CBSE Sample Paper Class 12 Daily Practice Questions

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Yamini
Yamini Saini
Assistant Manager - Content
Updated on May 13, 2022 17:39 IST

CBSE Class 12 exams will be conducted from April 26 to June 15, 2022 for Term 2. You can practice questions here daily.

CBSE Sample Paper Class 12: The Central Board of Secondary Education (CBSE) will be conducting the CBSE Class 12 exams from April 26 to June 15, 2022. The exams commenced with the Entrepreneurship, Beauty and Wellness exam on April 26, while the last exam will be of Psychology on June 15. These CBSE 12th preparations tips 2022 are helpful to organize the study time, marking a CBSE Class 12 study plan and other important things. This year, the CBSE class 12 syllabus 2022 has been reduced by 30% and divided into two terms. Each term contains 50% syllabus. The Term 1 exams were conducted from November 16 to December 30 for minor as well as major subjects. The result for the same was released offline on March 19.

CBSE has released CBSE Sample Paper Class 12 on its official website along with the marking scheme for Term 2 examinations. Students must practicse from them daily to assss their preparations and thereafter focus on weaker areas.

Also Read: CBSE Term 2 exam guidelines released for classes 10 and 12

Shiksha.com brings to you daily practice questions that will help you in better preparation of the exams. This blog is a compilation of subject questions from the CBSE Sample paper Class 12 2022. Get started with the CBSE practice test by solving the CBSE question bank below.

LATEST UPDATE
  • 5:39 PM IST13 May 2022

    CBSE Class 12 Biology Syllabus

     

    Unit

    Chapters

    Marks

    Unit - VIII

    Biology and Human Welfare

    Human Health and Disease

    14

    Microbes in Human Welfare

    Unit IX - Biotechnology and Its Applications

    Biotechnology: Principles and Processes

    11

    Biotechnology and its Applications

    Unit X - Ecology and Environment

    Organisms And Population

    10

    Biodiversity and its Conservation

     

  • 4:05 PM IST12 May 2022

    CBSE Class 12 Physics Syllabus Term 2

    Unit

    Chapters

    Marks

    Unit – V

    Electromagnetic Waves

    17

    Chapter–8: Electromagnetic Waves

    Unit – VI

    Optics

    Chapter–9: Ray Optics and Optical Instruments

    Chapter–10: Wave Optics

    Unit - VII

    Dual Nature of Radiation and Matter

    11

     

    Chapter–11: Dual Nature of Radiation and Matter

    Unit–VIII

    Atoms and Nuclei

    Chapter–12: Atoms

    Chapter–13: Nuclei

    Unit IX

    Electronic Devices

    7

    Chapter–14: Semiconductor -Electronics: Materials, Devices and Simple Circuits

     

    Total

    35

     

  • 4:09 PM IST11 May 2022

    CBSE Class 12 Sample Questions Biology- May 11

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    (b) In Bt cotton a cry gene has been introduce from bacterium Bacillus thuringiensis (Bt) which causes synthesis of a toxic This protein becomes active in the alkaline gut of bollworm feeding on cotton, punching holes in the lining causing death of the insect. However; a Non Bt crop will have no effect on the cotton bollworm/ the yield of cotton will decrease / non Bt will succumb to pest attack.
  • 2:06 PM IST10 May 2022

    CBSE Class 12 Sample Questions Physics - May 10

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    A. (iii) 

     

     

  • 6:35 PM IST6 May 2022
  • 6:06 PM IST6 May 2022

    CBSE Class 12 Sample Questions Chemistry- May 6

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    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
  • 5:07 PM IST6 May 2022

    CBSE Class 12 Sample Questions Chemistry- May 6

     

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    (a) Manganese is having lower melting point as compared to chromium , as it has highest number of unpaired electrons , strong interatomic metal bonding , hence no delocalisation of electrons . (b) There is much more frequent metal – metal bonding in compounds of the heavy transition metals i.e 4d and 5d series , whixh accounts for lower melting point of 3d series. (c) Tungsten
  • 2:16 PM IST6 May 2022

    CBSE Class 12 Sample Questions Chemistry- May 6

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    (a) The ability of fluorine to stabilize the highest oxidation state is attributed to the higher lattice energy or high bond enthalpy. (b) Co2+ has three unpaired electrons so it would be paramagnetic in nature, hence Co2+ ion would be attracted to magnetic field. (c) The transition elements of 5d series have intervening 4f There is greater effective nuclear charge acting on outer valence electrons due to the weak shielding by 4f electrons. Hence first ionisation energy of 5 d series of transition elements are higher than that of 3d and 4d series.
  • 7:31 PM IST5 May 2022

    CBSE Class 12 Sample Questions Chemistry- May 5

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    Al(s) /Cd2+  (0.1M) // Al3+ (0.01M) /Cd(s) 2Al(s) + 3Cd2+  (0.1M)  à 3Cd (s) + 2Al3+ (0.01M) Ecell = Eocell -0.059/n log [Al3+]2 n          [Cd2+]3 Ecell= 1.26 – .059/6 log (0.01)2        (0.1)3 = 1.26 – 0.059 (-1) 6 = 1.26+0.009 = 1.269 V
  • 12:08 PM IST5 May 2022

    CBSE Class 12 Sample Questions Physics - May 5

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    12. (a) Refraction, Total internal reflection

     

  • 7:41 PM IST4 May 2022

    CBSE Class 12 Sample Questions Biology- May 4

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    13. (a) The two different DNA molecules will have compatible ends to recombine.
  • 6:30 PM IST4 May 2022

    CBSE Class 12 Sample Questions Chemistry- May 4

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    (a). N-Ethyl-N-methylbenzenamine or N-Ethyl-N-ethylaniline (b). Refer to the official website of CBSE for the diagram based answer
  • 4:07 PM IST4 May 2022

    CBSE Class 12 Sample Questions Physics - May 4

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    (a). The total intensity at a point where the phase difference is ?, is given by ???? = ????1 + ????2 + 2?????1????2 ???????????? ?. Here ????1 and ????2 are the intensities of two individual sources which are equal. When ? is 0, I = 4????1. When ? is 90o, I = 0 Thus intensity on the screen varies between 4????1 and 0.
    (b). Refer to the official website of CBSE for this diagram based answer
  • 5:22 PM IST2 May 2022

    CBSE Class 12 Sample Questions Biology- May 2

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    (a) The two different DNA molecules will have compatible ends to (½ mark) (b) Restriction enzyme cuts the DNA of the vector and then ligates the gene of interest into the DNA of the (1 mark) (c) 2 fragments (½ mark) 5’ ATTTTGAG 3’5’GATCCGTAATGTCCT 3’ 3’ TAAAACTCCTAG 5’.3’GCATTACAGGA 5’                       (1 mark) (d) BamH1 site will affect tetracycline antibiotic resistance gene, hence the recombinant plasmids will lose tetracycline resistance due to inactivation of the resistance gene. (1 mark) Recombinants can be selected from non recombinants by plating into a medium containing tetracycline, as the recombinants will not grow in the medium because the tetracycline resistance gene is cut. (1 mark)
  • 4:49 PM IST2 May 2022

    CBSE Class 12 Sample Questions Chemistry- May 2

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    9. (a) When N-ethylethanamine reacts with benzenesulphonyl chloride, N,N-diethylbenzenesulphonamide is (b) When benzylchloride is treated with ammonia, Benzylamine is formed which on reaction with Chloromethane yields a secondary amine, N-methylbenzylamine. (c) When aniline reacts with chloroform in the presence of alcoholic potassium hydroxide, phenyl isocyanides or phenyl isonitrile is formed.
  • 12:45 PM IST2 May 2022

    CBSE Class 12 Sample Questions Physics - May 2

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    (a). If the Earth did not have atmosphere, then there would be absence of greenhouse effect of the atmosphere. Due to this reason, the temperature of the earth would be lower than what it is now. (b). An e.m. wave carries momentum with itself and given by P = Energy of wave(U)/ Speed of the wave(c) = U/c when it is incident upon a surface it exerts pressure on it.
  • 4:58 PM IST29 Apr 2022

    CBSE Class 12 Sample Questions Biology - April 29

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    A. a. Band III corresponds to 2500 base pairs, and Band IV corresponds to 100bp. (½ + ½ mark)
    b. The fragments will resolve according to their size. The shorter sequence fragments would move farthest from well as seen in Band IV (100 bp) which is lighter as compared to Band III which is heavier being 2500 base pairs. (1 mark)
    The significance of electrophoresis is to purify the DNA fragments for use in constructing recombinant DNA by joining them with cloning vectors. 
  • 3:42 PM IST29 Apr 2022

    CBSE Class 12 Sample Questions Chemistry - April 29

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    A. (a)electrodialysis
    (b) purification of colloidal solution
    (c) Yes. Dialysis is a very slow process to increase its speed electric field is applied 
  • 3:38 PM IST29 Apr 2022

    CBSE Class 12 Sample Questions Physics - April 29

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    A. Microwaves are suitable for the radar system used in aircraft navigation. Range of frequency of microwaves is 108 Hz to 1011 Hz.
  • 4:03 PM IST28 Apr 2022

    CBSE Class 12 Sample Questions Biology - April 28

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    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.
  • 2:28 PM IST28 Apr 2022

    CBSE Class 12 Sample Questions Chemistry - April 28

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    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
  • 12:11 PM IST28 Apr 2022

    CBSE Class 12 Sample Questions Physics - April 28

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    Given aµg = 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 aµm = 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
  • 3:40 PM IST27 Apr 2022

    CBSE Class 12 Sample Questions Biology - April 27

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    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)
  • 3:06 PM IST27 Apr 2022

    CBSE Class 12 Sample Questions Chemistry - April 27

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    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.
  • 1:20 PM IST27 Apr 2022

    CBSE Class 12 Sample Questions Physics - April 27

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    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.
  • 6:38 PM IST26 Apr 2022

    CBSE Class 12 Sample Questions Biology - April 26

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    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)
  • 4:29 PM IST26 Apr 2022

    CBSE Class 12 Sample Questions Chemistry - April 26

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    5. (a). As the fourth electron enters one of the eg orbitals giving the configuration t2g3 e g1
    , which indicates ?o < P hence forms high spin complex
    (b). Refer to the official website of CBSE for this diagram based answer
  • 12:50 PM IST26 Apr 2022

    CBSE Class 12 Sample Questions Physics - April 26

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    A. 8. (a) Refer to the official website of CBSE for this diagram based answer. (b) mo = 30, fo = 1.25 cm, fe = 5 cm when image is formed at least distance of distinct vision, D = 25 cm Angular magnification of eyepiece me = (1 + D/fe ) = 1 + 25/5  = 6 Total Angular magnification, m = mo me  ? mo = m/me   = 30/6 = 5 As the objective lens forms the real image, mo = vo/uo  = -5 ? vo   = -5uo using lens equation, uo = -1.5 cm, vo = ?5 × (-1.5)cm = +7.5 cm Given ve = -D = -25 cm, fe = +5 cm, ue = ? using again lens equation ue   = 25/6 Thus, object is to be placed at 1.5 cm from the objective and separation between the two lenses should be L = vo + Iuel = 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 = fo/fe = (140/5) = 28 (II) When the final image is formed at the least distance of distinct vision (25 cm) :   m = fo/fe(1 + fe/D) = (28 x 1.2) = 33.6
  • 12:04 PM IST25 Apr 2022

    CBSE Class 12 Sample Questions Biology - April 25

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    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) 
  • 1:02 AM IST25 Apr 2022

    CBSE Class 12 Sample Questions Chemistry - April 25

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    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
  • 12:57 AM IST25 Apr 2022

    CBSE Class 12 Sample Questions Physics - April 25

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    A. 6. Number of atoms present in 2 g of deuterium = 6 × 1023
    Number of atoms present in 2.0 Kg of deuterium = 6 × 1026
    Energy released in fusion of 2 deuterium atoms = 3.27 MeV
    Energy released in fusion of 2.0 Kg of deuterium atoms
     = 3.27/2 × 6 × 1026 MeV
     = 9.81 × 1026 MeV
    = 15.696 × 1013 J
    Energy consumed by bulb per sec = 100 J
    Time for which bulb will glow = 15.696 × 1013 /100
    s = 4.97 × 104 year
    7. A locus of points, which oscillate in phase is called a wavefront.
     OR
    A wavefront is defined as a surface of constant phase.
  • 4:14 PM IST22 Apr 2022

    CBSE Class 12 Sample Questions Biology - April 22

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    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). 
  • 2:27 PM IST22 Apr 2022

    CBSE Class 12 Sample Questions Chemistry - April 22

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    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. 
  • 1:54 PM IST22 Apr 2022

    CBSE Class 12 Sample Questions Physics - April 22

     

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    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 nth orbit. f= v/2?r= 4? 2mk2e4/n3h3
  • 12:03 PM IST22 Apr 2022

    CBSE Class 12 Sample Questions Biology - April 22

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    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. 
  • 6:08 PM IST21 Apr 2022

    CBSE Class 12 Sample Questions Chemistry - April 21

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    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. 
  • 5:40 PM IST21 Apr 2022

    CBSE Class 12 Sample Questions Physics - April 21

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    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)
About the Author
Yamini Saini
Assistant Manager - Content

Yamini has over 4 years of working experience in the editorial field. She has been working with Shiksha for the past two years. She covers current news and other information on school board exams. With a post-gradua... Read Full Bio

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