datavorous/entrance-exam-dataset huggingface dataset:
The controlled aerobic combustion of wastes inside chambers at a temperature of 900-1300°C is known as

Options:
A Incineration
B Recycling
C Pyrolysis
D Sanitary dumping

The correct answer is:
Incineration

Solution:
Incineration is controlled aerobic combustion of wastes inside chambers at temperature of 900 -1300°C. Incineration produces some pollutants. To check these pollutants incinerators are filled with scrubbers and electrostatic precipitators to prevent release of smoke and toxic chemicals directly to the environment.

Topic: Biology, Environmental Issues, NEET
The conversion:
Can be effected by

Options:
A LiAlH4 reduction
B Clemmensen’s reduction
C NaBH4 reduction
D H2/Ni reduction

The correct answer is:
NaBH4 reduction

Solution:
The NaBH4 is mild reducing agent which reduces carbonyl into alcohol but do not reduce ester. The clemmenson’s reduction converts carbonyl group into alkane.

Topic: Chemistry, Aldehydes and Ketones, NEET
The conversion of ADP into ATP by the electron transport system is called ___.

Options:
A Respiratory quotient
B Oxidation
C Oxidative phosphorylation
D none of the above

The correct answer is:
Oxidative phosphorylation

Solution:
The conversion of ADP into ATP by the electron transport system is called oxidative phosphorylation

Topic: Biology, Respiration in Plants, NEET
The conversion of ethyl chloride into diethyl ether takes place by

Options:
A Williamson's synthesis
B Perkin's reaction
C Wurtz reaction
D Grignard reaction

The correct answer is:
Williamson's synthesis

Solution:
Williamson's synthesis
C2H5ONa + ClC2H5 → C2H5OC2H5 + NaCl
diethyl ether

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
The conversion of fructose-6-phosphate to fructose-1,6-diphosphate is catalysed by:

Options:
A Phosphofructose kinase
B Aldolase
C Hexokinase
D None of the above

The correct answer is:
Phosphofructose kinase

Solution:
Fructose 6-phosphate + ATP â¶ Phosphofructokinase, Mg 2 + Fructose 1, 6 – diphosphate + ADP.

Topic: Biology, Respiration in Plants, NEET
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Options:
A The rate will increase 4 times
B The rate will increase 12 times
C The rate will increase 6 times
D The rate will increase 9 times

The correct answer is:
The rate will increase 9 times

Solution:
Correct Option is : (D)
The rate will increase 9 times

Topic: Chemistry, Chemical Kinetics, NEET
The conversion of NO3- to NO2- occurs in

Options:
A Cytosol of plant cells
B Chloroplast
C Mitochondria
D Peroxisome

The correct answer is:
Cytosol of plant cells

Solution:
Reactions throughout the course of the nitrate assimilation pathway occur in separate cells and cellular compartments; nitrate is first transferred from root epidermal cells to the cytosol of leaf cells via vascular tissues, and the subsequently reduced nitrogen species move, also via vascular tissues, from chloroplasts to sink organs.

Topic: Biology, Mineral Nutrition, NEET
The conversion of nitrate to nitrogen is called

Options:
A Nitrification
B Denitrification
C Ammonification
D Nitrogen fixation

The correct answer is:
Denitrification

Solution:
Denitrification is the reduction of nitrate to gaseous compounds of nitrogen. Under anaerobic conditions, some microorganisms use nitrate and other oxidized ions as a source of oxygen, e.g., Pseudomonas denitrificans, Thiobacillus denitrificans, Micrococcus denitrificans, etc.

Topic: Biology, Mineral Nutrition, NEET
The cook Mary Mallon was a carrier of which of the following diseases?

Options:
A The disease caused by Salmonella typhi.
B The disease caused by Streptococcus pneumoniae.
C The disease caused by Staphylococcus aureus.
D The disease caused by Treponema pallidum.

The correct answer is:
The disease caused by Salmonella typhi.

Solution:
The cook Mary Mallon was a carrier of the disease typhoid, caused by the bacterium Salmonella typhi. Being a typhoid carrier, she continued to spread typhoid for years together through the food she cooked and came to be known as Typhoid Mary.

Topic: Biology, Human Health and Diseases, NEET
The coordinates of centre of mass of the following quarter circular arc is -

Options:
A r2,r2
B 2r3,2r3
C 2rπ,2rπ
D 4rπ,4rπ

The correct answer is:
2rπ,2rπ

Solution:
dm=Î»RdÎ¸ =2mÏdÎ¸
â´xcm =1mâ«xdm
=1mâ«0Ï/2Rcosâ¡Î¸2mÏdÎ¸
Similarly ycm=1mâ«0Ï2R sin.â¡Î¸2mÏdÎ¸

Topic: Physics, Center of Mass Momentum and Collision, NEET
The coordinates of the center of mass of the following quarter circular arc are

Options:
A r2,r2
B 2r3,2r3
C 2rπ,2rπ
D 4rπ,4rπ

The correct answer is:
2rπ,2rπ

Solution:
Consider a small element as shown in the diagram

dm=λRdθ=2mπdθ
xcm=1m∫0π2Rcos⁡θ2mπdθ
xcm=2Rπ∫0π2cos⁡θdθ=2Rπsin⁡π/2-sin⁡0=2Rπ
Similarly
ycm=1m∫0π2Rsin⁡θ2mπdθ
ycm=2Rπ∫0π2sin⁡θdθ=2Rπ-cos⁡π/2-(-cos⁡0)=2Rπ

Topic: Physics, Center of Mass Momentum and Collision, NEET
The coordinates of the centre of mass of a system of three particles of masses 1 mg, 2 mg and 3 mg are 2, 2, 2. The coordinates of the fourth particle of mass 4 mg to be positioned so that the centre of mass of the four particles system is at the origin of the three-dimensional rectangular coordinate system are

Options:
A 1, 1, 1
B 2, 2, 2
C 3, 3, 3
D -3, -3, -3

The correct answer is:
-3, -3, -3

Solution:
Treat the three masses of 1 mg, 2 mg and 3 mg as a single mass of 6 mg at 2, 2, 2
Now, 0=4×x+6×210 or 4x+12=0
Or x=-3,
and, 0=4×y+6×210 or y=-3
Similarly, z=-3

Topic: Physics, Center of Mass Momentum and Collision, NEET
The coordinates of the centre of mass of a uniform flag-shaped lamina (thin flat plate) of mass 4 kg. (The coordinates of the same are shown in the figure) are:

Options:
A 1.25 m,1.50 m
B 0.75 m,1.75 m
C 0.75 m,0.75 m
D 1 m,1.75 m

The correct answer is:
0.75 m,1.75 m

Solution:
r → cm = 1× i ^ 2 + j ^ +1× i ^ + 5 j ^ 2 2
r→cm=34i^+74j^

Topic: Physics, Center of Mass Momentum and Collision, NEET
The co–ordination number of a central metal atom in a complex is determined by

Options:
A the number of mono dentate ligands around a metal ion bonded by sigma bonds
B the number of ligands around a metal ion boned by pi–bonds
C the number of ligands around a metal ion bonded by sigma and pi–bonds both
D the number of only anionic ligands bonded to the metal ion

The correct answer is:
the number of mono dentate ligands around a metal ion bonded by sigma bonds

Solution:
Co–ordination number is the maximum covalency shown by a metal or metal ion. It is the maximum number of ligands attached to metal by sigma bonds or co–ordinate bonds.
AlH2O6+3.3Cl- it has co-ordination number 6.

Topic: Chemistry, Coordination Compounds, NEET
The coordination number of a central metal atom in a complex is given by

Options:
A The number of Ï- bonds fromed by the ligands
B The number of ligands around a metal ion bonded by Ï-bonds
C The number of ligands around a metal ion bonded by Ï -and Ï- bonds both
D The number of only anionic ligands bonded to the metal ion

The correct answer is:
The number of Ï- bonds fromed by the ligands

Solution:
The coordination number is the maximum covalency shown by a metal or metal ion.
Coordination number = ∑ number of ligands   × denticity

Topic: Chemistry, Coordination Compounds, NEET
The co-ordination number of metal crystallizing in a hexagonal close packed structure is

Options:
A 12
B 4
C 8
D 6

The correct answer is:
12

Solution:
The co-ordination number of metal crystallizing in a hexagonal close packed structure is 12.

Topic: Chemistry, Solid State, NEET
The coordination number of the Ba2+ ions is barium fluoride is 8. The coordination number of the fluoride ion is

Options:
A 8
B 4
C 1
D 2

The correct answer is:
4

Solution:
BaF2 is a Â­ AX2 type salt. In its crystal, Ba2+ ions are arranged in ccp arrangement. Since there are two tetrahedral sites for each Ba2+ ion and F- ions occupy all the tetrahedral sites, hence one F- is surrounded by four Ba2+, i.e., the coordination number of F- is 4.

Topic: Chemistry, Solid State, NEET
The corm is used as a means of vegetative reproduction in

Options:
A members of Fabaceae family.
B members of Liliaceae family.
C members of Solanaceae family.
D members of Brassicaceae family.

The correct answer is:
members of Liliaceae family.

Solution:
Vegetative propagules are the plant parts or organs that are involved in the process of reproduction even though their primary function is not it. The reproductive process is called as vegetative propagation as the non-reproductive parts/organs (leaf/stem/roots) are involved.
Liliaceae commonly called the âLily familyâ is a characteristic representative of monocotyledonous plants. It is distributed worldwide. Vegetative characters: Perennial herbs with underground bulbs/corms/rhizomes.

Topic: Biology, Morphology of Flowering Plants, NEET
The correct arrangement of blood cells is

Options:
A In terms of quantity: erythrocytes > lymphocytes > monocytes > thrombocytes
B In terms of life span: thrombocytes > erythrocytes > monocytes > lymphocytes
C In terms of size: monocytes > lymphocytes > erythrocytes > thrombocytes
D In terms of motility: erythrocytes > thrombocytes > lymphocytes > monocytes

The correct answer is:
In terms of size: monocytes > lymphocytes > erythrocytes > thrombocytes

Solution:
Blood consists of plasma and cells. Cells are formed elements. Agranulocytes are the largest in size. Monocytes and lymphocytes are agranulocytes. Platelets or thrombocytes are smallest in size.

Topic: Biology, Body Fluids and Circulation, NEET
The correct arrangement of cardiac valves from the right side to the left sides is

Options:
A bicuspid valve - pulmonary semilunar valve - aortic semilunar valve - tricuspid valve
B tricuspid valve - aortic semilunar valve - pulmonary semilunar valve - bicuspid valve
C tricuspid valve - bicuspid valve - pulmonary semilunar valve - aortic semilunar valve
D tricuspid valve - pulmonary semilunar valve - aortic semilunar valve - bicuspid valve

The correct answer is:
tricuspid valve - pulmonary semilunar valve - aortic semilunar valve - bicuspid valve

Solution:
The valve present on the extreme right is the right atrioventricular valve also called the tricuspid valve it has three cusps. Just besides this valve, the right ventricle opens into the pulmonary trun and this opening is guarded by a pulmonary semilunar valve. To the left of this valve, the left semilunar valve guards the opening between left ventricle and aorta. The valve present on the extreme left is the left atrioventricular valve also called the bicuspid valve it has two cusps.

Topic: Biology, Body Fluids and Circulation, NEET
The correct arrangement of events in the cardiac cycle according to their duration is:

Options:
A Atrial systole < Ventricular systole < Ventricular diastole < Atrial diastole
B Atrial systole < Ventricular systole < Atrial diastole < Ventricular diastole
C Atrial systole < Atrial diastole < Ventricular systole < Ventricular diastole
D Ventricular systole < Atrial systole < Ventricular diastole < Atrial diastole

The correct answer is:
Atrial systole < Ventricular systole < Ventricular diastole < Atrial diastole

Solution:
The cardiac cycle includes systole and diastole in a cyclic manner. It includes joint diastole, ventricular systole and atrial systole. Duration of the total cardiac cycle is 0.8 sec. The durations are atrial diastole lasts for 0.7 sec. Ventricular diastole is for 0.5 sec. Ventricular diastole for 0.3 sec and ventricular systole for 0.3 and atrial systole for 0.1 sec.

Topic: Biology, Body Fluids and Circulation, NEET
The correct arrangement of these biomolecules in increasing order of their molecular weight is

Options:
A ATP, ADP, AMP, DNA
B ADP, ATP, AMP, DNA
C AMP, ADP, ATP, DNA
D DNA, ATP, ADP AMP

The correct answer is:
AMP, ADP, ATP, DNA

Solution:
AMP is Adenosine monophosphate. When one phosphate gets added it becomes ADP i.e. Adenosine diphosphate. Similarly when two phosphates get added it is ATP i.e. Adenosine triphosphate. Hence as phosphates get added, the weight increases. Therefore molecular weight of AMP ˂ ADP ˂ ATP. AMP consists of Phosphate + Sugar + Nitrogen bases = Nucleotides and DNA consists of millions nucleotides , hence molecular weight DNA is highest

Topic: Chemistry, Biomolecules, NEET
The correct bond order in the following species is:

Options:
A O22+<O2+<O2-
B O22+<O2-<O2+
C O2+<O2-<O22+
D O2-<O2+<O22+

The correct answer is:
O2-<O2+<O22+

Solution:
B.O=Nb -Na2
O 2 − ⇒σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <σ2 p z 2 <π2 p x 2 =π2 p y 2 < π * 2 p x 2 = π * 2 p y 1
B.O for O2-=1.5⇒10 - 72=1.5
O 2 ⇒σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <σ2 p z 2 <π2 p x 2 =π2 p y 2 < π * 2 p x 1 = π * 2 p y 1
B.O for O2=2.0⇒10 – 62=2
O 2 + ⇒σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <σ2 p z 2 <π2 p x 2 <π2 p y 2 < π * 2 p x 1 < π * 2 p y 0
B.O for O2+=2.5⇒10 - 52=2.5

O22+⇒σ1s2<σ*1s2<σ2s2<σ*2s2<σ2pz2<π2px1=π2py1<π*2px0=π*2py0
O22+⇒B.O=10-42=3.0
O2-<O2+<O22+

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2015 (Phase 1)
The correct Boolean operation represented by the circuit diagram drawn is,

Options:
A AND
B OR
C NAND
D NOR

The correct answer is:
NAND

Solution:
The truth table for the circuit can be given as:

A	B	Y

0	0	1

1	0	1

0	1	1

1	1	0

The above truth table clearly shows result of a NAND gate.

Topic: Physics, Semiconductors, NEET, NEET 2019 (All India)
The correct $\mathrm{C}-\mathrm{O}$ bond length among $\mathrm{CO}, \mathrm{CO}_3^{-2}, \mathrm{CO}_2$ is:

Options:
A $\mathrm{CO} < \mathrm{CO}_3^{2-} < \mathrm{CO}_2$
B $\mathrm{CO}_3^{2-} < \mathrm{CO}_2 < \mathrm{CO}$
C $\mathrm{CO} < \mathrm{CO}_2 < \mathrm{CO}_3^{2-}$
D $\mathrm{CO}_2 < \mathrm{CO}_3^{2-} < \mathrm{CO}$

The correct answer is:
$\mathrm{CO} < \mathrm{CO}_2 < \mathrm{CO}_3^{2-}$

Solution:
More is the single bond character more will be the bond length.
Thus, correct order is
$\mathrm{CO} < \mathrm{CO}_2 < \mathrm{CO}_3^{--}$
Related Theory
Compounds with formal $\mathrm{C} \equiv \mathrm{O}$ triple bonds do not exist except for carbon monoxide, which has a very short, and strong bond (112.8 pm).

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2007
The correct characterstics of chemical equilibrium for any reaction is

Options:
A Mutual opposite reactions undergoes
B Concentration of reactants and resulting products are equal
C Velocity of mutual reactions become equal
D The temperature of mutual opposite reactions become equal

The correct answer is:
Velocity of mutual reactions become equal

Solution:
When rate of forward reaction is equal to the rate of backward reaction then equilibrium is supposed to be established.

Topic: Chemistry, Chemical Equilibrium, NEET
The correct charge and co-ordination number of âFeâ in K3FeCN6 is

Options:
A + 2,4
B +3,6
C +2,6
D +3,3

The correct answer is:
+3,6

Solution:
K3FeCN6âFeCN63-
x+6-1=-3
â´ x=+3
Coordination number of Fe is 6 as 6 CN- are attached.

Topic: Chemistry, Coordination Compounds, NEET
The correct chronological arrangement of the following event is

A. In India, the Air (Prevention and Control of Pollution) Act was amended to include noise as an air pollutant.
B. The government of India passed the Water (Prevention and Control of Pollution) Act to safeguard our water resources.
C. Recognising the deleterious effects of ozone depletion, the Montreal Protocol became effective.
D. The National Forest Policy of India recommended 33 per cent forest cover for the plains and 67 per cent for the hills.

Options:
A AÂ âÂ BÂ âÂ DÂ âÂ C
B B â A â D â C
C BÂ âÂ AÂ âÂ CÂ âÂ D
D AÂ âÂ BÂ âÂ CÂ âÂ D

The correct answer is:
B â A â D â C

Solution:
The government of India has passed the Water (Prevention and Control of Pollution) Act, 1974 to safeguard our water resources. In India, the Air (Prevention and Control of Pollution) Act was amended in 1987 to include noise as an air pollutant. The National Forest Policy (1988) of India has recommended 33 per cent forest cover for the plains and 67 per cent for the hills. Recognising the deleterious effects of ozone depletion, an international treaty, known as the Montreal Protocol, was signed at Montreal (Canada) in 1987 (effective in 1989).

Topic: Biology, Environmental Issues, NEET
The correct combination of characters regarding slime molds is

Options:
A parasitic, plasmodium with true walls, spores dispersed by air currents
B parasitic, plasmodium without walls, spores dispersed by water
C saprophytic, plasmodium without walls, spores dispersed by water
D saprophytic, plasmodium without walls, spores dispersed by air currents

The correct answer is:
saprophytic, plasmodium without walls, spores dispersed by air currents

Solution:
Slime molds are saprophytic protists, this feature is similar to fungus so they are also called fungus-like Protists. Under favorable conditions, they form an aggregation called plasmodium which may grow and spread over several feet. During adverse conditions, this plasmodium differentiates and forms fruiting bodies bearing spores at their tips. They are highly resistant and survive for many years, even under adverse conditions. Their spores are dispersed by air currents.

Topic: Biology, Biological Classification, NEET
The correct corresponding order of names of four aldoses with configuration given below respectively, is

Options:
A L-erythrose, L-threose, L-erythrose, D-threose.
B D-threose, D-erythrose, L-threose, L-erythrose.
C L-erythrose, L-threose, D-erythrose, D-threose.
D D-erythrose, D-threose, L-erythrose, L-threose.

The correct answer is:
D-erythrose, D-threose, L-erythrose, L-threose.

Solution:
If the alcoholic group is present on the left side of the last chiral carbon then it is D- sugar and if it is on the left side then it is L- sugar.

All the four given molecules with their names are given below.

Topic: Chemistry, Biomolecules, NEET, NEET 2016 (Phase 2)
The correct corresponding order of names of four aldoses with configuration given below

respectively, is

Options:
A L-erythrose, L-threose, L-erythrose, D-threose
B D-threose, D-erythrose, L-threose, L-erythrose
C L-erythrose, L-threose, D-erythrose, D-threose
D D-erythrose, D-threose, L-erythrose, L-threose

The correct answer is:
D-erythrose, D-threose, L-erythrose, L-threose

Solution:

Topic: Chemistry, Biomolecules, NEET
The correct decreasing order of atomic radii $(\mathrm{pm})$ of $\mathrm{Li}, \mathrm{Be}, \mathrm{B}$ and C is

Options:
A $\mathrm{Be}\gt\mathrm{Li}\gt\mathrm{B}\gt\mathrm{C}$
B $\mathrm{Li}\gt\mathrm{Be}\gt\mathrm{B}\gt\mathrm{C}$
C $\mathrm{C}\gt$ B $\gt$ Be $\gt\mathrm{Li}$
D $\mathrm{Li}\gt\mathrm{C}\gt\mathrm{Be}\gt\mathrm{B}$

The correct answer is:
$\mathrm{Li}\gt\mathrm{Be}\gt\mathrm{B}\gt\mathrm{C}$

Solution:
As the atomic number in a period increases, the effective nuclear charge also increases hence, atomic radii along the period decreases.
Correct order of atomic radii
$\mathrm{Li}\gt\mathrm{Be}\gt\mathrm{B}\gt\mathrm{C}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2024 (Re-NEET)
The correct decreasing order of electropositive character among the following elements Fe, Sc, Rb, Br, Te, F and Ca is:

Options:
A Fe>Sc>Rb>Br>Te>F>Ca
B Ca>Rb>Sc>Fe>Te>F>Br
C Rb>Ca>Sc>Fe>Br>Te>F
D Rb>Ca>Sc>Fe>Te>Br>F

The correct answer is:
Rb>Ca>Sc>Fe>Te>Br>F

Solution:
(i) Alkali metals are more Electropositive than alkaline earth metals.
(ii) d-Block Elements which are near to alkaline earth metals are more Electropositive than other P-Block Elements.
(iii) Metalloids are more Electropositive than non-metals.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The correct decreasing order of priority for the functional groups of organic compound in the IUPAC system nomenclature is:

Options:
A -CHO, -COOH, -SO3H, -CONH2
B -CONH2, -CHO, -SO3H, -COOH
C -COOH, -SO3H, -CONH2, -CHO
D -SO3H, -COOH, -CONH2, -CHO

The correct answer is:
-COOH, -SO3H, -CONH2, -CHO

Solution:
Correct Option is : 3
-COOH, -SO3H, -CONH2, -CHO

Topic: Chemistry, General Organic Chemistry, NEET
The correct description of the sporophyte of Marchantia and gametophyte of a fern is

Options:
A Former is haploid while latter is diploid
B Both do not have vascular tissue
C Both are free living structures
D Both represent the dominant generation in their respective life cycle

The correct answer is:
Both do not have vascular tissue

Solution:
Bryophyta was the first land plant and both gametophyte and sporophyte in them do not have vascular tissue. Pteridophytes, on the other hand, were the first successful land plants with vascular tissue but in the gametophytes of pteridophytes, the vascular tissue is not present.

Topic: Biology, Plant Kingdom, NEET
The correct difference between first and second order reactions is that

Options:
A A first-order reaction can catalyze; a second-order reaction cannot be catalyzed
B The half-life of a first-order reaction does not depend on [A]0 ; the half-life of a second-order reaction does depend on [A]0
C The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
D The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations

The correct answer is:
The half-life of a first-order reaction does not depend on [A]0 ; the half-life of a second-order reaction does depend on [A]0

Solution:
For first order reaction, t1/2=0.693k, which is independent of initial concentration of reactant. (t1/2∝a)
For second order reaction, t1/2=1k[A0], which depends on initial concentration of reactant t1/2∝1a

Topic: Chemistry, Chemical Kinetics, NEET
The correct difference between the embryonic development of monocot and dicot plants is

Options:
A In monocots, suspensor is single celled while in dicots it contains 6-10 cells
B In monocots endosperm is formed before embryo while in dicots embryo is formed before endosperm
C In dicots, coleoptile and coleorhiza are seen while they are not seen in monocots
D In dicot the growth of embryo is lateral but in monocots it is vertical

The correct answer is:
In monocots, suspensor is single celled while in dicots it contains 6-10 cells

Solution:
In both monocots and dicots endosperm is formed before embryo. Coleoptile covering the plumule and coleorhiza covering the radicle is seen in the monocot embryo and not in the dicot embryo. In monocots, a single cotyledon is formed the upper cell of the embryo which the lower cell forms radicle and plumule. Thus the growth of the monocot embryo is lateral. In monocots, the basal cell enlarges and forms a large vesicular suspensor. In dicots, the basal cell divides horizontal for form a layer of 6 â 10 cells called. The cell touching the embryo is hypophysis while the cell on its extreme opposite is haustoria.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The correct difference between the ovary of rose and china rose is

Options:
A Rose has superior ovary while China rose has inferior ovary
B Rose has inferior ovary while China rose has superior ovary
C Rose has semi-inferior ovary while China rose has inferior ovary
D Rose has semi-inferior ovary while China rose has superior ovary

The correct answer is:
Rose has semi-inferior ovary while China rose has superior ovary

Solution:
Rose has perigynous flowers in which the gynoecium is present in the center and other whorls of the flower are located on the rim of the thalamus almost at the same level. Thus, it has a semi-inferior ovary. China rose has hypogynous flowers in which all the floral whorls are attached below the gynoecium. Thus, it has superior ovary.

Topic: Biology, Morphology of Flowering Plants, NEET