Test Booklet Code
DATE : 06/05/2018
CC ACHLA Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472
Time : 3 hrs.
Answers & Solutions
Max. Marks : 720
for NEET (UG) - 2018 Important Instructions : 1.
The test is of 3 hours duration and Test Booklet contains 180 questions. Each question carries 4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response, one mark will be deducted from the total scores. The maximum marks are 720.
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1
NEET (UG) - 2018 (Code-CC) ACHLA
1.
5.
Niche is (1) the functional role played by the organism where it lives
Secondary consumer : 120 g
(2) the range of temperature that the organism needs to live
Primary consumer : 60 g Primary producer : 10 g
(3) the physical space where an organism lives
(1) Upright pyramid of biomass (2) Upright pyramid of numbers
(4) all the biological factors in the organism's environment
(3) Pyramid of energy (4) Inverted pyramid of biomass
Answer ( 1 )
Answer ( 4 )
S o l . Ecological niche was termed by J. Grinnel. It refers the functional role played by the organism where it lives. 2.
What type of ecological pyramid would be obtained with the following data?
Sol. •
Which of the following is a secondary pollutant?
•
Pyramid of energy is always upright
(1) O3
•
Upright pyramid of biomass and numbers are not possible, as the data depicts primary producer is less than primary consumer and this is less than secondary consumers.
(2) SO2 (3) CO2 (4) CO Answer ( 1 )
6.
S o l . O3 (ozone) is a secondary pollutant. These are formed by the reaction of primary pollutant.
3.
Natality refers to (1) Number of individuals entering a habitat
CO – Quantitative pollutant
(2) Number of individuals leaving the habitat
CO2 – Primary pollutant
(3) Birth rate
SO2 – Primary pollutant
(4) Death rate Answer ( 3 )
In stratosphere, which of the following elements acts as a catalyst in degradation of ozone and release of molecular oxygen? (1) Oxygen
(2) Fe
(3) Cl
(4) Carbon
S o l . Natality refers to birth rate.
Answer ( 3 ) S o l . UV rays act on CFCs, releasing Cl atoms, chlorine reacts with ozone in sequential method converting into oxygen
7.
•
Death rate
– Mortality
•
Number of individual entering a habitat is
– Immigration
•
Number of individual leaving the habital
– Emigration
Offsets are produced by (1) Parthenogenesis
Carbon, oxygen and Fe are not related to ozone layer depletion 4.
The given data depicts the inverted pyramid of biomass, usually found in aquatic ecosystem.
(2) Parthenocarpy
World Ozone Day is celebrated on
(3) Mitotic divisions
(1) 22nd April
(4) Meiotic divisions
(2) 16th September
Answer ( 3 )
(3) 21st April
S o l . Offset is a vegetative part of a plant, formed by mitosis.
(4)
5th
June
Answer ( 2 ) S o l . World Ozone day is celebrated on 16 th September.
–
Meiotic divisions do not occur in somatic cells.
–
Parthenogenesis is the formation of embryo from ovum or egg without fertilisation.
–
Parthenocarpy is the fruit formed without fertilisation, (generally seedless)
5th June - World Environment Day 21st April - National Yellow Bat Day 22nd April - National Earth Day 2
NEET (UG) - 2018 (Code-CC) ACHLA
8.
11.
The experimental proof for semiconservative replication of DNA was first shown in a
Which of the following pairs is wrongly matched?
(1) Virus
(1) T.H. Morgan
: Linkage
(2) Plant
(2) XO type sex
: Grasshopper
determination
(3) Bacterium
(3) ABO blood grouping
(4) Fungus
(4) Starch synthesis in pea : Multiple alleles
Answer ( 3 )
Answer ( 4 )
S o l . Semi-conservative DNA replication was first shown in Bacterium Escherichia coli by Matthew Meselson and Franklin Stahl. 9.
S o l . Starch synthesis in pea is controlled by pleiotropic gene. Other options (1, 2 & 3) are correctly matched.
Select the correct match (1) Francois Jacob and - Lac operon
12.
Jacques Monod (2) Matthew Meselson
- Pisum sativum
and F. Stahl (3) Alfred Hershey and
Answer ( 1 ) S o l . Francois Jacob and Jacque Monod proposed model of gene regulation known as operon model/lac operon.
10.
–
Matthew Meselson and F. Stahl – Semiconservative DNA replication in E. coli.
–
Alfred Hershey and Martha Chase – Proved DNA as genetic material not protein
13.
(2) Oil content
(3) Cellulosic intine
(4) Pollenkitt
(3) Jackfruit
(4) Bamboo species
Select the correct statement (1) Transduction was discovered by S. Altman (2) Spliceosomes take part in translation (3) Punnett square was developed by a British scientist (4) Franklin Stahl coined the term ‘‘linkage’’
Answer ( 3 ) S o l . Punnett square was developed by a British geneticist, Reginald C. Punnett.
Which of the following has proved helpful in preserving pollen as fossils? (1) Sporopollenin
(2) Mango
Jackfruit, papaya and mango are polycarpic i.e., produce flowers and fruits many times in their life-time.
pneumoniae
Alec Jeffreys – DNA fingerprinting technique.
(1) Papaya
S o l . Bamboo species are monocarpic i.e., flower generally only once in its life-time after 50100 years.
- Streptococcus
–
Which of the following flowers only once in its life-time?
Answer ( 4 )
- TMV
Martha Chase (4) Alec Jeffreys
: Co-dominance
Answer ( 1 ) 14.
S o l . Sporopollenin cannot be degraded by enzyme; strong acids and alkali, therefore it is helpful in preserving pollen as fossil.
–
Franklin Stahl proved semi-conservative mode of replication.
–
Transduction was discovered by Zinder and Laderberg.
–
Spliceosome formation is part of posttranscriptional change in Eukaryotes
The Golgi complex participates in (1) Activation of amino acid (2) Respiration in bacteria
Pollenkitt – Help in insect pollination.
(3) Formation of secretory vesicles
Cellulosic Intine – Inner sporoderm layer of pollen grain known as intine made up cellulose & pectin.
(4) Fatty acid breakdown Answer ( 3 ) S o l . Golgi complex, after processing releases secretory vesicles from their trans-face.
Oil content – No role is pollen preservation. 3
NEET (UG) - 2018 (Code-CC) ACHLA
15.
S o l . ATP, NADPH and oxygen are products of light reaction, while NADH is a product of respiration process.
The stage during which separation of the paired homologous chromosomes begins is (1) Zygotene
19.
Which of the following is true for nucleolus?
(2) Diakinesis (3) Diplotene
(1) It is a site for active ribosomal RNA synthesis
(4) Pachytene
(2) It takes part in spindle formation
Answer ( 3 )
(3) It is a membrane-bound structure
S o l . Synaptonemal complex disintegrates. Terminalisation begins at diplotene stage i.e. chiasmata start to shift towards end.
(4) Larger nucleoli are present in dividing cells Answer ( 1 )
16.
Stomatal movement is not affected by (1) CO2 concentration
S o l . Nucleolus is a non membranous structure and is a site of r-RNA synthesis.
(2) O2 concentration
20.
(3) Light
Which among the following is not a prokaryote? (1) Oscillatoria
(4) Temperature (2) Nostoc Answer ( 2 ) (3) Mycobacterium S o l . Light, temperature and concentration of CO2 affect opening and closing of stomata while they are not affected by O2 concentration. 17.
(4) Saccharomyces Answer ( 4 )
Stomata in grass leaf are
S o l . Saccharomyces i.e. yeast is an eukaryote (unicellular fungi)
(1) Barrel shaped (2) Rectangular
Mycobacterium – a bacterium
(3) Kidney shaped
Oscillatoria and Nostoc are cyanobacteria. 21.
(4) Dumb-bell shaped Answer ( 4 )
(1) Carbonyl and hydroxyl
S o l . Grass being a monocot, has Dumb-bell shaped stomata in their leaves. 18.
The two functional groups characteristic of sugars are
(2) Carbonyl and phosphate (3) Carbonyl and methyl
Which of the following is not a product of light reaction of photosynthesis?
(4) Hydroxyl and methyl
(1) Oxygen
Answer ( 1 )
(2) NADPH
S o l . Sugar is a common term used to denote carbohydrate.
(3) NADH
Carbohydrates are polyhydroxy aldehyde, ketone or their derivatives, which means they have carbonyl and hydroxyl groups.
(4) ATP Answer ( 3 ) 4
NEET (UG) - 2018 (Code-CC) ACHLA
22.
Match the items given in Column I with those in Column II and select the correct option given below: Column I
24.
(1) Saccharomyces
Column II
a. Herbarium
b. Key
(2) Agaricus
(i) It is a place having a collection of preserved plants and animals
(3) Alternaria (4) Neurospora
(ii) A list that enumerates methodically all the species found in an area with brief description aiding identification
Answer ( 2 ) Sol.
In Agaricus (a genus of basidiomycetes), basidiospores or meiospores are produced exogenously.
c. Museum
(iii) Is a place where dried and pressed plant specimens mounted on sheets are kept
Neurospora (a genus of ascomycetes) produces ascospores as meiospores but endogenously inside the ascus.)
d. Catalogue
(iv) A booklet containing a list of characters and their alternates which are helpful in identification of various taxa.
Alternaria (a genus of deuteromycetes) does not produce sexual spores.
Saccharomyces (Unicellular ascomycetes) produces ascospores, endogenously.
25.
Winged pollen grains are present in
a
b
c
d
(1) Pinus
(1)
(iii)
(iv)
(i)
(ii)
(2) Mango
(2)
(ii)
(iv)
(iii)
(i)
(3)
(iii)
(ii)
(i)
(iv)
(4)
(i)
(iv)
(iii)
(ii)
(3) Cycas (4) Mustard Answer ( 1 )
Answer ( 1 ) Sol. •
23.
After karyogamy followed by meiosis, spores are produced exogenously in
Herbarium
–
Dried and pressed plant specimen
•
Key
–
Identification of various taxa
•
Museum
–
Plant and animal specimen are preserved
•
Catalogue
–
Alphabetical listing of species
S o l . In Pinus, winged pollen grains are present. It is extended outer exine on two lateral sides to form the wings of pollen. It is the characteristic feature, only in Pinus. Pollen grains of Mustard, Cycas & Mango are not winged shaped. 26.
Which one is wrongly matched? (1) Unicellular organism – Chlorella (2) Gemma cups
– Marchantia
Which of the following is commonly used as a vector for introducing a DNA fragment in human lymphocytes? (1) pBR 322
(2) phage
(3) Ti plasmid
(4) Retrovirus
(3) Biflagellate zoospores – Brown algae
Answer ( 4 )
(4) Uniflagellate gametes – Polysiphonia
S o l . Retrovirus is commonly used as vector for introducing a DNA fragment in human lymphocyte.
Answer ( 4 ) Sol. •
•
Polysiphonia is a genus of red algae, where asexual spores and gametes are non-motile or non-flagellated.
Gene therapy : Lymphocyte from blood of patient are grown in culture outside the body, a functional gene is introduced by using a retroviral vector, into these lymphocyte.
Other options (1, 2 & 3) are correctly matched 5
NEET (UG) - 2018 (Code-CC) ACHLA
27.
A ‘new’ variety of rice was patented by a foreign company, though such varieties have been present in India for a long time. This is related to (1) Basmati
(3) Annealing, Extension, Denaturation (4) Extension, Denaturation, Annealing Answer ( 1 ) S o l . This technique is used for making multiple copies of gene (or DNA) of interest in vitro.
(2) Lerma Rojo
(3) Sharbati Sonora (4) Co-667 Answer ( 1 )
Each cycle has three steps
S o l . In 1997, an American company got patent rights on Basmati rice through the US patent and trademark office that was actually been derived from Indian farmer’s varieties.
(i) Denaturation (ii) Primer annealing (iii) Extension of primer 31.
The diversity of rice in India is one of the richest in the world, 27 documented varieties of Basmati are grown in India. Indian basmati was crossed with semi-dwarf varieties and claimed as an invention or a novelty.
28.
(1) Genetic Engineering Appraisal Committee (GEAC)
Sharbati Sonora and Lerma Rojo are varieties of wheat.
(2) Research Committee Manipulation (RCGM)
Use of bioresources by multinational companies and organisations without authorisation from the concerned country and its people is called
(3) Council for Scientific and Industrial Research (CSIR)
Genetic
(1) Bioexploitation
Answer ( 1 )
(2) Biodegradation
S o l . Indian Government has setup organisation such as GEAC (Genetic Engineering Appraisal Committee) which will make decisions regarding the validity of GM research and safety of introducing GM-organism for public services. (Direct from NCERT).
(4) Bio-infringement Answer ( 3 ) S o l . Biopiracy is term used for or refer to the use of bioresources by multinational companies and other organisation without proper authorisation from the countries and people concerned with compensatory payment (definition of biopiracy given in NCERT).
32.
What is the role of NAD + in cellular respiration? (1) It is the final electron acceptor for anaerobic respiration. (2) It is a nucleotide source for ATP synthesis.
Select the correct match (1) G. Mendel
- Transformation
(3) It functions as an electron carrier.
(2) T.H. Morgan
- Transduction
(4) It functions as an enzyme.
(3) F2 × Recessive parent - Dihybrid cross
Answer ( 3 )
(4) Ribozyme
S o l . In cellular respiration, NAD+ act as an electron carrier.
- Nucleic acid
Answer ( 4 )
33.
S o l . Ribozyme is a catalytic RNA, which is nucleic acid. 30.
on
(4) Indian Council of Medical Research (ICMR)
(3) Biopiracy
29.
In India, the organisation responsible for assessing the safety of introducing genetically modified organisms for public use is
The correct order of steps in Polymerase Chain Reaction (PCR) is (1) Denaturation, Annealing, Extension (2) Denaturation, Extension, Annealing 6
Which one of the following plants shows a very close relationship with a species of moth, where none of the two can complete its life cycle without the other? (1) Viola
(2) Banana
(3) Yucca
(4) Hydrilla
NEET (UG) - 2018 (Code-CC) ACHLA
Answer ( 2 ) S o l . Potassium helps in maintaining turgidity of cells.
Answer ( 3 ) S o l . Yucca have an obligate mutualism with a species of moth i.e. Pronuba. 34.
39.
Pollen grains can be stored for several years in liquid nitrogen having a temperature of
(1) Submerged hydrophytes
(1) –160°C
(2) –196°C
(2) Carnivorous plants
(3) –80°C
(4) –120°C
(3) Free-floating hydrophytes (4) Halophytes
Answer ( 2 )
Answer ( 4 )
S o l . Pollen grains can be stored for several years in liquid nitrogen at –196°C (Cryopreservation) 35.
Pneumatophores occur in
Sol.
In which of the following forms is iron absorbed by plants? (1) Both ferric and ferrous 40.
(2) Free element (4) Ferric S o l . Iron is absorbed by plants in the form of ferric ions. (According to NCERT)
Apogeotropic (–vely geotropic) roots having lenticels called pneumathodes to uptake O2.
Select the wrong statement :
(3) Mushrooms belong to Basidiomycetes (4) Cell wall is present in members of Fungi and Plantae
*Plants absorb iron in both form i.e. Fe++ and Fe+++. (Preferably Fe++)
Answer ( 2 )
Double fertilization is (1) Syngamy and triple fusion
S o l . Pseudopodia are locomotory structures in sarcodines (Amoeboid)
(2) Fusion of two male gametes with one egg
41.
(3) Fusion of one male gamete with two polar nuclei
Secondary xylem and phloem in dicot stem are produced by (1) Axillary meristems (2) Phellogen
(4) Fusion of two male gametes of a pollen tube with two different eggs
(3) Vascular cambium (4) Apical meristems
Answer ( 1 )
Answer ( 3 )
S o l . Double fertilization is a unique phenomenon that occur in angiosperms only.
Sol. •
Vascular cambium is partially secondary
Syngamy + Triple fusion = Double fertilization
•
Oxygen is not produced during photosynthesis by
Form secondary xylem towards its inside and secondary phloem towards outsides.
•
4 – 10 times more secondary xylem is produced than secondary phloem.
(1) Chara 42.
(2) Cycas
Sweet potato is a modified
(3) Nostoc
(1) Rhizome
(4) Green sulphur bacteria
(2) Tap root
Answer ( 4 )
(3) Adventitious root
S o l . Green sulphur bacteria do not use H 2O as source of proton, therefore they do not evolve O2.
(4) Stem
38.
have
(2) Pseudopodia are locomotory and feeding structures in Sporozoans
Answer ( 4 * )
37.
mangrooves
(1) Mitochondria are the powerhouse of the cell in all kingdoms except Monera
(3) Ferrous
36.
Halophytes like pneumatophores.
Answer ( 3 ) S o l . Sweet potato is a modified adventitious root for storage of food
Which of the following elements is responsible for maintaining turgor in cells? (1) Calcium
(2) Potassium
(3) Sodium
(4) Magnesium 7
•
Rhizomes are underground modified stem
•
Tap root is primary root directly elongated from the redicle
NEET (UG) - 2018 (Code-CC) ACHLA
43.
Which of the following statements is correct?
(3) Glycolysis occurs in cytosol
(1) Stems are usually unbranched in both Cycas and Cedrus
(4) Enzymes of TCA cycle are present in mitochondrial matrix
(2) Horsetails are gymnosperms
Answer ( 1 )
(3) Selaginella is heterosporous, while Salvinia is homosporous
S o l . Oxidative phosphorylation takes place in inner mitochondrial membrane.
(4) Ovules are not enclosed by ovary wall in gymnosperms
48.
Answer ( 4 ) Sol. • • 44.
Gymnosperms have naked ovule. Called phanerogams without womb/ovary
Casparian strips occur in (1) Endodermis
(2) Cortex
(3) Pericycle
(4) Epidermis
• 45.
(1) Nucleosome
(2)
Plastidome
(3) Polyhedral bodies
(4)
Polysome
Answer ( 4 ) S o l . The phenomenon of association of many ribosomes with single m-RNA leads to formation of polyribosomes or polysomes or ergasomes.
Answer ( 4 ) Sol. •
Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as
Endodermis have casparian strip on radial and inner tangential wall. It is suberin rich.
49.
Plants having little or no secondary growth are
Which of the following terms describe human dentition?
(1) Cycads
(1) Pleurodont, Diphyodont, Heterodont
(2) Conifers
(2) Pleurodont, Monophyodont, Homodont
(3) Deciduous angiosperms
(3) Thecodont, Diphyodont, Heterodont
(4) Grasses
(4) Thecodont, Diphyodont, Homodont
Answer ( 4 )
Answer ( 3 )
S o l . Grasses are monocots and monocots usually do not have secondary growth.
S o l . In humans, dentition is
Palm like monocots secondary growth. 46.
have
anomalous
Nissl bodies are mainly composed of
Thecodont : Teeth are present in the sockets of the jaw bone called alveoli.
Diphyodont : Teeth erupts twice, temporary milk or deciduous teeth are replaced by a set of permanent or adult teeth.
Heterodont dentition : Dentition consists of different types of teeth namely incisors, canine, premolars and molars.
(1) Free ribosomes and RER (2) Nucleic acids and SER (3) DNA and RNA (4) Proteins and lipids Answer ( 1 ) S o l . Nissl granules are present in the cyton and even extend into the dendrite but absent in axon and rest of the neuron.
50.
(1) Phospholipid synthesis
Nissl granules are in fact composed of free ribosomes and RER. They are responsible for protein synthesis. 47.
Which of the following events does not occur in rough endoplasmic reticulum?
(2) Cleavage of signal peptide (3) Protein glycosylation
Which of these statements is incorrect?
(4) Protein folding
(1) Oxidative phosphorylation takes place in outer mitochondrial membrane
Answer ( 1 )
(2) Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atoms
S o l . Phospholipid synthesis does not take place in RER. Smooth endoplasmic reticulum are involved in lipid synthesis. 8
NEET (UG) - 2018 (Code-CC) ACHLA
51.
Select the incorrect match : (1) Polytene chromosomes
54.
– Oocytes of amphibians
(1) Parathyroid hormone and Prolactin
(2) Submetacentric – L-shaped chromosomes chromosomes
(2) Estrogen and Parathyroid hormone
(3) Allosomes
– Sex chromosomes
(3) Progesterone and Aldosterone
(4) Lampbrush
– Diplotene bivalents
(4) Aldosterone and Prolactin
chromosomes
Answer ( 2 )
Answer ( 1 )
S o l . Estrogen promotes the activity of osteoblast and inhibits osteoclast. In an ageing female osteoporosis occurs due to deficiency of estrogen. Parathormone promotes mobilisation of calcium from bone into blood. Excessive activity of parathormone causes demineralisation leading to osteoporosis.
S o l . Polytene chromosomes are found in salivary glands of insects of order Diptera. 52.
Which of the following hormones can play a significant role in osteoporosis?
Which of the following is an amino acid derived hormone? (1) Estriol (2) Estradiol
55.
(3) Ecdysone (4) Epinephrine
(1) smooth muscles attached to the ciliary body
Answer ( 4 ) S o l . Epinephrine is derived from tyrosine amino acid by the removal of carboxyl group. It is a catecholamine. 53.
(2) smooth muscles attached to the iris (3) ligaments attached to the iris
Which of the following structures or regions is incorrectly paired with its functions? (1) Corpus callosum
(2) Hypothalamus
(3) Limbic system
The transparent lens in the human eye is held in its place by
(4) ligaments attached to the ciliary body Answer ( 4 )
: band of fibers connecting left and right cerebral hemispheres.
S o l . Lens in the human eye is held in its place by suspensory ligaments attached to the ciliary body.
: production of releasing hormones and regulation of temperature, hunger and thirst.
56.
In a growing population of a country, (1) pre-reproductive individuals are less than the reproductive individuals. (2) reproductive and pre-reproductive individuals are equal in number.
: consists of fibre tracts that interconnect different regions of brain; controls movement.
(3) reproductive individuals are less than the post-reproductive individuals. (4) pre-reproductive individuals are more than the reproductive individuals.
(4) Medulla oblongata : controls respiration and cardiovascular reflexes.
Answer ( 4 ) S o l . Whenever the pre-reproductive individuals or the younger population size is larger than the reproductive group, the population will be an increasing population.
Answer ( 3 ) S o l . Limbic system is emotional brain. It controls all emotions in our body but not movements. 9
NEET (UG) - 2018 (Code-CC) ACHLA
57.
Match the items given in Column I with those in Column II and select the correct option given below : Column-I
(3) Sacred groves (4) Wildlife safari parks Answer ( 3 )
Column-II
a. Eutrophication
i.
b. Sanitary landfill
ii. Deforestation
c. Snow blindness
iii. Nutrient enrichment
Sol.
UV-B radiation
61.
d. Jhum cultivation iv. Waste disposal d
(1) Parietal cells
(1) i
ii
iv
iii
(2) Goblet cells
(2) iii
iv
i
ii
(3) Mucous cells
(3) i
iii
iv
ii
(4) Chief cells
(4) ii
i
iii
iv
S o l . a. Eutrophication
Answer ( 1 )
iii.
Nutrient enrichment
b. Sanitary landfill
iv.
Waste disposal
c. Snow blindness
i.
UV-B radiation
d. Jhum cultivation ii.
S o l . Parietal or oxyntic cell is a source of HCl and intrinsic factor. HCl converts iron present in diet from ferric to ferrous form so that it can be absorbed easily and used during erythropoiesis. Intrinsic factor is essential for the absorption of vitamin B 12 and its deficiency causes pernicious anaemia.
Deforestation
Which part of poppy plant is used to obtain the drug “Smack”? (1) Leaves
(2) Roots
(3) Latex
(4) Flowers
62.
Answer ( 3 )
Which one of the following population interactions is widely used in medical science for the production of antibiotics? (1) Amensalism
(2) Parasitism
(3) Mutualism
(4) Commensalism
S o l . Amensalism/Antibiosis (0, –)
Column II
a. Fibrinogen
(i) Osmotic balance
b. Globulin
(ii) Blood clotting
c. Albumin
(iii) Defence mechanism
a
Answer ( 1 )
Match the items given in Column I with those in Column II and select the correct option given below : Column I
S o l . ‘Smack’ also called as brown sugar/Heroin is formed by acetylation of morphine. It is obtained from the latex of unripe capsule of Poppy plant.
60.
Which of the following gastric cells indirectly help in erythropoiesis?
c
Answer ( 2 )
59.
Represent pristine forest patch as protected by Tribal groups.
b
a
58.
Sacred groves – in-situ conservation.
b
c
(1) (ii)
(iii)
(i)
(2) (i)
(iii)
(ii)
(3) (i)
(ii)
(iii)
(4) (iii)
(ii)
(i)
Answer ( 1 )
Antibiotics are chemicals secreted by one microbial group (eg : Penicillium) which harm other microbes (eg : Staphylococcus)
S o l . Fibrinogen forms fibrin strands during coagulation. These strands forms a network and the meshes of which are occupied by blood cells, this structure finally forms a clot.
It has no effect on Penicillium or the organism which produces it.
Antibodies are derived from -Globulin fraction of plasma proteins which means globulins are involved in defence mechanisms.
All of the following are included in ‘ex-situ conservation’ except (1) Seed banks
Albumin is a plasma responsible for BCOP.
(2) Botanical gardens 10
protein
mainly
NEET (UG) - 2018 (Code-CC) ACHLA
63.
Calcium is important in skeletal muscle contraction because it
(3) UGGTUTCGCAT (4) AGGUAUCGCAU
(1) Prevents the formation of bonds between the myosin cross bridges and the actin filament.
Answer ( 4 ) S o l . Coding strand and mRNA has same nucleotide sequence except, ‘T’ – Thymine is replaced by ‘U’–Uracil in mRNA.
(2) Detaches the myosin head from the actin filament.
66.
(3) Activates the myosin ATPase by binding to it. (4) Binds to troponin to remove the masking of active sites on actin for myosin.
(1) Both sons and daughters (2) Only grandchildren
Answer ( 4 ) Sol.
64.
A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by
(3) Only sons
Signal for contraction increase Ca++ level many folds in the sarcoplasm.
(4) Only daughters
Ca++ now binds with sub-unit of troponin (troponin "C") which is masking the active site on actin filament and displaces the sub-unit of troponin.
Answer ( 1 ) Sol. •
Once the active site is exposed, head of the myosin attaches and initiate contraction by sliding the actin over myosin.
67.
Which of the following is an occupational respiratory disorder? (1) Emphysema
(2) Botulism
(3) Silicosis
(4) Anthracis
Woman is a carrier
•
Both son X–chromosome
&
daughter inherit
•
Although only son be the diseased
Match the items given in Column I with those in Column II and select the correct option given below : Column I
Column II
a. Proliferative Phase i. Breakdown of endometrial
Answer ( 3 )
lining
S o l . Silicosis is due to excess inhalation of silica dust in the workers involved grinding or stone breaking industries.
ii. Follicular Phase
c. Menstruation
iii. Luteal Phase
a
Long exposure can give rise to inflammation leading to fibrosis and thus causing serious lung damage. Anthrax is a serious infectious disease caused by Bacillus anthracis. It commonly affects domestic and wild animals. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased.
b
c
(1) iii
i
ii
(2) ii
iii
i
(3) i
iii
ii
(4) iii
ii
i
Answer ( 2 ) S o l . During proliferative phase, the follicles start developing, hence, called follicular phase.
Botulism is a form of food poisoning caused by Clostridium botulinum. 65.
b. Secretory Phase
Secretory phase is also called as luteal phase mainly controlled by progesterone secreted by corpus luteum. Estrogen further thickens the endometrium maintained by progesterone.
AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA?
Menstruation occurs due to decline in progesterone level and involves breakdown of overgrown endometrial lining.
(1) UCCAUAGCGUA (2) ACCUAUGCGAU 11
NEET (UG) - 2018 (Code-CC) ACHLA
68.
71.
Match the items given in Column I with those in Column II and select the correct option given below : Column I Column II a. Tricuspid valve i. Between left atrium and left ventricle b. Bicuspid valve ii. Between right ventricle and pulmonary artery c. Semilunar valve iii. Between right atrium and right ventricle a b c (1) ii i iii (2) i ii iii (3) i iii ii (4) iii i ii Answer ( 4 ) S o l . Tricuspid valves are AV valve present between right atrium and right ventricle. Bicuspid valves are AV valve present between left atrium and left ventricle. Semilunar valves are present at the openings of aortic and pulmonary aorta. 72. Match the items given in Column I with those in Column II and select the correct option given below: Column I Column II a. Tidal volume i. 2500 – 3000 mL b. Inspiratory Reserve ii. 1100 – 1200 mL volume c. Expiratory Reserve iii. 500 – 550 mL volume d. Residual volume iv. 1000 – 1100 mL a b c d (1) iv iii ii i (2) i iv ii iii (3) iii i iv ii (4) iii ii i iv Answer ( 3 ) S o l . Tidal volume is volume of air inspired or expired during normal respiration. It is approximately 500 mL. Inspiratory reserve volume is additional volume of air a person can inspire by a forceful inspiration. It is around 2500 – 3000 mL. Expiratory reserve volume is additional volume of air a person can be expired by a forceful expiration. This averages 1000 – 1100 mL. Residual volume is volume of air remaining in lungs even after forceful expiration. This averages 1100 – 1200 mL.
According to Hugo de Vries, the mechanism of evolution is (1) Minor mutations (2) Phenotypic variations (3) Saltation (4) Multiple step mutations
Answer ( 3 ) S o l . As per mutation theory given by Hugo de Vries, the evolution is a discontinuous phenomenon or saltatory phenomenon/ saltation. 69.
All of the following are part of an operon except (1) a promoter (2) an enhancer (3) structural genes (4) an operator
Answer ( 2 ) Sol. •
• 70.
Enhancer sequences are present in eukaryotes. Operon concept is for prokaryotes.
Which of the following options correctly represents the lung conditions in asthma and emphysema, respectively? (1) Decreased respiratory Inflammation of bronchioles
surface;
(2) Increased respiratory Inflammation of bronchioles
surface;
(3) Increased number of bronchioles; Increased respiratory surface (4) Inflammation of bronchioles; Decreased respiratory surface Answer ( 4 ) S o l . Asthma is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased. 12
NEET (UG) - 2018 (Code-CC) ACHLA
73.
Answer ( 1 )
Hormones secreted by the placenta to maintain pregnancy are (1) hCG, progestogens, glucocorticoids
S o l . Spermiogenesis is transformation of spermatids into spermatozoa whereas spermiation is the release of the sperms from sertoli cells into the lumen of seminiferous tubule.
estrogens,
(2) hCG, hPL, progestogens, estrogens (3) hCG, hPL, estrogens, relaxin, oxytocin
76.
(4) hCG, hPL, progestogens, prolactin
The amnion of mammalian embryo is derived from
Answer ( 2 )
(1) ectoderm and endoderm
S o l . Placenta releases human chorionic gonadotropic hormone (hCG) which stimulates the Corpus luteum during pregnancy to release estrogen and progesterone and also rescues corpus luteum from regression. Human placental lactogen (hPL) is involved in growth of body of mother and breast. Progesterone maintains pregnancy, keeps the uterus silent by increasing uterine threshold to contractile stimuli.
(2) mesoderm and trophoblast
74.
(3) endoderm and mesoderm (4) ectoderm and mesoderm Answer ( 3 ) S o l . The extraembryonic or foetal membranes are amnion, chorion, allantois and Yolk sac. Amnion is formed from mesoderm on outer side and ectoderm on inner side. Chorion is formed from trophoectoderm and mesoderm whereas allantois and Yolk sac membrane have mesoderm on outerside and endoderm in inner side.
The contraceptive ‘SAHELI’ (1) is a post-coital contraceptive. 77.
(2) is an IUD. (3) increases the concentration of estrogen and prevents ovulation in females. (4) blocks estrogen receptors in the uterus, preventing eggs from getting implanted.
(1) Starfish
(2) Moth
(3) Tunicate
(4) Earthworm
Answer ( 4 )
Answer ( 4 )
S o l . Metamorphosis refers to transformation of larva into adult.
S o l . Saheli is the first non-steroidal, once a week pill. It contains centchroman and its functioning is based upon selective Estrogen Receptor modulation. 75.
Which of the following animals does not undergo metamorphosis?
Animal that perform metamorphosis are said to have indirect development. In earthworm development is direct which means no larval stage and hence no metamorphosis.
The difference between spermiogenesis and spermiation is 78.
(1) In spermiogenesis spermatozoa are formed, while in spermiation spermatozoa are released from sertoli cells into the cavity of seminiferous tubules.
Which one of these animals is not a homeotherm? (1) Psittacula (2) Camelus
(2) In spermiogenesis spermatozoa from sertoli cells are released into the cavity of seminiferous tubules, while in spermiation spermatozoa are formed.
Answer ( 3 )
(3) In spermiogenesis spermatozoa are formed, while in spermiation spermatids are formed.
S o l . Homeotherm are animals that maintain constant body temperature, irrespective of surrounding temperature.
(3) Chelone (4) Macropus
Birds and mammals are homeotherm.
(4) In spermiogenesis spermatids are formed, while in spermiation spermatozoa are formed.
Chelone (Turtle) belongs to class reptilia which is Poikilotherm or cold blood. 13
NEET (UG) - 2018 (Code-CC) ACHLA
79.
Column I
Column II
Which of the following features is used to identify a male cockroach from a female cockroach?
a. Glycosuria
i.
(1) Presence of anal cerci
b. Gout
ii. Mass of crystallised salts within the kidney
c. Renal calculi
iii. Inflammation in glomeruli
Answer ( 3 )
d. Glomerular nephritis
iv. Presence of glucose in urine
S o l . Males bear a pair of short, thread like anal styles which are absent in females.
a
b
c
d
(1)
iv
i
ii
iii
Anal/caudal styles arise from 9th abdominal segment in male cockroach.
(2)
ii
iii
i
iv
(3)
i
ii
iii
iv
Which of the following organisms are known as chief producers in the oceans?
(4)
iii
ii
iv
i
(2) Forewings with darker tegmina (3) Presence of caudal styles (4) Presence of a boat shaped sternum on the 9th abdominal segment
80.
(1) Euglenoids
(2) Cyanobacteria
(3) Diatoms
(4) Dinoflagellates
Answer ( 1 ) S o l . Glycosuria denotes presence of glucose in the urine. This is observed when blood glucose level rises above 180 mg/100 ml of blood, this is called renal threshold value for glucose. Gout is due to deposition of uric acid crystals in the joint.
Answer ( 3 ) S o l . Diatoms are chief producers of the ocean. 81.
Ciliates differ from all other protozoans in
Renal calculi are precipitates of calcium phosphate produced in the pelvis of the kidney.
(1) having two types of nuclei (2) using pseudopodia for capturing prey (3) having a contractile vacuole for removing excess water
Glomerular nephritis is the inflammatory condition of glomerulus characterised by proteinuria and haematuria.
(4) using flagella for locomotion 84.
Answer ( 1 ) S o l . Ciliates differs from other protozoans in having two types of nuclei. eg. Paramoecium have two types of nuclei i.e. macronucleus & micronucleus. 82.
Identify the vertebrate group of animals characterized by crop and gizzard in its digestive system (1) Osteichthyes
(2) Aves
(3) Reptilia
(4) Amphibia
Accumulation of uric acid in joints
Match the items given in Column I with those in Column II and select the correct option given below: Column I
Column II
(Function)
(Part of Excretory system)
a. Ultrafiltration
i.
Henle's loop
b. Concentration
ii. Ureter
of urine c. Transport of
iii. Urinary bladder
urine
Answer ( 2 )
d. Storage of
S o l . The digestive tract of Aves has additional chambers in their digestive system as crop and Gizzard.
iv. Malpighian
urine
corpuscle v.
Proximal convoluted tubule
Crop is concerned with storage of food grains. Gizzard is a masticatory organ in birds used to crush food grain. 83.
Match the items given in Column I with those in Column II and select the correct option given below : 14
a
b
c
d
(1) v
iv
i
iii
(2) v
iv
i
ii
(3) iv
i
ii
iii
(4) iv
v
ii
iii
NEET (UG) - 2018 (Code-CC) ACHLA
Answer ( 3 )
Answer ( 3 )
S o l . Ultrafiltration refers to filtration of very fine particles having molecular weight less than 68,000 daltons through malpighian corpuscle.
Sol.
Concentration of urine refers to water absorption from glomerular filtrate as a result of hyperosmolarity in the medulla created by counter-current mechanism in Henle's loop.
88.
Urine is carried from kidney to bladder through ureter.
85.
Dominant–recessive relationship
IAIB
-
Codominance
IA, IB
-
3-different allelic forms of a gene (multiple allelism)
&
IO
Which of the following is not an autoimmune disease? (1) Vitiligo (2) Alzheimer's disease
Among the following sets of examples for divergent evolution, select the incorrect option :
(4) Psoriasis
(3) Rheumatoid arthritis Answer ( 2 ) S o l . Rheumatoid arthritis is an autoimmune disorder in which antibodies are produced against the synovial membrane and cartilage.
(2) Brain of bat, man and cheetah (3) Heart of bat, man and cheetah
Vitiligo causes white patches on skin also characterised as autoimmune disorder.
(4) Forelimbs of man, bat and cheetah
Psoriasis is a skin disease that causes itchy or sore patches of thick red skin and is also autoimmune whereas Alzheimer's disease is due to deficiency of neurotransmitter acetylcholine.
Answer ( 1 ) S o l . Divergent evolution occurs in the same structure, example - forelimbs, heart, brain of vertebrates which have developed along different directions due to adaptation to different needs whereas eye of octopus, bat and man are examples of analogous organs showing convergent evolution.
89.
(1) Vitamin E
(2) Vitamin B12
(3) Vitamin A
(4) Vitamin D
Sol.
The similarity of bone structure in the forelimbs of many vertebrates is an example of (1) Adaptive radiation
Conversion of milk to curd improves its nutritional value by increasing the amount of
(2) Convergent evolution (3) Analogy (4) Homology Answer ( 4 )
Answer ( 2 )
87.
-
Urinary bladder is concerned with storage of urine.
(1) Eye of octopus, bat and man
86.
IAIO, IBIO
S o l . In different vertebrates, bones of forelimbs are similar but their forelimbs are adapted in different way as per their adaptation, hence example of homology.
Curd is more nourishing than milk. It has enriched presence of vitamins specially Vit-B12.
Which of the following characteristics represent ‘Inheritance of blood groups’ in humans?
90.
In which disease does mosquito transmitted pathogen cause chronic inflammation of lymphatic vessels?
a. Dominance
(1) Amoebiasis
b. Co-dominance
(2) Ringworm disease
c. Multiple allele
(3) Ascariasis
d. Incomplete dominance
(4) Elephantiasis Answer ( 4 )
e. Polygenic inheritance (1) a, c and e
(2) b, d and e
(3) a, b and c
(4) b, c and e
S o l . Elephantiasis is caused by roundworm, Wuchereria bancrofti and it is transmitted by Culex mosquito. 15
NEET (UG) - 2018 (Code-CC) ACHLA
91.
Answer ( 4 )
A carbon resistor of (47 ± 4.7) k is to be marked with rings of different colours for its identification. The colour code sequence will be
Sol. I
n nr r
So, I is independent of n and I is constant. I
(1) Green – Orange – Violet – Gold (2) Yellow – Green – Violet – Gold (3) Yellow – Violet – Orange – Silver (4) Violet – Yellow – Orange – Silver Answer ( 3 )
O
S o l . (47 ± 4.7) k = 47 × 103 ± 10% Yellow – Violet – Orange – Silver 92. A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is (1) 9 (2) 20 (3) 11 (4) 10 Answer ( 4 )
94.
E ...(i) nR R E 10 I ...(ii) R R n Dividing (ii) by (i), (n 1)R 10 1 n 1 R After solving the equation, n = 10 A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?
n The power radiated by a black body is P and it radiates maximum energy at wavelength, 0. If the temperature of the black body is now changed so that it radiates maximum energy 3 at wavelength 0 , the power radiated by it 4 becomes nP. The value of n is (1)
81 256
(2)
256 81
(3)
4 3
(4)
3 4
Sol. I
93.
I
Answer ( 2 ) S o l . We know, max T constant (Wien's law)
So, max1 T1 max2 T2 0 T
T
95.
(2)
O
O
n
I
n
I
(3)
n
4
P2 T 256 4 P1 T 81 3
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by l on applying a force F, how much force is needed to stretch the socond wire by the same amount? (1) F
(2) 4 F
(3) 6 F
(4) 9 F
Answer ( 4 ) S o l . Wire 1 :
(4)
O
4 T 3 4
So,
I
(1)
3 0 T 4
O
A, 3l
n 16
F
NEET (UG) - 2018 (Code-CC) ACHLA
Wire 2 :
98. 3A, l
F
The moment of the force, F 4iˆ 5 ˆj 6kˆ at (2, 0, –3), about the point (2, –2, –2), is given by (1) 7iˆ 4 ˆj 8kˆ
For wire 1,
F l 3l AY
(2) 7iˆ 8ˆj 4kˆ
…(i)
(3) 4iˆ ˆj 8kˆ
For wire 2,
(4) 8iˆ 4 ˆj 7kˆ
F l Y 3A l
Answer ( 1 )
F l l 3AY
Sol.
Y
…(ii)
F
From equation (i) & (ii),
A
F F l 3l l AY 3AY 96.
r0
F 9F
(1) 84.5 J
(2) 42.2 J
(3) 208.7 J
(4) 104.3 J
ˆi ˆj kˆ 0 2 1 7iˆ 4 ˆj 8kˆ 4 5 6
54 × 4.18 = U + 1.013 × 105(167.1 × 10–6 – 0) U = 208.7 J A small sphere of radius 'r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to
r2
r3
(3)
(4)
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is (1) 0.529 cm (2) 0.053 cm (3) 0.525 cm (4) 0.521 cm
Answer ( 1 )
Answer ( 2 )
S o l . Diameter of the ball
2 S o l . Power = 6 rVT iVT 6 rVT
= MSR + CSR × (Least count) – Zero error = 0.5 cm + 25 × 0.001 – (–0.004)
VT r 2 Power r
...(i)
0iˆ 2 ˆj kˆ
99.
(2) r5
X
ˆ (2iˆ 2 ˆj 2k) ˆ r r0 (2iˆ 0ˆj 3k)
S o l . Q = U + W
(1) r4
P
r
O (r r0 ) F
A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 105 Nm–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is
Answer ( 3 )
97.
r r0
= 0.5 + 0.025 + 0.004 5
= 0.529 cm 17
NEET (UG) - 2018 (Code-CC) ACHLA A
100. A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E . Due to the force q E , its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively
m a C
B
(1) a = g tan (3) a
(2) a = g cos
g sin
(4) a
g cosec
Answer ( 1 )
(1) 1.5 m/s, 3 m/s
Sol.
N cos
(2) 1 m/s, 3.5 m/s
N
(3) 1 m/s, 3 m/s
(4) 2 m/s, 4 m/s
ma (pseudo)
Answer ( 3 ) Sol. t = 0 A
a
–a
t=1 v = 6 ms C t=3
v=0
–1
mg
60 6 ms2 1
1 6(1)2 = 3 m 2
...(i)
1 6(1)2 3 m 2
...(i)
N cos = mg
...(ii)
a g
a = g tan 102. An em wave is propagating in a medium with
V Viˆ . The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along a
...(ii)
For t = 2 s to t = 3 s, S3 0
N sin = ma
tan
For t = 1 s to t = 2 s,
1 S2 6.1 6(1)2 3 m 2
a
In non-inertial frame,
–1
For t = 0 to t = 1 s, S1
t=2 B v=0
–a
v = –6 ms
Acceleration a
N sin
...(iii)
velocity
(1) –x direction (2) –y direction
Total displacement S = S1 + S2 + S3 = 3 m
(3) +z direction
3 Average velocity 1 ms 1 3
(4) –z direction
Total distance travelled = 9 m Average speed
Answer ( 3 )
9 3 ms 1 3
Sol. E B V
101. A block of mass m is placed on a smooth inclined wedge ABC of inclination as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and for the block to remain stationary on the wedge is
ˆ (B) Viˆ (Ej)
So, B Bkˆ Direction of propagation is along +z direction. 18
NEET (UG) - 2018 (Code-CC) ACHLA
103. The refractive index of the material of a
L
prism is 2 and the angle of the prism is 30°. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is (1) Zero
(2) 30°
(3) 45°
(4) 60°
25 2 106 10–3 3600 500 36
= 13.89 H 105. An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be (1) 36 cm towards the mirror
Answer ( 3 )
(2) 30 cm towards the mirror
S o l . For retracing its path, light ray should be normally incident on silvered face.
(3) 36 cm away from the mirror (4) 30 cm away from the mirror
Answer ( 3 )
30°
i
M
Sol.
60°
f = 15 cm
30°
O
40 cm
2 1 1 1 f v1 u
Applying Snell's law at M,
1 1 1 – 15 v1 40
sin i 2 sin30 1
–
1 sin i 2 2
sin i
1 2
v1 = –24 cm i.e. i = 45°
When object is displaced by 20 cm towards mirror.
104. The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance (1) 13.89 H
(2) 1.389 H
(3) 138.88 H
(4) 0.138 H
Now, u2 = –20 1 1 1 f v2 u2
1 1 1 – –15 v2 20
Answer ( 1 )
1 1 1 – v2 20 15
S o l . Energy stored in inductor U
1 2 Ll 2
25 10 –3
1 1 1 v1 –15 40
v2 = –60 cm
1 L (60 10 –3 )2 2
So, image shifts away from mirror by = 60 – 24 = 36 cm. 19
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Divide (ii) by (i),
106. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is (1) 1 : –2
(2) 2 : –1
(3) 1 : –1
(4) 1 : 1
eE0 t 1 mV 0 108. When the light of frequency 20 (where 0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v 1 . When the frequency of the incident radiation is increased to 5 0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is
Answer ( 3 ) S o l . KE = –(total energy) So, Kinetic energy : total energy = 1 : –1 107. An electron of mass m with an initial velocity
V V0 ˆi (V 0 > 0) enters an electric field
(1) 2 : 1
(2) 4 : 1
(3) 1 : 4
(4) 1 : 2
E –E0 ˆi (E0 = constant > 0) at t = 0. If 0 is its
Answer ( 4 )
de-Broglie wavelength initially, then its deBroglie wavelength at time t is
S o l . E W0
(1) 0
h 0
eE0 (3) 0 1 t mV0 eE0 1 mV 0
1 mv22 2
4h0
h mV0
v1 1 v2 2
...(i)
109. For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
E0 V0
F
…(ii)
1 v12 4 v22
S o l . Initial de-Broglie wavelength
Acceleration of electron eE0 a m
(1) 15
(2) 30
(3) 10
(4) 20
Answer ( 4 ) S o l . Number of nuclei remaining = 600 – 450 = 150
Velocity after time ‘t’
n
N 1 N0 2
eE0 V V0 t m
…(i)
1 mv22 2 Divide (i) by (ii),
t
Answer ( 4 )
h So, mV
1 mv12 2
1 mv12 2
h(50 ) h0
0
0
1 mv2 2
h(20 ) h0
(2) 0t
(4)
0
h eE m V0 0 m h
eE0 mV0 1 mV 0
t
t
150 1 t 1/2 600 2
t
2
0 eE0 1 mV 0
t
t
1 1 t 1/2 2 2 t = 2t1/2 = 2 × 10
…(ii)
= 20 minute 20
NEET (UG) - 2018 (Code-CC) ACHLA
110. The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is
l
4l 2l 32 3 2 20 13.33 cm 3
112. The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is (1) 12.5% (2) 6.25% (3) 20% 2 (1) 7
(3)
1 (2) 3
2 3
(4)
(4) 26.8% Answer ( 4 )
2 5
T S o l . Efficiency of ideal heat engine, 1 2 T1 T2 : Sink temperature
Answer ( 4 )
T1 : Source temperature
S o l . Given process is isobaric
dQ n Cp dT
T % 1 2 100 T1
5 dQ n R dT 2
273 1 100 373
dW P dV = n RdT
100 100 26.8% 373
dW nRdT 2 Required ratio dQ 5 5 n R dT 2
113. At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere?
111. The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is
(Given : Mass of oxygen molecule (m) = 2.76 × 10–26 kg Boltzmann's constant kB = 1.38 × 10–23 JK–1)
(1) 16 cm
(2) 12.5 cm
(1) 1.254 × 104 K
(3) 8 cm
(4) 13.2 cm
(2) 5.016 × 104 K
Answer ( 4 )
(3) 8.360 × 104 K
S o l . For closed organ pipe, third harmonic
(4) 2.508 × 104 K
Answer ( 3 )
3v 4l
S o l . Vescape = 11200 m/s
For open organ pipe, fundamental frequency
Say at temperature T it attains Vescape
v 2l
So,
Given,
3kB T 11200 m/s mO2
On solving,
3v v 4 l 2l
T = 8.360 × 104 K 21
NEET (UG) - 2018 (Code-CC) ACHLA
114. Unpolarised light is incident from air on a plane surface of a material of refractive index ''. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?
0.21
d
…(ii)
0.20 d Dividing we get, 0.21 2 mm
d = 1.9 mm
1 (1) i tan1
116. An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of
1 (2) i sin1 (3) Reflected light is polarised with its electric vector perpendicular to the plane of incidence
(1) Small focal length and small diameter (2) Large focal length and large diameter
(4) Reflected light is polarised with its electric vector parallel to the plane of incidence
(3) Large focal length and small diameter (4) Small focal length and large diameter Answer ( 2 )
Answer ( 3 ) S o l . When reflected light rays and refracted rays are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.
S o l . For telescope, angular magnification =
So, focal length of objective lens should be large.
i
Angular resolution =
117. In the circuit shown in the figure, the input voltage Vi is 20 V, VBE = 0 and VCE = 0. The values of IB, IC and are given by
Also, tan i = (Brewster angle) 115. In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same and D) the separation between the slits needs to be changed to (1) 1.7 mm
(2) 2.1 mm
(3) 1.9 mm
(4) 1.8 mm
20 V
Vi
RB 500 k B
RC 4 k C E
(1) IB = 40 A, IC = 5 mA, = 125
Answer ( 3 )
0.20 2 mm
D should be large. 1.22
So, objective should have large focal length (f0) and large diameter D.
S o l . Angular width
f0 fE
(2) IB = 20 A, IC = 5 mA, = 250
d
(3) IB = 25 A, IC = 5 mA, = 200 (4) IB = 40 A, IC = 10 mA, = 250 …(i)
Answer ( 1 ) 22
NEET (UG) - 2018 (Code-CC) ACHLA
S o l . VBE = 0
A
Sol. A
VCE = 0
B
Vb = 0
B A
20 V IC Vi
RB Ib
500 k
AB
RC = 4 k
Y AB
B
Y (A B A B)
Vb
120. A metallic rod of mass per unit length 0.5 kg m–1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
(20 0) 4 103 IC = 5 × 10–3 = 5 mA IC
Vi = VBE + IBRB Vi = 0 + IBRB
(1) 11.32 A
20 = IB × 500 × 103
(2) 14.76 A
20 40 A IB 500 103
(3) 5.98 A (4) 7.14 A
I 25 103 C 125 Ib 40 106
Answer ( 1 ) S o l . For equilibrium,
118. In a p-n junction diode, change in temperature due to heating
B
mg sin30 Il Bcos 30
(1) Affects the overall V - I characteristics of p-n junction
I
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance (4) Affects only reverse resistance
mg tan30 lB
0.5 9.8 0.25 3
0°
11.32 A
3 n si g m 30°
S o l . Due to heating, number of electron-hole pairs will increase, so overall resistance of diode will change.
(1) 1.13 W
Due to which forward biasing and reversed biasing both are changed.
(2) 2.74 W (3) 0.43 W
119. In the combination of the following gates the output Y can be written in terms of inputs A and B as
(4) 0.79 W Answer ( 4 )
A
2
V S o l . Pav RMS R Z
Y
2
1 Z R2 L 56 C (1) A B (3) A B A B Answer ( 3 )
° 30
llB 30° llB
121. An inductor 20 mH, a capacitor 100 F and a resistor 50 are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is
Answer ( 1 )
B
s co
(2) A B A B (4) A B
23
Pav
2
50 0.79 W 2 56 10
NEET (UG) - 2018 (Code-CC) ACHLA
125. The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is
122. A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from
(1) Inversely proportional to the distance between the plates (2) Proportional to the square root of the distance between the plates
(1) The induced electric field due to the changing magnetic field
(3) Linearly proportional to the distance between the plates
(2) The lattice structure of the material of the rod
(4) Independent of the distance between the plates
(3) The magnetic field
Answer ( 4 )
(4) The current source
S o l . For isolated capacitor Q = Constant
Answer ( 4 )
Q2 2A0
S o l . Energy of current source will be converted into potential energy of the rod.
Fplate
123. Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is
F is Independent of the distance between plates.
(1) 500
(2) 250
(3) 25
(4) 40
126. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s2 at a distance of 5 m from the mean position. The time period of oscillation is
Answer ( 2 )
(1) 1 s
S o l . Current sensitivity IS
NBA C
(2) 2 s (3) s
Voltage sensitivity
(4) 2 s
NBA VS CRG
Answer ( 3 ) S o l . |a| = 2y
So, resistance of galvanometer
20 = 2(5)
I 51 5000 RG S 250 VS 20 103 20
= 2 rad/s
124. A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27ºC two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27ºC is (1) 300 m/s
(2) 350 m/s
(3) 339 m/s
(4) 330 m/s
T
127. An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is (1) Equal
Answer ( 3 )
(2) 10 times greater
S o l . v = 2 () [L2 – L1] = 2 × 320 [73 – 20] × = 339.2
2 2 s 2
10–2
(3) 5 times greater
ms–1
(4) Smaller
= 339 m/s
Answer ( 4 ) 24
NEET (UG) - 2018 (Code-CC) ACHLA
Sol. h
1 eE 2 t 2 m
130. If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?
2hm eE
t
t m as ‘e’ is same for electron and proton.
(1) ‘g’ on the Earth will not change (2) Time period of a simple pendulum on the Earth would decrease
∵ Electron has smaller mass so it will take smaller time.
(3) Walking on the ground would become more difficult
128. The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then
(4) Raindrops will fall faster Answer ( 1 ) S o l . If Universal Gravitational constant becomes ten times, then G = 10 G So, acceleration due to gravity increases.
B A
i.e. (1) is wrong option.
C
S
(1) KB > KA > KC
(2) KB < KA < KC
(3) KA > KB > KC
(4) KA < KB < KC
131. A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere? (1) Angular momentum
Answer ( 3 )
(2) Rotational kinetic energy
B
Sol. perihelion A
S VA
VC
(3) Moment of inertia
C aphelion
(4) Angular velocity Answer ( 1 ) S o l . ex = 0
Point A is perihelion and C is aphelion. So, VA > VB > VC
dL 0 dt i.e. L = constant
So,
So, KA > KB > KC 129. A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (K t ) as well as rotational kinetic energy (K r ) simultaneously. The ratio Kt : (Kt + Kr) for the sphere is (1) 2 : 5
(2) 10 : 7
(3) 5 : 7
(4) 7 : 10
So angular momentum remains constant. 132. A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to
Answer ( 3 ) S o l . Kt
1 mv 2 2
Kt Kr
h
1 1 1 1 2 v mv2 I2 mv2 mr 2 2 2 2 25 r
2
A
(1)
7 mv2 10 So,
B
Kt 5 Kt Kr 7
5 D 4
(3) D 25
(2)
7 D 5
(4)
3 D 2
NEET (UG) - 2018 (Code-CC) ACHLA
Answer ( 1 )
134. Which one of the following statements is incorrect?
Sol.
h
(1) Coefficient of sliding dimensions of length.
B A
friction
has
(2) Frictional force opposes the relative motion.
vL
As track is frictionless, so total mechanical energy will remain constant
(3) Limiting value of static friction is directly proportional to normal reaction.
T.M.EI =T.M.EF
(4) Rolling friction is smaller than sliding friction.
1 0 mgh mvL2 0 2
Answer ( 1 )
v2 h L 2g
S o l . Coefficient of sliding friction has no dimension.
For completing the vertical circle, vL 5gR
f = sN
5 5gR 5 h R D 2g 2 4
f N
s
133. Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation
135. A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be
(1) WA > WC > WB
(1) 0.4
(2) WB > WA > WC (3) WA > WB > WC
(2) 0.8
(4) WC > WB > WA
(3) 0.25
Answer ( 4 )
(4) 0.5
S o l . Work done required to bring them rest Answer ( 3 )
W = KE
S o l . According to law of conservation of linear momentum,
1 W I2 2
mv 4m 0 4mv 0
W I for same
v
2 1 WA : WB : WC MR2 : MR2 : MR2 5 2
=
v 4
v Relative velocity of separation 4 e v Relative velocity of approach
2 1 : :1 5 2
= 4 : 5 : 10 WC > WB > WA
e 26
1 0.25 4
NEET (UG) - 2018 (Code-CC) ACHLA
136. Iron carbonyl, Fe(CO)5 is (1) Dinuclear
(2) Trinuclear
(3) Mononuclear
(4) Tetranuclear
138. Which one of the following ions exhibits d-d transition and paramagnetism as well? (1) MnO42– (2) MnO4–
Answer ( 3 )
(3) Cr2O72–
S o l . • Based on the number of metal atoms present in a complex, they are classified into mononuclear, dinuclear, trinuclear and so on.
(4) CrO42– Answer ( 1 ) S o l . CrO42– Cr6+ = [Ar]
eg: Fe(CO)5 : mononuclear Co2(CO)8 : dinuclear
Unpaired electron (n) = 0; Diamagnetic
Fe3(CO)12: trinuclear
Cr2O72– Cr6+ = [Ar] Unpaired electron (n) = 0; Diamagnetic
Hence, option (3) should be the right answer.
MnO42– = Mn6+ = [Ar] 3d1
137. Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code : Column I
Unpaired electron (n) = 1; Paramagnetic MnO4– = Mn7+ = [Ar] Unpaired electron (n) = 0; Diamagnetic
Column II
139. The geometry and magnetic behaviour of the complex [Ni(CO)4] are
a. Co3+
i.
8 BM
b. Cr3+
ii.
35 BM
(1) Tetrahedral geometry and paramagnetic
c. Fe3+
iii.
3 BM
(2) Square planar paramagnetic
d. Ni2+
iv.
24 BM
(3) Tetrahedral geometry and diamagnetic
v.
15 BM
(4) Square planar geometry and diamagnetic
geometry
a
b
c
d
Answer ( 3 )
(1)
iii
v
i
ii
S o l . Ni(28) : [Ar]3d8 4s2
(2)
iv
i
ii
iii
∵ CO is a strong field ligand
(3)
i
ii
iii
iv
Configuration would be :
(4)
iv
v
ii
i
and
3
sp -hybridisation
Answer ( 4 ) S o l . Co3+ = [Ar] 3d6, Unpaired e–(n) = 4 Spin magnetic moment =
4(4 2) 24 BM
3(3 2) 15 BM
5(5 2) 35 BM
CO CO CO
Ni
Ni2+ = [Ar] 3d8, Unpaired e–(n) = 2 Spin magnetic moment =
CO
CO
Fe3+ = [Ar] 3d5, Unpaired e–(n) = 5 Spin magnetic moment =
×× ×× ××
For, four ‘CO’-ligands hybridisation would be sp 3 and thus the complex would be diamagnetic and of tetrahedral geometry.
Cr3+ = [Ar] 3d3, Unpaired e–(n) = 3 Spin magnetic moment =
××
CO
OC CO
2(2 2) 8 BM 27
NEET (UG) - 2018 (Code-CC) ACHLA
140. The type of isomerism shown by the complex [CoCl2(en)2] is
Now,
CH3
(1) Linkage isomerism
CH – CH3
(2) Ionization isomerism
O2
CH3 – CH – CH3
(3) Coordination isomerism (4) Geometrical isomerism
(P)
Answer ( 4 )
CH3
S o l . In [CoCl2(en)2], Coordination number of Co is 6 and this compound has octahedral geometry.
HC –C –O– O –H 3
OH
O
+
H /H2O
CH3 – C – CH3 + (R)
(Q)
Hydroperoxide Rearrangement
142. Which of the following compounds can form a zwitterion? • As per given option, type of isomerism is geometrical isomerism.
OH ,
CH3 – CO – CH3
CHO ,
COOH ,
CH2CH2CH3
CHO
(4)
,
pKa = 2.34
MnO4
, CH3CH(OH)CH3
,
(3)
C2 O24
H+
(1) 5
16
2
(2) 2
16
5
(3) 2
5
16
(4) 16
5
2
Answer ( 3 )
Reduction
CH3CH2 – OH
,
+7
+3
S o l . MnO4– + C2O42– + H+
Answer ( 1 )
+
CH3 – CH – CH3
n-factor of
Al Cl
1, 2–H Shift
2+
+4
Mn + CO2 + H2O
Oxidation
Cl S o l . CH CH CH – Cl + 3 2 2
(Zwitterion form)
The correct coefficients of the reactants for the balanced equation are
OH
CH2CH2CH3
–
MnO4 C2 O24 H Mn2 CO2 H2 O
R
CH(CH3)2
(2)
H3N – CH2 – COO
143. For the redox reaction
Q+R
(ii) H3O+/
,
(1)
H3N – CH2 – COOH pKa = 9.60
Q CH(CH3)2
(4) Aniline
H2N – CH2 – COO–
(i) O2
P
(3) Acetanilide
Sol.
Anhydrous AlCl3
P
(2) Benzoic acid
Answer ( 1 )
141. Identify the major products P, Q and R in the following sequence of reactions:
+ CH3CH2CH2Cl
(1) Glycine
MnO4
5
n-factor of C2 O24 2
Cl +
CH3CH2CH2
Cl
(Incipient carbocation)
Ratio of n-factors of MnO4 and C2 O24 is 5 : 2 So, molar ratio in balanced reaction is 2 : 5
–
AlCl3
The balanced equation is
Cl
2MnO4 5C2 O24 16H 2Mn2 10CO2 8H2 O
–
AlCl3 28
NEET (UG) - 2018 (Code-CC) ACHLA
S o l . The reaction for fH°(XY)
144. The correction factor ‘a’ to the ideal gas equation corresponds to
1 1 X2 (g) Y2 (g) XY(g) 2 2
(1) Forces of attraction between the gas molecules (2) Electric field present between the gas molecules
Bond energies of X2, Y2 and XY are X,
X , X 2
respectively
(3) Volume of the gas molecules (4) Density of the gas molecules
Answer ( 1 )
X X H X 200 2 4
On solving, we get
2 S o l . In real gas equation, P an (V nb) nRT V2
X X 200 2 4
X = 800 kJ/mole
van der Waal’s constant, ‘a’ signifies intermolecular forces of attraction.
147. When initial concentration of the reactant is doubled, the half-life period of a zero order reaction
145. Which one of the following conditions will favour maximum formation of the product in the reaction,
(1) Remains unchanged (2) Is tripled
X2 (g) r H X kJ? A2 (g) B2 (g)
(3) Is doubled (1) High temperature and low pressure
(4) Is halved
(2) High temperature and high pressure
Answer ( 3 )
(3) Low temperature and low pressure
S o l . Half life of zero order
(4) Low temperature and high pressure t 1/2
Answer ( 4 ) X2 (g); H x kJ S o l . A2 (g) B2 (g)
t 1/2 will be doubled on doubling the initial concentration.
On increasing pressure equilibrium shifts in a direction where pressure decreases i.e. forward direction.
148 Which of the following molecules represents the order of hybridisation sp2, sp2, sp, sp from left to right atoms?
On decreasing temperature, equilibrium shifts in exothermic direction i.e., forward direction.
(1) CH3 – CH = CH – CH3
So, high pressure and low temperature favours maximum formation of product.
(2) CH2 = CH – CH = CH2 (3) CH2 = CH – C CH
146. The bond dissociation energies of X2, Y2 and XY are in the ratio of 1 : 0.5 : 1. H for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X2 will be
(4) HC C – C CH Answer ( 3 )
(1) 400 kJ mol–1 (2) 800 kJ
[A0 ] 2K
sp2
sp2
sp
sp
S o l . CH2 CH – C CH
mol–1
(3) 100 kJ mol–1
Number of orbital require in hybridization
(4) 200 kJ mol–1
= Number of -bonds around each carbon atom.
Answer ( 2 ) 29
NEET (UG) - 2018 (Code-CC) ACHLA
149. Which of the following carbocations is expected to be most stable?
NO2 H
(1) Y
Sol.
NO2
which is independent concentration of reactant.
NO2
NO2
(4)
H
initial
1 , k[A0 ]
152. Among CaH2, BeH2, BaH2, the order of ionic character is
Y
For second order reaction, t 1/2
0.693 , k
which depends on initial concentration of reactant.
Y
of
(2) H Y
(3)
For first order reaction, t 1/2
(1) BaH2 < BeH2 < CaH2
H
(2) BeH2 < BaH2 < CaH2
Answer ( 2 )
(3) CaH2 < BeH2 < BaH2
S o l . –NO2 group exhibit –I effect and it decreases with increase in distance. In option (2) positive charge present on C-atom at maximum distance so –I effect reaching to it is minimum and stability is maximum.
(4) BeH2 < CaH2 < BaH2 Answer ( 4 ) S o l . For 2nd group hydrides, on moving down the group metallic character of metals increases so ionic character of metal hydride increases.
150. Which of the following is correct with respect to – I effect of the substituents? (R = alkyl)
Hence the option (4) should be correct option.
(1) – NR2 > – OR > – F (2) – NH2 > – OR > – F
153. Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :
(3) – NR2 < – OR < – F (4) – NH2 < – OR < – F Answer ( 4 * )
– BrO4
S o l . –I effect increases on increasing electronegativity of atom. So, correct order of –I effect is –NH2 < – OR < – F.
– Br
*Most appropriate Answer is option (4), however option (3) may also be correct answer.
– BrO3
1.82 V
1.0652 V
1.5 V
Br2
HBrO
1.595 V
Then the species disproportionation is
151. The correct difference between first and second order reactions is that
undergoing
(1) HBrO
(1) The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations
(2) Br2
(2) A first-order reaction can catalyzed; a second-order reaction cannot be catalyzed
Answer ( 1 )
(3) BrO4 (4) BrO3 1
0
o S o l . HBrO Br2 , EHBrO/Br 1.595 V 2
(3) The half-life of a first-order reaction does not depend on [A] 0 ; the half-life of a second-order reaction does depend on [A]0
1
5
HBrO BrO3 , Eo
BrO3 /HBrO
1.5 V
o for the disproportionation of HBrO, Ecell
(4) The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
o o Ecell EHBrO/Br Eo 2
BrO3 /HBrO
= 1.595 – 1.5 = 0.095 V = + ve Hence, option (1) is correct answer.
Answer ( 3 ) 30
NEET (UG) - 2018 (Code-CC) ACHLA
154. In which case is number of molecules of water maximum? (1) 10–3 mol of water
(1) CH4
(2) CH3 – CH3
(3) CH2 CH2
(4) CH CH
Answer ( 1 )
(2) 0.00224 L of water vapours at 1 atm and 273 K
Br2/h
S o l . CH4 (A)
(3) 0.18 g of water
CH3Br Na/dry ether Wurtz reaction
(4) 18 mL of water Answer ( 4 )
CH3 — CH3
S o l . (1) Molecules of water = mole × NA = 10–3 NA (2) Moles of water =
Hence the correct option is (1) 157. The compound C7H8 undergoes the following reactions:
0.00224 = 10–4 22.4
3Cl /
Br /Fe
Zn/HCl
2 2 C7H8 A B C
Molecules of water = mole × NA = 10–4 NA
The product 'C' is (3) Molecules of water = mole × NA =
(1) p-bromotoluene
0.18 NA 18
(2) 3-bromo-2,4,6-trichlorotoluene (3) o-bromotoluene
= 10–2 NA
(4) m-bromotoluene
(4) Mass of water = 18 × 1 = 18 g Molecules of water = mole × NA =
Answer ( 4 )
18 NA 18
CH3
= NA
3Cl 2
Sol.
155. The compound A on treatment with Na gives B, and with PCl 5 gives C. B and C react together to give diethyl ether. A, B and C are in the order
CCl3
(C7H8)
CCl3 Br2 Fe
(B)
(A)
Br
Zn HCl CH3
(1) C2H5OH, C2H5ONa, C2H5Cl (2) C2H5Cl, C2H6, C2H5OH (3) C2H5OH, C2H5Cl, C2H5ONa
(C)
(4) C2H5OH, C2H6, C2H5Cl
So, the correct option is (4)
Answer ( 1 ) S o l . C2H5OH (A)
Na
158. Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?
C2H5O Na+ (B)
PCl5 C2H5Cl (C)
C2H5O Na+ + C2H5Cl (B) (C)
Br
(1) NO
(2) N2O
(3) NO2
(4) N2O5
Answer ( 4 ) SN2
S o l . Fact
C2H5OC2 H5
159. A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
So the correct option is (1) 156. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is 31
(1) 4.4
(2) 2.8
(3) 3.0
(4) 1.4
NEET (UG) - 2018 (Code-CC) ACHLA
Answer ( 2 )
162. Nitration of aniline in strong acidic medium also gives m-nitroaniline because
Conc.H2 SO4 S o l . HCOOH CO(g) H2 O(l) 1 1 mol 2.3 g or mol 20 20
COOH
Conc.H2SO4
COOH
(1) In acidic (strong) medium aniline is present as anilinium ion. (2) In absence of substituents nitro group always goes to m-position.
CO(g) + CO2 (g) + H2O(l) 1 mol 20
1 mol 20
(3) In electrophilic substitution reactions amino group is meta directive.
1 4.5 g or mol 20
Gaseous mixture formed is CO and CO2 when it is passed through KOH, only CO 2 is absorbed. So the remaining gas is CO.
(4) Inspite of substituents nitro group always goes to only m-position. Answer ( 1 )
So, weight of remaining gaseous product CO is
NH2
2 28 2.8 g 20
NH3 H
Sol.
Anilinium ion
So, the correct option is (2)
–NH3 is m-directing, hence besides para and
(51%) and ortho (2%), meta product (47%) is also formed in significant yield.
(1) Amylose is made up of glucose and galactose
163. Regarding cross-linked or network polymers, which of the following statements is incorrect?
160. The difference amylopectin is
between
amylose
(2) Amylopectin have 1 4 -linkage and 1 6 -linkage
(1) They contain strong covalents bonds in their polymer chains.
(3) Amylose have 1 4 -linkage and 1 6 -linkage
(2) Examples are bakelite and melamine. (3) They are formed from bi- and tri-functional monomers.
(4) Amylopectin have 1 4 -linkage and 1 6 -linkage
(4) They contain covalent bonds between various linear polymer chains.
Answer ( 4 ) S o l . Amylose and Amylopectin are polymers of D-glucose, so -link is not possible. Amylose is linear with 1 4 -linkage whereas Amylopectin is branched and has both 1 4 and 1 6 -linkages.
Answer ( 1 ) S o l . Cross linked or network polymers are formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains, e.g. bakelite, melamine etc. Option (1) is not related to cross-linking.
So option (4) should be the correct option. 161. Which of the following oxides is most acidic in nature? (1) CaO
(2) BaO
(3) BeO
(4) MgO
So option (1) should be the correct option. 164. Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations :
Answer ( 3 ) S o l . BeO < MgO < CaO < BaO Basic character increases. So, the most acidic should be BeO. In fact, BeO is amphoteric oxide while other given oxides are basic. 32
a. 60 mL
M M HCl + 40 mL NaOH 10 10
b. 55 mL
M M HCl + 45 mL NaOH 10 10
NEET (UG) - 2018 (Code-CC) ACHLA
Answer ( 4 )
M M c. 75 mL HCl + 25 mL NaOH 5 5
d. 100 mL
S o l . Solubility of BaSO4, s =
M M HCl + 100 mL NaOH 10 10
= 1.04 × 10–5 (mol L–1)
Ba2 (aq) SO 24(aq) BaSO 4 (s)
pH of which one of them will be equal to 1? (1) c
(2) d
(3) a
(4) b
s
s
Ksp = [Ba2+] [SO42–]= s2 = (1.04 × 10–5)2
Answer ( 1 ) Sol. •
2.42 103 (mol L–1) 233
= 1.08 × 10–10 mol2 L–2
1 Meq of HCl = 75 1 = 15 5
167. Given van der Waals constant for NH3, H2, O2 and CO2 are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?
•
1 Meq of NaOH = 25 1 = 5 5
•
Meq of HCl in resulting solution = 10
(1) CO2
(2) O2
•
Molarity of [H+] in resulting mixture
(3) H2
(4) NH3
=
10 1 100 10
Answer ( 4 )
1 pH = –log[H+] = log = 1.0 10 165. On which of the following properties does the coagulating power of an ion depend?
Sol. •
van der waal constant ‘a’, signifies intermolecular forces of attraction.
•
Higher is the value of ‘a’, easier will be the liquefaction of gas.
168. Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s2 2s2 2p3, the simplest formula for this compound is
(1) The sign of charge on the ion alone (2) Both magnitude and sign of the charge on the ion (3) Size of the ion alone
(1) Mg3X2
(2) Mg2X
(4) The magnitude of the charge on the ion alone
(3) MgX2
(4) Mg2X3
Answer ( 1 ) S o l . Element (X) electronic configuration
Answer ( 2 ) Sol. •
•
Coagulation of colloidal solution by using an electrolyte depends on the charge present (positive or negative) on colloidal
1s2 2s2 2p3
particles as well as on its size.
Formula of compound formed by Mg and X will be Mg3X2.
So, valency of X will be 3. Valency of Mg is 2.
Coagulating power of an electrolyte depends on the magnitude of charge present on effective ion of electrolyte.
169. Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is
166. The solubility of BaSO4 in water is 2.42 × 10–3 gL–1 at 298 K. The value of its solubility product (Ksp) will be (Given molar mass of BaSO4 = 233 g mol–1) (1) 1.08 × 10–8 mol2L–2
(1)
(2) 1.08 × 10–14 mol2L–2 (3) 1.08 × 10–12 mol2L–2 (3)
(4) 1.08 × 10–10 mol2L–2 33
1 2 4 3 3 2
(2)
(4)
3 3 4 2 3 2
NEET (UG) - 2018 (Code-CC) ACHLA
Answer ( 3 ) S o l . NO : (1s) 2 , ( 1s) 2 , (2s) 2,( 2s) 2,(2p z) 2 , (2px)2 = (2py)2,(2px)1 = (2py)0
Answer ( 2 ) S o l . For BCC lattice : Z = 2, a
4r 3
10 5 2.5 2 CN– : (1s)2, (1s)2, (2s)2,(2s)2, (2px)2 = (2py)2,(2pz)2 BO =
For FCC lattice : Z = 4, a = 2 2 r
d25C d900C
ZM N a3 A BCC
10 4 3 2 CN : (1s)2, (1s)2, (2s)2,(2s)2, (2px)2 = (2py)2,(2pz)1 BO =
ZM N a3 A FCC
9 4 2.5 2 CN+ : (1s)2, (1s)2, (2s)2,(2s)2, (2px)2 = (2py)2
3
3 3 2 2 2 r 4r 4 2 4 3
BO =
170. Which one is a wrong statement?
8 4 2 2 Hence, option(3) should be the right answer. BO =
(1) The value of m for dz2 is zero (2) The electronic configuration of N atom is 1s2
1
2s2
2px
1
2py
1
2pz
172. Which of the following statements is not true for halogens? (1) Chlorine has the highest electron-gain enthalpy
(3) An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers
(2) All but fluorine show positive oxidation states
(4) Total orbital angular momentum of electron in 's' orbital is equal to zero
(3) All are oxidizing agents
Answer ( 2 )
(4) All form monobasic oxyacids
S o l . According to Hund's Rule of maximum multiplicity, the correct electronic configuration of N-atom is
1s2
2s2
Answer ( 2 ) S o l . Due to high electronegativity and small size, F forms only one oxoacid, HOF known as Fluoric (I) acid. Oxidation number of F is +1 in HOF.
2p3
173. Which one of the following elements is unable to form MF63– ion?
OR
2
1s
2s
2
2p
3
(1) In
(2) B
(3) Al
(4) Ga
Answer ( 2 ) S o l . ∵ 'B' has no vacant d-orbitals in its valence shell, so it can't extend its covalency beyond 4. i.e. 'B' cannot form the ion like MF6 3(–) i.e. BF63(–).
Option (2) violates Hund's Rule. 171. Consider the following species : CN+, CN–, NO and CN Which one of these will have the highest bond order?
Hence, the correct option is (2). 174. In the structure of ClF 3, the number of lone pair of electrons on central atom ‘Cl’ is
(1) CN (2) CN+ (3) CN–
(1) Three
(2) Four
(4) NO
(3) Two
(4) One
34
NEET (UG) - 2018 (Code-CC) ACHLA
Answer ( 3 )
5
0
–3
Hence, the correct option is (4). 178. In the reaction
Cl F
–
OH
F
S o l . The structure of ClF3 is
F
2
S o l . H N O , N O, N2 , NH Cl 3 4
O Na
CHO
+ CHCl3 + NaOH
The number of lone pair of electrons on central Cl is 2.
+
The electrophile involved is
175. Considering Ellingham diagram, which of the following metals can be used to reduce alumina?
(1) Dichlorocarbene : CCl2
(2) Dichloromethyl anion CHCl2
(1) Cu (2) Mg
(3) Formyl cation CHO
(3) Zn
(4) Fe
(4) Dichloromethyl cation CHCl2
Answer ( 2 )
Answer ( 1 )
S o l . The metal which is more reactive than 'Al' can reduce alumina i.e. 'Mg' should be the correct option.
S o l . It is Reimer-Tiemann reaction. The electrophile formed is :CCl2 (Dichlorocarbene) according to the following reaction
176. The correct order of atomic radii in group 13 elements is
.–. CCl3 H2 O CHCl3 OH–
(1) B < Ga < Al < In < Tl
.–.
CCl3 : CCl2 Cl–
(2) B < Ga < Al < Tl < In
Electrophile
(3) B < Al < Ga < In < Tl
179. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their
(4) B < Al < In < Ga < Tl Answer ( 1 )
(1) Formation of intermolecular H-bonding
Sol. Elements Atomic radii (pm)
B 85
Ga 135
Al 143
In 167
(2) More extensive association of carboxylic acid via van der Waals force of attraction
Tl 170
(3) Formation of carboxylate ion
177. The correct order of N-compounds in its decreasing order of oxidation states is
(4) Formation of intramolecular H-bonding Answer ( 1 )
(1) NH4Cl, N2, NO, HNO3
S o l . Due to formation of intermolecular H-bonding in carboxylic acid, association occurs. Hence boiling point increases and become more than the boiling point of aldehydes, ketones and alcohols of comparable molecular masses.
(2) HNO3, NH4Cl, NO, N2 (3) HNO3, NO, NH4Cl, N2 (4) HNO3, NO, N2, NH4Cl Answer ( 4 )
35
NEET (UG) - 2018 (Code-CC) ACHLA
180. Compound A, C8H10O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell.
Answer ( 2 ) S o l . Option (2) is secondary alcohol which on oxidation gives phenylmethyl ketone (Acetophenone). This on reaction with I2 and NaOH form iodoform and sodium benzoate.
A and Y are respectively
CH3
2NaOH I2 NaOI NaI H2 O
(1) CH3
(2)
OH and I2
CH – CH3 OH (A)
CH – CH3 and I2 OH
(3)
(4) H3C
NaOI
C – CH3 O Acetophenone
COONa + CHI3 Sodium benzoate
CH2 – CH2 – OH and I2
CH2 – OH and I2
36
Iodoform (Yellow PPt)
I2 NaOH
