datavorous/entrance-exam-dataset huggingface dataset:
The colour of sky is due to

Options:
A transmission of light
B wavelength of scattered light
C absorption of light by atomspheric gases
D All of the above

The correct answer is:
wavelength of scattered light

Solution:
The blue colour of sky is due to Tyndall effect the colloidal particles absorb light, become self luminous and then scatter light of different wavelengths in all possible directions.

Topic: Chemistry, Surface Chemistry, KCET, KCET 2008
The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O(l) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ∆fH⊖of benzene. (Standard enthalpies of formation of CO2(g) and H2O(l) are –393.5 kJ mol–1and – 285.83 kJmol–1respectively)

Options:
A –33.25 kJ mol–1
B –56.43 kJ mol–1
C –48.51 kJ mol–1
D –65.78 kJ mol–1

The correct answer is:
–48.51 kJ mol–1

Solution:
Correct Option is : (C)
–48.51 kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
The common bottle cork is a product of

Options:
A dermatogen
B phellogen
C xylem
D vascular cambium

The correct answer is:
phellogen

Solution:
The cork cambium or phellogen cells divide periclinally cutting off cells towards the outside and inside. The cells cut off towards the out side become suberized and dead. These are compactly packed in radial rows without intercellular spaces and form cork or phellem. Cork is impervious to water due to suberin and provides protection to underlying tissues. The cells cut off from cork cambium towards inside add to the cortex and are called secondary cortex cells or phelloderm cells.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2012 (Screening)
The common roots of the equations $z^{3}+2 z^{2}+2 z+1=0$ and $\mathrm{z}^{2017}+\mathrm{z}^{2018}+1=0$ are

Options:
A $-1, \omega$
B $1, \omega^{2}$
C $-1, \omega^{2}$
D $\omega, \omega^{2}$

The correct answer is:
$\omega, \omega^{2}$

Solution:
Given equation, $\mathrm{z}^{3}+2 \mathrm{z}^{2}+2 \mathrm{z}+1=0$
$\Rightarrow \mathrm{z}^{3}-\mathrm{z}^{2}+\mathrm{z}+\mathrm{z}^{2}-\mathrm{z}+1+2 \mathrm{z}^{2}+2 \mathrm{z}=0$
$\Rightarrow(\mathrm{z}+1)\left(\mathrm{z}^{2}-\mathrm{z}+1\right)+2 \mathrm{z}(\mathrm{z}+1)=0$
$\Rightarrow(\mathrm{z}+1)\left(\mathrm{z}^{2}-\mathrm{z}+1+2 \mathrm{z}\right)=0$
$\Rightarrow(\mathrm{z}+1)\left(\mathrm{z}^{2}+\mathrm{z}+1\right)=0$
$\Rightarrow \mathrm{z}=-1, \omega, \omega^{2}$
$\mathrm{z}^{2017}+\mathrm{z}^{2018}+1=\omega+\omega^{2}+1=0$
$\therefore$ Common roots are $\omega, \omega^{2}$

Topic: Mathematics, Complex Number, NDA, NDA 2019 (Phase 1)
The common tangent to the parabola \(y^2=32 x\) and \(x^2=256 y\) will be

Options:
A \(2 x+4 y+64=0\)
B \(x+2 y-32=0\)
C \(2 x+4 y+32=0\)
D \(4 x+2 y+64=0\)

The correct answer is:
\(2 x+4 y+64=0\)

Solution:
Given parabolas are,
\(y^2=32 x \text { and } x^2=256 y\)
We use a standard result to find equation of common tangent. Equation of tangent common to \(y^2=4 a x\) and \(x^2=4 b y\) is,
\(b^{1 / 3} y+a^{1 / 3} x+\left(a^2 b^2\right)^{1 / 3}=0\)
Here,
\(\begin{aligned}
& 4 a=32 \Rightarrow a=8 \\
& 4 b=256, \quad b=64
\end{aligned}\)
So tangent is,
\(\begin{array}{ll}
& (64)^{1 / 3} y+(8)^{1 / 3} x+\left(8^2 \times 64^2\right)^{1 / 3}=0 \\
\Rightarrow \quad & 4 y+2 x+64=0
\end{array}\)

Topic: Mathematics, Parabola, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 2)
The complete solution of $3 \tan ^{2} x=1$ is given by:

Options:
A $x=n \pi \pm \frac{\pi}{3}$
B $x=n \pi+\frac{\pi}{3}$ only
C $x=n \pi \pm \frac{\pi}{6}$
D $x=n \pi+\frac{\pi}{6}$ only
where $n \in Z$

The correct answer is:
$x=n \pi \pm \frac{\pi}{6}$

Solution:
$3 \tan ^{2} x=1$
$\tan x=\pm \frac{1}{\sqrt{3}}$
$\tan x=\tan \left(\pm \frac{\pi}{6}\right)$
$x=n \pi \pm \frac{\pi}{6}$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2014 (Phase 1)
The compound on dehydrogenation gives a ketone. The original compound is

Options:
A primary alcohol
B secondary alcohol
C tertiary alcohol
D carboxylic acid

The correct answer is:
secondary alcohol

Solution:
Secondary alcohols gives ketones on dehydrogenation e.g.,

Topic: Chemistry, Aldehydes and Ketones, COMEDK, COMEDK 2017
The compound on dehydrogenation gives a ketone. The original compound is

Options:
A primary alcohol
B secondary alcohol
C tertiary alcohol
D carboxylic acid

The correct answer is:
secondary alcohol

Solution:
Secondary alcohols on dehydrogenation with $\mathrm{Cu}$ at $573 \mathrm{~K}$ give ketones.

Topic: Chemistry, Alcohols Phenols and Ethers, KCET, KCET 2008
The compound on dehydrogenation gives a ketone. The original compound is

Options:
A Primary alcohol
B Secondary alcohol
C Tertiary alcohol
D Carboxylic acid

The correct answer is:
Secondary alcohol

Solution:
Secondary alcohols on dehydrogenation with Cu at 573 K give ketones.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The compound that forms a yellow oily liquid with nitrous acid is

Options:
A 2 -methyl aniline
B methyl amine
C Benzyl amine
D diethyl amine

The correct answer is:
diethyl amine

Solution:
Secondary amines (-NH group containing amines), when reacts with nitrous acid $\left(\mathrm{HNO}_{2}\right)$, a yellow oily nitroso compound is formed.

Topic: Chemistry, Amines, MHT CET, MHT CET 2009
The compound which on reaction with aqueous nitrous acid at low temperature produces an oily nitrosoamine is

Options:
A methyl amine
B ethyl amine
C diethyl amine
D triethyl amine

The correct answer is:
diethyl amine

Solution:
Secondary $\left(2^{\circ}\right)$ amines (aliphatic as well as aromatic) react with nitrous acid $\left(\mathrm{HNO}_2\right)$ to form N-nitrosoamines.

Topic: Chemistry, Amines, AIIMS, AIIMS 2010
The condition where protoxylem lies towards the pith and metaxylem towards periphery is called as (i) while the condition where protoxylem lies towards periphery and metaxylem towards pith is known as (ii)

Options:
A (i) endarch xylem (ii) exarch xylem.
B (i) exarch xylem (ii) endarch xylem
C (i) diarch xylem (ii) polyarch xylem
D (i) hexarch xylem (ii) tetrarch xylem

The correct answer is:
(i) endarch xylem (ii) exarch xylem.

Solution:
The primary xylem tissue is of two types : protoxylem and metaxylem. The protoxylem is formed first and the metaxylem is formed later. The primary xylem is called endarch when the protoxylem lies towards the centre(pith) and metaxylem lies towards the periphery. E.g in stem. When the protoxylem lies towards periphery and metaxylem towards the centre it is called exarch xylem, eg : in roots.

Topic: Biology, Anatomy of Flowering Plants, NEET
The connective tissue having the presence of abundant fibroblasts, macrophages and mast cells is:

Options:
A

Type of connective tissue	Location

Adipose tissue	Beneath the skin

B

Type of connective tissue	Location

Areolar tissue	Beneath the skin

C

Type of connective tissue	Example

Dense connective tissue	Tendons

D

Type of connective tissue	Location

Specialised connective tissue	Cartilage

The correct answer is:

Type of connective tissue	Location

Areolar tissue	Beneath the skin

Solution:
Areolar connective tissue: A type of connective tissue consisting of a gel-like matrix incorporating strands of protein fibres (collagen and elastin) and such cells as fibroblasts, mast cells, macrophages, and fat cells. This tissue is found throughout the body under the skin and linking organs and other tissues.

Topic: Biology, Structural Organisation in Animals, NEET
The constant term in the expansion of \( \left|\begin{array}{ccc}3 x+1 & 2 x-1 & x+2 \\ 5 x-1 & 3 x+2 & x+1 \\ 7 x-2 & 3 x+1 & 4 x-1\end{array}\right| \) is

Options:
A \( 12 \)
B \( 06 \)
C \( 00 \)
D \( -10 \)

The correct answer is:
\( 06 \)

Solution:
(B)
Take \( x=0 \)
\[
\begin{array}{c}
\left|\begin{array}{ccc}
1 & -1 & 2 \\
-1 & 2 & 1 \\
-2 & 1 & -1
\end{array}\right| \\
=1(-3)+1(3)+2(3)=6
\end{array}
\]

Topic: Mathematics, Determinants, KCET, KCET 2019
The co-ordinates of foci of the ellipse $16 x^{2}+9 y^{2}=144$ are

Options:
A $(\pm 7,0)$
B $(0, \pm \sqrt{7})$
C $(\pm \sqrt{7}, 0)$
D $(0, \pm 7)$

The correct answer is:
$(0, \pm \sqrt{7})$

Solution:
Center is $(0,0)$, focii are $(0, \sqrt{7})$ and $(0,-\sqrt{7})$, major axis is along $y$-axis and is 8 , and minor axis is along $x$-axis and is 6 .
Explanation:
Note that equation $16 x^{2}+9 y^{2}=144$ by dividing each term by 144 can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$
As the standard equation is of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, center is $(0,0)$.
In this equation major axis is along $y$-axis and is $2 b=8$, and minor axis is along $x$ axis and is $2 a=6$.
As $a=3$ and $b=4$, focii are given by $\pm \sqrt{4^{2}-3^{2}}=\pm \sqrt{7}$ i.e. $(0, \sqrt{7})$ and $(0,-\sqrt{7})$

Topic: Mathematics, Ellipse, MHT CET, MHT CET 2020 (13 Oct Shift 1)
The coordinates of $\mathrm{P}$ and $\mathrm{Q}$ are $(-3,4)$ and $(2,1)$, respectively. If $P Q$ is extended to $R$ such that $P R=2 Q R$, then what are the coordinates of $\mathrm{R} ?$

Options:
A $\quad(3,7)$
B $(2,4)$
C $\left(-\frac{1}{2}, \frac{5}{2}\right)$
D $(7,-2)$

The correct answer is:
$(7,-2)$

Solution:
As given :
Coordinates of $\mathrm{P}$ and $\mathrm{Q}$ are $(-3,4)$ and $(2,1)$ respectively. Let coordinates of $\mathrm{R}$ be $(\mathrm{x}, \mathrm{y})$. As given : $\mathrm{PR}=2 \mathrm{QR}$
$\Rightarrow \quad \mathrm{PR}-\mathrm{QR}=\mathrm{QR} \Rightarrow \mathrm{PQ}=\mathrm{QR}$
So, Q is the mid point of $P$ and $R$
$\Rightarrow \quad 2=\frac{-3+\mathrm{x}}{2}$ and $1=\frac{4+\mathrm{y}}{2}$
$\Rightarrow \quad \mathrm{x}=7$ and $\mathrm{y}=-2$
$\therefore$ Coordinates of $\mathrm{R} \equiv(7,-2)$

Topic: Mathematics, Straight Lines, NDA, NDA 2007 (Phase 2)
The coordinates of the vertices $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ of a triangle $\mathrm{POR}$ are $(1,-1,1),(3,-2,2)$ and $(0,2,6)$ respectively. If $\angle \mathrm{ROP}=\theta$. then what is $\angle P R Q$ equal to?

Options:
A $30^{\circ}+\theta$
B $45^{\circ}-\theta$
C $60^{\circ}-\theta$
D $90^{\circ}-\theta$

The correct answer is:
$90^{\circ}-\theta$

Solution:
Given, $\mathrm{P}=(1,-1,1)$
$\mathrm{Q}=(3,-2,2)$
$\mathrm{R}=(0,2,6)$
Direction ratios of $\mathrm{PQ}\left(\mathrm{a}_{1}, \mathrm{~b}_{1}, \mathrm{c}_{1}\right)=(3-1,-2+1,2-1)$
$=(2,-1,1)$
Direction ratios of $\mathrm{PR}\left(\mathrm{a}_{2}, \mathrm{~b}_{2}, \mathrm{c}_{2}\right)=(0-1,2+1,6-1)$
$=(-1,3,5)$
Let us calculate, $\mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}$
$=(2)(-1)+(-1)(3)+(1)(5)$
$=-2-3+5$
$=0$
$\therefore P Q \perp$ PRi.e., $\angle \overline{Q P R}=90^{\circ}$
$\operatorname{In} \angle \mathrm{PQR}, \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180^{\circ} \Rightarrow 90^{\circ}+\theta+\angle \mathrm{R}=180^{\circ}$
$\Rightarrow \angle \mathrm{R}=90^{\circ}-\theta$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2018 (Phase 1)
The coordination number of $F e$ and $C o$ in the complex ions, $\left.\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}$ and $\left[\mathrm{Co}(\mathrm{SCN})_{4}\right]^{2-}$ are respectively

Options:
A 3 and 4
B 6 and 8
C 4 and 6
D 6 and 4

The correct answer is:
6 and 4

Solution:
The coordination number of $\mathrm{Fe}$ in $\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3+}$ is 6 and Co in $\left[\mathrm{Co}(\mathrm{SCN})_{4}\right]^{2-}$ is 4 . It is because $\mathrm{C}_{2} \mathrm{O}_{4}$ is a bidentate ligand. So, $3 \times 2=6$
SCN is a monodendate ligand $4 \times 1=4$
So, depending upon the number of the ligands, we can calculate the coordination number of central metal atom.

Topic: Chemistry, Coordination Compounds, KCET, KCET 2020
The core of a transformer is laminated because:

Options:
A ratio of voltage in primary and secondary may be increased
B energy losses due to eddy currents may be minimised
C the weight of the transformer may be reduced
D rusting of the core may be prevented

The correct answer is:
energy losses due to eddy currents may be minimised

Solution:
The core of a transformers is laminated to minimise the energy losses due to eddy currents.

Topic: Physics, Alternating Current, NEET, NEET 2006
The core of any transformer is laminated so as $\mathrm{to}$

Options:
A reduce the energy loss due to eddy currents
B make it light weight
C make it robust \& strong
D increase the secondary voltage

The correct answer is:
reduce the energy loss due to eddy currents

Solution:
The energy loss due to eddy currents is reduced by using laminated core in a transformer.

Topic: Physics, Alternating Current, AIIMS, AIIMS 2010
The cork cambium, cork and secondary cortex are collectively called

Options:
A phellogen
B periderm
C phellem
D phelloderm

The correct answer is:
periderm

Solution:
The periderm is a secondary protective structure and is made up of cork cambium (phellogen), cork (phellem) and secondary cortex (phelloderm).

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2011 (Screening)
The cork cambium, cork and secondary cortex are collectively called

Options:
A phelloderm
B phellogen
C periderm
D phellem.

The correct answer is:
periderm

Solution:
During secondary growth in extra stellar region hypodermis or outer cortical cells, a layer becomes meristematic which is known as cork cambium or phellogen. This phellogen also cuts off cells both on its outer side and inner side. The cells cut off on outer side are phellem or cork cells and cells cut off on inner side are phelloderm or secondary cortex. Phellem, phellogen and phelloderm collectively constitute periderm.

Topic: Biology, Anatomy of Flowering Plants, NEET
The correct option representing the structures forming phellogen, phellem and phelloderm is

Options:
A Phellogen: Cork cambium, Phellem: Secondary cortex and Phelloderm: Cork
B Phellogen: Cork cambium, Phellem: Cork and Phelloderm: Secondary cortex
C Phellogen: Cork, Phellem: Secondary cortex and Phelloderm: Cork cambium
D Phellogen: Secondary cortex, Phellem: Cork and Phelloderm: cork cambium

The correct answer is:
Phellogen: Cork cambium, Phellem: Cork and Phelloderm: Secondary cortex

Solution:
As the stem continues to increase in girth due to the activity of vascular cambium, the outer cortical and epidermis layers get broken and need to be replaced to provide new protective cell layers. Hence, sooner or later, another meristematic tissue called cork cambium or phellogen develops, usually in the cortex region. Phellogen is a couple of layers thick. It is made of narrow, thin-walled and nearly rectangular cells. Phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm.

Topic: Biology, Anatomy of Flowering Plants, NEET
The correct order of boiling point for primary $\left(1^{\circ}\right)$, secondary $\left(2^{\circ}\right)$ and tertiary $\left(3^{\circ}\right)$ alcohols is

Options:
A $1^{\circ}>2^{\circ}>3^{\circ}$
B $3^{\circ}>2^{\circ}>1^{\circ}$
C $2^{\circ}>1^{\circ}>3^{\circ}$
D $2^{\circ}>3^{\circ}>1^{\circ}$

The correct answer is:
$1^{\circ}>2^{\circ}>3^{\circ}$

Solution:
Alcohols with same molecular weight are expected to have almost same boiling point however two more factors other than molecular weight are important, they are namely H-bonding and surface area of molecule. Both these factors are least in $3^{\circ}$ alcohols and maximum in $1^{\circ}$ alcohols. Hence, $3^{\circ}$ alcohols have least boiling point while $1^{\circ}$ alcohols have maximum boiling point.

Topic: Chemistry, Alcohols Phenols and Ethers, MHT CET, MHT CET 2008
The correct order of boiling points of the following isomeric amines is
C4H9NH2,(C2H5)2NH,C2H5N(CH3)2

Options:
A C2H5N(CH3)2>(C2H5)2NH>C4H9NH2
B (C2H5)2NH>C2H5N(CH3)2>C4H9NH2
C (C4H9)NH2>(C2H5)2NH>C2H5N(CH3)2
D (C2H5)2NH>C4H9NH2>C2H5N(CH3)2

The correct answer is:
(C4H9)NH2>(C2H5)2NH>C2H5N(CH3)2

Solution:
Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. Therefore, the order of boiling points of amines is as follows.
(C4H9)NH2>(C2H5)2NH>C2H5N(CH3)2

Topic: Chemistry, Amines, NEET
The correct order of rates of $\mathrm{C}-\mathrm{Br}$ bond ionisation of the following bromides is

Options:
A (i) $>$ (ii) $>$ (iii)
B (ii) $>$ (iii) $>$ (i)
C (i) $>$ (iii) $>$ (ii)
D (ii) $>$ (i) $>$ (iii)

The correct answer is:
(ii) $>$ (i) $>$ (iii)

Solution:
The more stables carbocation, the more is the
rate of $\mathrm{C}-\mathrm{Br}$ ionisation.
Among the given species,
Specie (ii) shows aromatic nature along with resonance, thus gives most stable carbocation.
Specie (i) shows less stable carbocation due to absence of aromatic nature, it only shows resonance effect.
Specie (iii) is least stable because it is secondary carbocation which is stabilised by hyper conjugation. Hence, the correct order is $\mathrm{ii}>\mathrm{i}>\mathrm{iii}$

Topic: Chemistry, General Organic Chemistry, TS EAMCET, TS EAMCET 2019 (06 May Shift 1)
The correct order of the basis strength of methyl substituted amines in aqueous solution is:

Options:
A CH32NH>CH3NH2>CH33N
B CH33N>CH3NH2>CH32NH
C CH33N>CH32NH >CH3NH2
D CH3NH2>CH32NH>CH33N

The correct answer is:
CH32NH>CH3NH2>CH33N

Solution:
Combined effect of the inductive effect of Alkyl group +I, steric hindrance and the salvation of Amines causes the basicity order to be Secondary Amine > Primary Amine > Tertiary Amine

CH32NH>CH3NH2>CH33N

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2019 (All India)
The correct sequence of base strengths in aqueous solution is

Options:
A $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}>\mathrm{CH}_{3} \mathrm{NH}_{2}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}$
B $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}>\mathrm{CH}_{3} \mathrm{NH}_{2}$
C $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}>\mathrm{CH}_{3} \mathrm{NH}_{2}=\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$
D $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N} \mathrm{H}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}>\mathrm{CH}_{3} \mathrm{NH}_{2}$

The correct answer is:
$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}>\mathrm{CH}_{3} \mathrm{NH}_{2}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}$

Solution:
It is expected that the basic nature of amines should be in order tertiary $>$ secondary $>$ primary but the observed order in the case of lower members is found to be as secondary $>$ Primary $>$ tetriary. This anomalous behaviour of tetriary amines is due to steric factors i.e crowding of alkyl groups cover nitrogen atom from all sides thus makes the approach and bonding by a proton relatively difficult which results the maximum steric strain in tetiary amines. The electrons are there but the path is blocked, resulting the reduction in basicity. Thus the correct order is $\mathrm{R}_{2} \mathrm{NH}>\mathrm{R} \mathrm{NH}_{2}>\mathrm{R}_{3} \mathrm{~N}$

Topic: Chemistry, Amines, VITEEE, VITEEE 2007
The correct sequence of organisms during hydrarch succession is:

Options:
A Phytoplankton-submerged plant-submersed free floating plant stage-reed swamp stage-marsh- meadow stage-scrub stages-forest
B Scrub stages-forest-reed swamp stage-marsh-meadow stage-submersed free floating plant stage-submerged plant-phytoplankton
C Submerged plant-marsh-meadow stage-scrub stages-forest-phytoplankton-submersed free floating plant stage-reed swamp stage
D Phytoplankton -red swamp stage-scrub stages-forest-submerged plant-submersed free floating plant stage-marsh-meadow stage

The correct answer is:
Phytoplankton-submerged plant-submersed free floating plant stage-reed swamp stage-marsh- meadow stage-scrub stages-forest

Solution:
Ecological succession is defined as the process of gradual and predictable change in the species composition of an area. Primary succession occurs in those areas where no organism ever existed whereas secondary succession takes place in those areas where organisms once existed but now have lost them due to some reason. In a primary succession of wetland into the forest the first to appear are the phytoplanktons. They are replaced by submerged plants and later submerged and floating plants both appear. Gradually it passes through the reed swamp and marsh meadow stages and lastly scrubs appear before forming it into a forest land.

Topic: Biology, Ecosystem, NEET
The correct sequence of processes represented by A, B, and C are:-

Options:
A Diffusion → Synthesis → Active transport
B Digestion → Excretion → Cellular respiration
C Synthesis → Active transport → Excretion
D Digestion → Diffusion → Cellular respiration

The correct answer is:
Digestion → Diffusion → Cellular respiration

Solution:
The whole concept of nutrition is to obtain energy, food is digested and broken down into smaller molecules and stored. When there is need of energy these molecules are broken down through processes such as glycolysis, lipolysis Krebs cycle, and electron carrier carry these electrons and enter into cellular respiration or electron transfer chain where electron transfer generate ATP. ATP is used to carry out different tasks.

Topic: Biology, Digestion and Absorption, NEET
The correct sequence of spermatogenetic stages leading to the formation of sperms in a mature human testis is

Options:
A spermatocyte-spermatogonia-spermatid - sperms
B spermatogonia-spermatocyte-spermatid - sperms
C spermatid-spermatocyte-spermatogonia -sperms
D spermatogonia-spermatid-spermatocyte -sperms

The correct answer is:
spermatogonia-spermatocyte-spermatid - sperms

Solution:
In spermatogenesis, during growth phase some spermatogonia either due to growth or due to food storage become 2 or 3 times larger of their original size and are known as primary spermatocytes, which undergo meiosis-I and as a result 2 haploid secondary spermatocytes are formed. Further, meiosis-II takes place that results in the formation of 4-spermatids. Then, these round, non-motile and haploid spermatids are transformed into thread-like motile, haploid sperms (four).

Topic: Biology, Human Reproduction, NEET, NEET 2009 (Screening)
The CORRECT sequence of the pathway of air from outside to inside the lungs is

Options:
A Larynx-trachea-alveoli-bronchus-pharynx
B Pharynx-larynx-trachea-bronchus-alveoli
C Trachea-larynx-bronchus-pharynx-alveoli
D Alveoli-larynx-pharynx-bronchus-trachea

The correct answer is:
Pharynx-larynx-trachea-bronchus-alveoli

Solution:
Air enters the air through the nose. The pathway followed is External nares- Nasal chambers- Pharynx- Larynx- Trachea- Primary Bronchi- Secondary bronchi- Tertiary bronchi- Bronchioles- Alveolar ducts- Atria- Alveoli.

Topic: Biology, Breathing and Exchange of Gases, NEET