datavorous/entrance-exam-dataset huggingface dataset:
The haploid content of human DNA is ________.

Options:
A 1.3 × 109 bp
B 3.3 × 109 bp
C 3.3 × 106 bp
D 5.3 × 106 bp

The correct answer is:
3.3 × 109 bp

Solution:
The nucleic acid is a polymer of nucleotide. Nitrogen base attaches to pentose sugar by a glycosidic bond to form nucleoside. Phosphate attaches to 5' carbon of sugar to form nucleotide. A bacteriophage Ï 174 has 5386 nucleotides, bacteriophage lambda has 48502 base pairs (bp), Escherichia coli has 4.6 Ã 106 bp, and haploid content of human DNA is 3.3 Ã 109 bp.

Topic: Biology, Molecular Basis of Inheritance, NEET
The hardest part of the animal body is

Options:
A Bone
B Hair
C Dentine
D Enamel

The correct answer is:
Enamel

Solution:
The hardest part of the animal body is enamel. which is found on the upper layer of teeth.

Topic: Biology, Digestion and Absorption, NEET
The hardness of a water sample containing 10-3M MgSO4 expressed as CaCO3 equivalents (in ppm) is
(molar mass of MgSO4 is 120.38 g/mol and molar mass of CaCO3 is 100.08 g/mol)

Options:
A 50
B 75
C 90
D 100

The correct answer is:
100

Solution:
Given: 10-3M MgSO4
i.e. 10-3 mole MgSO4 present in 1 litre solution
nMgSO4=nCaCO3
Mass of CaCO3=10-3Ã100
ppm (in terms of CaCO3 ) =mass of CaCO3 weight of solutionÃ106
ppm (in terms of CaCO3 ) =10-3Ã1001000Ã106=100 ppm

Topic: Chemistry, s Block Elements, NEET
The harmful effects of air pollution depends on the

Options:
A concentration of pollutants.
B duration of exposure.
C type of organism.
D All of these

The correct answer is:
All of these

Solution:
Air pollutants deleteriously affect the respiratory system of humans and animals. The harmful effect of air pollution depends on the concentration of pollutants, duration of exposure and type of organism.

Topic: Biology, Environmental Issues, NEET
'The Haryana Kisan Welfare Club' is run by

Options:
A Ramesh Chandra Dagar
B Ahmed Khan
C Swaminathan
D Ramdeo Mishra

The correct answer is:
Ramesh Chandra Dagar

Solution:
Integrated organic farming is a cyclical zero waste procedure, where waste products from one process are cycled in as nutrients for other processes. To spread information and help with the practice of integrated organic farming Ramesh Chandra Dagar has created the Haryana Kisan Welfare club.

Topic: Biology, Environmental Issues, NEET
The heart of a man pumps 5 litres of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury be 13.6×103 kg m-3 and g=10 m s-2, then the power of heart in watt is:

Options:
A 2.35
B 3.0
C 1.50
D 1.70

The correct answer is:
1.70

Solution:
\(\begin{aligned} & \text { Pressure }=\mathrm{P}=150 \mathrm{~mm} \mathrm{Hg} \\ & \text { Pumping rate }=\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{5 \times 10^{-3}}{60} \mathrm{~m}^3 / \mathrm{s} \\ & \text { Power of heart }=\text { P. } \frac{\mathrm{dV}}{\mathrm{dt}}=\rho \mathrm{gh} \times \frac{\mathrm{dV}}{\mathrm{dt}} \\ & =\left(13.6 \times 10^3\right)(10) \times(0.15) \times \frac{5 \times 10^{-3}}{60} \\ & =\frac{13.6 \times 5 \times 0.15}{6}=1.70 \mathrm{watt}\end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2015 (Phase 2)
The heart sound heard on the stethoscope when the blood pressure in the artery is maximum is due to

Options:
A Closure of semi-lunar valves
B Closure of atrio-ventricular valves
C Opening of semi-lunar valves
D Opening of atrio-ventricular valves

The correct answer is:
Closure of atrio-ventricular valves

Solution:
The blood pressure in the artery is maximum during the systole of the heart. There are two heart sounds lub and dub. Lub is a systolic heart sound which is produced due to the closure of atrioventricular valves (bicuspid and tricuspid valves). Dub is a diastolic heart sound which is produced due to closure of semilunar valves (aortic and pulmonary valves)

Topic: Biology, Body Fluids and Circulation, NEET
The heat capacity of a bomb calorimeter is 500 J/oC. When 0.1 g of a methane was burnt in this calorimeter, the temperature rose by 2oC. The value of Î E per mole will be

Options:
A + 10 kJ
B - 10 kJ
C + 160 kJ
D - 160 kJ

The correct answer is:
- 160 kJ

Solution:
Δ E = - M w × C × Δ t
= - 1 6 0.1 × 5 0 0 × 2 = - 1 6 0 × 1 0 3 J = - 160 kJ (minus sign is used as heat is evolved.)

Topic: Chemistry, Thermodynamics (C), NEET
The heat generated in a circuit is given by Q=I2 Rt, where I is current, R is resistance, and t is time. If the percentage errors in measuring I,R, and t are 2%, 1% and 1% respectively, then the maximum error in measuring heat will be

Options:
A 2%
B 3%
C 4%
D 6%

The correct answer is:
6%

Solution:
Q=I2Rt

Required error is 2×2%+1%+1%=6%

Topic: Physics, Units and Dimensions, NEET
The heat of combustion of carbon to CO2 is -393.5 kJ/mol. The heat released upon the formation of 35.2 g of CO2 from carbon and oxygen gas is

Options:
A -315 kJ
B +315 kJ
C -630 kJ
D -3.15 kJ

The correct answer is:
-315 kJ

Solution:
Formation of CO2 from carbon and dioxygen gas can be represented as
Cs+O2(g)→CO2(g) ;∆fH=-393.5 kJ mol-1
(1 mole=44 g)
The heat released on formation of 44 g CO2
=-393.5 kJ mol-1
=-393.5 kJ mol-144 g×35.2 g
=-315 kJ

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2015 (Phase 2)
The heat of formation of NH3(g) is -46 kJmol-1 . The ÎH (in kJ) of the reaction is
2NH3(g)âN2(g)+3H2(g)

Options:
A 46
B - 46
C 92
D - 92

The correct answer is:
92

Solution:
12N2(g)+32H2(g)→NH3(g)
ΔHr=-46kJ/mol
2NH3(g)→N2(g)+3H2(g)
ΔHr=-2ΔHr
=-2(-46)=92kJ

Topic: Chemistry, Thermodynamics (C), NEET
The heat of neutralisation of a strong base and a strong acid is 13.7 kcal. The heat released when 0.6 mole HCl solution is added to 0.25 mole of NaOH is

Options:
A 3.425 kcal
B 8.22 kcal
C 11.645 kcal
D 13.7 kcal

The correct answer is:
3.425 kcal

Solution:
For the neutralization between a strong and a strong base, equal number of moles of H+ (from acid) and O-H (from base) come out. For one mole of following equation can be drawn
So, 0.25 mole of H2O is produced.
So, heat released =  0.25 × 13.7 kcal = 3.425 kcal

Topic: Chemistry, Thermodynamics (C), NEET
The heat of neutralisation will be highest for which of the following neutralisation reactions?

Options:
A NH4OH and HCl
B NH4OH and CH3COOH
C NaOH and HCl
D NaOH and CH3COOH

The correct answer is:
NaOH and HCl

Solution:
Heat of neutralisation between strong acid and a strong base is highest and it is about -13.7Kcal.

Topic: Chemistry, Thermodynamics (C), NEET
The heat of neutralization of NaOH with HCl is 57.3 KJ and with HCN is 12.1 KJ. The heat of ionization of HCN is

Options:
A +69.4 KJ
B +45.2 KJ
C -45.2 KJ
D -69.4 KJ

The correct answer is:
+45.2 KJ

Solution:
'HCN' is a weak acid
â´57.3-12.1=45.2 KJ

Topic: Chemistry, Ionic Equilibrium, NEET
The heat of solution of one mole of Na2SO4.10H2O is -78.7 kJ, and that of dehydration, -81.6 kJ. The heat of solution (kJ) of anhydrous sodium sulphate is

Options:
A +2.9 kJ
B +160.3 kJ
C -2.9 kJ
D -160.3 kJ

The correct answer is:
+2.9 kJ

Solution:
From Hess's law, Q=− −81.6 kJ + −78.7 kJ
=81.6 kJ−78.7 kJ=+2.9 kJ

Topic: Chemistry, Thermodynamics (C), NEET
The heat produced (in calories) in a resistance R when a current I amperes flows through it for t seconds is given by the expression

Options:
A I2Rt4.2
B IR2t4.2
C 4.2IRt2
D IRt24.2

The correct answer is:
I2Rt4.2

Solution:
H=I2RtJ=I2Rt4.2 cal

Topic: Physics, Current Electricity, NEET
The heating of carbonyl compounds with H2NNH2 and strong base is called

Options:
A Wolf-Kishner reduction
B Clemmensen reduction
C Rosenmund reduction
D Catalytic reduction

The correct answer is:
Wolf-Kishner reduction

Solution:
The heating of carbonyl compounds with H2NNH2 and strong base is called Wolf-Kishner reduction.
RCH=O+H2NNH2→RCH=NNH2→Strong baseRCH3+N2

Topic: Chemistry, Aldehydes and Ketones, NEET
The heating of phenyl-methyl ether with HI produces,

Options:
A ethyl chloride.
B iodobenzene.
C phenol.
D benzene.

The correct answer is:
phenol.

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET, NEET 2017
The heating of phenyl-methyl ethers with HI produces

Options:
A Iodobenzene
B Phenol
C Benzene
D Ethyl chlorides

The correct answer is:
Phenol

Solution:
Anisole reacts with HI to give methyl iodide & phenol.
CH3-O-C6H5→HICH3I+C6H5OH

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The heating rods of iron embedded in charcoal powder are used to make railway wagon axles. The process is known as

Options:
A Case hardening
B Sheradizing
C Annealing
D Tempering

The correct answer is:
Case hardening

Solution:
Case hardening : The process of hardening the surface of wrought iron by depositing a surface layer of steel on it is called case-hardening it is done by heating wrought iron in contact with potassium ferrocyanide.
Alternatively, case hardening can also be done by heating wrought iron with charcoal and then plunging it a suitable oil.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be

Options:
A Re4
B Re2
C 3Re4
D Re

The correct answer is:
Re

Solution:
g′=gR2R+h2
g4=gR2R+h2
14=R2R+h2
R+h2=4R2
R+h=2R
h=R=Re

Topic: Physics, Gravitation, NEET
The height above the surface of the earth at which the acceleration due to gravity becomes g9 (where g is the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is

Options:
A 2R
B R3
C R2
D 2R

The correct answer is:
2R

Solution:
gâ²=GMR+h2 (acceleration due to gravity at height h)
â¹ g9=GMR2âR2R+h2
=gRR+h2
â¹ 19=RR+h2
â¹ RR+h=13
â¹ 3R=R+h
â¹ 2R=h

Topic: Physics, Gravitation, NEET
The height at which the weight of a body becomes $1 / 16$ th, its weight on the surface of earth (radius $R$ ), is

Options:
A $5 R$
B $15 R$
C $3 R$
D $4 R$

The correct answer is:
$3 R$

Solution:
According to question $\frac{G M m}{(R+h)^2}=\frac{1}{16} \frac{G M m}{R^2}$
$\begin{array}{l}
\frac{1}{(R+h)^2} =\frac{1}{16 R^2} \\
\text {or } \frac{R}{R+h} =\frac{1}{4} \\
\text {or } \frac{R+h}{R} =4 \\
h =3 R
\end{array}$

Topic: Physics, Gravitation, NEET, NEET 2012 (Screening)
The height of a HCP unit cell is 5.715 Å. What is the volume of the unit cell in Å3?

Options:
A 91
B 182
C 273
D 82.5

The correct answer is:
182

Solution:
The height of the HCP unit cell =423.r=5.715
r= 5.715Ã 3 2 Ã4 =1.75 â«
The base area of the unit cell =63r2
The volume of the unit cell =base area Ãheight=24 2 r 3 =181.8 â«3

Topic: Chemistry, Solid State, NEET
The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ratio of density of mercury to that of air is 104. Then the height of the hill is

Options:
A 250Â m
B 2.5 km
C 1.25Â km
D 750Â m

The correct answer is:
2.5 km

Solution:
Difference of pressure between sea level and the top of hill
ÎP=h1-h2ÃÏHgÃg=75-50Ã10-2ÃÏHgÃg .....(i)
and pressure difference due to h meter of air
ÎP=hÃÏairÃg .....(ii)
By equating (i) and (ii) we get
hÃÏairÃg=75-50Ã10-2ÃÏHgÃg
â´ h=25Ã10-2ÏHgÏair=25Ã10-2Ã104=2500 m
â´ Height of the hill =2.5 km

Topic: Physics, Mechanical Properties of Fluids, NEET
The height of a plant is controlled by three genes. The maximum height in the species is 24 inches, whereas the minimum height in the species is 6 inches. What would be the height of a plant of the same species whose genotype is AaBBcc?

Options:
A 12 inches
B 15 inches
C 18 inches
D 21 inches

The correct answer is:
15 inches

Solution:
This is an example of polygenic or quantitative inheritance. Let the height contributed by each dominant allele be 'x' inches and that by each recessive allele be 'y' inches. AABBCC will have maximum height. Thus, 6x = 24 and x = 4. aabbcc will have a minimum height. Therefore, 6y = 6 and y = 1. The given genotype AaBBcc has 3 dominant alleles and 3 recessive alleles. Thus 3x+3y=34+3(1)=15

Topic: Biology, Principles of Inheritance and Variation, NEET
The height of a plant is controlled by three genes. The Maximum height in the species is 24 inches while the minimum height in the species is 6 inches. What would be the height of a plant of the same species whose genotype is AaBBcc?

Options:
A 12 inches
B 15 inches
C 18 inches
D 21 inches

The correct answer is:
15 inches

Solution:
Polygenic inheritance is defined as characters which are determined by two or more gene pairs, and they have additive or cumulative effect. Such genes are called polygenes.
In such inheritance, each polygene contribution = (Maximum height - Minimum height)/No. of polygene =24-6/6=3 inches. Thus, 3inches height is contributed by each dominant allele of the polygene.
Contribution of dominant polygene in plant having genotype AaBBCc=3×3=9
Total height of plant with genotype AaBBCc = Minimum height + Contribution of dominant allele = 6+9=15 inches

Topic: Biology, Principles of Inheritance and Variation, NEET
The Henry's law constant $\left(\mathrm{K}_H\right)$ values of three gases $(\mathrm{A}, \mathrm{B}, \mathrm{C})$ in water are $145,2 \times 10^{-5}$ and 35 kbar , respectively. The solubility of these gases in water follow the order:

Options:
A B $\gt$ C $\gt$ A
B A $\gt$ C $\gt$ B
C A $\gt$ B $\gt$ C
D B $\gt$ A $\gt$ C

The correct answer is:
B $\gt$ C $\gt$ A

Solution:
Value of Henry's law constant $\propto \frac{1}{\text { Solubility of gas }}$
Higher the value of $\mathrm{K}_H$ at a given pressure, lower is the solubility of the gas in the liquid.
$\mathrm{K}_H$ value of gases (given) : $\mathrm{A}\gt\mathrm{C}\gt\mathrm{B}$
$\therefore \quad$ Order of solubility of gases in water : $\mathrm{B}\gt\mathrm{C}\gt\mathrm{A}$

Topic: Chemistry, Solutions, NEET, NEET 2024
The hepatic portal vein drains blood to the liver from

Options:
A Heart
B Stomach
C Kidney
D Intestine

The correct answer is:
Intestine

Solution:
The hepatic portal system is found in all vertebrates. The flow of deoxygenated blood from the digestive and absorptive parts of the alimentary canal (especially from the intestine) to the liver before returning to the systemic circulation is called hepatic portal circulation.

The blood which comes from the alimentary canal contains absorbed food such as glucose, amino acids. The excess of glucose is converted into glycogen and is stored in the liver for later use.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2017
The hepatitis-B vaccine has been produced on a large scale through

Options:
A Recombinant DNA technology using Claviceps
B Recombinant DNA technology using E. coli
C Recombinant DNA technology using yeast
D Recombinant DNA technology using mice

The correct answer is:
Recombinant DNA technology using yeast

Solution:
Recombinant DNA technology has allowed the production of antigenic polypeptides of pathogens in bacteria or yeast. Vaccines produced using this approach allow large scale production and hence greater availability for immunization, e.g., hepatitis B vaccine (Recombivax HB) produced from yeast.

Topic: Biology, Human Health and Diseases, NEET
The Hershey and Chase's experiment using bacteriophage demonstrates:

Options:
A Semi-autonomus nature of DNA
B All viruses are DNA viruses
C DNA is genetic material
D DNA show repair mechanism

The correct answer is:
DNA is genetic material

Solution:
Bacteriophage experiment was conducted by Hershey and Chase, (1952). They selected T2 type phages for experimentation. From this experiment, they concluded that only DNA (and not proteins) passes from one generation to another.

Topic: Biology, Molecular Basis of Inheritance, NEET
The heterocysts present in cyanobacteria are sites of

Options:
A Photosynthesis
B Nitrogen fixation
C Carbon dioxide assimilation
D Gaseous exchange

The correct answer is:
Nitrogen fixation

Solution:
In cyanobacterium, the heterocysts are sights of nitrogen fixation. These specialized cells are larger as compared to cells and are distinctly visible under microscope.

Topic: Biology, Biological Classification, NEET
The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Which of the following explains this?

Options:
A The hexaquo manganese(II) ion undergoes oxidation to form the hexacyanoion, resulting in a change in the number of unpaired electrons.
B The hexacyanoion has a higher oxidation state for manganese, resulting in a different electron configuration compared to the hexaquo complex.
C Water, being a weak field ligand, was not able to pair up the electrons of the 3d subshell, but whereas in the case of hexacyano ion the CN−is a strong field ligand and it causes pairing of electrons in the d orbitals
D The coordination number of the manganese(II) ion in the hexacyano complex is higher, causing the change in the number of unpaired electrons.

The correct answer is:
Water, being a weak field ligand, was not able to pair up the electrons of the 3d subshell, but whereas in the case of hexacyano ion the CN−is a strong field ligand and it causes pairing of electrons in the d orbitals

Solution:
Correct Option is : (C)
Water, being a weak field ligand, was not able to pair up the electrons of the 3d subshell, but whereas in the case of hexacyano ion the CN−is a strong field ligand and it causes pairing of electrons in the d orbitals

Topic: Chemistry, Coordination Compounds, NEET
The high value of BOD (Biochemical Oxygen Demand) indicates that:

Options:
A Water is pure
B Water is highly polluted
C Water is less polluted
D Consumption of inorganic matter in the water is higher by the microbes

The correct answer is:
Water is highly polluted

Solution:
Biochemical Oxygen Demand (BOD) is the amount of oxygen that would be consumed if all the organic matter in one litre of water were oxidized by bacteria. The BOD test measures the rate of uptake of oxygen by micro-organisms in a sample of water and thus, indirectly, BOD is a measure of the organic matter present in the water. The greater the BOD of waste water, more is its polluting potential.

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2015 (Phase 1)
The highest boiling point is expected for

Options:
A iso-octane
B n-octane
C 2, 2, 3, 3-teramethylbutane
D n-butane

The correct answer is:
n-octane

Solution:
The highest boiling point is expected for n-octane and least for n-butane.
Boiling point of alkane increases with the increase in carbon atoms. In isomeric alkanes, the molecule becomes more and more compact with the increase in branching and therefore, surface area of the molecule decreases and consequently van dar Waal's forces, which depends upon surface area, decrease and hence, boiling point decreases with the increase in branching.

Topic: Chemistry, Hydrocarbons, NEET
The highest DDT concentration in aquatic food chain shall occur in

Options:
A Phytoplankton
B Seagull
C Crab
D Eel

The correct answer is:
Seagull

Solution:
Biomagnification is also called biological amplification. It is a phenomenon in which certain harmful chemicals get accumulated in the tissues of organisms in higher concentrations through the food chain. Biocides including pesticides, fungicides, insecticides, rodenticides, and herbicides show biomagnification.

Thus, in a food chain beginning at the marine ecosystem, harmful chemicals (DDT) that are deposited into the oceans, which are transferred to the plankton. These zooplanktons are fed on by smaller fish, which are in turn consumed by larger fish. These larger fish serve as the food for birds of prey like seagulls and hawks. Due to biomagnification, the highest quantity of DDT will be found in these birds.

Topic: Biology, Environmental Issues, NEET, NEET 2016 (Phase 2)
The highest factor contributing to the dry weight of organisms is

Options:
A Water
B Carbon
C Hydrogen
D Nitrogen

The correct answer is:
Carbon

Solution:
Dry weight of the organism is the weight after water is removed. Carbon constitutes 49% of dry weight in organisms. Thus, it contributes maximally to dry weight of the organism.

Topic: Chemistry, Biomolecules, NEET
The highest lattice energy corresponds to

Options:
A SrO
B BaO
C MgO
D CaO

The correct answer is:
MgO

Solution:
Lattice energy = −( Z + Z − )e 2 r
Here r is the interionic distance and it is minimum in case of MgO so, MgO has the highest lattice energy here.

Topic: Chemistry, s Block Elements, NEET
The highest number of helium atoms is in

Options:
A 4 u of helium
B 4 g of helium
C 2.271098 L of helium at STP
D 4 mol of helium

The correct answer is:
4 mol of helium

Solution:
(1) $4 u$ of $\mathrm{He}=\frac{4 u}{4 u}=1 \mathrm{He}$ atom
(2) 4 g of Helium $=\frac{4 \mathrm{~g}}{4 \mathrm{~g}}$ mole $=1$ mole $=\mathrm{N}_{\mathrm{A}} \mathrm{He}$ atom
(3)
$\begin{aligned}
2.2710982 \text { of He at STP } & =\frac{2.271}{22.710982} \text { mole } \\
& =0.1 \mathrm{~mole} \\
& =0.1 \mathrm{~N}_{\mathrm{A}} \mathrm{He} \text { atom }
\end{aligned}$
(4) 4 mol of $\mathrm{He}=4 \mathrm{~N}_{\mathrm{A}} \mathrm{He}$ atoms

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2024
The highest number of species in the world is represented by

Options:
A fungi
B mosses
C algae
D lichens

The correct answer is:
fungi

Solution:
Fungi represent the highest number of species in the world. Around 100000 species of fungi have been formally described by taxonomists, but the global biodiversity of kingdom fungi is not fully understood.

Topic: Biology, Biological Classification, NEET, NEET 2012 (Screening)
The highest oxidation no of iron in its oxide form is:

Options:
A +5
B +3
C +2
D +4

The correct answer is:
+3

Solution:
Correct Option is : 2
+3

Topic: Chemistry, d and f Block Elements, NEET
The highest oxidation state of Mn that combines with halogen to form halides is:

Options:
A +7
B +5
C +4
D +6

The correct answer is:
+4

Solution:
Correct Option is : 3
+4

Topic: Chemistry, d and f Block Elements, NEET
The hilum represents the junction between

Options:
A the two integuments
B ovule and funicle
C micropyle and chalaza
D ovule and placenta

The correct answer is:
ovule and funicle

Solution:
The ovule is a small structure attached to the placenta by means of a stalk called funicle. The body of the ovule fuses with funicle in the region called hilum. Thus, hilum represents the junction between ovule and funicle. Each ovule has one or two protective envelopes called integuments. Integuments encircle the nucellus except at the tip where a small opening called the micropyle is organised. Opposite the micropylar end, is the chalaza, representing the basal part of the ovule.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The historic convention on Biological Diversity held in Rio de Janeiro in 1992 is known as:

Options:
A CITES Convention
B The Earth Summit
C G-16 Summit
D MAB Programme

The correct answer is:
The Earth Summit

Solution:
The Earth Summit: This historical convention on biological diversity held in Rio-de-Janeiro, Brazil in 1992. Attending nations take appropriate measure for conservation of biodiversity and sustainable utilization of its benefits.

Topic: Biology, Biodiversity and its Conservation, NEET
The historic convention on Biological diversity held in Rio de Janeiro in 1992 is known as ____.

Options:
A The World Summit
B The Earth Summit
C G-16 Summit
D MAB Programme

The correct answer is:
The Earth Summit

Solution:
The historic Convention on Biological Diversity (‘The Earth Summit’) held in Rio de Janeiro in 1992, called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilisation of its benefits.

Topic: Biology, Biodiversity and its Conservation, NEET
The HNH angle value is higher than HPH, HAsH, and HSbH angles. Why? [Hint Can be explained on the basis of sp3hybridization in NH3and only s–p bonding between hydrogen and other elements of the group].

Options:
A NH3forms stronger bonds with hydrogen.
B NH3has a higher electronegativity.
C NH3undergoes sp3hybridization.
D NH3forms a larger number of bonds.

The correct answer is:
NH3undergoes sp3hybridization.

Solution:
Correct Option is : (C)
NH3undergoes sp3hybridization.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET