NWT-11

NEET WEEKEND TEST - 11 (22-07-2025)

SR ELITE (SET – I)
SYLLABUS
Botany : Principles of inheritance & variations (30Q) and Molecular basis of inheritance(15Q) : Nucleic acids, Chemical
composition, Deoxyribo nucleic acid (DNA), Numerical problems in DNA and Packaging of DNA helix
Zoology : Principles of Inheritance and Variation (10Q) , Molecular Basis of Inheritance (25Q): Complete chapter and
Evolution (10Q): Introduction to Theory of Biogenesis , Chemical Evolution, Experimental Verification of
Chemical Origin of Life
Physics : Moving charges and magnetism (33Q): Magnetization and Magnetic Intensity, Magnetic induction due to
SR ELITE, AIIMS S60 & MPL
Solenoid, The bar magnet as an equivalent solenoid , Torroid, Motion of charge in a magnetic field and
applications, Lorentz force, Force on a current carrying conductor placed in a magnetic field & Flemings left
hand rule, Force between two long straight parallel conductors carrying current , Torque on current loop in an
uniform magnetic field, construction and working of Moving coil galvanometer, Converting moving coil
galvanometer into ammeter and voltmeter, Tangent Galvanometer Comparison of M.C.G and T.G,
Galvanometer – Half deflection method and Electro Magnetic induction(12Q): Introduction, Faraday's
experiments, Magnetic flux, Change in flux, Faraday’s law - Induced E.M.F, Lenz law and related numerical
Chemistry : Track-1: Expt Skills (6Q): The chemistry involved in the titrimetric exercises - Acids. Bases and the use of
indicators, Theory involved in Permanganometry: Oxalic acid vs KMnO4.Mohr's salt vs KMnO4 and Kinetic
study of the reaction of iodide ions with hydrogenperoxide at room temperature, IIIA group (6Q) and IVA group
(11Q): Complete chapter as per updated NEET syllabus
Track-2:Organic chemistry: Alcohols and Phenols (5Q): preparations and properties and uses, Ethers (15Q)
:Nomenclature, preparation, Preparations of ether, properties of ether and Aldehydes and ketones(2Q):
Nomenclature
Guidelines:
1. Duration of the Exam is 3 Hours.
2. The question paper consists of 180 questions, comprising of 45 questions in each subject.
3. Each question carries four marks. One mark will be deducted for every incorrect response
from the total score. Maximum marks are 720

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 SR ELITE (SET – I) NWT-11 (22-07-25)
BOTANY 5. Four statements are listed below. How many
1. If “X” represents the number of double of them are pertinent to Law of dominance ?
heterozygous genotypes in F2 generation of A) Expression of only parental character in
dihybrid cross, then what would be the F1 when two pure line parents are crossed.
number of double homozygous genotypes ? B) Both alleles of a gene pair in a
(1) X (2) X/2 heterozygote are fully expressed resulting in
a phenotype displaying both traits
(3) 2X (4) 4X simultaneously.
2. Read the following statements and pick the C) 3 : 1 phenotypic ratio observed in F2
correct pair generation of a monohybrid cross.
A) DNA and RNA are two types of nucleic D) Gametes carry either dominant (or)
acids found in living systems recessive alleles for a particular trait not both.
B) DNA is long polymer of (1) A & B (2) B & C
ribonucleotides (3) C & D (4) A & C
C) Length of DNA is usually defined as 6. Which type of bonds get constructed when a
number of nucleotides present in it pentose sugar transforms into a nucleotide ?
D) RNA is the most abundant genetic A) N glycosidic bonds
material in cellular organisms B) Phosphoester bonds
(1) A & B (2) C&D C) Hydrogen bonds in between bases
(3) B & D (4) A&C D) Phosphodiester linkages
(1) A & B (2) B & C
3. Match column I with column II with respect
to DNA in different organisms (3) A & D (4) C & D
7. Given below are two statements.
Column - I Column - II
Statement I: The character that was not seen
A. Linear DNA I. Viruses
in F1 generation is expressed in F2
B. ss DNA II. Bacteria generation but in a lesser proportion in a
C. Circular DNA III. Mitochondria monohybrid cross.
D. DNA without IV. Humans Statement II: In codominance, the
histones heterozygote shows intermediate (or) blended
phenotype of two homozygous parents.
A B C D
In light of the above statements, choose the
(1) I II III IV most appropriate answer from the options
(2) III II IV I given below.
(3) IV I II III (1) Both statement I and statement II are
correct.
(4) II IV I III
(2) Both statement I and statement II are
4 Read the following statements and pick the incorrect.
incorrect one
(3) Statement I is correct but statement II is
(1) Backbone of a nucleic acid consists of incorrect.
alternating phosphates and pentose sugars
(4) Statement I is incorrect but statement II
(2) DNA is a type of nucleic acid acting as is correct.
genetic material in all cellular organisms
8. “It is a 5 membered cyclic compound with
(3) Phosphoric acid uses two of its three only carbon, hydrogen and oxygen and form
hydroxyl groups in the formation of 3'5' a building block of a nucleotide”. What is it?
diester linkages in between two nucleotides (1) Adenine base (2) Ribose sugar
(4) The sugar of RNA possess one less (3) Thymine base (4) Phosphate group
oxygen atom to that of DNA
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 SR ELITE (SET – I) NWT-11 (22-07-25)
9. If both father and mother is heterozygous for Column - I Column - II
A and B blood groups, then children could A. Monohybrid test I. 9 : 3 : 3 : 1
have cross
P) A group Q) B group
B. F2 complete II. 1 : 1
R) AB group S) O group
(1) P & Q only (2) R & S only dominance
(3) Q & S only (4) P, Q, R, S C. F2 incomplete III. 1 : 2 : 1
10. RNA in cellular organisms functions as all dominance
except
D. F2 dihybrid cross IV. 3 : 1
(1) Messenger (2) Adapter
(3) Catalyst (4) Genetic material A B C D
11. Pick the incorrect match (1) II IV III I
(1) Morgan : Linkage (2) II III IV I
(2) Correns, Hugo De Vries & Tschermak :
(3) IV I II III
Rediscovery of chromosomal theory of
inheritance (4) II IV I III
(3) SturtEvant : Genetic map 16. Any two nucleotides in a DNA could vary in
(4) Mendel : Principles of inheritance (1) Type of sugar (2) Phosphate group
12. “Complementary base pairing, hydrogen (3) Nitrogen base (4) Amino acid
bonding between opposite bases, antiparallel
nature of strands, right handed double helix 17. Length of DNA is 510A0 . It possess 20%
are features commonly seen in DNA. But in of Adenine. How many hydrogen bonds are
one organism listed below DNA does not seen in between Guanine and Cytosine ?
show any of the above attribute. What is it ? (1) 270 (2) 180
(1) E.Coli (3) 90 (4) 120
(2)  phage 18. Point mutations exemplify a direct and
(3) 174 bacteriophage immediate change in
(4) Human (1) Nucleotide sequence in DNA
13. An individual with genotype LlMm produce
(2) Chromosome structure
four types of gametes i.e., LM, Lm, lM & lm
in same proportion. The above result implies (3) Protein synthesized on ribosome
that (4) Chromosome number
(1) Genes are tightly linked 19. Given below are two statements. One is
(2) Genes are loosely linked without labelled Assertion (A) and the other is
crossing over labelled Reason (R).
(3) Genes are present on non homologous Assertion (A) : The pattern of inheritance for
chromosomes and alleles of these genes polygenic trait exemplify non Mendelian
assort independently during gamete inheritance pattern.
formation Reason (R) : Polygenic trait is an example for
(4) Genes underwent mutations quantitative inheritance unlike Mendel‟s
14. “The sequence of nucleotide bases if on one inheritance pattern which is a qualitative
strand is known, then the sequence of inheritance.
nucleotides in opposite strand is predicted”. In light of the above statements, choose the
What does it signify ? correct answer from the options given below.
(1) Central dogma of molecular biology (1) Both (A) and (R) are true and (R) is the
(2) Antiparallel orientation of both strands correct explanation of (A).
(3) Complementary nature of base pairing (2) Both (A) and (R) are true but (R) is not
(4) Right handed clockwise helix nature the correct explanation of (A).
15. Match column – I (cross) with column – II (3) (A) is true but (R) is false.
(phenotypic ratio) (4) Both (A) and (R) are false.
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 SR ELITE (SET – I) NWT-11 (22-07-25)
20. Which aspect given below is associated with 24. Phosphorylation of cytidine results in A and
phenylketonuria ? methylation of uracil results in B. What is A
(1) Pleiotropy and B ?
(2) Frame shift mutation A B
(3) Substitution of purine (A) with a
(1) Cytidylic acid Thymine
pyrimidine (T) in gene producing  globin
(4) Polygenic inheritance (2) Cytosine Thymine
21. F. Meischer identified acidic substance (3) Thymine Uridylic acid
namely DNA present inside
(1) Cytoplasm (2) Nucleus (4) Guanine Thymine
(3) Mitochondria (4) Chloroplast 25. If a pure tall plant (TT) is 50 cm height then
22. Read the following statements and pick the what is the height of F1 (Tt) plant if it follows
correct pair Mendel‟s inheritance pattern ?
A) DNA is scattered throughout the cells in
(1) 50 cm (2) 25 cm
prokaryotes due to the absence of defined
nucleus (3) 150 cm (4) 75 cm
B) Chromatin represents thread like stained 26. Pick the incorrect statement from the
bodies inside nucleus of eukaryotes following
C) The positively charged DNA is (1) All three laws of inheritance holds good
wrapped around negatively charged histone with respect to dihybrid cross involving
octomer to form a nucleosome unlinked genes
D) The packaging of chromatin at higher
(2) Phenotypic and genotypic ratios in F2
level require additional set of proteins
namely non histone chromosomal proteins in generation is 1 : 2 : 1 in both incomplete
eukaryotes dominance and codominance
(1) A & C (2) B & C (3) Genes always occur in pairs in gametes
(3) C & D (4) B & D (4) Besides the involvement of multiple
23. Given below are two statements. genes, polygenic inheritance also takes into
Statement A : The results of hybridization account the influence of environment
experiments done by Mendel pointed to 27. “Effect of each allele is additive”. The
general rules of inheritance rather than statement holds good for
unsubstantiated ideas. (1) Human skin colour
Statement B : Mendel‟s experiment on pea
has large sampling data and he confirmed his (2) Height of stem in garden pea
inferences from experiments on pea plant for (3) Size of starch grains in pea
many generations and recorded the data using (4) ABO blood grouping
statistical and mathematical logic.
28. Place the following in increasing order with
In light of the above statements, choose the
respect to their number in one complete
most appropriate answer from the options
spiral/turn of DNA.
given below.
(1) Both statement A and statement B are A) Nucleotide pairs
correct. B) Hydrogen bonds if all 10 base pairs are
(2) Both statement A and statement B are GC pairs
incorrect. C) Phosphodiester bonds
(3) Statement A is correct but statement B D) Glycosidic bonds
is incorrect.
(1) A C D B (2) C D A D
(4) Statement A is incorrect but statement
(3) D C B A (4) C D A B
B is correct.
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 SR ELITE (SET – I) NWT-11 (22-07-25)
29. Read the following statements and pick the (1) Both statement A and statement B are
incorrect one correct.
Statement Genes Chrom (2) Both statement A and statement B are
osome incorrect.
(3) Statement A is correct but statement B
(1) Occur in pairs √ √
is incorrect.
(2) Segregate at the (4) Statement A is incorrect but statement
time of gamete √ √ B is correct.
formation
33. Former is mutations involving changes in
(3) Independent pairs whole set of chromosomes and latter involve
segregate √ √ change in a subset of chromosome number.
independent of each What are they called respectively ?
other (1) Aneuploidy, Polyploidy
(4) One pair always
(2) Polyploidy, Aneuploidy
segregate √ √
independent of (3) Translocation, Aneuploidy
other pair (4) Polyploidy, Inversion

30. Read the following statements and pick the

A+G
34. If on one DNA strand is 0.80 then
incorrect one T+C
(1) Histones are rich in basic amino acids A+G
what is on opposite DNA strand ?
namely Lysine and Arginine T+C
(2) Histones are organized to form a unit of (1) 0.80 (2) 1
eight molecules called histone octomer in the
(3) 1.25 (4) 0.20
core region of nucleosome
(3) Euchromatin is transcriptionally 35. Match the following column I with column II
inactive chromatin, whereas heterochromatin Column - I Column - II
is with active genes A. Pleiotropy I. Single gene
(4) The length of DNA in a typical influencing
mammalian cell is far greater than the multiple unrelated
dimensions of a typical nucleus traits
31. Recombination frequencies among few genes B. Polygenic II. Two (or) more
is given below. Which gene is most likely to inheritance genes influencing a
occupy the centre in the order ? single trait
A – B = 34% B – C = 12 % C. Multiple allelism III. Gene that exist in
C – D = 10% D – E = 7% more than two
A – E = 5% allelic forms in
(1) A (2) E population
(3) D (4) B
D. Codominance IV. Both alleles of a
32. Given below are two statements. gene pair in a
Statement A : Stability of a helical structure heterozygote are
in DNA is due to proper stacking of base fully expressed
pairs in addition to hydrogen bonds between without blending
base pairs. Choose the correct answer from the options
Statement B : If each strand of parent DNA given below.
act as template for synthesis of new strand, A B C D
then the two daughter DNA‟s produced
would be identical to parent DNA molecule. (1) I II III IV
In light of the above statements, choose the (2) II III IV I
most appropriate answer from the options (3) IV I II III
given below. (4) II IV I III
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 SR ELITE (SET – I) NWT-11 (22-07-25)
36. Given below are two statements. One is 41. In a dihybrid test cross (AaBb  aabb) the
labelled Assertion (A) and the other is expected phenotypic ratio if genes assort
labelled Reason (R). independently is “X” and the same if the
Assertion (A) : Distance (diameter) between gene pair is completely linked is “Y”. What
two polynucleotide chains in B-DNA is is “X” and “Y”?
almost constant. X Y
Reason (R) : Always a purine with two ring (1) 9 : 3 : 3 : 1 3:1
structure pairs with pyrimidine with one ring (2) 1 : 1 : 1 : 1 1:1
structure. (3) 1 : 1 1:1:1:1
(4) 3 : 1 9:3:3:1
In light of the above statements, choose the
correct answer from the options given below. 42. Which feature in Drosophila makes it ideal
for studying linkage ?
(1) Both (A) and (R) are true and (R) is the (1) They cannot be grown on simple
correct explanation of (A). synthetic medium in a laboratory
(2) Both (A) and (R) are true but (R) is not (2) They complete the life cycle in about
the correct explanation of (A). two years
(3) (A) is true but (R) is false. (3) Single mating produce large number of
progeny flies
(4) Both (A) and (R) are false.
(4) Clear differentiation of sexes in the
37. Which does not exemplify blending male and female flies is absent
inheritance ? 43. What proportion of F2 offspring from a
(1) Antirrhinum (snapdragon) flowers dihybrid cross will be homozygous dominant
(2) Starch grain size in pea for both traits ?
(3) Mirabilis (4 o clock) flowers 1 2
(1) (2)
(4) ABO blood grouping 16 16
4 9
38. How many different genotypes appear in F2 (3) (4)
16 16
generation of a dihybrid cross ?
44. In garden pea large sized grains with
 AaBb  AaBb  effective synthesis of starch is in
(1) 9 (2) 4 (1) Heterozygous round seed plants
(2) Homozygous round seeded plants
(3) 8 (4) 16
(3) Wrinkled seeded plants
39. Nucleosomes are seen in genetic material of (4) Constricted pod containing plants
(1) 174 bacteriophage 45. Which among the following phenomenon
leads to variations in DNA ?
(2)  phage
(1) Linkage, mutation
(3) E. coli
(2) Recombination, linkage
(4) Human chromosomes (3) Mutation, recombination
40. Which condition favours higher chances of (4) Linkage, crossing over
crossing over between two genes ? ZOOLOGY
(1) Genes far apart on the same 46. Find out the percentage of colourblind
chromosome children but without PKU when a
(2) Genes located close together colourblind woman who is suffering from
PKU married a non-colourblind man who is
(3) Genes located on different non
carrier for PKU.
homologous chromosomes
(1) 75% (2) 25%
(4) Genes located close together in hetero
(3) 50% (4) 0%
chromatin region
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 SR ELITE (SET – I) NWT-11 (22-07-25)
47. Select the correct statements about sickle cell B. Turner‟s syndrome is due to aneuploidy.
anaemia. C. Cystic fibrosis is an autosomal recessive
A. There is a change in gene for beta globin gene disorder.
on chromosome 12 D. Sickle cell trait is represented as
B. In the beta globin, there is glutamate in the Hb Hbs
place of valine. (1) B, C and D are correct
C. It is an example of point mutation. (2) A, B and C are correct
D. In the mutated gene, CAC is present in the (3) A and D are correct
template strand (4) Only B and D are correct
Choose the correct answer from the options 51. Louis Pasteur proved that
given below. (1) Life comes from decaying and rotting
(1) Only B, C and D (2) Only C and D matter
(2) Life comes from pre existing life
(3) Only A, B and D (4) Only A and C
(3) Life comes spontaneously from spores
48. Given below are two statements: (4) Life comes due to special creation
Statement-I: The first form of life arose 52. Thalassemia is an inherited blood disorder
slowly through evolutionary forces from non- that affects haemoglobin synthesis. Select the
living molecules is accepted majority. correct statements and mark the correct
Statement-II: The first non-cellular forms of option
life could have originated 3 billion years ago. a) Thalassemia is a qualitative problem in
In the light of the above statements, choose which the body makes adequate but
the correct answer from the options given abnormal haemoglobin.
below: b) Beta thalassemia is caused by the
substitution of glutamic acid by valine of the
(1) Both Statement I and Statement II are
beta globin chain.
true
c) Alpha thalassemia is controlled by two
(2) Statement I is false but Statement II is closely linked genes.
true
d) Beta thalassemia is controlled by HB-B
(3) Statement I is true but Statement II is gene located on 11th chromosome.
false (1) c and d (2) b and d
(4) Both Statement I and Statement II are (3) a and c (4) b and c
false 53. Study the following statements with regard to
49. If a clourblind boy has normal brother and pedigree analysis
clourblind sister, then their parents are a) Since controlled crosses can be performed
A) Normal father and clourblind mother in human beings, it provides an alternative to
B) Both mother and father are normal understand the inheritance pattern of a trait.
C) Father is clourblind and mother is carrier b) It proves that Mendelian ratios are always
D) Both parents are clourblind observed in all families regarding
E) Father is colourblind and mother is chromosomal syndromes
homozygous normal c) It provides a tool to trace the inheritance of
Choose the most appropriate answer from the a disease.
options given below. d) It helps to predict the probability of the
(1) Only A and B (2) Only B and C occurrence of a disease in future generations.
(3) Only D and E (4) Only C Which of the above are correct?
50. Find out the correct statements. (1) a and b (2) c and d
A. DMD is a sex-linked dominant disorder. (3) a and c (4) a and d
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 SR ELITE (SET – I) NWT-11 (22-07-25)
54. Study the following statements w.r.t human 57. Study the following statements
genome a) The technique of DNA Fingerprinting was
a) The human genome contains 3164.7 initially developed by Alec Jeffreys.
million bp. b) Alec Jeffreys used a satellite DNA as
b) 99.9 per cent of base sequence among probe that shows very high degree of
humans is the same. polymorphism.
c) the functions are unknown for over 50 per c) A small DNA sequence is arranged
cent of the discovered genes. tandemly in many copy numbers. The copy
number remains same from chromosome to
d) Repetitive sequences make up very small
chromosome in an individual.
portion of the human genome.
d) DNA from a single cell is enough to
Which of the above are incorrect? perform DNA fingerprinting analysis.
(1) b and d (2) a and c Which of the above are correct?
(3) b and c (4) Only d (1) Only a and d (2) Only a, b and d
55. Match the following Column I with Column (3) Only b and c (4) Only a, c and d
II 58. Study the following statements
Column I Column II a) DNA fingerprinting technology has
A) Chromosome 1 i) 1.4 million application in forensic science.
locations b) DNA fingerprinting technology has no any
application in determining population and
B) Single – base ii) 3000 bases
genetic diversities.
DNA differences c) During the early years of the HGP, the
C) Average gene iii) 2968 genes Wellcome Trust (U.K.) became a major
D) Codes for iv) Less than 2 per partner
Proteins cent of the d) HGP was closely associated with the rapid
genome development of a new area in biology called
(1) A–ii, B–iii, C – i, D – iv Bioinformatics.
(2) A–iii, B–i, C – iv, D – ii Which of the above are incorrect?
(1) b and d (2) a and c
(3) A–iii, B–i, C – ii, D – iv
(3) b (4) c and d
(4) A–i, B–iv, C – ii, D – iii
59. Given below are two statements: One is
56. Study the following statements labeled as Assertion (A) and the other is
a) DNA polymorphism (variation at genetic labeled as Reason (R).
level) arises due to mutations. Assertion(A): Male honey bees do not have
b) Allelic sequence variation has traditionally father and thus can not have sons, but have a
been described as a DNA polymorphism if grand father and can have grand sons.
more than one variant (allele) at a locus Reason(R): An unfertilised egg develops as a
occurs in human population with a frequency male bee by means of parthenogenesis and
lesser than 0.01 fertilised egg develops as female bee.
c) There is a variety of different types of In the light of above statements, choose the
DNA polymorphisms ranging from single most appropriate answer from the options
nucleotide change to very large scale given below.
changes.
(1) Both A and R are correct and R is the
d) For evolution and speciation, DNA correct explanation of A
polymorphisms do not have any role
(2) Both A and R are correct but R is NOT
Which of the above are correct? the correct explanation of A
(1) c and d (2) a and b (3) A is not correct but R is correct
(3) a and d (4) a and c (4) A is correct but R is not correct
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 SR ELITE (SET – I) NWT-11 (22-07-25)
60. Match the following Column I with Column (1) Both A and R are correct and R is the
II correct explanation of A
Column I Column II (2) Both A and R are correct but R is NOT
Detection of the correct explanation of A
Automated DNA
A) hybridized i) (3) A is not correct but R is correct
sequencers
DNA fragments (4) A is correct but R is not correct
Frederck 63. The gases that are condensed under
B) ii) Autoradiography
Sanger gravitation and formed the galaxies of the
Separation of Restriction present day universe are
C) iii)
DNA fragments endonucleases
(1) Hydrogen and Ammonia
Digestion of
D) iv) Electrophoresis (2) Hydrogen and Helium
DNA
(3) Methane and Ammonia
(1) A–ii, B–i, C – iv, D – iii
(2) A–iii, B–i, C – ii, D – iv (4) Ammonia and Water vapour
(3) A–ii, B–i, C – iii, D – iv 64. Given below are two statements: One is
labeled as Assertion (A) and the other is
(4) A–iv, B–i, C – ii, D – iii
labeled as Reason (R).
61. Given below are two statements:
Assertion (A): Mutations in non – coding
Statement-I: The origin of life is considered a
DNA sequence keep on accumulating
unique event in the history of universe.
generation after generation, and form one of
Statement-II: According to Oparin and the basis of variability/polymorphism.
Haldane the first form of life could have
Reason (R): Mutations in non – coding DNA
come from pre-existing non-living organic
sequence may not have any immediate
molecules.
effect/impact in an individual‟s reproductive
In the light of the above statements, choose ability.
the correct answer from the options given
In the light of above statements, choose the
below:
most appropriate answer from the options
(1) Both Statement I and Statement II are given below.
true
(1) Both A and R are correct and R is the
(2) Statement I is false but Statement II is correct explanation of A
true
(2) Both A and R are correct but R is NOT
(3) Statement I is true but Statement II is the correct explanation of A
false
(3) A is not correct but R is correct
(4) Both Statement I and Statement II are
(4) A is correct but R is not correct
false
65. Match the following Column I with Column
62. Given below are two statements: One is
II
labeled as Assertion (A) and the other is
labeled as Reason (R). Column I Column II
Assertion (A): DNA taken from blood, hair- A) HGP was launched i) 1866
follicle, skin, etc., become very useful B) HGP was completed ii) May 2006
identification tool in forensic applications. C) sequence of chromosome 1 iii) 2003
Reason (R): DNA from every tissue (such as was completed
blood, hair-follicle, skin, bone, saliva, sperm D) Langdon Down described iv) 1990
etc.), from an individual shows the different Down‟s syndrome
degree of polymorphism.
(1) A–i, B–ii, C – iii, D – iv
In the light of above statements, choose the
(2) A–i, B–ii, C – iv, D – iii
most appropriate answer from the options
given below. (3) A–iii, B–ii, C – i, D – iv
(4) A–iv, B–iii, C – ii, D – i
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 SR ELITE (SET – I) NWT-11 (22-07-25)
66. Study the following statements (2) Both A and R are correct but R is NOT
a) Repetitive sequences of DNA are thought the correct explanation of A
to have direct coding functions but they do (3) A is not correct but R is correct
not shed light on chromosome structure, (4) A is correct but R is not correct
dynamics and evolution. 69. Which of the following statements are
b) One of the greatest impacts of having the correct about syndrome with trisomy of 23rd
HG sequence may well be enabling a chromosome
radically new approach to biological A. Secondary sexual characters are
research. developed normally
c) DNA fingerprinting is a very quick way to B. Such an individual has overall masculine
compare the DNA sequences of any two development. However, the feminine
individuals. development is also expressed.
d) During density gradient centrifugation, the C. The affected individual is short statured.
bulk DNA forms a major peak and the other D. Physical, psychomotor and mental
small peaks are referred to as satellite DNA. development is retarded.
Which of the above are correct? E. Such individuals are sterile.
(1) Only a and d (2) Only c and d Choose the correct answer from the options
given below
(3) Only b, c and d (4) Only a and c
(1) A and E only (2) A and B only
67. Match the following Column I with Column (3) C and D only (4) B and E only
II
70. Non-disjunction of autosomes during gamete
Column I Column II formation causes
A) Mini-satellites i) A:T rich (1) Sickle-cell anaemia
B) Micro-satellites ii) 0.1 to 20 kb (2) Turner‟s syndrome
C) VNTR iii) Mini-satellite (3) Down‟s syndrome
D) Size of VNTR iv) G:C rich (4) Haemophilia
(1) A–ii, B–iv, C – i, D – iii 71. Given below are two statements: One is
(2) A–i, B–iv, C – ii, D – iii labeled as Assertion (A) and the other is
labeled as Reason (R).
(3) A–iv, B–i, C – iii, D – ii
Assertion(A): The possibility of a female
(4) A–i, B–iv, C – iii, D – ii becoming a haemophilic is extremely rare.
68. Given below are two statements: One is Reason (R): Mother of such a female has to
labeled as Assertion (A) and the other is be at least carrier and the father should be
labeled as Reason (R). haemophilic (unviable in the later stage of
Assertion (A): Males are more likely to be life).
affected than females regarding X-linked In the light of above statements, choose the
recessive disorders. most appropriate answer from the options
Reason (R): Males have only one X- given below.
chromosome whereas females have two X- (1) Both A and R are correct and R is the
chromosomes. correct explanation of A
In the light of above statements, choose the (2) Both A and R are correct but R is NOT
most appropriate answer from the options the correct explanation of A
given below
(3) A is not correct but R is correct
(1) Both A and R are correct and R is the
correct explanation of A (4) A is correct but R is not correct

10
 SR ELITE (SET – I) NWT-11 (22-07-25)
72. Given below are two statements: One is (4) He studied “mutations” for the first time
labeled as Assertion (A) and the other is in organisms
labeled as Reason (R). 77. Given below are two statements:
Assertion (A): A father will pass the gene for Statement-I: Currently, many different
PKU to his son or daughter probes are used to generate DNA
Reason (R): PKU is an autosomal disorder fingerprints.
Statement-II: DNA polymorphisms play
In the light of above statements, choose the
very important role for evolution and
most appropriate answer from the options
speciation.
given below.
In the light of the above statements, choose
(1) Both A and R are correct and R is the the correct answer from the options given
correct explanation of A below:
(2) Both A and R are correct but R is NOT (1) Both Statement I and Statement II are
the correct explanation of A true
(3) A is not correct but R is correct (2) Statement I is false but Statement II is
(4) A is correct but R is not correct true
(3) Statement I is true but Statement II is
73. Sequencing the whole set of genome that false
contained all the coding and non-coding
sequence and later assigning different regions (4) Both Statement I and Statement II are
in the sequence with functions are referred to false
as 78. Polymorphism in DNA sequence is the basis
of
(1) Sequence annotation
(1) Only genetic mapping
(2) SNPs
(2) Genetic equilibrium
(3) ESTs (3) Only DNA fingerprinting
(4) RFLPs (4) 1 and 3
74. The fewest number of genes are present on 79. Choose the incorrect statement with respect
(1) Chromosome 15 to satellite DNA
(2) X-chromosome (1) It forms some portion of human genome
(3) Chromosome 9 (2) It is not inheritable
(4) Y-chromosome (3) It does not code any protein
75. A man with normal vision whose father was (4) It was used by Alec Jeffreys as probe
colourblind married a woman whose father 80. Which of the following was not a goal of
was also colourblind. Suppose their first HGP?
child is daughter then what are the chances of (1) Improve tools for data analysis
this child to be colourblind?
(2) Transfer related technologies to other
(1) 100% (2) 0% sectors
(3) 50% (4) 25% (3) Improve the technique of DNA
76. Choose the correct statement regarding fingerprinting
Frederick Sanger
(4) Address ELSI that may arise from the
(1) He started HGP in 1986 project
(2) He is credited for developing method for
determination of amino acid sequences in 81. The largest human gene is located on
proteins (1) 1st chromosome (2) X – chromosome
nd
(3) He developed DNA fingerprinting (3) 2 chromosome (4) 13th chromosome
technique
11
 SR ELITE (SET – I) NWT-11 (22-07-25)
82. Study the following statements regarding 86. Study the following syndromes
Phenylketonuria a) Down‟s syndrome
a) Phenylalanine cannot be converted into
tyrosine b) Klinefelter‟s syndrome
b) Phenylalanine cannot be absorbed from c) Edward‟s syndrome
the small intestine d) Turner‟s syndrome
c) Tyrosine accumulates in brain
Which of the above are trisomic?
d) Phenylpyruvate is efficiently reabsorbed
by kidneys (1) c and d (2) b and d
Which of the above are correct? (3) a, b and c (4) a and d
(1) a and c (2) only a 87. Which is an incorrect match?
(3) a and b (4) c and d (1) Alanine, Glycine, Aspartic acid – Miller
83. One of the following is not true with Miller‟s
experiment (2) Biogenesis theory – Louis Pasteur
(1) Miller created similar conditions like that (3) Greek thinkers – Panspermia
of primitive atmosphere in a laboratory scale. (4) Term “organic evolution” was coined by
(2) CH4, H2, NH3 and water vapour were – Darwin
kept in spark chamber
88. Stanley Miller conducted his experiment to
(3) Maintained the temperature of spark support chemical theory of origin of life in
chamber at 8000C
(1) 1981 (2) 1990
(4) Miller observed the formation of nitrogen
bases in the experiment (3) 1997 (4) 1953
84. Given below are two statements: One is 89. Most accepted theory of Origin of Life is
labeled as Assertion (A) and the other is (1) Theory of catastrophism
labeled as Reason (R).
(2) Theory of abiogenesis
Assertion (A): DNA fingerprinting is the
basis of paternity testing, in case of disputes. (3) Theory of chemical origin of life
Reason (R): DNA polymorphisms are (4) Theory of spontaneous generation
inheritable from parents to children. 90. Given below are two statements: One is
In the light of above statements, choose the labeled as Assertion (A) and the other is
most appropriate answer from the options labeled as Reason (R).
given below.
Assertion (A): When both parents are
(1) Both A and R are correct and R is the heterozygous regarding thalassemia, their
correct explanation of A children may be affected by the disorder
(2) Both A and R are correct but R is NOT Reason (R): Thalasemia is an autosomal
the correct explanation of A dominant disorder.
(3) A is not correct but R is correct
In the light of above statements, choose the
(4) A is correct but R is not correct most appropriate answer from the options
85. All the following theories were given for given below
explaining origin of life except (1) Both A and R are correct and R is the
(1) Big bang theory correct explanation of A
(2) Both A and R are correct but R is NOT
(2) Panspermia theory the correct explanation of A
(3) Spontaneous generation theory (3) A is not correct but R is correct
(4) Theory of chemical origin of life (4) A is correct but R is not correct
12
 SR ELITE (SET – I) NWT-11 (22-07-25)
PHYSICS (3) ˆ
30iN (4)  30iˆN
91. A charged particle moving along positive x- 94. A semicircular wire of radius „r‟ is kept with
direction with a velocity „ vi ‟ enters a region half of it in uniform magnetic field of
where there is a uniform magnetic field induction B and remaining half kept in u.m.f
B   Bkˆ, from x = 0 to x = d. The particle of induction 2B carrying a current I as
gets deflected gets deflected at an angle  shown. Net magnetic force experienced by
from its initial path. The specific charge of the wire is
the particle is
Bd v tan 
(1) (2)
v cos  Bd
B sin  v sin 
(3) (4)
vd Bd
92. The segment AB of wire carrying current I1
is placed perpendicular to a long straight wire (1) 3 BIr (2) 2 BIr
carrying current I 2 as shown in figure. The (3) 8 BIr (4) 10 BIr
magnitude of force experienced by the
straight wire AB is 95. Magnetic field in a region is given by
B   B0 x  kˆ . Two loops each of side „a‟ is
placed in this magnetic region in the x-y
I1 I2 plane with one of its sides on x-axis. If F1 is
A B the force on loop 1 and F2 be the force on
a loop 2, then
2a

0 I1I 2 0 I1I 2
(1) ln 3 (2) ln 2
2 2
2  0 I1 I 2 0 I1I 2
(3) (4)
2 2
93. A wire carrying a current of 3 A is bent in the (1) F1  F2  0 (2) F1  F2
form of a parabola y 2  4  x as shown in (3) F2  F1 (4) F1  F2  0
figure, where x and y are in metre. The wire
96. A charged particle with specific charge „s‟
is placed in a uniform magnetic field B  5kˆ
moves undeflected through a region of space
tesla. The force acting on the wire is
containing mutually perpendicular and
uniform electric and magnetic fields E and B.
When the electric field is switched off, the
particle will move in a circular path of radius
E Es
(1) (2)
Bs B
Es E
(3) 2
(4)
B B2 s

(1) ˆ
60iN (2)  60iˆN
13
 SR ELITE (SET – I) NWT-11 (22-07-25)
97. Equal currents are flowing in four infinitely 99. The magnetic flux threading a metal ring
long wires. Distance between any two varies with time „t‟ according to
adjacent wires is same and directions of
currents are shown in figure. Match the
 
B  3 at 3  bt 2 T  m2 with a = 2.00 s
3

following two columns. and b = 6.00 s 2 . The resistance of the ring

is 3.0  . Determine the maximum current
induced in the ring during the interval from
t = 0 to t = 2.0 s.
(1) 3.0 A (2) 12.0 A
(3) 6.0 A (4) 4.5 A
100. A uniform magnetic field B exists in a
cylindrical region of radius 10 cm as shown
in figure. A uniform wire of length 80 cm
and resistance 4.0  is bent into a square
frame and is placed with one side along a
Column I Column II
diameter of the cylindrical region. If the
A. Force on wire-1 P. Inwards magnetic field increases at a constant rate of
0.010 T/s, find the current induced in the
B. Force on wire-2 Q. Leftwards frame
C. Force on wire-3 R. Rightwards
D. Force on wire-4 S. Zero
(1) A-Q ; B-P ; C-R ; D-S
(2) A-S ; B-R ; C-Q ; D-P
(3) A-R ; B-P ; C-Q ; D-S
(4) A-Q ; B-R ; C-Q ; D-R
98. A proton is moving with constant speed
towards a conducting ring placed in the same (1) 3.9 105 A (2) 5 105 A
plane along the path OP. The direction of
path of proton ahead of „P‟ is (3) 3.9 105 A (4) 1106 A
101. A conducting wire frame is placed in a
magnetic field which is directed into the
paper. The magnetic field is increasing at a
constant rate. The directions of induced
currents in wires AB and CD are

(1) Along A
(2) Along B
(3) Along C (1) B to A and D to C
(4) Along any path depending upon speed of (2) A to B and C to D
proton (3) A to B and D to C
(4) B to A and C to D
14
 SR ELITE (SET – I) NWT-11 (22-07-25)
102. Two coaxial solenoids 1 and 2 of the same 106. An infinitely long, straight conductor AB is
length are set so that one is inside the other. fixed and an upward current is passed
The number of turns per unit length are n1 through it. Another movable straight wire CD
and n 2 . The currents i1 and i2 are flowing in of finite length and carrying current is held
opposite directions. The magnetic field inside perpendicular to it and released. Neglect
the inner coil is zero. This is possible when weight of the wire
a) i1  i2 and n1  n 2
b) i1  i2 and n1  n 2
c) i1  i2 and n1  n 2
d) i1 n1  i2 n 2
(1) only „a‟ (2) only „c‟ (1) The rod CD will move upwards parallel to
itself
(3) only „d‟ (4) only „c‟ and „d‟
(2) The rod CD will move downward parallel
103. An ionized gas contains both positive and
to itself
negative ions. If it is subjected
(3) The rod CD will move upward and turn
simultaneously to an electric field along the
clockwise at the same time
negative X-direction and a magnetic field
(4) The rod CD will move upward and turn anti
along the +Z direction, then
–clockwise at the same time
(1) Positive ions deflect towards +y direction
and negative ions towards –y direction 107. A uniform conducting wire ABC has a mass
of 10g. A current of 1A flows through it. The
(2) All ions deflect towards +y direction
wire is kept in a uniform magnetic field
(3) All ions deflect towards –y direction
B  2T . The acceleration of the wire will be
(4) Positive ions deflect towards –y direction
and negative ions towards +y direction
104. A particle of mass m and charge q moves
with a constant velocity v along the positive x
direction. It enters a region containing a
uniform magnetic field B directed along the
(1) Zero
negative z direction, extending from x = a to
x = b. The value of v required so that the (2) 12 ms 2 along y-axis
particle can enter the region x  b is (3) 6 ms 2 along y-axis
q (b  a ) B q (b  a ) B
(1) v  (2) v  (4) 0 .6  10 3 ms 2 along y - axis
m m
q (b  a ) qB 108. A proton (Q) accelerated by P.D. and enter
(3) v  (4) v  with velocity „v‟ though a transverse
mB (b  a)m
magnetic field of „B‟ as shown in figure, if
105. A long wire AB is placed on a table. Another 2mV
wire PQ of mass 1.0 g and length 50 cm is length of the field d  , the angle 
Bq
set to slide on two rails PS and QR. A current
of 50A is passed through the wires. At what through which the proton deviates from the
distance above AB, will the wire PQ be in initial direction of its motion is
equilibrium x x x x
x x x x
B
x x x x
x x x x
Q
x x x x
(1) 25 mm (2) 50 mm x x x x
(3) 75 mm (4) 100 mm (1) 30° (2) 180°
(3) 90° (4) 60°
15
 SR ELITE (SET – I) NWT-11 (22-07-25)
109. A particle of mass M and charge Q moving 10–4 A (Anticlockwise)

with velocity v describes a circular path of (4) 2  10–4 A (Anticlockwise),
radius R when subjected to a uniform 10–4 A (Anticlockwise)
transverse magnetic field of induction B. The 112. A current-carrying, straight wire is kept
work done by the field when the particle along the axis of a circular loop carrying a
completes one half circle is current. The straight wire
 Mv 2  (1) will exert an inward force on the
(1) B Qv  R (2)  R  2R circular loop
 
(2) will exert an outward force on the
(3) Zero (4) BQv.2R
circular loop
110. A highly conducting ring (viewed from left)
(3) will not exert any force on the circular
of radius R is perpendicular to and concentric
loop
with the axis of a long solenoid as shown in
fig. The ring has a narrow gap of width d in (4) will exert a force on the circular loop
its circumference. The solenoid has cross parallel to itself
sectional area A and a uniform internal field 113. The phosphor bronze string used in Moving
of magnitude B0. Now beginning at t = 0, the coil galvanometer to suspend the coil in it
solenoid current is steadily increased to so should posses. (Y-young‟s modulus, n –
that the field magnitude at any time t is given rigidity modulus)
by B(t) = (B0 + t), where   0 . Assuming (1) high Y, high n (2) low Y, low n
that no charge can flow across the gap. At (3) high Y, low n (4) low Y, high n
which end of ring which has excess of
114. A magnetic field of 2 102 T acts at right
positive charge ? Find the magnitude of
induced e.m.f. in the ring are respectively angles to coil of area 100cm2 with 50 turns.
If the average emf induced in the coil in 0.1V
when it is removed from the field in time t.
The value of „t‟
(1) 0.4S (2) 0.3S
(3) 0.2S (4) 0.1S
(1) X, A (2) X, R2
115. For the given figure, a charged particle of
(3) Y, AB0 (4) Y, R2 mass m and charge q starts sliding from rest
111. Two circular loops made of thin wires of on a vertical fixed perfectly smooth circular
resistance R = 50 milli ohm/metre are located track of radius R from the position shown
in a uniform magnetic field perpendicular below. There exists a uniform and constant
into the plane of the figures and which horizontal magnetic field of induction B
decrease at the rate dB/dt = 0.1 m T/s. Then acting outward as shown in the figure. The
currents in the inner and outer boundary are. maximum force exerted by the track on the
(The inner radius a = 10 cm and outer radius particle is
b = 20 cm)

(1) 10–4 A (Clockwise),

2  10–4 A (Clockwise)
(2) 10–4 A (Anticlockwise), (1) mg (2) 3mg  qB 2 gR
2  10–4 A (Clockwise)
(3) 2  10–4 A (clockwise), (3) 3mg  qB 2 gR (4) mg  qB 2 gR
16
 SR ELITE (SET – I) NWT-11 (22-07-25)
116. Three long wires of resistances in the ratio 121. A coil having 20 turns is suspended in a
3 : 4 : 5 are connected in parallel to each magnetic field of 0.2 wb/m2 of a
other as shown in figure. If net force on Galvanometer by a suspension wire of couple
d1 per unit twist 10-6 N-m/degree. If the least
middle wire is zero then will be measurable deflection is 0.1° then the least
d2
measurable current is (area of coil is 103 m2 )
(1) 25 mA (2) 50 mA
(3) 25 µA (4) 50 µA
122. When a magnet is moved quickly towards a
closed coil the induced emf and charge are
„e‟ and „q‟. When the same magnet is moved
slowly towards the same coil, the induced
emf and charge are e1 and q1 respectively.
(1) 9.25 (2) 5:3
Then
(3) 5: 3 (4) 1 : 1
(1) e  e1, q  q1 (2) e  e1, q  q1
117. The magnetic flux passing through a coil

placed in magnetic field   5t 2  2t  10  (3) e  e1, q  q1 (4) e  e1 , q  q1
123. A moving coil galvanometer has a resistance
weber. The emf induced in coil at t = 2 sec is
of 900  . In order to convert into ammeter
(1) 22 volt (2) 10 volt and send only 10% of the current through
(3) 30 volt (4) 50 volt the galvanometer, a shunt resistance is
connected across it, then resistance of
118. A solenoid of length 1m and having 200
ammeter is nearly
turns of wire carries a current of 2A. A thin
(1) 0.9  (2) 100 
coil having 10 turns of wire and of radius
(3) 405  (4) 90 
0.1m carries a current of 2A. Calculate the
torque required to hold the coil in the middle 124. A galvanometer of 25  resistance can read a
of the solenoid with its axis perpendicular to maximum current of 6 mA. It can be used as
the axis of the solenoid voltmeter to measure maximum of 6V by
connecting a resistance to the galvanometer.
(1) 6 106 N (2) 32 105 Nm Identify the correct choice in the given
answers.
(3) 9.54 106 Nm (4) 5.9 108 Nm
(1) 1025  in series
119. A tangent galvanometer is connected directly
to an ideal battery. If the number of turns in (2) 1025  in parallel
the coil is doubled, the deflection will (3) 975  in series
(1) Increase (4) 975  in parallel
125. The deflection in a galvanometer falls to
(2) Decrease
1
(3) Remain unchanged th when a resistance of 5  is connected
10
(4) Either increase or decrease in parallel with it. If an additional resistance
120. Two identical charged particles enter a of 3  is connected in parallel to
uniform magnetic field with same speed but galvanometer, the deflection falls to ____ of
at angles 30° and 60° with field. Let a, b and intial
c be the ratio of their time periods, radii and 1 2
(1) th (2) th
pitches of the helical paths than 6 25
(1) abc = 1 (2) abc > 1 1 1
(3) th (4) th
(3) abc < 1 (4) a = bc 25 12

17
 SR ELITE (SET – I) NWT-11 (22-07-25)
126. Lenz‟s law is in accordance with law of 130. A vertical circular coil of one turn and radius
conservation of 12.56 cm is placed with its plane in the
magnetic meridian and short magnetic needle
(1) charge (2) energy
is pivoted at the centre of the coil so that it
(3) mass (4) momentum can freely rotate in the horizontal plane. If a
current of 8A is passed through the coil, then
127. A square loop of side „a‟ hangs from an
insulating hanger of spring balance. The 
the needle deflects by BH  4 105T 
magnetic field of strength B occurs only at
the lower edge. It carries a current I. Find (1) 30° (2) 45°
the change in the reading of the spring (3) 60° (4) 90°
balance if the direction of current is reversed.
131. A galvanometer, having a resistance of 50 
gives a full scale deflection for a current of 0.05
A. The length in meter of a resistance wire of
area of cross-section 2.97× 10–2 cm2 that can be
used to convert the galvanometer into an
ammeter which can read a maximum of 5 A
current is (Specific resistance of the wire = 5 ×
10 7 m)

(1) 9 (2) 6
(3) 3 (4) 1.5
132. Some magnetic flux is changed from a coil of
(1) IaB (2) 2 IaB resistance 10 . As a result an induced
current is developed in it, which varies with
IaB 3 time as shown in figure. The magnitude of
(3) (4) IaB
2 2 change in flux through the coil in webers is
128. A galvanometer of resistance 50  is
connected to a battery of 3 V along with a
resistance of 2950  in series. A full scale
deflection of 30 divisions is obtained in the
galvanometer. In order to reduce this
deflection to 20 divisions the resistance in
series should be
(1) 4450  (2) 5050 
(3) 5550  (4) 6050 
(1) 2 (2) 4
129. A galvanometer of resistance, G is shunted
by a resistance S ohm. To keep the main (3) 6 (4) 8
current in the circuit unchanged, the 133. A moving coil galvanometer has 30 turns and
resistance to be put in series with the
galvanometer is area 1.5 x 103 m 2 . It is placed in radial field
0.25T using suspension wire of C =
S2 SG 103 Nm / deg ree. The current sensitivity of
(1) (2)
 S  G S  G the galvanometer is
(1) 41.25 degree/A (2) 31.25 degree/A
G2 G
(3) (4) (3) 21.25 degree/A (4) 11.25 degree/A
 S  G S  G
18
 SR ELITE (SET – I) NWT-11 (22-07-25)

134. A square loop ACDE of area 20 cm 2 and (4) A – III, B – I, C – II

resistance 5  is rotated in a magnetic field 138. Incorrect statement(s) among the following
B  2T through 180° in 0.02 s is/are
I) Potassium dichromate and potassium
permanganate are primary standards in
volumetric analysis.
II) Sodium hydroxide is a second standard
in volumetric analysis.
III) Quinonoid form of phenolphthalein is
pink coloured.
Find the magnitude of induced current
(1) 0.04A (2) 0.08A IV) KMnO4 acts as self indicator in a
(3) 0.32A (4) 0.16A redox titration.
135. In half deflection method, a galvanometer of (1) I only (2) I, II only
resistance 200  is connected to a battery of (3) I, III only (4) II, III only
potential difference „V‟ with a series 139. Volume of 0.1 M KMnO4 solution required
resistance 50  the deflection observe is  . to completely oxidise 20 mL of 0.05 M
When a shunt resistance „S‟ is connected oxalic acid solution in acidic medium is
across galvanometer, the deflection found to
(1) 10 mL (2) 4 mL

be , the value of „S‟ is (3) 20 mL (4) 8 mL
2
140. Given below are two statements. One is
(1) 40  (2) 50 
labeled as assertion (A) and other is labeled
200 50
(3)  (4)  as reason (R)
3 3 Assertion (A) : Hydrochloric acid can not be
CHEMISTRY used as medium in the titration of potassium
136. According to Ostwald‟s theory, methyl permanganate solution and Mohr‟s salt
orange and phenolphthalein are solution.
(1) Weak acid and strong acid respectively Reason (R) : Hydrochloric acid can reduce
(2) Weak base and salt of weak base potassium permanganate.
respectively In the light of the above statements, choose
(3) Weak base and strong base respectively the correct answer from the options given
below:
(4) Weak base and weak acid respectively
(1) Both A and R are true, and R is the
137. Match list – I with list – II
correct explanation of A.
List – I List – II
(2) Both A and R are true, but R is not the
(Titration) (Suitable indicator) correct explanation of A.
Methyl orange (3) A is true, but R is false
HCOOH vs
A. I or
NaOH (4) A is false, but R is true
phenolphthalein
B. HBr vs KOH II. Methyl red 141. During the preparation of Mohr‟s salt
solution (Ferrous ammonium sulphate),
HNO3 vs
C. III. Thymol blue which of the following acid is added to
NH 4OH prevent hydrolysis of Fe2 ion?
Choose the correct answer from the options (1) Dilute nitric acid
given below.
(2) Dilute sulphuric acid
(1) A – II, B – I, C – III
(2) A – I, B – III, C – II (3) Dilute hydrochloride acid
(3) A – I, B – II, C – III (4) Concentrated sulphuric acid
19
 SR ELITE (SET – I) NWT-11 (22-07-25)
142. The elements having highest and the lowest (2) C > Ge > Si > Pb > Sn
melting points in group 13 are respectively (3) C > Pb > Si  Ge  Sn
(1) B, Al (2) Tl, B (4) Sn > Pb > Si > Ge > C
(3) B, Ga (4) B, Tl 147. Hybridisation of boron in BCl3 .NH 3 is
143. Given below are two statements:
(1) sp 2 (2) sp3
Statement – I : Aluminium dissolves in
dil.HCl as well as in aqueous solution of (3) sp 3 d (4) sp3d 2
NaOH evolving H 2 . 148. Match list – I with list – II
Statement – II : Aluminium forms List – I List – II
amphoteric Al2O3 on heating in air. (Ore) (Formula)
In the light of the above statements, choose A. Kernite I. Na3 AlF6
the correct answer from the options given B. Cassiterite II. Na2 B4O7 .4 H 2O
below:
(1) Both the statements I and II are correct C. Cryolite III. SnO2
(2) Both the statements I and II are D. Galena IV. PbS
incorrect Choose correct answer from the options
(3) Only statement I – is correct given below
(4) Only statement II – is correct (1) A – I, B – II, C – IV, D – III
144. Match list – I with list – II (2) A – II, B –I, C – III, D – IV
List – I List – II (3) A – II, B – III, C – I, D – IV
(Element) (Property) (4) A – I, B – II, C – III, D – IV
A. B I. Least electronegativity 149. Correct statements among the following are
B. Al II. Positive value of E 0 3 I) IE1 of C > B
M /M

C. Tl III. Non metal II) Covalent radius C > B

D. In IV. Least value of IE1 III) E.N of C > B
IV) Density of diamond > graphite
Choose correct answer from the options
given below (1) I, III only (2) I, II, III only
(1) A – III, B – I, C – II, D – IV (3) II, IV only (4) I, III, IV only
(2) A – II, B – III, C – I, D – IV 150. Set of only amphoteric oxides in the options
(3) A – III, B – II, C – I, D – IV given below are
(4) A – II, B – IV, C – III, D – I (1) Ga2O3 , SnO, PbO2 (2) Al2O3 , CO, CO2
145. Incorrect statement among the following is
(3) SnO2 , In2O3 , CO2 (4) B2O3 , CO2 , SiO2
(1) Ultrapure form of germanium and
silicon are used to make semiconductor 151. Correct combinations with respect to
devices. hybridization of central atom is/are
(2) Covalent radius of Pb is slightly less I) SiF62  d 2 sp 3
than that of Sn
(3) Silicon is the most abundant element in II) SiCl4  sp 3
14th group
 Sn  OH  
2
III)  sp 3d 2
(4) IE1 of Si is slightly greater than that of 6

Ge. IV) CO2  sp

146. Corect order of electronegativity of group 14
(1) I, II, IV, V only (2) II, III, IV only
elements is
(1) C > Si > Ge  Sn  Pb (3) II, IV only (4) I, V only

20
 SR ELITE (SET – I) NWT-11 (22-07-25)
152. Correct order of bond enthalpies is 158. Maximum covalence of carbon is limited to 4
(1) C – C > Si – Si > Ge – Ge > Sn – Sn because of
(2) C – C < Si – Si < Ge – Ge < Sn – Sn (1) availability of s and p orbitals only
(3) C – C > Ge – Ge > Si – Si > Sn – Sn (2) its non metallic nature
(4) C – C > Sn – Sn > Ge – Ge > Si – Si (3) having low bond energy
153. Given below are two statements. One is (4) having high electronegativity
labeled as assertion (A) and other is labeled 159. The products A & B in the following reaction
as reason (R) is
Assertion (A) : PbI 4 does not exist. O  C2 H 5
Reason (R) : Pb – I bond initially formed  
    A B
conc . HBr Excess
during the reaction does not release enough
energy to unpair 6s 2 electrons and excite one
of them to higher orbital.
In the light of the above statements, choose CH  CH 2
the correct answer from the options given (1)
below:
Br
(1) Both A and R are true, and R is the
correct explanation of A.  C 2 H 5O H
(2) Both A and R are true, but R is not the
correct explanation of A.
(3) A is true, but R is false.
CHBr  CH 3
(4) A is false, but R is true .
(2)
154. Correct order with respect to stability of
oxidation states is OH
(1) Sn2  Pb2 (2) Sn4  Pb4  C2 H 5 I
3 3 2 4
(3) Ga  In (4) Si  Si
155. Which of the following species is not stable ?
 SiF6  GeCl6 
2 2
(1) (2) CH 2  CH 2 Br

 SiCl6 
2
 Sn  OH 6 
2 (3)
(3) (4)
OH
156. Which of the following is incorrect
statement?  C2 H 5 I
(1) PbF4 is covalent in nature
(2) SiCl4 is easily hydrolysed
CHBr  CH 3
(3) GeX 4  X  F , Cl , Br  is more stable
(4)
than GeX 2
OH
(4) SnF4 is ionic in nature
157. Hybridisation of Al in monomeric form of  C2 H 5OH
AlCl3 and dimeric form are respectively
(1) sp 2 , sp 2 (2) sp 2 , sp 3
CH 2  CH 2 Br
(3) sp 3 , sp 3 (4) sp 3 , sp 2
21
 SR ELITE (SET – I) NWT-11 (22-07-25)
160. Statement – I : Both Reimer – Tiemann (1) A, B only correct (2) Only A correct
reaction and Kolbe‟s reaction of phenol are
(3) Only C correct (4) A, B & C correct
carried out in alkaline medium.
Statement – II : Phenoxide is more reactive 164. When CH 2  CH  O  CH 2  CH 3 heated
than phenol towards electrophilic aromatic strongly with one mole of HI, one of the final
substitution reaction. product formed is
(1) Both the statements I and II are correct (1) Ethane (2) Ethanol
(2) Both the statements I and II are (3) Ethanal (4) Iodoethene
incorrect 165. Oxilation of phenol with chromic acid
(3) Only statement I – is correct produces product (A). The number of sigma
and pi bonds present in the product A
(4) Only statement II – is correct respectively are
161. P  Q  Anisole  HI

 R  S; correct (1) 8 & 4 bonds (2) 8 & 2 bonds
statement among the following is
(3) 12 & 2 bonds (4) 12 & 4 bonds
(1) P & Q are C6 H 5  ONa & C2 H 5Cl
166. Which of the following alcohols gives the
(2) R & S are C6 H 5Cl and CH 3OH best yield of dialkyl ether on being heated
with traces of sulphuric acid?
(3) R & S are C6 H 5OH & CH 3 I
(1) 2 – propanol
(4) P & Q are C6 H 5Cl and CH 3ONa (2) 1 – propanol
162. Major product of the given reaction are (3) 2 – butanol
CH 3 (4) 2 – methyl – 2 – propanol
CH 3  C  O  CH 2 CH 3 

HI 1eq
167. Tetra hydro furan on heating with excess of
con HI, the final product formed is
CH 3
(1) CH2OH CH2 2  CH2 I
(1) CH3 3 CI  C2 H5OH
(2) CH2 I CH2 3  CH2 I
(2) CH3 3 C  OH  C2 H5  I
(3) CH2OH CH2 2  CH2OH
(3) CH3 3 COH  CH2  CH2
(4) CH2 I CH2 2  CH2 I
(4) CH3 3 CI  CH3  CH3  CH CO  O
168. Phenol 
NaOH
 A 
1) CO2
2) H 
 B 
3
H
2
C
163. Which of the following statement (s) are
CORRECT regarding Williamson‟s synthesis  CH 3COOH
of ether
INCORRECT statement among the
A) Better result obtained if the alkyl halide following is
used is primary without branching at  – C
– atom (1) B is steam volatite

B) Better result obtained if the alkyl halide (2) C can be used as anti inflammatory
used is tertiary & alkoxide is primary. (3) C has free OH group on benzene ring
C) Beter result obtained if the alkyl halide (4) Formation of B from phenol is Kolbe‟s
is tertiary and alkoxide is tertiary reaction
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 SR ELITE (SET – I) NWT-11 (22-07-25)
169. Total number of possible cyclic ether isomers
(both structural and stereo) of molecular
formula C4 H 8O is

(1) 6 (2) 8
CH 3O
(3) 10 (4) 12 (3) – 3,3 – diethyl – 1
– methoxy cyclo hexane (IUPAC name)
170. Which of the following compounds is
aromatic alcohol? (4) C2 H5  O  CH2  CH CH3 2 –
OH 1 – ethoxy – 2 – methyl propane
OH 174. Given below are two statements. One is
labeled as assertion (A) and other is labeled
as reason (R)
Assertion (A) : Preparation of ethers by
CH 3 acidic dehydration of secondary and tertiary
A) B)
alcohols is not suitable method
CH 2  OH CH 2  OH
Reason (R) : Secondary and tertiary alcohols
to give ethers is unsuccessful as elimination
competes over substitution to form alkenes

CH 3 (1) Both A and R are true, and R is the

C) D) correct explanation of A
(1) A & B only (2) C & D only (2) Both A and R are true, but R is not the
(3) A, B, C and D (4) A only correct explanation of A

CH3  O  C CH3 3 
171. A  B  HI

 X Y (3) A is true, but R is false
(4) A is false, but R is true
Correct statement among the following is
O
(1) A & B are CH3  O Na & CH3 3 CBr 
C 6 H 5  C  OH  C 2 H 5  OH  H

A
175.
(2) X & Y are CH 3 I & CH3 3  COH
O
X & Y are CH3OH &CH3 3 CI

(3) CH 3  CH 2  C  OH  C6 H 5  OH 
H

B

(4) A & B are CH3OH & CH3 3 COH Consider the above reactions, INCORRECT
statements are
172. Which of the following reagents can
distinguish ethyl methyl ether and isopropyl A) Both A & B are metamers
alcohol? B) A & B readily undergo electrophilic
(1) Br2 / CCl4 (2) I 2 / NaOH aromatic substitution compared with benzene
C) A & B are same compounds
(3) Neutral FeCl3 (4) AgNO3 / NH 4OH
D) Both A & B are esters
173. Wrong match among the following is
(1) Only A & B correct
(1) C6 H5  O  CH2  CH2  CH CH3 2 -
(2) Only B & C correct
3-methyl butoxy benzene (IUPAC name)
(3) Only A & D correct
(2) C6 H 5  O  CH 2  CH 3 - Phenetole
(4) Only A, B and D are correct
(Common name)
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 SR ELITE (SET – I) NWT-11 (22-07-25)
176. Statement – I : Ethanol is known as wood C) C6 H 5  CH  CH  CHO
spirit.
(1) A & C only (2) A & B only
Statement – II : Sugar in molasses is
converted to glucose and fructose in presence (3) B & C only (4) A, B & C
of enzyme zymase .
180. Incorrect match among the following is
(1) Both the statements I and II are correct
Formula IUPAC name
(2) Both the statements I and II are
incorrect CH 3 CHO
3 – methyl
(3) Only statement I – is correct (1) cyclohexane
carbaldehyde
(4) Only statement II – is correct
177. Which of the following pair of reagents could
be used to prepare 2 – phenyl – 2 - butanol CHO
Benzene – 1, 2 –
(1) CH 3  CH 2  MgBr & C6 H 5  CO  CH 3 (2)
dicarbaldehyde
CHO
(2) CH 3  CH 2  MgBr  C6 H 5  CH 2  CHO

(3) CH 3  MgBr & C6 H 5  CH 2  CO  CH 3 4 – Methyl pent –

(3) CH3 2 C  CH  COCH3
3 – en – 2 – one
(4) C6 H 5  MgBr & CH 3  CO  CH 3

178. How many primary alcohols (only structural) CHO  CH 2  CH  CH 2  CHO 2 – formyl propane
are possible for C5 H12O ? (4) – 1, 3 -
CHO dicarbaldehyde
(1) 2 (2) 3
(3) 4 (4) 5
179. Which of the following has pleasant
fragrance?
A)

OH

OCH 3

CHO

B)
CHO

OH

24