datavorous/entrance-exam-dataset huggingface dataset:
BOD of waste water is estimated by measuring the amount of:

Options:
A Total organic matter
B Biodegradable organic matter
C Oxygen evolution
D Oxygen consumption

The correct answer is:
Oxygen consumption

Solution:
Biochemical Oxygen demand is the amount of oxygen, present in water, taken up by bacteria and other microorganisms to degrade organic matter present in water. BOD estimates the organic matter in water. If there is a high amount of organic matter in water, there will be increased consumption of oxygen by microbes for the degradation of organic matter. Hence BOD is determined by the amount of consumption of oxygen by microbes. An increase in BOD results in a decrease in oxygen in the water.

Topic: Biology, Microbes in Human Welfare, NEET
BOD refers to:

Options:
A The amount of the oxygen that would be consumed to oxidize all the inorganic matter in one liter of water
B The amount of the oxygen that would be consumed if all the organic matter in one liter of water were oxidised by bacteria.
C The amount of the oxygen that would be consumed if all the organic and inorganic matter in one liter of water were oxidised by bacteria.
D The amount of the oxygen that would be produced if all the organic matter in one liter of water were oxidised by bacteria.

The correct answer is:
The amount of the oxygen that would be consumed if all the organic matter in one liter of water were oxidised by bacteria.

Solution:
The measure of oxygen used by microorganisms to decompose the waste is called biological oxygen demand. To decompose waste a large number of bacteria are present and for that demand for oxygen will be high. So the BOD will be high. It will decrease when waste is completely consumed. So the correct answer will be C.

Topic: Biology, Microbes in Human Welfare, NEET
BOD values of different samples of water are given:

Sample A BOD=5, Sample B BOD=12, Sample C BOD=2, Sample D BOD=7

Which of the above-given sample will have the least number of Daphnia?

Options:
A A
B B
C C
D D

The correct answer is:
B

Solution:
Daphnia is an indicator of freshwater thus it is not found in the polluted water body. Biochemical oxygen demand is the indirect measurement of organic pollutant present in the water body. Higher the value of BOD, higher is its polluting potential.

Topic: Biology, Environmental Issues, NEET
Body A of mass 4m moving with speed u collides with another body B of mass 2m at rest. The collision is head on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is

Options:
A 19
B 89
C 49
D 59

The correct answer is:
89

Solution:
4mu+0=4mv1+2mv2
⇒2u=2v1+v2 …(A)
For elastic collision,
e=1
⇒u=v2-v1 …(B)

From A and B,
4u=3v2⇒v2=43u
and v1=u3
Fraction of energy lost by (A),
=124mu2 - 124mv12124mu2
=u2 - u29u2=89

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2019 (All India)
Body centred cubic and face centred cubic unit cells have n1 and n2 effective number of atoms. Which one the following (n1, n2) combination is correct?

Options:
A (4, 1)
B (4, 2)
C (1, 4)
D (2, 4)

The correct answer is:
(2, 4)

Solution:
Number of atoms in
BCC unti cell =1+8×18=1+1=2
FCC unit cell =8×18+6×12=1+3=4

Topic: Chemistry, Solid State, NEET
Body having meshwork of cells, internal cavities lined with food filtering flagellated cells and indirect development are the characteristics of phylum:

Options:
A Porifera
B Mollusca
C Protozoa
D Coelenterata

The correct answer is:
Porifera

Solution:
In Poriferans, the body is a loose aggregate of cells (meshwork of cells). Internal cavities and canals are lined with food filtering flagellated cells i.e. choanocyte/collar cell. Choanocytes help in filter-feeding. In sponges, the development of adult sponge from a diploid zygote is indirect which means it is mediated by an intermediate free-swimming flagellated larval stage for dispersal of species. The larval stage is morphologically distinct from an adult.

Topic: Biology, Animal Kingdom, NEET, NEET 2015 (Phase 2)
Body is segmented in

Options:
A Arthropoda
B Cnidaria
C Annelida
D Both (a) and (c)

The correct answer is:
Both (a) and (c)

Solution:
The phylum Annelida typically has numerous segments, known as somites, and are covered with the locomotory structures called setae. The phylum Arthropoda has a body segmented into head thorax and abdomen. The body is covered by an exoskeleton that protects their organs and muscles.

Topic: Biology, Animal Kingdom, NEET
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

Options:
A 411.02 g
B 258.46 g
C 354.12 g
D 121.67 g

The correct answer is:
121.67 g

Solution:
Correct Option is : (D)
121.67 g

Topic: Chemistry, Solutions, NEET
Bond angle in PH3 is closer to 90° while that in NH3 is 104.5°. Which of the following best explains this structural feature?

Options:
A Due to larger size of the lone pair electron cloud, there is larger lone pair – bond pair repulsion in PH3 compared to NH3 .
B Higher electronegativity of nitrogen concentrates the bond pair electron cloud near the central atom which increases the bond pair – bond pair repulsion which in turn decreases the bond angle in NH3 .
C Energy difference between 3s and 3p orbitals is quite high and hence the lone pair on phosphorous prefers to occupy unhybridized s- orbital rather than hybridized sp3 hybridized orbital which causes its s- orbital energy to increase.
D Phosphorous forms pπ-dπ bonds while nitrogen does not.

The correct answer is:
Energy difference between 3s and 3p orbitals is quite high and hence the lone pair on phosphorous prefers to occupy unhybridized s- orbital rather than hybridized sp3 hybridized orbital which causes its s- orbital energy to increase.

Solution:
In the hydrides of group 15 and group 16 (except NH3 and H2O ) the energy difference between 3s and 3p orbital’s is quite high. Hybridization increases the energy of 3s orbital so much that lone pair rather prefers to occupy unhybridized s orbital. For example, in PH3 , 600kJmol-1 of energy is required to hybridize the central atom. So, to avoid such energy demanding hybridization P forms bonds with unhybridized p orbitals leaving the lone pair in the spherical s orbital which leads to a bond angle close to 90° .

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Bond angle in PH4+is higher than that in PH3. Why?

Options:
A Due to increased repulsion between lone pairs.
B Due to the presence of a positive charge on phosphorus.
C Due to decreased repulsion between lone pairs.
D Due to the presence of a negative charge on phosphorus.

The correct answer is:
Due to decreased repulsion between lone pairs.

Solution:
Correct Option is : (C)
Due to decreased repulsion between lone pairs.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Bond dissociation enthalpy of $\mathrm{H}_2, \mathrm{Cl}_2$ and $\mathrm{HCl}$ Enthalpy of formation of $\mathrm{HCl}$ is

Options:
A $93 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-245 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $-93 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $245 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$-93 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
Key Idea:
$\Delta H_{\text {reaction }}=\Sigma$ Bond energy of reactant $-\Sigma$ Bond energy of product
$\begin{array}{ll}
\text {Here, } & \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}=434 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{Cl}-\mathrm{Cl}}=242 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{H}_{\mathrm{H}-\mathrm{Cl}}=431 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
\because & \frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{HCl}
\end{array}$
$\begin{aligned}
\Delta H_{\text {reaction }} & =\frac{1}{2} \Delta H_{\mathrm{H}-\mathrm{H}}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{Cl}-\mathrm{Cl}}-\Delta \mathrm{H}_{\mathrm{H}-\mathrm{Cl}} \\
& =\frac{1}{2} \times 434+\frac{1}{2} \times 242-431 \\
& =217+121-431 \\
& =-93 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2008 (Screening)
Bond enthalpy is defined as the amount of energy required for what specific action?

Options:
A Forming one mole of bonds between two atoms
B Breaking one mole of bonds between two atoms
C Changing the state of one mole of bonds
D Bonding two moles of atoms

The correct answer is:
Breaking one mole of bonds between two atoms

Solution:
The bond enthalpy is the amount of energy required to break one mole of bonds between two atoms in a gaseous state.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Bond order depends on the number of electrons in the bonding and antibonding molecular orbitals. Which of the following statements about bond order is correct?

Options:
A Can have a negative quantity
B Has always an integral value
C Can assume any positive or integral or fractional value including zero
D Is a non-zero quantity

The correct answer is:
Can assume any positive or integral or fractional value including zero

Solution:
bond order of a molecule is given by Half the difference between the number of electrons in bonding and antibonding electrons, it can have any positive or integral or fractional value including zero

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Bond order of 1.5 is shown by

Options:
A $\mathrm{O}_2^{+}$
B $\mathrm{O}_2^{-}$
C $\mathrm{O}_2^{2-}$
D $\mathrm{O}_2$

The correct answer is:
$\mathrm{O}_2^{-}$

Solution:
$\begin{aligned}
& \text{(a) } \text {MO configuration of } \mathrm{O}_2^{+}(8+8-1=15) \\
& =\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \stackrel{*}{\sigma} 2 s^2, \sigma 2 p_z^2, \\
& \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi 2 p_x^1 \approx \pi 2 p_y^0 \\
& \mathrm{BO}=\frac{N_b-N_a}{2}
\end{aligned}$
(where, $N_b=$ number of electrons in bonding molecular orbital
$N_a=$ number of electrons in anti-bonding molecular orbital
$\therefore \mathrm{BO}=\frac{10-5}{2}=2.5$
Similarly,
(b) $\mathrm{O}_2^{-}(8+8+1=17)$
so $\mathrm{BO}=\frac{N_b-N_a}{2}=\frac{10-7}{2}=1.5$
(c) $\mathrm{O}_2^{2-}(8+8+2=18)$
$\mathrm{BO}=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1$
(d) $\mathrm{O}_2(8+8=16)$
$\mathrm{BO}=\frac{10-6}{2}=2$
Thus, $\mathrm{O}_2^{-}$shows the bond order 1.5.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2012 (Screening)
Bond order of N2+ and N2- are same. Which relation is correct for N2+ and N2- ?

Options:
A Bond energy of N2+= bond energy of N2-
B Bond energy of N2+> bond energy of N2-
C Bond energy of N2+< bond energy of N2-
D Bond energy of N2+≥ bond energy of N2-

The correct answer is:
Bond energy of N2+> bond energy of N2-

Solution:
By MOT, the bond orders of both are 2.5. Now acc to MOT only, the compound having more electrons in antibonding orbitals will form a weaker bond and a longer bond. If their bond orders are equal. N2- has 5e- in antibonding while N2+ has 4. So, N2+ will make a stronger bond and will have a shorter bond length.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Bonding between deoxyribose and base in pyrimidine nucleoside molecule is

Options:
A 1'-1' glycosidic linkage
B 1'-6' glycosidic linkage
C 1'-9' glycosidic linkage
D 1'-4' glycosidic linkage

The correct answer is:
1'-1' glycosidic linkage

Solution:
Bonding between carbon number 1 of deoxyribose and nitrogen number 1 of the base in pyrimidine nucleoside molecule occurs, forming a 1'-1' glycosidic linkage.

Topic: Biology, Molecular Basis of Inheritance, NEET
Bones become fragile in

Options:
A Osteoporosis
B Gout
C Arthritis
D None of these

The correct answer is:
Osteoporosis

Solution:
Bones become fragile in osteoporosis, i.e., reduction in bone tissue mass causing weakness of skeletal strength. It is characterized by pain in the bone, specifically in the back and vertebral crush, usually in weight-bearing vertebrae.
Arthritis: Inflammation of joints.
Gout: Inflammation of joints due to the accumulation of uric acid crystals.

Topic: Biology, Locomotion and Movement, NEET
Boric acid is an acid because its molecule

Options:
A contains replaceable H+ ion.
B gives up a proton.
C accepts OH- from water releasing proton.
D combines with proton from water molecule.

The correct answer is:
accepts OH- from water releasing proton.

Solution:
Boric acid is an acid because its molecule accepts OH(-) from water releasing proton. It acts as a Lewis acid because of the electron deficient nature of boron in boric acid.
BOH3+H2O→BOH4-aq+H+(aq)

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET, NEET 2016 (Phase 2)
Boric acid is polymeric due to

Options:
A its acidic nature
B the presence of hydrogen bonds
C its monobasic nature
D its geometry

The correct answer is:
the presence of hydrogen bonds

Solution:
Correct Option is : (B)
the presence of hydrogen bonds

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Boron does not form B3+ ion because

Options:
A it has small size and high ionization energy
B it has high electronegativity
C it has high charge density (charge/radius ratio)
D none of the above

The correct answer is:
it has small size and high ionization energy

Solution:
The total energy to produce a B3+ ion (or the sum of the first three ionization energies) is too large to be compensated by lattice energies (in case of solids) or by hydration energies (in aqueous solution).

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Boron fluoride exists as BF3but boron hydride doesn’t exist as BH3. Give reason. In which form does it exist? Explain its structure.

Options:
A BH3is highly unstable and exists only in a dimerized form, (BH3)2.
B BH3is highly unstable and exists only in a polymeric form.
C BH3is highly stable and exists as a monomer, BH3.
D BH3is highly stable and exists as a dimer, (BH3)2.

The correct answer is:
BH3is highly unstable and exists only in a polymeric form.

Solution:
Correct Option is : (B)
BH3is highly unstable and exists only in a polymeric form.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Boron has an exceptionally high melting point in the group 13th elements, because–

Options:
A boron has the smallest size in the group
B boron atoms are joined by Vander Waals force
C boron is covalent solid
D boron has higher ionisation energy

The correct answer is:
boron is covalent solid

Solution:
Boron exists in giant covalent structure. Boron due to small size forms strong covalent bonds with the neighbouring atoms.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Boron plays role in all except-

Options:
A Cell elongation
B Cell differentiation
C Membrane functioning
D None of the above

The correct answer is:
None of the above

Solution:
Boron is required for uptake and utilisation of Ca²⁺, membrane functioning, pollen germination, cell elongation, cell differentiation and carbohydrate translocation.

Topic: Biology, Mineral Nutrition, NEET
Botanical gardens and zoological parks are collection of

Options:
A exotic living species only
B endemic living species only
C only local plants and animals
D endemic and exotic living species

The correct answer is:
endemic and exotic living species

Solution:
Botanical gardens are place where living plants are grown and in zoological parks living animals are kept. It includes both endemic species of that area and exotic species which are brought from different place.

Topic: Biology, Biodiversity and its Conservation, NEET
Botanical gardens and zoological parks have

Options:
A Collection of endemic living species only
B Collection of exotic living species only
C Collection of endemic and exotic living species
D Collection of only local plants and animals

The correct answer is:
Collection of endemic and exotic living species

Solution:
They are used to collect different species of plants & animals.

Topic: Biology, The Living World, NEET
Both antacids and antiallergic drugs are antihistamines but they cannot replace each other. Explain why?

Options:
A Antacids are stronger antihistamines than antiallergic drugs.
B Because both antiallergic and antacid drugs work on different receptors.
C Antacids are specifically designed to treat allergies.
D Antiallergic drugs are specifically designed to treat acid reflux.

The correct answer is:
Because both antiallergic and antacid drugs work on different receptors.

Solution:
Correct Option is : (B)
Because both antiallergic and antacid drugs work on different receptors.

Topic: Chemistry, Chemistry in Everyday Life, NEET
Both, autogamy and geitonogamy are prevented in

Options:
A papaya
B cucumber
C castor
D maize

The correct answer is:
papaya

Solution:
Autogamy involves pollination within the same flower, while geitonogamy involves transfer of pollen grains from the anther of one flower to the stigma of another flower of the same plant. Both the processes are prevented in papaya because it is a dioecious plant (i.e., male and female sex organs are borne on separate plants) and it always needs cross-pollination.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2012 (Screening)
Both centrioles in a centrosome lie ..........(i)........ to each other and each of them contains ....(ii)..... peripheral fibril so ........(iii)......... array is present.

Options:
A (i) Parallel (ii) Triplet (iii) 9+0
B (i) Perpendicular (ii) Doublets (iii) 9+0
C (i) Parallel (ii) Doublet (iii) 9 + 0
D (i) Perpendicular (ii) Triplet (iii) 9+0

The correct answer is:
(i) Perpendicular (ii) Triplet (iii) 9+0

Solution:
Centrosome consists of two centrioles which lies perpendicular to each other and a pair of centriole is called as diplosome. Each centriole consists of nine peripheral fibril and a central protein hub. It thus shows 9+0 arrangement.

Topic: Biology, Cell - The Unit of Life, NEET
Both centrioles in a centrosome lie .......(i)....... to each other and each of them contains .......(ii)....... peripheral fibril with .......(iii)....... arrangement of microtubules.

Options:
A (i) parallel (ii) triplet (iii) 9 + 0
B (i) perpendicular (ii) doublet (iii) 9 + 2
C (i) parallel (ii) doublet (iii) 9 + 0
D (i) perpendicular (ii) triplet (iii) 9 + 0

The correct answer is:
(i) perpendicular (ii) triplet (iii) 9 + 0

Solution:
A centrosome consists of two centrioles that lie perpendicular to each other. It shows a cart wheel-like arrangement having 9 + 0 arrangement in which each peripheral fibril is a triplet.

Topic: Biology, Cell - The Unit of Life, NEET
Both deoxyribose and ribose belong to a class of sugars called:

Options:
A trioses
B hexoses
C pentoses
D polysaccharides

The correct answer is:
pentoses

Solution:
Both deoxyribose and ribose belong to a class of sugars called pentoses which contain 5 carbons.

Topic: Biology, Molecular Basis of Inheritance, NEET
Both lithium and magnesium are known to be harder and lighter than other elements in their respective groups. Which property contributes to their hardness?

Options:
A High electronegativity
B High atomic radius
C High density
D Strong metallic bonding

The correct answer is:
Strong metallic bonding

Solution:
The hardness of lithium and magnesium is due to strong metallic bonding in their crystal structures.

Topic: Chemistry, s Block Elements, NEET
Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect is:

Options:
A Both form soluble bicarbonates
B Both form nitrides
C Nitrates of both Li and Mg yield NO2 and O2 on heating
D Both form basic carbonate

The correct answer is:
Both form basic carbonate

Solution:
LiC O 3 is acidic while MgC O 3 is basic carbonate

Topic: Chemistry, s Block Elements, NEET
Both Sickle cell anaemia and Huntington's chorea are:

Options:
A virus-related disease
B bacteria-related diseases
C congenital disorders
D pollutant-induced disorders

The correct answer is:
congenital disorders

Solution:
Sickle cell disease is a group of inherited red blood cell disorders that affects haemoglobin, the protein that carries oxygen through the body.
Huntington's disease is a progressive brain disorder caused by a defective gene. This disease causes changes in the central area of the brain, which affect movement, mood and thinking skills. Both sickle cell anaemia and Huntington's chorea are congenital genetic disorders.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2006
Both the lobes of thyroid gland are interconnected with

Options:
A Connective tissue called the stromal tissue.
B Connective tissue called the thyroid tissue.
C Connective tissue called follicle tissue
D Connective tissue called the isthmus tissue.

The correct answer is:
Connective tissue called the isthmus tissue.

Solution:
The thyroid gland is composed of two lobes which are located on either side of the trachea. Both the lobes are interconnected with a thin flap of connective tissue called the isthmus.

Topic: Biology, Chemical Coordination and Integration, NEET
Both the strings, shown in the figure are made of the same material and have the same cross-section. The pulleys are light. The wave speed of a transverse wave in the string AB is v1 and in the CD it is v2. Then v1v2 is

Options:
A 1
B 2
C 2
D 12

The correct answer is:
12

Solution:
We know that v=Tμ
Therefore v1v2=T1T2
v1v2=T2T
v1v2=12

Topic: Physics, Waves and Sound, NEET
Bovine spongiform encephalopathy is caused by

Options:
A an agent having RNA without protein coat
B an agent having RNA with protein coat
C an agent consisting of abnormally folded protein
D an agent having ill-defined nucleus wth cell wall

The correct answer is:
an agent consisting of abnormally folded protein

Solution:
Bovine spongiform encephalopathy or mad cow disease is caused by prions which are abnormally folded proteins that can transmit their abnormality to normal proteins. RNA without protein coat is seen in a viroid. RNA with a protein coat is seen in plant viruses and certain animal viruses. An ill-defined nucleus with a cell wall is seen in bacteria.

Topic: Biology, Biological Classification, NEET
Bowman's glands are found in:

Options:
A juxtamedullary nephrons
B olfactory epithelium
C external auditory canal
D cortical nephrons only

The correct answer is:
olfactory epithelium

Solution:
Bowman's glands are situated in the olfactory mucosa, beneath the olfactory epithelium, in the lamina propria, a connective tissue also containing fibroblasts, blood vessels, and bundles of fine axons from the olfactory neurons.
Related Theory
Bowman's capsule is a part of the nephron that forms a cup-like sack surrounding the glomerulus. It serves as a filter to remove organic wastes, excess inorganic salts, and water.

Topic: Biology, Neural Control and Coordination, NEET, NEET 2006
Bowman's glands are located in the:

Options:
A anterior pituitary
B female reproductive system of cockroach
C olfactory epithelium of our nose
D proximal end of uriniferous tubules.

The correct answer is:
olfactory epithelium of our nose

Solution:
Bowman's glands occur below the olfactory epithelium which secretes mucus over the epithelium to keep it moist.
Caution
Bowman's capsule is a cup-like sac at the beginning of the tubular component of $a$ nephron in the mammalian kidney that performs the first step in the filtration of blood to form urine.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2007
Brainstem of human brain consists of:

Options:
A Thalamus, Hypothalamus and Corpora quadrigemina
B Amygdala, Hippocampus and Corpus Callosum
C Mid-brain, Pons and Medulla Oblongata
D Forebrain, Cerebellum and Pons

The correct answer is:
Mid-brain, Pons and Medulla Oblongata

Solution:
Option (3) is the answer because mid brain, pons and medulla oblongata are the three major regions to make up the brain stem of human brain. Amygdala and hippocampus form a complex structure called the limbic system. Hindbrain comprises pons, cerebellum and medulla.

Topic: Biology, Neural Control and Coordination, NEET, NEET 2023 (Manipur)