine. This generates approximately uniform between the two strands of the helix.
4. The plane of one base pair stacks over the other in double helix. This, in addition to H-bonds, confersstability of the helical structure.

Answer 12.
The product of fertilization is the zygote. The mitotic division starts as the zygote moves through the isthmus of the oviduct called cleavage towards the uterus and forms 2,4,8,16 daughter cells called blastomeres. The embryo with 8 to 16 blastomeres is called a morula. The morula continues to divide and transforms into blastocyst as it moves further into the uterus. The blastomeres in the blastocyst are arranged into an outer layer called trophoblast and an inner group of cells attached to trophoblast called the inner cell mass. The trophoblast layer then gets attached to the endometrium and the inner cell mass gets differentiated as the embryo. After attachment the uterine cells divide rapidly and cover the blastocyst. As a result, the blastocyst becomes embedded in the endometrium of the uterus. This is called implantation and it leads to pregnancy.

Answer 13.
Har Govind Khorana, Marshall Nirenberg and Severo Ochoa elucidated the genetic code. Genetic Code directs the sequence of amino acids during protein synthesis of proteins. The chemical method was developed by Har Gobind Khorana to synthesise RNA molecules with defined combinations of bases (homopolymers and copolymers). Marshall Nirenberg’s cell-free system for protein synthesis finally helped the code to be deciphered. Further, Severo Ochoa enzyme (polynucleotide phosphoryalse) was also helpful in polymerizing RNA with defined sequences in a template independent manner (enzymatic synthesis of RNA). This finally gave rise to the checker-board, for genetic code. The salient features of genetic code are as follows:
(a) The codon is triplet 4³ = 64. (61 codons code for amino and 3 codons do not code for any amino acids, hence they function as stop codons.)
(b) One codon codes for only one amino acid, hence, it is unambiguous and specific.
(c) Some amino acids are coded by more than one codon, hence the code is degenerate.
(d) The codon is read in mRNA in a contiguous fashion. There are no punctuations.
(e) The code is nearly universal: for example, from bacteria to human UUU would code for Phenylalanine (phe).
(f) AUG has dual functions. It codes for Methionine (met), and it also as initiator codon

Answer 14.
p² + 2pq + q² = 1, equation represents the gene frequency of a given population as stated by Hardy-Weinberg law. It states that the frequencies of allele in a given population are stable and constant.
Where p² = frequency of homozygous dominant alleles (AA).
q² = frequency of homozygous recessive alleles (aa).
2pq = frequency of heterozygous alleles (Aa).
This is also referred to as genetic equilibrium and is a binomial expansion of (p + q)².

Answer 15.
Gene regulation could be exerted at

1. Transcriptional level (formation of primary transcript).
2. Processing level (regulation of splicing).
3. Transport of mRNA from nucleus to the cytoplasm.
4. Translational Level.

Answer 16.
The primary lymphoid organs are bone marrow and thymus where immature lymphocytes differentiate into antigen-sensitive lymphocytes. Both provide micro-environments for the development and maturation of T-lymphocytes.The bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced.

The thymus is a lobed organ located near the heart and beneath the breastbone. The thymus is quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained, it reduces to a very small size.

Answer 17.
Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in child/embryo. Here, genes are inserted into a person’s cells and tissues to treat a disease. Correction of a genetic defect involves delivery of a normal gene into the individual or embryo to take over the function of and compensate for the non-functional gene.

The first clinical gene therapy was given to a 4-year-old girl with adenosine deaminase (ADA) deficiency. This disorder is caused due to the deletion of the gene for adenosine deaminase. In some children, ADA deficiency can be cured by bone marrow transplantation; in others it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. Both of these approaches are not completely curative.

Answer 18.

Answer 19.
The Bt toxin is coded by a gene named cry protein. The proteins encoded by the genes crylAc and crylAb control the cotton bollworms that of crylAb control com borer. B. thuringiensis  forms protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein. The Bt toxin protein exist as inactive protoxin but once an insect ingest 1 the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the gut which solubilise the crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually cause death of the insect.

OR

Carbon cycling occurs through atmosphere, ocean and through living and dead organisms. A considerable amount of carbon returns to the atmosphere as CO₂ through respiratory activities of the producers and consumers. Decomposers also contribute substantially to CO₂ pool by their processing of waste materials and dead organic matter of land oceans. Some amount of the fixed carbon is lost to sediments and removed from circulation. Burning of wood, forest fire and combustion of organic matter, fossil fuel, volcanic activity are additional sources for releasing CO₂ in the atmosphere.

Answer 20.
Two reasons for the loss of biodiversity are:
1. Habital Loss and Fragmentation: The degradation of many habitats by pollution also threatens the survival of many species. When large habitats are broken up into small fragments due to various human activities, mammals and birds requiring large territories and certain animals with migratory habits are badly affected, leading to population declines.

2. Co-extinctions: When a species becomes extinct, the plant and animal species associated with it in an obligatory way also become extinct. When a host fish species becomes extinct, its unique assemblage of parasites also meets the same fate. Another example is the case of a coevolved plant-pollinator mutualism where Extinction of one invariably leads to the extinction of the other.

Answer 21.
Some adaptive features that the desert plants have are:

1. Thick cuticle on their leaf surfaces.
2. Their stomata arranged in deep pits to minimize water loss through transpiration.
3. Special photosynthetic pathway (CAM) enables their stomata to remain closed during day time.
4. Some plants like opuntia, have no leaves-they are reduced to spines-and the photosynthetic function is taken over by the flattened stems

Answer 22.
The following picture represents the global biodiversity

Fishes have the maximum global biodiversity in the phylum vertebrates followed by birds, reptiles, amphibians and mammals.

SECTION-D

Answer 23.
(a) Vijay was alert, curious, clever and a responsible child.
(b) According to the teacher, Sunita lacks the concept of colours. When she could not identify red colour, it proved to be a case of colour blindness. It is a sex linked inherited disorder.
(c) This is a human disease which causes the loss of ability to differentiate between red and green colour. The gene for this red-green colour blindness is present on X chromosome. Colour blindness is recessive to normal vision. If a colour blind man (XcY) marries a girl with normal vision (XX), the daughters would have normal vision but would be a carrier, while sons would also be normal. If the carrier girl (heterozygous for colour blindness, XcX) now marries a colour blind XcY the off spring would show 50% females and 50% males. Of the females, 50% would be the carrier for colour blindness and the rest 50% would be colour blind. Of the males, 50% would have normal vision and the 50% would be colour blind.

SECTION-E

Answer 24.
Hershey and Chase grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but not protein. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur. Radioactive phages were allowed to attach to E. coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge. Bacteria which were infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.

Pollination By Wind:
(a) Wind pollination requires that the pollen grains are light and non-sticky so that they can be transported in wind currents.
(b) Possess well-exposed stamens so that the pollens are easily dispersed into wind currents.
(c) Ltirgc and feathery stigma to easily trap air-bome pollen grains.
(d) Have a single ovule in each ovary and numerous flowers packed into an inflorescence.
(e) Quite common in grasses.

Pollination By Water: 
(a) Water pollination is quite rare in flowering plants and is limited to mostly monocotyledons.
(b) Water is a regular mode of transport for the male gametes among the lower plant groups such as algae, bryophytes and pteridophytes.
(c) Some examples of water pollinated plants are Vallisneria and Hydrilla which grow in fresh water and several marine sea-grasses such as Zostera.
(d) In Vallisneria, the female flower reach the surface of water by the long stalk and the male flowers or pollen grains are released on the surface of water. They are carried passively by water currents some of them eventually reach the female flowers and the stigma.
(e) In sea grasses, female flowers remain submerged in water and the pollen grains are released inside the water.
(f) Pollen grains in many such species are long, ribbon like and they are carried passively inside the water; some of them reach the stigma and achieve pollination. In most of the water- pollinated species, pollen grains are protected from wetting by a mucilaginous covering.

Answer 25.
In testis, the immature male germ cells (spermatogonia) produce sperms by spermatogenesis that begins at puberty. The spermatogonia (sing, spermatogonium) present on the inside wall of seminiferous tubules multiply by mitotic division and increase in numbers. Each spermatogonium is diploid and contains 46 chromosomes. Some of the spermatogonia called primary spermatocytes periodically undergo meiosis. A primary spermatocyte completes the first meiotic division (reduction division) leading to formation of two equal, haploid cells called secondary spermatocytes, which have only 23 chromosomes each. The secondary spermatocytes undergo the second meiotic division to produce four equal, haploid spermatids. The spermatids are transformed into spermatozoa (sperms) by the process called spermiogenesis. After spermiogenesis, sperm heads become embedded in Sertoli cells, and are finally released from the seminiferous tubules by the process called sperimation.

The hormones involved are:

1. GnRH – Gonadotropin Releasing Hormone
2. LH – Luteinising Hormone
3. FSH – Follicle Stimulating Hormone.

OR

DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times.

• These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.

• The bulk DNA forms a major peak and the other small peaks are reffered to as satellite DNA.
• Depending on base composition (A : T rich or G:C rich), length of segment, and number of repetitive units, the satellite DNA is classified into many categories, such as micro- satellities, mini-satellites etc.
• These sequences normally do not code for any proteins, but they form a large portion of human genome.
• These sequence show high degree of polymorphism and form the basis of DNA fingerprinting.

The technique of DNA fingerprinting involves the following steps: 

1. Isolation of DNA, i.e., extraction from the nuclei of the different possible cells.
2. The DNA molecules are digested with the help of enzyme restriction endonuclease (called chemical knife) that cuts them into fragments. The fragments of DNA also contains the VNTRs.
3. The fragments are separated according to the size by the gel electrophoresis.
4. Multiplication of fragments of a particular size having VNTRs through PCR technique.
   Here they are treated with alkaline chemicals to split them into single stranded DNAs.
5. Blotting/transferring of separated fragments of a single stranded DNA to a nylon membrane.
   
6. Radioactive DNA probes having repeated base sequences complementary to possible VNTRs are poured over the nylon membrane. Some of them will bind to the single stranded VNTRs. Hybridisation of DNA with probes is called Southern Blotting. The nylon membrane is washed to remove extra probes.
7. Autoradiography: An X-ray film is exposed to the nylon membrane to mark the place where the radioactive DNA probes have bound to the DNA fragments. These places are marked as dark bands when X-ray film is developed.
8. The dark bands so formed on X-ray film represent the DNA fingerprints (= DNA profiles).

Answer 26.
Parasitism is an interspecies relationship in which one organism gets benefitted and the other is harmed. Parasites evolve special adaptations such as the loss of unnecessary sense organs, presence of adhesive organs or suckers to cling on to the host, loss of digestive system and high reproductive capacity. The life cycles of parasistes are often complex, involving one or two intermediate hosts or vectors to facilitate parasitisation of its primary host. The human liver fluke (a trematode parasite) depends on two intermediate hosts (a snail and a fish) to complete its life cycle. The malarial parasite needs a vector (mosquito) to spread to other hosts.

There are three types of parasitism:

(1) Ecoparasitsm: Parasites feeding on the external surface of the host organism. For example the lice on humans and ticks on dogs, marine fish are infested with ectoparasitic copepods and, cuscuta found growing on hedge plants, to derive its nutrition from the host plant which it parasitizes.

(2) Endoparasitism: Parasites living inside the host body at different sites, /.<?., liver, kidney, lungs, red blood cells etc. The life cycles of endoparasites are more complex because of their extreme specialisation. Their morphological and anatomical features are greatly simplified while emphasising their reproductive potential.

(3) Brood parasitism: Parasitisism in which birds lays its eggs in the nest of its host and lets the host incubate them. The eggs of the parasitic bird have evolved to resemble the host’s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest.

OR

The benefits that people obtain from ecosystems are termed as ecosystem services. The four categories of ecosystem services are supporting, provisioning, regulating and cultural. These services forms a base for a wide range of economic, environmental and aesthetic goods and services. Out of the total cost of various ecosystem services, the soil formation accounts for about 50 percent, and contributions of other services like recreation and nutrient cycling, are less than 10 per cent each. The cost of climate regulation and habitat for wildlife are about 6 per cent each. Examples of such services are as follows:
(a) Healthy Forest ecosystems.
(b) Purify air and water.
(c) Mitigate droughts and floods.
(d) Cycle nutrients.
(e) Generate fertile soils.
(f) Provide wildlife habitat.
(g) Maintain biodiversity.
(h) Pollinate crops.
(i) Provide storage site for carbon.

We hope the CBSE Sample Papers for Class 12 Biology Paper 5 help you. If you have any query regarding CBSE Sample Papers for Class 12 Biology Paper 5, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 12 Biology. Here we have given CBSE Sample Papers for Class 12 Biology Paper 1.

CBSE Sample Papers for Class 12 Biology Paper 1

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 1 of Solved CBSE Sample Paper for Class 12 Biology is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. There are total 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 Questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer Type Questions of three marks each.
5. Section D contains question number 23, Value Based Question of four mark.
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 examiner is to attempt any one of the question out of the two given in the question paper with the same question number.
8. No. of printed pages are three.

SECTION-A

Question 1.
Recently chikungunya cases were reported from various parts of the country. Name the vector responsible for it.

Question 2.
Why cashew is referred to as false fruit?

Question 3.
Name two animals that have become extinct due to over-exploitation.

Question 4.
What forms the backbone of a polynucleotide strand of a nucleic acid?

Question 5.
How is the presence of cyanobacteria in the paddy fields beneficial to rice crop?

SECTION-B

Question 6.
Why should a bisexual flower be emasculated and then bagged prior to artificial pollination in hybridisation programmes?

Question 7.
How does CuT act as an effective contraceptive for human females?

Question 8.
How and why is the bacterium, Thermus aquaticus, useful in rDNA technology?

Question 9.
Differentiate between predation and competition.

Question 10
Highlight any four advantages of genetically modified or transgenic animals.

SECTION-C

Question 11.
Draw a diagram of a mature embryo of grass and label six parts in it.

OR

Describe the structure of a microsporangium of an angiosperm along with its diagram.

Question 12.
Who proposed chromosome theory of inheritance? Point structure of Microsporangium
out any two similarities between the behaviour of genes and that of chromosomes.   

Question 13.
A person is suffering from Amoebiasis. Mention the pathogen that causes it and one organ of the body that gets affected. Give three symptoms and one mode of its transmission.

Question 14.
How do organisms manage with the stressful conditions prevailing in their habitat for short duration? Explain with the help of one example each.

Question 15.
Mention three uses of PCR in molecular diagnostics.   

Question 16.
Explain the three ways in which natural selection operates on different traits in nature.

Question 17.
Explain the antibiotics resistance observed in bacteria in light of Darwinian selection theory.

Question 18.
What is the cause of adenosine deaminase deficiency in a person? What is the importance of this enzyme? Why is it that even after infusion of genetically engineered lymphocytes into the patient suffering from ADA deficiency, the cure is not permanent?

Question 19.
(a) What is the objective of Water Act?
(b) Mention any four problems caused due to deforestation.

Question 20.
Draw well labelled diagram of a magnified view of seminiferous tubule.    

Question 21.
(a) List two essential roles of ribosomes during translation.
(b) Who discovered nuclei acid? What was it called then?

Question 22.
Fill in the spaces a, b and c in the table showing certain events and their sites in human
(male / female) reproductive system

SECTION-D

Question 23.
In many villages, people do not go for vaccination/immunization because of some fear or religious beliefs; they feel they are healthy and they do not have the disease.

1. How can we explain to them that vaccination will help to prevent the diseases?
2. How can this idea be made to reach them?

SECTION-E

Question 24.
(a) What are

1. introns, and
2. exons?

(b) Represent schematically the steps in transcription in Eukaryotes.

OR

(a) Expand BAC and YAC.
(b) What is a promoter in a transcription unit?
(c) Differentiate between template strand and coding strand.

Question 25.
(a) What is meant by contact inhibition?
(b) Why is using tobacco in any form injurious to health? Explain.
(c) Why do sports persons often fall a victim to cocaine addiction?

OR

How are morphological and biochemical/’physiological characteristics of plants associated with resistance to certain pests? Explain with the examples

Question 26.
(a) Why pollen grains are known as living fossils?
(b) Study the population growth curves in the graph given below and answer the questions which follow:

• Identify the growth curves ‘a’ and ‘b’
• Which one of them is considered a more realistic one and why?
• If \(\cfrac { dN }{ dt } =rN\left( \cfrac { K-N }{ K } \right) \) is the equation of one of the
  growth curves, what does K stands for?
• What is symbolized by N?

OR

(a) Name any two human activities that influence carbon cycle.
(b) Describe the particular type of agriculture which is also responsible for deforestation.

Answers

SECTION-A

Answer 1.
Vector responsible for chikungunya is Aedes aegyptii.

Answer 2.
Cashew is referred to as false fruit because the thalamus also contributes to the fruit formation.

Answer 3.
Two animals that have become extinct due to over-exploitation are:

1. Dodo bird
2. Caspian Bali (species of Tiger)
3. Quagya (Any two)

Answer 4.
Phosphodiester linkage forms the backbone of a polynucleotide strand of a nucleic acid.

Answer 5.
The presence of cyanobacteria in the paddy fields is beneficial to rice crop because these bacteria have the property of nitrogen fixation due to the presence of heterocysts (nitrogen fixing cells).

SECTION-B

Answer 6.
A bisexual flower is emasculated to prevent self pollination and then bagged to further prevent its stigma from contamination by unwanted pollen. This is all done to get the desired result in the artificial hybridisation.

Answer 7.
CuT acts as an effective contraceptive for human females due to the following reasons:

1. It releases Cu ions which suppresses sperm motility.
2. Cu ions also reduces the fertilising capacity of the sperms.

Answer 8.
In PCR reaction, in the process of repeated DNA replication, the segment of DNA is amplified to approximately 1 billion time which is achieved by a thermostable DNA polymerase isolated from bacterium Thermus aquaticus.

The bacterium Thermus aquaticus (Taq) is useful in rDNA technology because it contains thermostable enzyme – DNA polymer as (Taq-DNA polymerase) which can withstand high temperature.

Answer 9.

Answer 10.
(1) Biological Products: Transgenic animals produces useful biological products only by the introduction of the portion of DNA or genes, which codes for a particular product like human protein (α-1-antitrypsin) is produced to treat emphysema.The first transgenic cow, Rosie produced human protein-enriched milk (2.4 gms per litre).

(2) Vaccine Safety: Transgenic mice are being used in testing the safety of vaccines before they are used on human beings.

(3) Chemical Safety Testing: It is also called as toxicity/safety testing. Transgenic animals are developed with genes exposed to toxic substance and are used to study their effects.

(4) Growing of Spare Parts: Spare parts (e.g., heart, pancreas) of pig for human use can be grown through the formation of transgenic animals

SECTION-C

Answer 11.

OR

1. In a transverse section, a typical microsporangium appears circular in outline.
2. It is generally surrounded by four wall layers—the epidermis, endothecium, middle layers, and the tapetum.
3. The outer three wall layers perform the function of protection and help in dehiscence of anther to release the pollen.
4. The innermost wall layer is the Tapetum. It nourishes the developing pollen grains. Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus.

Answer 12.
Chromosomal theory of inheritance was proposed by Sutton and Boveri.

Answer 13.
Pathogen – Entamoeba histolytica
Organ affected – Large intestine
Symptoms:

1. Constipation
2. Abdominal pain
3. Cramps, stools are covered with mucous and blood clots.

Mode of Transmission: Houseflies act as mechanical carriers which transmit the parasite from the faces of infected person to the food and drinking water.

Answer 14.
Some organisms possess phenotypic adaptations which help them to respond quickly to an unfavourable situation.
For example, at high altitudes due to the low atmospheric pressure the human body does not get sufficient oxygen which causes altitude or mountain sickness characterised by nausea, vomiting and heart palpitation. The body compensates this low oxygen availability by increasing red blood cells, which decreases the binding-capacity of haemoglobin thereby increasing breathing rate.

Answer 15.
1. PCR (Polymerase Chain Reaction) is used to detect the presence of gene sequences of the infectious agents.
2. It is used to detect specific microbes from the samples of soil, sediments and water.
3. It is also used to detect HIV in AIDS patient.

Answer 16.
The three types of natural selection are as follows:
(1) Stabilizing Selection: The stabilizing influence of natural selection -in an environment changes little in space and time. It is also called maintenance evolution.

(2) Directional or Progressive Selection: Directional selection produces a regular change within a population in one direction in respect to certain characteristics.

(3) Disruptive Selection: In this original population splits into two groups which later evolved into two different species.

Answer 17.
In the original population of bacteria, there were some bacteria which possessed antibiotic resistant genes. When antibiotic was not used, these bacteria had no advantage with the use of antibiotics, these bacteria survived, reproduced and spread throughout the population.

Answer 18.
1. ADA is caused by the deletion of the gene responsible for the enzyme adenosine deaminase.
2. The enzyme adenosine deaminase plays crucial role in the functioning of the immune system.
3. The infusion of genetically engineered lymphocytes into the patient suffering from ADA deficiency is not a permanent cure because these cells are not immortal.

Answer 19.
(a) India has passed the Water Act in 1974. The objective of Water Act is the prevention and control of water pollution.
(b) Four problems caused due to deforestation are:       

1. Deforestation has enhanced CO₂ level which has contributed to the greenhouse effect or global warming.
2. It has caused soil erosion.
3. It has disturbed the hydrological cycle or water cycle.
4. It has lead to the loss of diversity.

Answer 20.

Answer 21.
(a) Two essential roles of ribosomes during translation are:

1. The larger subunit of ribosome has a groove for pushing out the newly formed polypeptide and protects the same from cellular enzymes.
2. The smaller unit of ribosome has a point for recognising mRNA and binding area for initiation factors.

(b) Nucleic acids were first identified by Friedrich Meischer. It was named as nucleic acids because it was isolated from the nuclei of cells.

Answer 22.
a → Ovary
b → Testis
c → Ampullary Isthmic Junction(fallopian tube)

Answer 23.
1. We can explain the villagers that vaccination is not given to cure any disease at the given time. Now they may be healthy, but when an epidemic of a disease occurs, or when they happen to encounter an infectious disease, the body does not have enough of antibodies to fight against. By vaccination, the body produces antibodies and also memory cells that can produce antibodies during an encounter with the same pathogen.

2. Advertisement propaganda can be made:

• By giving examples of cases where death has occurred due to failure of vaccination.
• Advertisement in newspapers about vaccination and its benefits.

SECTION-E

Answer 24.
(a) Coding segments in DNA are known as Exons whereas non-coding segments are known as introns.

(a) BAC → Bacterial Artificial Chromosome
YAC → Yeast Artificial Chromosome
(b) Promoter initiates the process of transcription.
(c)

Answer 25.
(a) Contact inhibition is a property by virtue of which normal cells by contact with other cells inhibit their uncontrolled growth.
(b) Harmful effects of Tobacco:

1. Tobacco contains nicotine which stimulates the adrenal gland to release adrenaline and noradrenaline into blood circulation, both of which raise blood pressure and increase , heart rate.
2. It is associated with increased incidence of cancers of lung, throat, urinary bladder, bronchitis, emphysema, gastric ulcer and coronary heart disease.
3. Tobacco chewing is associated with increase risk of cancer of the orai cavity.

(c) Sports person often take cocaine to increase their energy level for the game because it has a potent stimulating action to provide them instant energy. With regular intake they fall victim to cocaine addiction as it acts on their nervous system producing a sense of euphoria and interferes with the transport of the nano-transmitter dopamine.

OR

Morphological and biochemical characteristics of plants associated with resistance to certain pests has been seen in many host crop plants such as:

• Hairy leaves in cotton are resistant to Jassids and cereals leaf beetle in wheat.
• In wheat solid stems lead to non-preference by the stem sawfly.
• In cotton, smooth leaved and nectar-less varieties do not attract bollworms.
• In maize, high aspartic acid, low nitrogen and sugar content leads to resistance to maize stem borers.
• Thoms in Acacia, Cactus are the most common morphological means of defence in plants

Answer 26.
(a) The outer hard layer of pollen grain called the exine is made up of sporopollenin. It can withstand the worst condition of fossilization such as high temperatures and action of strong acids and alkali.

No enzyme is so far known that can degrade sporopollenin. Pollen grains are well preserved as fossils because of the presence of sporopollenin.

(b)
1. a → exponential curve
2. b→ logistic curve
3. Since resources for growth for most animal populations are finite and will become limiting sooner or later, therefore the logistic growth model/curve is considered to be a more realistic one.
4. K — carrying capacity
5. N — Population density at time “t”

OR

(a)
1. Combustion of fossil fuels
2. Deforestation

(b) Slash and Burn cultivation or Jhum cultivation or Jhuming is a type of agriculture which is responsible for deforesation. It is technically known as shifting cultivation. In India about five lakh hectares of land is cleared every year through hopping, burning the remainder, mixing the ash with soil & sowing the cleared land with crop seeds.

We hope the CBSE Sample Papers for Class 12 Biology Paper 1 help you. If you have any query regarding CBSE Sample Papers for Class 12 Biology Paper 1, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 12 Biology. Here we have given CBSE Sample Papers for Class 12 Biology Paper 3.

CBSE Sample Papers for Class 12 Biology Paper 3

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 3 of Solved CBSE Sample Paper for Class 12 Biology is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. There are total 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 Questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer Type Questions of three marks each.
5. Section D contains question number 23, Value Based Question of four mark.
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 examiner is to attempt any one of the question out of the two given in the question paper with the same question number.
8. No. of printed pages are three.

SECTION-A

Question 1.
Identify the types of immunization in case of injection of ready made antibodies for a tetanus case.

Question 2.
What are Homologous organs?

Question 3.
What is the sequence of nitrogen bases of the coding strand of DNA in a transcription unit?

Question 4.
Explain the term emasculation.

Question 5.
What is Adaptive radiation?

SECTION-B

Question 6.
How do sweet potato and potato differ in terms of evolution?

Question 7.
Identify the diagram and label the parts.

Question 8.
What are the barriers that comprise innate immunity?

Question 9.
What are the parts of the fallopian tube?

Question 10.
What is out-crossing?

OR

What is succession?

SECTION-C

Question  11.
What is the principle of Genetic equilibrium?

Question  12.
Draw a labelled diagram of a human sperm

Question  13.
List the salient features of DNA double helix model.

OR

Explain the parts of an ovule with a diagram

Question  14.
Explain convergent evolution with examples

Question  15.
What do you mean by withdrawal syndrome? Write the side effects of the use of anabolic steroids in males.

Question  16.
Explain some methods of Molecular Diagnosis.

Question  17.
Explain with examples, how do the plant-animal interactions involve co-evolution.

Question  18.
Cancer is one of the most dreaded diseases of human beings and is a major cause of death all over the globe. Explain the
(1) Causes of cancer
(2) Techniques of detection and diagnosis
(3) Treatment and cure

Question  19.
The rate of decomposition of detritus is affected by the abiotic factors like availability of oxygen, pH of the soil substratum, temperature etc. Discuss.

Question  20.
What are the different methods of breeding?

Question  21.
When is insulin fully functional?

Question  22.
You have identified a useful gene in a bacteria. Make a flow chart of the steps that you would follow to transfer this gene to a plant.

SECTION-D

Question  23.
A young couple quarrelled with the hospital authority on suspicion that their child had been exchanged after birth. The couple based their argument on the fact that their child is O blood group whereas they are A and B blood groups respectively. The doctor smiled and explained.
(a) What values of the doctor is reflected here?
(b) How can the child be O blood group as explained by the doctor?
(c) Which test method can be considered authentic to identify the biological parents of the child?
(d) Name the other blood group(s) which the child could have inherited.

SECTION-E

Question  24.
(a) What may be the probable reasons for greater biodiversity of tropics?
(b) Explain the importance of species diversity in reference to the “rivet popper hypothesis”.

OR

Give the journey of sperm formation with diagram. What are the hormones involved

Question  25.
In a medium where E.coli was growing, lactose was added, which induced the lac operon. Then why does the lac operon shut down after some time after the addition of lactose in the medium. Explain.

OR

Answer the following:
(a) Represent schematically the independent assortment of chromosomes.
(b) What are the requisites for a molecule to be a genetic material?

Question 26.
How do you represent the food and energy relationships between organisms?

OR

What are biogeochemical cycles? Explain the carbon cycle.

Answers

SECTION-A

Answer 1.
Passive immunization.

Answer 2.
The same structure developed along different directions due to adaptations to different needs. This is divergent evolution and these structures are homologous. Homology indicates common ancestry.

Answer 3.
5′ – AUGAAUG – 3′

Answer 4.
If the female parent bears bisexual flowers, removal of anthers from the flower bud before the anther dehisces using a pair of forceps is necessary. This step is referred to as emasculation.

Answer 5.
This process of evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography (habitats) is called adaptive radiation.

SECTION-B

Answer 6.
Sweet potato is a root modification and potato is a stem modification. Different structures evolving for same function are considered analogous structures and are examples of convergent evolution.

Answer 7.
Structure of microsporangium showing wall layers

Answer 8.
Innate immunity consists of four types of barriers. These are:

1. Physical barriers
2. Physiological barriers
3. Cellular barriers
4. Cytokine barriers

Answer 9.
Infundibulum with fimbriae, Ampulla, Isthmus that joins with the uterus.

Answer 10.
The practice of mating of animals within the same breed, but having no common ancestors on either side of their pedigree up to 4-6 generations. The offspring of such a mating is known as an out-cross.

OR

The gradual and fairly predictable change in the species composition of a given area is called Ecological succession. Example: hydrarch and xerarch succession.

SECTION-C

Answer 11.
Hardy-Weinberg principle states that allele frequencies in a population are stable and is constant from,generation to generation. The gene pool (total genes and their alleles in a population) remains constant. This is called genetic equilibrium. Sum total of all the allelic frequencies is 1. Individual frequencies, for example, can be named/), q, etc. In a diploide P and q represent the frequency of allele A and allele a. The frequency of AA individuals in a population is simply p₂ The probability that an allele A with a frequency of p appear on both the chromosomes of a diploid individual is simply the product of the probabilities, i.e.,p₂ Similarly of aa is q₂, of  Aa 2pq. Hence, p₂ + 2pq + q₂ = 1. This is a binomial expansion of (p + qf. When frequency measured differs from expected values, the difference (direction) indicates the extent of evolutionary change. Disturbance in genetic equilibrium, or Hardy-Weinberg equilibrium, i.e., change of frequency of alleles in a population would then be interpreted as resulting in evolution.

Answer 12.

Answer 13.
The salient features of the Double-helix structure of DNA are as follows:

1. It is made of two polynucleotide chains, where the backbone is constituted by sugarphosphate, and the bases project inside.
2. The two chains have anti-parallel polarity. It means, if one chain has the polarity 5′ – >3′, the other has 3′ – >5′.
3. The bases in two strands are paired through hydrogen bond (H-bonds) forming base pairs. Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa. Similarly, Guanine is bonded with Cytosine with three H-bonds. As a result, always a purine comes opposite to a pyrimidine. This generates approximately uniform distance between the two strands of the helix.
4. The two chains are coiled in a right-handed fashion. The pitch of the helix is 3.4 nm (a nanometre is one billionth of a metre, that is 10-9 m) and there are roughly 10 bp in each turn. Consequently, the distance between a bp in a helix is approximately equal to 0.34 nm.
5. The plane of one base pair stacks over the other in double helix. This, in addition to H-bonds, confers stability of the helical structure.

OR

The ovule is a small structure attached to the placenta by means of a stalk called funicle. The body of the ovule fuses with funicle in the region called hilum. Each ovule has one or two protective envelopes called integuments. Integuments encircle the ovule except at the tip where a small opening called the micropyle is organised. Opposite the micropylar end, is the chalaza, representing the basal part of the ovule. Enclosed within the integuments is a mass of cells called the nucellus. Cells of the nucellus have abundant reserve food materials. Located in the nucellus is the embryo sac or female gametophyte.

Answer 14.
Convergent evolution: Evolution of different structures for the same function and hence having similarity. The similar habitat that has resulted in selection of similar adaptive features in different groups of organisms but toward the same function. Examples are the eye of the octopus and of mammals, the flippers of Penguins and Dolphins. Sweet potato (root modification) and potato (stem modification). Similarities in proteins and genes performing a given function among diverse organisms give clues to common ancestry.

Answer 15.
The tendency of the body to manifest a characteristic and unpleasant withdrawal syndrome if regular dose of drugs/alcohol is abruptly discontinued. This is characterised by anxiety, shakiness, nausea and sweating, which may be relieved when use is resumed again. In some cases, withdrawal symptoms can be severe.

The side effects of the use of anabolic steroids in males are: Reduction of the size of testicles, decreased sperm production, potential for kidney and liver dysfunction, breast enlargement, premature baldness, enlargement of the prostate gland, and premature closure of the growth centres of the long bones which result in stunted growth, acne, increased aggressiveness, mood swing, depression.

Answer 16.
Some methods of Molecular Diagnosis are:

(1) Recombinant DNA Technology: It includes isolating the antibiotic resistance gene by cutting out a piece of DNA from a plasmid which was responsible for conferring antibiotic resistance.

(2) Polymerase Chain Reaction (PCR): In this a single stranded DNA or RNA is tagged with a radioactive molecule which is allowed to hybridise to its complementary DNA in a clone of cells followed by using autoradiography.

(3) ELISA: It is based on the principle of antigen-antibody interaction. Infection by pathogen can be detected by the presence of antigens (proteins, glycoroteins, etc.) or by detecting the antibodies synthesised against the pathogen

Answer 17.
Plant-animal interactions often involve co-evolution of the mutualists. For example, the evolutions of the flower and its pollinator species are tightly linked with one another.
In many species of fig trees, there is a tight one-to-one relationship with the pollinator species of wasp. It means that a given fig species can be pollinated only by its ‘partner’ wasp species and no other species. The female wasp uses the fruit’not only as on oviposition (egg-laying) site but uses the developing seeds within the fruit for nourishing its larvae. The wasp pollinates  the fig inflorescence while searching for suitable egg-laying sites. In return for the favour of pollination, the fig offers the wasp some of its developing seeds, as food for the developing wasp larvae.

Another example, is of the Mediterranean orchid Ophrys which employs ‘sexual deceit’ to get pollination done by a species of bee. One petal of its flower bears an uncanny resemblance to the female of the bee in size, colour and markings. The male bee is attracted to what it perceives as a female, ‘pseducopulates’ with the flower, and during that process is dusted with pollen from the flower. When this same bee ‘pseudocopulates’ with another flower, it transfers pollen to it and thus, pollinates the flower. 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

Answer 18.
(1) Causes of Cancer:

• Ionizing radiations like X-rays and gamma rays and non-ionizing radiations like UV.
• The chemical carcinogens present in tobacco smoke have been identified as a major cause of lung cancer.
• Oncogenic viruses have genes called viral oncogenes.

(2) Techniques of detection and diagnosis

• Biopsy and histopathological studies of tissues and blood.
• Bone marrow tests for increased cell counts in the case of leukemias.
• Use of antibodies against cancer-specific antigens.
• Techniques of molecular biology is used for detection of genes in individuals with inherited susceptibility to certain cancers.
• Radiography, CT and MRI for cancer of internal organs.

(3) Treatment and Cure:

• Surgery
• Radiation therapy: In radiotherapy, tumor cells are irradiated lethally, taking proper care of the normal tissues surrounding the tumor mass.
• Immunotherapy
• Chemotherapeutic drugs are used to kill cancerous cells. Some of these are specific for particular tumors.
• Use of a-interferons, which activate immune system to destroy tumor.

Answer 19.
Decomposition is largely an oxygen-requiring process. The rate of decomposition is controlled by chemical composition of detritus and climatic factors. In a particular climatic condition, decomposition rate is slower if detritus is rich in lignin and chitin, and quicker, if detritus is rich in nitrogen and water-soluble substances like sugars. Temperature and soil moisture are the most important climatic factors that regulate decomposition through their effects on the activities of soil microbes. Warm and moist environment favour decomposition whereas low temperature inhibits decomposition resulting in build up of organic materials.

Answer 20.
The different methods of breeding are:

1. Inbreeding
2. Out-breeding
3. Out-crossing
4. Cross-breeding and
5. Interspecific hybridisation

Answer 21.
Insulin is synthesised as a pro-hormone which like a pro-enzyme also needs to be processed before it becomes a fully mature and functional hormone and contains an extra stretch called the C peptide. This C peptide is not present in the mature insulin and is removed during maturation into insulin

Answer 22.

SECTION-D

Answer 23.
(a) The doctor was assertive, patient and pragmatic.
(b) A child can be of blood group O if the parents are heterozygotes, i.e., Ai x Bi. If the child receives i from both the parents, it becomes, ii, and expresses O blood group.
(c) DNA finger printing
(d) A or B or AB.

SECTION-E

Answer 24.
(a) Reasons for the greater biodiversity of tropics.

1. Speciation is generally a function of time, unlike temperate regions subjected to frequent glaciations in the past, latitudes have remained relatively undisturbed for millions of years and thus, had a long evolutionary time for species diversification.
2. Tropical environments, unlike temperate ones, are less seasonal, relatively more constant and predictable. Such constant environments promote niche specialisation and lead to a greater species diversity.
3. There is more solar energy available in the tropics, which contributes to higher productivity; this in turn might contribute indirectly to greater diversity.

(b) The rivet popper hypothesis was used by Paul Ehrlich. It states that in an airplane (ecosystem) all parts are joined together using thousands of rivets (species). If every passenger travelling in it starts popping a rivet to take home (causing a species to become extinct), it may not affect flight safety (proper functioning of the ecosystem) initially, but as more and more rivets are removed, the plane becomes dangcrousl y- weak over a period of time. Furthermore, which rivet is removed may also be critical. Loss of rivets on the wings (key species that drive major ecosystem functions) is obviously a more serious threat to flight safety than loss of a few rivets on the seats or windows inside the plane.
OR

In testis, the immature male germ cells (spermatogonia) produce sperms by spermatogenesis that begins at puberty.
The spermatgonia (sing, spermatogonium) present on the inside wall of seminiferous tubules multiply by mitotic division and increase in numbers. Each spermatogonium is diploid and contains 46 chromosomes. Some of the spermatogonia called primary spermatocytes periodically undergo meiosis.

A primary spermatocyte completes the first meiotic division (reduction division) leading to formation of two equal, haploid cells called secondary spermatocytes, which have only 23 chromosomes each. The secondary spermatocytes undergo the second meiotic division to produce four equal, haploid spermatids. The spermatids are transformed into spermatozoa (sperms) by the process called spermiogenesis. After spermiogenesis, sperm heads become embedded in the Sertoli cells, and are finally released from the seminiferous tubules by the process called spermiation.

The hormones involved are:

• Gonadotropin Releasing Hormone (GnRH)
• Luteinising Hormone (LH)
• Follicle Stimulating Hormone (FSH

Answer 25.
The lac operon consists of one regulatory gene (the i gene) and three structural genes (z, y, and a).
The (i gene) codes for the repressor of the lac operon. The z gene codes for beta-galactosidase (β-gal), which is primarily responsible for the hydrolysis of the disaccharide, lactose into its monomeric units, galactose and glucose. The y gene codes for permease, which increases permeability of the cell to
β-galactosides. The a gene encodes a transacetylase. Hence, all the three gene products in lac operon are required for metabolism of lactose.

Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. Hence, it is termed as inducer. In the absence of a preferred carbon source such as  glucose, if lactose is provided in the growth medium of the bacteria, the lactose is transported into the cells through the action of permease. The lactose then induces the operon in the following manner. The repressor of the operon is synthesised (all-the-time – constitutively) from the i gene.

The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon. In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase access to the promoter and transcription proceeds. Essentially, regulation of lac operon can also be visualised as regulation of enzyme synthesis by its substrate. Glucose or galactose cannot act as inducers for lac operon. Thus, regulation of lac operon by repressor is referred to as negative regulation.

(b) The requisites for a molecule to be a genetic material are:

• It should be able to generate its replica (Replication).
• It should chemically and structurally be stable.
• It should provide the scope for slow changes (mutation) that are required for evolution.
• It should be able to express itself in the form of ‘Mendelian Characters’.

Answer 26.
The food or energy relationship is expressed in terms of number, biomass or energy. The base of each pyramid represent the producer or the first trophic level while the apex represents tertiary or top level consumer. The three ecological pyramids that are usually studied are:
(a) pyramid of number;
(b) pyramid of biomass and
(c) pyramid of energy
Calculations of any energy content, biomass, or numbers must include all organisms at that trophic level. No generalisations can be true with a few individuals at any trophic level into account. Also a given organism may occupy more than one trophic level simultaneously. One must remember that the trophic level represents a functional level, not a species as such. A given species may occupy more than one trophic level in the same ecosystem at the same time; for example, a sparrow is a primary consumer when it eats seeds, fruits, peas, and a secondary consumer when it eats insects and worms.

In most ecosystems, all the pyramids, of number, of energy and biomass are upright, i.e., producers are more in number and biomass than the herbivores, and herbivores are more in number and biomass than the carnivores. Also energy at a lower trophic level is always more than at a higher level. .

The pyramid of biomass in sea is also generally inverted because the biomass of fishes far exceeds that of phytoplankton. Pyramid of energy is always upright, can never be inverted, because when energy flows from a particular trophic level to the next trophic level, some energy is always lost as heat at each step. Each bar in the energy pyramid indicates the amount of energy present at each trophic level in a given time or annually per unit area. There are certain limitations of ecological pyramids such as it does not take into account the same species belonging to two or more trophic levels. It assumes a simple food chain, something that almost never exists in nature; it does not accommodate a food web. Moreover, saprophytes are not given any place in ecological pyramids even though they play a vital role in ecosystem.


The movement of nutrient elements through the various components of an ecosystem is called nutrient cycling or biogeochemical cycles.
Nutrient cycles are of two types:
(a) gaseous and
(b) sedimentary.
Carbon cycling occurs through atmosphere, ocean and through living and dead organisms. A considerable amount of carbon returns to the atmosphere as CO₂ through respiratory activities of the producers and consumers. Decomposers also contribute substantially to CO₂ pool by their processing of waste materials and dead organic matter of land or oceans. Some amount of the fixed carbon is lost to sediments and removed from circulation. Burning of wood, forest fire and combustion of organic matter, fossil fuel, volcanic activity are additional sources for releasing CO₂ in the atmosphere.

We hope the CBSE Sample Papers for Class 12 Biology Paper 3 help you. If you have any query regarding CBSE Sample Papers for Class 12 Biology Paper 3, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 12 Physics Paper 3 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 3.

CBSE Sample Papers for Class 12 Physics Paper 3

Time Allowed : 3 Hours

Max. Marks : 70

General Instructions 

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

Questions :
SECTION : A

Question 1.
In which orientation, a dipole placed in a uniform electric field is in
(i) stable,
(ii) unstable equilibrium?

Question 2.
Which part of electromagnetic spectrum has largest penetrating power?

Question 3.
A plot of magnetic flux (Φ) versus current (I) is shown in the figure  for two inductors A and B. Which of the two has larger value of self inductance.

Question 4.
Figure shows three point charges, + 2q, -q and +3q. Two charges + 2q and -q are enclosed within a surface ‘S’. What is the electric flux due to this configuration through the surface ‘S’?

Question 5.
A glass lens of refractive index 1.45 disappears  when immersed in a liquid. What is the value of refractive index of the liquid.

SECTION : B

Question 6.
What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom?

Question 7.
A wire of resistance 8R,is bent in the form of a circle. What is the effective resistance between the ends of a diameter AB?

Question 8.
State the conditions for the phenomenon of total internal reflection to occur.

Question 9.
Explain the function of a repeater in a communication system.

Question 10.
(i) Write two characteristics of a material used for making permanent magnets.
(ii) Why is core of an electromagnet made of ferromagnetic materials?

OR

Draw magnetic field lines when a
(i) diamagnetic,
(ii) paramagnetic substance is placed in an external magnetic field. Which magnetic property distinguishes this behaviour of the field lines due to the two substances?

SECTION : C

Question 11.
Draw a circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to measure light intensity?

Question 12.
An electric lamp having coil of negligible inductance connected in series with a capacitor and an AC source is glowing with certain brightness. How does the brightness of the lamp change on reducing the
(i) capacitance, and
(ii) the frequency? Justify your answer.

Question 13.
Arrange the following electromagnetic radiations in ascending order of their frequencies :

1. Microwave
2. Radio wave
3. X-rays
4. Gamma rays

Write two uses of any one of these.

Question 14.
The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length of the lens is 12 cm, find the refractive index of the material of the lens.

Question 15.
An electron is accelerated through a potential difference of 100 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?

Question 16.
A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 MeV per nucleon. Calculate the energy Q released per fission in MeV.

Question 17.

1. The bluish colour predominates in clear sky.
2. Violet colour is seen at the bottom of the spectrum when white light is dispersed by a prism. State reasons to explain these observations.

Question 18.
Plot a graph showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions W₁ and W₂ (W₁ > W₂).
On what factors does the
(i) slope and
(ii) intercept of the lines depend?

Question 19.
A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will
(i) the capacitance of the capacitor,
(ii) potential difference between the plates and
(iii) the energy stored in the capacitor be affected?
Justify your answer in each case.

Question 20.
Write the principle of working of a potentiometer. Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a given cell.

Question 21.

Write the expression for the magnetic moment ( m ) due to a planar square loop of side l carrying a steady current I in a vector form. In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying a steady current I₁ at a distancel as shown.
Given reasons to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop.

Question 22.
(a) Depict the equipotential surfaces for a system of two identical positive point charges placed at a distance d apart.
(b)

SECTION : E

Question 23.
For the last four months, the English teacher of Komal had erratic body movement, lack of coordination etc in her daily activities. Occasionally she had headache also. Komal suggested to get a MRI and other complete medical check up. After that she was diagnosed a brain tumour.

1. What values are displayed by Komal?
2. How can radioisotopes help a doctor to diagnose brain tumour?

Question 24.
Describe briefly, with the help of a labelled diagram, the basic elements of an A.C. generator. State its underlying principle. Show diagrammatically how an alternating emf is generated by a loop of wire rotating in a magnetic field. Write the expression for the instantaneous value of the emf induced in the rotating loop

OR

A series LCR circuit is connected to an ac source having voltage V = V_m sin cot. Derive the expression for the instantaneous current T and its phase relationship to the applied voltage. Obtain the condition for resonance to occur. Define ‘power factor’. Sate the conditions under which it is
(i) maximum and
(ii) minimum.

Question 25.
State Huygens principle. Show, with the help of a suitable diagram, how this principle is used to obtain the diffraction pattern by a single slit. Draw a plot of intensity distribution and explain clearly why the secondary maxima become weaker with increasing order (n) of the secondary maxima.

OR

Draw a ray diagram to show the working of a compound microscope. Deduce an expression for the total magnification when the final image is formed at the near point. In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, estimate the magnifying power of the microscope.

Question 26.
(a) Explain the formation of depletion layer and potential barrier in a p-n junction output.
(b) In the figure given below the input waveform is converted into the output waveform by a device ‘X’. Name the device and draw its circuit diagram.

(c) Identify the logic gate represented by the circuit as shown and write its truth table.

OR

(a) Draw the circuit diagram of a common emitter transistor as an amplifier, define current gain and voltage gain of it.
(b) Distinguish between a conductor, a semiconductor and an insulator on the basis of energy band diagrams.

CBSE Sample Papers for Class 12 Physics Paper 4 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 4.

CBSE Sample Papers for Class 12 Physics Paper 4

Time Allowed : 3 Hours

Max. Marks : 70

General Instructions 

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

Questions
SECTION : A

Question 1.
Show graphically, the variation of the de-Broglie wavelength (λ) with the potential (V) through which an electron is accelerated from rest.

Question 2.
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 V in terms of ε, V and R.

Question 3.
A plane electromagnetic wave travels in vacuum along Z-direction. What can you say about the direction of electric and magnetic field vector?

Question 4.
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 Q is
(i) positive
(ii) negative?

Question 5.
The permeability of magnetic material is 0.9983. Name the type of magnetic materials it represents.

SECTION : B

Question 6.
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

Question 7.
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 8.
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 9.
Figure shows two identical capacitors, C₁ and C₂, each of 1μF 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 10.
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 11.
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.

Question 12.
(i) Define modulation index.
(ii) Why is the amplitude of modulating signal kept less than the amplitude of carrier wave?

Question 13.
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.

Question 14.
Define the terms
(i) drift velocity,
(ii) relaxation time. A conductor of length L is connected to a dc source of emf ε. 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?

Question 15.
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.

Question 16.
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 17.
How are infrared waves produced? Why are these referred to as ‘heat wave’? Write their one important use.

Question 18.
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.

Question 19.
An electron and a photon each have a wave lenght of 1.00 nm.
Find
(a) their momenta,
(b) the energy of the photon and
(c) the kinetic energy of electron

Question 20.
Describe Young’s 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.

Question 21.
(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 unpolarized light passes from air to a transparent medium, under what condition does the refracted light get polarized?

Question 22.
The energy levels of a hypothetical atom are shown. 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?

SECTION : D

Question 23.
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.

SECTION : E

Question 24.
(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.

Question 25.
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 axis. 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. Explain 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.

Question 26.
(a) Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism.Deduce thee xpression 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.

OR

(a) Obtain Lens Maker formula using the expression

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.

 Answers :
SECTION : A

Answer 1.

Answer 2.

Answer 3.
In electromagnetic wave, the electric field vector E and magnetic field vector B 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.

Answer 4.
If Q is positively charged, V_A-V_B = positive. If Q is negatively charged, V_A-V_B = negative.

Answer 5.
Paramagnetic material (0 < μ < 1).

SECTION : B

Answer 6.
Downward force due to gravity is balanced by upward force due to electric field.


When this dipole is placed in an electric field along + X-axis, it will experiences a torque.

According to Right Hand Thumb Rule, torque will be into the plane of the paper.

Answer 7.

∴  The telescope with objective of aperture A₁ should be preferred for viewing as this would :

1. give a better resolution.
2. have a higher light gathering power of telescope.

Answer 8.
Given,

Answer 9.
When S is closed: Potential difference across C₁  = Potential difference across C₂ = 6V
V₁ = V₂ = 6V
∴ q₁ = q₂ = 1μF x 6V = 6μC
When S is open and dielectric slab (K = 3) are put in between plates of capacitors,
C₁ = 3 x 1μF = 3μF
C₁ = 3 x 1μF = 3μF
When S is open, potential difference across C₁ , V₁= 6V
∴ q₁ = 3μF x 6V = 18μC
Potential difference across C₂

Answer 10.

The logic operation performed by this circuit is OR.

SECTION : C

Answer 11.

In the active region, a (small) increase of V_i results in a (large, almost linear) increase in I_c. This results in an increase in the voltage drop across R.

Answer 12.
(i) Modulation index is defined as the ratio of amplitude of modulating signal to amplitude of carrier wave. It is given by

A_m = amplitude of modulation wave
A_c = amplitude of carrier wave
(ii) The amplitude of modulating signal is kept less than the amplitude of carrier wave to avoid noise.

Answer 13.
(a) Any two ways to obtain large deflection in G :

• Moving C₂ faster towards C₁.
• Insertion of soft iron core in C₁
  

(b) Alternative device that can be used in place of galvanometer is LED.

Answer 14.
(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.
The drift velocity will be inversely proportional to L, i.e.,

Answer 15.
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/ decrease of the voltage drop across series resistance without any change in the voltage across Zener diode,

Answer 16.
(i) Standard AM Broadcast : 540-1600 kHz
(ii) Television : 54-72 MHz VHF (very high frequencies); 76-88 MHz TV; 174-216 MHz UHF (ultra high frequencies); 420-890 MHz TV
(iii) Satellite communication : 5.925-6.425 GHz uplink and 3.7-4.2 GHz downlink.

Answer 17.
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 earth’s warmth, in physiotherapy, remote switches etc.

Answer 18.
Electric field at point P at a distance r out side the spherical shell.

According to gauss law,

image

Answer 19.
λ_e =λ_photon = 1.00 nm= 10^-9 m
(a) For electron or photon, momentum

Answer 20.

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 slits 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 superimpose on each other, dark fringes appear on the screen when the crest of one wave 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.
Light waves spread out from S and fall on both S₁ and S₂. The spherical waves emanating from S₁ and S₂ will produce interference fringes on the screen MN.


Now, the intensity at point P is maximum or minimum accordingly 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 (second law)
First Law, incident wavefront, refracted wave front, normals all lie in the same plane.

Answer 21.

(a) When a polaroid P₁ is rotated in the path of an unpolarised light, there is no change in transmitted intensity. 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.

Answer 22.
If a photon of wavelength λ = 275nm is to be emited , they energy of photon is given by

hence transition B would result in the emission of a photon of wavelenght 275nm.
(i) Transition A corresponds to max wavelenght due to least energy change transition.
(ii) Transition D corresponds to min wavelenght due to least energy change transition.

SECTION : D

Answer 23.
R₂  R₃  R₄  are in parallel combination

SECTION : E

Answer 24.
(i) Step-Up Transformer :

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 (Φ) be the flux in each turn in the core at any time t due to current in the primary when a voltage V_p is applied to it.
Then the induced emf or voltage E_ç, in the secondary with N_s, turns is

The alternating flux also induces an emf, called back emf in the primary given by

But E_p = V_p (since resistance of primary is small) and E_s. = V_s (since secondary current is small)

(i) Sources of energy loss in transformer are : Heat loss due to resistance of windings, loss due to eddy currents.
(ii) A step up transformer steps up the voltage while it steps down the current. So the input and output power remain same (provided there is no loss). Hence there is no violation of the principle of energy conservation.

OR

The effective resistance offered by a series LCR circuit to the flow of current is called its impedance. It is denoted by Z. Suppose an inductance L, capacitance C and resistance R are connected in series to a source of alternating emf, V = V₀ sin ωt
Let I be the instantaneous value of current in series circuit. Then voltages across the three components are

• V_L = X_L It is ahead of current I in phase by 90°
• V_c = X_C It lags behind the current I in phase by 90°
• V_R = RI. It is in phase with current I.

These voltages are shown in the phasor diagram given below :

As V_L and V_c are in opposite direction, their resultant is OD = V_L – V_c, in the positive y-direction.

Here, V/ I is the effective resistance of the series LCR circuit and is called its impedance (Z).

The given graph showing variation of current 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 matches 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.

Answer 25.
Biot-Savart Law states that the magnitude of magnetic Y field dB due to current element is directly proportional to the current I,


Explanation :
A circular current loop produces magnetic field and its magnetic moment is the product of current and its area.
M = IA

OR

Principle :
Acharged particle can be accelerated to high energy by making it cross the same electric field again and again using a perpendicular magnetic field.

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 traces 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, needed by the charged particle to move in circular track, is provided by the magnetic field.

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

Answer 26.
(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.

OR

(a)

The point, where image of an object, located at infinity is formed, is called the focus F, of the lens and the distance f gives its focal length.

We hope the CBSE Sample Papers for Class 12 Physics Paper 4 help you. If you have any query regarding CBSE Sample Papers for Class 12 Physics Paper 4, drop a comment below and we will get back to you at the earliest.

These Sample papers are part of CBSE Sample Papers for Class 12 Biology. Here we have given CBSE Sample Papers for Class 12 Biology Paper 2.

CBSE Sample Papers for Class 12 Biology Paper 2

Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 2 of Solved CBSE Sample Paper for Class 12 Biology is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

1. There are total 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 Questions of two marks each.
4. Section C contains question number 11 to 22, Short Answer Type Questions of three marks each.
5. Section D contains question number 23, Value Based Question of four mark.
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 examiner is to attempt any one of the question out of the two given in the question paper with the same question number.
8. No. of printed pages are three.

SECTION-A

Question 1.
Observe the pedigree chart and answer the following questions:
(a) Identity whether the trait is sex linked or autosomal.
(b) Give an example of a disease in human beings which shows such a pattern of inheritance.

Question 2.
Identify the reason for the selection of DNA polymerase from Thermus aquaticus for Polymerase Chain Reaction

Question 3.
Govt, of India has raised the marriageable age of females to 18 yrs and of males 21 yrs. Suggest any two measures adopted by the government for the purpose.

Question 4.
Thymus of a new born child was degenerating right from birth due to genetic disorder. Predict its two impacts on the health of the child.

Question 5.
Give an example of a chromosomal disorder caused due to non-disjunction of autosomes

SECTION-B

Question 7.
A single pea plant in your kitchen garden produces pods with seeds, but the individual papaya plant does not. Explain.

Question 8.
Suggest four important steps to produce a disease resistant plant through conventional plant breeding technology.

Question 9.
Name a genus of baculovirus. Why are they considered good biocontrol agents?

Question 10.
Explain the relationship between CFCs or Ozone in the stratosphere.

OR

Why are sacred groves highly protected?

SECTION-C

Question 11.
Why is breast feeding recommended during the initial period of an infant’s growth? Give reasons.

Question 12.
Give an example of an autosomal recessive trait in humans. Explain its pattern of inheritance with the help of a cross.

Question 13.
Describe the experiment that helped Louis Pasteur to dismiss the theory of spontaneous generation of life

Question 14.
Plant breeding technique has helped sugar industry in North India. Explain how.

Question 15.
Suggest and describe a technique to obtain multiple copies of a gene of interest in vitro.

Question 16.
What is GMO? List any five possible advantages of a GMO to a farmer.

Question 17.
During a school trip to ‘Rohtang Pass’, one of your classmate suddenly developed “altitude sickness”. But, she recovered after sometime.
(a) Mention one symptom to diagnose the sickness.
(b) What caused the sickness?
(c) How could she recover by herself after sometime?

Question 18.
How has RNAi technique help to prevent the infestation of roots into tobacco plants by nematode Meloidegyne incognitia?

Question 19.
“In a food chain, a trophic level represents a functional level, not a specie.” Explain. 3

OR

(a) Name any two places where it is essential to install electrostatic precipitators. Why is it required to do so?
(b) Mention one limitation of the electrostatic precipitator

Question 20.
Prior to a sports event, blood and urine sample of sports person are collected for drug tests.3
(a) Why is there a need to conduct such tests?
(b) Name the drugs the authorities usually look for.
(c) Write the generic names of two plants from which these drugs are obtained.

Question 21.
Describe the experiment that helped to demonstrate the semi conservative mode of DNA replication.

Question 22.
Given below is a list of six micro-organisms. State their usefulness to humans:
(a) Nucleopolyhedrovirus
(b) Saccharomycs cerebisiae
(c) Monascus purpureus
(d) Trichoderma polysporum
(e) Penicillium notatum
(f) Propionibacterium sharmanii

SECTION-D

Question 23.
Reproductive and Child Flealthcare (RCH) programme are currently in operation. One of the majof tasks of these programmes is to create awareness amongst people about the wide range of reproduction related aspects. As this is important and essential for building a reproductively healthy society.
(a) “Providing sex education in schools is one of the ways to meet this goal.” Give four points in support of your opinion regarding this statement.
(b) List any two ‘indicators’ that indicate a reproductively healthy society.

SECTION-E

Question 24.
(a) Explain the post-pollination events leading to seed production in angiosperms.
(b) List the different types of pollination depending upon the source of pollen grains.

OR

(c) Briefly explain the events of fertilisation and implantation in an adult human female.
(d) Comment on placenta as an endocrine gland.

Question 25.
(a) How are the following formed and involved in DNA packing in a nucleus of a cell? 5

• Histone octomer
• Nucleosome
• Chromatin

(b) Differentiate between Euchromatin and Heterochromatin.

OR

Explain the role of lactose as an inducer in a lac operon

Question 26.
(a) Why should we conserve biodiversity? How can we do it?
(b) Explain the importance of biodiversity hot spots and sacred groves.

OR

(a) Represent diagrammatically three kinds of age-pyramids for human populations.
(b) How does an age pyramid for human population at given point of time help the policy makers in planning for future

Answers

SECTION-A

Answer 1.

(a) Sex-linked.

(b) Haemophilia/Colourblindness.

Answer 2.
Identify the reason for the selection of DNA polymerase from Thermus aquaticus for Polymerase Chain Reaction.

Answer 3.
Incentives given to couples with small families/media publicity – posters of happy couples with two children (slogan-Hum Do Humare Do)/Motivate to promote smaller families by using contraceptive methods. (Any two)

Answer 4.
Thymus provides micro-environment for the development and maturation of T-lymphocytes; its degeneration will weaken the immune system so the child will be prone to frequent infections.

Answer 5.
Down’s Syndrome.
SECTION-B

Answer 6.
Pea: Flowers of pea plants are bisexual, monoecious / self pollinated (to produce pods with viable seeds.

Papaya: Dioecious plant/unisexual plant bearing male and female flowers on separate plants, unable to produce viable seeds as there is no cross pollination/it could be a male plant which is unable to produce fruit and seeds.

Answer 7.
Stop codon: does not code for any amino acid/terminates the synthesis of polypeptide chain. Unambiguous codon: one codon codes for one amino acid only
Degenerate codon: some amino acids are coded by more than one codon
Universal codon: genetic code is same for all organisms (bacteria to humans)

Answer 8.
Steps for producing disease resistant plants:

1. Screening of germplasm (for resistance sources)
2. Hybridization of selected parents !
3. Selection and evaluation of hybrids
4. Testing and release of new varieties

Answer 9.
Nucleopolyhedrovirus.
Species specific, narrow spectrum insecticidal application, no negative impact on non target organisms.

Answer 10.
UV rays act on CFC’s, release Cl atoms, which act on ozone to release O₂, resulting in ozone layer depletion/causing ozone hole.

 OR

Sacred groves are highly protected because of religious and cultural traditions, refuge for large number of rare and threatened plants/ecologically unique and biodiversity rich regions.

SECTION-C

Answer 11.
Colostrum, rich in nutrients, rich in antibodies/rich in IgA/provide passive immunity/provides :immunity to new born/helps to develop resistance in new born/readily available for new born? hygienic/develops a bond between mother and child.

Answer 12.
Sickle cell anaemia/Phenylketonuria/Thalassemia/O Blood group /Non-rolling of tongue / fused or attached ear lobes / inability to taste PTC (phenyl thiocarbamide).

Similar cross can be considered for any other trait mentioned above.

Answer 13.

Two pre sterilised flasks with killed yeast, one sealed, other open to air, differential growth of life in two flasks/life was found only in open flask. The given diagram can be considered in lieu of above explanation.
Life comes from pre-existing life (it came from air entering the flask)/proved the theory of biogenesis.

Answer 14.
Two species (Saccharum barberi and Saccharum officinarum) were crossed to get sugarcane varieties with high yield/thick stem/high sugar content/ability to-grow in North India.

Answer 15.
PCR/Polymerase Chain Reaction.
Separation/ denaturation of two strands of two dsDNA, using two sets of primers / small chemically synthesised oligonucleotides complementary to regions of DNAand (thermostable) DNA polymerase/Taq polymerase, extension of the primers, by enzyme using nucleotides replicates the DNA and if the process of replication is repeated many times, multiple copies of DNA are produced.
The following diagram can be considered in lieu of the explanation.

Answer 16.
1. Plants/bacteria/ fungi/ animals whose genes have been altered by manipulation.
2. Tolerance to abiotic stresses/like cold/drought/salt/heat, reduced reliance on chemical pesticides/pest resistant crops, reduce post harvest losses, increased efficiency of mineral usage by plants, enhanced nutritional value, to create tailor made plant.

Answer 17.
(a) Nausea/fatigue/heart palpitation
(b) Low atmospheric pressure at high altitude, body deprived of O₂.
(c) Increase in RBC, decreases binding capacity of haemoglobin, increased breathing rate, get acclimatised.

Answer 18.
Using Agrobacterium vectors, nematode specific genes introduced into host plant, produced sense antisense RNA in host cells, ds RNA – initiated RNAi, silenced specific mRNA of nematode, parasite could not survive in transgenic host.

Answer 19.
Position of a species in any trophic level is determined by the function performed by that mode of nutrition of species in a particular food chain. A given species may occupy more than one trophic level in the same ecosystem (in different food chains) at the given time. If the function of the mode of nutrition of species changes its position shall change in the trophic levels, same species can be at primary consumer level in one food chain and at secondary consumer level in another food chain in the same ecosystem at the given time.

OR

(a) Thermal power plants/smelters/ other particulate matter releasing industries.
To remove particulate matter (Any two)
(b) Very small particulate matter/less than 2.5 micrometres are not removed/velocity of air between plates must be low enough to allow the dust to fall/cannot work without electricity.

Answer 20.
(a) To detect drug abuse/use of banned drugs/use of cannabinoids/anabolic steroids/narcotic analgesic/diuretics/hormones/drugs used to accelerate performance / increase muscle strength/bulk/promote aggressiveness/to ensure fair game.
(b) Cannabinoids/cocaine/coca alkaloid / coke /crack/ hashish /charas/ganj a /hemp plant extract.
(c) Cannabis/Atropa/Erythroxylum/Datura. (Any two)

Answer 21.
Matthew Meselson and Franklin Stahl grew E.coil in ¹⁵NH₄C1 for many generations to get ¹⁵N incorporated into DNA. Then the cells are transferred into ¹⁴NH₄C1. The extracted DNA are centrifuged in CsCl and measured to get their densities, DNA extracted from the culture after one generation (20 minutes), showed intermediate hybrid density, DNA extracted after two generations (40 minutes) showed light DNA and hybrid DNA.

A correctly labelled diagrammatic representation in lieu of the above explanation of experiment to be considered.

Answer 22.
(a) As bio control agents / species specific / narrow spectrum insecticidal application / no negative impacts on plants / mammals / birds / fish / non target insects / Integrated Pest Management.
(b) Used in bread making/brewing industry/ethanol/CO, production
(c) Cholesterol lowering agent/competitively inhibiting the enzyme responsible for synthesis of cholestrol
(d) Produces cyclosporin-A/immuno suppressive agent
(e) Produces antibiotic penicillin
(f) Produces large holes in swiss cheese/products large amount of CO₂ in swiss cheese.

SECTION-D

Answer 23.
(a)

• Provide right information to the young so as to discourage children from believing in myths and
• misconception about sex related aspects.
• Proper information about reproductive organs
• Proper information about adolescence and related changes
• Safe hygienic practices
• STDs/AIDS
• Available birth control options
• Care of pregnant mothers
• Post natal care
• Importance of breast feeding
• Equal opportunities for male and female child
• Awareness of problems due to uncontrolled population growth
• Sex abuse
• Sex related crimes (Any four)

(b) Better awareness about sex related matters/increase in number of assisted deliveries/better post natal care/decrease in IMR (Infant Mortality Rate)/decrease MMR(Matemal Mortality Rate)/increase in number of couples with small families/better detection and cure of STDs/ overall increased medical facilities for sex related problems/total well being in all aspects of reproduction/physical-behavioural-social /physically and functionally normal reproductive organs/normal emotional and behavioural interaction among all sex related aspects. (Any two)

SECTION-E

Answer 24.
(a) Pollen pistil interaction, germination of pollen tube that carries two male gametes, double
fertilization / syngamy and triple fusion, development of endosperm, development of embryo, maturation of ovule into seed.
(b) Autogamy/self pollination/Geitonogamy Xenogamy/cross pollination

OR

(a) Fertilization:

• Sperm comes in contact and enters the secondary oocyte
• Activates / induces secondary oocyte to complete meiosis II leads to formation of ovum/ootid
• The haploid nucleus of sperm and that of ovum fused to form a diploid zygote completing the process of fertilization

Implantation:

• Trophoblast layer of blastocyst attaches to the endometrium (of the uterus)
• The uterine cells divide rapidly and cover the blastocyst,
• The blastocyst becomes embedded in the endometrium and the implantation is

Answer 25.
(a)
1. Eight molecules of (positively charged basic proteins called) histones are organised to form histone octamer.
2. Negatively charged DNA wrapped around positively charged histone octamer to give rise to nucleosome
3. Nucleosome constitute the repeating unit of a structure called chromatin

(b)

OR

Lactose/inducer binds with repressor protein, inactivates it, frees operator gene, RNA polymerase freely move over structural genes/RNA polymerase access to the promoter, transcribing to, lac mRNA, which on translation, produce transacetylase, permease, β-galactosidase.
The following diagram to be considered in lieu of above explanation.

Answer 26.
(a)
(1) Narrowly utilitarian-related examples like derive economic benefits from nature food (cereals, pulses, fruits)/firewood/fibre/construction materials/industrial products (tannins, lubricants, dyes, resi’ns, perfumes) / product of medicinal importance/drugs.

• Broadly utilitarian — 20% of total 0₂ from Amazon forests / pollination / aesthetic pleasures.
• Ethical — millions of species (plants, animals, microbes) share this planet/we need to realise that every species has an intrinsic value (even if it may not have any economic value to us)/we have a moral duty to care for their well being and pass on our biological legacy to future generations.

(2)

• In situ conservation / biosphere reserves/ national parks / sanctuaries / sacred groves.
• Ex situ conservation/zoological parks /botanical gardens/wild life safari parks/cryopreservation/seed banks/tissue culture (eggs in vitro).

(b) Hot spots – regions with high level of species richness, high degree of endemism Sacred groves – tracts of forest containing tree/wildlife were venerated, and given total protection/to protect last refuges for a large number of rare and threatened plants.

OR

(b) Planning of health/ education/transport/ infrastructure/finance/food/employment can depend on the age-pyramid analysis of a population. (Any two explanations)

We hope the CBSE Sample Papers for Class 12 Biology Paper 2 help you. If you have any query regarding CBSE Sample Papers for Class 12 Biology Paper 2, drop a comment below and we will get back to you at the earliest.

CBSE Sample Papers for Class 12 Physics Paper 3 are part of CBSE Sample Papers for Class 12 Physics. Here we have given CBSE Sample Papers for Class 12 Physics Paper 3.

CBSE Sample Papers for Class 12 Physics Paper 3

Time Allowed : 3 Hours

Max. Marks : 70

General Instructions 

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

Questions :
SECTION : A

Question 1.
If the angle between the pass axis of polarizer and the analyser is 45°, write the ratio of the ‘ intensities of original light and the transmitted light after passing through the analyser.

Question 2.
You are given following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope?

Question 3.
Write the following electromagnetic waves in ascending order with respect to their frequencies: X-rays, microwaves, UV rays and radio waves.

Question 4.
What is sky wave propagation?

Question 5.
Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why?

SECTION : B

Question 6.
The figure shows a plot of three curves a, b, c, showing the variation of photocurrent v/s collector plate potential for three different intensities I₁ I₂ and I₃ having frequencies v₁ v₂ and v₃ respectively incident on a photosensitive surface. Point out the two curve for which the incident radiations have same frequency but different intensities.

Question 7.
What type of wavefront will emerge from a
(i) point source, and
(ii) distant light source?

Question 8.
Two nuclei have mass numbers in the ratio 1 : 2. What is the ratio of their nuclear densities?

Question 9.
A cell of emf ‘E’ and internal resistance ‘V’ is connected across a variable resistor ‘R’ Plot a graph showing the variation of terminal potential ‘V’ with resistance R. Predict from the graph the condition under which ‘V’ becomes equal to ‘E’.

Question 10.
(i) Can two equipotential surfaces intersect with each other? Give reasons.
(ii) Two charges -q and +q are located at points A (0,0, -a) and B (0,0, +a) respectively. How much work is done in moving a test charge from point P (7, 0, 0) to Q (-3, 0, 0)?

SECTION : C

Question 11.
A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme :

The mass number and atomic number of A are 180 and 72 respectively. What are these numbers for A_{4 ?}

Question 12.
An electron and a proton are accelerated through the same potential. Which one the two has
(i) greater value of de Broglie wavelength associated with it and
(ii) less momentum? Justify your answer.

Question 13.
By what percentage will the transmission ranges of TV tower be affected when the height of the tower is increased by 21 %?

Question 14.
Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time.

Question 15.
In a single slit diffraction experiment, when tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why? State two points of difference between the interference pattern obtained in Young’s double slit experiment and the diffraction pattern due to a single slit.

Question 16.
A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell. Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∝.

Question 17.
Why are high frequency carrier waves used for transmission?

OR

What is meant by term modulation’? Draw a block diagram of a simple modulator for obtaining an AM signal.

Question 18.
Three identical capacitors C₁, C₂ and C₃ of capacitance 6 μF each are connected to a 12 V battery as shown :

Find
(i) charge on each capacitor
(ii) equivalent capacitance of the network
(iii) energy stored in the network of capacitors

Question 19.
(a) The energy levels of an atom are as shown below. Which of them will result in the transition of a photon of wavelength 275 nm?

(b) Which transition corresponds to emission of radiation of maximum wavelength?

Question 20.
(a) Define self inductance. Write its S.I. units.
(b) Derive an expression for self inductance of a long solenoid of length l, cross-sectional area A having N number of turns.

Question 21.
The figure shows experimental set up of a meter bridge. When the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance of 10Ω is connected in series with X, the null point shifts by 10 cm.Find the position of the null point when the 10Ω resistance is instead connected in series with resistance ‘Y’. Determine the values of the resistances X and Y.

Question 22.
Derive the expression for force per unit length between two long straight parallel current carrying conductors. Hence define one ampere.

OR

Explain the principle and working of a cyclotron with the help of a schematic diagram. Write the expression for cyclotron frequency.

SECTION : D

Questin 23.
Shivani was interested to gift something to her mother on her birthday. But she could not decide what to give her mother.She asked her brother. Her brother advised to gift a microwave oven to the mother. But their father said it is a wastage of money. They convinced him for a microwave and ultimately her father was agreed for microwave.

1. hat qualities Shivani and her brother possess?
2. State how a microwave oven works?

SECTION : E

Question 24.
(i) Draw a circuit diagram to study the input and output characteristics of an n-p-n transistor in its common emitter configuration. Draw the typical input and output characteristics.
(ii) Explain, with the help of a circuit diagram, the working of n-p-n transistor as a common emitter amplifier.

OR

How is a zener diode fabricated so as to make it a special purpose diode? Draw I-V characteristics of zener diode and explain the significance of breakdown voltage. Explain briefly, with the help of a circuit diagram, how a p-n junction diode works as a half wave rectifier

Question 25.
Trace the rays of light showing the formation of an image due to a point object placed on the axis of a spherical surface separating the two media of refractive indices n₁ and n₂. Establish the relation between the distances of the object, the image and the radius of curvature from the central point of the spherical surface. Hence derive the expression of the lens makers formula.

OR

Draw the labelled ray diagram for the formation of image by a compound microscope. Derive the expression for the total magnification of a compound microscope. Explain why both the objective and the eyepiece of a compound microscope must have short focal lengths.

Question 26.
(a) Derive an expression for the average power consumed in a series LCR circuit connected to a.c. source in which the phase difference between the voltage and the current in the circuit is Φ.
(b) Define the quality factor in an a.c. circuit. Why should the quality factor have high value in receiving circuits? Name the factors on which it depends.

OR

(i) Derive the relationship between the peak and the rms value of current in an a.c. circuit.
(ii) Describe briefly, with the help of labelled diagram, working of a step-up transformer. A step-up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain.

 Answers :

SECTION : A 

Answer 1.

Answer 2.
For constructing an astronomical telescope, the objective should have the maximum diameter. Of the three lenses given, L₁ has the maximum diameter so, use lens L₁ as an objective. The eyepiece should have the highest power for better magnification. Therefore, we use lens L₃ as an eyepiece.

Answer 3.
The given electromagnetic waves can be arranged in ascending order with respect to their frequencies as :
Radio waves < Microwaves < UV rays < X-rays

Answer 4.
The type of propagation in which radio waves are transmitted towards the sky and are reflected by the ionosphere towards the desired location on earth is called sky wave propagation.

Answer 5.
Magnetic field lines form closed loops around a current – carrying wire. The geometry of a straight solenoid is such that magnetic field lines cannot loop around circular wires without spilling over to the outside of the solenoid. The geometry of a toroid is such that magnetic field lines can loop around electric wires without spilling over to the outside. Hence, magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid.

SECTION : B

Answer 6.

The stopping potential increases with increasing frequency of radiations. Thus curves a and b have the same frequency but different intensities.

Answer 7.
(i) For a point source, wavefront will be spherical,
(ii) For a distant light source, the wavefronts will be plane.

Answer 8.
Nuclear density is independent of mass number. Hence, both the atoms have the same nuclear density.
i.e.,  d₁ : d₂ = 1 : 1

Answer 9.

V becomes equal to E when no current flows through the circuit.

Hence, can say that when resistance increases from R = 0 to R = ∝ then V_T increases from minimum to maximum and current decreases from
maximum to minimum value.

Answer 10.
(i) Two equipotential surfaces cannot intersect with each other because when they will intersect,then at a point the electric field will have two directions, which is not possible.
(ii) Potential at P (7, 0, 0)

SECTION : C

Answer 11.

∴ The mass number of A₄ = 172 and the atomic number of A₄= 69

Answer 12.
(i) The de Broglie wavelength associated with potential V is

Answer 13.

Answer 14.
If there are N electrons and the velocity of the i^th electron at a given time is v_i where, i = (1, 2, 3,…N), then

For all the electrons in the conductor, average value of v. is zero. The average of v_t is v_d or drift velocity. This is the average velocity experienced by an electron in an external electric field. There is no fixed time after which each collision occurs. Therefore, we take the average time after which one collision takes place by an electron.
Let this time, also known as relaxation time, be t. Substituting this in equation (1)

Then, Negative sign shows that electrons drift opposite to the applied field.

Answer 15.
A bright spot is observed when a tiny circular object is placed in the path of light from a distant source in a single slit diffraction experiment because light rays flare into the shadow region of the circular object as they pass the edge of the tiny circular object. The light from all the edges of the tiny circular object are in phase with each other. Thus they form a bright spot at the centre of the shadow of the tiny circular object. The two differences between the interference patterns obtained in Young’s double slit experiment and the diffraction pattern due to a single slit are as follows: The fringes in the interference pattern obtained from diffraction are of varying width, while in case of interference, all are of the same width. The bright fringes in the interference pattern obtained from diffraction have a central maximum followed by fringes of decreasing intensity, whereas in case of interference, all the bright fringes are of equal intensity.

Answer 16.
According to Gauss law,

Where, q is the point charge,
E is electric field due to the point charge,
dA is a small area on the Gaussian surface at any distance,
e₀ is the proportionally constant.
For a spherical shell at distance r from the point charge, the integral § dA is merely the sum of all differential of dA on the sphere.
Therefore,

Therefore, for a thin conducting spherical shell of radius R and charge Q, spread uniformly over its surface, the electric field at any point outside the shell is


The graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∝.

Answer 17.
For transmitting a signal, the antenna should have a size comparable to the wavelength of the signal (atleast λ/4 in dimension), where λ is the wavelength. If the frequency of the signal is small, then its wavelength becomes very large and it is impractical to make those large antennas for the corresponding large wavelengths. For higher frequencies, wavelength is smaller, which is the reason why high frequency carrier waves are used for transmission.

OR

The process of superimposing information containing in a low frequency signal on a high frequency signal is called modulartion

Answer 18.
Existing diagram can be redrawn as follows :

Answer 19.
(a) Energy transitions for A, B, C and D are :


(b)

For maximum wavelength, energy transition should be minimum. A undergoes minimum energy transition.
A = 2 eV
Thus, photon in A will have the maximum wavelength.

Answer 20.
(a) The phenomenon in which emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil is called self inductance. S.I. unit of inductance is henry (H).
(b) Magnetic field B inside a solenoid carrying a current i is B = μ₀ n i
Let n be the number of turns per unit length.

Answer 21.
For meter bridge :

Answer 22.

1 ampere is the value of that steady current which when maintained in each of the two very long, straight, parallel conductors of negligible cross-section and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 x 10^-7 Newton per metre of length.

OR

Cyclotron is a machine used to accelerate charged particles or ions to high energies. It uses both electrical and magnetic fields in combination to increase the speed of the charged particles. The particles move in two semi-circular containers D₁ and D₂, called Dees. Inside the metal box, the charged particle is shielded from external electric fields. When the particle moves from one dee to another, electric field is acted on the particle. The sign of the electric field is changed alternately, in tune with the circular motion of the particle. Hence, the particle is always accelerated by the electric field. As the energy of the particle increases, the radius of the circular path increases.


The above expression is the expression for cyclotron frequency. The oscillator applies an ac voltage across the dees and this voltage must have a frequency equal to that of cyclotron frequency.

SECTION : D

Answer 23.

1. (a) Caring, love, affection and respect towards her parents.
   (b) General awareness about scientific gadgets and their utilities.
2. In microwave oven the frequency of microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of molecules. In this way temperature of any food item containing water increases.

SECTION : E

Answer 24.
(i)

Input characteristics :

Output characteristics :

(ii) The input (base-emitter) circuit is forward and the output (collector-emitter) circuit is reversed biased. When no a.c. signal is applied, the potential difference Vcc between the collector and emitter is given by    V_cc = V_ce +I_c R_L
When an a.c. signal is fed to the input circuit, the forward bias increase during  the positive half cycle of the input. This result in increase in I_c and decrease inV_cc. Thus during the positive half cycle of the input, the collector becomes less positive.

During the negative half cycle of the input, the forward bias is decreased resulting in decrease in I_E and hence I_c. Thus V_cc would increase making the collector more positive. Hence in a common-emitter amplifier, the output voltage is 180° out of phase with the input voltage.

• Input signal voltage V_in = I_BR_in
• Output signal voltage V_out = I_CR_L
  
• Voltage gain (A_v) of the amplifier is

OR

Zener diode is fabricated such that both the p-type and the n-type are highly doped. This makes the depletion region thin. When an electric field is applied, a high electric field appears across the thin depletion region. When the electric field becomes very high, it knocks off electrons from the host atoms to create a large number of electrons. This results in a large value of current inside the circuit.

Zener has a sharp breakdown voltage and this property of zener is used for voltage regulation.

An a.c. current has a positive half cycle and a negative half cycle. A p-n junction allows current to pass only in one direction and that is when it is forward biased. When a positive half-cycle occurs, the p-side has a lower potential. Therefore, the diode is now forward biased and therefore, conducts and this positive cycle is available for the load. When a negative half cycle occurs, the «-side has a higher potential than the p-side. Hence, the diode is now reverse biased and thus, does not conduct. As a result, this negative half cycle does not conduct. Therefore, it does not appear at the load and is cut-off. We obtain a waveform, which has only positive half cycles and therefore it is called halfwave rectifier.

Answer 25.
In the given figure, image is I and object is denoted as O. The centre of curvature is X. The rays are incident from a medium of refractive index n₁ to another of refractive index n₂ We consider NM to-be perpendicular to the principal axis.

For a thin lens , BI₁  = DI₁
Adding  (ii) and (iii) , we obtain

Answer 26.
(a) The rate of dissipation of energy in an electrical circuit is called the ‘power’. It is equal to the product of the emf and the current. The power of an alternating-current depends upon the phase difference between the emf and the current. The instantaneous values of the emf and the current in an a.c. circuit are given by

(b) Quality Factor :
It is defined as the ratio of the voltage drop across inductor or capacitor to the applied voltage.

The quality factor has high value in receiving circuits in order to get a sharp gain for the desired channel frequency. The quality factor depends on the following values :

• Inductance
• Resistance
• Capacitance

OR

(a) The instantaneous power dissipated in the resistor is


No, a step-up transformer steps up the voltage while steps down the current. So input and output power remains same, provided there is no loss. Hence, there is no violation of principle of conservation of energy.

We hope the CBSE Sample Papers for Class 12 Physics Paper 2 help you. If you have any query regarding CBSE Sample Papers for Class 12 Physics Paper 2, drop a comment below and we will get back to you at the earliest.