datavorous/entrance-exam-dataset huggingface dataset:
MALT constitutes about _______ percent of the lymphoid tissue in human body.

Options:
A 50%
B 20%
C 70%
D 10%

The correct answer is:
50%

Solution:
The mucous membranes lining the digestive, respiratory and urogenital systems have a combined surface area of about 400 m2, which is equal to the size of a standard basketball court. They are the major sites of entry for most pathogens. These vulnerable membrane surfaces are defended by a group of organized lymphoid tissues collectively known as mucosal-associated lymphoid tissue (MALT). It constitutes about 50 per cent of the lymphoid tissue in human body.

Topic: Biology, Human Health and Diseases, NEET, NEET 2017
Mark the correct statement about Selaginella.

Options:
A Macrophylls are a characteristic feature of Selaginella.
B The spores germinate to give rise free-living, multicellular thalloid gametophytes.
C The sporophytes bear sporangia, which produce spores mitotically in the spore mother cells.
D It is a homosporous plant, meaning it can produce only one type of spores.

The correct answer is:
The spores germinate to give rise free-living, multicellular thalloid gametophytes.

Solution:
The leaves in Pteridophyta are small (microphylls) as in Selaginella or large (macrophylls) as in ferns. The sporophytes bear sporangia that are subtended by leaf-like appendages called sporophylls. In some cases, sporophylls may form distinct compact structures called strobili or cones. The sporangia produce spores by meiosis in spore mother cells. The spores germinate to give rise to inconspicuous, small but multicellular, free-living, mostly photosynthetic thalloid gametophytes called prothallus.
Heterospory is observed as both small and large spores are produced.

Topic: Biology, Plant Kingdom, NEET
Mark the INCORRECT statement regarding connective tissue.

Options:
A Most abundant and widely distributed body tissue
B Highly vascular
C Arise from the endoderm
D Abundant intercellular matrix

The correct answer is:
Arise from the endoderm

Solution:
Connective tissues arise from mesenchyme - an embryonic tissue which in turn arises from mesoderm, an embryonic germ layer. Cells of connective tissues secrete collagen and elastin fibres. They help in connecting, binding and supporting different structures of the animal body. Ligaments connect the bone to muscles whereas tendon connects bone to bone. They are highly vascular in nature.

Topic: Biology, Structural Organisation in Animals, NEET
Mark the incorrect statement regarding the head of the cockroach.

Options:
A It is hypognathous.
B Formed by the fusion of six embryonic segments.
C Head bears appendages forming biting and chewing type of mouth parts.
D A pair of broad antennae arise from membranous sockets lying in front of eyes.

The correct answer is:
A pair of broad antennae arise from membranous sockets lying in front of eyes.

Solution:
The head of the cockroach is hypognathous i.e. the lower jaw is longer than the upper jaw. Head is triangular in shape and lies anteriorly at right angles to the longitudinal body axis. It is formed by the fusion of six segments and shows great mobility in all directions due to the flexible neck. The head capsule bears a pair of compound eyes. A pair of thread-like antennae arise from membranous sockets lying in front of eyes. Antennae have sensory receptors that help in monitoring the environment. Anterior end of the head bears appendages forming biting and chewing type of mouth parts.

Topic: Biology, Structural Organisation in Animals, NEET
Marks obtained by 7 students in a subject are $30,55,75,90$, $50,60,39 .$ The number of students securing marks less than the mean marks is $\quad$

Options:
A 7
B 6
C 5
D 4

The correct answer is:
4

Solution:
Given marks are $30,55,75,90,50,60,39$.
Mean marks $=y$ $\bar{x}=\frac{30+55+75+90+50+60+39}{7}=\frac{399}{7}=57$
Hence, 4 students secured marks less than the mean marks

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 1)
Mass of cell enclosed within the integuments in ovule is called

Options:
A Chalaza
B Endosperm
C Nucellus
D Sporogenous tissue

The correct answer is:
Nucellus

Solution:
Each ovule has one or two protective envelopes called integuments. Enclosed within the integuments is a mass of cells called the nucellus which has abundant reserve food materials. It provides nutrition to developing female gametophyte or embryo sac.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Match column- I with column - II and select the CORRECT option from the codes given below.

Column â I	Column â II

(1) Bulliform cells	(i) Regulate the opening and closing of stomata

(2) Guard cells	(ii) Aerating pores in the bark of the plant

(3) Lenticels	(iii) Rolling in and out of leaves

(4) Subsidiary cells	(iv) Accessory cells

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

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

Solution:
On the upper surface of the isobilateral leaves of monocots large, bubble-shaped cells are present called bulliform cells. These cells help in rolling leaves to prevent transpiration and loss of water. Guard cells surround the stomatal pore and help in the opening and closing of the stomata. Pores formed by rupture of the epidermal region of root and stem at certain locations which help in aeration is called lenticels. These are commonly seen on the bark after secondary growth. Epidermal cells found around the guard cells are called subsidiary cells. The cells are also referred to as accessory cells.

Topic: Biology, Anatomy of Flowering Plants, NEET
Match List I and List II and select the correct answer with the help of the codes given below the lists

List I	List II

A. Somatotropin	1. Neurohypophysis

B. Oxytocin	2. Cretinism

C. Thyroxine	3. Rickets

D. Vitamin D	4. Gigantism

Options:
A A = 3, B = 1, C = 4, D = 2
B A = 4, B = 1, C = 2, D = 3
C A = 2, B = 3, C = 4, D = 1
D A = 1, B = 4 C = 3, D = 2

The correct answer is:
A = 4, B = 1, C = 2, D = 3

Solution:
Somatotropin is growth hormone and gigantism is medical condition in which there is abnormal growth. Posterior part of pituitary is called as neurohypophysis. Neurohypophysis is structural foundation of neurohumoral system coordinating fluid balance and reproductive function. Parturition and lactation is done by oxytocin. Cretinism is seen in newborns. Cretinism is a condition due to deficiency of thyroid hormones. There are 2 types of thyroid hormones a Triiodothyronine and b Thyroxine. Vitamin D is a fat soluble vitamin. Rickets is a condition found in children's in which there is softening of bones due to deficiency of vitamin D.

Topic: Biology, Chemical Coordination and Integration, NEET
Match List I with List II and select the correct answer using the code given below the lists:

Options:
A $A-1\quad B-4\quad C-5\quad D-3$
B $A-1\quad B-3\quad C-5\quad D-4$
C $A-5\quad B-4\quad C-2\quad D-3$
D $A-5\quad B-3\quad C-2\quad D-4$

The correct answer is:
$A-5\quad B-4\quad C-2\quad D-3$

Solution:
(A) Graph of $f(x)=\cos x$ cuts $x$ -axis at infinite number of points. ( 5 of list II)
(B) Graph of $\mathrm{f}(\mathrm{x})=\operatorname{In} \mathrm{x}$ cuts $\mathrm{x}$ -axis in only one point. (4 of list II)
(C) Graph of $f(x)=x^{2}-5 x+4$ cuts $x$ axis in two points (2 of list II)
(D) Graph of $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$ cuts $\mathrm{y}$ -axis in only one point. (3 of list II)

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 1)
Match List I with List II and select the correct answer using the code given below the lists
\begin{array}{llll}
\hline List I & & List II \\
\hline A. A cube root of unity & 1. & -2(1+i) \\
B. A square root of -1 & 2. & 2 i \\
C. Cube of 1-\mathrm{i} & 3. & -i \\
D. Square of 1+\mathrm{i} & 4. & -\frac{1}{2}(1+i \sqrt{3})
\end{array}

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

The correct answer is:
4 3 1 2

Solution:
\begin{array}{|l|l|}
\hline A cube root of unity & -\frac{1}{2}(1+i \sqrt{3}) \\
\hline A square root of -1 & -i \\
\hline Cube of 1-i & -2(1+i) \\
\hline Square of 1+i & 2 i \\
\hline
\end{array}

Topic: Mathematics, Complex Number, NDA, NDA 2008 (Phase 2)
Match List I with List II and select the correct answer using the code given below the lists

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

The correct answer is:
A-4 B-2 C-1

Solution:
(A) $\tan 15^{\circ}=\tan \left(45^{\circ}-30^{\circ}\right)$
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3}$
(B) $\tan 75^{\circ}=\tan \left(45^{\circ}+30^{\circ}\right)=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}}$
(C) $\tan \left(105^{\circ}\right)=\tan \left(60^{\circ}+45^{\circ}\right)$
$=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$ (By Rationaliziang)
$=2+\sqrt{3}$
$=\frac{\sqrt{3}+1}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}}=\frac{(\sqrt{3}+1)^{2}}{1-3}$
$=\frac{4+2 \sqrt{3}}{-2}=-2-\sqrt{3}$
Hence, option (b) is correct.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
Match List I with List II.
\begin{array}{|l|l|l|l|}\hline & {|c|}{\begin{array}{c} List I \\(Spectral Lines of \\Hydrogen for \\transitions from)\end{array}} & & {|c|}{\begin{array}{l} List II \\(Wavelengths (nm))\end{array}} \\\hline A. & n_2=3 to m_1=2 & I. & 410.2 \\\hline B. & n_2=4 to m_1=2 & II. & 434.1 \\\hline C. & n_2=5 to n_1=2 & III. & 656.3 \\\hline D. & n_2=6 to n_1=2 & IV. & 486.1 \\\hline\end{array}
Choose the correct answer from the options given below:

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

The correct answer is:
A-III, B-IV, C-III, D-I

Solution:
Energy difference $\Delta E=\frac{h c}{\lambda}$
$\begin{aligned}
& \therefore \lambda \propto \frac{1}{\Delta E} \\
& (\Delta E)_{6-2}\gt(\Delta E)_{5-2}\gt(\Delta E)_{4-2}\gt(\Delta E)_{3-2} \\
& \lambda_{6-2} \lt \lambda_{5-2} \lt \lambda_{4-2} \lt \lambda_{3-2}
\end{aligned}$
A-III, B-IV, C-II, D-I

Topic: Physics, Atomic Physics, NEET, NEET 2024
Match List I with List II.
Choose the correct answer from the options given below:

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

The correct answer is:
A-IV, B-I, C-II, D-III

Solution:
It is reductive ozonolysis
It is Friedel-Crafts acylation reaction.
Secondary alcohols are oxidised to ketones by $\mathrm{CrO}_3$

Topic: Chemistry, Carboxylic Acid Derivatives, NEET, NEET 2024
Match List - I with List.-II.

List-I	List-II

(a)	Lenticels	(i)	Phellogen

(b)	Cork cambium	(ii)	Suberin deposition

(c)	Secondary cortex	(iii)	Exchange of gases

(d)	Cork	(iv)	Phelloderm

Select the correct answer from the options given below.

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

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

Solution:
List-I	List-II

(a)	Lenticels	(iii)	Exchange of gases

(b)	Cork cambium	(i)	Phellogen

(c)	Secondary cortex	(iv)	Phelloderm

(d)	Cork	(ii)	Suberin deposition

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2021
Match List-I with List-II and select the correct answer using the code given below the lists:

Options:
A $A-2\quad B-3\quad C-1\quad D-4$
B $A-2\quad B-3\quad C-4\quad D-1$
C $A-3\quad B-2\quad C-1\quad D-4$
D $A-3\quad B-2\quad C-4\quad D-1$

The correct answer is:
$A-2\quad B-3\quad C-4\quad D-1$

Solution:
$\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\cos \mathrm{x}$.
We know, the maximum value of a $\sin x+b \cos x$ is
$\sqrt{a^{2}+b^{2}}$
$\therefore$ Maximum value of $\sin x+\cos x$ is $\sqrt{1+1}=\sqrt{2}$.
$\mathrm{A} \rightarrow(2)$
Maximum value of $3 \sin x+4 \cos x$ is $\sqrt{9+16}=5$
$\mathrm{B} \rightarrow(3)$
Maximum value of $2 \sin x+\cos x$ is $\sqrt{4+1}=\sqrt{5}$
$\mathrm{C} \rightarrow(4)$
Maximum value of $\sin x+3 \cos x$ is $\sqrt{1+9}=\sqrt{10}$
$\mathrm{D} \rightarrow(1)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
Match List-I with List-II :
\begin{array}{llll} & List-I & & List-II \\A. & China rose & I. & Free central \\B. & Mustard & II. & Basal \\C. & Primrose & III. & Axile \\D. & Marigold & IV. & Parietal\end{array}
Choose the correct answer from the options given below :

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

The correct answer is:
A-III, B-IV, C-I, D-II

Solution:
China rose exhibits axile placentation.
Mustard exhibits parietal placentation.
Primrose exhibits free central placentation.
Marigold exhibits basal placentation.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2024 (Re-NEET)
Match List-I with List-II.
\begin{array}{|l|l|l|l|}\hline & List-I & & List-II \\\hline A. & Primary structure of protein & I. & Human haemoglobin \\\hline B. & Secondary structure of protein & II. & Disulphide bonds \\\hline C. & Tertiary structure of protein & III. & Polypeptide chain \\\hline D. & Quaternary structure of protein & IV. & Alpha helix and \beta sheet \\\hline\end{array}
Choose the correct answer from the options given below :

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

The correct answer is:
A-III, B-IV, C-II, D-I

Solution:
$\begin{array}{|l|l|r|l|}\hline & \text{List-I} & & \text{List-II} \\ \hline A. & \text{Primary structure of protein} & III. & \begin{array}{l} \text{Polypeptide chain with positional information of amino} \\ \text{acids} \end{array} \\ \hline B. & \text{Secondary structure of protein} & IV. & \text{Alpha helix and } \beta \text{ sheet structure} \\ \hline C. & \text{Tertiary structure of protein} & II. & \begin{array}{l} \text{Hollow woolen ball like structure with hydrogen and} \\ \text{disulphide bonds} \end{array} \\ \hline D. & \text{Quaternary structure of protein} & I. & \begin{array}{l} \text{Assembly of more than one polypeptides, seen in adult} \\ \text{human haemoglobin} \end{array} \\ \hline \end{array}$
Hence, A-III, B-IV, C-II, D-I is the correct match.

Topic: Biology, Biomolecules (B), NEET, NEET 2024 (Re-NEET)
Match List-I with List-II.
Choose the correct answer from the options given below :

Options:
A $(A) -\mathrm{I},(\mathrm{B})-\mathrm{IV},(\mathrm{C})-\mathrm{II},(\mathrm{D})-\mathrm{III}$
B $(A) -\mathrm{IV},(\mathrm{B})-\mathrm{III},(\mathrm{C})-\mathrm{I},(\mathrm{D})-\mathrm{II}$
C $(A) -\mathrm{IV},(\mathrm{B})-\mathrm{II},(\mathrm{C})-\mathrm{III},(\mathrm{D})-\mathrm{I}$
D $(A) -\mathrm{III},(\mathrm{B})-\mathrm{I},(\mathrm{C})-\mathrm{IV},(\mathrm{D})-\mathrm{II}$

The correct answer is:
$(A) -\mathrm{IV},(\mathrm{B})-\mathrm{III},(\mathrm{C})-\mathrm{I},(\mathrm{D})-\mathrm{II}$

Solution:
(i) In hydrarch succession pioneer species phytoplanktons, with series of seral community leads to mesic condition.
(ii) In xerarch succession pioneer species lichens, with series of seral community leads to mesic condition.
(iii) Ecological succession leads to gradual change in the species composition.
(iv) Secondary succession is faster and reach to the climax quickly.
Therefore, option (2) is correct.

Topic: Biology, Ecosystem, NEET, NEET 2023 (Manipur)
Match List-I with List-II :
$\begin{array}{|l|l|}
\hline \text{List-I } & \text{List-II } \\
\text{(Reaction) } & \text{(Product formed) } \\
\hline \text{(a) Gabriel synthesis} & \text{(i) Benzaldehyde} \\
\text{(b) Kolbe synthesis} & \text{(ii) Ethers} \\
\text{(c) Williamson synthesis} & \text{(iii) Primary amines} \\
\text{(d) Etard synthesis} & \text{(iv) Salicyclic acid} \\
\hline
\end{array}$
Choose the correct answer from the options given below :

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

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

Solution:
(a) Gabriel synthesis $\longrightarrow$ (iii) Primary amines
Pthalimide is alkylated with primary or unbranched secondary alkyl halides or sulphonate. The product thus obtained is hydrolysed to give primary amines.
Treatment of toluene with chromyl chloride in an inert solvent like $\mathrm{CS}_2$ or $\mathrm{CCl}_4$ forms an insoluble complex $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}\left(\mathrm{OCrCl}_2 \mathrm{OH}\right)_2$ which further on hydrolysis produces benzaldehyde.

Topic: Chemistry, Amines, NEET, NEET 2022 (Phase 2)
Match the column A with column B and choose the correct option

Column I

Column II

A. Porifera

i. Canal System

B. Aschelminthes

ii. Water-vascular system

C. Annelida

iii. Muscular pharynx

D. Arthropods

iv. Jointed appendages

E. Echinodermata

v. Metameres

Options:
A A-ii, B-iii, C-v, D-iv, E-i
B A-ii, B-v, C-iii, D-iv, E-i
C A-i, B-iii, C-v, D-iv, E-ii
D A-i, B-v, C-iii, D-iv, E-ii

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

Solution:
Porifera has Canal System, Aschelminthes has Muscular pharynx, Annelida has Metameres, Arthropods has Jointed appendages and Echinodermata has Water-vascular system.

Topic: Biology, Animal Kingdom, NEET
Match the column I with column II and choose the correct option

Column-I	Column- II

A. Inguinal canal	1. Network of seminiferous tubules

B. Rete testes	2. Secondary sexual characters

C. Leydig’s cells	3. Far descending of the testis

D. Prepuce	4. Dorsal bundles of muscles

E. Corpora cavernosa	5. Terminal skin of the penis

​

Options:
A

A	B	C	D	E

1	2	3	5	4

​
B

A	B	C	D	E

3	1	4	2	5

​
C

A	B	C	D	E

3	1	2	5	4

​
D

A	B	C	D	E

2	4	3	5	1

​

The correct answer is:

A	B	C	D	E

3	1	2	5	4

​

Solution:
Column I	Column IIâ

Inguinal canal	Far descending of testis

Rete testes	Network of seminiferous tubules

Leydig’s cells	Secondary sexual characters

Prepuce	Terminal skin of penis

Corpora cavernosa	Dorsal bundles of muscles

Topic: Biology, Human Reproduction, NEET
Match the columns

Column I	Column II

i. Butyric acid	p. Streptomyces venezuelae

ii. SCP	q. Saccharomyces cerevisiae var. ellipsoides

iii. Chloromycetin	r. Clostridium

iv. Wine	s. Chlorella

Options:
A i- s, ii-r, iii-q, iv-p
B i-r, ii-p, iii-s, iv-q
C i-r, ii- s, iii-p, iv-q
D i-q, ii-p, iii-s, iv-r

The correct answer is:
i-r, ii- s, iii-p, iv-q

Solution:
Ethanol is a type of alcohol that is a product of the fermentation process. Ethanol is widely consumed as a mix in varying degree of concentration (Wine, beer, whisky). Fermentation is an anaerobic respiratory process where many strains of Saccharomyces are used.
Clostrodium is a microbial source for commercial production of butyric acid.
Secondary metabolites like antibiotics are produced by the microbes which inhibit growth of other microbes. Chloromycetin is a type of antibiotics produced by Streptomyces venezuelae.
Microbes are used a direct source of food as they are rich in protein. Single cell protein denotes dead and dried cells of microbes where each cell is like a source of protein. Chlorella, a type of algae is used as SCP.

Topic: Biology, Microbes in Human Welfare, NEET
Match the columns:

Column-I	Column-II

A.	Nitrous oxide	(i)	Secondary pollutant from car exhausts

B.	Chlorofluorocarbons	(ii)	Combustion of fossil fuels

C.	Methane	(iii)	Denitrification

D.	Ozone	(iv)	Refrigerators aerosols sprays

E.	Carbon dioxide	(v)	Cattle, rice fields, toilets

Options:
A A = (ii), B = (iv), C = (v), D = (i), E = (ii)
B A = (v), B = (i), C = (iii), D = (iv), E = (ii)
C A = (iv), B = (v), C = (i), D = (ii), E = (iii)
D A = (ii), B = (iii), C = (iv), D = (v), E = (ii)

The correct answer is:
A = (ii), B = (iv), C = (v), D = (i), E = (ii)

Solution:
The nitrogen oxides are released into the atmosphere when fossil fuels are burnt and that contributes to the formation of smog and acid rain.
Chlorofluorocarbons (CFCs) are the compounds released into the atmosphere due to the usage of air-conditioning, refrigeration, blowing agents in foams, insulations and packing materials, propellants in aerosol cans and as solvents.
The digestion activity of the methanogenic bacteria, the decomposition of fecal matter in the toilets and rice fields release methane into the environment.
Ozone in the troposphere is the secondary pollutant formed by the photooxidation of the car exhausts pollutants.
Oxygen combines with carbon to form CO2 and with hydrogen to form water (H2O) when fossil fuels are burned.

Topic: Biology, Environmental Issues, NEET
Match the columns given below correctly.

Types of Wood	Description

1. Early wood	i. Has narrow vessels.

2. Late wood	ii. Dead elements with highly lignified walls.

3. Heartwood	iii. Vessels with wider cavities.

4. Sapwood	iv. Lighter in colour.

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

Solution:
Heartwood does not conduct water or transport it but it is hard and gives plant mechanical support and thus consist of dead elements with lignified walls. Early wood or Spring wood is formed during spring season and known to have wide vessels and Late wood have narrow vessels and cambium is less active in it. Sapwood is a part of secondary xylem in peripheral region which is lighter in colour.

Topic: Biology, Anatomy of Flowering Plants, NEET
Match the columns I, II and III and choose the CORRECT combination from the options given.

Column I	Column II	Column III

a.	Tall and slender cells	1.	Cuboidal epithelium	K.	Bronchioles

b.	Flattened cells with irregular boundaries	2.	Columnar epithelium	L.	Tubular parts of nephrons

c.	Cilia on their free surface	3.	Squamous epithelium	M.	Lining of intestine

d.	Cube like cells	4.	Ciliated epithelium	N.	Wall of blood vessels

Options:
A a - 2 - M, b - 3 - N, c - 4 - K, d - 1 - L
B a - 2 - K, b - 1 - L, c - 4 - M, d - 3 - N
C a - 4 - M, b - 3 - L, c - 1 - K, d - 1 - N
D a - 2 - M, b - 3 - N, c - 4 - L, d - 1 - K

The correct answer is:
a - 2 - M, b - 3 - N, c - 4 - K, d - 1 - L

Solution:
Epithelial tissue is divided into two types: Simple and compound based on number of layers resting on the basement membrane. Simple epithelium has only one layer resting on the basement membrane and such epithelium brings abo transport of substances. This epithelium is further divided into various types depending on the shape of cells. These are as follows. A) Squamous epithelium: This consists of flat cells and is responsible for diffusion of substances. The borders are irregular sometimes to increase the surface area and this is tesallated squamous epithelium. This is seen in the inner lining of blood vessels. B) Cuboidal epithelium: This epithelium consists of cube shaped cells. This is seen in the tubular parts of nephrons and is responsible for absorption and secretion of cells. C) Columnar epithelium: This epithelium consists of tall column like cells and is seen lining the intestine. D) Ciliated epithelium: This is cuboidal or columnar cells with cilia that bring about movement of substances. This is seen in the bronchioles.

Topic: Biology, Structural Organisation in Animals, NEET