NEET
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Model Grand Test 2017
Instructions :
This paper contains 180 objective questions; 90 from Biology, 45 from Physics and
45 from Chemistry
2.
Each question caries 4 marks. Candidate will get 4 marks for a correct response.
For each incorrect response 1 mark will be deducted from the total score
3.
Maximum time is 3 hours and maximum marks is 720
4.
To darken the appropriate circle of the correct option against each question use blue
or black ball point pen only. Answer marked with pencil would not be evaluated.
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BIOLOGY :
Column I
A. Wheat
4. The given below figure shows a generalized
life cycle of a fungus. The suitable terms for
A, B, C, D and E are
2. Diptera
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B. Mango
Column II
(Taxonomic
category order)
1. Primata
en
(Common name)
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1. Match the following and choose the correct
combination from the options given
3. Chrysophytes, Euglenoids, Dinoflagellates
and Slime moulds are included in the
kingdom
1) Monera
2) Protista
3) Fungi
4) Animalia
C. Housefly
3. Sapindales
D. Man
4. Poales
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1) A-1, B-2, C-4, D-3 2) A-4, B-3, C-2, D-1
3) A-2, B-4, C-1, D-3 4) A-3, B-4, C-2, D-1
A
B
C
D
E
Fertilizat
ion
Spore
Dikaryoti
c phase
Mitosis
Dikaryot
ic phase
Fertilizat
ion
Amitosis
1)
Meiosis
Mitosis
Spore
1) Meiosis is arrested
2)
Fertiliza
tion
Meiosis
2) The two haploid cells do not fuse
immediately
3)
Dikaryo
tic
phase
Fertiliza
tion
Myceliu
m
Meiosis
Mitosis
Mitosis
Meisis
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2. A dikaryon is formed when
3) Cytoplasm does not fuse
4) None of the above
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4)
Spore
5. ‘Viroids’ differ from ‘viruses’ in being
9. In the diagram of lenticel identify the parts as
A, B, C, D
1) Naked RNA molecules only
2) Naked DNA molecules only
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3) Naked DNA packaged with viral genome
4) Satellite RNA packaged with viral genome
6. Consider the following statements regarding
the major pigments and stored food in the
different groups of algae and select the
correct options given below
2) A-phellem, B-complementary cells, Cphelloderm, D-periderm
3) A-complementary cells, B-phellogen, Cphelloderm, D-periderm
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B. In phaeophyceae, laminarin is the stored
food and major pigments are chlorophylla and b
1) A-phellem, B-complementary cells, Cphellogen, D-phelloderm
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A. In chlorophyceae the stored food material
is starch and the major pigments are
chlorophyll-a and d
4) A-complementary cells, B-phellem, Cperiderm, D-phelloderm
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C. In rhodophyceae, floridean starch is the
stored food and the major pigments are
chlorophyll-a, d and phycoerythrin
1) A is correct, but B and C are wrong
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2) A and B are correct, but C is wrong
10. Trees near sea do not have annual rings
because
1) Soil is sandy
3) A and C are correct, but B is wrong
2) There is climatic variation
4) C is correct, but A and B are wrong
3) There is no marked climatic variation
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7. Cells of this tissue are living and show
angular wall thickening. They also provide
mechanical support. The tissue is
1) Xylem
2) Sclerenchyma
3) Collenchyma
4) Epidermis
8. Which of the following represents the floral
characters of Liliaceae
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1) Six tepals, zygomorphic, six stamens,
bilocular, ovary, axile placentation
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2) Tetramerous, actinomorphic, polyphyllous, unilocular ovary, axile placentation
3) Trimerous, actinomorphic, polyandrous,
superior ovary, axile placentation
4) Bisexual, zygomorphic, gamophyllous,
inferior ovary, marginal placentation
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4) There is enough moisture in the atmosphere
11. Which one of the following pairs is not
correctly matched
Cristae
The “Shelves” formed by
the folding of the inner
membrane
of
the
mitochondrion
2) Plasmodes The membrane surrounding
mata
the vacuole in plants
3) Grana
Membrane bound discs in
chloroplasts that contain
chlorophylls
and
carotenoids
4) Middle
Layer between adjacent
lamella
cell wall in plants derived
from cell plate
A
B
Vmax
Km
Km
Km
C
Vmax/2
Vmax/2
Ki
Vmax
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A
Km
Vmax
Vmax/2
Ki
X
[S]
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B
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14. ‘G0’ state of cells in eukaryotic cell cycle
denotes
1) Check point before entering the next
phase
2) Pausing in the middle of a cycle to cope
with a temporary delay
3) Death of a cell
4) Exit of cells from cell cycle
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15. Observe the following figure
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4) The water level is higher in side A than in
side B
16. The given diagram illustrates a Potato plant
forming new tubers. Which route would be
taken by most of the food at this time
1) 6 → 5 → 4 → 1
2) 1 → 4 → 5 → 6
3) 6 → 5 → 2 → 3
4) 1 → 4 → 2 → 3
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C
1)
2)
3)
4)
3) The water level is higher in side B than in
side A
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2) No change is observed
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13. The given adjacent graph depicts the change
in conc. Of substrate on enzyme activity.
Identify A, B and C
1) First the level of water is high in tube A
and then water level is decreased
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12. Which one of the following biomolecules is
correctly characterised
1) Lecithin – a phosphorylated glyceride
found in cell membrane
2) Palmitic acid – an unsaturated fatty acid
with 18 carbon atoms
3) Adenylic acid – adenosine with a glucose
phosphate molecule
4) Alanine amino acid – Contains an amino
group and an acidic group anywhere in the
molecule
After the system reaches equilibrium, Which
of the following changes will have occrred
17. Which one of the following statements can
best explain the term critical concentration of
an essential element.
1) Essential element concentration below
which plant growth is retarded
2) Essential element concentration below
which plant growth becomes stunted
3) Essential element concentration below
which plant remains in the vegetative
phase
4) None of the above
18. The correct sequence of flow of electrons in
the light reaction is
1 PSII, plastoquinone, cytochromes, PSI,
ferredoxin
2) PSI, plastoquinone, cytochromes, PSII,
ferredoxin
3) PSI, ferredoxin, PSII
4) PSI, plastoquinone, cytochromes, PSII,
ferredoxin
20.
The end product of oxidative phosphorylation is
1) NADH
2) Oxygen
3) ADP
4) ATP + H2O
24. Geitonogamy involves
1) Fertilization of a flower by the pollen from
a flower of another plant in the same
population
2) Fertilization of a flower by the pollen
from a flower of another plant belonging
to a distant population
3) Fertilization of a flower by the pollen
from another flower of the same plant
4) Fertilization of a flower by the pollen
from the same flower
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21. Apical dominance in plant is caused by
1) High concentration of auxins in the
terminal bud
2) High concentration of gibberellins in the
apical bud
3) Low concentration of auxins in the lateral
bud
4) Absence of auxins and gibberellins in the
apical bud
23. Which of the following statements is not
correct
1) Pollen grains of many species can
germinate on the stigma of a flower, but
only one pollen tube of the same species
grows into the style
2) Insects that consume pollen or nectar
without bringing about pollination are
called pollen/nectar robbers
3) Pollen germination and pollen tube
growth are regulated by chemical components of pollen interacting with those of
the pistil
4) Some reptiles have also been reported as
pollinators in some plant species
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Photolysis of each water molecule in light
reaction will yield
1) 2 electrons and 4 protons
2) 4 electrons and 4 protons
3) 4 electrons and 3 protons
4) 2 electrons and 2 protons
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19.
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22. See the following experiment and observe the
result
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Now identify plants (P-I, II and III)
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1) P-I = Long day plant; P-II = Long day
plant; P-III = Day neutral plant
2) P-I = Short day plant; P-II = Short day
plant; P-III = Day neutral plant
3) P-I = Short day plant; P-II = Long day
plant; P-III = Day neutral plant
4) P-I = Long day plant; P-II = Short day
plant; P-III = Day neutral plant
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25. Read the following statements and choose the
correct option
A. Nitrogenous base is linked to the pentose
sugar through a N-glycosidic linkage
B. Phosphate group is linked to 5’- OH of a
nucleoside through phosphoester linkage
C. Two nucleosides are linked through 3’-5’
N-glycosidic linkage
D. Negatively charged DNA is wrapped
around positively charged histone octamer
to form nucleosome
E. The chromatin that is more densely
packed and stains dark is called
euchromatin
1) A, B and C alone are wrong
2) D alone is wrong
3) C and E alone are wrong
4) A, B and D alone are wrong
29. Refer the given flowchart of sewage
treatment, accordingly match Column I with
Column II and select the correct answer from
the codes given below
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26 Which of the given statements is correct in
the context of visualizing DNA molecules
separated by agarose gel electrophoresis
1) DNA can be seen in visible light
2) DNA can be seen without staining in
visible light
3) Ethidium bromide stained DNA can be
seen in visible light
4) Ethidium bromide stained DNA can be
seen under exposure to UV light
27. Given below are a few statements regarding
somatic hybridization. Choose the correct
statements
A. The stage in
which physical
treatment
of
sewage is done
B. The stage in
which biological treatment of
sewage is done
C. Name of the
sediment
in
primary
treatment
D. It is carried to
aeration tanks
from primary
settling
E. Name of the
sediment
in
secondary
treatment
F. Site of flocs
growth
G. Function
of
sludge digester
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ii. Protoplasts from cells of different species
can be fused
Column I
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i. Protoplasts of different cells of the same
plant are fused
up
iii. Treatment of cells with cellulase and
pectinase is mandatory
iv. The hybrid protoplast contains characters
of only one parental protoplast
2) i and ii
3) i and iv
4) ii and iii
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1) i and iii
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28. Match the microbes in column I with their
commercial/industrial products in column II
and choose the correct answer
Column I
Column II
1. Ethanol
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A. Aspergillus niger
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B. Clostridium
2.
butylicum
C. Saccharomyces
3.
cerevisiae
D. Trichoderma
4.
polysporum
E. Monascus
5.
purpureus
1) A-4, B-5, C-2, D-1, E-3
2) A-5, B-4, C-1, D-2, E-3
3) A-3, B-4, C-1, D-5, E-2
4) A-3, B-4, C-5, D-1, E-2
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Statins
Citric acid
Butyric acid
Cyclosporin A
Column II
i.
ii.
Anaerobic digestion of activated
sludge and production of biogas
Activated sludge
iii.
Aeration tanks
iv.
Primary effluent
v.
Primary sludge
vi.
Secondary
treatment
Primary
treatment
vii.
1) A-vii, B-vi, C-v, D-iv, E-ii, F-iii, G-i
2) A-i, B-iii, C-v, D-vii, E-ii, F-iv, G-vi
3) A-i, B-ii, C-iii, D-iv, E-v, F-vi, G-vii
4) A-vii, B-vi, C-i, D-ii, E-iii, F-iv, G-v
31. Which of the following is not a feature of the
plasmids
1) Independent replication
2) Circular structure
3) Transferable
4) Single – stranded
→ Malic
OAA + NADPH + H + 
Dehydrogenase
acid + NADP +
3)
Malic
→ Pyruvic
Malic acid + NADPH 2 
enzyme
acid + CO2 + NADP +
4) Both 1 and 2
36. Bacterial structure
respectively:
1) Simple, Simple
3) Simple, Complex
and
behaviour
are
2) Complex, Simple
4) Complex, Complex
37. In pea plant, genotype (Bb) shows 1) Round seed with large sized starch grain
2) Round seed with intermediate sized starch
grain
3) Round seed with small sized starch grain
4) Wrinkled seed with small sized starch
grain
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32. What is true about Bt toxin
1) The inactive protoxin gets converted into
active form due to alkaline ph of the insect
gut
2) Bt protein exists as active toxin in the
Bacillus
3) The activated toxin enters the ovaries of
the pest to sterilize it and thus prevent its
multiplication
4) The concerned Bacillus has antitoxins
2)
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1) Exonuclease enzyme removes nucleotides
from site within DNA
2) Endonuclease
enzyme
removes
nucleotides from the ends of DNA
3) Endocuclease enzyme cut long palyndromic DNA strand
4) Exonuclease enzyme removes nucleotides
from ends of DNA
35. Which of the following equation holds true
for acidification reactions of CAM pathway ?
1) PEP + CO2 + H 2O 
→ OAA + H 2 PO4
PEP case
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30. Which of the following option is correct for
recombinant DNA technology
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33. When peacock eats snakes which eat insects
thriving on green plants, the peacock is
1) A primary consumer
2) A primary decomposer
3) Final decomposer
4) The apex of food pyramid
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34. Which one of the following regarding
ecological pyramid is not correct
1) In most ecosystems, the pyramid of
numbers and biomass are upright
2) In tree-dominated ecosystem the pyramid
of numbers is inverted
3) The pyramid of energy expresses mainly
the rate of food production
4) In deep water ecosystem, the pyramid of
biomass is upright
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38. In plant cells during cytokinesis wall
formation :1) Starts at periphery of the cell and grows
inward
2) Starts at centre of the cell and grows
inward
3) Starts at centre of the cell and grows
outward
4) Starts at periphery of the cell and grows
outward
39. The life cycle in Bryophyta is
haplo – diplontic type, it means that 1) Haploid generation is multicellular and
diploid generation is single celled
2) Haploid generation is dominant and multicellular, diploid generation is also multicellular but comparatively less developed.
3) Haploid and diploid generations are
equally developed.
4) Haploid generation is highly reduced.
40. Tap roots of carrot, turnip and adventitious
roots of sweet potato become swollen for :1) Support
2) Respiration
3) Storage
4) Nitrogen fixation
44. Enzymes, which catalyse joining of C – O,
C – S, C – N, P – O etc. Bonds, are included
in which of the following classes ?
1) Oxidoreductases
2) Ligases
3) Isomerases
4) Lyases
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41. Identify the correct match from the column, I,
45. Double fertilization is the fusion of :II and III and select the correct option.
1) Two eggs
ColumnColumnColumn-I
2) Two eggs and polar nuclei with pollen
II
III
nuclei
A Catkin
i Modified a Neem
3)
One male gamete with egg and other with
calyx
synergid
B Pinnately ii Additional b Mulberry
4) One male gamete with egg and other with
compound
support
secondary nucleus
leaf
46. The term ‘Systematics’ is related to
C Pappus
iii Unisexual c Maize
1) Systematic arrangement of organisms
flowers
2) It gives relationships among organisms
D Stilt root
iv Rachis
d Sunflower
3) It gives evolutionary differences among
1) A – iii - b; B – iv – a; C – i - d; D – ii – c
2) A – i – b; B – ii – a; C – iii – d; D – iv – c
animals only.
3) A – ii – a B – iii – b; C – iv – c; D – i – d
4) Both 1 and 2
4) A – iii – c; B – iv – d; C- i – b; D – ii – a
47. The given figure shows mouth parts of
42. In monocot stems, water canals are present
cockroach .Identify the parts labeled as A to
they are formed by :E and select the correct option
1) Disintegration of protoxylem elements
2) Fusion of metaxylem elements
3) The dissoulution of phloem parenchyma
4) Disintegration of rhytidome layers.
43. Fill up the blanks.
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I. Ethanol is produced by ......A.......
II. Large scale production of beverages in
industries is done in very large vessels
called .....B.....
III. Pencillin is discovered by ......C......
IV. LAB checks disease causing microbes in
....D.....A to D in the above statements
refers to
1) A-virus,
B-fermentors,
C-Alexander
Fleming, D-intesine
2) A-yeast, B-Fermentors, C-Alexander
Fleming, D-stomach
3) A- bacteria, B-fermentors,
C-S Waksman, D-chest
4) A-bacteria,B-fermentors,C-S Waksman,
D-liver
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1) A–Labrum, B–Hypopharynx, C– Mandible,
D– Labium, E– Maxilla
2) A– Labrum, B– Mandible,
C– Hypopharynx,D– Maxilla,E– Labium
3) A– Hypopharynx, B–Labrum, C– Labium,
D– Mandible, E– Maxilla
4) A– Maxilla, B– Mandible, C– Labium,
D– Labrum, E– Hypopharynx
48. Which of the following is deviation from the
rules of nomenclature
1) The names are generally in ‘Latin’ and
written in italics.
2) Mangifera indica Linn. indicates Linnaeus
first discovered mango tree
3) The first word, genus and the second
component species epithet.
4) ICZN for animal nomenclature.
52. Match the name of the animal Column - I
with Column – II and find out the incorrect?
49. Which one of the following figure shows
pseudocoelomate condition?
53. The figures (A-D ) show four animals.
Select the correct option with respect to a
common characterstic of two of these
animals
Column - II
1
Limulus
Living fossil arthropod
2
Adamsia
Radially symmetrical ctenophore
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Column - I
3 Petromyzon
Ichthyophis
Terrestrial Limbless amphibian
2) c
3) b
4) b& c
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4
Ectoparasitic cyclostome
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50. Which of the following is incorrect combination
1) Mast cells
- Heparin
2) Macrophages - Phagocytosis
3) Plasma cells
- Antibodies
4) Adipocytes
- Heat storage
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51. Which type of tissue correctly matches with
its location?
Tissue
Location
Cuboidal
epithelium
Lining of
intestine
2
Areolar tissue
Tendons
3
Skeletal muscle
Diaphragm
4
Transitional
epithelium
Ear pinna
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1
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1) A and D respire mainly through body wall
2) B and C are aquatic marine
3) A and B have cnidoblasts for self-defence
4) C and D have haemocoel
54. All these members of the following class are
with internal fertilization and oviparous
1) Prototherians
2) Reptilia
3) Vertebrata
4) Aves
55. Drowsiness after a heavy meal occurs due
to
1) Increased blood pressure in the brain
2) Decreased pulse rate
3) Reduced blood pressure in the brain
4) Increased pulse rate
2) A-IV, B- III, C and D-I, E-III
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3) A-II, B and C- III, D-I, E-IV
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1) A-IV, B and C- III, D-I, E-II
4) A-III, B and C- I, D-II, E-IV
The graph shows the percentage saturation
of haemoglobin with oxygen at different
partial pressures of oxygen. Which range
of partial pressure of oxygen produces the
greatest change of percentage saturation of
haemoglobin per unit oxygen tension?
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58.
60. Observe the abnormality in the given ECG
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1) Statement-I & Statement-II are correct
2) Statement-I is correct & Statement-II is
incorrect
3) Statement-I is incorrect but Statement-II is
correct
4) Both Statement-I andStatement-II are
incorrect
57. Match the following correctly –
Respiratory Organs
Animals
A
Earthworms
I
Lungs
B
Most aquatic II Trachea
arthropods
C
Fishes
III Gills
D
Birds/Reptiles IV Moist cuticle
E
Insects
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Statement-II: Movement ofwater molecule
requires energy
59. Which one of the following is matching
pair?
1) Lubb – Sharp closure of AV valves at the
beginning of ventricular systole
2) Dup – Sudden opening of semilunar valves
at the beginning of ventricular diastole
3) Pulsation of the radial artery – Valves in
the blood vessels
4) Initiation of the heartbeat – Purkinje fibres
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56. Statement-I: At the time of absorption water
moves into the cells by diffusion
1) Tachycardia
3) Arythmia
61. Match the following
Excretory organ
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Animal
A Antennal glands
1 Pheretima
B Protonephridia
2 Limulus
C Nephridia
3 Palaeomon
D
Malpighian
tubules
4 Amphioxus
5 Periplaneta
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1) A
2) Hyperkalemia
4) Myocardial infarction
2) B
3) C
4) D
1)
2)
3)
4)
A
2
3
5
3
B
4
4
1
2
C
1
1
4
1
D
5
5
3
5
62. Human urine is usually acidic, because
1) Sodium transporter exchanges
one
hydrogen ion for each sodium ion, in
peritubular capillaries
2) Excreted plasma proteins are acidic
65. Release of a neurotransmitter in a synapse
results in the activation of ligand gated ionic
channels, leading to the generation of IPSPs
(Inhibitory post synaptic potentials) in the
post synaptic neuron
The neurotransmitter must be
1) Glycine or GABA
4) Sodium and potassium exchange generates acidity
3) Oxytocin or serotonin
I) Actin is the contractile protein
II) Myosin is the major constituent of thick
filaments
4) Acetyl choline or glycine
66. Select the answer with correct matching of
the structure, its location and function.
Structure Location
Function
1 Blind
spot
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III) Tn I subunit of troponin has a binding site
(active site) for myosin
2) Epinephrine or Norephiner phrine
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63. Which of the following statements are
true/false about the skeletal muscle proteins.
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3) Hydrogen ions are actively secreted into
the filtrate
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1) I, II are true & III, IV are false
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IV)Actin and myosin,which generate force
during contraction are called structural
proteins
2) III, IV are true & I, II are false
3) All are true except III
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4) All are true except IV
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64. Statement-I : Myasthenia gravis is an
autoimmune disorder
Statement-II : In myasthenia gravis, neuromuscular junctions are affected leading to
fatigue, weakening & paralysis of smooth
muscles
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1) Both statement-I & statement-II are true
2) Statement-I is true but statement-II is false
3) Both statement-I & statement -II both are
false
4) Statement-I is false but statement-II is
true.
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10
2 Hippo
campus
3 Macula
Near
the
place
where
optic nerve
leaves the
eye
Cerebellar
cortex
Recall
memory
otolith
organ
angular
acceleration
4 Eustachai connects
n tube
middle ear
cavity with
pharynx
Rods
and
cones are
present but
inactive here
of
Equalizes the
presure of air
on either side
of ear drum
67. A pregnant female delivers a
baby who
suffers from stunted growth, mental retardation, low intelligence quotient and abnormal
skin. This is a result of
1) Deficiency of iodine in diet
2) Cancer of the thyroid gland
3) Over secretion of thyroxine
4) Over secretion of TSH
72. The first movements of the fetus and
appearance of hair on its head are usually
observed during which month of pregnancy?
69. Why do all copulations not lead to
fertilization and pregnancy? The root cause
is ______
1) Due to numerous sperms and one ovum
1) Third month
2) Fourth month
3) Fifth month
4) Sixth month
73. Match the following in relation to contraceptive methods
LIST-I
LIST-II
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68. Damage to thymus in a child may lead to
1) a reduction in RBC of blood.
2) a reduction in stem cell production.
3) loss of cell mediated immunity.
4) loss of antibody mediated immunity.
2) Due to less progesterone
A Vasectomy
3) Ovum and sperms are not transported
simultaneously tothe ampullary- isthmic
junction
B Coitus interrptus 2 Fallopian tube
C Condom
D Tubectomy
70. Mark the incorrect statement
4 Natural method
5 Vas deferens
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1) Polyspermy is prevented bydepolarization of the membrane of ovum is called
as fast block
3 Barrier method
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4) Due to non – formation of corpus luteum
1 vaginal rings
3)A-4, B-5, C-3, D-2 4)A-2, B-4, C-3, D-5
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2) Entry of sperm into ovum restarts the cell
cycle by breaking down MPF and turning
on the APC
1)A-5, B-4, C-3, D-2 2)A-5, B-4, C-2, D-3
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3) If fertilization occurs anywhere else other
than uterus, it is called tubal pregnancy
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4) Ability to reproduce is lost in female
primate after menopause
75. Pedigree chart for pattern baldness in
humans, is given below
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71. Statement-I: Lack of menstruation may be
indicative of pregnancy
Statement-II: Menstruation only occurs if
the released ovum is fertilized
74. Find the incorrect match
1) Genital warts - Human papilloma virus
2) Gonorrhoea - Neisseria gonorrhoea
3) Syphilis - Treponema pallidum
4) Genital herpes - Trichomonas vaginalis
1) Both Statement-I and Statement-II are
correct.
Statement-II is the correct
explanation of Statement-I.
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2) Both Statement-I and Statement-II are
correct. Statement-II is not the correct
explanation of Statement-I.
3) Statement-I is true, but Statement-II is
false
4) Statement-I is false, but Statement-II is
true
Page
11
Identify the
respectively
genotypes
of
A,
1) Bb, bb, Bb
2) Bb, Bb, BB
3) Bb, Bb, Bb
4) Bb, BB, Bb
B,
C
80. What is the principle of vaccination?
1) Feeding of antibodies in our body
2) Feeding of immunological memory
76. Find out the incorrect match.
Type of sex Character
Determinatio
n
Example
3) To give active pathogens is our body
4) To provide passive immunity
Sex of offspring Fumea
depends
of
fertilizing ovum
ZW - ZZ
81. The following drug shows effects on
cardiovascular system of the body.
ha
.n
et
2)
ZO -ZZ
Sex of offspring Butterflies
depends
of
fertilizing sperm
XX-XY
Sex of offspring Man
depends
of
fertilizing sperm
4)
XX-XO
Sex of offspring Bugs
depends
of
fertilizing sperm
2) Coca alkaloid
3) Cannabinoids
4) LSD
82. The equation for the population growth
depicted in the following graph is
ra
3)
1) Opioids
tib
1)
I II
b
a
c
b
c
c
+
1)
b
b
+
2)
ca
c
+
a
b
4)
+
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.e
3)
a
en
a
ad
up
77. This segment of DNA has restrictions sites I
and II which create restriction fragments a, b
and c. Which of the following gel (s) produced by electrophoresis would represent the
separation and identify of these fragments?
78. In a small population the Genetic drift is the
change in the frequency of gene that occurs
merely by
w
1) Selection but not by chance
2) Chance but not by selection
w
3) neither by selection nor chance
4) special creation but not by biological
evolution
79. The cranial capacity waslargest among the
1) Pekingman
2) Java ape man
3) African man
4) Neanderthal man
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12
1)
dN
= rN
dt
dN
dt
2)
dN
3) Nt = N0ert
dt
4)
 K-N 

 K 
= rN 
 K-N 

 N 
= rN 
83. Match the items in column I with those in
column II and choose the correct option using
the codes given below:
Column I
(a)
Ophrys
Column II
(i) Roots of higher plants
(b) Fig tree
(ii)
(c) Mycorrhizae (iii)
(d)
Bee
(iv)
Code:
(b)
(a)
(1) (ii)
(iv)
(2) (iii)
(ii)
(3) (iv)
(ii)
(4) (iv)
(ii)
Wasp
Pseudocopulation
Sexual deceit
(c)
(i)
(i)
(iii)
(i)
(d)
(iii)
(iv)
(i)
(iii)
Read the following
ha
.n
et
84. What is true for marine animals?
89.
1) Because of exosmosis they drink Sea
water
2) As they drink sea water exosmosis
occurs
3) They have glomerular kidneys
4) Salt gaining mechanism is evolved in
them to a maintain homoeostasis
85. As we move away from the equator towards
the poles, the species diversity
1) Increases
2) Decreases
3) First increases then decreases
4) First decreases then increases
In the above ‘A’ and ‘B’ respectively are
1) Dissolved O2 and BOD
2) BOD and dissolved O2
3) Dissolved O2 and COD
4) COD and dissolved O2
ra
tib
86. Given below are pie diagrams A, B and C
related to proportionate number of species
of major taxa of invertebrates, vertebrates
and plants respectively. Critically study and
fill in the blank I, II, III and IV
Statement(1) : High concentration of DDT
in birds causes thinning of egg shell and
their premature breaking.
up
90.
III –
2) I- Molluscs, II – Amphibians,
Angiosperms, IV - Fungi
III –
3) I- Hexapoda, II – Amphibians,
Fungi, IV - Angiosperms
III –
en
ad
1) I- Molluscs, II – Amphibians,
Fungi, IV – Angiosperms
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4) I- Turtles, II – Amphibians, III – Fungi,
IV – Angiosperms
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87. Find out the incorrect match
1. Dodo - Mauritius
2. Reserpine - Rauwolfia
3. Botanical gardens - In-situ conservation
4. Sacred forests - Khasi and Jaintia hills
88. Bioindicator of SO2 pollution
1) Pseudomonas putida
2) Lichens
4) Corals
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13
3) Escherichia coli
Statement(2) : DDT disturbs
metabolism
calcium
1) Statement 1 and 2 are correct
2) Statement 1 and 2 are in correct
3) Statement 1 is correct but 2 in correct
4) Statement 1 is incorrect and statement 2
is correct
PHYSICS :
αz
α −
91. In the relation P = e k θ , P is pressure, z
β
is distance, k is Boltzmann constant and θ is
the temperature . The dimensional formula
of β will be
1)  M 0 L2T 0 


2)  ML2T 


3)  ML0T −1 


4)  M 0 L2T −1 


1) 2
2) 0.5
3)
3
2
4) 1
α 2 + β2
2) t 2 α 2 + β2
3) 3t α 2 + β2
4) 2mg
1) µ = 2 tan θ
1
3) µ =
tan θ
2) µ = tan θ
2
4) µ =
tan θ
97. The work done by a force acting on a body
is as shown in the graph. The total work
done in covering an initial distance of 20m is
up
1)
3) zero
96. The upper half of an inclined plane of
inclination θ is perfectly smooth while
lower half is rough. A block starting from
rest at the top of the plane will again come to
rest at the bottom, if the coefficient of
friction between the block and lower half of
the plane is given by
ra
93. The co-ordinates of a moving particle at any
time t are given by x = αt 3 and y = βt 3 . The
speed of the particle at time t is given by :-
1) 3mg 2) 6mg
tib
At the same time, another body is projected
vertically upwards from B with velocity v2 .
The point B lies vertically below the highest
v
point. For both the bodies to collide, 2
v1
should be
ha
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et
92. A body is projected with velocity v1 from
the point A as shown in figure.
4) 3t 2 α 2 + β2
en
ad
94. Two blocks A and B of masses 2m and m,
respectively are connected by a massless and
inextensible string. The whole system is
suspended by a massless spring as shown in
the figure. The magnitudes of accelerations
of A and B, immediately after the string is
cut, are respectively
1) 225 J
g
g
2) , g
2
2
3) g , g
1) 100
4)
g g
,
2 2
95. Three blocks with masses m, 2m and 3m are
connected by strings, as shown in the figure.
After an upward force F is applied on block
m, the masses move upward at constant
speed v . The net force on the block of mass
2m is (g is the acceleration due to gravity)
Page
14
3) 400 J
4) 175 J
98. In the non-relativistic regime, if the momentum, is increased by 100%, the percentage
increase in kinetic energy is
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.e
w
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1) g ,
2) 200 J
2) 200
3) 300
4) 400
99. An engine pumps water continuously
through a hose. Water leaves the hose with a
velocity v and m is the mass per unit length
of the water jet. The rate at which kinetic
energy imparted to water is
1
2) mv 2
1) mv 3
2
3)
1 2 2
m v
2
4)
1 3
mv
2
2
3)
n(n + 1)l
2
4)
2l
105. Two wires are made of the same material
and have the same volume. However wire 1
has cross-sectional area A and wire 2 has
cross-sectional area 3A. If length of wire 1
increased by ∆x on applying force F, how
much force is needed to stretch wire 2 by
the same amount?
1) 4F
ha
.n
et
100. Particles of masses m, 2m,3m,..., nm grams
are placed on the same line at distances
l , 2l ,3l ,...., nl cm from a fixed point. The
distance of centre of mass of the particles
from the fixed point in centimeters is
(2n + 1)l
l
1)
2)
3
n +1
2) 6F
3) 9F
4) F
106. A copper rod of length l1 and an iron rod of
2
n(n + 1)
l2 always maintained at the same
common temperature T. If the difference
( l2 − l1 ) is 15 cm and is independent of the value
length
of T, the l1 and l2 have the values (given the
coefficient of linear expansion for copper and
iron are 2.0 × 10 −6 C −1 and 1.0 × 10 −6 C −1
respectively) :1) l1 = 15cm, l2 = 30cm
tib
101. A T joint is formed by two identical rods A
and B each of mass m and length L in the
XY plane as shown. Its moment of inertia
about axis coinciding with A is
2mL
3
2)
mL
12
4) None of these
2
3)
mL
6
en
ad
102. A thin rod of length L and mass M is bent at
its midpoint into two halves so that the angle
between them is 90°. The moment of inertia
of the bent rod about an axis passing through
the bending point and perpendicular to the
plane defined by the two halves of the rod is
2ML2
24
ML2
3)
24
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.e
ML2
2)
1)
6
ML2
4)
12
w
103. The height of the point vertically above the
earth’s surface at which the acceleration due
to gravity becomes 1% of its value at the
surface is (R is the radius of the earth)
1) 8R
2) 9R
3) 10R
3) l1 = 10cm, l2 = 25cm
4) l1 = 25cm, l2 = 10cm
up
1)
2
ra
2) l1 = 30cm, l2 = 15cm
2
107. Copper of fixed volume V is drawn into
wire of length l . When this wire is subjected to a constant force F, the extension
produced in the wire is ∆l . Which of the
following graphs is a straight line?
1) ∆l versus 1 / l
2) ∆l versus l 2
3) ∆l versus 1 / l 2
4) ∆l versus l
108. A thermodynamic system undergoes cyclic
process ABCDA as shown in figure. The
net work done by the system is
4) 20R
w
104. Infinite number of bodies, each of mass 2kg
are situated on x-axis at distances 1m, 2m,
4m, 8m,…., respectively, from the origin.
The resulting gravitational potential due to
this system at the origin will be
1) −
Page
15
4
G
3
2) −4G
3) −G
8
3
4) − G
1) zero
2) 2 P0V0
3) P0V0
4)
3
P0V0
2
113. A thick metallic spherical shell of inner
radius r1 and outer radius r2 has a charge
+ Q . A charge + q is placed at the centre of
the shell. The charge per unit area on the
outer surface is
r2
ha
.n
et
109. The oscillation of a body on a smooth
horizontal surface is represented by the
where
equation,
X is
X = A cos(ωt )
displacement at time t and ω is frequency
of oscillation. Which one of the following
graphs shows correctly the variation of a
with t ?
r
•+1q
+Q
1)
2)
1)
(Q − q )
4π(r22
a
O
4)
4πr22
2)
4)
(Q − q )
4πr22
(Q + q )
4π(r22 + r12 )
114. A parallel plate condenser with plate area A and
separation ‘d’ is filled with two dielectric
materials as shown in the figure. The dielectric
constants are K1 and K2 respectively. The
capacitance will be:-
up
ra
110. Two sound waves with wavelengths 5.0m
and 5.5m respectively, each propagate in a
gas with velocity 330m/s. We expect the
following number of beats per second.
1) 6
2) 12
3) 0
4) 1
(Q + q )
tib
3)
3)
− r12 )
111. A train emiting a sound of frequency
ad
1000 Hz , is moving at a speed of 220 ms −1
en
towards a stationary object. Some of the
sound reaching the object gets reflected
back to the train as echo. The frequency of
the echo as detected by the driver of the
w
.e
train is (Speed of sound in air is 330 ms −1 )
1) 3500 Hz
2) 4000 Hz
4) 3000 Hz
3) 5000 Hz
ε0 A
( K1 + K 2 )
d
2)
ε 0 A  K1 + K 2 


d  K1 K 2 
3)
2ε 0 A  K1K 2 


d  K1 + K 2 
4)
2ε 0 A  K1 + K 2 


d  K1K 2 
115. A network of six identical capacitors, each
of value C is made as shown in the figure.
Equivalent capacitance between points A
and B is
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112. In the circuit shown in figure, charge stored
in the capacitor of capacitance 5µF is
1)
1) 20 µC
3) 60 µC
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16
2) zero
4) 30 µC
1)
C
4
2)
3C
4
3)
4C
3
4) 3C
116. In the network shown below, the ring has
zero resistance. The equivalent resistance
between the points A and B is
121. What is the value of inductance L for which
the current is maximum in a series LCR
circuit with C = 10 µF and ω = 1000 s −1 ?
1) 1 mH
2) 10 mH
3) 100 mH
1) 2R
2) 4R
3) 7R
4) 10R
117. A uniform wire of resistance 36 ohm is bent
in the form of a circle. The effective
resistance across the points A and B is
ha
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et
4) cannot be calculated unless R is known
122. A conducting circular loop is placed in a
uniform magnetic field 0.04T with its
plane perpendicular to the magnetic field.
The radius of the loop starts shrinking at
2mm/s. The induced emf in the loop when
the radius is 2 cm is
2) 0.8 πµV
1) 4.8 πµV
4) 3.2 πµV
123. A small square loop of wire of side l is
placed inside a large square loop of wire of
side L(>> l ) . The loops are coplanar and
their centres coincide. The mutual
inductance of the system is
ad
up
ra
1) 5Ω
2) 15Ω
3) 7.2 Ω
4) 30 Ω
118. A galvanometer has resistance of 5 ohm and
a full scale deflection is produced by 15
milliamperes. What resistance may be
connected in series with it to enable it to
read maximum 1.5 volt?
2) 100 Ω
1) 95Ω
tib
3) 1.6 πµV
3) 105 Ω
4) 110 Ω
2
en
119. An arc of a circle of radius R subtends an
angle π at the centre. It carries a current I .
The magnetic field at the centre will be
µ0 I
2R
2)
µ0 I
µ I
3) 0
8R
4R
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.e
1)
4)
2µ0 I
5R
120. P, Q and R are long straight wires in air,
carrying currents as shown in the figure.
The force on Q is directed
Q
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w
P
20A
40A
R
60 A
1) to the left
2) to the right
3) perpendicular to the plane of the
diagram
4) along the current in Q
Page
17
µ0 l 2
1) 2 2
π L
µ0 l 2
2) 8 2
π L
µ l2
3) 2 2 0
µ L
4) 2 2 0
2
2π L
124. A
plane
πl
electromagnetic
wave
8
E z = 100cos(6 × 10 t + 4 x ) V / m propagates
in a medium. The refractive index of the
medium is
1) 1.5
2) 2.0
3) 2.4
4) 4.0
125. The energy of the em waves is of the order
of 15 keV. To which part of the spectrum
does it belong?
1) Ultraviolet rays 2) γ - rays
3) X - rays
4) Infra-red rays
126. If one face of a prism of prism angle 30° and
µ = 2 is silvered, the incident ray retraces
its initial path. The angle of incidence is
1) 60°
2) 30°
3) 45°
4) 90°
127. A plane mirror is placed along the x-axis
facing negative y-axis. The mirror is fixed. A
point object is moving with 3iˆ + 4 ˆj in front
of the plane mirror. The relative velocity of
image with respect to its object is
132. Half life of a radioactive substance A is two
times the half life of another radioactive
substance B. Initially, the number of nuclei
of A and B are N A and NB respectively.
After three half lives of A, number of nuclei
NA
is
NB
1
1
1
1
2)
3)
4)
1)
3
6
8
4
238
133. A radioactive reaction is 92U
→ 82Pb 206
. The number of α and β particles emitted
respectively are
1) 10, 6
2) 8, 8
up
1) A.B
ad
1) a single convergent beam
2) two different convergent beams
3) two different divergent beams
4) a convergent and divergent beam
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.e
en
129. The magnifying power of an astronomical
telescope is 8 and the distance between the
two lenses is 54cm. The focal lengths of eye
lens and objective lens will be respectively
1) 6cm and 48cm
2) 48 cm and 6cm
3) 8cm and 64 cm
4) 64cm and 8cm
w
w
130. A particle of mass M at rest decays into two
masses m1 and m2 with non-zero velocities.
The ratio of de-Broglie wavelength of the
particles λ1 / λ 2 is
1)
m1
m2
2)
m2
m1
3) 1:1
4)
m1
m2
131. The number of photons of wavelength of
540nm emitted per second by an electric
bulb of power 100W is
(Given h = 6 × 10 −34 J s )
1) 100
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18
2) 1000
3) 3 × 1020
4) 3 × 1018
3) 12, 6
4) 8, 6
134. The Boolean expression for the given circuit
is
ra
liquids of refractive indices µ2 and µ3 , as
shown in the figure. µ 2 > µ1 > µ3 . A wide,
parallel beam of light is incident on the lens
from the left. The lens will give rise to
tib
1) −8 ˆj
2) 8 ˆj
3) 3iˆ − 4 ˆj 4) − 6iˆ
128. A double convex lens, made of a material of
refractive index µ1 , is placed inside two
ha
.n
et
of both are equal. Then the ratio
2) A + B
3) A + B
4) A + B
135. Two ideal diodes are connected to a battery
as shown in the circuit. The current supplied by the battery is
1) 0.75A
2) zero
3) 0.25A
4) 0.5A
CHEMISTRY :
136. In the Schrodinger’s wave equation the term
Ĥ represents
−h2  ∂ 2
∂2
∂2 
+
+
v
8π 2 m  ∂x 2 ∂y 2 ∂z 2 
−h 2  ∂ 2
∂2
∂2 
2) 2  2 + 2 + 2  + v
8π m  ∂x ∂y ∂z 
1)
h2  ∂ 2
∂2
∂2 
+
+
+v
8π 2 m  ∂x 2 ∂y 2 ∂z 2 
−h2  ∂ 2
∂2
∂2 
4) 2  2 + 2 + 2  + Ψ
8π  ∂x ∂y ∂z 
3)
138.Which of the following is an incorrect match?
1) Z=103…Unniltrium
2) Thorium…f-block element
3)ns1-2(n-1)d1-10…general outer electronic
configuration of d-block
4)Fluorine….highest electron gain enthalpy
139. Bond polarity is higher in which of the
following molecule
1) NF3
2) NCl3
3) CCl4
4) BeF2
140.Removal of electron is easiest in
2) O2+( g )
3) N 22(−g )
4) O22+
M
4)
1
ra
M
3)
4
en
ad
142. A 50L vessel contain 36g water at 27 0 C . The
vessel is saturated with nitrogen gas. After
some time the loss in water is found to be
1.20g. Assuming ideal behavior to water,
what will be the aqueous tension at 27 0 C ?
2) 0.1256 atm
4) 0.03284 atm
w
.e
1) 0.0625 atm
3) 0.6275 atm
w
144. “X” on hydroformylation followed
reduction gave n-butyl alcohol. “X” is
1) Ethene
2) Propene
3) 1-Butene
4) 2-Butene
w
degree of decomposition of PCl5 is 0.5. The
total number of moles at equilibrium will be:
1) 1.5
2) 3
3) 4
4) 6
150. 100ml of 0.1M NH4OH is mixed with 50ml of
0.1M HCl. pH of the mixture is
1) 4.8
2) 4.5
3) 9.2
4) 9.5
151.Herbicide among the following is
2) Na3 AsO3
1) NaClO3
3) NaCl
143. Gas with higher rate of diffusion is with
atomic number
1) Z = 7
2) Z= 8
3) Z = 9
4) Z = 10
by
145.Na2CO3 and NaHCO3 can be distinguished
using
1) Phenolpthalein
2) Methyl orange
3) Methyl red
4) All of these
Page
19
establish an equilibrium
PCl5( g ) ⇌ PCl3( g ) + Cl2( g ) . At equilibrium the
up
M
2)
3
148. Incorrect statement is
1) All carbon-carbon bond lengths in
fullerene are equal
2) Graphite is a lubricant
3) Fullerene is aromatic
4) Diamond has 3D network structure
149. 4 mole of PCl5 is heated in a 2L vessel to
141.Equivalent weight of ferrous oxalate in its
reaction with acidified KMnO4 is
[M = molecular weight FeC2O4 ]
M
1)
2
147. B2O3 is the dehydration product of
1) H 3 BO3
2) HBO2
3) BF3 .NH 3
4) Both 1 and 2
tib
1) N 2+( g )
146. Thermodynamic stability is least for
1) BeCO3
2) NaHCO3
3) CaCO3
4) Na2CO3
ha
.n
et
137.For an element with Z=38 the number of
electrons present in orbitals with n+l value 5
are
1) 14
2) 16
3) 18
4) 20
4) Both 1 and 2
152. An organic compound on analysis produces
1.88g of AgBr from 2g of the sample. The
weight percentage of Bromine (atomic
weight=80) in the organic compound is
1) 40%
2) 62%
3) 50%
4) 34%
153. IUPAC name of
CH 3 − CH 2 − CH 2 − CH 2 − N ( CH 3 )2
1) N,N-dimethyl-1-butanamine
2) N,N-dimethyl-2-butanamine
3) N,N-dimethyl-3-butanamine
4) n-Butyl dimethylamine
154.Which of the following will not show
optical isomerism:
1) Cl-CH=C=CH-CI
2) Br-CH=C=CH-Br
H
H
OH
OH
H
159.An optically active compound ‘X’ having
molecular formula C4H4O3 it evolves CO2 with
NaHCO3. ‘x’ on treatment with LiAIH4 gives
achiral compound then ‘x’ is:
H
Me
Cl
COOH
Me
COOH
4)
3)
155.Ethane on oxidation in presence of manganous acetate gives “X”. Sodium salt of “X” on
kolbe’s electrolysis gives n-butane. “X” is
1) methyl alcohol
2) formaldehyde
3) acetic acid
4) formic acid
1)
O
2)
OH
OH
OH
3)
O
156.
Br
1)
en
ad
2)
4)
OH
157.
w
.e
OH
Hg ( OAc )2 , THF , H 2O
X.
Here 'X' is
NaBH 4 , OH
w
OH
1)
2)
w
OH
OH
HO
OH
3)
4)
OH
OH
Page
20
OH
O
OH
X

→
H 2O
. “X” is
2) NaBH4
4) B2H6
1) LiAlH4
3) both 1 and 2
161.Most acidic is
1) Methanol
2) Ethanol
3) n – Propyl alcohol 4) n – Butyl alcohol
up
Undergoes hydrolysis by SN1
mechanism. Major product is
OH
OH
4)
OH
ra
160.
3)
COOH
OH
tib
COOH
2) Antimicrobial
4) Antacid
ha
.n
et
Cl
158. Prontosil is
1) Antimalarial
3) Antihistamine
162.Glucose is converted to ethanol by using
1) Maltase
2) Diastase
3) Zymase
4) Invertase
163.In
the
conversion
(C2 H5 )2 OH
+
of
(C2 H 5 )2 O
to
HSO4− , the role of ether is
1) Lewis acid
2) Bronsted acid
3) Lewis base
4) Dehydrating agent
X
164. RCOOR 
→ RCH 2OH + ROH . ‘X’ is
1) H 2 / Catalyst
2) R2Cd
3) KMnO4
4) R Mg X
165.An organic compound can react with 2,4D.N.P. but not
I2/NaOH. The organic
compound is
1) Ethanal
2) Acetone
3) 3–Pentanone
4) Acetyl chloride
166.Acetamide is converted to methyl amine by
using
2) NaBH4
4) Br2/NaOH
1) Only oxidation
3) Both oxidation and reduction
4) Neither oxidation nor reduction
174.The chemical used in the estimation of
carbon monoxide is:
2) I 2O5
3) I 2O7
4) IO2
1) I 2O4
175.The complex Cr ( en )2 Cl2  Br can exhibit the
isomerism
A) Ionisation
C) Geometrical
1) Only A
3) A and C
B) Optical
D) All of these
2) A and B
4) A, B and C
176.Wrong match is
1) CO…Pi acceptor ligand
2) Co2(CO)8… bridged carbonyl group
3) Li[AlH4]…cationic complex
4) Ni(CO)4…EAN is 36
ra
168.The halide of a metal MX crystalizes as ccp
structure (NaCl type). If all the atoms situated at one of the tetrad axis are removed
then what will be the simplest molecular
formula of halide
3) MX2 4) MX3
1) MX
2) M3X4
2) Only reduction
ha
.n
et
167. The osmotic pressure is exhibited By 450ml
of solution containing 1g of polymer of
molecular weight 1,85,000 at 370 C is
1) 30.9 Pascals
2) 420 Pascals
3) 19.8 Pascals
4) 16.25 Pascals
tib
1) LiAlH 4
3) PCC
∆
173. 2 KMnO4 
→ X + Y + O2 . X and Y are
compounds of Mn. In the reaction KMnO4
undergoes:
ad
up
169.The number of faradays with 50% current
efficiency needed to reduce one mole
potassium permanganate in acidic medium
are
1) 2.5
2) 15
3) 10
4) 2
w
.e
en
170. During electrophoresis of a colloidal sol the
dispersed phase migrates towards cathode.
Which of the following electrolyte has the
maximum coagulation power to coagulate
the sol?
1) Potassium ferrocyanide
2) Barium chloride
3) Calcium phosphate
4) Aluminium fluoride
171.For the reaction N 2 + 3H 2 → 2 NH 3 . If the rate
w
w
of disappearance of nitrogen is 1.12g/min
then the rate of disappearance of hydrogen
will be
1) 0.6g/min
2) 0.24g/min
3) 2.4g/min
4) 6g/min
172.Moist NH 3 can be dried by using
1) H 2 SO4
2) P4O10
3) CaCl2
Page
21
4) CaO
177. Which of the following contains a alpha
glycosydic linkage between C-1 of glucose
and C-4 of glucose
1) Sucrose
2) Galactose
3) Maltose
4) Cellulose
178.Nylon – 2 – Nylon – 6 is an example of
A) Polyamide polymer
B) Biodegradable polymer
C) Condensation polymer
1) A and B
2) A and C
3) B and C
4) A, B and C
179.Reaction of which of the following with
water is a disproportionation reaction?
2) XeF4
3) XeF6
4) XeO3
1) XeF2
180. Monds process is used in the refining of
1) Ni
2) NiO
4) Zr
3) Ni ( CO ) 4
KEY SHEET
BIOLOGY
2
2)
2
3)
2
4)
2
5)
1
6)
4
7)
3
8)
3
9)
1
10)
3
11)
2
12)
1
13)
2
14)
4
15)
4
16)
3
17)
1
18)
1
19)
1
20)
4
21)
1
22)
4
23)
1
24)
4
25)
3
26)
4
27)
4
28)
3
29)
1
30)
4
31)
4
32)
1
33)
4
34)
4
35)
4
36)
3
37)
2
38)
3
39)
2
40)
3
41)
1
42)
1
43)
2
44)
2
45)
4
46)
4
47)
2
48)
2
49)
3
50)
4
51)
3
52)
2
53)
2
54)
4
55)
3
56)
2
57)
1
58)
2
59)
1
60)
4
61)
2
62)
3
63)
1
64)
2
65)
1
66)
4
67)
1
68)
3
69)
3
70)
3
71)
3
72)
3
73)
1
74)
4
75)
3
76)
2
77)
2
78)
2
79)
4
80)
2
81)
3
82)
2
83)
4
84)
1
85)
2
86)
1
87)
3
88)
2
89)
1
90)
1
1
92)
2
93)
4
94)
2
101)
2
102)
4
103)
2
104)
2
111)
3
112)
2
113)
3
121)
3
122)
4
123)
1
131)
3
132)
3
136)
146)
4
tib
1
97)
2
98)
3
99)
4
100)
1
105)
3
106)
1
107)
2
108)
1
109)
3
110)
1
3
115)
3
116)
1
117)
1
118)
1
119)
2
120)
1
124)
2
125)
3
126)
3
127)
1
128)
4
129)
1
130)
3
en
133)
ra
96)
134)
1
135)
4
2
137)
3
138)
4
139)
4
140)
3
141)
2
142)
4
143)
4
144)
2
145)
1
1
147)
4
148)
1
149)
4
150)
3
151)
4
152)
1
153)
1
154)
3
155)
3
2
157)
2
158)
2
159)
3
160)
3
161)
1
162)
3
163)
3
164)
1
165)
3
171)
2
172)
4
173)
2
174)
2
175)
4
w
156)
3
114)
w
.e
CHEMISTRY
95)
ad
91)
up
PHYSICS
4
167)
1
168)
1
169)
3
170)
1
176)
3
177)
3
178)
4
179)
2
180)
1
w
166)
This model test paper was prepared and verified by Senior faculty
Sri Chaitanya Medical Academy
Telangana & Andhra Pradesh.
Page
22
ha
.n
et
1)
NEET MODEL GRAND TEST PHYSICS SOLUTIONS
αz
should be dimensionless ,
kθ
91.
=  MLT −2 


−2 

α
 α   MLT   0 2 0 
= M L T Ans.
P = = [β] =   =

β
 P   ML−1T −2  


tib
−2 

α
 α   MLT   0 2 0 
= M L T Ans.
P = = [β] =   =

β
 P   ML−1T −2  


ha
.n
et
 2 −2 −1

K θ  ML T K V K 
∴ [α] =
=
Z
[ L]
Two bodies will collide G + highest point if both cover same vertical height in the same time. So,
92.
dx
dy
= 3xt 2 and
= 3βt 2
dt
dt
93.
dx
dt
2
+
dy
dx
2
en
t=
up
V2
= sin 30 = 0.5 Ans.
V1
ad
⇒
ra
V12 sin 2 30 V22
=
⇒
2g
2g
94.
After string is cut
w
.e
2 ma∆ = 3mg − 2mg ⇒ G∆ =
g
Ans.
2
maB = mg or aB = g Ans.
95.
Conceptual.
w
96.
As all blocks are moving with constant velocity. So a =0 so net force on each block is zero.
w
97.
F
20
10
0
Page
23
B
C
D
15
E
20
Work done = Area under F-S graph
= Area at trapezium ∆BCD + Area
At trapezium CEFD =
1
1
× (10 + 15) × 10 + (10 + 20) × 5
2
2
= 125 + 75 = 200J Ans.
P2
⇒ K α P2
2m
2
ha
.n
et
K=
98.
2
 P 
K1  P1 
1
=  = 1  =
⇒ K 2 = 4 K1
K 2  P2 
4
 2P 1 
% increase is
99.
Velocity of water is V, mass flowing per unit length is ‘m’
ra
∴Mass flowing per second = mv
up
∴Rate at K.E or K.E per second
1
1 
=  mv  v 2 = mv3
2
2 
ml + 2m × 2l + 3m.3l + ......
m + 2m + 3m + ......
l n(n + 1)(2n + 1)
l (2n + 1)
6
=
n(n + 1)
3
2
w
w
.e
en
ml (1 + 4 + 3 + ......)
=
=
m(1 + 2 + 3 + ......)
ad
100. X CM =
101.
tib
 K + K1 
KE =  2
 × 100 = 300% Ans.
 K1 
w
I = I A + IB
= 0+
Page
24
mL2 mL2
=
Ans.
12
12
102.
O
L
2
A
ha
.n
et
L
2
B
Total Max = M total length = L
2
2
 M  L  1 ML
M.I.total = 2 × 
Ans.
  × =
 2  2  3 12
gR 2
g
gR 2
⇒
=
100 ( R + h) 2
( R + h) 2
tib
103. g h =
104. V =
−G × 2
1
−G × 2
2
−G × 2
4
−G × 2
.....
8
en
ad


 1 
= −2G 
1
1 − 
 2
up
 1 1 1

= −2G 1 + + + + ......
 2 4 8

ra
⇒ R + h = 10 R or h = 9 R Ans.
2
= −2G   = −4G Ans.
1
w
.e
2
2
2
105. Y = FL ⇒ F = YA∆L = Y A ∆L = YA ∆L = YA ∆x
A∆L
L
AL
V
V
2
w
F 1 A12 ( 3A )
= 2 =
= g ⇒ F 1 = gF Ans.
2
F
A
A
106. Conceptual.
 F / ∆   Fl 
=

 ∆l / l   Α∆l 
w
107. V = Al , Y = 
Fl Fl 2
⇒ ∆l =
=
∆y YV
∴ ∆l α l 2 , hence graph between ∆l and l 2 is a straight line.
Page
25
PV
108. WBCOB = − area of ∆BCO = − 0 0
2
P0V0
2
WAOD∆ = + area of ∆AOD∆ = +
2
ha
.n
et
∴ WNet = WBOB + WAOD∆ = − P0V0 + P0V0 = 0
2
109. Conceptual.
V V 330 330
−
=
−
λ1 λ 2
5
5.5
110.
330 −330
= 6 Ans.
5
5.5
ra
= n1 − n2 =
tib
∴ Number of beats per second
111. Frequency of the echo detected by the driver of the train is

( 330 + 220 ) = 1000 × 550 = 5000
 = 1000
(330 − 220)
110

up
 V + VT
n1 = n 
 V − VT
112. 5µF and 3µF capacitors are short circuted .
ad
∴ charge on 5µF will be zero.
113. Charge per unit area on outer surface
Q+q
4πr22
114. Q = q =
10 × 10−6
2 × 8.85 × 10
−12
= 0.56 × 106 ≈ 6 × 105 Mm2C −1
w
.e
26
en
=
115. Redraw the Ckt
On simple fraction we will
w
w
A
⇒ Ceq =
Page
26
C
C
C
C
C
C
2C 2C 4C
+
=
Ans.
3
3
3
B
116. Eq. Ckt diagram is
3R
R
⇒ Req = 2 R
ha
.n
et
3R
B
3R
117. Resistance of smaller arc A13,
R1 = R / 6 = 36 / 6 = 6Ω
Resistance of bigger arc,
5R
36
= 5 × = 30Ω
6
6
tib
R2 =
Coefficient resistance between
R+5=
1.5
1500
=
= 100Ω
0.015
15
∴ R = 95Ω Ans.
µ0 I
µ I
( π / 2) = 0 Ans.
4πR
8R
en
119. B =
ra
V
1.5
or 0.015 =
R+G
R+G
ad
118. I g =
6 × 30
= 5Ω
6 + 30
up
A& B =
w
.e
120. Parallel currents attract and antiparallel currents
both due to P(20A) and Q(60A). then force on Q is towards left.
1
1
1
⇒ ω2 =
or L = 2
LC
LC
ωC
w
121. ω =
w
∴L =
1
1000 × 1000 × 10 × 10−6
122. Emf = E =
= 0.1H ≃ 100mH
dθ
dr
= Bπ(2r )
dt
dt
= 0.04 × π× 2 × 2 × 10−2 × 2 × 10−3 = 3.2πµV
Page
27
123.
L
l
ha
.n
et
I
I
I
Let current I be flowing in the larger loop
B = 4×
µ Ι
2
= 2 2 0
π L
2
P0 2 I
4π L
Flux linked with smaller loop
W 6 × 108 3
124. V =
=
= × 108
K
4
2
C
3 × 108
=
= 2 Ans.
V (3 / 2) × 108
ad
∴µ=
ra
µ l2
Q2
> M =2 2 0
Ans.
π L
I
up
M=
µ0 I 2
×l
π L
tib
Q2 = BA2 = 2 2 ×
hc 1240 evnm
=
= 0.083nm
E 15 × 103 ev
X-rays range are from 1nm − 10−3 nm
en
125. λ =
w
.e
126.
A
µ
30°
60°
w
P
>
• >
l
R
r = 30°
µ
w
C
B
It is clear from fig that ray will retrace the path when incident ray QR, incident normally
on
µ=
the
surface
sin l
1
⇒ sin l = 2 sin 30 = 2 ×
sin r
2
∴ Q = 45°
Page
28
polished
AC,
this
angled
refraction
r = 30° .
(
) (
)
127. Vrel = V ima − V ob = 3iˆ − 4 ˆj − 3iˆ + 4 ˆj = −8 ˆj
128. As µ 2 > µ1 the upper half of lense because diverging as µ1 > µ3 the lower half of lens become
converging.
fe
∴ f e = 6cm, f 0 = 48cm
130. MXO = m1v1 + m2v2 ⇒ m1v1 = − m2v2 = m2v2 ( is magnitude)
λ1 h / m1v1
=
=1
λ 2 h / m2v2
∴
M =
100 × 540 × 10−9
18 × 10
−26
= 3 × 1020
ra
Nhc
M × (6 × 10 − 34 ) (3 × 108 )
⇒ 100 =
λ
540 × 10 −5
up
=
tib
131. Power
ha
.n
et
f
129. M = 0 = 8, L = f0 + f e = 54cm
132. Three half lines of B and equivalentto six half lines of B
3
6
3
ad
N
(1 / 2)6  1 
1
1
1
N A   = NB   ⇒ A =
=  =
3
N B (1 / 2)
8
 2
2
2
en
133. Conceptual.
134. Y = A.B + A + B = A + B + A . B
w
.e
= A + B + A. B = A + B (using A.B. B. A )
= AB
135. D1 is formed biased and D2 is reverse biased . So
5V
1
= = 0.5 A
10Ω 2
w
w
I=
current supplied by the battery is
This model test paper solutions were prepared and verified by Senior faculty
Sri Chaitanya Medical Academy
Telangana & Andhra Pradesh.
Page
29