CBSE Sample Paper Class 12 Daily Practice Questions

12 (a) Ratio of neutrons to protons is 2.3: 1 which is not the stable ratio of 1:1                                 (b) Age of fossils can be estimated by C-14 All living organisms have C-14  which decays without being replaced back once the organism dies. (c) carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to  stable calcium (d) t = 2.303/ k log (Co/Ct)                                                                                                    Co = 20 g Ct = ? t = 10320 years k = 0.693/6000 (half-life given in passage) substituting in equation: 10320 = 2.303 / (0.693/6000) log 20/ Ct                                                                            0.517 = log 20 / Ct  anlilog (0.517) = 20/Ct 3.289 = 20/Ct                                                                                                                        Ct = 6.17 g                                                                                                                            OR t = 2.303/ k log (Co/Ct)                                                                                                    Co = 32 g Ct = 12 t = ? k = 0.693/6000 (half life given in passage) substituting in equation: t = 2.303 / (0.693/6000) log 32/ 12                                                                                     t = 2.303 x 60000 /0.693 log 2.667                                                                                     t = 2.303x6000x0.4260 /0.693

12. (a) Refraction, Total internal reflection

a. India’s history of religious and cultural traditions emphasized the protection of In many cultures, tracts of forest are set aside, all the trees and wildlife within are venerated and given total protection. Sacred groves in many states are the last refuges for a large number of rare and threatened plants. (2 marks) b. Area A will have more species richness and a steeper slope.

Compound A is an alkene, on ozonolysis it will give carbonyl compounds. As both  B and C have >C=O group, B gives positive Fehling’s test so it is an aldehyde and it gives iodoform test so it is so it has CH3C=O group. This means the aldehyde is acetaldehyde C does not give Fehling’s test, so it is a ketone. It gives positive iodoform test so it is a methyl ketone means it has CH3C=O group Compound A (C5H10) on ozonlysis gives B (CH3CHO) + C (CH3COR) So “C” is CH3COCH3 CH3CH=C(CH3)2 (i)O3 (ii) Zn/H3O?⁺             CH3CHO + CH3COCH3 CH3CHO + 2Cu²⁺ + 5OH^-  ? CH3COO^- + Cu2O (red ppt) + 3H2O CH3COCH3   + 2Cu²⁺ + 5OH^- ?     No reaction CH3CHO + 3I2 + 3 NaOH? CHI3 (yellow ppt) + 3HI + HCOONa CH3COCH3 + 3I2 + 3 NaOH ?     CHI3 (yellow ppt) + 3HI + CH3COONa A = CH3CH=C(CH3)2 B = CH3CHO C = CH3COCH3

Given a_µg = 1.5 Focal length of the given convex lens when it is placed in air is f = + 20 cm Refractive index of the given medium with respect to air is a_µm = 1.25 New focal length of the given convex lens when placed in a medium is f ^' 1/f= a/µg  - 1) [(1/R  ) + (1/R  )]--------- (A)
1/fF  = M/µg  ? 1) [(1/R  ) + (1/R  )]-------- (B)
Dividing (A) by (B), we get f ^' /f = (aµg  ? 1)/ (mµ  -1) = (1.5 ? 1)/(1.2 ? 1)  =  0.5/ 0.2= 5/2= 2.5 f ^' = 2.5f = (2.5 × 20)cm = +50cm as m _µg = µg _µm = 1.5/1.25 = 1.2 New focal length is positive. The significance of the positive sign of the focal length is that given convex lens is still converging in the given medium

8. The flow chart shows the three steps involved in the process of PCR showing the following -              Denaturation The DNA strands are treated with a temperature of 940C (Heat) and the strands are separated. -              Annealing The primers anneal to the complementary strands -              Extension The DNA polymerase facilitates the extension of the strands.  (1x3=3 marks) OR Refer to the official website of CBSE for the diagram 10. a. When a large habitat is broken into small fragments due to various activities, mammals and birds requiring large territories and certain animals with migratory habitats are badly affected, leading to population decline.      (1 mark) b.

• Nile perch introduced in Lake Victoria eventually led to the extinction of an ecologically unique assemblage of more than 200 species of child fish. 3
• Parthenium/Lantana/water hyacinth caused environmental damage and threat to our native species
• African catfish-Clarias gariepinus introduced for aquaculture purposes is posing a threat to the indigenous catfishes in our rivers. (Any one) (1 mark)

C. Yes; Humans have overexploited natural resources for their ‘greed’ rather than ‘need’ leading to extinction of these animals. Sustainable harvesting could have prevented extinction of these species. (1 mark)

4. (a) Ti is having electronic configuration [Ar] 3d2 4s2. Ti (IV) is more stable as 1 Ti4+ acquires nearest noble gas configuration on loss of 4 e-. (b) In case of transition elements, ions of the same charge in a given series show 1 progressive decrease in radius with increasing atomic number. As the new electron enters a d orbital each time the nuclear charge increases by unity. The shielding effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and the ionic radius decreases. (c) Iron and Chromium are having high enthalpy of atomization due to the presence of unpaired electrons, which accounts for their hardness. However, Zinc has low 1 enthalpy of atomization as it has no unpaired electron. Hence zinc is comparatively a soft metal.

9 (a). ? = 2000 Å = (2000 × 10–10)m Wo = 4.2eV h = 6.63 × 10–34JS Using Einstein's photoelectric equation K. E. = (6.2 – 4.2) eV = 2.0 eV (b). The energy of the emitted electrons does not depend upon intensity of incident light; hence the energy remains unchanged. (c). For this surface, electrons will not be emitted as the energy of incident light (6.2 eV) is less than the work function (6.5 eV) of the surface.

7. Transformation of normal cells into cancerous neoplastic cells may be induced by following physical, chemical or biological agents causing DNA damage:

• Ionising radiations like X-rays and gamma rays
• Non-ionizing radiations like UV.
• Chemical carcinogens present in tobacco smoke
• Cellular oncogenes (c-onc) or proto-oncogenes, when activated under certain conditions cause cancer. Viruses with oncogenes can transform normal cells to cancerous cells. (any 3; 1 x 3 marks)

OR If the person has sustained high fever (39° to 40°C), weakness, stomach pain, constipation, headache and loss of appetite, it is Typhoid.   (1 mark) If the person has fever, chills, cough and headache; and the lips and fingernails turn gray to bluish, it is Pneumonia.                (1 mark) If the person has chills and high fever recurring every three to four days then, it is Malaria.            (1 mark) 8. When our body encounters an antigenic protein or a pathogen for the first time it produces a response which is of low intensity and our body retains memory of the first encounter. (1 mark) The subsequent encounter with the same pathogen elicits a highly intensified response carried out with the help of two special types of lymphocytes present in our blood, B-              3 lymphocytes, and T-lymphocytes.            (1 mark) The B-lymphocytes produce an army of proteins in response to these pathogens into our blood to fight with them. These proteins are called antibodies. The T-cells themselves do not secrete antibodies but help B-cells produce them.                (1 mark)

A. 8. (a) Refer to the official website of CBSE for this diagram based answer. (b) m_o = 30, f_o = 1.25 cm, f_e = 5 cm when image is formed at least distance of distinct vision, D = 25 cm Angular magnification of eyepiece m_e = (1 + D/f_e ) = 1 + 25/5  = 6 Total Angular magnification, m = m_o m_e  ? m_o = m/m_e   = 30/6 = 5 As the objective lens forms the real image, m_o = v_o/u_o  = -5 ? v_o   = -5u_o using lens equation, u_o = -1.5 cm, v_o = ?5 × (-1.5)cm = +7.5 cm Given v_e = -D = -25 cm, f_e = +5 cm, u_e = ? using again lens equation u_e   = 25/6 Thus, object is to be placed at 1.5 cm from the objective and separation between the two lenses should be L = v_o + Iu_el = 11.67 cm Or 8.  (a) Refer to the official website of CBSE for this diagram based answer. (b) (I) In normal adjustment: Magnifying m = f_o/f_e = (140/5) = 28 (II) When the final image is formed at the least distance of distinct vision (25 cm) :   m = f_o/f_e(1 + f_e/D) = (28 x 1.2) = 33.6

A. 5. This interaction will lead to competition between the individuals of population A,B and C for resources. Eventually the ‘fittest’ individuals will survive and reproduce. (1 mark)
The resources for growth will become finite and limiting, and population growth will become realistic. (1 mark)
6. The relationship between the plant and pollinator is called mutualism. Fig depends on wasp for pollination, and wasp depends on fig for
food and shelter. (1 mark)
With the decline in population of figs, wasp loses its source of food and shelter. (1 mark)
OR
Regulators;
Thermoregulation, Osmoregulation
Birds/mammals (any one) 

A. (a) The colour of coordination compound depends upon the type of ligand and dd transition taking place .
H2O is weak field ligand , which causes small splitting , leading to the d-d
transition corresponding green colour , however due to the presence of ( en )
which ia strong field ligand , the splitting is increased . Due to the change in
t2g -eg splitting the colouration of the compound changes from green to blue.
(b)Formula of the compound is [Co(H2NCH2CH2NH2 )3 ]2 (SO4 )3
The hybridisation of the compound is: d2sp3

A. 3. It is Morphine. Physically it appears as a white, odourless, crystalline compound.
4. At collection points A and B, the BOD level is high due to high organic pollution caused by sugar factory and sewage discharge. 
At the collection point C, the water was released after secondary treatment/ biological treatment (where vigorous growth of useful aerobic microbes into flocs consume the major part of the organic matter present in the river water or effluent due to sugar factory and sewage discharge). 

A. a) In case of chlorobenzene, the C—Cl bond is quite difficult to break as it acquires
a partial double bond character due to conjugation.
So Under the normal conditions, ammonolysis of chlorobenzene does not yield aniline.
b) Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. Due to the presence of three hydrogen atoms, the intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it.
c) During the acylation of aniline, stronger base pyridine is added. This done in order to remove the HCl so formed during the reaction and to shift the equilibrium to the right hand side. 

A. From Bohr’s theory, the frequency f of the radiation emitted when an electron de – excites from level n₂ to level n₁ is given as f = 2? ²mk²z²e⁴/h³ (1/n₁²-n₂²) Given n₁= n-1, n2 = n, derivation of it f = 2? ²mk²z²e⁴/h³ (2n-1)/ (n-1)² n² For large n, 2n ? 1 = 2n, n ? 1 = n and z = 1 Thus, f= 4?²mk²e⁴/n³h³ which is same as orbital frequency of electron in n^th orbit. f= v/2?r= 4? ²mk²e⁴/n³h³

A. 1. Microbial pathogens enter the gut of humans along with food:
Physical barriers: Mucus coating of the epithelium lining the
gastrointestinal tract helps in trapping microbes entering our body.
Physiological barriers: Acid in the stomach, saliva in the mouth prevent microbial growth. 
2. Streptokinase (produced by the bacterium Streptococcus) is used as a
‘clot buster’ for removing clots from the blood vessels of patients who have undergone myocardial infarction. 
Statins (produced by the yeast Monascus purpureus) act as bloodcholesterol lowering agents.
OR
Eradication of pests will disrupt predator-prey relationships, where beneficial predatory and parasitic insects which depend upon flora and fauna as food or hosts, may not be able to survive.
Holistic approach ensures that various life forms that inhabit the field, their life cycles, patterns of feeding and the habitats that they prefer are extensively studied and considered. 

A. 1. (a)Picric acid < salicylic acid < benzoic acid (b)Methyl tert – butyl ketone < acetone< Acetaldehyde
(c)ethanol increase in molar mass)  2. B is a strong electrolyte. The molar conductivity increases slowly with dilution as there is no increase in number of ions on dilution as strong electrolytes are
completely dissociated. 
3.(a) The alpha hydrogen atoms are acidic in nature due to presence of electron withdrawing carbonyl group. These can be easily removed by a base and the carbanion formed is resonance stabilized.
(b) Tollen’s reagent is a weak oxidizing agent not capable of breaking the C-C bond in ketones . Thus ketones cannot be oxidized using Tollen’s reagent itself gets
reduced to Ag. 

A. 1. As given in the statement antimony is added to pure Si crystal, then a n -type extrinsic semiconductor would be so obtained, Since antimony(Sb) is a pentavalent impurity. 
2. No, because according to Bohr's model, En = -13.6/n² and electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as L = mvr = nh/2? (i) The increase in the frequency of incident radiation has no effect on photoelectric current. This is because of incident photon of increased energy cannot eject more than one electron from the metal surface. (ii) The kinetic energy of the photoelectron becomes more than the double of its original energy. As the work function of the metal is fixed, so incident photon of higher frequency and hence higher energy will impart more energy to the photoelectrons.
3. Photodiodes are used to detect optical signals of different intensities by changing current flowing through them. 
Applications of photodiodes:
In detection of optical signals, demodulation of optical signals, light operated switches, speed reading of computer punched cards, electronic counters
(any two out of these or any other relevant application)