SRI CHAITANYA EDUCATIONAL INSTITUTIONS

A.P,TELANGANA,KARNATAKA,TAMILNADU,MAHARASHTRA,DELHI,RANCHI,CHANDIGARH

NEET GRAND TEST-

TEST - 5
Name :........................................... Hall Ticket No:

INSTRUCTIONS TO CANDIDATES

1. The Model NEET- 2020 is of 3 Hrs duration. Time: 10.00 AM – 1.00

PM.
2. The question paper for NEET-2020 consists of 180 questions
comprising 45 questions in Botany, 45 in Zoology, 45 in Physics
and 45 in Chemistry for NEET.
3. All questions are of objective type (Multiple choices only)
4. Each question carries four marks.
5. Negative marking: one mark will be deducted for every wrongly
answered question.
6. Total Marks 720.
7. The candidates are prohibited from carrying any paper to the
examination hall except HALL TICKET.
8. No Calculators, Mini-Cards, Watches with Calculators, Pager, Cell
Phone, Slide rules or outer aids to calculation will be allowed in
the examination hall.
9. Candidates once admitted will not be allowed to leave the hall till
half an hour before the closing of the test.
10. A separate sheet is attached in the middle of this booklet for rough
work, you can detach and use it.
11. A detachable answer sheet with 180 question blocks, with 4 circles
corresponding to 4 multiple choice for each question will be
provided. Use HB Pencil to darken the appropriate circle against
the question number provided in the sheet. Answer should be
marked only on the answer sheet, but not on the question paper
booklet.

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 SRI CHAITANYA EDUCATIONAL INSTITUTIONS,INDIA.
A.P,TELANGANA,KARNATAKA,TAMILNADU,MAHARASHTRA,DELHI,RANCHI,CHANDIGARH
SEC : SR ELITE, AIIMS S60, NEET MPL & MEDICON DATE : 31-01-2020
SUB: BOTANY NEET GRAND TEST - 5 Max. Marks : 720

01. Systematics is the study of (1) chromosome number

(1) Kinds of organisms (2) chromosome structure
(2) Their diversities (3) chromosome behaviour
(3) Relationship among them (4) all the above
(4) All the above 08.
02. Select the incorrect match
(1) Herbarium – quick referral system
(2) Botanical gardens – live collection of
plants
(3) Flora – information on any one taxa
(4) Manuals – identification of names of
species of any area
03. Which of the following group show most Which of the following is not associated with
extensive metabolic diversity given organism
(1) Bacteria (2) Fungus (1) chl – c
(3) Algae (4) Bryophytes (2) Laminarin
04. Which of the following is heterotrophic (3) Lack of flagella/cilia
protist (4) Cellulose & Algin
(1) diatoms (2) desmids 09. (i) Tracheophyte
(3) slime moulds (4) dinoflagellates (ii) Sporophitic main plant body
05. Which of the following is coenocytic fungi (iii) Haplontic life cycle
(1) Saccharomyces (iv) Sessile sex organs
(2) Mucor (v) Thallus body
(3) Claviceps (vi) Protonema
(4) Usatilago (vii) Independent sporophyte
06. Potato spindle tuber disease caused by (viii) Elaters
(1) Viruses How many are associated with pteridophytes.
(2) Viroids (1) 3
(3) Prions (2) 4
(4) Virusoids (3) 5
07. Cytotaxonomy is based on (4) 6

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10. Most of the algae genera are (1) 2
(1) Haplontic (2) 3
(2) Diplontic (3) 4
(3) Haplodiplontic (4) 5
(4) Diplohaplontic 16. Presence of phloem only toward outer side of
11. Which of the following modified root found xylem is feature of which type of vascular
in Rhizophora bundle
(1) Prop root (1) Radial
(2) Stilt root (2) Conjoint
(3) Pneumatophore (3) Collateral
(4) Assimilatory roots (4) Bicollateral
12. Rachis is absent in 17. Annual rings can be observe in
(1) Neem (1) Heart wood
(2) Rose (2) Sap wood
(3) Dalbergin (3) Periderm
(4) Alstonia (4) Both 1 & 2
13. Parietal placentation present in 18. Glogi complex participate in
(1) chinarose (1) Formation of secretory vesicles
(2) Argenone (2) Respiration in bacteria
(3) Dianthus (3) Fatty acid breakdown
(4) Tomato (4) Activation of amino acids
19. Given diagram represent to
14.
Floral formula is applicable on
(1) Gloriosa
(2) Makoi
(3) Sesbania
(4) Indigofera
15. (i) Deposition of cellulose, hemicelluloses &
pectin
(ii) Often contain chloroplast
(1) Basal body
(iii) Lack of inter cellular space
(2) Basal plate
(iv) Presence of pits
(3) Axoneme
(v) Universal occurance in plants
(4) All the above
How many are true for collenchyma
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20. Which of the following is true for nucleolus (1) High C : N
(1) It is a membrane bound structure (2) low N : C ratio
(2) It take part in spindle formation (3) High N : C ratio
(3) Large nucleoli present in dividing cells (4) Both 1 & 3
(4) It is a site for active r – RNA synthesis 26. Which of thefollowing is final electron
COOH acceptor of ‘Z’ Scheme
|
21. H− C − NH 2 structure which of the (1) CO2
|
CH 2OH (2) NAD +
following is incorrect
(1) It is serine (3) NADP +

(2) It is substituted methame (4) PS-I

(3) It is basic amino acid 27. During aerobic respiration, which of the

(4) It is hydroxy amino acid following ETC compex doesnot pump any

22. S − G + S' → S + S' − G . This type of proton

reaction can be catalysed by (1) Complex – I

(1) Calss – 1 enzymes (2) Complex – II

(2) Class – 2 enzymes (3) Complex – III

(3) Class – 5 enzymes (4) Complex – IV

(4) both 1 & 2 28. Select out the incorrect match

23. Split of centromeres, chromatid separation (1) Auxin – induction of Parthenocarpy

and movement of chromatids to opposite (2) Gibberellins – bolting

poles observed in (3) Cytokinins – herbicide

(1) Anaphase (4) Ethylene – Synchronising fruit set in

(2) Anphase – I pineapple

(3) anaphase – II 29. Which of the following is dioecious plant.

(4) Both 1 & 3 (1) Cucurbits

24. In which of the following plant cell value of (2) Coconuts

(3) Papaya
ψ w (water potential) is minimum
(4) Solanum
(1) Root hair cell
30. Sporopollenin is absent in
(2) Root Pericycle
(1) Exine
(3) Vessels of main stem
(2) Intine
(4) Midrib xylem cells
(3) Germpore
25. As compare to amino acids, amides are
(4) Both 2 and 3
suitable for translocation in plant body
because they have
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31. Which of the following gene/character can (2) Regulator (i) gene
assort independently (3) Operator
(1) genes of same chromosome
(4) Structural genes
(2) genes present on homologens
36. DNA replication is discontinuous because
chromosomes
(3) genes present on non homologens (1) Fork moves in both direction
chromosomes (2) Same DNA polymerase molecule works
(4) all the above on both template
32. If gene A & b showing 30% recombination (3) DNA polymerase can polymerise only in
then what would be type & proportion of 5' → 3' direction
gametes in AaBb plant (4) Replication is semiconservative in nature
AB ab Ab aB 37. Which of the following variety of Brassica is
(1) 35 35 15 15 white rust resistant
(1) Pusa Gaurav
(2) 15 15 35 35
(2) Pusa Swarnim
(3) 20 20 30 30
(3) Pusa Shubra
(4) 35 15 15 35
(4) Pusa Komal
33.
38. Hairy leaves are associated with
(1) resistance to Jassids in cotton
(2) resistence to cereal leaf beetles in wheat
(3) resistance to bollworms
(4) Both 1 & 2
39. Major part of activated sludge is pumped in
to
Species (3) is the result of (1) aeration tank
(1) Autopolyploidy (2) anaerobic sludge digeter
(2) Allopolyploidy (3) aerobic sludge digester
(4) tertiary treatment
(3) Autoallopolyploidy
40. The ladybird used to get rid of
(4) Aneuploidy
(1) aphids
34. Lac selection method of screening is
applicable on which of the following vector (2) mosquitoes
(1) P BR322 (3) both 1 & 2
(4) bollworms
(2) PUC8
(3) Cosmid 41. In P BR322 , BamHI is present in
(4) All the above (1) amp R gene
35. In Lac operon which of the following gene is (2) tet R gene
constitutive in function (3) rop region
(1) Promoter
(4) Kanamycin R gene
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42. Gel electrophoresis is used for separation of (4) Genital herpes
(1) DNA 48. The ancestor of elephant
(2) RNA (1) Protylopus
(3) Protein (2) Jamoytius
(4) All the above (3) Moeritherium
43. For PCR how many types of primers are (4) Sahelanthropus
required 49. Sustained high fever, headache, stomach pain
(1) One and constipation are the symptoms of
(1) Malaria
(2) Two
(2) Typhoid
(3) Three
(3) Ascariasis
(4) Not fixed
(4) Amoebiasis
44. Out of the critical research areas of
50. Choose correct pair
biotechnology best catalyst usually provide
(1) Frog – ectothermic animal
inform of
(2) Hedge hog – osmoconformer
(1) microbe
(3) Pacific salmon – stenohaline animal
(2) pure enzyme
(4) Desert lizard – thermoregulator
(3) pure m – RNA
51. Multiple genes influence this trait in a human
(4) either 1 or 2
(1) ABO blood types
45. Cotton bollworms controlled by
(2) Sickle cell anaemia
(1) Cry I Ac
(3) Skin colour
(2) Cry II Ab
(4) Phenylketonuria
(3) Cry I Ab
52. Match the contents in column – I with column
(4) both 1 & 2
– II and identify correct answer.
46. Which part of male reproductive system of
human helps in storage, nutrition, Column – I Column – II
physiological maturation of sperms and also (A) Kupffer cells (i) Secretion of HCl
aids in attaining motility of sperms? (B) Leydig cells (ii) Secretion of mucus
(1) Vagina
(C) Oxyntic cells (iii) Secretion of renin
(2) Epididymis
(D) JG cells (iv) Phagocytosis
(3) Seminal vesicle
(E) Brunner’s glands (v) Secretion of
(4) Sertoli cells
androgens
47. Identify the sexually transmitted viral
(1) A – iv, B – v, C – i, D – ii, E – iii
infection affecting the sex organs
(1) AIDS (2) A – iv, B – v, C – i, D – iii, E – ii
(2) Hepatitis - B (3) A – v, B – iv, C – iii, D – i, E – ii
(3) Syphilis (4) A – iv, B – v, C – ii, D – iii, E – i

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53. Statement A: The binding affinity of (3) Star fish and oyster
haemoglobin for O2 decreases at tissue (4) Clown fish and sea anemone
level. 58. Choose incorrect statement about rods and
CO2
Statement B: P is relatively higher in cones of human eye
tissues than P O2 . (1) Photopigments are found in the
(1) Both statement A and statement B are
membrane discs of outer segment.
true.
(2) Dissociation of retinal from opsin alters
(2) Statement A is true and statement B is
membrane permeability
false.
(3) They are absent in blind spot.
(3) Statement A is false and statement B is
(4) They are a type of bipolar neurons.
true.
59. Which of the following set is not a correct
(4) Both statement A and statement B are
match ?
false.
(1) Male Ascaris – Copulatory spicules
54. Which of the following joint connects axial
(2) Female cockroach – boat shaped 7th
skeleton with appendicular skeleton?
sternum
(1) Joint between skull and atlas
(3) Female sea horse – brood pouch
(2) Joint between heads of ribs and vertebrae
(4) Male frog – Vocal sacs
(3) Joint between acetabulum and femur.
60. Hermaphrodite with internal fertilisation and
(4) Joint between sacrum and coxal bone
indirect development.
55. Conditional reabsorption of Na+ and water
(1) Fasciola
occurs in this part of human nephron.
(2) Ctenplana
(1)DCT
(3) Hirudinaria
(2) Descending limb of loop of Henle
(4) Aedes
(3) PCT
61. Cells of areolar tissue which secrete
(4) Collecting duct
immunoglobulins
56. Lipolytic enzymes play a role in the
(1) Fibroblast
breakdown of dietary lipids. Which gland
does not secrete such enzymes? (2) Mast cells
(1) Gastric glands (3) Plasma cells
(2) Liver (4) Macrophages
(3) Intestinal glands 62. Which of the following is closest to Homo
(4) Pancreas sapiens ?
57. The type of association between barnacles (1) Dryopithecus
and whale is similar to that of (2) Homo erectus
(1) Ticks and dogs (3) Gorilla
(2) Myxcorrhizae and trees (4) Ramapithecus
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63. ‘Water Act’ was passed by the Government (4) pancreas
of India in 68. Choose incorrect pair.
(1)1986 (1) Sea hare - radula
(2) 1974 (2) Sea pen – cnidoblasts
(3) 1997 (3) Sea lion – gills
(4) 1982 (4) Sea urchin – spines
64. Identify the incorrect statement about tissue 69. The primary function of epithelium lining
grafting. the proximal convoluted tubule of nephron
(1) Cell mediated immune response causes (1) filtration and absorption
graft rejection (2) absorption and secretion
(2) Cyclosporins are used to suppress graft (3) protection and secretion
rejection (4) acid base balance and reabsorption of
(3) Autografts are not generally rejected urea
(4) It requires blood group and tissue 70. Open circulatory system is found in
incompatability. (1) Apis
65. Choose the set of chromosomal disorders (2) Nereis
resulting from trisomy of autosomes (3) Ichthyophis
(a) Down’s syndrome
(4) Bungarus
(b) Turner’s syndrome
(c) Cri-du-chat syndrome 71. Gonadotropins secreted by ___A___ act on
(d) Klinefelter’s syndrome ___B__. “A’ and ‘B’ represent
(e) Edward’s syndrome
(1) A- hypothalamus, B – adenohypophysis
(f) Patau syndrome
(1) a, b, d (2) A – adenohypophysis, B – ovary
(2) a, d, e, f (3) A – Ovary, B – uterus
(3) a, e, f (4) A – hypothalamus, B – testis
(4) d only 72. If a couple with normal vision has two
66. The primary producers of aquatic ecosystem colorblind sons and daughter with normal
(1) Phytoplankton vision the genotypes of the couple will be
(2) Zooplankton
(1) - X+ Xc , - XcY
(3) Molluscs
(4) Small fishes
(2) - X+ Xc , - X+Y
67. Which of the following endocrine gland of
human has a duct ?
(3) - Xc Xc , - X+Y
(1) liver
(2) thyroid
(4) - X+ X+ , - XcY
(3) pituitary
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73. Which of the following statements is not (1) Calotropis – Cardiac glycosides
related to the ‘Y’ chromosome of human? (2) Cinchona - Quinine
(1) It has 231 genes
(3) Rauwolfia - reserpine
(2) It has gene for dystrophin
(4) Vinca - taxol
(3) It determines male sex
80. Sexually transmitted bacterial infection which
(4) It has Y linked and XY linked genes affects new born through transplacental
74. The insect which produces sperms with ‘X’ or transmission
‘Y’ chromosome (1) AIDS
(1) Human (2) Syphilis
(2) Honey bee (3) Genital warts
(3) Grass hopper (4) Trichomoniliasis
(4) Drosophila 81. Semelparity is a reproductive strategy in
75. Which contraceptive method has high risk of (1) Salmon
breast cancer?
(2) Beaver
(1) CuT
(3) Pigeon
(2) Tubectomy
(4) Human
(3) Implants
82. Choose correct combination with reference to
(4) Vaults
sarcomere of skeletal muscle fibre.
76. Enterokinase is secreted by
(1) Pancreas Structure Character Function
Polymer of
(2) Stomach Thick Hydrolysis
(1) globular
filament of ATP
(3) Crypts of Lieberkhun proteins
(4) Liver Provides
Elastic fibre
attachment
77. The maximum volume of air expelled after (2) Z – line of
for actin
deepest inhalation A – band
filaments
(1) TLC
Active sites
(2) VC Pulling of
Thin for
(3) thick
(3) IC filament interaction
filaments
with myosin
(4) ERV
Myosin Binding site Cross bridge
78. The increase in blood volume causes high (4)
head for actin formation
venous pressure. The effect of this could be
(1) release of renin by JG cells 83. Which hormone level is relatively low during

(2) secretion of ADH ovulation in a woman?

(3) release of ANF (1) Progesterone

(4) secretion of aldosterone (2) Estrogen

79. Choose incorrect pair with reference to (3) FSH

chemical defences in plants (4) LH
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84. SO2 pollution in thermal power stations can (2) B - Pyramid numbers of a big tree

be checked by using (3) C - Pyramid of biomass in sea

(1) Incinerator (4) A - Pyramid of energy in parasitic food

(2) Scrubber chain

(3) Bag filter 90. Choose incorrect statement about Malpighian

(4) Catalytic converter tubules of cockroach.

85. Sacred groves of Rajasthan (1) terrestrial adaptation for water

(1) Khasi and Jaintia conservation

(2) Sarguja (2) help in osmoregulation and water

(3) Aravalli Hills conservation.

(4) Bastar (3) whitish filamentous structures which

86. The Himalayan musk deer is an endangered excrete uric acid

species which is protected at (4) lined by glandular and ciliated cells.
(1) Kedarnath wild life sanctuary



91. A , B and C are three orthogonal vectors
(2) Keoladeo Ghana national park with magnitudes 3, 4 and 12 respectively. The




(3) Kaziranga national park value of A − B + C will be
(4) Kanha national park (1) 11
87. Which of the following examples does not (2) 19
explain convergent evolution?
(1) eyes of octopus and cat (3) 13

(2) flipper of penguins and dolphins (4) can't be determined

(3) sweet potato and potato 92. If the system is released, then the acceleration
of the centre of mass of the system is
(4) Forelimbs of whale and horse
88. Cytokine barriers used in treatment of cancer
(1) NK cells
(2) Monoclonal antibodies
(3) α - interferons
(4) Taxol
89. Which of the following ecological pyramids
A, B and C is correctly matched with the
given options. g
(1)
4
g
(2)
2
(3) g
(1) A - Pyramid of biomass in forest (4) 2g

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93. A particle is kept at rest at the top of a sphere energy per unit volume of the two wires will
of diameter 42 m. When disturbed slightly, it be in the ratio
slides down. At what height 'h' from the (1) 1 : 2
bottom, the particle will leave the sphere (2) 4 : 1
(1) 14 m (3) 2 : 1
(2) 28 m (4) 16 : 1
(3) 35 m 97. Two pendulums of time period 3 s and 7s
(4) 7 m respectively start oscillating simultaneously
94. Velocity-time graph of a particle moving in a from two opposite extreme positions. After
straight line is as shown in figure. Mass of the how much least time they will be in phase
particle is 2 kg. Work done by all the forces 21
(1) s
acting on the particle in time interval between 8
t = 0 to t = 10 s is 21
(2) s
4
21
(3) s
2
21
(4) s
10
98. The equation of a wave on a string of linear
(1) 300 J
mass density 0.04 kg m–1 is given by
(2) –300 J
  t x 
(3) 400 J y = 0.02 ( m ) sin  2π  − .
(4) –400 J   0.04 ( s ) 0.50 ( m )  
95. A thin hollow cylinder is free to rotate about The tension in the string is
its geometrical axis. It has a mass of 8 kg and (1) 6.25 N
a radius of 20 cm. A rope is wrapped around (2) 4.0 N
the cylinder. What force must be exerted (3) 12.5 N
along the rope to produce an angular (4) 62.5 N
acceleration of 3 rad/sec2 ? 99. The average kinetic energy of a gas molecule
(1) 8.4 N at t 0C is directly proportional to
(2) 5.8 N
(1) t
(3) 4.8 N
(2) t
(4) None of these
1
96. Two wires of the same material and same (3)
t
length but diameters in the ratio 1:2 are
(4) t + 273
stretched by the same force. The potential
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100. An observer starts moving with uniform
acceleration 'a' towards a stationary sound
source of frequency f. As the observer
approaches the source, the apparent frequency
f' heard by the observer varies with time t as:
(1) Increase of 12.5 gm-wt
(1)
(2) Increase of 6.25 gm-wt
(3) Decrease of 12.5 gm-wt
(4) Decrease of 6.25 gm-wt
102. In the indicator diagram fig. shown of carnot
cycle Ta, Tb, Tc, Td represent temperature of
gas at A, B, C, D respectively. Which of the
(2) following is correct relation

(3)
(1) Ta = Tb = Tc = Td
(2) Ta = Tc, Tb = Td
(3) Ta = Td, Tc = Tb
(4) Ta = Tb, Tc = Td
103. One end of a copper rod of length 1.0 m and
(4) area of cross-section 10-3m2 is immersed in
boiling water and the other end in ice. If the
coefficient of thermal conductivity of copper
is 92 cal / m-s - oC and the latent heat of ice is

8 × 10 4 cal / kg , then the amount of ice which

will melt in one minute is

101. A cylinder containing water upto a height of
(1) 9.2 × 10 −3 kg
1 2
25 cm has a hole of cross-section cm in
4 (2) 8 × 10 −3 kg
the bottom. It is counterpoised in a balance.
(3) 6.9 × 10 −3 kg
What is the initial change in the balancing
weight when water begins to flow out? (4) 5.4 × 10 −3 kg
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104. One mole of a an ideal gas is heated at (d) The energy of the mass m remains
constant pressure through 1K work done by constant
the gas is (1) a, b
(1) 1 Joule (2) b, d
1 (3) a, c
(2) Joule
R (4) a, d
(3) R Joule 107. In the given figure each plate of capacitance C
(4) 2.5 R Joule has charge
105. A pipe of length 85 cm is closed from one
end. Find the number of possible natural
oscillations of air column in the pipe whose
frequencies lie below 1250 Hz. The velocity
of sound in air is 340 m/s.
(1) 6
(2) 4 (1) CE
(3) 12 CER 1
(2)
(4) 8 R2 − r
106. Two bodies, each of mass M are kept fixed CER 2
(3)
with a separation 2L. A particle of mass m is R2 + r
projected from the mid-point of the line
CER 1
joining their centres, perpendicular to the line. (4)
R2 + r
The gravitational constant is G. The correct
108. Full scale deflection current for galvanometer
statement(s) is (are)
is 1 Amp. What should be the value of shunt
(a) The minimum initial velocity of the mass
resistance so that galvanometer shows half
‘m’ to escape the gravitational field of the
scale deflection.
GM
two bodies is 4
L
(b) The minimum initial velocity of the mass
‘m’ to escape the gravitational field of the

GM
two bodies is 2 (1) 10 Ω
L
(c) The minimum initial velocity of the mass (2) 1 Ω

‘m’ to escape the gravitational field of the (3) 12Ω

2GM (4) 2Ω
two bodies is
L
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109. A bar magnet is held at right angle to a IV) height of the satellite from the surface of
uniform magnetic field. The couple acting on the earth.
the magnet is to be halved by rotating it from
Which of the following option is correct?
this position. The angle of rotation is
(1) 60° (1) Only I
(2) 45° (2) I and II
(3) 30° (3) I, II and III
(4) 75° (4) II, III and IV
110. Which of the following loop is in stable 112. For a ring having radius R of uniform linear
equilibrium charge density λ (As shown in figure) its
(1) intensity of electric field at its centre equals to

(2)
2K λ
(1)
R
4K λ
(2)
R
Kλ
(3) (3)
R
(4) Zero
113. Which bulb will be glow brightest in the
following circuit diagram?

(4)

111. The orbit of a geostationary satellite is

circular, the time period of satellite depends
(1) B1
on
I) mass of the satellite. (2) B2
II) mass of the earth. (3) B3
III) radius of the orbit. (4) B4
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114. In the potentiometer circuit shown in the Neglecting source resistance, the voltmeter
figure, AB is a uniform wire of length 100 cm and ammeter readings will be
and resistance 2.4 Ω . The length AC of the
wire for which the galvanometer G shows no
deflection is 40 cm. The emf of the test cell,
ε is

(1) 0 V, 2.0 A
(2) 0 V, 1.4 A
(3) 5.6 V, 1.4 A
(1) 1.56 V (4) 8 V, 2.0 A
(2) 0.96 V 117. An electron is moving through a field. It is
(3) 2.77 V moving
(i) opposite an electric field
(4) 1.44 V
(ii) perpendicular to a magnetic field as
115. A uniform magnetic field B existas in a shown. For each situation the de-Broglie
cylindrical region of radius 10 cm as shown wave length of electron
in figure. A uniform wire of length 80 cm
and resistance 4.0 Ω is bent into a square
frame and is placed with one side along a
diameter of the cylindrical region. If the
magnetic field increases at a constant rate of
0.010 T/s, find the current induced in the
frame
(1) Increasing, increasing
(2) Increasing, decreasing
(3) Decreasing, same
(4) Increasing, Same
118. A ray of light incident on an equilateral glass
prism shows minimum deviation of 30°.
(1) 3.9 × 10 −5 A
Calculate the speed of light through the
−5
(2) 0.2 × 10 A prism.

(4) 0.8 × 10 −5 A (1) 2.12 × 108 m/sec

(2) 1.5 × 108 m/sec
−5
(4) 10 × 10 A
(3) 3.1 × 108 m/sec
116. In the circuit shown in figure, the AC source
(4) 5 × 108 m/sec
gives a voltage V = 20 cos (2000t).
Sri Chaitanya Page Hyderabad
15
119. A radioactive nucleus is being produced at a
constant rate α per second. Its decay
constant is λ . If N0 are the number of nuclei
at time t = 0, then maximum number of nuclei
possible are
Bω l 2
α (1) VA − V0 =
(1) 2
λ
7
α (2) V0 − VC = Bω l 2
(2) N 0 + 2
λ
(3) VA − VC = 4Bω l 2
(3) N 0
9
λ (4) VC − VO = Bω l 2
(4) + N 0 2
α
124. An astronomical telescope is able to resolve
120. Electromagnetic wave of intensity 1400 W/m2
two stars at an angular separation of 10–3
falls normally on metal surface of area 4.5 m2
degree. If wavelength of light used is 500nm,
and is completely reflected by it. Find out
then the diameter of the objective of the
force exerted by beam
telescope is
(1) 14 × 10–5 N
(1) 3.5 cm
(2) 14 × 10–6 N
(3) 21 × 10–5 N (2) 0.35cm
(4) 42 × 10–6 N (3) 3.5 m
121. A telescope has an objective lens of focal
(4) None
length 150 cm and an eye piece of focal
length 5 cm. If this telescope is used to view a 125. The following diagram indicates the energy
100 m high tower 3 km away, find the height levels of a certain atom when the system
of the final image when it is formed 25 cm moves from 4E level to E. A photo of
away from the eye piece. wavelength λ 1 is emitted. The wavelength of
(1) 11 cm photon produced during it's transition from
(2) 15 cm 7 λ
E level to E is λ 2 . The ratio 1 will be
(3) 22 cm 3 λ2
(4) 30 cm
122. The mobility of free electron is greater than
that of free holes because
(1) The carry negative charge
(2) They are light
(3) They mutually collide less 9
(1)
(4) They require low energy to continue their 4
4
motion (2)
9
123. A conducting rod AC of length 4l is rotated 3
about a point O in a uniform magnetic field (3)

 2
B directed into the paper. AO = l and OC = 7
(4)
3l. Then 3
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126. In hydrogen atom, electron makes transition (1) A common base amplifier circuit
from n = 4 to n = 1 level. Recoil momentum (2) A common emitter amplifier circuit
of the H atom will be
(3) A common collector amplifier circuit
(1) 3.4 × 10–27 N-sec
(4) Neither of the above
(2) 6.8 × 10–27 N-sec
130. In the circuit below, A and B represents two
(3) 3.4 × 10–24 N-sec
inputs and C represents the output. The circuit
(4) 6.8 × 10–24 N/sec
represents
127. Two coherent sources of intensity ratio x2
interfere that in interference pattern

I −I 1+ X2
(1) max . min . =
I max . + I min . 2 X
I + I min . 1 + X
(2) max . =
I max . − I min . 2 X
I −I 2X (1) AND gate
(3) max . min . =
I max . + I min . 1+ X2 (2) NOR gate
I + I min . 2X (3) OR gate
(4) max . =
I max . − I min . 1+ X 2
(4) NAND gate
128. A lens made up of different material forms 131. One centimetre on the main scale of vernier
two images of a point object 'O' as shown in
calipers is divided into ten equal parts. If 10
figure. Then
divisions of vernier scale coincide with 8
small divisions of the main scale, the least
count of the callipers is :-
(1) 0.01 cm
(2) 0.02 cm
(3) 0.05 cm
(4) 0.005 cm
(1) µ1 > µ2
132. An object moving with a speed of
(2) µ1 = µ2
(3) µ2 > µ1 6.25 ms −1 , is decelerated at a rate given by
dv
(4) Data insufficient = −2.5 v . Where v is the instantaneous
dt
129. An NPN-transistor circuit is arranged as
speed. The time taken by the object, to come
shown in the figure. It is
to rest, would be
(1) 1 s
(2) 2 s
(3) 4 s
(4) 8 s

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17
133. A bullet of mass 0.01 kg, is travelling at a (1) 1.66 m/s2
speed of 500 m/sec, strikes a block of 2 kg (2) 2.33 m/s2
which is suspended by a string of length 5m. (3) 3.66 m/s2
The centre of gravity of the block is found to (4) 4.66 m/s2
rise a vertical distance of 0.1 m. What is the 136. The energy of second Bohr orbit of the
approximate speed of the bullet after it hydrogen atom is −328 kJ mol −1 ; hence the
emerges from the block ?
energy of fourth Bohr orbit would be
(1) 200 m/s
(1) − 82 kJ mol −1
(2) 220 m/s
(3) 204 m/s (2) −41 kJ mol −1
(4) 284 m/s
(3) − 1312 kJ mol −1
134. A hoop rolls on a horizontal ground without
slipping with linear speed v. Speed of a (4) − 164 kJ mol −1
particle P on the circumference of the hoop at
137. The set of quantum numbers, n = 3, l = 2,
angle θ is :
ml = 0
(1) Describes an electron in a 2s orbital
(2) is not allowed
(3) describes an electron in a 3p orbital
(4) describes one of the five orbitals of same
θ energy.
(1) 2v sin
2 138. At low pressures, van der Waals equation is
θ
(2) v sin  a 
2 written as  P +  V = RT . The
V2
θ
(3) 2v cos compressibility factor is equal to
2
θ  a 
(1)  1 −
(4) v cos  RTV 
2
135. Two blocks, 4 kg and 2 kg are sliding down  RTV 
(2)  1 −
an incline plane as shown in the figure. The  a 
acceleration of 2 kg block is  a 
(3)  1 +
 RTV 

 RTV 
(4)  1 +
 a 

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18
139. 20 mL of methane is completely burnt using 143. If the enthalpy change for the transition of

liquid water to steam is 30 kJ mol −1 at

50 mL of oxygen. The volume of the gas left
after cooling to room temperature is
(1) 80 mL
27 0 C , the entropy change for the process
would be
(2) 40 mL
(3) 60 mL (1) 10 J mol −1 K −1
(4) 30 mL
(2) 1.0 J mol −1 K −1
140. Which of the following is a buffer solution?
(3) 0.1J mol −1 K −1
(1) 500 mL of 0.1 N CH 3COOH + 500 mL

of 0.1 N NaOH (4) 100 J mol −1 K −1

(2) 500 mL of 0.1 N CH 3COOH + 500 mL 144. Elements of which of the following group(s)

of 0.1 N HCl of periodic table do not form hydrides?

(1) Groups 7, 8, 9
(3) 500 mL of 0.1 N CH 3COOH + 500 mL
(2) Group 13
of 0.2 N NaOH
(3) Groups 15, 16, 17
(4) 500 mL of 0.2 N CH 3COOH + 500 mL
(4) Group 14
of 0.1 N NaOH 145. The vapour pressure of two liquids P and Q
141. Which of the following is most soluble in are 80 and 60 torr respectively. The total
water? vapour pressure of solution obtained by

(
(1) Ba3 ( PO4 ) K sp = 6 × 10 −39
2 ) mixing 3 mol of P and 2 mole of Q would be
(1) 68 torr
(
(2) ZnS K sp = 7 × 10 −16
) (2) 20 torr

( )
(3) 140 t orr
(3) Fe (OH ) 3 K sp = 6 × 10 −38
(4) 72 torr

(
(4) Ag3 ( PO4 ) K sp = 1.8 × 10 −18 ) 146. The oxidation potential of hydrogen electrode
at pH = 10 and PH = 1 is
2
142. In an exothermic equilibrium
(1) 0.51 V
A + 3B ⇌ AB3 all the reactants and
products are in the gaseous state. The (2) 0.00 V

formation of AB3 is favoured at (3) + 0.59 V

(1) low temperature and low pressure (4) 0.059 V

(2) low temperature and high pressure
(3) high temperature and high pressure
(4) high temperature and low pressure

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19
 (1) k
147. E 0 for the cell Zn | Zn 2 + ( aq ) || Ch 2 + ( aq ) | Cu is

1.10 V at 25 o C , the equilibrium constant for (2) log k

the reaction Zn + Cu 2+ ( aq ) ⇌ Cu + Zn 2+ ( aq ) (3) In k

is of the order of
(4) 1/n
(1) 10 −28
152. The method of zone refining of metals is
(2) 10 −37
based on the principle of
(3) 10 +18

(4) 10 +17
(1) Greater mobility of the pure metal than

148. The rate constant of the reaction A → B is that of impurity

0.6 × 10 −3 mol L−1 s −1 . If the concentration (2) Higher melting point of the impurity than
of A is 5 M, then concentration of B after 20
that of the pure metal
minutes is
(3) Greater noble character of the solid metal
(1) 3.60 M
than that of the impurity
(2) 0.36 M
(4) Greater solubility of the impurity in the
(3) 0.72 M
molten state than in the solid
(4) 1.08 M
153. Which of the following polymers is not
149. If AgI crystallises in zinc blende structure
with I − ions at lattice points. What fraction correctly matched?
of tetrahedral voids is occupied by Ag + (1) Formation of dacron – Step growth
ions?
(1) 25 % polymerisation
(2) 50 % (2) Formation of polytetrafluoroethene – Step
(3) 100 %
growth polymerisation
(4) 75 %
150. The number of carbon atoms per unit cell of (3) formation of polythene – Chain growth
diamond unit cell is polymerisation in presence of benzoyl
(1) 1
peroxide
(2) 4
(3) 8 (4) formation of Polyacrylonitrile - Chain
(4) 6 growth polymerisation in presence of
151. For adsorption of gas on solid surface, the
peroxide
plots of log x / m vs.log P is linear with a
slope equal to

Sri Chaitanya Page Hyderabad

20
 (3) A + + B → A + B +
154. If one strand of DNA has the sequence
ATGCTTGA, the sequence in the
(4) cannot be predicted
complimentary strand would be
158. The photochemical bromination of
(1) TCCGAACT
ethylbenzene with excess of bromine
(2) TACGTAGT
followed by hydrolysis with aqueous KOH
(3) TACGAATC
gives
(4) TACGAACT (1) PhCHO

155. Which of the following is not a correct (2) PhCH 2CHO

statement? (3) PhCOCH 3

(1) Transparent soaps are made by dissolving (4) PhCHOHCH 3
the soap in ethanol and then evaporating
159. C6 H 5 14COOH on heating with Na2CO3
excess solvent.
(2) Soaps that float in water are made by releases

beating tiny air bubbles before their (1) CO2

hardening.
(2) 14 CO2
(3) Medicated soaps contain alcohol to
(3) CO
prevent rapid drying.
(4) H 2
(4) Potassium soaps are soft to the skin than
sodium soaps. 160. The correct order of basicities of the

156. The vapour pressure of an aqueous solution of following compounds is

sucrose at 373 K is found to be 750 mm Hg.

The molarity of the solution at the same
temperature will be (I)
(1) 0.26 (II) CH 3 − CH 2 − NH 2
(2) 0.73
(III) (CH 3 ) NH
(3) 0.74 2

(4) 0.039 O
||
157. A hypothetical electrochemical cell is shown
(IV) CH 3 − C − NH 2
below: A | A + ( XM ) || B + (YM ) | B . The
(1) II > I > III > IV
e.m.f. measured is + 0.20 V. The cell reaction
(2) I > III > II > IV
is

(1) A + B + → A + + B
(3) III > I > II > IV

(2) A + + B → A + B +
(4) I > II > III > IV

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21
161. 163. Which of the following is with square
pyramidal structure?
(1) XeO4

(2) XeF6

would be (3) XeO2 F2

(1) (4) XeOF4
164. The inner transition elements with

4 f 7 5d 1 6s 2 and 5 f 7 6d 17s 2
configuration are respectively.
(1) Gd, Cm
(2)
(2) Gd, Lu
(3) Lu, Cm
(4) Gd, Lw
165. Which of the following is responsible for the
high reducing nature of phosphinic acid?
(3)
(1) 3 P – OH bonds
(2) 2 P – H bonds
(3) 1 P = O bond
(4) 2 P – OH bonds
166. Regarding sulphur dioxide the incorrect
(4)
statement is
(1) SO2 is used as an antichlor, disinfectant
and food preservative
(2) Decolourisation of acidified potassium
permanganate is the convenient identification
162. Which of the following can oxidise H 2S to test for SO2
S?
(3) Liquid SO2 is as good solvent for both
−2
(1) Cr2O7 organic and inorganic compounds.

(2) MnO4− (4) In the reaction with water and alkalies the
behaviour of SO2 is very similar to that of
(3) Cl2
carbon monoxide.
(4) All

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22
167. Statement I: When HCl reacts with finely (4) π 2Py > σ 2Pz < π * 2Px = π * 2Py
powdered iron it forms FeCl2 but not 171. The low solubility of LiF in water is due to
FeCl3 . __and the low solubility of CsI is due to ___

Statement II: Cl2 oxidises H 2 S to SF6 . (1) Low lattice enthalpy, high lattice enthalpy
(2) Low hydration enthalpy, high lattice
(1) Both statement - I and statement - II are
enthalpy
true
(3) High lattice enthalpy, low hydration
(2) Statement - I is true, but statement - II is
enthalpy
false
(4) High lattice enthalpy, high hydration
(3) Both statement - I and statement - II are
enthalpy
false
172. The dehydration of chlorides of which of the
(4) Statement - I is false, but statement - II is
following alkaline earth metal cannot be
true
achieved on heating?
168. Methyl halides with highest and least dipole
(1) Ba
moment values are
(2) Ca
(1) CH 3 F ,CH 3 I
(3) Sr
(2) CH 3Cl,CH 3 I (4) Mg
(3) CH 3Cl,CH 3 F 173. Colourless metal metaborate is

(4) CH 3 F ,CH 3 Br (1) Co ( BO2 )

2
169. Considering the elements B, C, N, F and Si (2) Cu ( BO2 )
2
the correct order of their non metallic
(3) Cr ( BO2 )
character is 3

(1) B > C > Si > N > F (4) Al ( BO2 )

3
(2) Si > C > B > N > F 174. Incorrect statement among the following is
(1) Graphite is thermodynamically most
(3) F > N > C > B > Si
stable allotrope of carbon
(4) F > N > C > Si > B
(2) Tridymite and Cristobalite are the
170. Which of the following order of energies of amorphous forms of silica
molecular orbitals of N 2 is correct? (3) Kieselghur an amorphous form of silica is
used in filtration plants.
(1) π 2Py < σ 2Pz < π * 2Px = π * 2Py
(4) A substantial amount of CO2 is used in
(2) π 2Py > σ 2Pz > π * 2Px = π * 2Py
the manufacture of urea.
(3) π 2Py < σ 2Pz > π * 2Px = π * 2Py

Sri Chaitanya Page Hyderabad

23
175. The best and latest technique for isolation, (2)
purification and separation of organic
compounds is
(1) Crystallisation
(2) Distillation
(3) Sublimation
(4) Chromatography (3)

176. The extent of crystal field splitting in

octahedral complexes of the given metals
with a particular weak field ligand are such
that
(1) Ir (III) > Th (III) > Fe (III) > Cr (III)
(2) Ir (III) > Th (III) > Cr (III) > Fe (III)
(4)
(3) Cr (III) > Fe (III) > Rh (III) > Ir (III)
(4) Ir (III) > Rh (III) > Cr (III) ≡ Fe (III)

177. Co ( NH 3 ) ( NO2 ) 2  Cl exhibits

 4
(1) Linkage isomerism, ionization isomerism
and geometrical isomerism
(2) Ionization isomerism, geometrical 179. Statement I: Dipole moment of NH 3 is

isomerism and optical isomerism greater than the dipole moment of NF3
(3) Linkage isomerism, geometrical Statement II: N – F bond is more polar than
isomerism and optical isomerism N – H bond.
(4) Linkage isomerism, ionization isomerism (1) Both statement - I and statement - II are
and optical isomerism true
(2) Statement - I is true, but statement - II is
false
(3) Both statement - I and statement - II are
false
178.
(4) Statement - I is false, but statement - II is
Which of the following is X?
(1) true
180. Which of the following is optically inactive?
(1) Lactic acid
(2) Meso tartaric acid
(3) Alanine
(4) 2 – butanol
Sri Chaitanya Page Hyderabad
24
 SRI CHAITANYA EDUCATIONAL INSTITUTIONS,INDIA
A.P,TELANGANA,KARNATAKA,TAMILNADU,MAHARASHTRA,DELHI,RANCHI
SR ELITE, AIIMS S60, NEET MPL & MEDICON NEET GRAND TEST – 5 KEY Date : 31-01-2020

BOTANY
1) 4 2) 3 3) 1 4) 3 5) 2 6) 2 7) 4 8) 3 9) 2 10) 1
11) 3 12) 4 13) 2 14) 2 15) 2 16) 3 17) 4 18) 1 19) 3 20) 4
21) 3 22) 4 23) 4 24) 4 25) 3 26) 3 27) 2 28) 3 29) 3 30) 4
31) 3 32) 2 33) 2 34) 2 35) 2 36) 3 37) 2 38) 4 39) 2 40) 1
41) 2 42) 4 43) 2 44) 4 45) 4

ZOOLOGY
46) 2 47) 4 48) 3 49) 2 50) 1 51) 3 52) 2 53) 1 54) 4 55) 1
56) 2 57) 4 58) 4 59) 3 60) 1 61) 3 62) 2 63) 2 64) 4 65) 3
66) 1 67) 4 68) 3 69) 2 70) 1 71) 2 72) 2 73) 2 74) 4 75) 3
76) 3 77) 2 78) 3 79) 4 80) 2 81) 1 82) 4 83) 1 84) 2 85) 3
86) 1 87) 4 88) 3 89) 2 90) 3

PHYSICS
91) 3 92) 1 93) 3 94) 1 95) 3 96) 4 97) 1 98) 1 99) 4 100) 1
101) 3 102) 4 103) 3 104) 3 105) 1 106) 2 107) 3 108) 4 109) 1 110) 4
111) 4 112) 1 113) 3 114) 2 115) 1 116) 3 117) 3 118) 1 119) 1 120) 4
121) 4 122) 4 123) 3 124) 1 125) 2 126) 2 127) 3 128) 1 129) 2 130) 3
131) 2 132) 2 133) 2 134) 1 135) 2

CHEMISTRY
136) 1 137) 4 138) 1 139) 4 140) 4 141) 4 142) 2 143) 4 144) 1 145) 4
146) 3 147) 2 148) 3 149) 2 150) 3 151) 4 152) 4 153) 2 154) 4 155) 3
156) 3 157) 1 158) 3 159) 1 160) 2 161) 3 162) 4 163) 4 164) 1 165) 2
166) 4 167) 2 168) 2 169) 3 170) 1 171) 3 172) 4 173) 4 174) 2 175) 4
176) 2 177) 1 178) 1 179) 1 180) 2
 SOLUTIONS
SOLUTIONS
PHYSICS



 


91. Say A = 3iˆ , B = 4 ˆj and C = 12kˆ A − B + C = 3iˆ − 4 ˆj + 12kˆ



⇒ A − B + C = 3 2 + 4 2 + 12 2 = 13

g mg
3m − m g 3m × −
92. a= = Acceleration of centre of mass = 2 2 =g
3m + m 2 3m + m 4

mv 2
93. At B, mg cos θ = ∴ mv 2 = mgR cos θ
R

Applying COME from A to B ∆ PE = ∆ KE

1 1
mgR ( 1 − cos θ ) = mv 2 ⇒ mgR ( 1 − cos θ ) = mgR cos θ
2 2
cos θ 3 2
⇒ 1 − cos θ = ⇒ 1 = cos θ ⇒ cos θ =
2 2 3
 2 R
∴ Height dropped = R ( 1 − cos θ ) = R  1 −  =
 3 3
R 5R 5
Height from bottom = 2R − = = × 21 = 35 m
3 3 3
94. From work energy theorem
1
2
(
1
)
W = ∆ KE = K f − K i = m v f 2 − vi 2 = × 2 ( 400 − 100 ) = 300 J
2

95. I = MR 2 = 0.32 kg × m 2 τ = Iα = 0.96 N − m

τ 0.96
But F= = = 4.8 N
R 0.2
2
1 1 Stress 1  F  1
96. Strain energy density ( µ ) = × Stress × Strain = Stress × =   ×
2 2 Y 2  A Y
 1
⇒ µ∝
r4
97. Let they will meet in phase after time 't'.
2π 2π 21
Then φ1 − φ 2 = π t− t =π ⇒t = s
T1 T2 8

T λ
98. v= =
µ T

99. KE ∝ T ( Kelvine) = ( t + 273 )

 V + V0   V + at   a 
100. f' =  f =  f = f  1 + t
 V   V   V 

101. F = ρav 2 = 12.5 gm − ω t

102. AB & CD are iso thermal Ta = Tb & Tc = Td

∆Q KA ∆Q 92 × 10 −3
103.
∆t
= (T − TL ) ⇒ ∆t = 1 [ 100 − 0 ]
L H
∆Q 92
∴ = ⇒ ∆Q = 92 × 6 = 552 cal
60 10
552
∆Q = mL ⇒ 552 = m × 8 × 10 4 ⇒ m = = 6.9 × 10 −3 kg
4
8 × 10
104. W = p (v2 – v1) = nR (T2 – T1) = R Joule
 v
105. 2P +   ≤ n
 4l 

−GMm GMm 1 2
106. − + mv = 0
L L 2
E
107. In steady state current drawn from the battery i =
( R2 + r )
 In steady state capacitor is fully charged hence. No current will flow through line (2)
E
Hence potential difference across line (1) is V = × R the same potential
( R2 + r ) 2
ER2
difference appears across the capacitor, so charge on capacitor Q = C ×
( R2 + r )
GA 38 38
108. S= = = = 2Ω
i 10 39
−1 −1
ig 0.5

109. τ = MB sin 90 0 = MB

MB 1
= MB sin θ ⇒ = sin θ ⇒ θ = 30 0
2 2
Angle of rotation 90 0 − 30 0 = 60 0
110. In stable equilibrium, M and B must be in the same direction.

GMm 2 GM r3
111. = mrω ⇒ω = ⇒ T = 2π
r2 r3 GM

dq λ R dθ sin θ
112. dE = 2dE1 sin θ = 2 × K . sin θ = 2K ×
R2 R2

π/2
2K λ 2K λ
E= ∫ R
sin θ dθ =
R
0

V2
113. R=
P

R1 =
( 200 )
2
= 800 Ω ; R2 =
( 200 )
2
= 400 Ω ; R3 =
( 100 )
2
= 100 Ω ; R4 =
( 100 )
2
= 50 Ω
50 100 100 200
100 1
Current through R1 & R2 = = A
200 12
100 2
Current through R3 & R4 = = A
50 3
 Power dissipated = Pmax = i 2 R ( ) max

 Ep R 5 2.4
114. e=  ×l = × 40 = 0.96 V
 r + Rs + R  L 5 100

dB π r 2 dB π ( 0.1)
2
dφ d π
115. Eind = = ( BA ) = A = × = × 0.01 = × 10 −4 V
dt dt dt 2 dt 2 2

E π / 2 × 10 −4
∴ i = ind = = 3.9 × 10 −5 A
R 4
1 1
116. X L = ω L = 2000 × 5 × 10 −3 = 10 Ω XC = = = 10 Ω
ωC 2000 × 50 × 10 −6
X L = XC ⇒ Z = R = 4 + 6 = 10 Ω

V 20 2
(A) reading = rms = = 1.4 A
Z 10
(V) reading = irms × 4 Ω ∴ XC and X L cancel out = 1.4 × 4 = 5.6 V

h
117. λ= since v is increasing in case(i), but it is not changing in case (ii). Hence, in the
mv
first case de-Broglie wavelength will change, but it second case, it remain the same.

 0 0
 A + δ m  sin  60 + 30 
sin 
 2   2 
118. µ= = = 2
 A  60 0 
sin   sin 
 2 
 2 

C C 3 × 10 8
µ= ⇒v= = = 2.12 × 10 8 m / sec
v µ 2
119. Maximum number of nuclei will be present when rate of decay = rate of formation
α
⇒ λN = α ⇒ N =
λ
F 2I 2I 2 × 1400
120. P= = ⇒F= ×A= × 4.5 = 42 × 10 −6 N
8
A C C 3 × 10
121. Angle subtended by the 100 m tall tower at 3 km away
100 1
∝= = rad
3
3 × 10 30
 h is the height of image of tower formed by objective lens
h h 1
∝= = = ⇒ h = 5 cm
f 0 150 30

D 25
magnification produced by the eye piece me = 1 + = 1+ =6
fe 5
height of final image = h × me = 5 × 6 = 30 cm
122. Conceptual
1
123. For any rotating conducting rod in a field B, ∆V = Bω l 2
2
1 1
Here V0 − VA = Bω ( l ) , V0 − VC = Bω ( 3l ) ⇒ VA − VC = 4Bω l 2
2 2
2 2

1.22 λ
124. Limit of resolution of telescope =
a

π 1.22 × 500 × 10 −9
10 −3 × = ⇒ a = 3.5 cm
180 a
hc hc
125. Transition from 4E to E ( 4E − E ) = ⇒ λ1 = …………(i)
λ1 3E

7 7  hc 3hc
Transition from E to E  3 E − E  = λ ⇒ λ2 = …….(ii)
3 2 4E

λ1 4
From equation (i) and (ii) =
λ2 9
h
126. Recoil momentum = momentum of photon =
λ
 1 1  hR × 15
= hR  − = = 6.8 × 10 −27 N × sec
2 2 16
 n1 n2 

I1

127.
I max − I min
=
( I1 + I 2 ) −(
2
I1 − I 2 )=
2
4 I 1I 2
=
I2
=
2
2x
2 2( I + I )
I max + I min
( I1 + I2 ) + (
2
I1 + I 2 ) 1 2 I
1+ 1 1+ x
I2
2

1
128. f ∝ f1 < f 2 ⇒ µ1 > µ 2
( µ − 1)
129. Conceptual
130. Conceptual
131. One verner division = 0.8/10=0.08 cm; L.C = 1 MSD – 1 VSD = 0.1– 0.08 = 0.02 cm
dv 1
132. = −2.5 v or dv = −2.5 v
dt v

On integrating, within limit (u 1 = 6.25 ms

−1
to u2 = 0 )
v2 − 0 t
−1/ 2
∴ ∫ υ dυ = − 2.5 ∫ dt
v1 = 6.25 ms −2 0

−2 × ( 6.25 )
0 1/ 2
2 × υ 1/ 2  = − ( 2.5 ) t or t = = 2s
  6.25 −2.5
133. Suppose the velocity of the bullet of mass m is u and it strikes the block of mass M. After
collision, the linear velocity of the block is V and that of the bullet is v'.
Applying law of conservation of linear momentum, we get.
mv = MV + mv' or 500 × 0.01 = 2V = 0.01 v' or 5 = 2V + 0.01 v' .......(i)
1
By conservation of energy, we get. MV 2 = Mgh or
2
V = 2gh = 2 × 9.8 × 0.1 = 1.4 m / s
Putting value of V in eqn. (i), we get 5 = 2 × 1.4 + 0.01 v' or v' = 220 m/s.

134.
 180 − θ  θ
VP = 2V cos   = 2V sin
 2  2

135. a=
( m1 + m2 ) g sinθ − µ1m1g cos θ − µ 2m2 cos θ On solving a = 2.33 m/sec2
m1 + m2
CHEMISTRY

E 2 ( n4 )
2
136. =
E4 ( n ) 2
2

PV a
138. =Z ∴ Z = 1−
RT VRT
141. For most soluble slat, solubility should be maximum.
142. Exothermic reactions are favoured at low temperature. As reaction is accompanied by
decrease in the number of moles, it is favoured by high pressure.
143. We know that ∆G = ∆ H − T ∆ S ∆ H = T ∆S [∵ ∆G = 0 ]
∆H 30 × 10 3
∆S = = = 100 J mol −1 K −1
T 300
144. It is called hydride gap.
3 2
145. PM = P1o .X 1 + P2o .X 2 = 80 × + 60 × = 72torr
3+2 3+2
H + 
log  
0.059
146. 0
EOP = EOP − ∵  H +  = 10 −10 ; PH 2 = 1atm EOP = 0.59V
1 PH  
2

0.059 1.10 × 2
147. o
Ecell = log K ⇒ log K = = 37.2881 or K = 10 −37
n 0.059
148. Reaction is zero order as the unit of rate constant.

∴ conc. of B = K .t = 0.6 × 10 −3 × 20 × 60 = 0.72 M

149. In AgI crystal, number of Ag + ions is equal to I − ions. However, the number of tetrahedral
voids are twice the number of atoms forming the cubic lattice.

∴ Number of tetrahedral voids occupied by Ag + ion = 50%.

150. In the diamond cubic unit cell, there are eight corner atoms, six face centered atoms and four
more atoms inside the structure.
∴ Number of atoms present in a diamond cubic unit cell = 1 + 3 + 4 + 8 atoms.
x 1 1
151. log = log k + log P ; this is Freundlich isotherm. Thus slope =
m n n
153. Polytetrafluoroethene (Teflon) is a chain growth polymer formed by polymerisation of
Tetrafluoroethene in presence of persulphate catalyst.
156. Given PA = 750 mm Hg ∵ 373 K is boiling point of water.
 Po − P  1000 10 1000
Thus, PAo = 760 mm Hg m= × ⇒ × ⇒ 0.74
 P  M solvent 750 18

o
157. Ecell = + ve, then A → A + + e B+ + e → B