datavorous/entrance-exam-dataset huggingface dataset:
In humans, at the end of the first meiotic division, the male germ cells differentiate into the

Options:
A primary spermatocytes
B secondary spermatocytes
C spermatids
D spermatogonia

The correct answer is:
secondary spermatocytes

Solution:
At sexual maturity the germ cells in human male divide several times by mitosis to produce a large number of spermatogonia. In growth phase each spermatogonium actively grows to a larger primary spermatocyte by obtaining a nourishment from the nurshing cells. In maturation phase each primary spermatocyte undergoes two successive divisions called maturation divisions. The primary spermatocyte undergoes first maturation division (meiotic) to producing two haploid daughter cells called secondary spermatocytes. Both secondary spermatocytes undergo second maturation division to form four haploid spermatids. The transformation of spermatids into spermatozoa is called spermiogenesis or spermateliosis.

Topic: Biology, Human Reproduction, NEET, NEET 2008 (Screening)
In humans, blood passes from the post caval to the diastolic right atrium of heart due to

Options:
A pushing open of the venous valves
B suction pull
C stimulation of the sino auricular node
D pressure difference between the caval and atrium

The correct answer is:
pressure difference between the caval and atrium

Solution:
The action of heart includes contractions and relaxations of the atria and ventricles. Contraction of the heart is called systole and relaxation is called diastole. The dynamics of blood flow in blood vessels is no exception and blood flows through the blood vessels along a pressure gradient, always moving from higher to lower pressure areas. Fundamentaly, the pumping action of heart generates blood flow. The contraction of atria is initiated and activated by sinoatrial node (SA node-Pacemaker), which spreads waves of contraction across the wall of atria via muscle fibres at regular intervals.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2008 (Screening)
In Kjeldahl's method of estimation of nitrogen, nitrogen is quantitatively converted to ammonium sulphate. It is then treated with standard solution of alkali. The nitrogen which is present is estimated as

Options:
A N2
B NO2
C NH3
D (NH4)2SO4 ppt.

The correct answer is:
NH3

Solution:
When treated with standard solution of alkali nitrogen is liberated as NH3 which is absorbed in H2SO4 to get percentage of nitrogen.

Topic: Chemistry, General Organic Chemistry, NEET
In malignant tumours, the cells proliferate, grow rapidly and move to other parts of the body to form new tumours. This stage of disease is called

Options:
A
metagenesis

B
metastasis

C
teratogenesis

D
mitosis

The correct answer is:
metastasis

Solution:
Cancer is defined as an uncontrolleddivision or proliferation of cells without any differentiation. Repeated division of cells form a large mass of tissue called tumours. These are of two types i.e., benign (non-cancerous) and malignant (cancerous).
The invasion of malignant tumour from primary site of cancer to the other secondary growth is called metastasis, while metagenesis is the alteration of generation.
Teratogenesis is a prenatal toxicity, that is characterised by structural, functional defects in the developing embryo or foetus. Mitosis is a type of cell division that results in two daughter cells.

Topic: Biology, Human Health and Diseases
In mammals growing oocytes are surrounded by special nutritive cells called

Options:
A Follicle cells
B Nurse cells
C Follicle cells and nurse cells
D None of the above

The correct answer is:
Follicle cells

Solution:
Graffian follicle is about 2.5 cm in diameter and has a secondary oocyte surrounded by a few layers of follicular cells, those are nourishing cells formed from undifferentiated oogonia.

Topic: Biology, Human Reproduction, NEET
In Meghalaya, sacred groves are found in

Options:
A Khasi and Jaintia Hills
B Aravalli Hills
C Western Ghat regions
D Sarguja, Chanda and Bastar areas

The correct answer is:
Khasi and Jaintia Hills

Solution:
Sacred groves are forests which are generally worshipped by certain cultures. Such sacred groves are found in Khasi and Jaintia Hills in Meghalaya, Aravalli Hills of Rajasthan, Western Ghat regions of Karnataka and Maharashtra and the Sarguja, Chanda and Bastar areas of Madhya Pradesh.

Topic: Biology, Biodiversity and its Conservation, NEET
In metre bridge experiment, the null point is obtained at $20 \mathrm{~cm}$ from left end of the wire, when resistance $X$ is balanced against another resistance $Y(X

PhysicsCurrent ElectricityMHT CETMHT CET 2022 (05 Aug Shift 1)

Options:

A $40 \mathrm{~cm}$
B $80 \mathrm{~cm}$
C $60 \mathrm{~cm}$
D $50 \mathrm{~cm}$

Solution:

2082 Upvotes

Verified Answer

The correct answer is:
$50 \mathrm{~cm}$

The correct option is (D).
Concept: Meter bridge experiment is based on two facts:
(1) the resistance across a wire is directly proportional to its length
(2) balanced Wheatstone bridge has null deflection when the ratio of resistances is equal, and no current flows through the circuit.
The current through galvanometre is zero, when the ratio of resistances, i.e., $\frac{P}{Q}=\frac{R}{S}$.
As given in the problem, $P=X, Q=Y, R \propto 20 \mathrm{~cm}$ and $S \propto 80 \mathrm{~cm}$.
Therefore,
$\frac{X}{Y}=\frac{20}{80}=\frac{1}{4}$
Similarly, to balance a resistance $P=4 X$ against $Q=Y$, let us assume a length $l$ is required. Then, $R=l$
$\frac{4 X}{Y}=\frac{l}{100-l}$
On solving, we get $l=50 \mathrm{~cm}$.

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Options:
A $40 \mathrm{~cm}$
B $80 \mathrm{~cm}$
C $60 \mathrm{~cm}$
D $50 \mathrm{~cm}$

The correct answer is:
$50 \mathrm{~cm}$

Solution:
The correct option is (D).
Concept: Meter bridge experiment is based on two facts:
(1) the resistance across a wire is directly proportional to its length
(2) balanced Wheatstone bridge has null deflection when the ratio of resistances is equal, and no current flows through the circuit.
The current through galvanometre is zero, when the ratio of resistances, i.e., $\frac{P}{Q}=\frac{R}{S}$.
As given in the problem, $P=X, Q=Y, R \propto 20 \mathrm{~cm}$ and $S \propto 80 \mathrm{~cm}$.
Therefore,
$\frac{X}{Y}=\frac{20}{80}=\frac{1}{4}$
Similarly, to balance a resistance $P=4 X$ against $Q=Y$, let us assume a length $l$ is required. Then, $R=l$
$\frac{4 X}{Y}=\frac{l}{100-l}$
On solving, we get $l=50 \mathrm{~cm}$.

Topic: Physics, Current Electricity, MHT CET, MHT CET 2022 (05 Aug Shift 1)
In metre bridge experiment, with a standard resistance in the right gap and a resistance coil
dipped in water (in a beaker) in the left gap, the balancing length obtained is '1'. If the
temperature of water is increased, the new balancing length is

Options:
A \(>l \)
B \( < l \)
C \( =l \)
D \( =0 \)

The correct answer is:
\(>l \)

Solution:
Given, balancing length of bridge with standard resistance in right gap and a resistance coil dipped in water in left gap is
\( I=1 \). If temperature of water is increased the new balancing length is greater than \( 1 \).
Unknown resistance is calculated as \( X=R \frac{l}{(1-l)} \)
If temperature increases then resistance increases. Thus, option (1) is correct.

Topic: Physics, Ray Optics, KCET, KCET 2017
In monocot roots which types of vascular bundles are found?

Options:
A Collateral, conjoint and closed
B Radial with exarch xylem
C Bicollateral, conjoint and closed
D Radial with endarch xylem

The correct answer is:
Radial with exarch xylem

Solution:
When xylem and phloem groups are located on different radii, the bundles are said to be radial e.g. root. The protoxylem element lie at the periphery and metaxylem in the centre, this condition is called exarch e.g. root.

Topic: Biology, Anatomy of Flowering Plants, JIPMER, JIPMER 2005
In old trees, the central dark-coloured, non-conducting part of secondary xylem is referred to

Options:
A heartwood
B sapwood
C softwood
D hardwood

The correct answer is:
heartwood

Solution:
Over time the centre of the innermost layers of the stem gets deposited with organic compounds like tannins, raisins, oils, gums and other aromatic substances. This deposition gives the secondary xylem a dark brownish colour and makes it durable and resistant to the microorganisms and insects. This dead inner wood with highly lignified walls is known as the hardwood. They are not involved in the conduction and transport of water but gives mechanical support to the stem. The sapwood is the outermost portion of the secondary xylem lighter in colour and involved in the conduction of water and minerals.

Topic: Biology, Anatomy of Flowering Plants, NEET
In old trees the greater part of secondary xylem is dark brown and resistant to insect attack due to:
(a) secretion of secondary metabolites and their deposition in the lumen of vessels.
(b) deposition of organic compounds like tannins and resins in the central layers of stem.
(c) deposition of suberin and aromatic substances in the outer layer of stem.
(d) deposition of tannins, gum, resin and aromatic substances in the peripheral layers of stem.
(e) presence of parenchyma cells, functionally active xylem elements and essential oils.
Choose the correct answer from the options given below:

Options:
A (c) and (d) Only
B (d) and (e) Only
C (b) and (d) Only
D (a) and (b) Only

The correct answer is:
(a) and (b) Only

Solution:
In old trees, the greater part of secondary xylem is dark brown due to deposition of organic compounds like tannins, resins, oils, gums, aromatic substances and essential oils in the central or innermost layers of the stem. These substances make it hard, durable and resistant to the attacks of microorganisms and insects. This region comprises dead elements with highly lignified walls and is called heartwood.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2022 (Phase 1)
In Oppenauer's oxidation,

Options:
A secondary alcohol is oxidised to carboxylic acid in acetone solvent using aluminium tertiary butoxide
B secondary alcohol is oxidised to carboxylic acid without affecting the $\mathrm{C}=\mathrm{C}$ or $\mathrm{C} \equiv \mathrm{C}$ bond by aluminium tertiary butoxide in acetone solvent
C secondary alcohol is oxidised to ketone without affecting $\mathrm{C}=\mathrm{C}$ or $\mathrm{C} \equiv \mathrm{C}$ bond by aluminium tertiary butoxide
D secondary alcohol is oxidised to ketone by chromic acid - pyridine complex.

The correct answer is:
secondary alcohol is oxidised to ketone without affecting $\mathrm{C}=\mathrm{C}$ or $\mathrm{C} \equiv \mathrm{C}$ bond by aluminium tertiary butoxide

Solution:
In Oppenauer's oxidation, secondary alcohol is oxidised to corresponding ketone in the presence of aluminium tertiary butoxide. Other oxidisable groups are not affected.

Topic: Chemistry, Aldehydes and Ketones, AIIMS, AIIMS 2009
In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to $\frac{3}{4}$ th of the original length and the tension is changed. The factor by which the tension is to be changed is

Options:
A $\frac{3}{8}$
B $\frac{2}{3}$
C $\frac{8}{9}$
D $\frac{9}{4}$

The correct answer is:
$\frac{9}{4}$

Solution:
Frequency of stretched string.
$\mathrm{n}=\frac{1}{2 l} \frac{\sqrt{T}}{m}$
$\begin{aligned} & n \propto \frac{\sqrt{T}}{l} \\ & \frac{n_1}{n_2}=\frac{l_2}{I_1} \sqrt{\frac{T_1}{T_2}} \Rightarrow \frac{n}{2 n}=\frac{\frac{3}{4}}{l} \sqrt{\frac{T_1}{T_2}} \\ & \frac{1}{2}=\frac{3}{4} \sqrt{\frac{T_1}{T_2}} \Rightarrow \sqrt{\frac{T_1}{T_2}}=\frac{2}{3} \\ & \frac{T_1}{T_2}=\frac{4}{9} \Rightarrow T_2=\frac{9}{4} T_1\end{aligned}$
$\therefore$ Tension should be changed by the factor $\frac{9}{4}$.

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2001
In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to $\frac{3}{4}$ th of the original length and the tension is changed. The factor by which the tension is to be changed is

Options:
A $\frac{3}{8}$
B $\frac{2}{3}$
C $\frac{8}{9}$
D $\frac{9}{4}$

The correct answer is:
$\frac{9}{4}$

Solution:
Frequency of stretched string.
$\mathrm{n}=\frac{1}{2 l} \frac{\sqrt{T}}{m}$
$\begin{aligned} & n \propto \frac{\sqrt{T}}{l} \\ & \frac{n_1}{n_2}=\frac{l_2}{I_1} \sqrt{\frac{T_1}{T_2}} \Rightarrow \frac{n}{2 n}=\frac{\frac{3}{4}}{l} \sqrt{\frac{T_1}{T_2}} \\ & \frac{1}{2}=\frac{3}{4} \sqrt{\frac{T_1}{T_2}} \Rightarrow \sqrt{\frac{T_1}{T_2}}=\frac{2}{3} \\ & \frac{T_1}{T_2}=\frac{4}{9} \Rightarrow T_2=\frac{9}{4} T_1\end{aligned}$
$\therefore$ Tension should be changed by the factor $\frac{9}{4}$.

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2001
In oxidation pond, once BOD from the sewage is reduced significantly, the 'flocs' are allowed to sediment and it is known as:

Options:
A Activated sludge
B Inactivated sludge
C Primary sludge
D Secondary sludge

The correct answer is:
Activated sludge

Solution:
Once BOD of sewage water is reduced significantly, the effluent is the passed into the setteling tank where the bacterial ' flocs' are allowed to sediment. This sediment is called as activated sludge.

Topic: Biology, Microbes in Human Welfare, NEET
In photosystem-I, the first electron acceptor is:

Options:
A an iron-sulphur protein
B ferredoxin
C cytochrome
D plastocyanin

The correct answer is:
an iron-sulphur protein

Solution:
In photosystem I, the primary electron acceptor is probably a Fe-S protein. The reduced primary acceptor transfers the electrons to the secondary electron acceptor $\mathrm{P}_{430}$. The reduced $\mathrm{P}_{430}$ passes its electrons to ferredoxin $(\mathrm{Fd})$ present at the outer surface of the thylakoid membrane.
Related Theory
In light reaction of photosynthesis two types of photosystems are involved. PS-I consists of plenty of chlorophyll- $a$ and very less quantity of chlorophyll-b. These pigments absorb light energy and transfer it to the reaction centre - $P_{700}$ After absorbing adequate amount of light energy electron gets excited from $P_{700}$ molecule and moves to iron-sulphur protein complex, designated as $A$ (Fe-S). It gets reduced after accepting electrons. It later gives this electron to ferredoxin and gets oxidized again.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2006
In plants towards the centre of the root, next to endodermis lies a few layers of thick-walled parenchymatous cells referred to as

Options:
A pith.
B epidermis.
C phloem.
D pericycle.

The correct answer is:
pericycle.

Solution:
In the root of angiospermic plant, the tissue group adjacent to endodermis and towards the central part is called as pericycle. Pericycle generally is composed of parenchymatous cell and in the centre of it lies the pith which is mostly inconspicuous. Formation of lateral root and vascular cambium during the secondary growth occurs at this section.
Epidermis forms the outermost layer of the root having cells giving out root hair in some cells. Phloem is a type of complex vascular tissue which helps in transportation of food throughout the plant.

Topic: Biology, Anatomy of Flowering Plants, NEET
In $\Delta \mathrm{PQR}, \angle \mathrm{R}=\frac{\pi}{2}$. If than $\left(\frac{1}{2}\right)$ and
$\left(\frac{Q}{2}\right)$ are the roots of the equation $a x^{2}+b x+c=0$, then which one of the following is correct?

Options:
A $a=b+c$
B $\mathrm{b}=\mathrm{c}+\mathrm{a}$
C $\mathrm{c}=\mathrm{a}+\mathrm{b}$
D $\mathrm{b}=\mathrm{c}$

The correct answer is:
$\mathrm{c}=\mathrm{a}+\mathrm{b}$

Solution:
$\angle \mathrm{R}=\frac{\pi}{2}$
$\therefore \angle \mathrm{P}+\angle \mathrm{Q}=\frac{\pi}{2}$
$\Rightarrow \frac{\angle \mathrm{P}+\angle \mathrm{Q}}{2}=\frac{\pi}{4}$...(1)
$\tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2} \tan \frac{Q}{2}}$...(2)
Given, $\tan \frac{P}{2}$ and $\tan \frac{Q}{2}$ are roots of $a x^{2}+b x+c=0$. $\tan \frac{P}{2}+\tan \frac{Q}{2}=\frac{-b}{a} ; \tan \frac{P}{2} \cdot \tan \frac{Q}{2}=\frac{c}{a}$
$(2) \Rightarrow \tan \left(\frac{\mathrm{P}}{2}+\frac{\mathrm{Q}}{2}\right)=\frac{\frac{-\mathrm{b}}{\mathrm{a}}}{1-\frac{\mathrm{c}}{\mathrm{a}}}$
$\Rightarrow \tan \frac{\pi}{4}=\frac{\frac{-b}{a}}{1-\frac{c}{a}} \quad($ from $(1))$
$\Rightarrow 1=\frac{-b}{a-c} \Rightarrow-b=a-c \Rightarrow a+b=c$
$11-x+z^{2}=5$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2017 (Phase 2)
In resonance tube of length $0.8 \mathrm{~m}$, air column vibrates with a source of frequency
$375 \mathrm{~Hz}$ for a certain height of water from bottom of the tube. Water level corresponding to fundamental frequency is (Neglect end correction, speed of sound in air = $330 \mathrm{~m} / \mathrm{s}$ )

Options:
A $0.45 \mathrm{~m}$
B $0.58 \mathrm{~m}$
C $0.8 \mathrm{~m}$
D $0.65 \mathrm{~m}$

The correct answer is:
$0.58 \mathrm{~m}$

Solution:
$\ell=0.8 \mathrm{~m} \quad \mathrm{f}=375 \mathrm{~Hz} \quad \mathrm{v}=330 \mathrm{~m}$
$\mathrm{n}_{0}=\frac{\mathrm{v}}{4 \ell}=375$
$\therefore \ell=\frac{330}{4 \times 375}=0.22 \mathrm{~m}$
$\therefore \mathrm{L}-\ell=0.8-0.22 \mathrm{~m}=0.58 \mathrm{~m}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (20 Oct Shift 1)
In respect of the equation
$\left[\begin{array}{ll}
2 & 3 \\
4 & 6
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y}
\end{array}\right]=\left[\begin{array}{c}
5 \\
\mathrm{c}-5
\end{array}\right]$
correctly match List I with List II and select the correct answer using the code given below the lists:

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

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

Solution:
$\left[\begin{array}{ll}2 & 3 \\ 4 & 6\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}5 \\ c-5\end{array}\right] \Rightarrow\left[\begin{array}{l}2 x+3 y \\ 4 x+6 y\end{array}\right]=\left[\begin{array}{l}5 \\ c-5\end{array}\right]$
$\Rightarrow 2 x+3 y=5$ ...(1)
and $4 x+6 y=c-5$ ...(2)
$\Rightarrow$ Solving equation (1) and (2) $\mathrm{c}=15$ Now, if $\mathrm{c}=15$, equations 1 and 2 becomes $\Rightarrow 2 x+3 y=5$ and $4 x+6 y=10$
This shows that the equation has an infinite set of solutions. For other values of $\mathrm{c}$, equation has no solution.

Topic: Mathematics, Determinants, NDA, NDA 2006 (Phase 2)
In secondary proteins H-bonding occurs between :

Options:
A CO and CO
B NH and NH
C CO and NH
D None of these

The correct answer is:
CO and NH

Solution:
Correct Option is : 3
CO and NH

Topic: Chemistry, Biomolecules, NEET
In situ conservation refers to:

Options:
A Conserve only high risk species
B Conserve only endangered species
C Conserve only extinct species
D Protect and conserve the whole ecosystem

The correct answer is:
Protect and conserve the whole ecosystem

Solution:
In the in situ conservation, the protection of an organism will automatically take place, since the natural habitat of the organisms is protected in this sort of conservation. For example, announcing Kanha forest as a tiger reserve. It is nothing but conservation ‘at home’. This helps in conserving biodiversity hotspots. India has three of the world’s biodiversity hotspots, Western ghats, Indo Burma and Eastern-Himalayas.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2022 (Phase 1)
In solving a problem that reduces to a quadratic equation, one student makes a mistake in the constant term and obtains 8 and 2 for roots. Another student makes a mistake only in the coefficient of first-degree term and finds $-9$ and $-1$ for roots. The correct equation is

Options:
A $x^{2}-10 x+9=0$
B $x^{2}-10 x+9=0$
C $x^{2}-10 x+16=0$
D $x^{2}-8 x-9=0$

The correct answer is:
$x^{2}-10 x+9=0$

Solution:
Letcorrect equation is $a x^{2}+b x+c=0$ According to first student, equation is $a x^{2}+b x+c_{1}=0$ and roots are 8 and
$8+2=\frac{-b}{a} \Rightarrow \frac{b}{a}=-10$
Quadratic equation according to second student $a x^{2}+b_{1} x+c=0$ and roots are $-9$ and $-1$
$(-9) \times(-1)=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{\mathrm{c}}{\mathrm{a}}=9$
Putting value in original equation
$x^{2}+\frac{b}{a} x+\frac{c}{a}=0$
$x^{2}-10 x+9=0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2015 (Phase 1)
In spermatogenesis reduction division of chromosome occurs during the conversion of

Options:
A primary spermatocytes to secondary spermatocytes
B spermatogonia to primary spermatocytes
C spermatids to sperms
D secondary spermatocytes to spermatids

The correct answer is:
primary spermatocytes to secondary spermatocytes

Solution:
The germ cells in the seminiferous tubules giving rise to the haploid spermatozoa are known as spermatogenesis. The three stages involved in this process are mitosis, meiosis, and spermiogenesis. The first step begins when the stem cells start multiplying and produce duplicate cells(spermatogonia). When these cells mature they are known as primary spermatocytes which undergo meiotic-I cell division to form secondary spermatocytes.

Topic: Biology, Human Reproduction, NEET
In stcp-up transformer, relation between number of turns in primary $\left(N_{P}\right)$ and number of turns in secondary $\left(N_{S}\right)$ coils is

Options:
A $N_{S}$ is greater than $N_{P}$
B $N_{P}$ is greater than $N_{S}$
C $N_{S}$ is equal to $N_{P}$
D $N_{P}=2 N_{S}$

The correct answer is:
$N_{S}$ is greater than $N_{P}$

Solution:
In step-up transformer, number of turns in secondary coil is greater than the number of turns in the jxrimary coil.
1.e. $\quad N_{g}>N_{\mathrm{P}}$

Topic: Physics, Alternating Current, COMEDK, COMEDK 2020
In stems, the protoxylem lies towards the center (pith) and the metaxylem lies towards the periphery of the organ. This type of primary xylem is called

Options:
A Exarch
B Endarch
C Both (A) and (B)
D None of these

The correct answer is:
Endarch

Solution:
In stems, the protoxylem lies towards the centre (pith) and the metaxylem lies towards the periphery of the organ. This type of primary xylem is called endarch. In roots, the protoxylem lies towards periphery and metaxylem lies towards the centre. Such an arrangement of primary xylem is called exarch.

Topic: Biology, Anatomy of Flowering Plants, NEET
In temperate regions, cambium is less active during the winter season and forms fewer xylary elements that have narrow vessels, this wood is called ____________.

Options:
A early wood
B late wood
C heart wood
D sapwood

The correct answer is:
late wood

Solution:
In the spring season, cambium is very active and produces a large number of xylary elements having vessels with wider cavities. The wood formed during this season is called spring wood or early wood. In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels, and this wood is called autumn wood or late wood.
In old trees, the greater part of secondary xylem is dark brown due to deposition of organic compounds like tannins, resins, oils, gums, aromatic substances and essential oils in the central or innermost layers of the stem. These substances make it hard, durable and resistant to the attacks of microorganisms and insects. This region comprises dead elements with highly lignified walls and is called heartwood. The peripheral region of the secondary xylem, is lighter in colour and is known as the sapwood.

Topic: Biology, Anatomy of Flowering Plants, NEET