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RE-AIPMT EXAMINATION - 2015 QUESTION WITH SOLUTION PAPER CODE- B Fastest Growing Institute of Kota (Raj.) FOR JEE Advanced (IIT-JEE) | JEE Main (AIEEE) | AIPMT | CBSE | SAT | NTSE | OLYMPIADS

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RE-AIPMT EXAMINATION - 2015

QUESTION WITH SOLUTION

PAPER CODE- B

Fastest Growing Institute of Kota (Raj.)

FOR JEE Advanced (IIT-JEE) | JEE Main (AIEEE) | AIPMT | CBSE| SAT | NTSE | OLYMPIADS

RE-AIPMT Examination- 2015 (Code-B) (Page # 1)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)

: 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

[BIOLOGY]

1. Read the different components from (a) to(d) in the list given below and tell the cor-rect order of the components with referenceto their arrangement from outer side to innerside in a woody dicot stem:(a) Secondary cortex(b) Wood(c) Secondary phloem(d) PhellemThe correct order is :(1) (c), (d), (b), (a)(2) (a), (b), (d), (c)(3) (d), (a), (c), (b)(4) (d), (c), (a), (b)

Sol. [3]

2. Chromatophores take part in :(1) Photosynthesis(2) Growth(3) Movement(4) Respiration

Sol. [1]

3. Which of the following joints would allow nomovement?(1) Fibrous joint(2) Cartilaginous joint(3) Synovial Joint(4) Ball and Socket joint

Sol. [1]

4. The wheat grain has an embryo with onelarge, shield-shaped cotyledon known as:(1) Epiblast (2) Coleorrhizia(3) Scutellum (4) Coleoptile

Sol. [3]

5. A gene showing codominance has :(1) one allele dominant on the other(2) alleles tightly linked on the same chromosome(3) alleles that are recessive to each other(4) both alleles independently expressed in theheterozygote

Sol. [4]

6. Which of the .following structures is not foundin a prokaryotic cell?(1) Nuclear envelope(2) Ribosome(3) Mesosome(4) Plasma membrane

Sol. [1]

7. The term "linkage" was coined by:(1) T.H. Morgan(2) T. Boveri(3) G. Mendel(4) W. sutton

Sol. [1]

8. the imperfect fungi which are decomposer oflitter and help in mineral cycling belong to(1) Deuteromycetes (2) Basidiomycetes(3) Phycomycetes (4) Ascomvcetes

Sol. [1]

9. Match the columns and identify the correctoption.Column-I Column-II(a) Thylakoids (i) Dis-shaped sacs in

Golgi apparatus C(b) Cristae (ii) Condensed structure

of DNA(c) Cisternae (iii) flat membranous

sacs in stroma(d) Chromatin (iv) Infoldings

mitochondrial(a) (b) (c) (d)

(1) (iv) (iii) (i) (ii)(2) (iii) (iv) (i) ( ii)(3) (iii) (i) (iv) (ii)(4) (iii) (iv) (ii) (i)

Sol. [2]

10. Select the wrong statement(1) the viroids were discovered by D.J.Ivanowski(2) W.M. Stanley showed that viruses couldbe crystallized(3) The term 'contagium vivum fluidum' wasconined by M.W. Beijerinek(4) Mosaic disease in tobacco and AIDS inhuman being are caused by viruses

Sol. [1]

11. During biological nitrogen fixation inactiva-tion of nitrogenase by oxygen poisoning isprevented by(1) Leghaemogolobin(2) Xanthophyll(3) Carotene(4) Cytochrome

Sol. [1]

RE-AIPMT Examination- 2015 (Code-B)(Page # 2)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

12. The species confined lo a particular Regionand not found elsewhere is termed as:(1) Keystone (2) Alien(3) Endemic (4) Rare

Sol. [3]

13. Which Which one of the following hormonesis not involved in sugar metabolism?(1) Cortisone (2) Aldosterone(3) Insulin (4) Glucagon

Sol. [2]

14. Which of the following is not a function ofthe skeletal system?(1) Production of erythrocytes(2) Storage of minerals(3) Production of body heat(4) Locomotion

Sol. [3]

15. Which one of the following is not applicableto RNA?(1) Complementary base pairing(2) 5, phosphoryl and 3' hydroxyl ends(3) Heterocyclic nitrogenous bases(4) Chargaff's rule

Sol. [4]

16. Which one is a wrong statement?(1) Archegonia are found in Bryophyta,Pteridophyta and Gymnosperms(2) Mucor has biflagellate zoospores(3) Haploid endosperm is typical feature ofgymnosperms(4) Brown algae have chlorophyll a and c,and fucoxanthin

Sol. [2]

17. A childless couple can be assisted to have achild through a technique called GIFT. Thefull form of this technique is:(1) Gamete inseminated fallopian transfer(2) Gamete intra fallopian transfer(3) Gamete internal fertilization and transfer(4) Germ cell internal fallopian transfer

Sol. [2]

18. The wings of a bird and the wings of an in-sect are:(1) homologous structures and represent di-vergent evolution(2) analogous structures and represent con-vergent evolution .(3) phylogenetic structures and representdivergent evolution(4) homologous structures and representconvergent evolution

Sol. [2]

19. Golden rice is a genetically modified crop plantwhere the incorporated gene i's meant forbiosynthesis of:(1) Vitamin B (2) Vitamin C(3) Omega 3 (4) Vitamin A

Sol. [4]

20. Outbreeding is an important strategy of ani-mal husbandry because it:(1) helps in accumulation of superior genes.(2) is useful in producing purelines of ani-mals.(3) is useful in overcoming inbreeding de-pression.(4) exposes harmful recessive genes that areeliminated by selection.

Sol. [3]

21.  Which one of the following hormones thoughsynthesised elsewhere, is stored and releasedby the master gland?(1) Antidiuretic hormone(2) Luteinizing hormone(3) Prolactin(4) Melanocyte stimulating hormone

Sol. [1]

22.  An association of individuals of different spe-cies living in the same habital and havingfunctional interactions is:(1) Ecological niche (2) Biotic community(3) Ecosystem (4) Population

Sol. [2]

RE-AIPMT Examination- 2015 (Code-B) (Page # 3)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)

: 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

23.  In which of the following both pairs have cor-rect combination?

Gaseous nutrient cycle Carbon and Nitrogen

Sedimentary nutrient cycle Sulphur and Phosphorus

Gaseous nutrient cycle Carbon and sulphur

Sedimentary nutrient cycle Nitrogen and Phosphours

Gaseous nutrient cycle Nitrogen and sulphur

Sedimentary nutrient cycle Carbon and Phosphorus

Gaseous nutrient cycle Sulphur and Phosphorus

Sedimentary nutrient cycle Carbon and Nitrogen

(1)

(2)

(3)

(4)

Sol. [1]

24.  Identify the correct order of organisation ofgenetic material from largest to smallest:(1) Chromosome, gene, genome, nucleotide(2) Genome, chromosome, nucleotide, gene(3) Genome, chromosome, gene, nucleotide(4) Chromosome, genome, nucleotide, gene

Sol. [3]

25.  A jawless fish, which lays eggs in fresh wa-ter and whose ammocoetes larvae aftermetamorphosis retum to the ocean is:(1) Eptatretus (2) Myxine(3) Neomyxine (4) Petromyzon

Sol. [4]

26. Industrial melanism is an example of :(1) Neo Darwinism (2) Natural selection(3) Mutation (4) Neo Lamarckism

Sol. [2]

27. Cell wall is absent in :(1) Aspergillus (2) Funaria(3) Mycoplasma (4) Nostoc

Sol. [3]

28. The chitinous exoskeleton of arthropods isformed by the polymerisation of :(1) Keratin sulphate and chondraitin sulphate(2) D-glucosamine(3) N-acetyl glucosamine(4) lipoglycans

Sol. [3]

29. Filiform apparatus is characteristic featureof :(1) Generative cell (2) Nucellar embryo(3) Aleurone cell (4) Synergids

Sol. [4]

30. In angiosperms, microsprogenesis and me-gasporogenesis :(1) occur in anther(2) form gametes without further divisions(3) Involve meiosis(4) occur in ovule

Sol. [3]

31. Metagenesis refers to :(1) Presence of different morphic forms(2) Alternation of generation between asexualand sexual phases of an organism(3) Occurrence of a drastic change in formduring post-embyonic development(4) Presence of a segmented body and par-thenogenetic mode of reproduction

Sol. [2]

32. Which of the following immunoglobulins doesconstitule the largest percentage in humanmilk ?(1) lgD (2) lgM(3) lgA (4) lgG

Sol. [3]

33. Destruction of the anterior horn cells of thespinal cord would result in loss of :(1) Sensory impulses(2) Voluntary motor impulses(3) Commissural impulses(4) integrating impulses

Sol. [2]

34. The cutting of DNA at specific locations be-come possible with the discovery of :(1) Restricition enzymes(2) Probes(3) Selectable markers(4) Ligases

Sol. [1]

35. In the following human pedigree, the filledsymbols represent the affected individuals.Identify the type of given pedigree.

(i)

(ii)

(iii)

(iv)

RE-AIPMT Examination- 2015 (Code-B)(Page # 4)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

(1) Autosomal dominant(2) X-linked resessive(3) Autosomal resessive(4) X-linked dominant

Sol. [2]

36. A colour blind man marries a woman with nor-mal sight who has no history of colour blind-ness in her family. What is the probability oftheir grandson being colour blind ?(1) 0.5 (2) 1(3) Nil (4) 0.25

Sol. [3]

37. Flowers are unisexual in :(1) Pea (2) Cucumber(3) China rose (4) Onion

Sol. [2]

38. Roots play insignificant role in absorption ofwater in :(1) Sunflower (2) Pistia(3) Pea (4) Wheat

Sol. [2]

39. Balbiani rings rigs are sites of :(1) Lipid synthesis(2) Nucleotide synthesis(3) Polysaccharide synthesis(4) RNA and protein synthesis

Sol. [4]

40. Which of the following pairs .is not correctlymatched?

Mode of reproduction Example

(1) Offset Water hyacinth

(2) Rhizome Banana

(3) Binary fission Sargassum

(4) Conidia PenicilliumSol. [3]

41. Ectopic pregnancies are referred to as :

(1) Pregnancies with genetic abnormality.

(2) Implantation of embryo at site other than uterus

(3) Implantation of defective embryo in the uterus

(4) Pregnancies terrninated due to hormonalimbalance

Sol. [2]

42. Choose the wrong statement :(1) Penicillium is multicelluar and produces

antibioties(2) Neurospora is used in the study of biochemi-

cal genetics(3) Morels and truffles are poisonous mushrooms

(4) Yeast is unicellular and useful in fermenta-tion

Sol. [3]

43. The function of the gap junction is to :(1) performing cementing to keep neighbouring

cells together.(2) facilitate communication between adjoining

cells by connecting the cytoplasm for rapidtransfer of ions, small molecules and somelarge molecules.

(3) separate two cells from each other.

(4) stop substance from leaking across a tissue.Sol. [2]

44. Axile placentation is present in :(1) Dianthus (2) Lemon(3) Pea (4) Argemone

Sol. [2]

45. Which of the following are not membrane-bound?

(1) Vacuoles (2) Ribosomes(3) Lysosomes (4) Mesosomes

Sol. [2]

46. In his classic experiments on pea plants,Mendel did not use :(1) Seed colour (2) Pod length(3) Seed shape (4) Flower position

Sol. [2]

47. During ecological succession :(1) the gradual and predictable change in spe-

cies composition occurscin a given area.(2) the establishment of a new biotic community

is very fast in its primary phase.(3) the number and types of animals remain

constant

(4) the changes lead to a community that is innear equilibrium with the envirounment andis called pioneer community

Sol. [1]

RE-AIPMT Examination- 2015 (Code-B) (Page # 5)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)

: 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

48. The body cells in cockroach discharge theirnitrogenous waste in the haemolymph mainlyin the form of

(1) Ammonio

(2) Potassium urate

(3) Urea

(4) Calcium carbonateSol. [2]

49. Which of the following biomolecules does havea phosphodiester bond?

(1) Fatty acids in a diglyceride

(2) Monosaccharides in a polysaccharide

(3) Amino acids in a polypeptide

(4) Nucleic acids in a nucleotideSol. [4]

50. The UN conference of Parties on climatechange in the year 2012 was held at

(1) Durban (2) Doha

(3) Lima (4) WarsawSol. [2]

51. Arrange the following events of meiosis incorrect sequence:

(a) Crossing over

(b) Synapsis

(c) Terminalisation of chiasmata

(d) Disappearance of nucleolus

(1) (b), (a,) (d), (c)

(2) (b), (a) (c), (d)

(3) (a), (b) (c), (d)

(4) (b), (c) (d), (a)Sol. [2]

52. Root pressure develops due to:

(1) Active absorption

(2) Low osmotic potential in soil

(3) Passive absorption

(4) Increase in tranpirationSol. [1]

53. Which one of the following animals has twoseparate circulatory pathways?

(1) Frog (2) Lizard(3) Whale (4) Shark

Sol. [3]

54. Which of the following events is not associ-ated with ovulation in human female?(1) Decrease in estradiol(2) Full development of Graafian follicle

(3) Release of secondary oocyte(4) LH surge

Sol. [1]

55. Most animals that live in deep oceanic wa-ters are:(1) primary consumers(2) secondary consumers

(3) tertiary consumers(4) detritivores

Sol. [4]

56. If you suspect major deficiency of antibod-ies in a person, to which of the followingwould you look for confirmatory evidence?(1) Fibrinogin in plasma(2) Serum albumins(3) Haemocytes

(4) Serum globulinsSol. [4]

57. The structures that help some becteria toattach to rock and/or host tissues are:(1) Rhizoids (2) Fimbriae(3) Mesosomes (4) Holdfast

Sol. [2]

58. Increase in concentration of the toxicant atsuccessive tropic levels is known as:

(1) Biomagnification(2) Biodeterioration(3) Biotransformation(4) Biogechemical cycling

Sol. [1]

59. Body having meshwork of cells, internal, cavi-ties lined with food filtering flagellated cellsand indirect development are the character-istics of phylum:

(1) Coelenterate (2) Porifera(3) Mollusca (4) Protozoa

Sol. [2]

RE-AIPMT Examination- 2015 (Code-B)(Page # 6)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

60. The oxygen evolved during photosynthesiscomes from water molecules. Which one ofthe following pairs of elements is involved inthis reaction?

(1) Manganese and Chlorine

(2) Manganese and Potassium

(3) Magnesium and Molybdenum

(4) Magnesium and ChlorineSol. [1]

61. The primary denition in human differs frompermanent dentition in not having one of thefollowing type of teeth:

(1) Canine (2) Premolars

(3) Molars (4) IncisorsSol. [2]

62. Coconut water from a tender coconut is:

(1) Immature embryo

(2) Free nuclear endosperm

(3) Innermost layers of the seed coat

(4) Degenerated nucellusSol. [2]

63. Which of the following layers in an antral fol-licle is acellular?

(1) Granulosa (2) Theca interna

(3) Stroma (4) Zona pellucidaSol. [4]

64. The introduction of t-DNA into plates involves :(1) Infection of the plant by agrobacteriumtumefaciens(2) Altering the pH of the soil, then heat-shocking the plants(3) Exposing the plates to cold for a briefperiod(4) Allowing the plant roots to stand in water

Sol. [1]

65. In which group of organisms the cell wallsform two thin overlapping shells which fittogether ?(1) Chryosphytes (2) Euglenoids(3) Dinoflagellates (4) Slime moulds

Sol. [1]

66. Human urine is usually acidic because :(1) The sodium transporter exchanges onehydrogen ion for each sodium ion, inperitubular capillaries.(2) Excreted plasma proteins are acidic(3) potassium and sodium exchangegenerates acidity(4) hydrogen ions are actively secreted intothe filtrate.

Sol. [4]

67. In photosynthesis, the light-independentreactions take place at :(1) Thylakoid lumen (2) Photosystem I(3) Photosystem II (4) Stromal matrix

Sol. [4]

68. In mammalian eye, the 'fovea' is the centerof the visual field, where :(1) high density of cones occur, but has norods.(2) the optic nerve leaves the eye.(3) only rods are present(4) more rods than cones are found.

Sol. [1]

69. The DNA molecule to which the gene ofinterest is integrated for cloning is called :(1) Transformer (2) Vector(3) Template (4) Carrier

Sol. [2]

70. Pick up the wrong statement :(1) Cell wall is absent in Animalia(2) Protista have photosynthetic andheterotrophic modes of nutrition(3) Some fungi are edible(4) Nuclear membrane is present in Monera

Sol. [4]

71. Among china rose, mustard, brinjal, potato,guava, cucumber, onion and tulip, how manyplates have superior ovary ?(1) Five (2) Six(3) Three (4) Four

Sol. [2]

RE-AIPMT Examination- 2015 (Code-B) (Page # 7)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)

: 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

72. Name the pulmonary disease in which alveolarsurface area involved in gas exchange isdrastically reduced due to damage in thealveolar walls .(1) Pleurisy (2) Emphysema(3) Pneumonia (4) Asthma

Sol. [2]

73. A column of water within xylem vessels oftall trees does not break under its weightbecause of :(1) Dissolved sugars in water(2) Tensile strength of water(3) Lignification of xylem vessels(4) Positive root pressure

Sol. [2]

74. Acid rain is caused by increase in theatmospheric concentration of :(1) SO2 and NO2 (2) SO3 and CO(3) CO2 and CO (4) O3 and dust

Sol. [1]

75. The enzyme that is not present in succusentericus is :(1) maltase (2) nucleases(3) nucleosidase (4) lipase

Sol. [2]

76. In which of the following interatctions bothpartners are adversely affected ?(1) Competition (2) Predation(3) Parasitism (4) Mutualism

Sol. [1]

77. Match the following list of microbes and theirimportance :

(a) Sacharomyces cerevisiae

(i) Production of immunosuppressive agents

(b)Monascus purpureus (ii) Riperuing of Swiss cheese

(c) Trichoderma polysporum (iii)

Commercial production of ethanol

(d) Propionibacterium sharmanii

(iv) Production of blood cholesterol lowering angents

(a) (b) (c) (d)(1) (iii) (iv) (i) (ii)(2) (iv) (iii) (ii) (i)(3) (iv) (ii) (i) (iii)(4) (iii) (i) (iv) (ii)

Sol. [1]

78. A pleiotropic gene :(1) is expressed only in primitive plants.(2) is a gene evolved during Plicene(3) controls a trait only in combination withanother gene(4) controls multiple traits in an individual

Sol. [4]

79. A protoplast is a cell :(1) Without plasma membrane(2) without nucleus(3) undegoing division(4) without cell wall

Sol. [4]

80. Which of the following are most suitable in-dicators of SO2 pollution in the environment?(1) Lichens (2) Conifers(3) Algae (4) Fungi

Sol. [1]

81. Grafted kindney may be rejected in a patientdue to :(1) Humoral immune response(2) Cell-mediated immune respone(3) Passive immune respone(4) Innate immune respone

Sol. [2]

82. Which one of the following fruits is partheno-carpic ?(1) Brinjal (2) Apple(3) Jackfruit (4) Banana

Sol. [4]

83. Which of the following diseases is caused bya protozoan ?(1) Syphilis (2) Influenza(3) Babesiosis (4) Blastomycosis

Sol. [3]

84. In human females, meiosis-II is not completeduntil ?(1) puberty(2) fertilization(3) uterine implantation(4) birth

Sol. [2]

85. Mele gametophyte in angiosperms produces:(1) Two sperms and a vegetative cell(2) Single sperm and a vegetative cell(3) Single sperm and a two vegetative cells(4) Three sperms

Sol. [1]

RE-AIPMT Examination- 2015 (Code-B)(Page # 8)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

86. Doctors use stethoscope to hear the soundsproduced during each cardiac cycle. The sec-ond sound is heard when:(1) AB valves open up(2) Ventricular walls vibrate due to gushingin of blood from atria(3) Semilunar valves close down after theblood flows into vessels from ventricles(4) AV node receives signal from SA node

Sol. [3]

87. Auxin can be bioassayed by:(1) Avena coeoptile curvature(2) Hydroponics(3) Potometer(4) Lettuce hypocotyl elongation

Sol. [1]

88. Satellite DNA is important because it:(1) codes for proteins needed in cell cycle.(2) shows high degree of polymorphism inpopulation and also the same degree of poly-morphism in an individual, which is heritablefrom parents to children.(3) does not code for proteins and is same inall members of the population.(4) codes for enzymes needed for DNAreplicaton.

Sol. [2]

89. Cellular organelles with membranes are:(1) nuclei, ribosomes and mitochondria(2) chromosomes, ribosomes and endoplas-mic(3) endoplasmic reticulum, ribosomes andnuclei(4) Lysosomes, Golgi apparatus and mitochon-dria

Sol. [4]

90. Eutrophication of water bodies leading to kill-ing of fishes is mainly due to non-availabilityof:(1) food (2) light(3) essential minerals (4) oxygen

Sol. [4]

RE-AIPMT Examination- 2015 (Code-B) (Page # 15)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

[PHYSICS]91. The cylindrical tube of a spray pump has

radius R, one end of which has n fine holes,each of radius r. If the speed of the liquid inthe tube is V, the speed of the ejection ofthe liquid through the holes is :

(1) 22

2

rn

VR(2) 2

2

nr

VR

(3) 23

2

rn

VR(4)

2V Rnr

Sol. 2

Incompressible liquidA1V1 = A2V2R2V = nr2v

2 2

2 2

R v R vv

r nR

92. Point masses m1 and m2 are placed at theopposite ends of a rigid rod of length L, andnegligible mass. The rod is to be set rotatingabout an axis perpendicular to it. The positionof point P on this rod through which the axisshould pass so that the work required to setthe rod rotating with angular velocity 0 isminimum, is given by :

m1 P m2

(L–x)x

(1) x = 21

1

mmLm

(2) x = 2

1

mm

L

(3) x = 1

2

mm

L (4) x = 21

2

mmLm

Sol. 4

rm1 m2

L-xx

K = 21

(m1x2) 0

2 + 21

m2 (L – x)2 02

dxdk

= 0

m1x = m2 (L – x)m1x = m2L – m2x

x = 21

2

mmLm

93. A proton and an alpha particle both enter aregion of uniform magnetic field B, moving atright angles to the field B. If the radius ofcircular orbits for both the particles is equaland the kinetic energy acquired by proton is1 MeV, the energy acquired by the alphaparticle will be :(1) 4 MeV (2) 0.5 MeV(3) 1.5 MeV (4) 1 MeV

Sol. 4

q

2mEr

B

r = sameB = same

P P

P

m Em E

q q

m = 4mpq = 2qp

p

2 EE

2

Ep = E

94. A plank with a box on it at one end is graduallyraised about the other end. As the angle ofinclination with the horizontal reaches 30°,the box starts to slip and slides 4.0 m downthe plank in 4.0 s. The coefficients of staticand kinetic friction between the box and theplank will be, respectively.

mg

(1) 0.6 and 0.6 (2) 0.6 and 0.5(3) 0.5 and 0.6 (4) 0.4 and 0.3

RE-AIPMT Examination- 2015 (Code-B)(Page # 16)

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Sol. 4

mg sin 30° = s mg cos 30°

s = 30cos30sin

= 32

21

=

3

1

s = 732.11

= 0.6

30

fnet

fnet = mg sin 30 – mg cos

a = 210

– 2

310

a = 5 – 35

a = 5 (1 3 )

s = 21

at2

4 = 21

× 5 (1 – 3 )

101

= 1 – 3

3 = 1 – 101

= 109

= 10

33 = 0.3 × 1.932 = 0.5

95. An ideal gas is compressed to half its initialvolume by means of several processes. Whichof the process results in the maximum workdone on the gas ?(1) Adiabatic (2) Isobaric(3) Isochoric (4) Isothermal

Sol. 1Adiabatic

96. A ball is thrown vertically downwards from aheight of 20 m with an initial velocity 0. Itcollides with the ground, loses 50 percent ofits energy in collision and rebounds to thesame height. The initial velocity 0 is (take g= 10 ms–2)(1) 14 ms–1 (2) 20 ms–1

(3) 28 ms–1 (4) 10 ms–1

Sol. 2

V0

KE v'

21

mv'2 = mgh ...(1)

for return motion

21

mv'2 = 20 5 10 × m

21

mv'2 = 200 m ....(2)

KE when it collides = 2 × 21

mv'2 = 400 m

energy conservation for downfall

21

mv02 + 200 mg = 400 m

v02 = 400

v0 = 400 = 20 m/s

97. In the spectrum of hydrogen, the ratio of thelongest wavelength in the Lyman series to thelongest wavelength in the balmer series is :

(1) 94

(2) 49

(3) 527

(4) 275

RE-AIPMT Examination- 2015 (Code-B) (Page # 17)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.) : 0744-2209671, 08003899588 | url : www.motioniitjee.com, :[email protected]

Sol. 4

1

= Rz2

22

21 n

1

n

1

For lymenn1 = 1n2 = 2

1

= Rz2

22 21

11

= Rz2

4

14

1

= 43

Rz2

1

= Rz2

22

21 n

1

n

1

For barmer = n1 2 n2 = 3

1

= Rz2

22 31

21

1

= Rz2

91

41

= Rz2

36

49 = Rz2

365

= 2

2

Rz43

Rz365

= 365

× 34

= 275

98. A source of sound S emitting waves offrqeuency 100 Hz and an observer O arelocated at some distance from each other.The source is moving with a speed of 19.4ms-1 at an angle of 60° with the sourceobserver line as shown in the figure. Theobserver is at rest. The apparent frequencyobserved by the observer (velocity of soundin air 330 ms–1) is :

vs

60°

S O

(1) 100 Hz (2) 103 Hz(3) 106 Hz (4) 97 Hz

Sol. 2

n' = n

60cosVV

V

s

99. If dimensions of critical velocity c of a liquidflowing through a tube are expressed as[xyrz], when , and r are the coefficient ofviscosity of liquid, density of liquid and radiusof the tube respectively, then the values ofx, y and z are given by :(1) 1, –1, –1 (2) –1, –1, 1(3) –1, –1, –1 (4) 1,1,1

Sol. 1V xyrz

put valuesx + y = 0–x – 3y + z = 1 – x = – 1x = 1, y = –1, z = – 1

100. 4.0 g of a gas occupies 22.4 litres at NTP.The specific heat capacity of the gas atconstant volume is 5.0 JK–1 mol–1. If the speedof sound in thsi gas at NTP is 952 ms–1, thenthe heat capacity at constant pressure is(Take gas constant R = 8.3 JK–1 mol–1)(1) 8.0 JK–1 mol–1 (2) 7.5 JK–1 mol–1

(3) 7.0 JK–1 mol–1 (4) 8.5 JK–1 mol–1

Sol. 1

Vs = M

rRT

put values we getr = 1.6

p

V

C

C = r

Cp = rCv= 1.6 × 5 = 8

101. If vectors A

= cos t i + sin t j and B

=

cos 2t

i + sin 2t

j are functions of time,

then the value of t at which they areorthogonal to each other is :

(1) t =

4(2) t =

2

(3) t =

(4) t = 0

Sol. 3

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B.A = 0

2t

= 2

t =

102. In the given figure, a diode D is connectedto an external resistance R = 100 and ane.m.f. of 3.5 V. If the barrier potentialdeveloped across the diode is 0.5 V, thecurrent in the circuit will be :

D

R

3.5V

(1) 30 mA (2) 40 mA(3) 20 mA (4) 35 mA

Sol. 1Potential at R = 3.5 – 0.5 = 3V

23i .03 3 10

100

= 30 × 10–3

103. If potential (in volts) in a region is expressedas V(x,y,z) = 6xy – y + 2 yz, the electricfield (in N/C) at point (1,1,0) is :

(1) k3j5i3 (2) k2j5i6

(3) kj3i2 (4) kj9i6

Sol. 2

dvE 6y a 0 6y 0

dx

y

dvE 6x 1 2z

dy

z

dvE 2y

dz

Point (1,1,0)

ˆˆ ˆE 6i 5j 2k

104. A remote - sensing satellite of earth revolvesin a circular orbit at a height of 0.25 × 106 mabove the surface of earth. If earth's radiusis 6.38 × 106 m and g = 9.8 ms–2, then theorbital speed of the satellite is :(1) 7.76 kms–1 (2) 8.56 kms–1

(3) 9.13 kms–1 (4) 6.67 kms–1

Sol. 1h = 0.25 × 106 Re = 6.385 × 106

r = Re + h = 6.63 × 106 m

2

0

GM gRv

r r

gR

r

2

0

gR gRv

hR h 1R

6 6

0

9.8 6.38 10 62.54 10v

0.25 1.0416.38

= 7.9 × 103 m/s

105. Two metal wires of identical dimensions areconnected in series. If 1 and 2 are theconductivities of the metla wires respectively,the effective conductivity of the combinationis :

(1) 1 2

1 2

2 (2)

21

21

2

(3) 21

21

(4) 21

21

Sol. 1R = R1 + R2

1 2 1 1 2 1l l l lA A A

l1 = l2 = l

1 2

2

1 2

2 1 1

1 2

1 2

2

1 2

1 2

2

106. A satellite S is moving in an elliptical orbitaround the earth. The mass of the satelliteis very small compared to the mass of theaerth. Then,(1) The angular momentum of S about thecentre of the earth changes in direction, butits magnitude remains constant(2) The total mechanical energy of S variesperiodically with time.(3) the linear momentum of S remains constantin magnitude(4) The acceleration of S is always directedtowards the centre of the earth.

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Sol. 4Acceleration always towards the center

107. Two particles A and B, move with constantvelocities 1

and 2

. At the initial momentum

their position vectors are 1r

and 2r

respectively. The condition for particles A andB for their collisions is :

(1) 21

21

rrrr

= 22

12

(2) 2211 .r.r

(3) 2211 rr

(4) 2121 rr

Sol. 1Their relative velocity should remain indirection of relative position

108. Two stones of masses m and 2m are whirledin horizontal circles, the heavier one in a radius

2r

and the ligher one is radius r. The tangential

speed of lighter stone is an times that of thevalue of heavier stone when they experiencesame centripetal forces. The value of n is :(1) 2 (2) 3(3) 4 (4) 1

Sol. 1

1

211

rvm

= 2

222

rvm

n2 = 4 n = 2

109. A parallel plate air capacitor has capacity'C', distance of separation between plates is'd' and potential difference 'V' is appliedbetween the plates. Force of attractionbetween the plates of the parallel plate aircapacitor is :

(1) d2VC 22

(2) d2

CV2

(3) d

CV2

(4) d2VC 22

Sol. 2

F = 0

2

A2Q

F = 0

2

A2)CV(

+++++++

–––––––

FF

dF = ddA

2

VC

0

22

= d2

CV2

110. The position vector of a particle R

as a func-tion of time is given by:

ˆ ˆR 4sin 2 t i 4 cos 2 t i

Where R is in meters, t is in second and

ˆ ˆi and j denote unit vectors along x and y

direction, respectively. Which one of the fol-lowing statements is wrong for the motion ofparticle?

(1) Acceleration vector is along R

(2) Magnitude of acceleration vector is 2v

R.

(3) Magnitude of the velocity of particle is 8meter/second(4) Path of the particle is circle of radius 4meters

Sol. 3

r = 48 m t i + 4 cos t j

v = 4 cos t i – 4 sin t j

a = – 24 sin t i – 42 cos t j

= – w2 [4 sin t i + 4 cos t j ]

rwa 2

|a| = 42

|v| = 216 = 4

|r| = 4 a = 4

16 2 = 42

111. A series R–C circuit is connected to an alter-nating voltage source. Consider two situa-tions :(i) When capacitor is air filled.(ii) When capacitor is mica filled.Current through resistor is i and voltageacross capacitor is V then :(1) Va < Vb (2) Va > Vb(3) ia > ib (4) Va = Vb

Sol. 2

R C

V

i = ZV

Z = 222

c

1R

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G = cain C2 = CmedC2 > C1 z2 < z1 VC

2 = v2 – VR2

i2 > i1voltage across C = Vc = i xC

2C1

V = V2 – 2R1

V

2C2

V = V2 – 2R2

V

2C2

V < 1CV VA > Vb

112. A string is stretched between fixed pointsseparated by 75.0 cm. It is observed to haveresonant frequencies of 420 hz and 315 Hz.There are no other resonant frequencies be-tween these two. The lowest resonant fre-quency for this string is :(1) 155 Hz (2) 205 Hz(3) 10.5 Hz (4) 105 Hz

Sol. 4

2

= 75

= 1.5 m

2 2

=

22

32

=

= 32

n1 = 2

PV

n = 2

P (v)

105, 210, 315, 420

113. The coefficient of performance of arefrigerator is 5. If the temperature insidefreezer is –20°C, the temperatures of thesurroundings to which it rejects heat is :(1) 31°C (2) 41°C(3) 11°C (4) 21°C

Sol. 1Refrigerator

T1

Q1

T2

Q2

Performance factorT2 = 253

n = 2Q

= 5

= 21

2

QQQ =

21

2

TTT = 5

T1 = 56

T2 = 56

× 253 T2 = 5T1 – 5T2

= 303.6 = 30.6 31°C

114. A photoelectric surface is illuminatedsuccessively by monochromatic light of

wavelength and 2 . If the maximum kinetic

energy of the emitted photoelectrons in thesecond case is 3 times that in the first case,the work function of the surface of thematerial is :

(1) hc2

(2) hc

(3) 2hc

(4) hc3

Sol. 1

hck1

hc2k2

k2 = 3k1

hc2

= hc3

– 3

2 = hc

F = 2

hc

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115. In an astronomical telescope in normaladjustment a straight black line of length L isdrawn on inside part of objective lens. Theeye-piece from a real image of this line. Thelength of this image is I. The Magnificationof the telescope is :

(1) L

1I (2)

L1

I

(3) L IL I

(4) LI

Sol. 4Magnification (M) = L/I

116. Two slits in Youngs experiment have widthsin the ratio 1 : 25. ?The ratio of intensity atthe maxima and minima in the interference

pattern max

min

II is :

(1) 94

(2) 12149

(3) 49121

(4) 49

Sol. 1

2 21 2max

2min

1 2

I II 1 25I 25 1I I

2 26 3 94 2 4

117. Two vessels separately contain two idealgases A and B at the same temperature, thepressure of A being twice that of B. Undersuch conditions, the density of A is found tobe 1.5 times the density of B. The ratio ofmolecular weight of A and B is :

(1) 23

(2) 34

(3) 2 (4) 12

Sol. 2

RTP

M

T = samePA = 2PB

A A B

B A B

P MP M

A A B

B B A

MM

1 31.5

2 4

118. A circuit contains an ammeter, a battery of30 V and a resistance 40.8 ohm all connectedin series. If the ammeter has a coil ofresistance 480 ohm and a shunt of 20 ohm,the reading in the ammeter will be :(1) 0.5 A (2) 0.25 A(3) 2 A (4) 1 A

Sol. 1

20

480

m

RA = 480 20

500

9619.2

5

Req. = R + RA

= 40.8 + 19.2= 60 E = 30 v

30 1i 0.5

60 2

119. The value of coefficient of volume expansionof glycerin is 5×10–4 K–1. The fractionalchange in the density of glycerin for a rise of40°C in its temperature is :(1) 0.015 (2) 0.020(3) 0.025 (4) 0.010

Sol. 2

0 1 r t

0

r t

= 5 × 10–4 × 40

= 2 × 10–2 = .02

120. The heart of a man pumps 5 liters of bloodthrough the arteries per minute at a pressureof 150 mm of mercury. If the density ofmercury be 13.6 × 103 kg/m3 andg = 10 m/s2 then the power of heart in wattis :(1) 1.70 (2) 2.35(3) 3.0 (4) 1.50

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Sol. 1

energy work PvPower

time time t

P = hdg = 150 × 10–3 × 13.6 × 103 × 10= 2040 × 10

P = 2.04 × 104

4 32.04 10 5 10Power

60

= 1.7

121. A beam of light consisting of red, green andblue colours is incident on a right angled prism.The refractive index of the material of theprism for the above red, green and bluewavelengths are 1.39, 1.44 and 1.47,respectively.

BlueGreen

Red

A

B C

45°

The prism will:(1) separate the blue colour part from thered and green colours(2) separate all the three colours from oneanother(3) not-separate the three colours at all(4) separate the red colour part from thegreen and blue colours

Sol. 4critical angle for red light

(C)Red = sin–1

Rn1

= sin–1

39.11

= sin–1

39.11

Similarly criticle angle of green light

(C)Green = sin–1

44.11

(C)Blue = sin–1

47.11

angle of incidence for all colour is

45° = sin–1 2

1 = sin-1

414.11

This means that red light will emerge butgreen and blue will totaly reflect.

122. A rectangular coil of length 0.12 m and width0.1 m having 50 turns of wire is suspendedvertically in a uniform magnetic field ofstrength 0.2 Weber/m2. The coil carries acurrent of 2A. If the plane of the coil is inclinedat an angle of 30° with the direction of thefiled, the torque required to keep the coil instable equilibrium will be :(1) 0.15 Nm (2) 0.20 Nm(3) 0.24 Nm (4) 0.12 Nm

Sol. 2T = NIAB sin T = 50 × 2 × (0.12 × 0.1) 0.2 × sin 60

T = 50 × 2 × 12 × 1 ×2 × 10–4 × 32

4T 1200 10 3 T = 0.12 × 1.732T = 0.20

123. An electron moves on a straight line path XYas shown. The abcd is a coil adjacent to thepath of electron. What will be the directionof current, if any, induced in the coil?

c

db

a

X Yelectron

(1) abcd(2) adcb(3) The current will reverse its direction asthe electron goes past the coil(4) No current induced

Sol. 3

As electron comes near magnetic flux willincrease and then decrease hence inducedcurrent will change its direction.

124. A nucleus of uranium decays at rest into nu-clei of thorium and helium. Then :(1) The helium nucleus has more kinetic en-ergy than the thorium nucles.(2) The helium nucleus has less momentumthen the thorium nucleus.(3) The helium nucleus has more momentumthan the thorium nucleus.(4) The helium nucleus has less kinetic en-ergy than the thorium nucleus.

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Sol. 1momentum = same

2PKE

2m

1K.E.

mass

mHe < mth

HE thKE KE

125. A for F = a i +3 j +6 k is acting at a point r

= 2 i –6 j –12k . The value of for which an-

gular momentum about origin is conserved is:(1) –1 (2) 2(3) zero (4) 1

Sol. 1

ˆˆF i 3j 6k

ˆr 2i 6j 12k

angular momentum = conserved = 0

r F 0

r and F are parallel

= –1

ˆˆ ˆF i 3j 6k

ˆˆ ˆr 2 i 3j 6k

r 2F

126. Water rises to a height ‘h’ in capillary tube. Ifthe length of capillary tube above the sur-face of water is made less than ‘h’, then :(1) water rises upto the tip of capillary tubeand then starts overflowing like a fountain.(2) water rises upto the top of capillary tubeand stays there without overflowing.(3) water rises upto a point a little below thetop and stays there.(4) water does not rise at all.

Sol. 2h1R1 = h2R2water stays at upper height

127. A particle is executing a simple harmonicmotion. Its maximum acceleration is andmaximum velocity is . Then, its time periodof vibration will be :

(1) 2

2a

(2) (3)

2

(4) 2

Sol. 4vmax = a = max = 2a = a = 2 = (aw)(w) = ()

2T

T 2

128. The energy of the em waves is of the orderof 15 keV. To which part of the spectrumdoes it belong?(1) X-rays (2) Infra-red rays(3) Ultraviolet rays (4) - rays

Sol. 1O12400

0.82 A15 1000

wavelength Range of x-ray is approximate0.03 Å to 3 Å

129. Light of wavelength 500 nm is incident on ametal with work function 2.28 eV. The deBroglie wavelength of the emitted electron is:(1) <2.8x10-10m(2) <2.8x10-9m(3) 2.8x10-9m(4) 2.8x10-12m

Sol. 3

maxK E

12400ev 2.28ev

5000

= 2.48 – 2.28 ev= 0.20 ev

ma

h2m·K

min

ma

h

2m K

9min 2.8 10 m

130. At the first minimum adjacent to the centralmaximum of a single-slit diffraction pattern,the phase difference between the Huygen’swavelet from the edge of the slit and thewavelet from the midpoint of the slit is :

(1) 4

radian (2) 2

radian

(3) radian (4) 8

radian

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Sol. 3The path difference will be /2 hence phasedifference between wares reaching the

central maximum is 2

.

131. On a frictionless surface, a block of mas M.moving at speed collides elastically withanother block of same mass M which is ini-tially at rest. After collision the first blockmoves at an angle to its initial direction

and has a speed 3

. The second block’s

speed after the collision is :

(1) 2 23

(2) 34

(3) 32 (4)

32

Sol. 12

2 21 1 v 1mv m mv'

2 2 3 2

22 2 q

q' qg

22 8q

q'g

q' = gv8 2

= 2 2

q v3

132. A potentiometer wire of length L and a resis-tance r are connected in series with a bat-tery of e.m.f. E0 and a resistance r1. An un-known e.m.f. E is balanced at a length l ofthe potentiometer wire. The e.m.f. E will begiven by :

(1) 0

1

L E rl r (2)

0

1

E r(r r )

lL

(3) 0E lL

(4) 0

1

L E r(r+r )l

Sol. 4

E0 r1

Lr

0

1

E rx

r r L

0

1

E rx

r r

.

L1

E = xl

0

1 0

E rlr r L

133. The Young’s modulus of steel is twice that ofbrass. Two wires of same length and of samearea of cross section, one of steel and an-other of brass are suspended from the sameroof. If we want the lower ends of the wiresto be at the same level,. then the weightsadded to the steel and brass wires must bein the ratio of:(1) 1:2 (2) 2:1(3) 4:1 (4) 1:1

Sol. 2

mg wly l

A l A l

yA lw

l

l = samel = sameA = samew × y

s s

B B

w y 2w y 1

134. The input signal given to a CE amplifier hav-ing a voltage gain of 150 is Vi = 2 cos

15t3

. The corresponding output signal

will be :

(1) 300 cos 15t3

(2) 75 cos 2

15t3

(3) 2 cos 5

15t6

(4) 300 cos 4

15t3

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Sol. 4

0v

c

vA

v

vi = 2 cos (15t + /3)

v0 = (Av)vi

= 300 cos (15t + + /3)

135. An automobile moves on a road with a speed

of 5 km h-1. The radius of its wheels is 0.45

m and the moment of inertia of the wheel

about its axis of rotation is 3 kg m2. If the

vehicle is brought to rest in 15s, the magni-

tude of average torque transmitted by its

brakes to the wheel is :

(1) 6.66 kg m2 s-2

(2) 8.58 kg m2 s-2

(3) 10.86 kg m2 s-2

(4) 2.86 kg m2 s-2

Sol. 1

0 t

0

t

54km/hr–1

R = 45

I = 3 kg m2

qrt

554 100 2018

0.45 15 45 9

T = I

20 20T 3 6.66 Kg.

9 3

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[CHEMISTRY]136. In which of the following pairs, both the

species are not isostructural(1) XeF4, XeO4(2) SiCl4, PCl4

+

(3) diamond, silicon carbide(4) NH3, PH3

Sol. 1

1

These are having different structures.

137. Which one of the following esters getshydrolysed most easily under alkalineconditions?

(1)

OCOCH3

Cl (2)

OCOCH3

O N2

(3)

OCOCH3

H CO3

(4)

OCOCH3

Sol. 2– NO2 group having – M effect

138. Reaction of phenol with chloroform inpresence of dilute sodium hydroxide finallyintroduces which one of the followingfunctional group?(1) – CHO (2) – CH2Cl(3) –COOH (4) –CHCl2

Sol.. 1

OH

+ CHCl3Dil. NaOH

OH OH

CHO

CHOSalicylaldehyde

+

139. Which of the following reactions can beused for the preparation of alkyl halides ?I CH3CH2OH + HCl anh. ZnCl2

II CH3CH2OH + HCl

II CH32COH + HCl IV CH32CHOH + HCl anh. ZnCl2

(1) III and IV only(2) I, III and IV only(3) I and II only(4) IV only

Sol. 2I, III and IV Lucas reagent

140. In an SN1 reaction on chiral centres, thereis :(1) 100 % inversion(2) 100 % recemization(3) inversion more than retention leading topartial recemization(4) 100 % retention

Sol. 3Some degree of racemisation will take place.

141. Which of the following is not the product of

dehydration of OH ?

(1) (2)

(3) (4)

Sol. 1

OH

Hydride Shift

C

CH3

CH2 CH3

–H

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142. On heating which of the following releasesCO2 most easily ?(1) CaCO3(2) K2CO3(3) Na2CO3(4) MgCO3

Sol. 4

143. In the reaction with HCl, an alkene reactsin accordance with the Markovnikov’s rule,to give a product 1-chloro-1-methylcycclohexane. The possible alkene is:

(1)

CH2

I (2)

CH3

II

(3) I and II (4)

CH3

Sol. 2

CH3

+ HClMR

CH3 Cl

144. Number of possible isomers for the complexCoen2Cl2Cl will be : en = ethylenediamine(1) 4 (2) 2(3) 1 (4) 3

Sol. 4

145. A gas such as carbon monoxide would bemost likely to obey the ideal gas law at :(1) Low temperatures and low pressures.(2) High temperatures and low pressures(3) Low temperatures and high pressures(4) High temperatures and high pressures.

Sol. 2At high temperature & low pressure.

146. If Avogadro number NA, is changed form6.022 ×1023 mol–1 to 6.022 × 1020 mol–1,this would change :(1) The ratio of elements to each other in acompound .(2) The definition of mass in units of grams.(3) The mass of one mole of carbon.(4) The ratio of chemical species of eachother in a balanced equation.

Sol. 2As 6.022 × 1023 × massnuclear = 1.But 6.022 × 1020 × massnuclear 1

147. Gadolinium belongs to 4f series. It’s atomicnumber is 64. Which of the following is thecorrect electronic configuration of gado-linium ?(1) Xe 4f65d26s2

(2) Xe4f86d2

(3) Xe4f95s1

(4) Xe4f75d16s2

Sol. 4

148. What is the pH of the resulting solution whenequal volumes of 0.1M NaOH and 0.01 M HClare mixed ?(1) 1.04 (2) 12.65(3) 2.0 (4) 7.0

Sol. 2pH = –logH+

OH– = 0.1 y 0.01 y

2y

= 0.09

2M

pOH = –logOH–

= –log 0.09

2

= –1.346

pH = 14 – pOH= 14 – 1.345= 12.65

149. Decreasing order of stability of O2, O–2, O

+2

and 22O is :

(1) 22 2 2 2O O O O

(2) 22 2 2 2O O O O

(3) 22 2 2 2O O O O

(4) 22 2 2 2O O O O

Sol. 2

O2 kk 2s2 *2s2 2z2P 2

x2p = 2y2p ,

xx2p ' = '

y* 2p

B.O. = 2O2

+ B.O.= 2.5O2

– B.O. = 1.5O2

2– B.O. = 1 Stability order O2

+ > O2 > O2– > O2

2–

150. the correct statement regarding defects incrystalline solids is :(1) Frenkel defect is found in halides of alkalinemetals.(2) Schottky defects have no effect thedensity of crystalline solids.(3) Frenkel defects decreases the density ofcrystalline solids.(4) Frenkel defect is a dislocation defect.

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Sol. 4Frenkel defect is a dislocation defect

151. Which of the following statements is notcorrect for a nucleophile ?(1) Nucleophiles are not electron seeking(2) Nucleophile is a Lewis acid(3) Ammonia is a nucleophile(4) Nucleophiles attack low e– density sites

Sol. 2Nucleophiles is a lewis base.

152. The hybridization involved in complexNiCN42– is

: At. No. Ni = 28(1) d2sp3 (2) dsp2

(3) sp3 (4) d2sp2

Sol. 2

153. The stability of +1 oxidation state among Al,Ga, ln and Tl increase in the sequence :(1) In < Tl < Ga < Al (2) Ga < In < Al < Tl(3) Al < Ga < In < Tl (4) Tl < In < Ga < Al

Sol. 3

154. The sum of coordination number and oxidationnumber of the metal M in the complex Men2C2O4Cl where en is ethylenediamine is :(1) 8 (2) 9(3) 6 (4) 7

Sol. 2

155. Which of the statements given below isincorrect ?(1) OF2 is an oxide of fluorine(2) Cl2O7 is an anhydride of perchloric acid(3) O3 molecule is bent(4) ONF is isoelectronic with O2N

Sol. 1

156. In the extraction of copper from its sulphideore, the metal is finally obtained by thereduction of cuprous oxide with :(1) sulphur dioxide (2) iron II sulphide(3) carbon monoxide (4) copperI sulphide

Sol. 4

157. Which one of the following pairs of solutionis not an acidic buffer ?(1) H3PO4 and Na3PO4(2) HClO4 and NaClO4(3) CH3COOH and CH3COONa(4) H2CO3 and Na2CO3

Sol. 1Must be a conjugate acid base pair

158. Assuming complete ionization, same moles ofwhich of the following compounds will requirethe least amount of acidified KMnO4 forcomplete oxidation ?(1) FeNO22 (2) FeSO4(3) FeSO3 (4) FeC2O4

Sol. 27

24kMnO Mn

It is an oxidising agent for FeSO4 least amoung of FeSO4 is

required because sulphurc is present in its highestO.S., so it will only oxidise Fe2+ to Fe3+

159. The number of structural isomers possiblefrom the molecular formula C3H9N is :(1) 3 (2) 4(3) 5 (4) 2

Sol. 2C3H9N H C7 3 NH2 2

H C3 N H C3

H C3

1

H C3 NH C H2 5 1160. 20.0 g of a magnesium carbonate sample

decomposes on heating to give carbon dioxideand 8.0 g magnesium oxide. What will be thepercentage purity of magnesium carbonatein the sample ?(1) 84 (2) 75(3) 96 (4) 60

Sol. 1

MgCO MgO + CO 3 2

n = 51 moles

8 gm

n = 408

n = 51

W = 15

× 84 = 16.8 gm

present purity = 16.8

10020

= 84 %

161. Two possible stereo–structuers ofCH3CHOH.COOH, which are optically active,are called :(1) Mesomers (2) Diastereomers(3) Atropisomers (4) Enantiomers

Sol. 4Enantiomers

COOH COOH

CH3 CH3

OH HH HO

d - lactic acid l - lactic acid

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162. The heat of combustion of carbon to CO2 is–393.5 kJ/mol. the heat released uponformation of 35.2 g of CO2 from carbon andoxygen gas is :(1) –3.15 kJ (2) –315 kJ(3) +315 kJ (4)–630 kJ

Sol. 3C + O2 CO2 ; x = –393.5 kJ/mole for 2 mole of CO2 heat released –393.5 kJ

for 0.8 mole 35.244

of CO2

heat released = + 393.5 × 0.8= +314.8 kJ

163. The rate constant of the reaction A B is0.6× 10–13 mole per second. If theconcentration of A is 5 M, then concentrationof B after 20 minutes is :(1) 0.72 M (2) 1.08 M(3) 3.60 M (4) 0.36 M

Sol. 1CO – Ct = kT5 – Ct = 0.6 × 10–3 × 20 × 60Ct = 5 –0.72Ct = 4.28, ie the conc. of A left B formed = 0.72 M

164. the formation of the oxide ion, O2–g, fromoxygen atom requires first an exothermic andthen an endothermic step as shown below :Og + e– O–g ; fH

–= –141 kJ mol–1

O–g + e– O2–g ; fH–= +780 kJ mol–1

Thus process of formation of O2– in gas phaseis unfavourable even though O2– isisoelectronic with neon. It is due to the factthat,(1) addition of electron in oxygen results inlarger size of the ion(2) electron repulsion outweighs the stabilitygained by achieving noble gas configuration.(3) O–ion has comparatively smaller size thanoxygen atom.(4) oxygen is more electronegative.

Sol. 2This is due to the inner e– repulsion of O– toform O2-

165. What is the mass of the precipitate formedwhen 50 mL of 16.9% solution of AgNO3 ismixed with 50 mL of 5.8% NaCl solution ?Ag = 107.8, N = 14, O = 16, Na = 23, Cl =35.5(1) 14g (2) 28 g(3) 3.5 g (4) 7 g

Sol. 4AgNo3 + NaCl AgCl + NaNO316.9% 5.8%50 ml 50ml

166. Which is the correct orderof increasing energyof the listed orbitals in the atom of titaniumAt.no. Z = 22(1) 3s 3p 4s 3d (2) 3s 4s 3p 3d(3) 4s 3s 3p 3d (4) 3s 3p 3d 4s

Sol. 1using n + mole

167. Reaction of a carbonyl compound with oneof the following reagents involves nucleophilicaddition followed by elimination of water. Thereagent is :(1) sodium hydrogen sulphite(2) a Girgnard reagent(3) hydrazine in presence of feebly acidicsolution(4) hydrocyanic acid

Sol. 3C = O + H 2 N NH2

pH = 4.5 - 7

> C = N – NH2Hydrazone

168. The variation of the boiling points of thehydrogen halides is in the order HF > Hi > HBr> HCl.What explains the higher boiling point ofhydrogen fluoride ?(1) The effect of nuclear shielding is muchreduced in fluorine which polarises the HFmolecules.(2) The electronegativity of fluorine is muchhigher than for other elements in the group(3) There is strong hydrogen bondingbetween HF molecules.(4) The bond energy of HF molecules isgreater than in other hydrogen halides.

Sol. 3

169. The name of complex ion, FeCN63– is :

(1) Hexacyanidoferrate III ion(2) Hexacyanoiron III ion(3) Hexacyanitoferrate IIIion(4) Tricyanoferrate III ion

Sol. 1

170. Method by which Aniline cannot be preparedis :(1) potassium salt of phthalimide treated withchlorobenzene followed by hydrolysis withaqueous NaOH solution.(2) hydrolysis of phenylisocyanide with acidicsolution.(3) reduction of nitrobenzene with brominein alkaline solution.(4) reduction of nitrobenzene with H2/Pd inethanol.

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Sol. 1

C

C

O

O

NK +

Cl

[Resonance]

Hydrolysis

aq. NaOH

Aniline does not form

171. If the equilibrium constant for N2g + O2g 2Nogis K, the equilibrium constant for

12

N2g + 12

O2 gNOg will be :

(1) K2 (2) K1/2

(3) 12

K (4) K

Sol. 2K1/2

172. Strong reducing behaviour of H3PO2 is dueto:(1) Presence of two –OH groups and one P–H bond(2) Presence of one –OH group and two P–Hbonds(3) High electron gain enthalpy of phosphorus(4) High oxidation state of phosphorus

Sol. 2

173. 2,3–Dimethyl–2–butene can be prepared byheating which of the following compoundswith a strong acid ?(1) CH32 CH–CH2–CH=CH2

(2)

(3) CH33C—CH=CH2(4) CH32C=CH–CH2–CH3

Sol. 3

CH3 C CH CH2

CH3

CH3

CH3 C CH CH3

CH3

CH3

Methyl Shift

CH3 C CH CH3

CH3 CH3

CH3 C CH CH3

CH3 CH3

2, 3 – Dimethyl - 2 - butene

174. Aqueous solution of which of the followingcompounds is the best conductor of electriccurrent ?(1) Fructose, C6H12O6(2) Acetic acid, C2H4O2(3) hydrochloric acid, HCl(4) Ammonia, NH3

Sol. 3

175. The vacant space in bcc lattice unit cell is :(1) 32% (2) 26%(3) 48% (4) 23%

Sol. 2

packing efficiency= occupied

total

v

v or

Packing fraction

= 3

3

4 /3 r 2

4r3

= 0.74 vacant space = 26%

176. What is the mole fraction of the solute in a1.00 m aqueous solution ?(1) 0.0177 (2) 0.177(3) 1.770 (4) 0.0354

Sol. 1

M = solute

solventsolute

x 1000MM1 x

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1 = solute

solute

x 10001 x 18

1–xsolute = 55.55 xsolute

xsolute = 1

56.55= 0.01768

177. The oxidaiton of benzene by V2O5 in thepresence of air produces :(1) benzaldehyde(2) benzoic anhydride(3) maleic anhydride(4) benzoic acid

Sol. 3

V O2 5

H

H

C

C

C

C

O

O

O

Maleic anhydride

178. Caprolactam is used for the manufacture of(1) Nylon–6,6 (2) Nylon–6(3) Teflon (4) Terylene

Sol. 2Nylon - 6

179. The following reaction

is known by the name :(1) Schotten–Baumen reaction(2) Friedel–Craft's reacton(3) Perkin's reaction(4) Acetylation reaction

Sol. 1Schotten - baumen reaction

Ph

Ph

PhNH + 2

NH

C

C

Cl

Ph

O

NaOH

O

180. The number of water molecules is maximumin :(1) 18 moles of water(2) 18 molecules of water(3) 1.8 gram of water(4) 18 gram of water

Sol. 1

2H On = 18 moles no. of H2O molecules = 18 × NA