Category: CBSE Class 12

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.

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

CBSE Sample Papers for Class 12 Maths 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 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 100

General Instructions:

• All questions are compulsory.
• Questions 1-4 in section A are very short answer type questions carrying 1 mark each.
• Questions 5-12 in section B are short answer type questions carrying 2 marks each.
• Questions 13-23 in section C are long answer I type questions carrying 4 marks each.
• Questions 24-29 in section D are long answer II type questions carrying 6 marks each.

SECTION A

Question 1.
If A is a matrix of order 2 x 3 and B is of order 3 x 4, what is the order of (AB)’?

Question 2.
If \(\vec { P }\) is a unit vector and \(\left( \vec { x } -\vec { P } \right) \cdot \left( \vec { x } +\vec { P } \right) =80\), then find the value of \(\left| \vec { x } \right|\).

Question 3.
Evaluate \(\int { \sqrt { \frac { x }{ 1-{ x }^{ 3 } } } } dx\)

Question 4.
Find the point at which tangent to the curve y = x² makes an angle of 45° with x-axis.

SECTION B

Question 5.
Express the matrix \(A=\begin{pmatrix} 3 & 5 \\ 1 & -1 \end{pmatrix}\) as the sum of symmetric and skew symmetric matrix.

Question 6.
If x^y = y^x, find \(\frac { dy }{ dx }\)

Question 7.
Verify Rolle’s theorem for f(x) = sin 2x in [0, \(\frac { \pi }{ 2 }\)] and find the value of ]0, \(\frac { \pi }{ 2 }\)[

Question 8.
Discuss continuity of the function at x = 0

Question 9.
Find the coordinate of point of intersection of lines

Question 10.
If P(A) = \(\frac { 1 }{ 4 }\), P(A | B) = \(\frac { 1 }{ 2 }\), P(B | A) = \(\frac { 2 }{ 3 }\) then find P(B).

Question 11.
If 3 tan^-1x + cot^-1x = π then find the value of x.

Question 12.

SECTION C

Question 13.

Question 14.

Question 15.

Question 16.

Question 17.

Question 18.
Solve the differential equation (1 + y²) dx = (tan^-1 y – x) dy ; y(0) = 0

Question 19.
The scalar product of the vector \(\hat { i } +\hat { j } +\hat { k }\) with a unit vector along the sum of vectors \(2\hat { i } +4\hat { j } -5\hat { k }\) and \(\lambda \hat { i } +2\hat { j } +3\hat { k }\) is equal to one. Find the value of λ.

Question 20.

Question 21.
A and B throw a die alternately till one of them gets a 5 and wins the game. F ind their respective probabilities of winning if A starts the game. Why gambling is not a good way of earning money?

Question 22.
In a bolt factory, machine A, B and C manufacture respectively 25%, 35% and 40% of the bolts. of their output, 5%, 4% and 2% are respectively defective bolts. A bolt is drawn at random from the total production and is found to be defective. Find the probability that it is manufactured by machine B.

Question 23.
A binary operation * on the set {0, 1, 2, 3, 4, 5} is defined as

Show that zero is the identity element for. this operation and each non-zero element ‘a’ of the set is invertible with 6 – a being the inverse of a.

SECTION D

Question 24.

Question 25.
Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.
OR
An open box with square base is to be made out of a given quantity of sheet of area a² sq.units. Show that the maximum volume of the box is \(\frac { { a }^{ 3 } }{ 6\surd 3 }\) cubic units.

Question 26.
Using integration find the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.
OR
Find the area of the region in the first quadrant enclosed by the y-axis, the line y = x and the circle x² + y² = 32 using integration.

Question 27.
A dealer wishes to purchase a number of fans and sewing machines. He has only ₹ 5760 to invest and has space for at most 20 items. A fan cost him ₹ 360 and a sewing machine ₹ 240. On selling he get a profit of ₹ 22 on a fan and ₹ 18 on a sewing machine. Assuming that he can sell all the items that he store, how should he invest his money in order to maximize profit? Formulate this as L.P.P. and solve it graphically.

Question 28.
Find the image of the point (1, 2, 3) in the plane x + 2y + 4z = 38.
OR
Find the equation of the plane passing through the points A(3, -1, 2), B (5, 2, 4) and C(-1, -1, 6). Also find the distance of the point P(6, 5, 9) from the plane.

Question 29.
Show that the differential equation x dy – y dx = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } }\) dx is homogeneous and hence solve it.

Solutions

Solution 1.
A = [a_ij]_2×3
B = [b_ij]_3×4
Order of AB = 2 x 4
Order of (AB)’ = 4 x 2

Solution 2.

Solution 3.

Solution 4.
Let the point is (x, y)
y = x²

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

Solution 12.

Solution 13.

Solution 14.

Solution 15.
y = sinpt
x = sint

Solution 16.

Solution 17.

Solution 18.

Solution 19.

Solution 20.

Solution 21.

Solution 22.
Let E₁ is the event the bolt is manufactured by machine A
E₂ is the event the bolt is manufactured by machine B
E₃ is the event the bolt is manufactured by machine C
A is the event bolt drawn is defective

Solution 23.

Solution 24.

Solution 25.

Solution 26.
x + 2y = 2 ……. (1)
y – x = 1 …… (2)
2x + y = 7 ……. (3)
From (1) and (2), (0, 1)
From (2) and (3), (2, 3)
From (1) and (3), (4, -1)

Solution 27.
Let dealer purchases x fans and y sewing machines.
Objective function is maximize profit Z = 22x + 18y

Solution 28.

Solution 29.

We hope the CBSE Sample Papers for Class 12 Maths Paper 6 help you. If you have any query regarding CBSE Sample Papers for Class 12 Maths 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 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 Chemistry Paper 1 are part of CBSE Sample Papers for Class 12 Chemistry. Here we have given CBSE Sample Papers for Class 12 Chemistry Paper 1.

CBSE Sample Papers for Class 12 Chemistry 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 Chemistry is given below with free PDF download solutions.

Time Allowed : 3 Hours
Max. Marks : 70

General Instructions

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

Questions 

Question 1.
Why are aquatic species more comfortable in cold water as compared to warm water?

Question 2.
Write IUPAC name of

Question 3.
Which of the following complex is more stable, write its IUPAC name:
[CO(NH₃)₆]³⁺ or [Co(CO)₆]³⁺

Question 4.
Name the method used for the refining of Zirconium metal.

Question 5.
How are deltas formed?

Question 6.
Write two differences between Covalent and Molecular solids.

Question 7.
Calculate ΔG° for the following cell at 298 K.

Question 8.
An element with density 5.6 g cm 3 exists in bcc lattice with edge length 400 pm. Calculate atomic mass of this element.

Question 9.
Draw the structure of following molecules :
(a) IF^–₄
(b) XeF₆

Question 10.
Calculate mass of glucose (molar mass 180 g/mol) to be diluted in 60 g of H₂O to lower the freezing point of its solution by 0.93 K (K_f for H₂O is 1.86 K kg mol^-1)

Question 11.
(a) Which solution is used for the leaching of gold metal in the presence of air in the metallurgy of gold?
(b) Out of C and CO, which is better reducing agent at lower temperature for extraction of iron from its oxide ore.
(c) Write the principle involved in zone refining.

Question 12.
Complete the following reactions:

1. XeF₂ + PF5 →
2. H₃PO₃ \(\underrightarrow { Heat } \)
3. ClO^–₃ + Cl \(\underrightarrow { { H }^{ + } } \)

Question 13.
(a) Chlorobenzene is less soluble in water as compared to chloroethane. Explain.
(b) Which will show faster S_N1 reaction and why?

(c) Which of the following will be optically active: 2-chlorobutane or 1-chlorobutane?

Question 14.
Explain the mechanism of the following reactions :
(a) C₂H₅OH \(\xrightarrow [ { Conc.H }_{ 2 }{ SO }_{ 4 } ]{ { 170 }^{ 0 }C } \) CH₂ = CH₂
(b) CH₃ – OH \(\xrightarrow [ { H }^{ + } ]{ 413K } \) CH₃ – O – CH₃

Question 15.
(a) Distinguish between lyophilic and lyophobic colloids.
(b) Name the absorbent used for controlling of humidity in operation theatres.

Question 16.
(a) Define the limiting molar conductivity.
(b) State Kohlrausch’s law.
(a) Define corrosion.

Question 17.

1. State Henry’s law.
2. Define azeotropes. What type of azeotropes is formed by positive deviation from Raoult’s law? Give an example.

Question 18.
How would you account for the following?

1. Transition metal exhibits variable oxidation state.
2. Zr(Z = 40) and Hf(Z = 72) have almost identical radii.
3. Transition metals and their compounds act as complexing agents.

OR
Complete the following chemical equations:

1. Cr₂O^2-₇ + Fe²⁺ + H⁺ →
2. CrO^2-₄ + H⁺ →
3. MnO^–₄ + I^– + H₂O →

Question 19.

1. What type of isomerism is exhibited by the complex [Pt(NH₃)₆] [NiCl₆]?
2. Write the formula for potassium hexacyanoferrate (II).
3. Write electronic arrangement of [Co(NH₃)₆]³⁺ and discuss its geometry and magnetic behaviour.

Question 20.
Give reason for the following:

1. (CH₃)₃ P = O exists but (CH₃)₃ N = O does not.
2. Phosphorus has more negative electron gain enthalpy than nitrogen.
3. H₂Se is more acidic than H₂O.

Question 21.
Give the structure of A, B and C in the following reactions:

Question 22.
Write the name and structure of the monomers of the following polymers
(a) Buna-N
(b) Nylon-2, 6
(c) PVC

Question 23.
Read the following passage and answer the questions that follows:
Shivani has recently started facing difficulty in visualising in dark. Prema, her best friend wonders why Shivani cannot see properly in the dark, on observation she found that Shivani often carries junk foods in her lunch. Prema has started sharing her lunch full of green vegetables like carrots, tomatoes, coloured fruits, milk and other dairy products. Shivani is now better and has also started taking a balanced diet.

1. Name the disease Shivani is suffering from.
2. Name the Vitamin and its source to cure this disease.
3. What values are shown by Prema?

Question 24.
Account for the following:
(a) pK_b of aniline is more than that of methylamine.
(b) Methylamine in water reacts with FeCl₃ from hydrated ferric oxide.
(c) Why do primary amines have higher boiling point than tertiary amines?
(d) Complete the following equations:
(i) C₆H₅N₂Cl + C₂H₅OH →
(ii) C₆H₅NH₂ + (CH₃CO)₂O →
OR
(a) Give one chemical test to distinguish between:
(i) Methylamine and dimethylamine
(ii) Aniline and Benzyl amine

(b) How will you convert:
(i) Ethanoic acid to Methanamine
(ii) Aniline to p-bromoaniline
(c) Arrange according to increasing basic strength C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH, CH₃NH₂

Question 25.
(a) The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its l/16th value?
(b) Differentiate between order and molecularity of reaction.
(c) Write unit of rate constant (K) for
(i) First order reaction
(ii) 2nd order reaction
OR

1. The rate of a reaction quadruples when the temperature changes from 290 K to 310 K, calculate the activation energy for this reaction.
2. A reaction is first order wrt. P and wrt. 2nd order wrt. Q, write the differential rate equation for it.
3. Write factors (any two) that affects the rate of a chemical reaction.
4. Define specific rate constant.

Question 26.

1. Phenol is more acidic than cyclohexanol. How?
2. Which is steam volatile, 2-nitrophenol or 4-nitrophenol and why?
3. Explain following name reactions with a suitable chemical reaction.
   1. Sandmeyer’s reaction
   2. Cannizaro’s reaction

OR
(a) Write the reaction of glucose with HI.
(b) Name the force responsible for stabilisation of secondary structure of protein.
(c) Arrange according to the given instructions in increasing order:
(i) Phenol, Methoxyphenol, 2,4-Dinitrophenol (acidic strength)

Answers

Answer 1.
In cold water the concentration of dissolved oxygen gas is more and hence respiration for aquatic species will be easier so it will be comfortable as compared to warm water.

Answer 2.
2-methyl pent-4-enal

Answer 3.
[Co(CO)₆]³⁺ complex is more stable. It is due to synergic bonding between metal and carbonyl group.
IUPAC name → Hexacarbonyl cobalt (III) ion.

Answer 4.
van Arkel method

Answer 5.
Coagulation of silt present in river water due to electrolytes present in sea water leads to formation of delta.

Answer 6.

Answer 7.

= 0.34 – (- 2.37) = 2.71 V
Here change in number of electrons (n) = 2
ΔG° = – nFE_cell° = – 2 × 96500 × 2.71 = – 523.03 kJ/mole

Answer 8.
Since d = \(\frac { ZM }{ { a }^{ 3 }{ N }_{ A } } \)
From question → bcc arrangement i.e. Z = 2
a = 400 pm = 400 × 10^-10 cm = 4 × 10^-8 cm
d = 5.6 g/cm³

Answer 9.

Answer 10.

i.e. 54 g glucose needs to be added.

Answer 11.

1. Aqueous solution of NaCN or KCN is used for leaching.
2. CO at lower temperature is a better reducing agent than C for the extraction of iron from its oxide ore.
3. Impurities are more soluble in molten metal as compared to its solid state.

Answer 12.

1. XeF₂ + PF₅ → [XeF]⁺ [PF₆]^–
2. H₃PO₃ \(\underrightarrow { Heat } \) PH₃ + H₃PO₄
3. ClO^–₃ + Cl^– \(\underrightarrow { { H }^{ + } } \) Cl₂ + H₂O

Answer 13.
(a) Phenyl group is more hydrophobic as compared to smaller ethyl group, hence chloroethane is more soluble than chlorobenzene.

Due to more stable carbocation formation which is there because of more number of resonance structures.
(c) 2-chlorobutane.

Answer 14.
(a) H₂SO₄ + H₂O → H₃O⁺ + HSO^–₄

Answer 15.
(a)

(b) Silica gel

Answer 16.

1. Limiting molar conductivity : The value of molar conductivity when the concentration of electrolytes approaches zero is known as limiting molar conductivity or molar conductivity at infinite dilution. It is the maximum possible molar conductivity for any electrolyte. Represented by Δ°_m.
2. Kohlrausch’s law : It states that the limiting molar conductivity of on electrolyte is the sum of the individual contribution of the anion and cation of the electrolyte, e.g.
   Δ°_m(CH₃COOH) = λ°_CH3COO^– + λ^°_H+
3. Corrosion : The tarnishing of upper layer of any active metal due to environmental factors such as water vapours, CO₂, O₂, etc. in known as corrosion.
   For example-rusting of iron, blackening of silver, deposition of greenish layer on copper etc.

Answer 17.

(a) Henry’s law : It states that the partial pressure (p) of a gas in vapour phase is proportional to the mol fraction of the gas (x) in the solution. It is expressed as
p = k_Hx
where kH is known as Henry’s law constant and x is mol fraction.

(b) Azeotropes : It is a binary mixture, having same composition in liquid state as well as in vapour phase and boils at constant temperature like a pure liquid. If there is positive deviation, then minimum boiling azeotrope will form. In case of negative deviations, maximum boiling azeotrope will form.
Examples 95.5% Ethanol and 4.5% H₂O by volume.

Answer 18.

1. Transition metals exhibits variable oxidation state due to comparable energies of (n – 1)d and ns orbitals, so electrons from both can form bond.
2. Due to lanthanoid contraction.
3. Transition metals acts as complexing agent due to vacant d-orbitals and high charge density on metal ion.

OR

1. Cr₂O^2-₇ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
2. CrO^2-₄ + H⁺ → Cr₂O^2-₇
3. 2MnO^–₄ + 10 I^– + 8H₂O → 2Mn²⁺ + 5I₂ + 16 OH^–

Answer 19.

1. Co-ordination isomerism is exhibited by the complex [Pt(NH₃)₆] [NiCl₆]
2. K₄[Fe(CN)₆]
3. Electronic arrangement for [Co(NH₃)₆]³⁺ according to VBT.
   Co → 3d⁷4s²
   Co³⁺ → 3d⁶4s⁰
   

Answer 20.

1. Due to lack of d-orbital in N, dπ-pπ bonding in not possible, hence formation of (CH₃)₃ N = O is not possible.
2. Due to small size and high electron density in valence shell of N, its electron gain enthalpy is less than phosphorous.
3. Due to longer H-Se bond than H-O bond H₂Se is more acidic.

Answer 21.

Answer 22.

Answer 23.

1. Night blindness
2. Vitamin-A, Source – Carrot, cod liver oil, butter, milk etc.
3. Care for friend, knowledge of science, sincerity in her effort.

Answer 24.
(a) Aniline is a weaker base than methylamine due to positive charge on nitrogen, which is there because of resonance. Hence its pK_b value is more than methylamine.
(b) Methyl amine gives OH- ions in water, hence FeCl₃ get hydrolysed to give hydrated ferric oxide.

(c) There is intermolecular hydrogen bonding in primary amines, which is not possible in tertiary amines. Hence primary amines have higher boiling point.

OR
(a) (i) 

(ii)

(b) (i) Ethanoic acid to methanamine

(ii) Aniline to p-bromo aniline

(c) C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH

Answer 25.

(b) Difference between order and molecularity:

(c) (i) Second^-1
(ii) mol^-1 L s^-1
OR

1. from Arrhenius equation
   
2. rate = K[P]¹ [Q]²
3. Concentration, pressure, temperature, catalyst
4. It is the rate of reaction when concentration of all the reactants is unity.

Answer 26.

1. Since phenoxide ion is resonance stablised, hence phenol is more acidic than cyclohexanol.
2. 2-nitrophenol is steam volatile due to intramolecular hydrogen bonding whereas 4-nitrophenol undergoes intermolecular hydrogen bonding and thus connected to each other to become less volatile.
3. 1. Sandmeyer’s reaction
      
   2. Cannizaro’s reaction
      HCHO \(\underrightarrow { NaOH(Conc.) } \) CH₃OH + HCOONa

OR

(b) Intermolecular and intramolecular hydrogen bonding, coloumbic interactions, disulphide
linkage etc.
(c) (i) Methoxy phenol < phenol < 2,4-dinitrophenol

We hope the CBSE Sample Papers for Class 12 Chemistry Paper 1 help you. If you have any query regarding CBSE Sample Papers for Class 12 Chemistry Paper 1, 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 English Core Paper 2 are part of CBSE Sample Papers for Class 12 English Core. Here we have given CBSE Sample Papers for Class 12 English Core Paper 2.

CBSE Sample Papers for Class 12 English Core 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 Englsih Core is given below with free PDF download solutions.

Time Allowed: 3 hours
Maximum Marks: 100

General Instructions

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

SECTION A
READING (30 MARKS)

Question 1.
Read the following passage carefully. (12 Marks)

1. Maharana Pratap ruled over Mewar only for 25 years. However, he accomplished so much grandeur during his reign that his glory surpassed the boundaries of countries and time turning him into an immortal personality. He along with his kingdom became a synonym for valour, sacrifice and patriotism. Mewar had been a leading Rajput kingdom even before Maharana Pratap occupied the throne. Kings of Mewar, with the cooperation of their nobles and subjects, had established such traditions in the kingdom, as augmented their magnificence despite the hurdles of having a smaller area under their command and less population. There did come a few thorny occasions when the flag of the kingdom seemed sliding down. Their flag once again heaved high in the sky thanks to the gallantry and brilliance of the people of Mewar.

2. The destiny of Mewar was good in the sense that barring a few kings, most of the rulers were competent and patriotic. This glorious tradition of the kingdom almost continued for 1500 years since its establishment, right from the reign of Bappa Rawal. In fact only 60 years before Maharana Pratap, Rana Sanga drove the kingdom to the pinnacle of fame. His reputation went beyond Rajasthan and reached Delhi. Two generations before him, Rana Kumbha had given a new stature to the kingdom through victories and developmental work. During his reign, literature and art also progressed extraordinarily. Rana himself was inclined towards writing and his works are read with reverence even today. The ambience of his kingdom was conducive to the creation of high quality work of art and literature. These accomplishments were the outcome of a long standing tradition sustained by several generations.

3. The life of the people of Mewar must have been peaceful and prosperous during the long span of time; otherwise such extraordinary accomplishment in these fields would not have been possible. This is reflected in their art and literature as well as their loving nature. They compensate for lack of admirable physique by their firm but pleasant nature. The ambience of Mewar remains lovely thanks to the cheerful and liberal character of its people.

4. One niay observe astonishing pieces of workmanship not only in the forts and palaces of Mewar but also in public utility buildings. Ruins of many structures which are still standing tall in their grandeur are testimony to the fact that Mewar was not only the land of the brave but also a seat of art and culture. Amidst aggression and bloodshed, literature and art flourished and creative pursuits of literature and artists did not suffer. Imagine, how glorious the period must have been when the Vijaya Stambha which is the sample of our great ancient architecture even today, was constructed. In the same fort, Kirti Stambha is standing high, reflecting how liberal the then administration was which allowed people from other communities and kingdoms to come and carry out construction work. It is useless to indulge in the debate whether the Vijaya Stambha was constructed first or the Kirti Stambha. The fact is that both the capitals are standing side by side and reveal the proximity between the king and \ the subjects of Mewar.

5. The cycle of time does not remain the same. Whereas the reign of Rana Sanga was crucial in raising the kingdom to the acme of glory, it also proved to be his nemesis. 1“ History took a turn. The fortune of Mewar—the land of the brave, started waning. Rana tried to save the day with his acumen which was running against the stream and the glorious traditions for sometime. (Delhi, All India, Foreign 2016)

I. On the basis of your understanding of the passage, answer the following questions by choosing the most appropriate option. (1 × 4 = 4 Marks)

(a) Maharana Pratap became immortal because:

1. he ruled Mewar for 25 years.
2. he added a lot of grandeur to Mewar.
3. of his valour, sacrifice and patriotism.
4. both (ii) and (Hi).

(b) Difficulties in the way of Mewar were:

1. lack of cooperation of the nobility,
2. ancient traditions of the kingdom.
3. its small area and small population.
4. the poverty of the subjects.

(c) During thorny occasions:

1. the flag of Mewar seemed to be lowered.
2. the flag of Mewar was hoisted high.
3. the people of Mewar showed gallantry.
4. most of the rulers heaved a sigh of relief.

(d) Mewar was lucky because:

1. all of its rulers were competent.
2. most of its people were competent.
3. most of its rulers were competent.
4. only a few of its people were incompetent.

II. Answer the following questions as briefly as possible. (1 × 6 = 6 Marks)

(e) Who is the earliest king of Mewar mentioned in the passage?
(f) What was Rana Kumbha’s contribution to the glory of Mewar?
(g) What does the writer find worth admiration in the people of Mewar?
(h) How could art and literature flourish in Mewar?
(i) How did the rulers show that they cared for their subjects?
(j) What does the erection of Vijaya Stambha and Kirti Stambha in the same fort signify?

III. Find words from the passage which mean the same as: (1 × 2 = 2 Marks)

(a) surprising (para 4)
(b) evidence (para 4)

Question 2.
Read the following passage carefully. (10 Marks)

1. To ensure its perpetuity, the ground is well held by the panther both in space and in time. It enjoys a much wider distribution over the globe than its bigger cousins, and procreates sufficiently profusely to ensure its continuity for all time to come.

2. There seems to be no particular breeding season of the panther, although its sawing and caterwauling is more frequently heard during winter and summer. The gestation period is about ninety to hundred days (Whipsnade, ninety-two days). The litter normally consists of four cubs, rarely five. Of these, generally two survive and not more than one reaches maturity. I have never come across more than two cubs at.the heels of the mother. Likewise, graziers in the forest have generally found only two cubs hidden away among rocks, hollows of trees, and other impossible places.

3. Panther cubs are generally in evidence in March. They are born blind. This is a provision of Nature against their drifting away from the place of safety in which they are lodged by their mother, and exposing themselves to the danger of their being devoured by hyenas, jackals and other predators. They generally open their eyes in about three to four weeks.

4. The mother alone rears its cubs in seclusion. It keeps them out of the reach of the impulsive and impatient male. As a matter of fact the mother separates from the male soon after mating and forgets all about their tumultuous union. The story that the male often looks in to find out how the mother is progressing with her cubs has no foundation except in what we wish it should do at least.

5. The mother carries its cubs about by holding them by the scruff of their neck in its mouth. It trains them to stalk, and teaches them how to deliver the bite of death to the prey. The cubs learn to treat all and sundry with suspicion at their mother’s heels. Instinctively the cubs seek seclusion, keep to cover and protect their flanks by walking along the edge of the forest.

6. I have never had an opportunity to watch mother panther train its cubs. But in Pilibhit forests, I once saw a tigress giving some lessons to its little ones. I was sitting over its kill at Mala. As the sun set, the tigress materialised in the twilight behind my machan. For about an hour, it scanned and surveyed the entire area looking and listening with the gravest concern. It even went to the road where my elephant was awaiting my signal. The mahout spotted it from a distance and drove the elephant away.

7. When darkness descended upon the scene and all was well and safe, the tigress called its cubs by emitting a low haa-oon. The cubs, two in number and bigger than a full-grown cat, soon responded. They came trotting up to their mother and hurried straight to the kill in indecent haste. The mother spitted at them so furiously that they doubled back to its heels immediately. Thereafter, the mother and its cubs sat under cover about 50 feet (15 m) away from the kill to watch, wait, look, and listen. After about half an hour’s patient and fidgetless vigil the mother seemed to say ‘paid for’. At this signal, the cubs cautiously advanced, covering their flanks, towards the kill. No longer did they make a beeline for it, as they had done before.

8. The mother sat watching its cubs eat, and mounted guard on them. She did not partake of the meal. (Delhi, All India, Foreign, 2016)

I. On the basis of your understanding of the passage, answer the following questions by choosing the most appropriate option. (1 × 2 = 2 Marks)

(a) To protect its cubs the mother panther hides them :

1. among rocks
2. in the branches of the trees
3. behind the tree trunks
4. at its heels

(b) The male panther :

1. is protective of its cubs
2. trains its cubs
3. watches the progress of the mother
4. is impulsive and impatient

II. Answer the following questions as briefly as possible. (1 × 6 = 6 Marks)

(c) How many cubs does the mother panther rarely deliver?
(d) What may happen if the panther cubs are not born blind?
(e) Why did the mahout drive his elephant away?
(f) Why did the tigress spit at its cubs?
(g) From the narrator’s observation, what do we learn about the nature of the tigress?
(h) Why does the panther not face the risk of extinction?

III. Find words from the passage which are similar in meaning to the following. (1 × 2 = 2 Marks)

(a) moving aimlessly (para 3)
(b) came down / fell (para 7)

Question 3.
Read the following passage carefully. (8 Marks)

People tend to amass possessions, sometimes without being aware of doing so. They can have a delightful surprise when they find something useful which they did not know they owned. Those who never have to change house become indiscriminate collectors of what , can only be described as clutter. They leave unwanted objects in drawers, cupboards and attics for years in the belief that they may one day need them. Old people also accumulate belongings for two other reasons, lack of physical and mental energy, and sentiment. Things owned for a long time are full of associations with the past, perhaps with the relatives who are dead, and so they gradually acquire a sentimental value.

Some things are collected deliberately in an attempt to avoid wastage. Among these are string and brown paper, kept by thrifty people when a parcel has been opened. Collecting small items can be a mania. A lady cuts out from newspapers sketches of model clothes that she would like to buy if she had money. As she is not rich, the chances are that she will never be able to afford such purchases. It is a harmless habit, but it litters up her desk. Collecting as a serious hobby is quite different and has many advantages. It provides relaxation for leisure hours, as just looking at one’s treasure is always a joy. One doesn’t have to go out for amusement as the collection is housed at home. Whatever it consists of – stamps, records, first editions of books, china – there is always something to do in connection with it, from finding the right place for the latest addition to verifying facts in reference books. This hobby educates one not only in the chosen subject, but also in general matters which have some bearing on it.

There are other benefits also. One gets to meet like-minded collectors to get advice, compare notes, exchange articles, to show off one’s latest find. So one’s circle of friends grows. Soon the hobby leads to travelling, perhaps a meeting in another town, possibly a trip abroad in search of a rare specimen, for collectors are not confined to one country. Over the years one may well become an authority on one’s hobby and will probably be asked to give informal talks to little gatherings and then, if successful, to larger audiences. [Delhi, All India, Foreign 2016]

A. On the basis of your reading of the above passage make notes on it, using headings and sub-headings. Use recognizable abbreviations (wherever necessary—minimum four) and a format you consider suitable. Also supply an appropriate title to it. (5 Marks)
B. Write a summary of the passage in about 80 words. (3 Marks)

SECTION B
ADVANCE WRITING SKILLS (30 MARKS)

Question 4.
Draft an advertisement in about 50 words for a leading practising lawyer Karan Kumar who wishes to buy an independent house of about 1000 square feet at City Light Road to be used as an office- cum-residence. It should be located in a posh area with excellent facilities. Uninterrupted water & electricity. Nearby market must. His mobile No. is 45645678. (4 Marks)
OR
You have planned to organize a summer camp for the children of age group 10¬13 years for developing their creative skills in art—craft, clay modeling, music etc. in your school during the month of May. Draft a notice in not more than 50 words for your notice-board giving details. Mention the last date of registration of the students/ you are Reena/Roshan, Secretary Art and Craft Club, B.V. Public School, Chandigarh.

Question 5.
Your school, KKR Senior Secondary School, is situated in the heart of the city of Mangalore. Many stray animals roam on the school road causing traffic jams as well as accidents. Write a letter to the editor of The Indian Express drawing attention of the municipal authorities to this nuisance. Sign as Rukmani/Rakesh, Head Girl/Head Boy. (Word limit 120-150) (6 Marks)
OR
You are Saurabh/Shweta, 15, Shashigardens, Ghaziabad. Read the advertisement given below and write a letter to the advertiser, applying for the job. Also give your detailed resume which you would send along with your letter of application.

Question 6.
You are Ram/Devi. You recently attended a health camp on yoga. Write an article on the benefits of Yoga for holistic health and the need to practice daily. (Word limit 150-200) (10 Marks)
OR
Write a speech in 150-200 words on the topic ‘Discipline shapes the future of student’. It is to be delivered in the morning assembly. You are Karuna/Karan. (All India 2016)

Question 7.
You witnessed a street fight between a landlord and his tenant and felt bad about their uncouth behavior in a public place. Write a speech to be given in the school assembly, bringing out the need for good manners and polite behavior especially in public places. You are Lalita/Lalit. (Word Limit: 150-200) (10 Marks)
OR
You are Venu/Fatima of Sapphire International School, Delhi. Your school recently organised winter carnival which featured the famous band Rock On. Write a report on the event in150-200 words.

SECTION C
TEXTBOOKS AND EXTENDED READING TEXT (40 MARKS)

Question 8.
Read the following extract carefully and answer the questions that follow: (1 × 4 = 4 Marks)

Now we will count to twelve
and we will all keep still.
For once on the face of the Earth
let’s not speak in any language,
let’s stop for one second,
and not move our arms so much.

(а) What is the significance of the number ‘twelve’?
(b) Which two activities does the poet want us to stop?
(c) What does the poet mean by ‘let’s not speak in any language’?
(d) Describe the pun on the word, ‘arms’.

OR
…………. I saw my mother,
beside me,
doze, open mouthed, her face
ashen like that
of a cdrpse and realised with
pain ……….

(a) Who is “I”?
(b) What did ‘I’ realise with pain?
(c) Why was the realisation painful?
(d) Identify and name the figure of speech used in these lines.

Question 9.
Answer any four of the following questions in about 30-40 words each: (3 × 4 = 12 Marks)

(a) What does the reference to chappals in ‘Lost Spring’ tell us about the economic condition of the ragpickers? (All India 2016)
(b) What do we learn about the crofter’s nature from the story The Rattrap? (All India 2016)
(c) What is suggested by the ‘Massive weight of the uncle’s wedding band’?
(d) What does the poet want for the children of the slums? How can this change be affected?
(e) How did the dewan procure the hundredth tiger for the tiger king?
(f) How did the governor of Oxford Prison describe Evans to the secretary, examination board? (All India 2016)

Question 10.
Language is an essential part of one’s identity. How did M. Hamel and his students show their love for their language on the day of the last lesson? (6 Marks)
OR
“Sophie’s dreams and disappointments are all in her mind.” Do you agree? Why or Why not? (NCERT)

Question 11.
Both the units of ‘Memories of Childhood’ present autobiographical episodes from the lives of two women from ‘marginalised communities’. Describe the main issues raised as well as the common features highlighted in them. (6 Marks)
OR
Both Derry and Mr. Lamb are victims of physical impairments. Yet they are poles apart in their values and outlook to life. Comment. (Word limit 120 150)

Question 12.
How did Mrs. Hall show her excitement when Griffin came to stay at Coach and Horses Inn? ( Word Limit 120-150) [All India 2016] (6 marks)
OR
What are Godfrey’s arguments for adopting Eppie? (Word limit 120-150)

Question 13.
Marvel was a perfect choice to carry out Griffin’s objective. Comment. (Word limit 120-150) (6 Marks)
OR
Attempt a character sketch of Eppie. (Word limit 120-150) (All India 2016)

ANSWERS

Answer 1.
I. (a) (iv) both (ii) and (iii)
(b) (iii) its small area and small population
(c) (i) the flag of Mewar seemed to be lowered
(d) (iii) most of its rulers were competent

II. (e) Bappa Rawal is the earliest king of Mewar mentioned in the passage. His reign started around 1500 years ago.

(f) Rana Kumbha gave a new stature to the kingdon of Mewar. Literature and art flourished during his reign. He himself had penchant for writing. His works are read with reverence even today.

(g) The people of Mewar may be peaceful and prosperous. It is reflected in their art, literature and loving nature. The pleasant nature of people compensate for lack of their admirable physique.

(h) The enviroment of Mewar was conducive to the creation of high quality work of art and literature. They flourished amidst aggression and bloodshed.

(i) Besides the forts and palaces of Mewar, the surprising pieces of architecture are seen in the public utility buildings also. These structures are testimony to the fact that the rulers had care for their subjects.

(j) The erection of Vijaya Stambha and Kirti Stambha in the same fort signifies the proximity/close relationship between the king and the subjects of Mewar.

III.
(a) Astonishing (b) Testimony

Answer 2.
I. (a) – (i) among rocks (b) (iv) is impulsive and impatient

II. (c) The mother panther rarely delivers five cubs.
(d) If the panther cubs are not born blind, they may drift away from the place of safety. They may also be devoured by hyenas, jackals and the other predators.
(e) The tigress was keeping a vigil over her cubs. The mahout did not want to disturb her. Instead, he desired to assure her that there was no danger. So, he drove his elephant away.
(f) The tigress spat at its cubs to make them aware of dangers. She also wanted them to learn patience.
(g) The tigress is protective of her cubs. She raises them alone and does not allow the male to reach them. Also she trains them well.
(h) Panthers do not face the risk of extinction because they procreate quite profusely and are distributed much widely.

III. (a) Drifting away (b) Descended

Answer 3.
A. TITLE: The Habit of Collection

NOTES:

1. Sentimental value of collection

1. delightful purpose of collection
2. indiscriminate collectors
3. lack of physical and mental energy
4. association with past

2. Purpose of deliberate collection

1. avoidance of wastage
2. can be a mania
3. a harmless habit

3. Adv. of collection

1. relaxtn. for leisure hrs.
2. as verfiying facts in ref. books
3. source of educating people

4. Other benefits of collection

1. meeting of like minded collectors/people
2. source of travelling
3. as authority on one’s hobby

B. SUMMARY

Collection gives a delightful surprise to the people. The indiscriminate collectors only clutter their house. The people are sentimentally attached with the collections as they are closely linked with the past. But deliberate collection is merely a waste of time. The collection of small items bears no meaning as it can be said a mania. However, it can be treated as a harmless habit. On the other hand the habit of valuable collection is the source of relaxation. We get amusement at home. It also serves us as the source of knowledge. This hobby gives opportunity to meet like minded people also. We may apt for travelling to meet them in another town. We may become an authority on our hobby.

Answer 4.

Answer 5.

KKR Senior Secondary School
Kamal Enclave Mangalore
10th May, 20××
The Editor
The Indian Express
New Delhi

Sub: Nuisance Caused by Stray Animals Dear Sir,

Through the columns of your esteemed daily, I would like to express the difficulties caused by stray animals near our school. Ours is a newly developed school situated on the outskirts of the city.

Stray animals are a great threat to pedestrians and vehicle drivers regularly causing traffic jams and often accidents too. Only recently a cow drove its horns into a class nine student causing severe injuries in his abdomen. The dairy owners let loose their domestic animals to graze on the domestic waste and garbage. Many a time the life of many a student is endangered on account of accidents caused by these animals. Stray animals are ubiquitous in the street in front of our school. Often they sit on the road causing considerable inconvenience and danger to the passersby most of whom are students of our school.

Several complaints made to the authorities in the past year have been in vain.

You are requested to give some space to my letter in your prestigious newspaper so that it may reach the ears of the authorities. I hope that the authorities concerned would look into the matter and take immediate steps to solve the problems of the residents.

Thanking you
Yours truly
Rukmani Head Girl
KKR Senior Secondary School

OR

15, Shashi Gardens, Ghaziabad
Uttar Pradesh
19th Sept. 20××
Principal
Sapphire Senior Secondary School
Mira Enclave, Delhi

Sub: Application for the post of a librarian Dear Sir,

With reference to your advertisement in ‘The Times of India’ dated 7 th Sept. 20xx inviting applications for the position of a librarian, I hereby offer my candidature for the same.

As regards my qualifications and experience, I am enclosing my bio-data to enable you to make an assessment of my suitability for the given post. In case my application is considered, I am available for the interview at any time suitable to you.

If selected, I assure you that I shall work with utmost dedication and sincerity to your full satisfaction.

Hoping for a favourable response.
Yours faithfully
Shweta
Enel. : Bio-data and Testimonials

Answer 6.

Yoga: A Holistic Approach to Health
By Ram

You are truly healthy when you are not just physically fit but also mentally and emotionally balanced. Weight loss, a strong and flexible body, glowing beautiful skin, peaceful mind, good health – whatever you may be looking for, yoga has it on offer. Very often, yoga is only understood as being limited to asanas . As such, its benefits are only perceived to be at the body level and we fail to realize the immense benefits yoga offers in uniting the body, mind and breath.

Sun Salutations and Kapal Bhati pranayama are some ways to help you lose weight with yoga. Pranayama and meditation are effective techniques to release stress. Yoga is also one of the best ways to calm a disturbed mind.

Yoga poses massage organs and strengthen muscles. Breathing techniques and meditation release stress and improve immunity. Yoga and pranayama help you bring your mind back to the present moment, where you can stay happy and focused. Yoga can even help improve your relationship with your loved ones by keeping the mind happy . Multiple tasks through the day can be quite exhausting. Yoga provides the secret to feeling fresh and energetic even after a long day.

The deeper you move into your yoga practice, the more profound are its benefits.

OR

Discipline Shapes the Future of Student

Good morning, respected Principal, teachers and students! Today I wish to talk to you about the decisive role that discipline plays in a student’s life.

Discipline is the most fundamental of all qualities that shape a student. Discipline means self-control and never trespassing the rules of propriety.

The lack of discipline among the students is one of the chief social problems of the day. You should be a model student, diligent in studies, well behaved at home, in school and outside.

Elders who are indisciplined, set a bad example to the youngsters. They are unconsciously influenced by the elders. A healthy society which is composed of people who mind their business and who are not sources of botheration to others. Non-interference in others’ affairs is one of the most prized virtues .You should be obedient to your parents, teachers and elders.

You should get up early in the morning, do some exercises and keep yourself trim. Study diligently the lessons of the day. Our nation is often caught in a crisis as people become unruly and fight among themselves. From our young age we should train ourselves to be the ideal citizens of our nation.

It is said that the people of a nation are its wealth. Let us take a vow that we will be dutiful, honorable citizens of India.

Thank you

Answer 7.
Good morning, respected Principal, teachers and other staff! Today I wish to talk to you about the importance of good manners in public life

Man is considered as the most intelligent creation of God on the earth as he lives in the society as well as he has capacity to think, talk and act accordingly. So, he must know how to behave well and practise good manners to behave well with others in the society. Parents must teach their kids how should they behave with family members, neighbours, teachers, etc and respect the views of others. Being a well behaved person, one must be polite, gentle, disciplined, and sweet. Some people behave well with sweet words only at home but when they are outside or with strangers they behave badly.

Good manners help a person to show the same character, everywhere. Well mannered people know well how to show or explain the bitter truth with sweet words. They do not shout or quarrel in public. People, who lack good manners, generally have sharp tongue and behave roughly and abuse others without any reason.

They show that they lack a good culture, society and discipline, thus they are hated everywhere. Well mannered people are honoured and admired everywhere.

I urge you to always be polite and well mannered to all at home as well as outside.

Thank you.

OR

Rock On Performs at Sapphire International
By Venu

Delhi, 23rd Sept. 20xx: A grand winter carnival was organized on 22nd Sept. 20xx by Sapphire International School in the spacious school grounds. It was a warm and pleasant morning. The carnival was inaugurated by the Education Minister of Delhi Manish Sisodia.

The programme started at 11 am when the Education Minister arrived. He appreciated the beautifully organized carnival. About 100 stalls selling beautiful handicraft and utility items had been put up. There was a separate food court offering both Indian and foreign cuisines. The colourfully decorated game stalls were a great hit among the children.

The highlight of the carnival was the musical performance by the renowned band Rock On Himanshu, the lead singer regaled the audience with his melodious songs. The audience comprising students and their friends and family, enthusiastically applauded the performances.

The carnival ended at 6 pm. A student remarked, “This is the best carnival I have seen this year’. A part of the proceeds were donated to the NGO—Saathi.

Answer 8.

(a) The poet wishes to remind the readers the need to keep quiet all twelve months of the year. Twelve could also refer to the twelve hours of the clock.
(b) He wants us to stop speaking and keep our arms still.
(c) He wishes people to to refrain from speaking in any language and unite by the common bond of silence.
(d) Arms could refer to both arms of the human body and the destructive weapons that humans have devised.

OR

(a) Here T refers to the poet.
(b) She realized with pain that her mother was ageing and had become pale and ashen. She was slowly nearing death.
(c) The realization is painful as she fears the pain of separation from her mother.
(d) The figure of speech used here is ‘similie’. It occurs in the line ‘ashen like that of a corpse’.

Answer 9.
(a) The ragpickers move around on dirty roads barefoot. There are several reasons to explain their lack of chappals. One says that wearing chappals was not a tradition in their family. Another says that his mother had not brought it down. The fact is that they were too poor to be able to afford chappals.

(b) The crofter was a lonely and credulous man. He was a generous man as he offered the peddlar food, hospitality and shelter. He even entertained him by playing mjolis with him. In good faith he even told him where he had kept his 30 kronor.

(c) Adrienne Rich’s ‘Aunt Jennifer’s Tigers’ is very rich in imagery. ‘Massive weight of Uncle’s wedding’ is a suggestive imagery. It is symbolic of the weight of the harsh and difficult experiences of her married life. The wedding band represents the unbreakable bond of marriage between the husband and the wife.

(d) The poet has a keen desire that these slum children should break the bonds of living in a slum area. They should not remain dejected, depressed and isolated from the rest of the civilized world. So he urges the governors, teachers, inspectors, invigilators and visitors to come forward and educate the slum children. They should be taken to the horizons of the blue sky so that they can progress and cope with the other world. Through education their life can be made to change.

(e) Dewan’s tiger was an old tiger. It was procured from the People’s Park in Madras. It was not ferocious and agile. It was passive and exhausted. He was pushed down to the ground. He wandered into the Maharaja’s presence and stood as if in humble supplication. The tiger was kept hidden in Dewan’s house. At midnight when the town slept in peace, the Dewan and his wife dragged the tiger into the car drove straight to the forest.

(f) James Roderick Evans was quite a pleasant sort of chap. He had no record of violence. He was one of the stars at the Christmas concert. But he was ‘just a congenital kleptomaniac.’ The prison officers called him “Evans the Break.” Thrice he had escaped from prison.

Answer 10.
It is true that one’s language is a part of one’s cultural identity. It is inseparable part of one’s existence. In the story The Last Lesson, the villagers and M. Hamel became acutely aware of their love for French only when it was snatched from them. They regretted wasting their time in impulsive and futile pursuits and neglecting learning their own language. So the scene in the school on the day of the last lesson in French was different from that on other days. German was going to be imposed on them from the next day. Hence, all students and even the village elders had gathered there. The love for their native language French dominated all other things. Now it was too late as their opportunity had been taken away from them by the cruel Germans.

OR

Like every youth when he is about to leave the school, the struggle of thought begins, even in Sophie. On their way back home, Sophie and Jansie discuss possibilities of work to be started. Sophie’s flight took off from boutique to shop manager, to an actress cum boutique owner and to a fashion designer, though both the girls were earmarked for a biscuit factory.

Like every young girl, Sophie also fancied the world. She also had in her fancies a strong feeling of knowing the unknown, seeing the unseen and even meeting the welcoming strangers. Though her father forbade for taking her to unknown places and countries, yet Sophie expected Geoff to acquaint her to all new and strange things, people and places. She fancied the world greeting and applauding them.

Finally in her fantasizing she had Danny Casey in her mind. So she visited the perfect place for lovers and sat on the solitary wooden bench under the tree. She waited for her lover coming out of the shadows but there was no sign of him. She became burdened with sadness. Thus Sophie’s dreams and disappointments are all of her own creations.

Answer 11.
The lesson ‘Memories of Childhood’ presents autobiographical episodes from the lives of two women. They highlight the plight of ‘the marginalised communities’. The people of these communities never receive honour and dignity due to them. In ‘The Cutting of My Long Hair’ Bonnin describes how an Indian girl suffered extreme indignities. She tried to maintain her identity and her distinct culture. She liked to wear, long and thick hair. But the ‘paleface woman’ and others dragged her out. They tied her fast in a chair and gnawed off her long hair. In ‘We Too Are Human Beings’ Bama, a Tamil Dalit writer, presents the struggle of a girl of a low caste. The people of low castes are never respected and honoured. They have to carry a food packet by its string without touching it. The girl in ‘We Too Are Human Beings’ struggles, studies hard and stands first in the class. She attains equality, and honour.

The most common feature of both these episodes is the struggle and fight against racial and social discrimination. They don’t take oppression, exploitation and injustice meekly.

OR

Derry’s main problem is his burnt face. One side of his face was eaten up by acid. He suffers from a tremendous sense of inferiority complex. He is always conscious that his face is ‘bad’, ‘terrible’ and ‘the ugliest’. People are ‘afraid’ of him. He tries to escape people. He allows himself to be alienated from the world. On the other hand, Mr Lamb doesn’t allow his physical disability to come in his way. He accepts life as it comes. He has a positive attitude towards life, things and people. He doesn’t find solace in escapism. Children tease him by calling ‘Lamey-Lamb’, but he doesn’t mind it. One of his legs was blown off in the war. But he is full of life and enjoys it at his best. He enjoys sitting in the sun, reading books and growing weeds and flowers.

Derry is withdrawn and defiant. He doesn’t trust people. He thinks that no one will ever love and kiss him except his own mother. He can’t stand people staring at him or passing uncharitable remarks. He is touchy and hyper-sensitive. Mr Lamb is open-minded and open-hearted. For him all are welcome. He loves everybody and everything. He teaches Derry how to handle people and things.

Answer 12.
Griffin arrives in Bramblehurst railway station bundled from head to foot with only the tip of his nose showing. He enters the Coach and Horses Inn and demands a room and a fire. Mrs. Hall, the owner is quite excited and obliged as she feels lucky to have a customer in the chill of February. She was also delighted that he did not haggle over the rent. She prepares a supper for him and offers to take his coat and hat, but he refuses to take them off. When he finally removes the hat, his entire head is swathed in a bandage. Mrs. Hall thinks he has endured some accident that had disfigured his face. She tries to get him to talk about himself, but he is taciturn with her, although not particularly rude. When she went to clear his lunch dishes, the stranger finally requests her to have her luggage sent from Bramblehurst railway station. Her bid to initiate yet another conversation with the stranger is met with resistence as the stranger abruptly asks her to get some matches.

OR

Godfrey insists that he has a claim on Eppie and confesses that he is her father. Hitherto he had refrained from revealing this secret out of moral cowardice and guilt he felt for Molly’s death. However he did feel a sense of responsibility towards Eppie. When Godfrey is ready to claim Eppie as his own, he tells Silas that he is standing in the way of Eppie’s welfare. He persuades him that being wealthier he would be able to give a more luxurious upbringing to Eppie. Silas says that he will not argue anymore and leaves the decision up to Eppie. Eppie who dearly loves Silas and regards him as her father refuses Godfrey’s offer of taking care of her.

Answer 13.
Marvel is poor, homeless and jobless tramp. He wears shabby, old-fashioned clothes, like his ‘obsolete hat’, and he has buttons replaced by pieces of string. The narrator goes so far as to tell us that he does everything in a leisurely manner. He doesn’t seem to like work or excitement. The Invisible Man chooses him to carry out his ends as his own state of invisibility had necessitated the help of a normal person. Marvel was a destitute and a jobless man. Also he wasn’t as intelligent, strong willed and sharp as the invisible man. It was easy to manipulate him. Marvel was timid and quite terrified by the abusive and violent ways of the invisible man. He definitely doesn’t like working for the Invisible Man. So it’s no surprise when he decides to resign from his services.

OR

Eppie is Silas’ adopted daughter and the light of his life. Even as a little girl, she is adorable. She is affectionate and trusting—she latches on to Silas right away—and she is full of high spirits and lovable mischief. She loves flowers and butterflies, and birds and animals and loves to explore the world around her.

Not much changes are seen in her even after sixteen years. When we meet Eppie again, she is still high-spirited and full of mischief, talking about her desire for a garden just where Aaron Winthrop will hear, because she knows that he will volunteer to dig it for her. She has a little dog and cat that she frolics with, and she calls Silas by pet names. Eppie is completely devoted to Silas. She doesn’t even want to marry Aaron because she is afraid that she will have to leave her father. And when Godfrey, her biological father, offers to make her a lady, she refuses quickly and firmly thus showing her love and loyalty to Silas.

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CBSE Sample Papers for Class 12 Maths Paper 7 are part of CBSE Sample Papers for Class 12 Maths. Here we have given CBSE Sample Papers for Class 12 Maths Paper 7.

CBSE Sample Papers for Class 12 Maths Paper 7

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 7 of Solved CBSE Sample Paper for Class 12 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 100

General Instructions:

• All questions are compulsory.
• Questions 1-4 in section A are very short answer type questions carrying 1 mark each.
• Questions 5-12 in section B are short answer type questions carrying 2 marks each.
• Questions 13-23 in section C are long answer I type questions carrying 4 marks each.
• Questions 24-29 in section D are long answer II type questions carrying 6 marks each.

SECTION A

Question 1.
If A is a square matrix of order 3 and |adj A| = 144, then find the value of |3A| ?

Question 2.

Question 3.
Evaluate \(\int { \frac { { sec }^{ 2 }\left( logx \right) }{ x } } dx\)

Question 4.
Differentiate log (x + \(\sqrt { { x }^{ 2 }+{ a }^{ 2 } }\)) with respect to x.

SECTION B

Question 5.
Using elementary transformation find the inverse of the matrix \(\begin{pmatrix} 5 & 4 \\ 3 & 2 \end{pmatrix}\)

Question 6.
Using differentials find the approximate value of f(5.001) where f(x) = x³ – 7x² + 15.

Question 7.
Find the equations of tangents to the curve 3x² – y² = 8 which passes through the point (\(\frac { 4 }{ 3 }\) , 0)

Question 8.
If cos y = x cos (a + y) prove that \(\frac { dy }{ dx } =\frac { { cos }^{ 2 }\left( a+y \right) }{ sina }\)

Question 9.
Find the coordinate of the point where the line \(\frac { x+1 }{ 2 } =\frac { y+2 }{ 3 } =\frac { z+3 }{ 4 }\) meets the plane x + y + 4z = 6.

Question 10.
Let A and B are two events such that P (\(\bar { A\cup B }\)) = \(\frac { 1 }{ 6 }\), P(A ∩ B) = \(\frac { 1 }{ 4 }\) and P(\(\bar { A }\)) = \(\frac { 1 }{ 4 }\) .Prove that A and B are independent events.

Question 11.
Solve for x, tan^-12x + tan^-13x = \(\frac { \pi }{ 4 }\)

Question 12.
Evaluate \(\int { \frac { x-4 }{ \left( x-2 \right) ^{ 3 } } } { e }^{ x }dx\)

SECTION C

Question 13.

Question 14.

Question 15.

Question 16.

Question 17.

Question 18.
Form the differential equation of the family of circles in the first quadrant which touches the coordinate axes.

Question 19.

Question 20.

Question 21.
An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question.

Question 22.
An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01 and that of a motorcycle is 0.02. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle. How we can avoid accidents?

Question 23.

SECTION D

Question 24.
If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is \(\frac { \pi }{ 3 }\).
OR
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8m³. If building of tank cost ₹ 70 per square metre for the base and ₹ 45 per square metre for the sides, what is the cost of least expensive tank?

Question 25.
Find the area of the region included between the parabola y² = x and the line x + y = 2 using integration.
OR
Find the area of the smaller region bounded by the ellipse \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 4 } =1\) and the line \(\frac { x }{ 3 } +\frac { y }{ 2 } =1\) using integration.

Question 26.
If \(A=\left( \begin{matrix} 3 & 2 & -1 \\ -2 & 1 & 2 \\ 1 & -3 & 1 \end{matrix} \right)\) , find A^-1. Hence solve the system of linear equations
3x – 2y + z = 2
2x + y – 3z = – 5
-x + 2y + z = 6

Question 27.
A company manufactures two types of toys. Toys of Type A require 5 minutes each for cutting and 10 minutes each for assembling. Toys of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours available for assembling. The profit is ₹ 0.50 each for type A and ₹ 0.60 each for type B toys. How many toys of each type should be manufactured in order to maximize the profit?

Question 28.
The points A (4, 5, 10), B(2, 3, 4) and C(1, 2, -1) are three vertices of a parallelogram ABCD. Find the vector and cartesian equation of the sides AB and BC and find coordinates of D.
OR
Find the angle between lines whose direction cosines are given by the relations 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0

Question 29.
Solve the differential equation \(\frac { dy }{ dx }\) = sin (x + y) + cos (x + y)

Solutions

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

Solution 12.

Solution 13.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

Solution 18.
Equation of family of circle in the first quadrant which touches both the coordinate axes is

Solution 19.

Solution 20.

Solution 21.
Let E₁ is the event: it is an easy question
E₂ is the event: it is an M.C.Q (Multiple Choice Question)
True/False Easy Question = 300
True/False Difficult Question = 200
MCQ easy question = 500
Difficult MCQ = 400
Total Questions = 300 + 200 + 500 + 400 = 1400
Total easy questions = 300 + 500 = 800
Total difficult questions = 200 + 400 = 600
E₁ ∩ E₂ = easy M.C.Q = 500
E₂ = 500 + 400 = 900

Solution 22.
E₁ : Total insured scooters = 2000
E₂ : Total insured motorcycles = 3000
Total vehicles = 5000
A is the event insured vehicle meets with an accident.

Solution 23.

Solution 24.

Solution 25.

Solution 26.

Solution 27.
Let x type A and y type B toys manufactured.

Hence profit is maximum at the point (8, 20) means 8 toys of type A and 20 toys of type B should be manufactured.

Solution 28.
In parallelogram diagonal bisect to each other, so mid point of BD = Mid point of AC

Solution 29.

We hope the CBSE Sample Papers for Class 12 Maths Paper 7 help you. If you have any query regarding CBSE Sample Papers for Class 12 Maths Paper 7, drop a comment below and we will get back to you at the earliest.

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

CBSE Sample Papers for Class 12 Maths 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 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 100

General Instructions:

• All questions are compulsory.
• Questions 1-4 in section A are very short answer type questions carrying 1 mark each.
• Questions 5-12 in section B are short answer type questions carrying 2 marks each.
• Questions 13-23 in section C are long answer I type questions carrying 4 marks each.
• Questions 24-29 in section D are long answer II type questions carrying 6 marks each.

SECTION A

Question 1.
What is the principal value of \({ tan }^{ -1 }\left( tan\frac { 2\pi }{ 3 } \right)\)

Question 2.
If A is a square matrix of order 3 x 3 and |3A| = k |A|, then find the value of k.

Question 3.
What is the distance of the point (p, q, r) from the x-axis?

Question 4.
Let f: R → R be defined by f(x) = 3x² – 5 and g : R → R be defined by g(x) = \(\frac { x }{ { x }^{ 2 }+1 }\). Find fog.

SECTION B

Question 5.
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even} is an equivalence relation.

Question 6.
Let l_i, m_i, n_i For i = 1, 2, 3 be the direction cosines of three mutually perpendicular vectors in space. Show that AA’ = I₃.
Where

Question 7.
Form the differential equation corresponding to the function y² = a(b – x) (b + x) by eliminating a and b.

Question 8.
If e^y (x + 1) = 1, prove that \(\frac { dy }{ dx }\) = -e^y

Question 9.
Find the Cartesian and vector equations of line which passes through the point (-2, 4, -5) and parallel to the line given by
\(\frac { x+3 }{ 3 } =\frac { 4-y }{ -5 } =\frac { 8-z }{ -6 }\)

Question 10.
Solve the following L.R.R graphically
Maximise Z = 3x + 4y
Subject to constraints x + y ≤ 4, x ≥ 0, y ≥ 0

Question 11.
If P (E) = 0.35, P(F) = 0.45, P(E ∪ F) = 0.65. Find (i) P (E|F) (ii) P(E ∩ \(\bar { F }\)).

Question 12.
Find the points on the curve y = x³ at which the slope of the tangent is equal to y-coordinate of the point.

SECTION C

Question 13.
Find the area of the region bounded by the y-axis, y = cos x and y = sin x, 0 ≤ x ≤ \(\frac { \pi }{ 4 }\).

Question 14.
Evaluate \(\int _{ 0 }^{ 4 }{ \left( 2{ x }^{ 2 }+5x \right) } dx\) dx as a limit of sum.

Question 15.

Question 16.

Question 17.
Use differentials to find the approximate value of \(\frac { 1 }{ { \left( 2.002 \right) }^{ 2 } }\)
OR
Find the intervals for which the function f(x) = sin x – cos x is increasing or decreasing in [0, 2π].

Question 18.
Solve the differential equation y dx + (x – y²) dy = 0
OR
Solve the differential equation \(\frac { dy }{ dx } =\frac { x+y+1 }{ x+y }\)

Question 19.
A bag contains (2n + 1) coins. It is known that n of these coins have a head on both its sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is \(\frac { 31 }{ 42 }\). Find the value of n.

Question 20.
A plane meets the coordinate axis in A, B and C such that the centroid of ΔABC is the point (α, β, γ). Show that the equation of the plane is \(\frac { x }{ \alpha } +\frac { y }{ \beta } +\frac { z }{ \gamma } =3\).

Question 21.
Two cards are drawn simultaneously without replacement from a well shuffled pack of 52 cards. Find the mean and variance of number of red cards.

Question 22.

Question 23.
If a 20 year old girl drives her car at 25 km/h, she has to spend ₹ 4/km on petrol. If she drives her car at 40 km/h, the petrol cost increases to ₹ 5/km. She has ₹ 200 to spend on petrol and wishes to find the maximum distance she can travel within one hour. Express the above problem as L.P.P. Write any one value reflected in the problem.

SECTION D

Question 24.

Question 25.

OR
Let N be the set of all natural numbers and let R be a relation in N x N defined by (a, b) R (c, d) ⇒ ad = bc
For all (a, b), (c, d) ∈ N x N, show that R is an equivalence relation on N x N.

Question 26.

Question 27.
Define skew lines. Using only vector approach, find the shortest distance between the following two skew lines

Question 28.

Question 29.
Show that the semivertical angle of a right circular cone of given surface area and maximum volume is \({ sin }^{ -1 }\left( \frac { 1 }{ 3 } \right)\)

Solutions

Solution 1.

Solution 2.
A is a square matrix of order 3 x 3
|3A| = k|A| ⇒ |3A| = 3³ |A| = 27 |A| = k |A|
So k = 27.

Solution 3.

Solution 4.

Solution 5.
A= {1,2, 3, 4, 5}
R = {(a, b): |a- b| is even}
Reflexive: (a, a): |a – a| = 0 is even
So (a, a) ∈ R, ∀ a ∈A
So R is reflexive.
Symmetric: (a, b): |a – b| is even, ∀ (a, b) ∈ R
= |b – a| is even
= (b, a) ∈ R
So R is symmetric.
Transitive: Let (a, b), (b, c) ∈ R
(a, b): |a – b| is even and (b, c): |b – c| is even.
a – b = ± 2m, b – c = ± 2n, ∀ m, n ∈ N
a – c = ± 2m ± 2n = ± 2 (m + n)
|a – c| is even.
So (a, c) ∈ R.
So R is transitive.
Since R is reflexive, symmetric and transitive, so R is an equivalence relation.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

Solution 12.

Solution 13.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

Solution 18.

Solution 19.
Let E₁ is the event: Coin having head on both sides
E₂ is the event: Coin is fair.
A is the event: head comes upon tossing a coin

Solution 20.

Solution 21.
Let Random variable x denote the number of red cards drawn.
Possible values of x are 0, 1, 2
Two cards are drawn simultaneously without replacement.
Total cards = 52, Red cards = 26
Remaining cards = 26

Solution 22.

Solution 23.
Let distance covered with speed of 25 km/h is x km, distance covered with speed of 40 km/h is y km.
Objective function is maximize distance = Z = x + y,
Subject to constraints are
4x + 5y ≤ 200
x ≥ 0, y ≥ 0
Value: At slow speed consumption of petrol is low, so maintain normal speed and save petrol.

Solution 24.

Solution 25.

Solution 26.

Solution 27.
Skew lines: The lines which are neither intersecting nor parallel are called skew lines.

Solution 28.

Solution 29.

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

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.

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

CBSE Sample Papers for Class 12 Chemistry 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 Chemistry is given below with free PDF download solutions.

Time Allowed : 3 Hours
Max. Marks : 70

General Instructions :

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

Questions

Question 1.
Name the non-stoichiometric point defect responsible for colour in alkali metal halides.

Question 2.
Gelatin is generally added to ice cream. Give reason.

Question 3.
Give IUPAC name of the following organic compound :

Question 4.
Write the structure of Acetoxy salicylic acid.

Question 5.
Name the mono saccharides persent in lactose.

Question 6.
Differentiate between molarity and molality for a solution. How does a change in temperature influence their values?

Question 7.
A chemical reaction is of first order w.r.t. reactant A and of second order w.r.t. reactant B. How is the rate of this reaction affected when

1. the concentration of B is alone increased to three times.
2. the concentration of A as well as B are doubled?

Question 8.
Complete the following chemical equations:

1. Cu + Cone. HNO₃ →
2. S₈ + H₂SO₄ (conc.) →

Question 9.
Write complete chemical equations for

1. Oxidation of Fe²⁺ by Cr₂O^2-₇ in acidic medium
2. Oxidation of S₂O^2-₃ by MnO^–₄ in neutral aqueous medium

Or
Complete the following chemical equations:

1. MnO^–₄ (aq) + C₂O^2-₄ (aq) + H⁺ (aq) →
2. Cr₂O^2-₇ (aq) + Fe²⁺ (aq) + H⁺(aq) →

Question 10.
Which one of the following pair of sustances undergoes S_N2 substitution reaction faster and why?

Question 11.
Calcium crystallises in Face Centered Cubic structure. Radius of the atom in the metal is 141.4 pm.
(a) What is the length of the side of the unit cell?
(b) How many atoms of Calcium are there in its 5 cm³?

Question 12.
Calculate the freezing point of an aqueous solution containing 10.5 g of MgBr₂ in 200 g of water (given molar mass of MgBr₂ = 184 g mol^-1, K_f for water = 1.86 K kg mol^-1, also MgBr₂ is 80% dissociated).

Question 13.
The rate of a reaction becomes four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. (R = 8.314 JK^-1 mol^-1)

Question 14.

1. How electrophoresis leads to coagulation of gold sol?
2. Explain Tyndall effect briefy by taking an example.
3. Explain why coagulation of sols will occur.p-

Question 15.
(a) Briefly explain leaching process with reaction.
(b) With the help of reactions, explain how pure Zr is obtained from impure Zr?
(c) With the help of reactions, explain how pure gold is obtained?

Question 16.
How would you carry out the following conversions? Write complete equations in each case.
(a) 1-methylcyyclohexene to 1-iodo methylcyclohexanol
(b) Styrene to 1-bromo-l-phenylethane
(c) But-4-enal to 2-Bromobutane

Question 17.

1. Why in transition elements, the melting point first increases and then decreases?
2. Why scandium (Atomic no. 21) is considered a transition metal?
3. How would you account for the following-Actinoid contraction is greater than lanthanoid contraction?

OR
With the help of chemical reactions, explain how KMnO₄ is prepared from pyrolusite ore? Also, write the chemical reaction of KMnO₄ with SO₂ gas in acidic medium?

Question 18.
How will you bring about following conversions? Write complete equations in each case.
(a) Ethanal to 3-hydroxybutanal
(b) Benzaldehyde to benzophenone
(c) Propanone to propene

Question 19.
Give the structures of A, B and C in the following reactions:

Question 20.
Explain the following:
(a) Peptide Linkage
(b) Essential amino acid
(c) Pyranose Structure of β Glucose

Question 21.
Write the names and structures of the monomers of the polymers given below:
(a) Buna-S
(b) Neoprene
(c) Nylon-6,6

Question 22.

1. Give two macromolecules used as drug traget.
2. What are antiseptics?
3. What are anionic detergents?

Question 23.
Mamta, a housewife lives in Sonepat (Haryana). The tap water she gets is rich in dissolved salts. She uses ordinary washing soap for washing clothes and she finds that soaps are not very effective in cleaning her clothes. One of her neighbours, Shilpa is a student of science. Shilpa suggested to her to use synthetic detergents for washing clothes.
Answer the following questions:
(i) As a science student, why did Shilpa suggest to Mamta to use synthetic detergent?
(ii) Give one chemical reaction to justify why not to use ordinary soap.
(iii) What were the values associated with the above decision?

Question 24.
What type of a battery is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it. Calculate the emf of the following cell at 25°C:
Cu(s) | CuSO₄(10^-1 M)|| Ag₂SO₄(10^-3 M) | Ag
Given E_cell° = 0.46 V
OR
(a) Write the anode and cathode reactions occurring in a commonly used mercury cell. How is the overall reaction represented?
(b) Calculate the degree of dissociation and dissociation constant of 0.01 M acetic acid at 298 K, given that
Δ⁰_m (CH₃COOH) = 39 S cm² mol^-1
Δ⁰_m (CH₃COO^–) = 40.9 S cm² mol^-1
Δ⁰_m (H⁺) = 349.1 S cm² mol^-1

Question 25.
(a) Explain each of the following:

1. Molten PCl₅ conducts electricity
2. The electron gain enthalpy of Fluorine is less than Chlorines .
3. Fluorine forms only one oxoacid, that is HOF

(b) Draw the structure of
(i) XeOF
(ii) ClF^–₄
OR
Answer the following

1. Which neutral molecule would be isoelectronic with CIO^–?
2. Of Bi (V) and Sb(V) which may be stronger oxiding agent and why?
3. What is the basicity of H₃PO₃ and why?
4. Why do noble gases have very low boiling points?
5. Nitrogen is much less reactive than phosphorus at room temperature. Why?

Question 26.
(a) Arrange the following sets of compounds in increasing order according to the given instruction.

1. Phenol, 2-nitrophenol, Cyclohexanol (Acidic character)
2. pentan-l-ol, n-butane, pentanal, ethoxyethane (boiling points)

(b) Describe the mechanism of hydration of ethene to yield ethanol.
(c) Write Kolbe’s reaction with an example.
OR
(a) Explain the following name reactions with a suitable example.

1. Reimer-Tiemann reaction
2. Williamson’s synthesis

(b)

1. Haloalkanes are more soluble in organic solvent as compared to water, why?
2. Grignard’s reagent is prepared under anhydrous conditions, why?
3. Phenol is more acidic than ethanol, why?

Answers 

Answer 1.
F-centre or metal excess defect

Answer 2.
Gelatin acts as an emulsifying agent as well as prevents crystallisation of water.

Answer 3.
3-Hydroxyhex-4-enoic acid

Answer 4.

Answer 5.
β-D-galactose and β-D-glucose.

Answer 6.

• Molarity : It is the number of moles of solute per litre of solution.
• Molality : It is number of moles of solute per kilogram of solvent.

In case of molarity, its value will decrease with increasing temperature but molality remains unchanged with change in temperature.

Answer 7.
According to given information the rate law equation is
rate = k[A] [B]²

1. On increasing [B] three times
   rate₁ = k [A] [3B]²
   = 9 k[A] [B]²
   i.e. rate becomes nine times.
2. When [A] and [B] gets doubled, then
   rate₂ = K[2A] [2B]² = 8 K[A] [B]²
   i.e. rate becomes 8 times.

Answer 8.

1. Cu + 4HNO₃ (conc.) → Cu (NO₃)₂ + 2NO₂ + 2H₂O
2. S₈ + 16H₂SO₄ (conc.) → 24SO₂ + 16H₂O

Answer 9.
(a) Cr₂O^2-₇ + 6Fe²⁺ + 14⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(b) 8MnO^–₄ + 3S₂O^2-₃ + H₂O → 8MnO₂ + 6SO^2-₄ + 2OH^–
OR
(a) 2MnO^–₄ + 5S₂O^2-₄ + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂
(b) CrO^2-₇ + 3H₂S + 8H⁺ → 2Cr³⁺ + 3S + 7H₂O

Answer 10.

Answer 11.
Since calcium crystallises in fee structure, so

Answer 12.
From question

Answer 13.
Since K₂ = 4K₁
T₁ = 300 K, T₂ = 320 K, R = 8.314 JK^-1 mol.
from Arrhenius relation:

Answer 14.

1. The colloidal particles move towards oppositely charged electrode i.e. electropheoresis takes place and it will lead to coagulation of gold sol.
2. The path of light becomes visible due to scattering of light by particles of starch sol and tyndall effect will take place.
3. Coagulation of sol will occur due to neutralisation of charges on the colloidal particles.

Answer 15.
(a) The process of leaching with NaOH (aq) is used for leaching of bauxite ore to get pure alumina (Al₂O₃). The reaction involved is
Al₂O₃(g) + 2NaOH(aq) + 3H₂O(g) → Al₂O₃2CH₂O↓ + 2NaHCO₃ (aq)
Precipitate is filtered, washed and dried, followed by heating to get pure Al₂O₃.

Answer 16.

Answer 17.

1. It is due to strong metal metal bond which is there because of small size and large number of unpaired electrons in the valence shell.
2. Sc (21) → [Ar] 3d¹4s²
   It contains partially filled d-orbital in the ground state, i.e. 3d¹ configuration, hence it is d-orbital considered as a transition metal.
3. It is due to poorer shielding of 5f electrons in actinoids as compared to 4f electrons of lanthanoids.

OR
Preparation of KMnO₄ from pyrolusite ore :

Reaction with SO₂
2MnO^–₄ + 5SO₂ + 2H₂O \(\underrightarrow { { H }^{ + } } \) 2Mn²⁺ + 5SO^2-₄ + 4H⁺

Answer 18.

Answer 19. 

Answer 20.
(a) Peptide linkage : Proteins are connected to each other through

(amide linkage) bond which is called peptide bond or linkage in protein chemistry,
e.g.

(b) Essential amino acid : The amino acids which must be taken in diet regularly are known as essential amino acids.
(c) Pyranose structure of β-glucose is as follows:

Answer 21.

Answer 22.

1. Protein, enzymes, hormones etc.
2. Antiseptics : The class of compounds used to kill microorganisms on living cells are known as antiseptics.
3. These are sodium salts of sulphonated long chain alcohols or hydrocarbons, since large part of such detergent is anionic and cleansing action is also due to this part. Hence such detergents are known as anionic detergents. Most of the household detergents are anionic.

Answer 23.
(a) Synthetic detergents provide cleaning in soft water as well as in hard water.

(c) Values associated are knowledge of chemistry, concern for friend, cleanliness and environment protection, (any two)

Answer 24.
It is a secondary battery, i.e. rechargeable
Cell reaction during battery in use:
At anode : Pb(s) + SO^2-₄(aq) → PbSO₄(aq) + 2e^–
Cathode : Pb(s) + SO^2-₄(aq) + 4H⁺(aq) + 2e → PbSO₄(s) + 2H₂O (l)
Net reaction : Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄ + 2H₂O(l)

(b) Cell reactions at
Anode → Cu(s) Cu²⁺ + x  Here, n = 2
Cathode → (Ag⁺ + e^– → Ag) × 2
Net reaction :
Cu(s) + 2Ag⁺ → Cu²⁺ + 2Ag
At 298 K, Nemst equation for emf of cell:

OR
Reaction of mercury cell at
Anode : Zn (Hg) + 20H^– → ZnO(s) + H₂O + 2e^–
Cathode : HgO + 2H₂O + 2e^– → Hg(l) + 2OH^–
Overall reaction:
Zn(Hg) + HgO(s) → ZnO(s) + Hg(s)
[since no ion is involved during the overall reaction, so cell potential remains constant during the entire life of the cell]

(b) from question (Δ⁰_m) = 39 S cm² mol^-1
limiting molar conductivity (Δ⁰_m) for acetic acid
Δ⁰_m (CH₃COOH) = λ_CH3COO^– + λ°_H+
= 349.1 + 40.9 – 390 S cm² mol^-1
Now, Degree of dissociation,

α = 0.1
Dissociation constant (K_a) for fa acetic acid :
K = Cα² = (0.01) (0.1)² = 10^-4

Answer 25.
(a)

1. In molten state PCl₅ exist as [PCl₄]⁺ and [PCl₆]^– ions, hence it conducts electricity.
2. It is due to high electron density in valence shell of fluorine.
3. Due to absence of d orbitals in F, it cannot expand its covalency.

OR

1. OF₂ or ClF
2. Bi(V) is an stronger oxidising agent, reason being Bi(III) is more stable than Bi(V) because of inert pair effect.
3. Structure of H₃PO₃ –
   
   Its basicity is two and the reason being is, it consists of two OH groups, hence only 2H⁺ ions can be released.
4. Noble gases are monotomic gas molecules and hence only weak dispersion forces are there. As a result they have low boiling point.
5. Triple bond present between two nitrogen atoms is much stronger than P-P single bond, hence nitrogen is less reactive than phosphorous.

Answer 26.
(a)

1. Cyclohexanol < phenol < 2-nitrophenol
2. n-butane < ethoxyethane < pentanal < pentan-1-ol

OR
(a)

1. Reimer-Tiemann reaction:
   
2. Williamson’s synthesis:
   An alkyl halide reacts with sodium alkoxide to form ether.
   R-X + R’- ONa → R-O-R’ + NaX
   CH₃Cl + NaOC₂H₅ → CH₃-O-CH₂-CH₃ + NaCl

(b)

1. Haloalkanes are unable to form hydrogen bond with water but bonding of haloalkanes with organic solvent is as strong as among haloalkanes. Due to this reason haloalkanes are more soluble in organic solvent than water.
2. Grignard reagent reacts immediately with water to form alkane, hence they are prepared under anhydrous conditions.
   RMgX + H₂O → RH + MgX(OH)
3. Phenoxide ion is resonance stablised, hence phenol is more acidic than ethanol (ethoxide ion is destablised by +I effect of alkyl group).

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