datavorous/entrance-exam-dataset huggingface dataset:
Mark out the correct statement among the following.

Options:
A Aqueous AgNO3 solution can be stored in a copper bowl.
B Aqueous CuSO4 solution can be stored in a silver bowl.
C Cu , Ag can release hydrogen gas from dil HCl .
D H2 cannot reduce Cu2+ and Ag+ in the form of metals.

The correct answer is:
Aqueous CuSO4 solution can be stored in a silver bowl.

Solution:
Aq CuSO4 Solution can be stored in a silver bowl because Ag is less reactive than Cu

Topic: Chemistry, Electrochemistry, JEE Main
Mark out the number of incorrect statement/s:

[I] Some essential elements can alter the osmotic potential of a cell.
[II] Nitrogen is required by the meristematic tissues and the metabolically active cells only.
[III] Mg+2 is needed by several respiratory enzymes.
[IV] Mo is essential for the water-splitting reaction in photosynthesis, a reaction that leads to oxygen evolution.
[V] Sulphur is the main constituent of several coenzymes, vitamins (thiamine, biotin, coenzyme-A) and ferredoxin.

Options:
A Two
B Three
C Four
D Five

The correct answer is:
Two

Solution:
Statement II and IV are incorrect. Mn is essential for water splitting and Nitrogen is needed by all the cells. Nitrogen is required by all parts of a plant, particularly the meristematic tissues and the metabolically active cells.

Topic: Biology, Mineral Nutrition, NEET
Mark the correct order of decreasing acid strength of the following compounds.

Options:
A (V) > (IV) > (II) > (I) > (III)
B (II) > (IV) > (I) > (III) > (V)
C (IV) > (V) > (III) > (II) > (I)
D (V) > (IV) > (III) > (II) > (I)

The correct answer is:
(II) > (IV) > (I) > (III) > (V)

Solution:
The electron withdrawing group -NO2 increases the acidity of phenols and the electron donating group -OCH3 decreases the acidity of phenols. The effect oat p-position is greater than at m-position.
Hence the order of acidic nature is (II) > (IV) > (I) > (III) > (V).

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Mark the correct statement

Options:
A All mammals have placenta
B All Chondrichthyes are marine animals with a streamlined body with some exceptions
C Larva of all cyclostomates return to fresh water after metamorphosis
D Chelone and Testudo are able to maintain a constant body temperature

The correct answer is:
All Chondrichthyes are marine animals with a streamlined body with some exceptions

Solution:
Chondrichthyes : They are marine animals with a streamlined body and have cartilaginous endoskeleton.
Placental mammal (infraclass Eutheria), any member of the mammalian group characterized by the presence of a placenta, which facilitates exchange of nutrients and wastes between the blood of the mother and that of the fetus. The placentals include all living mammals except marsupials and monotremes
Cyclostomes are marine but migrate for spawning to fresh water. After spawning, within a few days, they die. Their larvae, after metamorphosis, return to the ocean.
Chelone and Testudo (reptiles) are not able to maintain a constant body temperature

Topic: Biology, Animal Kingdom, NEET
Mark the correct statement in a reversible reaction

Options:
A The catalyst catalyses the forward reaction
B The catalyst catalyses the backward reaction
C The catalyst influences the direct and the reverse reaction to the same extent
D The catalyst increases the rate of forward reaction and decreases the rate of backward reaction

The correct answer is:
The catalyst increases the rate of forward reaction and decreases the rate of backward reaction

Solution:
A catalyst is a substance, which increases the rate of reaction without itself being consumed at the end of the reaction, and the phenomenon is called catalysis. There are some catalysts which decrease the rate of reaction and such catalysts are called negative catalyst. Obviously, the catalyst accelerating the rate will be positive catalyst. However, the term positive is seldom used and catalyst itself implies positive catalyst. Catalysts are generally foreign substances but sometimes one of the product formed may act as a catalyst and such a catalyst is called "auto catalyst" and the phenomenon is called auto catalysis.

Topic: Chemistry, Surface Chemistry, JEE Main
Mark the correct statement

Options:
A Lenticels remain always open.
B Lenticels pores are larger than stomata.
C Lenticular transpiration is only 1% of the total transpiration.
D All the above.

The correct answer is:
All the above.

Solution:
A lenticel is a porous tissue consisting of cells with large intercellular spaces in the periderm of the secondarily thickened organs and the bark of woody stems and roots of dicotyledonous flowering plants. It functions as a pore, providing a pathway for the direct exchange of gases between the internal tissues and atmosphere through the bark, which is otherwise impermeable to gases. Though its contribution is very less in comparison to stomata, it is much needed for the plants.

Topic: Biology, Anatomy of Flowering Plants, NEET
Mark the correct statements about major abiotic factors.

I. Temperature on land varies seasonally, increases progressively from the equator towards the poles and from plains to the mountain tops.

II. Light is important in animals as they use the diurnal and seasonal variations in light intensity and duration as cues for timing their foraging, reproductive and migratory activities.

III. Various characteristics of the soil such as soil composition, grain size and aggregation determine the percolation and water holding capacity of the soils.

IV. The salt concentration is more than 5% in inland waters.

Options:
A I and III
B II and III
C I,II and IV
D I, III and IV

The correct answer is:
II and III

Solution:
Major Abiotic factors that affect the survival of species are: temperature, water, light and soil. When moving from equator to the polar region or from plain to mountain tops; there is progressive decreases in temperature. The temperature on land never remains constant but varies seasonally. Aquatic organisms are largely affected by concentration of salt in water. In hypersaline lagoons, it is > 100% whereas in the sea it is 30-35% and just 5% in inland waters.

Topic: Biology, Organisms and Populations, NEET
Mark the incorrect statement.

Options:
A Insect and pest infestation is one of the major causes for large scale destruction of crop plants.
B Breeding method for insect pest resistance involves the special steps that are not similar to use in other agronomic traits such as yield or quality.
C Malnutrition of micronutrients and vitamins can be termed as hidden hunger.
D Somaclones are genetically identical to original plants.

The correct answer is:
Breeding method for insect pest resistance involves the special steps that are not similar to use in other agronomic traits such as yield or quality.

Solution:
The breeding method for insect or pest-resistant plants is the same as used for other agronomic traits. It involves 5 steps as used for other agronomic traits.
The main steps in breeding a new genetic variety of a crop are:
(i) Collection of variability
(ii) Evaluation and selection of parents
(iii) Cross hybridization among the selected parents
(iv) Selection and testing of superior recombinants
(v) Testing, release, and commercialization of new cultivars

Topic: Biology, Strategies for Enhancement in Food Production, NEET
Marsh gas mainly contains –

Options:
A C2H2
B CH4
C H2S
D CO

The correct answer is:
CH4

Solution:
Methane is produced due to the decay of vegetables or animal organisms present in swamps and marsh, by the action of bacteria. Due to this method of formation, methane is also known as marsh gas.

Topic: Chemistry, Hydrocarbons, NEET
Mass $M$ is divided into two parts $x m$ and $(1-x) m$. For a given separation the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is

Options:
A $\frac{1}{2}$
B $\frac{3}{5}$
C $1$
D $2$

The correct answer is:
$\frac{1}{2}$

Solution:
Force of gravitation $F \propto m_1 m_2$
$\begin{aligned} & F=k m_1 m_2 \\ & F=k(x m)(1-x) \times m\end{aligned}$
For $F$ to be maximum $\frac{d F}{d x}=0$
$\begin{aligned} \frac{d}{d x}[k(x m)(1-x) \times m] & =0 \\ m^2-2 x m^2 & =0 \\ \therefore \quad x & =\frac{1}{2}\end{aligned}$

Topic: Physics, Gravitation, JEE Main
Mass numbers of two nuclei are in the ratio of 4:3. Their nuclear densities will be in the ratio of

Options:
A 4:3
B 3413
C 1:1
D 4313

The correct answer is:
1:1

Solution:
Radius of nucleus R=R0 A13

Density of nucleus ρ= Mass of nucleus volume of nucleus

ρ=m×A43πR3 Where m: mass of proton or neutron and A: mass number.

⇒ρ=m×A43πR0 3A

⇒ρ∝A0

Hence, density of nucleus is independent of mass number.

Therefore, the ratio is 1:1.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Mass of a person sitting in a lift is $50 \mathrm{~kg}$. If lift is coming down with a constant acceleration of $10 \mathrm{~m} / \mathrm{sec}^2$. Then the reading of spring balance will be $\left(g=10 \mathrm{~m} / \mathrm{sec}^2\right)$

Options:
A 0
B $1000 \mathrm{~N}$
C $100 \mathrm{~N}$
D $10 \mathrm{~N}$

The correct answer is:
0

Solution:
$R=m(g-a)=0$

Topic: Physics, Laws of Motion, JEE Main
Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is kg (Molar mass in gmol-1 for ethylene glycol 62, Kf of water =1.86 K kg mol-1)

Options:

The correct answer is:
15

Solution:
Given, ΔTf= -240C

Molar mass of Ethylene glycol= 62

ΔTf=i Kf×molality

24=(1)×1.86×W62×18.6

W=14880gm

=14.880 kg

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Mass of methane required to produce 22Â g of CO after complete combustion is g. (Given Molar mass in g mol-1, C=12.0, H=1.0, O=16.0)

Options:

The correct answer is:
8

Solution:
The molecular mass of methane is 16 g/mol.

The reaction is as follows,

CH4+2O2→CO2+2H2O

Moles of CO2=massmolar mass=2244=0.5

So, required moles of CH4=0.5

Mass of CH4 =0.5×16 =8 gm

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Mass of Urea NH2CONH2 required to be dissolved in 1000 g of water in order to reduce the vapour pressure of water by 25% is _______ g. (Nearest integer)

Given : Molar mass of N, C, O and H are 14, 12, 16 and 1 g mol-1 respectively.

Options:

The correct answer is:
1111

Solution:
Relative lowering of vapour pressure=Mole fraction of solute

Lowering of vapour pressureVapoure pressure of solution=moles of solutemoles of solvent

2575= moles of urea 100018

13×100018=moles of urea

Mass of urea=13×100018×60

=1111.11 gm

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Mass per unit area of a circular disc of radius a depends on the distance r from its centre as σr=A+Br . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is:

Options:
A 2πa4A4+aB5
B 2πa4aA4+B5
C πa4A4+aB5
D 2πa4A4+B5

The correct answer is:
2πa4A4+aB5

Solution:
σ=A+Br
∫dm=∫A+Br2πrdr
I=∫dmr2
∫0aA+Br2πr3dr
=2πAa44+Ba55
=2πa4A4+aB5

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Masses, M1, M2 and M3 are connected by strings of negligible mass which pass over massless and frictionless pulleys P1 and P2 as shown in the figure. The masses move such that the portion of the string between P1 and P2 is parallel to the inclined plane and the portion of the string between P2 and M3 is horizontal. The masses M2 and M3 are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37Â° with the horizontal and the mass M1 moves downwards with a uniform velocity. Find the tension in the horizontal portion of the string.
(Take g=9.8 m s-2, sin 37Â° â 35)

Options:

The correct answer is:
9.8

Solution:
Constant velocity means the net acceleration of the system is zero. Or the net pulling force on the system is zero. While calculating the pulling force, tension forces are not taken into consideration. Therefore,

M1g=M2g sin 37Â°+Î¼M2g cos37Â°+Î¼M3g
or M1=M2 sin 37Â°+Î¼M2 cos 37Â°+Î¼M3

Substituting the values

M1=(4)(35)+(0.25)(4)(45)+(0.25)(4) =4.2 kg
Since, M3 is moving with uniform velocity

T=Î¼M3g=(0.25) (4) (9.8)=9.8 N

Topic: Physics, Laws of Motion, JEE Main
Match animals in column I with their respective respiratory structures in column II.

Column I (Animals)	Column II (Respiratory structures)

A. Pigeon	(i) Book gills

B. Scorpion	(ii) Pharyngeal wall

C. Planaria	(iii) Lungs

D. Earthworm	(iv) Gills

E. Insects	(v) Tracheal tubes

F. King crab	(vi) Body surface

G. Prawn	(vii) Skin

H. Labeo

Options:
A A-(iii), B-(v), C-(vi),D-(vii), E-(v), F-(i), G-(iv), H-(iv)
B A-(v), B-(ii), C-(vi), D-(vii), E-(vi), F-(iv),G- (i), H-(iii)
C A-(vi), B-(iv), C-(vii), D-(v), E-(i), F-(ii), G-(iii), H-(vii)
D A-(i), B-(v), C-(vii), D-(ii),E-(vii), F-(ii),G-(iv), H-(vi)

The correct answer is:
A-(iii), B-(v), C-(vi),D-(vii), E-(v), F-(i), G-(iv), H-(iv)

Solution:
Mechanisms of breathing vary among different groups of animals depending mainly on their habitats and levels of the organisation. Lower invertebrates like sponges, coelenterates, flatworms, etc., exchange O2 with CO2 by simple diffusion over their entire body surface. Earthworms use their moist cuticle and insects have a network of tubes (tracheal tubes) to transport atmospheric air within the body. Special vascularised structures called gills (branchial respiration) are used by most of the aquatic arthropods and molluscs whereas vascularised bags called lungs (pulmonary respiration) are used by the terrestrial forms for the exchange of gases. Among vertebrates, fishes use gills whereas amphibians, reptiles, birds and mammals respire through lungs. Amphibians like frogs can respire through their moist skin (cutaneous respiration) also.

Topic: Biology, Breathing and Exchange of Gases, NEET
Match column A and B.

Column A	Column B

A. Catalytic converter	(i) Particulate matter

B. Electrostatic precipitator	(ii) CO2 and Nitrogen oxide

C. Earmuffs	(iii) Solid wastes

D. Landfills	(iv) High noise level

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

The correct answer is:
A - (ii), B - (i), C - (iv), D - (iii)

Solution:
Methods of removal of air pollutants before their discharge into the atmosphere:
a. Electrostatic precipitator: It is the most widely used way to remove particulate matter. It can remove over 99% particulate matter from the exhaust of a thermal power plant.
b. Scrubber: It can remove gases like sulfur dioxide.
Ways to reduce the emission of pollutants from automobiles:
a. Proper maintenance of automobiles.
b. Use of lead-free petrol or diesel.
c. Catalytic converters

Topic: Biology, Environmental Issues, NEET
Match Column A with B and choose the CORRECT option from below.

Column A	Column B

(1) Skeletal Muscles	(P) Interrupts dark band

(2) Z line	(Q) Autonomic nervous system

(3) Henson zone	(R) Striated

(4) Visceral muscles	(S) Seen in light band

Options:
A 1-R,Â 2-P,Â 3-Q,Â 4Â âS
B 1-R,Â 2-Q,Â 3-P,Â 4-S
C 1-Q,Â 2-P,Â 3-S,Â 4-R
D 1-R, 2-S, 3-P, 4-Q

The correct answer is:
1-R, 2-S, 3-P, 4-Q

Solution:
Skeletal muscles are different from visceral muscles as these muscles show striations and are voluntary in nature. These are made up of many bundles known as Fasciculi. Each bundle has many muscle fibres which in turn has many myofibrils. Each myofibril consists of repeating units called as sarcomere. This is the main functional unit of contraction and relaxation of muscles. One sarcomere is the portion between two Z lines. Z lines bisect the 'I' or Isotrophic or light bands. In the centre of Sarcomere or between two Z lines there are present Dark bands or 'A'or Anisotrophic bands. Hensons zone or H zone is also seen in the centre which is demarcated by M line in the middle.

Topic: Biology, Locomotion and Movement, NEET
Match Column - I with Column - II and select the CORRECT option from codes given below

Column-I	Column-II

(1)	Meristem	(i)	Photosynthesis, storage

(2)	Parenchyma	(ii)	Living mechanical tissue

(3)	Collenchyma	(iii)	Actively dividing cells

(4)	Sclerenchyma	(iv)	Stomata

(5)	Epidermal tissue	(v)	Sclereids

Options:
A 1-(i), 2-(iii), 3-(v), 4-(ii), 5-(iv)
B 1-(iii), 2-(i), 3-(ii), 4-(v), 5-(iv)
C 1-(ii), 2-(iv), 3-(v), 4-(i), 5-(iii)
D 1-(v), 2-(iv), 3-(iii), 4-(ii) 5-(i)

The correct answer is:
1-(iii), 2-(i), 3-(ii), 4-(v), 5-(iv)

Solution:
Collenchyma, part of the ground tissue, is a simple living tissue found in plants. The cell wall of collenchyma tissue cells is thickened at corner due to higher deposition of cellulose, pectin and hemicellulose. The cells are generally spherical or polygonal in shape. This tissue provides mechanical support to the young stem and leaf. The outermost covering of the whole plant body forms the epidermal tissue system that includes the epidermal cells and appendages with stomata. This system basically provides protection to the main plant body. The outermost layer of primary plant body called as epidermis have elongated compactly arranged cells. Stomata structures present in the epidermis help in regulating transpiration and gas exchange. These stomata are made of two guard cells each that encloses the pore.
Meristem are a type of tissue found in such plants which can divide indefinitely. These are generally considered undifferentiated as they can give develop into all other tissues and organs of the plant. The meristematic tissues that appear early in plant life and help in formation of the plant body is called primary meristem. The apical and intercalary meristem are included in it. Meristematic tissue occurring between the mature regions of root and shoots of angiosperms are called secondary meristem.
Parenchyma is a type of simple permanent tissues forming a major component of the tissues of plant. These cells are in various shapes like oval, round, polygonal etc. with a thin cell wall made of cellulose. These cells perform functions like storage, photosynthesis, secretion.
The simple permanent tissue consisting of dead cells without protoplast is called sclerenchyma. These cells are long and narrow with lignified cell wall. The thick-walled, elongated pointed cells form the sclerenchyma fibres and the spherical, oval shaped cell with highly thickened cell wall and narrow cavity form the sclerenchyma sclereids.

Topic: Biology, Anatomy of Flowering Plants, NEET
Match column I with column II and select the CORRECT option from the codes given below.

Options:
A (i) - a; (ii) - d ; (iii) - c; (iv) - b
B (i) - c; (ii) - a ; (iii) - b; (iv) - d
C (i) - a; (ii) - c ; (iii) - d; (iv) - b
D (i) - d; (ii) - a ; (iii) - b; (iv) - c

The correct answer is:
(i) - c; (ii) - a ; (iii) - b; (iv) - d

Solution:
Nitrogen is one of the essential macronutrients required by plants and absorbed in the form of nitrates and nitrites. It forms a major constituent of proteins, vitamins, chlorophyll, pyrimidines and hormones. Deficiency of nitrogen of leads to stunted growth, chlorosis, reduced cell division, decrease in protein content, wrinkling of cereal grains, purple coloration of the shoot axis.
Iron is another essential element necessary for plant growth and development. Chlorophyll production requires iron and it is a component of many enzymes which participate in energy transfer, nitrogen fixation, lignification. Iron deficiency is clearly indicated by chlorosis where leaf yellowing appears first in the interveinal regions of the young leaves. Chlorosis referes to production of insufficienct chlorophyll leading to yellowing of the leaves.
Chlorine is absorbed in the form of chloride ions and it helps in maintaining salt balance in cells of plant. Chlorine is important for opening and closing of stomata. Chlorine deficiency results in the wilting of the leaves which later turn in bronze colour.
Boron is another essential element which is required for membrane functioning, pollen germination, cell differentiation. It is also required for calcium uptake and utilization. Boron deficiency symptoms include leaves with yellowish or reddish cast, growing points stop functioning leading to stunted or rosette appearance. Its deficiency also causes formation of cracked fruits.

Topic: Biology, Mineral Nutrition, NEET
Match Column - I with Column - II and select the CORRECT option from the codes given below

Column â I	Column â II

P	Vessels	i	Cells are living, with thin cellulosic cell walls

Q	Tracheids	ii	Cells possess highly thickened walls with obliterated central lumen

R	Xylem fibres	iii	Individual members are interconnected Cells are living, with thin cellulosic cell walls

(T)	Xylem parenchyma	iv	Elongated tube-like cells with thick, lignified walls and tapering ends

Options:
A P-(iv), Q-(iii), R-(ii), S-(i)
B P-(iii), Q-(iv), R-(ii), S-(i)
C P-(ii), Q-(iv), R-(iii), S-(i)
D P-(iv), Q-(ii), R-(iii), S-(i)

The correct answer is:
P-(iii), Q-(iv), R-(ii), S-(i)

Solution:
Vascular bundles are components of the vascular tissue system that consist of two main parts i.e.xylem and phloem. In which xylem is called water-conducting tissue and phloem is called food conducting tissue. Xylem was proposed by Nageli in 1858.Xylem composed of several types of cells and out of which some are living and some are dead. The elements of xylem are tracheids, vessel, xylem fiber, and xylem parenchyma. Tracheids are tube-like cells and both the ends of the cells are tapered. Vessels are shorter than tracheids and they have perforations at the end wall through which they are interconnected. Xylem fibers have cells with the lignified secondary cell wall. So they possess cells with highly thickened walls. Xylem parenchyma is living cells with cytoplasm and prominent nucleus. So the correct answer is B.

Topic: Biology, Anatomy of Flowering Plants, NEET
Match Column I with Column II.

Column I	Column II

(1)	Xenogamy	(i)	Transfer of pollen grain from the anther to the stigma of another flower of the same plant

(2)	Autogamy	(ii)	Transfer of pollen grain from the anther to the stigma of the same flower

(3)	Cleistogamy	(iii)	Transfer of pollen grain from the anther to the stigma of different plant

(4)	Geitonogamy	(iv)	Flowers are invariably autogamous which do not open at all

Options:
A 1-iv, 2-iii, 3-i, 4-ii
B 1-iii, 2-ii, 3-i, 4-iv
C 1-iii, 2-ii, 3-iv, 4-i
D 1-i, 2-iii, 3-iv, 4-ii

The correct answer is:
1-iii, 2-ii, 3-iv, 4-i

Solution:
The different types of pollination taking place -
Xenogamy - pollen transfer to different flower of different plant
Autogamy - pollen transfer to same flower of same plant
Cleistogamy - Self pollination is maintained by keeping the flower closed.
Geitonogamy - Pollen transfer to different flower of same plant

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Match Column I with Column II:

Column I	Column II

(A) Bud in the axil of a leaf	(i) Pitcher plant and Venus flytrap

(B) Bud is absent in the axil of leaflets	(ii) Cacti

(C) Spines (modified leaves)	(iii) Compound leaf

(D) Leaves modified to catch insects	(iv) Simple leaf

(E) Fleshy leaves with stored food	(v) Garlic and onion

Options:
A A-(iv); B-(iii); C-(ii); D-(i); E-(v)
B A-(v); B-(iv); C-(iii); D-(ii); E-(i)
C A-(i); B-(ii); C-(iii); D-(iv); E-(v)
D A-(iv); B-(ii); C-(iii); D-(i); E-(v)

The correct answer is:
A-(iv); B-(iii); C-(ii); D-(i); E-(v)

Solution:
In plants mainly 2 types of leaves are present as simple leaves and compound leaves. In both types of leaves bud is present in the axil of leaf but in compound leaf, leaflets are also present which are devoid of bud. Leaves show various types of modification according to need of plant like spine for protection in cactus, fleshy leaves for storage of food in garlic and onion and for catching insects in some insectivorous plants like pitcher plant, venus fly trap etc.

Topic: Biology, Morphology of Flowering Plants, NEET