datavorous/entrance-exam-dataset huggingface dataset:
In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4%, respectively. Then the maximum percentage of error in the measurement X, where X=A2B12C13D3 will be

Options:
A 313%
B 16%
C -10%
D 10%

The correct answer is:
16%

Solution:
Given: Percentage error in A, B, C and D is 1%, 2%, 3% and 4%, respectively.

The maximum percentage error in X is given by,

∆XX×100=2∆AA×100+12∆BB×100+13∆CC×100+3∆DD×100

∆XX×100=2×1+12×2+13×3+3×4

∆XX×100=16%. will be the maximum percentage error in X.

Topic: Physics, Mathematics in Physics, NEET, NEET 2019 (All India)
In an experiment, to find the loss of energy with respect to time in case of swinging simple pendulum, the graph between (amplitude)2 and time is

Options:
A
B
C
D

The correct answer is:

Solution:
E=12Ka2
E=12mω2a2
So a2t=constant

Topic: Physics, Oscillations, NEET
In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a 4 Î© resistance and at 3 m when cell is shunted by a 8 Î© resistance. The internal resistance of cell is

Options:
A 12 Î©
B 8 Î©
C 16 Î©
D 1 Î©

The correct answer is:
8 Î©

Solution:
If Ï is the potential gradient then, VA-VB=ÏÃl
when the cell is shunted by a 4 Î© resistance, E-Err+4=ÏÃ2
E4r+4=ÏÃ2 ........ (1)
Similarly, when the cell is shunted by a 8 Î© resistance,
E8r+8=ÏÃ3 ......... (2)
From equation (1) and (2),
r+82r+4=23
3r+24=4r+16
r=8 Î©

Topic: Physics, Current Electricity, NEET
In an experiment tungsten cathode which has a threshold wavelength 2300 Å is irradiated by ultraviolet light of wavelength 1800 Å . The maximum energy of emitted photo-electron will be

Options:
A 1.2Â eV
B 1.5 eV
C 1.6Â eV
D 1.8Â eV

The correct answer is:
1.5 eV

Solution:
Kmax=hc1Î»-1Î»0
Kmax=6.62Ã10-34Ã3Ã1081.6Ã10-1910101800-10102300
Kmax=1.5 eV

Topic: Physics, Dual Nature of Matter, NEET
In an experiment, two long day plants belonging to the same species are selected. In plant A, all leaves are covered in such a way that no light can reach them but the floral buds are exposed and light can reach them. In plant B, all floral buds are covered in such a way that no light can reach them but the leaves are exposed and light can reach them. Both these plants are now subjected to light for a duration greater than their critical photoperiod. Which of these results will be obtained?

Options:
A Flowering will occur in plant A but not in plant B
B Flowering will occur in plant B but not in plant A
C Flowering will occur in both plant A as well as plant B
D Flowering will not occur in both plant A as well as plant B

The correct answer is:
Flowering will occur in plant B but not in plant A

Solution:
The site of perception of light/dark duration is the leaves. It has been hypothesized that there is a hormonal substance(s) that is responsible for flowering. This hormonal substance migrates from leaves to shoot apices for inducing flowering only when the plants are exposed to the necessary inductive photoperiod. In plant B, leaves are exposed to light and hence flowering occurs

Topic: Biology, Plant - Growth and Development, NEET
In an f.c.c. unit cell, atoms are numbered as shown below. The atoms not touching each other are (Atom numbered 3 is face centre of front face) .

Options:
A 3 & 4
B 1 & 3
C 1 & 2
D 2 & 4

The correct answer is:
1 & 2

Solution:
​Atoms along one edge or at corners do not touch each other in fcc cell.
​

Topic: Chemistry, Solid State, NEET
In an ideal transformer, the turns ratio is $\frac{N_P}{N_S}=\frac{1}{2}$. The ratio $V_S: V_P$ is equal to (the symbols carry their usual meaning) :

Options:
A 2:1
B 1:1
C 1:4
D 1:2

The correct answer is:
2:1

Solution:
According to transformer ratio,
$\frac{V_S}{V_P}=\frac{N_S}{N_P}=2: 1$

Topic: Physics, Alternating Current, NEET, NEET 2024
In an inducible operon, the genes are:

Options:
A usually not expressed unless a signal turns them "on".
B usually expressed unless a signal turns them "off".
C never expressed
D always expressed

The correct answer is:
usually not expressed unless a signal turns them "on".

Solution:
When $E$. coli is exposed to both lactose and glucose, the organism first metabolise glucose and then temporarily cease to grow until the genes of lac operon become included to provide the ability to metabolise lactose. Allolactose is called an inducer because in its presence it turns on, or induces the expression of the lac genes.
Related Theory
When lactose is present outside the cell. it crosses the cell membrane and acts as an inducer of the operon. It does so once lactose is broken down to create allolactose. The lac operon is then membrane facilitating lactose transport into the cell, and \& beta galactosidase, which eats up lactose to Make glucose molecules. And beta-galactosidase also makes allolactose. This leads to a positive feedback loop.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2013 (Karnataka)
In an inelastic collision of two bodies, the quantities which do not change after the collision are the of the system of two bodies.

Options:
A total kinetic energy
B total linear momentum
C total energy
D None of the above

The correct answer is:
total linear momentum

Solution:
Correct Option is : (B)
total linear momentum

Topic: Chemistry, Work Power Energy, NEET
In an irreversible process taking place at constant T and P and in which only P-V work is being done, the change in Gibbs energy (dG) and change in entropy (dS) satisfy the criteria

Options:
A (dS)V,U=0,(dG)T,P=0
B (dS)V,U=0,(dG)T,P=+ve
C (dS)V,U=-ve,(dG)T,P=-ve
D (dS)V,U=+ve,(dG)T,P=-ve

The correct answer is:
(dS)V,U=+ve,(dG)T,P=-ve

Solution:
For spontaneous change (work is being done by the system)
So (dG)T,P be –ve and (dS)V,U be +ve.

Topic: Chemistry, Thermodynamics (C), NEET
In an isobaric process, heat is supplied to a monoatomic ideal gas. The fraction of heat that goes into mechanical work is

Options:
A 1
B 23
C 35
D 25

The correct answer is:
25

Solution:
QP=nCP∆T=52nR∆TW=P∆V=nR∆T⇒WQP=nR∆T52nR∆T=25

Topic: Physics, Thermodynamics, NEET
In an isothermal and irreversible expansion of a gas, what is the relationship between q and w?

Options:
A q = w
B q < w
C q > w
D q is unrelated to w.

The correct answer is:
q = w

Solution:
In an isothermal expansion, both q (heat absorbed) and w (work done) are equal and have the same magnitude but opposite signs.

Topic: Chemistry, Thermodynamics (C), NEET
In an isothermal expansion of one mole of an ideal gas against vacuum from 5 litre to 50 litre at 300 K, the quantity of heat absorbed by the gas is

Options:
A zero
B 80 lit. atm
C 1380 cal
D -1380 cal

The correct answer is:
zero

Solution:
The quality of heat absorbed by the gas, Q=ΔE+W, here W is the work done by the gas. For free expansion i.e., expansion against vacuum, W=0 ; at constant temperature , the internal energy of an ideal gas is constant and so ΔE=0 . Thus Q=0 .

Topic: Chemistry, Thermodynamics (C), NEET
In an isothermal process at 300 K, 1 mole of an ideal gas expands from a pressure 100 atm against an external pressure of 50 atm. Then total entropy change (Cal K-1 ) in the process is â

Options:
A +0.39
B – 0.39
C +1.59
D –1.59

The correct answer is:
+0.39

Solution:
ÎSsys=1Ã2Ã2.303log10050â1.39 Cal K-1
ÎSsurr=-1Ã2Ã3001-50100300=-1.0 Cal K-1
ÎStotal=+0.39calK-1

Topic: Chemistry, Thermodynamics (C), NEET
In an n-p-n transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector

Options:
A The base current will be 1 mA
B The base current will be -1 mA
C The emitter current will be 9 mA
D The emitter current will be 15 mA

The correct answer is:
The base current will be 1 mA

Solution:
As IC=90% of IE (according to questions)
⇒ 10 mA=90100IE
Or IE=10×109≈11 mA
IB=IE-IC=11 mA-10 mA=1 mA

Topic: Physics, Semiconductors, NEET
In an n-p-n transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector

Options:
A The base current will be 1 mA

B The base current will be -1 mA

C The emitter current will be 9 mA

D The emitter current will be 15 mA

The correct answer is:
The base current will be 1 mA

Solution:
As IC=90% of IE (according to questions)
â 10 mA=90100IE
Or IE=10Ã109â11 mA
IB=IE-IC=11 mA-10 mA=1 mA

Topic: Physics, Semiconductors, NEET
In an NPN transistor circuit, the collector current is 10  mA. If 90% of the emitted electrons reach the collector, the emitter current IE and base current IB are given by

Options:
A IE=-1 mA, IB=9 mA
B IE=9Â mA,Â IB=-1Â mA
C IE=1Â mA,Â IB=11Â mA
D IE=11 mA, IB=1 mA

The correct answer is:
IE=11 mA, IB=1 mA

Solution:
We know that
IC=90100ÃIEâ10=0.9ÃIE
â´IE=100.9=11 mA
Also IE=IB+ICâIB=11-10=1 mA .

Topic: Physics, Semiconductors, NEET
In an n-type silicon, which of the following statement is true:

Options:
A Electrons are majority carriers and trivalent atoms are the dopants.
B Electrons are minority carriers and pentavalent atoms are the dopants.
C Holes are minority carriers and pentavalent atoms are the dopants.
D Holes are majority carriers and trivalent atoms are the dopants.

The correct answer is:
Holes are minority carriers and pentavalent atoms are the dopants.

Solution:
Correct Option is : (C)
Holes are minority carriers and pentavalent atoms are the dopants.

Topic: Chemistry, Semiconductors, NEET
In an octahedral complex, which set of orbitals experiences a higher energy level due to ligand-metal repulsions?

Options:
A dxz, dyz, dxy
B dz2, dx2-y2
C s, p
D All d orbitals experience equal energy levels.

The correct answer is:
dz2, dx2-y2

Solution:
In an octahedral complex, the dz2, dx2-y2 orbitals experience higher energy levels due to ligand-metal repulsions.

Topic: Chemistry, Coordination Compounds, NEET
In an octahedral structure, the pair of $d$ orbitals involved in $d^2 s p^2$ hybridisation is:

Options:
A $d_{x^2-y^2} \cdot d_{z^2}$
B $d_{x z} d_{x^2-y^2}$
C $d_{z 2} \cdot d_{x z}$
D $d_{x y^{\prime}} d_{y z}$

The correct answer is:
$d_{x^2-y^2} \cdot d_{z^2}$

Solution:
$s p^3 d^2$ or $d^2 s p^3$ are hybridisation for the octahedral geometry. In octahedron, the bonds are formed parallel to the $x, y$, and $z$-axes, hence $d x^2-d y^2$ and $d z^2$ will be used to form the hybrid orbitals.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2004
In an octahedral structure, the pair of d-orbitals involved in d2 sp3 hybridisation is

Options:
A dx2-y2,dz2
B dxz ,dx2-y2
C dz2,dxz
D dxy,dyz

The correct answer is:
dx2-y2,dz2

Solution:
In the formation of d2sp3 hybrid orbitals, two (n-1)d orbitals of eg set i.e., n-1 dz2and n-1dx2-y2orbitals, one ns and three npnpx, npy and npz orbitals combine together and form six d2sp3 hybrid orbitals.

Topic: Chemistry, Coordination Compounds, NEET
In an octahedral structure, the pair of d-orbitals involved in d2 sp3 hybridisation is

Options:
A dx2-y2,dz2
B dxz ,dx2-y2
C dz2, dxz
D dxy, dyz

The correct answer is:
dx2-y2,dz2

Solution:
In the formation of d2sp3 hybrid orbitals, two n-1d orbitals of eg set i.e., n-1 dz2and n-1dx2-y2orbitals, one ns and three npnpx, npy and npz orbitals combine together and form six d2sp3 hybrid orbitals.

Topic: Chemistry, Coordination Compounds, NEET
In an organic compound, the presence of halogen is detected by

Options:
A Iodoform test
B Silver nitrate test
C Beilstein’s test
D Millon’s test

The correct answer is:
Beilstein’s test

Solution:
Halogens are detected by Beilstein’s test. In this test, a copper wire is dipped in original solution and heated in a bunsen burner flame. Green colour is imparted to the flame, due to the formation of a volatile copper halide. This proves the presence of halogen.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
In an organism, enzymes generally have

Options:
A Same pH and temperature optima
B Same pH but different temperature optima
C Different pH but same temperature optima
D Different pH and different temperature optima

The correct answer is:
Different pH but same temperature optima

Solution:
In an organism, each enzyme has its own different favourable pH values but generally has the same optimum temperature.

Topic: Chemistry, Biomolecules, NEET
In an organism, if the normal diploid number of chromosomes is 8, how many chromatids are present in each daughter cell at the end of meiosis I

Options:
A 2
B 4
C 8
D 16

The correct answer is:
8

Solution:
Meiosis I ends when the chromosomes of each homologous pair arrive at opposing poles of the cell. The microtubules disintegrate, and a new nuclear membrane forms around each haploid set of chromosomes. The chromosomes uncoil, forming chromatin again, and cytokinesis occurs, forming two non-identical daughter cells. A resting phase known as interkinesis or interphase II happens in some organisms. Therefore, each daughter cell will have 4 chromosomes and 8 chromatids, at the end of the meiosis I.

Topic: Biology, Cell Cycle and Cell Division, NEET
In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is

Options:
A Q2
B Q3
C Q2
D Q

The correct answer is:
Q2

Solution:
Let Q denote maximum charge on the capacitor.
Let q denote charge when energy is equally shared
∴ 1212Q2C=12q2C⇒Q2=2q2
∴ q=Q2

Topic: Physics, Alternating Current, NEET
In an p-type silicon, which of the following statement is true:

Options:
A Electrons are majority carriers and trivalent atoms are the dopants.
B Electrons are minority carriers and pentavalent atoms are the dopants.
C Holes are minority carriers and pentavalent atoms are the dopants.
D Holes are majority carriers and trivalent atoms are the dopants.

The correct answer is:
Holes are majority carriers and trivalent atoms are the dopants.

Solution:
Correct Option is : (D)
Holes are majority carriers and trivalent atoms are the dopants.

Topic: Chemistry, Semiconductors, NEET
In an SN1 reaction on chiral centres, there is:

Options:
A 100% racemization
B Inversion more than retention leading to partial racemization
C 100% retention
D 100% inversion

The correct answer is:
Inversion more than retention leading to partial racemization

Solution:
SN1 reaction gives racemic mixture with slight predominance of that isomer which corresponds to inversion because SN1 also depends upon the degree of 'shielding' of the front side of the reacting carbon.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2015 (Phase 2)
In an unbiased p - n junction electrons diffuse from n-region to p-region because:

Options:
A electrons travel across the junction due to potential difference
B only electrons move from n to p region and not the vice - versa
C electron concentration in n - region is more as compared to that in p - region
D holes in p - region attract them

The correct answer is:
electron concentration in n - region is more as compared to that in p - region

Solution:
In a p-n junction diffusion occurs due to spontaneous movement of majority charge carrier from the region of high concentration to low concentration.

Topic: Physics, Semiconductors, NEET
In an unbiased $p-n$ junction holes from the p-region to $n$-region because of:

Options:
A The attraction of the free electrons of n-region.
B The higher hole concentration in p-region than that in $n$-region.
C The higher concentration of electrons in the $n$-region than that in the p-region.
D The potential difference across the $p-n$ junction.

The correct answer is:
The higher hole concentration in p-region than that in $n$-region.

Solution:
The higher hole concentration in P-region that in $n$-region.

Topic: Physics, Semiconductors, NEET, NEET 2013 (Karnataka)
In an unbiased p-n junction, holes diffuse from the p-region to n-region because

Options:
A free electrons in the n-region attract them.
B they move across the junction by the potential difference.
C hole concentration in p-region is more as compared to n-region.
D All the above

The correct answer is:
hole concentration in p-region is more as compared to n-region.

Solution:
Correct Option is : (C)
hole concentration in p-region is more as compared to n-region.

Topic: Chemistry, Semiconductors, NEET
In anaerobic glycolysis, 2 molecules of inorganic phosphate are used per molecule of glucose consumed. Which of the following enzyme catalyzes the uptake of inorganic phosphate?

Options:
A Hexokinase
B Phosphofructokinase
C Glyceraldehyde 3-phosphate dehydrogenase
D Pyruvate kinase

The correct answer is:
Glyceraldehyde 3-phosphate dehydrogenase

Solution:
Glyceraldehyde-3-phosphate dehydrogenase (GAPDH) is a very important enzyme in the production of energy. This enzyme catalyzes the sixth step in the process of breaking down glucose, also known as glycolysis which occurs in organisms of all phyla. The sixth step consists of the oxidation of Glyceraldehyde 3-phosphate by NAD and inorganic phosphate to yield 1,3 bisphosphoglycerate

Topic: Biology, Respiration in Plants, NEET
In angiosperm all the four microspores of tetrad are covered by a layer which is formed by:

Options:
A pectocellulose
B callose
C cellulose
D sporopollenin

The correct answer is:
callose

Solution:
Callose wall surrounds the sporocytes during the meiotic stage. It provides an isolation barrier sealing off one meiotic cell (megaspore mother cell) from another.
Related Theory
Tetrad is a group of four haploid and immature pollen grains in tetrahedral fashion produced by meiotic microsporogenesis. The tetrads are of five types namely tetrahedral (most common), isobilateral, $T$ shaped, decussate and linera. In dicots, the most common type of pollen tetrad is tetrahedral.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2002
In angiosperm, the haploid, diploid and triploid structures of a fertilized embryo sac sequentially are:

Options:
A Synergids, antipodals and Polar nuclei
B Synergids, Primary endosperm nucleus and zygote
C Antipodals, synergids, and primary endosperm nucleus
D Synergids, Zygote and Primary endosperm nucleus

The correct answer is:
Synergids, Zygote and Primary endosperm nucleus

Solution:
Synergids are pairs of haploid cells present at the micropylar end of embryo sac. Zygote which is a diploid structure undergoes divisions to form an embryo. Antipodal and polar nuclei are also haploid in nature. Primary endosperm nucleus is triploid in nature. In a fertilized embryo sac the haploid, diploid and triploid structures are synergid, zygote and primary endosperm respectively.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2023 (All India)
In angiosperms, female gametophyte is embryo sac which is formed by

Options:
A Reduction division in megaspore
B Equational division in megaspore
C Reduction division followed by equational division in megaspore mother cell
D Both B and C

The correct answer is:
Reduction division followed by equational division in megaspore mother cell

Solution:
The megaspore mother cell undergoes, reductional division to form four megaspore, three of which degenerates, one functional megaspore undergoes mitosis to form embryo sac or female gametophyte. The functional megaspore undergoes, three mitosis to form seven celled, eight nucleate embryosac.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In angiosperms, ___________ nuclei participate in fertilization.

Options:
A 2
B 4
C 5
D 7

The correct answer is:
5

Solution:
Angiosperms show double fertilization. In syngamy, there is a fusion of one male and one female gamete having one nucleus each. In triple fusion, there is a fusion of one male gamete having one nucleus and two polar nuclei. Thus, five nuclei (2 polar nuclei, 1 egg nuclei, 2 male gametes) participate in double fertilization.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In Angiosperms pollen tubes liberate their male gametes into the:

Options:
A Central cell
B Antipodal cells
C Egg cell
D Synergids

The correct answer is:
Synergids

Solution:
The male gametes in the pollen tube are discharged into the synergid cells from where the first male gamete is transferred to the egg cell of the embryo sac. The other male gamete is transferred to the central cell through the cytoplasmic current.
Related Theory
Towards the micropylar end, a mature ovule consists of embryo sac which has an egg apparatus (an egg cell and two synergids), centrally placed two polar nuclei and 3 antipodal cells at the chalazal end.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2002
In angiosperms, the endosperm is

Options:
A Short lived
B Persistent female gametophyte
C Formed before fertilisation
D Formed after fertilisation

The correct answer is:
Formed after fertilisation

Solution:
In angiosperms, the endosperm is triploid and is a post-fertilization product. It is formed by the fusion of second male gamete with two polar nuclei; this fusion is also called triple fusion.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In angiosperms, the first cell from which the development of male gametophyte takes place is:

Options:
A Pollen grain
B Zygote
C Microspore
D PEC

The correct answer is:
Microspore

Solution:
In angiosperm male gametophyte is pollen grain, it is formed from microspore which develops after reductional division of microspore mother cell. Microspore develops two layered wall, outer exine and inner intine to form pollen grain.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In animal and plant cell the division of cytoplasm is respectively

Options:
A Centripetal and centripetal
B Centrifugal and Centripetal
C Centripetal and centrifugal
D Centrifugal and centrifugal

The correct answer is:
Centripetal and centrifugal

Solution:
In animal cell cytokinesis occurs by cell furrow or groove formation in which a furrow develops in membrane which gradually grows towards the center thus shows centripetal growth. In plant cell cytokinesis occurs by cell plate which gradually grows from center to periphery thus shows centrifugal growth.

Topic: Biology, Cell Cycle and Cell Division, NEET
In animal cell, lipid-like steroidal hormones are synthesized in

Options:
A endoplasmic reticulum
B glyoxysome
C sphaerosome
D lysosome

The correct answer is:
endoplasmic reticulum

Solution:
In animal cells, the lipid-like steroid hormone is synthesized by the smooth endoplasmic reticulum. Glyoxysome causes the breakdown of stored fats during seed germination by the glyoxylate cycle. Spherosomes are present in the plant cells which are involved in synthesis and storage of lipids. Lysosomes are suicidal bags which degrade the macromolecules by pocessing the lytic enzymes.

Topic: Biology, Cell - The Unit of Life, NEET
In animals, specialized cells for osmoregulation and excretion was first seen in

Options:
A Annelida
B Coelenterata
C Mollusca
D Platyhelminthes

The correct answer is:
Platyhelminthes

Solution:
In Platyhelminthes, specialized cells called flame cells help in osmoregulation and excretion.

Topic: Biology, Animal Kingdom, NEET
In animals, which type of cells are known to exhibit mitotic cell division?

Options:
A Diploid somatic cells
B Haploid somatic cells
C Diploid reproductive cells
D Haploid reproductive cells

The correct answer is:
Diploid somatic cells

Solution:
In animals, mitotic cell division is only seen in the diploid somatic cells, with a few exceptions such as male honey bees.

Topic: Biology, Cell Cycle and Cell Division, NEET
In any fission process the ratio: mass of fission products mass of parent nucleus is:

Options:
A equal to 1
B greater than 1
C less than 1
D depends on the mass of the parent nucleus

The correct answer is:
less than 1

Solution:
The mass of fission products is always less than mass of parent nucleus.

Topic: Physics, Nuclear Physics, NEET, NEET 2005
In apomixis, how is the diploid egg cell formed?

Options:
A Through reduction division
B Without reduction division
C Through fertilisation
D None of the above

The correct answer is:
Without reduction division

Solution:
In some species, the diploid egg cell is formed without reduction division and develops into the embryo without fertilisation.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET