datavorous/entrance-exam-dataset huggingface dataset:
Dishwashing soaps are synthetic detergents. What is their chemical nature?

Options:
A Cationic detergent.
B Anionic detergent.
C Non-ionic detergents
D Hydrophobic detergents

The correct answer is:
Non-ionic detergents

Solution:
Correct Option is : (C)
Non-ionic detergents

Topic: Chemistry, Chemistry in Everyday Life, NEET
Disjunction is:

Options:
A Chromosome separation during mitosis
B Chromosome separation during prophase I
C Chromosome separation in anaphase I
D Chromosome separation during metaphase I

The correct answer is:
Chromosome separation in anaphase I

Solution:
The junction is the point at which two things meet. When these get separated from each other then it is called a disjunction. In mitosis, sister chromatids get separated so it is not found in mitosis. During meiosis, chromosome number gets reduced to half in progeny so it is called reductional division also. Two processes that occur in meiosis one is chromosome separation and second is sister chromatid separation. So the separation of homologous chromosome occurs in anaphase I of meiosis I and in Anaphase II sister chromatids get separate.

Topic: Biology, Cell Cycle and Cell Division, NEET
Disjunction is:

Options:
A Chromosomes separation during mitosis
B Chromosomes separation during prophase I
C Chromosomes separation in anaphase I
D Chromosomes separation during metaphase I

The correct answer is:
Chromosomes separation in anaphase I

Solution:
Normal separation of chromosomes towards the opposite poles during anaphase-I of cell division is called a disjunction.

Topic: Biology, Cell Cycle and Cell Division, NEET
Displacement between the maximum potential energy (P.E.) position and the maximum energy (K.E.) position for a particle executing simple harmonic motion is:

Options:
A $\pm \frac{a}{2}$
B $\pm \mathrm{a}$
C $\pm 2 \mathrm{a}$
D - 1

The correct answer is:
$\pm \mathrm{a}$

Solution:
The K.E. of pendulum
$\begin{aligned}
& =\mathrm{K} . \mathrm{E} . \\
& =\frac{1}{2} K\left(a^2-y^2\right)
\end{aligned}$
and P.E. of pendulum $=\frac{1}{2} K Y^2$
when $y=0$ (mean position) K.E. is maximum and PE is 0 .
When $y=0$, K.E. is 0 and P.E. is maximum.
$\therefore$ The displacement between position of maximum potential energy and maximum kinetic energy is $\pm a$.

Topic: Physics, Oscillations, NEET, NEET 2002
Displacement - time graphs for two waves, wave - 1 and wave - 2 are shown here. The ratio of the intensity of wave - 1 to that of wave - 2 is

Options:
A 1 : 1
B 9 : 4
C 16 : 9
D 9 : 1

The correct answer is:
9 : 4

Solution:
Average intensity along the string
<I> =2π2f2A2ρv
I1I2=f1A1f2A22=T2A1T1A22=12×32=94

Topic: Physics, Waves and Sound, NEET
Displacement - time graphs for two waves, wave - 1 and wave - 2 are shown here. The ratio of the intensity of wave - 1 to that of wave - 2 is

Options:
A 1 : 1
B 9 : 4
C 16 : 9
D 9 : 1

The correct answer is:
9 : 4

Solution:
Average intensity along the string
<I> =2π2f2A2ϱv
l1l2=f1A1f2A22=T2A1T1A22=12×32=94

Topic: Physics, Waves and Sound, NEET
Displacement x (in meters), of a body of mass 1 kg as a function of time t, on a horizontal smooth surface, is given as x=2t2. Then work done in the first one second by the external force is

Options:
A 2Â J
B 4Â J
C 8 J
D 16Â J

The correct answer is:
8 J

Solution:
Given displacement is,
x=2t2
⇒v=velocity=dxdt=4t
vinitial=v t=0
= 4×0=0 m s-1
vfinal=v t=1
= 4×1=4 m s-1
∆K.E= change in K.E of body
=12mvfinal2-vinitial2
=12×1×16-0=8 J
By work - kinetic energy theorem, work done =∆K.E=8 J

Topic: Physics, Work Power Energy, NEET
Dissociation constants of CH3COOH and NH4OH are 1.8×10–5 each at 25°C. The equilibrium constant for the reaction of CH3COOH and NH4OH will be -

Options:
A 1.8×1.8 10 4
B 1.810â9
C 1.8Ã1.8Ã 10 4
D 3.24Ã 10 â10

The correct answer is:
1.8×1.8× 10 4

Solution:
CH 3 COOH + NH 4 OH ⇌ CH 3 COO - + NH 4 + + H 2 O
K h = CH 3 COO – NH 4 + CH 3 COOH NH 4 OH = K a K b K w = 1.8 2 × 10 4

Topic: Chemistry, Ionic Equilibrium, NEET
Dissociation constants of CH3COOH and NH4OH in aqueous solution are are almost same (10-5) . If pH of a CH3COOH solution is 3, what will be the pH of NH4OH ?

Options:
A 3.0
B 4.0
C 10.0
D 11.0

The correct answer is:
11.0

Solution:
pH of CH3COOH
Sol =3
So [H3O+]=c×Ka=10-3
or 10-3=c×10-5
or c=10-1 M
Kb=Ka=10-5
pH of 10-3 M
Solution of NH4OH
[OH-]=c×Kb
=10-1×10-5=10-3
pOH =3
or pH=14-3=11

Topic: Chemistry, Ionic Equilibrium, NEET
Dissociation of phosphorus pentachloride is favoured by –

Options:
A High temperature and high pressure
B High temperature and low pressure
C Low temperature and low pressure
D Low temperature and high pressure

The correct answer is:
High temperature and low pressure

Solution:
PCl5→ΔPCl3+Cl2 ,
Number of molecules in products is higher than that of reactant thus low pressure condition favours reaction in forward direction.

Topic: Chemistry, Chemical Equilibrium, NEET
Dissolution of 1.5 g of non-volatile solute (Molecular weight = 60) in 250 g of a solvent reduces its freezing point by 0.01oC . Find the molal depression constant of the solvent.

Options:
A 0.01
B 0.001
C 0.0001
D 0.1

The correct answer is:
0.1

Solution:
Depression in freezing point,
ÎTf=kfÃm
Where,
m=Molality=Weight of soluteÃ1000Molecualr weight of solute ÃWeight of solvent
=1.5Ã100060Ã250
ÎTf=0.1oC molal-1
âÎTf=kfÃ0.1
0.01=kfÃ0.1
â´ kf=0.010.1
=0.1

Topic: Chemistry, Solutions, NEET
Dissolution of a solute is an exothermic process if

Options:
A hydration energy > Lattice energy
B hydration energy < Lattice energy
C hydration energy = Lattice energy
D None of the above

The correct answer is:
hydration energy > Lattice energy

Solution:
Dissolution of a solute is an exothermic process when the magnitude of hydration energy is more than that of lattice energy.
ÎHsolution=ÎHhydration+ÎHlattice energy
ÎHhy.=-ve
ÎHlat.=+ve

Topic: Chemistry, Solutions, NEET
Dissolution of the synaptonemal complex occurs during

Options:
A Zygotene
B Diplotene
C Leptotene
D Pachytene

The correct answer is:
Diplotene

Solution:
Diplotene:
• Synapotnemal complex is dissolved at this stage.
• The recombined homologous chromosomes of the bivalent separate from each other; except at the site of crossing over.
• The X-shaped structures; thus formed; are called chiasmata.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2020 (Phase 1)
Dissolved oxygen in water is very important for aquatic life. What processes are responsible for the reduction of dissolved oxygen in water?

Options:
A Photosynthesis and respiration by aquatic plants and animals.
B Aeration and wind action on the water surface.
C Bacterial decomposition of organic matter.
D Absorption of oxygen from the atmosphere.

The correct answer is:
Bacterial decomposition of organic matter.

Solution:
Correct Option is : (C)
Bacterial decomposition of organic matter.

Topic: Chemistry, Environmental Chemistry, NEET
Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/ml. The molarity of the solution is

Options:
A 2.00 M
B 2.22 M
C 1.78 M
D 2.05 M

The correct answer is:
2.05 M

Solution:
Total mass of solution =1000+120=1120g
Total volume of solution in (L)=11201.15Ã103
M=WMÃ1V(L)=12060Ã1.15Ã1031120=2.05 M

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Distance between a frog and an insect on a horizontal plane is 10 m. The frog can jump with a maximum speed of 10 m s-1. (g = 10âm s-2) The minimum number of jumps required by the frog to catch the insect is :

Options:
A 5
B 10
C 100
D 50

The correct answer is:
10

Solution:
For the minimum number of jumps, the range must be maximum.
maximum range =u2g=10210=1 m
Total distance to be covered = 10 m
So the minimum number of jumps = 10

Topic: Physics, Motion In Two Dimensions, NEET
Distance between gene and percentage of linkage shows:

Options:
A A direct relationship
B A parallel relationship
C An inverse relationship
D No relationship

The correct answer is:
An inverse relationship

Solution:
Distance between the two genes determine strength of linkage, closely located genes showed strong linkage and widely located genes showed weak linkage. Morgan and his group found that even when genes were grouped on the same chromosome, some genes were tightly linked and showed very low recombination, while others were loosely linked showed higher recombination.

Topic: Biology, Principles of Inheritance and Variation, NEET
Distance between tetrahedral and octahedral void in fcc lattice will be
(Given edge length of unit cell = a)

Options:
A 3a4
B 3a
C 3a2
D 3a3

The correct answer is:
3a4

Solution:
Octahedral void is found at centre of cube and edge centre while tetrahedral voids are present at 1/4th of the distance along each body diagonal.
2× distance between tetrahedral and octahedral voids =3a2
So distance between tetrahedral and octahedral void =3a4 .

Topic: Chemistry, Solid State, NEET
Distance between the genes and percentage of recombination shows:

Options:
A A direct relationship with each other.
B An inverse relationship with each other.
C A parallel relationship with each other.
D They are not related to each other.

The correct answer is:
A direct relationship with each other.

Solution:
Crossing over separates genes away from each other. So, the distance between the genes and percentage of recombination shows a direct relationship, i.e., when genes are close together they have high linkage and exhibit low recombination frequencies. Thus, the other option is wrong as it does not show parallel or inverse relationship.

Topic: Biology, Principles of Inheritance and Variation, NEET
Distance in free space at which intensity of 5 eV neutron beam reduces to half will be nearly: (Take half-life of the neutron =12.8 min )

Options:
A 6000 km

B 12000 km

C 18000 km

D 24000 km

The correct answer is:
24000 km

Solution:
The kinetic energy of the neutron beam 5 eV=12 mv2
121.67×10-27v2=5×1.6×10-19
⇒v=31 km s-1
distance travelled in one half-life d=vt=31×103×12.8×60 m≃24000 km

Topic: Physics, Nuclear Physics, NEET
Distinguish between Face-centred and end-centred unit cells.

Options:
A Face-centred unit cells have atoms at the corners and in the center of each face, while end-centred unit cells have atoms only at the corners.
B Face-centred unit cells have atoms only at the corners, while end-centred unit cells have atoms at the corners and in the center of each face.
C Face-centred unit cells have atoms at the corners and in the center of each face, while end-centred unit cells have atoms at the center of each face.
D Face-centred unit cells have atoms at the corners and in the center of each face, while end-centred unit cells have atoms only at the center of each face.

The correct answer is:
Face-centred unit cells have atoms at the corners and in the center of each face, while end-centred unit cells have atoms only at the corners.

Solution:
Correct Option is : (A)
Face-centred unit cells have atoms at the corners and in the center of each face, while end-centred unit cells have atoms only at the corners.

Topic: Chemistry, Solid State, NEET
Distinguish between Hexagonal and monoclinic unit cells.

Options:
A Hexagonal unit cells have all sides and angles equal, while monoclinic unit cells have unequal sides and angles.
B Hexagonal unit cells have unequal sides and angles, while monoclinic unit cells have all sides and angles equal.
C Hexagonal unit cells have three unequal sides and angles, while monoclinic unit cells have three equal sides and angles.
D Hexagonal unit cells have three equal sides and angles, while monoclinic unit cells have three unequal sides and angles.

The correct answer is:
Hexagonal unit cells have all sides and angles equal, while monoclinic unit cells have unequal sides and angles.

Solution:
Correct Option is : (A)
Hexagonal unit cells have all sides and angles equal, while monoclinic unit cells have unequal sides and angles.

Topic: Chemistry, Solid State, NEET
Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.

Options:
A Adsorption is the process of a substance penetrating into the bulk of another substance. An example is a sponge absorbing water. Absorption is the process where molecules adhere to the surface of a solid or a liquid material. An example is activated charcoal adsorbing impurities from a liquid.
B Adsorption is the process where molecules adhere to the surface of a solid or a liquid material. An example is activated charcoal adsorbing impurities from a liquid. Absorption is the process of a substance penetrating into the bulk of another substance. An example is a sponge absorbing water.
C Adsorption and absorption are synonymous terms with no practical distinction.
D Adsorption is a chemical reaction, while absorption is a physical process.

The correct answer is:
Adsorption is the process of a substance penetrating into the bulk of another substance. An example is a sponge absorbing water. Absorption is the process where molecules adhere to the surface of a solid or a liquid material. An example is activated charcoal adsorbing impurities from a liquid.

Solution:
Correct Option is : (A)
Adsorption is the process of a substance penetrating into the bulk of another substance. An example is a sponge absorbing water. Absorption is the process where molecules adhere to the surface of a solid or a liquid material. An example is activated charcoal adsorbing impurities from a liquid.

Topic: Chemistry, Surface Chemistry, NEET
Distribution of energy in the spectrum of a black body can be correctly represented by

Options:
A wien's law
B stefan's law
C planck's law
D kirchhoff's law

The correct answer is:
planck's law

Solution:
Because Planck's law explains the distribution of energy correctly at low temperature as well as at high temperature.

Topic: Physics, Thermal Properties of Matter, NEET
Diuresis is prevented by:

Options:
A Renin from JG cell via switching off the osmoreceptors
B ANF from atria of the heart
C Aldosterone from adrenal medulla
D Vasopressin from Neurohypophysis

The correct answer is:
Vasopressin from Neurohypophysis

Solution:
The correct answer is option (4) because,
- Diuresis is prevented by vasopressin or anti-diuretic hormone from neurohypophysis or posterior pituitary. When there is excessive loss of fluid from the body the secretion of, ADH facilitates water reabsorption from latter part of renal tubules.
- Option (1) is incorrect because renin from JG cells convert angiotensinogen in blood to angiotensin I and further to angiotensin II; which is a powerful vasoconstrictor and increases the glomerular blood pressure and thereby GFR.
- Option (2) is incorrect because ANF causes vasodilation and thereby decrease GFR.
- Option (3) is incorrect because aldosterone is released from adrenal cortex and causes reabsorption of $\mathrm{Na}^{+}$and water from distal parts of renal tubules.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2024 (Re-NEET)
Diversity shown by a single species at the genetic level over its distribution range is known as

Options:
A species diversity
B community diversity
C ecological diversity
D none of these

The correct answer is:
none of these

Solution:
Genetic diversity is the diversity shown by a single species at the genetic level over its distribution range. It is the diversity in the number and types of genes as well as chromosomes present in different species and the variations in the genes and their alleles in the same species.

Topic: Biology, Biodiversity and its Conservation, NEET
Division of centromere is absent in

Options:
A Telophase
B Anaphase-I
C Anaphase-II
D All of these

The correct answer is:
Anaphase-I

Solution:
During anaphase I chromosomal pairs separates and chromosomes starts moving towards opposite poles. This separation of chromosomal pair is called disjunction and it results in reduction in number of chromosome to half. Centromere splits and chromatids are separated during anaphase of mitosis and anaphase II of meiosis.

Topic: Biology, Cell Cycle and Cell Division, NEET
D-mannose is epimeric with D-glucose at C2 . Which of the following structure represents β -D-mannopyranose?

Options:
A
B
C
D

The correct answer is:

Solution:
In D-mannose -OH at C-2 and C-3 are cis position to each other so the structure of β -D-mannopyranose can be given as

Topic: Chemistry, Biomolecules, NEET
DNA and RNA are chiral molecules in nature, their chirality is due to

Options:
A L-sugar component
B D-sugar component
C chiral phosphate ester units
D chiral bases

The correct answer is:
D-sugar component

Solution:
The constituents of nucleic acids are nitrogenous bases, sugar and phosphoric acid. The sugar present in DNA is D(-)-2-deoxyribose and the sugar present in RNA is D(-) ribose. Due to these D(-)-sugar components, DNA and RNA molecules are chiral molecules in nature.

Topic: Chemistry, Biomolecules, NEET
DNA and RNA, both are types of nucleic acids. Figure out the correct statements concerning DNA and RNA from below.

I. Adenine and guanine are bases that have a double ring structure; cytosine, thymine and uracil are bases with a single ring structure.
II. An adenine nucleotide from DNA is the same as an adenine nucleotide from RNA; in DNA, adenine pairs with thymine and in RNA adenine pairs with uracil.
III. The base pairing that occurs in a double DNA helix and when RNA is synthesised during transcription, it follows the rule that a purine pairs with a pyrimidine.
IV. The two polynucleotide strands of a DNA molecule run in opposite directions so that the double helix formed has two strands that are parallel to each other.

Options:
A I, II and III only
B I, II and IV only
C II, III and IV only
D I, III and IV only

The correct answer is:
I, III and IV only

Solution:
Adenine and guanine are purines which have two-ring structure whereas thymine, uracil and cytosine are pyrimidines having single ring structure. The nucleotide of DNA and RNA are not same as the sugar molecule is different. Ribose in RNA and deoxyribose in DNA.

Topic: Biology, Molecular Basis of Inheritance, NEET
DNA can be introduced into any cell by:

Options:
A Injection
B Being complexed with calcium salts
C Being placed along with the cell into a gene gun
D Gel electrophoresis

The correct answer is:
Being complexed with calcium salts

Solution:
Bacterial wall is made permeable, and hence competent, by treatment with calcium chloride or lysozyme. The process is called as electroporation.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Dna can be visualized in agarose gel by applying:

Options:
A Ethidium bromide
B Plasmid
C Eco r1
D Protease

The correct answer is:
Ethidium bromide

Solution:
DNA is made visible by staining it with Ethidium bromide, which on exposure to UV rays, appears to be orange-coloured bands.

Topic: Biology, Biotechnology - Principles and Processes, NEET
DNA chemically is less reactive and structurally more stable when compared to RNA
stability as a genetic material?

Options:
A DNA has a reactive 2'-OH group at every nucleotide
B DNA is a better catalytic material than RNA
C The two strands of DNA are complementary
D DNA is less reactive and structurally more stable than RNA

The correct answer is:
DNA is less reactive and structurally more stable than RNA

Solution:
DNA is less reactive and structurally more stable when compared to RNA due to the presence of a reactive 2'-OH group at every nucleotide in RNA, making it labile and easily degradable.

Topic: Biology, Molecular Basis of Inheritance, NEET
DNA dependent RNA polymerase catalyzes the transcription of one strand of the DNA which is called the

Options:
A template strand.
B coding strand.
C alpha strand
D non-template strand.

The correct answer is:
template strand.

Solution:
Transcription is formation of RNA from DNA. RNA polymerases catalyses the formation of complementary RNA on one strand of DNA known as template strand. The strand which do not undergo transcription is called as coding strand or non-template strand.

Topic: Biology, Molecular Basis of Inheritance, NEET
DNA fingerprinting involves identifying differences in some specific regions in DNA sequence, called as:

Options:
A Satellite DNA
B Repetitive DNA
C Single nucleotides
D Polymorphic DNA

The correct answer is:
Repetitive DNA

Solution:
DNA fingerprinting is a very quick way to compare the DNA sequences of any two individuals. DNA fingerprinting involves identifying differences in some specific regions in a DNA sequence called repetitive DNA because, in these sequences, a small stretch of DNA is repeated many times.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2021
DNA fingerprinting is a technique used in generating a profile of an individual. This technique identifies the various satellite DNA found in the sequence and has high degree of

Options:
A Metamorphism
B Polymorphism
C Biomorphism
D Pathomorphism

The correct answer is:
Polymorphism

Solution:
DNA fingerprinting involves identifying differences in some specific regions in a DNA sequence called satellite DNA. These sequences of satellite DNA normally do not code for any proteins and they show a high degree of polymorphism which forms the basis of DNA fingerprinting. Polymorphism is the ability to take any form or change in sequence (viz., SNP). Metamorphism is the phenomenon of the transition of certain animal groups from the larval stage to the adult stage. (viz. tadpole to frog). Pathomorphism is a term given for abnormal morphology shown in any part of body and biomorphism is an art form imitating the living organisms.

Topic: Biology, Biotechnology - Principles and Processes, NEET
DNA fingerprinting recognizes the differences in

Options:
A satellite DNA
B bulk DNA
C repetitive DNA
D both a and c

The correct answer is:
both a and c

Solution:
DNA fingerprinting or DNA typing/genetic fingerprinting or DNA profiling is a technique to identify a person on the basis of person DNA specificity. The technique is based upon the fact that the DNA constitution of an individual carries some specific sequence of nucleotides, which do not carry any information for protein synthesis. These specific sequences are repeated many times and found at many places within the genome. These loci are known as the Variable number tandem repeats (VNTR`s) or minisatellites with repetitive sequence.

Topic: Biology, Biotechnology - Principles and Processes, NEET
DNA fingerprinting refer to:

Options:
A molecular analysis of profiles of DNA samples
B analysis of DNA samples using imprinting devices
C techniques used for molecular analysis of different specimens of DNA
D techniques used for identification of fingerprints of individuals.

The correct answer is:
molecular analysis of profiles of DNA samples

Solution:
DNA fingerprinting involves determining the genetic makeup of a person. Determining and analysing the repeated sequences (as in STR), form the basis of DNA fingerprinting.
Related Theory
Sir Alec Jeffreys is the father of DNA fingerprinting. DNA (deoxyribonucleic acid) represents the blueprint of the human genetic makeup. A DNA fingerprint, therefore, is a DNA pattern that has a unique sequence such that it can be distinguished from the DNA patterns of other individuals. DNA fingerprinting is also called DNA typing.
Caution
Earlier, VNTRs (Variable Number Tandem Repeats) sequence were analysed during the process. However, with advancement of technology, STRs (Short Tandem Repeats) are studied present, which provide better results than VNTR.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2004
DNA fingerprinting refers to

Options:
A Analysis of DNA samples using imprinting devices
B Techniques used for the amplification of different specimens of DNA
C Techniques used for identification of fingerprints of individuals
D Molecular analysis of profiles of DNA samples

The correct answer is:
Molecular analysis of profiles of DNA samples

Solution:
DNA fingerprinting is the technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual. DNA contains non-cistronic hypervariable repeat sequences called VNTR. DNA fingerprinting involves the identification of these VNTRs.

Topic: Biology, Molecular Basis of Inheritance, NEET