datavorous/entrance-exam-dataset huggingface dataset:
Arteries are best defined as the vessels which

Options:
A carry blood away from the heart to different organs
B break up into capillaries which reunite to form a vein
C carry blood from one visceral organ to another visceral organs
D supply oxygenated blood to the different organs

The correct answer is:
carry blood away from the heart to different organs

Solution:
Arteries are blood vessels that carry blood away from the heart towards different organs. They generally contain oxygenated blood (except pulmonary artery which contains deoxygenated blood). The blood flows in an artery under alternate increased pressure and with jerks.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2011 (Screening)
Artificial induction of roots on stems before it is separated from the parent plant for propagation is called

Options:
A Cutting
B Plant tissue culture
C Root-stem joint
D Layering

The correct answer is:
Layering

Solution:
In layering middle part of a soft basal branch is defoliated, slightly injured (tonguing, notching, ringing) and pegged in the soil to develop adventitious roots. Later on, the branch of the layer is separated and planted

Topic: Biology, Reproduction in Organisms, NEET
Artificial insemination means

Options:
A transfer of sperms of a healthy donor to a test-tube containing ova
B transfer of sperms of husband to a test-tube containing ova
C artificial introduction of sperms of a healthy donor into the vagina
D introduction of sperms of healthy donor directly into the ovary

The correct answer is:
artificial introduction of sperms of a healthy donor into the vagina

Solution:
Infertility due to inability of the male to inseminate the female or due to very low sperm counts in the ejaculates can be corrected by artificial insemination.
Ova from the wife/donor (female) and sperms from the husband/donor (male) are callected and are induced to form zygate under simulated conditions in the laboratory is called test-tube baby programme. Introduction of sperms of healthy donor directly into the ovary is called Intra Cytoplasmic Sperm Injection (|CSI).

Topic: Biology, Reproductive Health, NEET, NEET 2013 (All India)
Artificial selection to obtain cows yielding higher milk output represents

Options:
A Stabilizing selection as it stabilizes this character in the population
B Directional as it pushes the mean of the character in one direction
C Disruptive as it splits the population into two one yielding higher output and the other lower output
D Stabilizing followed by disruptive as it stabilizes the population to produce higher yielding cows

The correct answer is:
Directional as it pushes the mean of the character in one direction

Solution:
Artificial selection to obtain cow yielding higher milk output will shift the peak to one direction, hence, will be an example of Directional selection. In stabilizing selection, the organisms with the mean value of the trait are selected. In disruptive selection, both extremes get selected (more individuals acquire peripheral character value at both ends of the distribution curve).

Topic: Biology, Evolution, NEET, NEET 2017
Artificial silk is a

Options:
A polypeptide
B polysaccharide
C polythene
D polyvinyl chloride

The correct answer is:
polysaccharide

Solution:
Chemically-treated cellulose (a polysaccharide) is called artificial silk.

Topic: Chemistry, Polymers, NEET
Artificial sweetener which is stable under cold conditions only is:

Options:
A Saccharine
B Sucralose
C Aspartame
D Alitame

The correct answer is:
Aspartame

Solution:
Aspartame is stable at cold conditions but unstable at cooking temperature.

Aspartame consists of aspartic acid, phenylalanine and methanol. It is 180-200 times sweeter than sugar. They are used in frozen desserts, gelatins, chewing gums and beverages. At high temperature, it decomposed into constituent amino acids. Thus it is referred to as a baking sweetener. Its melting point is very low. Thus is stable at very cool conditions.

Topic: Chemistry, Chemistry in Everyday Life, NEET, NEET 2014
Artificial sweetener which is stable under cold conditions only is:

Options:
A Saccharine
B Sucralose
C Aspartame
D Alitame

The correct answer is:
Aspartame

Solution:
Aspartame is stable at cold conditions but unstable at cooking temperature.

Topic: Chemistry, Chemistry in Everyday Life, NEET
Artificial sweetner which is stable under cold conditions only is

Options:
A Alitame
B Saccharine
C Aspartame
D Sucralose

The correct answer is:
Aspartame

Solution:
Aspartame is Artificial sweetner which is stable under cold conditions only.

Topic: Chemistry, Biomolecules, NEET
Artificial system of classification gave equal weightage to

Options:
A morphological and anatomical characters
B vegetative and sexual characters
C vegetative and anatomical characters
D genetical and sexual characters

The correct answer is:
vegetative and sexual characters

Solution:
Artificial system of classification based on the vegetative and sexual characteristics and it gave equal weightage to these for classification. They were able to separate closely related species.

Topic: Biology, Biological Classification, NEET
Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidizing agent?

Options:
A Aryl iodides are more stable than aryl chlorides and bromides, so an oxidizing agent is needed to increase their reactivity.
B Because Iodination reactions are reversible in nature. To carry out the reaction in the forward direction, HI formed during iodination is removed by oxidation. HIO4is used as an oxidizing agent.
C Aryl iodides are less reactive than aryl chlorides and bromides, so an oxidizing agent is used to activate iodine molecules for substitution.
D Aryl iodides have a higher boiling point than aryl chlorides and bromides, so an oxidizing agent is used to reduce their boiling point.

The correct answer is:
Because Iodination reactions are reversible in nature. To carry out the reaction in the forward direction, HI formed during iodination is removed by oxidation. HIO4is used as an oxidizing agent.

Solution:
Correct Option is : (B)
Because Iodination reactions are reversible in nature. To carry out the reaction in the forward direction, HI formed during iodination is removed by oxidation. HIO4is used as an oxidizing agent.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution:

Options:
A II > III > I
B I > II > III
C II > I > III
D III > II > I

The correct answer is:
III > II > I

Solution:
Correct Option is : (D)
III > II > I

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Aryl halides do not undergo nucleophilic substitution reactions under ordinary conditions because
(1) Approach of nucleophile is retarded
(2) Carbon carrying halogen atom is sp3 hybridised
(3) The substrate molecule is destabilized due to resonance
(4) Partial double bond character between carbon and halogen

Options:
A 2 and 3 only
B 2, 3 and 4 only
C 1 and 4 only
D 1 and 3 only

The correct answer is:
1 and 4 only

Solution:
Aryl halides do not undergo nucleophilic substitution reactions under ordinary conditions because Approach of nucleophile is retarded and partial double bond character between carbon and halogen which is reso stabilized so removal of halide ions gets tough as more energy is needed to break C-X bond.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
As a result of change in the magnetic flux linked to the closed loop as shown in the figure, an e.m.f. $V$ volt is induced in the loop. The work done (joules) in taking a charge $Q$ coulomb once along the loop is:

Options:
A $Q V$
B $2 Q V$
C $\frac{Q V}{2}$
D zero

The correct answer is:
$Q V$

Solution:
Work done due to a charge $W=Q V$

Topic: Physics, Electrostatics, NEET, NEET 2005
As compared to a $\mathrm{C}_3$-plant, how many additional molecules of ATP are needed for net production of one molecule of hexose sugar by $\mathrm{C}_4$-plants?

Options:
A two
B $\operatorname{six}$
C twelve
D zero.

The correct answer is:
twelve

Solution:
18 ATP molecules are required by $\mathrm{C}_3$ plants for the synthesis of one molecule of hexose sugar while 30 ATP molecules are needed by the $\mathrm{C}_4$ plants for the same. Thus, $\mathrm{C}_4$ plants have a need of 12 ATP molecules extra than $\mathrm{C}_3$ plants for the synthesis of one molecule of hexose sugar.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2005
As compared to meiosis, in mitosis

Options:
A homologous chromosomes form pairs.
B daughters have half chromosome number.
C telophase stage is absent.
D the prophase is shorter.

The correct answer is:
the prophase is shorter.

Solution:
Mitosis is equational diffusion in which prophase is very short as compared to prophase of meiosis which is very long. Pairing of homologous chromosomes, reduction in number of chromosomes and absent of telophase I occurs only in meiosis.

Topic: Biology, Cell Cycle and Cell Division, NEET
As compared to the other cells in a mammalian testis, Leydig cells will have more amount of

Options:
A Rough Endoplasmic Reticulum
B Golgi complex
C Smooth Endoplasmic Reticulum
D Lysosomes

The correct answer is:
Smooth Endoplasmic Reticulum

Solution:
Smooth endoplasmic reticulum is a network of folded interconnected organelle that has no ribosomes attached. This organelle is involved in the synthesis of lipid including steroidal hormones. It is also involved in detoxification of lipid soluble drugs and harmful molecules formed during metabolism.

Topic: Biology, Cell - The Unit of Life, NEET
As each cell of the sporogenous tissue is capable of giving rise to ..................

Options:
A a megaspore diad
B a megaspore tetrad
C a microspore diad
D a microspore tetrad

The correct answer is:
a microspore tetrad

Solution:
Each cell of the sporogenous tissue is capable of giving rise to a microspore tetrad. Each one is potential pollen or microspore mother cell (PMC). The process of formation of microspores from a pollen mother cell through meiosis is called microsporogenesis. The microspores, as they are formed, are arranged in a cluster of four cells–the microspore tetrad

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
As far as flowering behavior of plant is concerned, Neelkuranji and Bambusa do not come along with

Options:
A Wheat, rice and marigold
B Henbane
C Radish and carrot
D Mango and apple

The correct answer is:
Mango and apple

Solution:
Neelakuranji and bambusa are perennial but monocarpic plants; they flower only once during their lifetime. Mango and apple are perennial trees that flower many times during their lifetime.

Topic: Biology, Reproduction in Organisms, NEET
As O2 (l) is cooled at 1 atm pressure, it freezes to form solid I at 54.5 K. At a lower temperature, solid I rearrange to solid II, which has a different crystal structure. Thermal measurements show that for the phase transition solid I to slid II, ÎH=-743.1 Jmol-1 and ÎS=-17.0 JK-1mol-1 . At what temperature are solids I and II in equilibrium?

Options:
A 2.06 K
B 31.6 K
C 43.7 K
D 53.4 K

The correct answer is:
43.7 K

Solution:
Since, ÎG=ÎH-TÎS
At equilibrium ÎG=0
â´ ÎH=TÎS
Given ÎH=-743.1 J/mol
ÎS=-17.0 JK-1mol-1
â´ -743.1-17.0=T
T=43.7 K
At this temperature solid, I and II are in equilibrium.

Topic: Chemistry, Thermodynamics (C), NEET
As observed in the laboratory system, a 6 MeV proton is incident on a stationary 12 C target. The velocity of center of mass of the system is (Take mass of proton to be 1amu )

Options:
A 2.6Ã106 m s-1
B 6.2×106 m s-1
C 10×106 m s-1
D 10 m s-1

The correct answer is:
2.6Ã106 m s-1

Solution:
vcm=m1v1+m2v2m1+m2
vcm=1vP+1201+12
vcm=vP13
K.E. of proton,
12mvP2=KE
121.67Ã10-27vP2=6Ã1061.6Ã10-19
vP2=12Ã1014
vP=23Ã107
â´ vcm=2313Ã107
vcm=2.6Ã106 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
As part of a student project, Sheela surgically made a full cut in the hypothalamo-hypophyseal tract (infundibulum) in pregnant rabbits and studied the physiological effect on different tissues/organs. Which of the following should be expected?

(i) Formation of a large volume of dilute urine
(ii) Increased blood sugar levels
(iii) Increased heart rate
(iv) Difficulty in parturition

Options:
A (i) and (ii)
B (ii) and (iii)
C (iii) and (iv)
D (i) and (iv)

The correct answer is:
(i) and (iv)

Solution:
Pituitary gland is the master endocrine gland of the body. It is connected to hypothalamus which is a part of the forebrain. Hypothalamus controls the pituitary gland and thus it is known as Master of master endocrine gland. Pituitary gland consists of two lobes: Adenohypophysis and Neurohypophysis. Neurohypophysis is also known as posterior pituitary and it is connected to the hypothalamus through infundibulum or pituitary stalk. Hypothalamic nuclei secretes two hormones which are stored in the posterior pituitary in hassall's corpuscles. These two hormones are ADH and Oxytocin. Hence if the stalk is cut these two hormones will not be stored in the posterior part and thus will not be able to perform its function. If ADH secretion decreases, water reabsorption will be affected and thus dilute urine will be formed. Oxytocin is the birth hormone as it plays an important role during parturition and milk production. In the question the rabbit is pregnant and hence if the infundibulum is cut there will be no oxytocin release and this will affect parturition.

Topic: Biology, Chemical Coordination and Integration, NEET
As per ABO blood grouping system, the blood group of father is $\mathrm{B}^{+}$, mother is $\mathrm{A}^{+}$and child is $\mathrm{O}^{+}$. Their respective genotype can be
A. $\quad{ }^{\mathrm{B}} / / \|^{A^A} / / i \mathrm{i}$
B. $\left.\quad I^{\mathrm{B}}\right|^{\mathrm{B}} /\left.\left.\right|^{\mathrm{A}}\right|^{\mathrm{A}} / \mathrm{ii}$
C. $|\mathrm{A}| \mathrm{B} /\left.i\right|^{\mathrm{A}} /\left.\right|^{\mathrm{B}} \mathrm{j}$
D. $\quad \mathrm{A}_{\mathrm{A}} / I^{\mathrm{B}} / /^{\mathrm{A}} \mathrm{i}$
E. $\quad \mathrm{il}^{\mathrm{B}} /\left.\mathrm{i}\right|^{\mathrm{A}} /\left.\mathrm{I}^{\mathrm{A}}\right|^{\mathrm{B}}$
Choose the most appropriate answer from the options given below :

Options:
A B only
B C \& B only
C D \& E only
D A only

The correct answer is:
A only

Solution:
Genotype of father with blood group $\mathrm{B}^{+}=І^{\mathrm{B}} / / \mathrm{i}^{\mathrm{B}}$
Genotype of mother with blood group $\mathrm{A}^{+}=\mid \mathrm{A}_{\mathrm{A} /\left.\right|^{\mathrm{A}}}$
Genotype of child with blood group $\mathrm{O}^{+}=\mathrm{ii}$
Hence only ' $A$ ' is correct.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2024
As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z=3) is

Options:
A 1.51
B 13.6
C 40.8
D 122.4

The correct answer is:
122.4

Solution:
For hydrogen and hydrogen like atoms
E n = - 1 3 . 6 Z 2 n 2 eV
Therefore,ground state energy of doubly ionized lithium atom (Z=3,n=1) will be
Eâ¡ n = - 1 3 Â· 6 3 2 3 2 1 2 = - 1 2 2 Â· 4 e V
â´ Ionization energy of an electron in ground state of doubly ionized lithium atom will be 122.4 eV

Topic: Physics, Atomic Physics, NEET
As per the diagram a point charge $+q$ is placed at the origin $\mathrm{O}$. Work done in taking another point charge $-\mathrm{Q}$ from the point $\mathrm{A}$ [coordinate $(0, a)$ ] to another point $\mathrm{B}$ [coordinates $(a, 0)$ ] along the straight path $\mathrm{AB}$ is:

Options:
A zero
B $\left(\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{a^2}\right) \cdot \sqrt{2} a$
C $\left(\frac{-q Q}{4 \pi \varepsilon_0} \frac{1}{a^2}\right) \cdot \sqrt{2} a$
D $\left(\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{a^2}\right) \cdot \frac{a}{\sqrt{2}}$

The correct answer is:
zero

Solution:
Work done is equal to zero because the potential of A and B are the same $=\frac{1}{4 \pi \epsilon_0} \frac{q}{a}$
No work done if a particle does not change its potential energy.
i.e. initial potential energy $=$ final potential energy.

Topic: Physics, Electrostatics, NEET, NEET 2005
As per the fluid mosaic model, how are phospholipid molecules arranged in a lipid bilayer?

Options:
A Scattered
B Series
C Alternate
D Amphiphatic

The correct answer is:
Amphiphatic

Solution:
According to the fluid mosaic model proposed by Singer and Nicolson, the major lipids are phospholipids that are arranged in a bilayer. Also, the lipids are arranged within the membrane with the polar head towards the outer sides and the hydrophobic tails towards the inner part. This ensures that the nonpolar tail of saturated hydrocarbons is protected from the aqueous environment.

Topic: Biology, Cell - The Unit of Life, NEET
As shown in figure, the tension in the horizontal cord is 30 N. The weight w and tension in the string OA in newton are

Options:
A 303, 30
B 303,60
C 603, 30
D None of the above

The correct answer is:
303,60

Solution:
Horizontal and vertical components of tension are shown in the figure.
From the figure, Tsinâ¡30o=30 .....(i)
Tcosâ¡30o=w .....(ii)
By solving Equations (i) and (ii), we get
w=303 N
and T=60 N

Topic: Physics, Laws of Motion, NEET
As shown in the figure, a mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take g = 10 m s-2). Neglect the mass of the rope.

Options:
A 37 °
B 53 °
C 60 °
D 40 °

The correct answer is:
40 °

Solution:
Correct Option is : (D)
40 °

Topic: Chemistry, Laws of Motion, NEET
As shown in the figure initially switch k1 is open and k2 is closed and deflection of galvanometer is Î¸0 . Now both switch are closed then the deflection in galvanometer is (given RH=5 Î© )

Options:
A 24 Ω
B 22 Ω
C 42 Ω
D 50 Ω

The correct answer is:
22 Ω

Solution:
θ0∝iG⇒Cθ0=iG
I-case Cθ0=v220+R ......(1)
II-case Cθ05v220+5R5+R×55+R ..........(2)
From (1) and (2) R=22 Ω

Topic: Physics, Experimental Physics, NEET
As shown in the figure, wheel A of radius rA=10 cm is coupled by a belt B to the wheel C of radius rC=25 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad s-2. Time after which wheel C reaches a rotational speed of 100 rpm, assuming the belt does not slip, is nearly

Options:
A 4 s
B 8 s
C 12 s
D 16 s

The correct answer is:
16 s

Solution:
at=αA rA=αCrC
αC= αA rArC= 1.6×1025=0.64 rad/s2
t=ωCαC=100×2π600.64=16.35 s

Topic: Physics, Rotational Motion, NEET
As the ........(A).......... molecules cannot pass through the ........(B).......... lipid bilayer, they require .....(C)........ to facilitate their transport across the membrane. Choose the correct combination respectively -

Options:
A

(A)	(B)	(C)

Non-polar	Polar	Transmembrane protein

B

(A)	(B)	(C)

Polar	Non-polar	Carrier protein

C

(A)	(B)	(C)

Polar	Polar	Transmembrane protein

D

(A)	(B)	(C)

Non-polar	Non-polar	Carrier protein

The correct answer is:

(A)	(B)	(C)

Polar	Non-polar	Carrier protein

Solution:
One of the most important functions of the plasma membrane is the transport of the molecules across it. The membrane is selectively permeable to some molecules present on either side of it. Many molecules can move briefly across the membrane without any requirement of energy and this is called the passive transport. Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient, i.e., from higher concentration to the lower. Water may also move across this membrane from higher to lower concentration. Movement of water by diffusion is called osmosis. As the polar molecules cannot pass through the nonpolar lipid bilayer, they require a carrier protein of the
membrane to facilitate their transport across the membrane. A few ions or molecules are transported across the membrane against their concentration gradient, i.e., from lower to the higher concentration.

Topic: Biology, Cell - The Unit of Life, NEET
As the temperature increases, the electrical resistance

Options:
A decreases for both conductors and semiconductors
B increases for conductors but decreases for semiconductors
C decreases for conductors but increases for semiconductors
D increases for both conductors and semiconductors

The correct answer is:
increases for conductors but decreases for semiconductors

Solution:
For conductors, on increasing temperature, thermal agitation of free electrons increases. Hence, resistance increases.

For Semiconductors, number of holes or charge carriers increases with increase in temperature.

Hence, resistance increases for conductors but decreases for semiconductors.

⇒ Answer is (b)

Topic: Physics, Current Electricity, NEET, NEET 2022 (Phase 1)
As the temperature is raised from 20 oC to 40 oC , the average kinetic energy of neon atoms changes by a factor of which of the following?

Options:
A 12
B 2
C 313293
D 313293

The correct answer is:
313293

Solution:
As average K.E =32RTNA
i.e., K.E∝T
Hence K.E313K.E293=313293

Topic: Chemistry, States of Matter, NEET
Ascaris is characterised by

Options:
A absence of true coelom but presence of metamerism
B presence of neither true coelom nor metamerism
C presence of true coelom but absence of metamerism
D presence of true coelom and metamerism (metamerisation)

The correct answer is:
presence of neither true coelom nor metamerism

Solution:
Ascaris is an endoparasite of man. It inhabits the small intestine more frequently of children than of adults. The body is elongate, cylindrical and gradually tapering at both ends. There is no metameric segmentation. In Ascaris between body wall and visceral organs is a spacious fluid filled cavity. This cavity is not true coelom as it is not lined by coelomic epithelium, has no relation with reproductive and excretory organs and develops from blastocoel. This body cavity is referred pseudocoel.

Topic: Biology, Human Health and Diseases, NEET, NEET 2008 (Screening)
Ascaris is characterized by

Options:
A Absence of true coelom but presence of metamerism
B Presence of neither true coelom nor metamerism
C Presence of true coelom and metamerism
D Presence of true coelom but the absence of metamerism

The correct answer is:
Presence of neither true coelom nor metamerism

Solution:
Body of Ascaris is elongate, cylindrical gradually tapering at both ends. There is no metameric segmentation. In Ascaris, between the body wall and visceral organs is a spacious fluid-filled cavity. They belong to phylum Aschelminthes that have pseudocoelom as there body cavity.
Segmentation, also called metamerism, or metameric segmentation, in zoology, the condition of being constructed of a linear series of repeating parts, each being a metamere (body segment, or somite) and each being formed in sequence in the embryo, from anterior to posterior. E.g. animals of phylum Annelida.

Topic: Biology, Animal Kingdom, NEET
Ascending limb of Henle's loop is ___ to water whereas the descending limb is ___ to it.

Options:
A impermeable, permeable
B permeable, impermeable
C semi-permeable, impermeable
D permeable, semi-permeable

The correct answer is:
impermeable, permeable

Solution:
Ascending limb of Henle's loop is impermeable to water whereas the descending limb is permeable to it.

Topic: Biology, Excretory Products and their Elimination, NEET
Asci are sexually reproductive structures formed in

Options:
A Morels
B Saccharomyces
C Claviceps
D All the above

The correct answer is:
All the above

Solution:
Sexual spores are called as ascospores which are produced endogenously in a sac-like structure called asci (singular: ascus). These asci are arranged in different types of fruiting bodies called ascocarps. Ascobolus, Saccharomyces, morels, truffles, Aspergillus, Claviceps and Penicillium belong to ascomycetes.

Topic: Biology, Biological Classification, NEET