datavorous/entrance-exam-dataset huggingface dataset:
Sand is to be piled up on a horizontal ground in the form of a regular cone of a fixed base of radius $R$. Coefficient of static friction between the sand layers is $\mu$. Maximum volume of the sand can be piled up in the form of cone without slipping on the ground is

Options:
A $\frac{\mu R^3}{3 \pi}$
B $\frac{\mu R^3}{3}$
C $\frac{\pi R^3}{3 \mu}$
D $\frac{\mu \pi R^3}{3}$

The correct answer is:
$\frac{\mu \pi R^3}{3}$

Solution:
Let $h=$ critical height of sand cone of radius $R$.
Then, for a sand particle to be in equilibrium (it must no slips to the ground)
$$
\begin{aligned}
& \quad f \sin \phi=N \cos \phi \\
& \text { where, } f=\text { friction, } \\
& \\
& \Rightarrow \quad N=\text { normal reaction and } f=\mu N \\
& \Rightarrow \quad \mu N \sin \phi=N \cos \phi \\
& \Rightarrow \quad \tan \phi=\frac{1}{\mu}=\frac{R}{h} \text { (from figure) }
\end{aligned}
$$
So, maximum volume of sand cone that can be formed over level ground is
$$
V_{\max }=\frac{1}{3} \pi R^2 h=\frac{1}{3} \pi R^2(\mu R)=\frac{\mu \pi R^3}{3}
$$

Topic: Physics, Laws of Motion, JEE Main
Saponification of ethyl benzoate with caustic soda as alkali gives

Options:
A Benzyl alcohol and ethanoic acid
B Sodium benzoate and ethanol
C Benzoic acid and sodium ethoxide
D Phenol and ethanoic acid

The correct answer is:
Sodium benzoate and ethanol

Solution:
\(\underset{\text { Ethylbenzoate }}{\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOC}_2 \mathrm{H}_5}+\mathrm{NaOH} \xrightarrow{\Delta} \underset{\text { Sod.benzoate }}{\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}}+\underset{\text { Ethanol }}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}\)

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Saponification of ethyl benzoate with caustic soda as alkali gives

Options:
A Benzyl alcohol and ethanoic acid
B Sodium benzoate and ethanol
C Benzoic acid and sodium ethoxide
D Phenol and ethanoic acid

The correct answer is:
Sodium benzoate and ethanol

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
s-block element which cannot be qualitatively confirmed by the flame test is

Options:
A Li
B Na
C Rb
D Be

The correct answer is:
Be

Solution:
The electrons in beryllium and magnesium are too strongly bound to get excited by flame. Hence, these elements do not impart any colour to the flame. The flame test for Ca, Sr and Ba is helpful in their detection in qualitative analysis and estimation by flame photometry.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Schottky defect defines imperfection in the lattice structure of a

Options:
A solid
B gas
C liquid
D plasma.

The correct answer is:
solid

Solution:
Schottky defect - If some of the lattice points are unoccupied, called vacancies or holes, the lattice becomes defective. Equal number of cations and anions usually absent so that the crystal remains neutral.

Topic: Chemistry, Solid State, AIIMS, AIIMS 2002
Schottky defect generally appears in

Options:
A $\mathrm{NaCl}$
B $\mathrm{KCl}$
C $\mathrm{CsCl}$
D All of these

The correct answer is:
All of these

Solution:
Schottky defects occurs in highly ionic compounds which have high colo-ordination number ex. $\mathrm{NaCl}, \mathrm{KCl}, \mathrm{CsCl}$.

Topic: Chemistry, Solid State, JEE Main
Schottky defect in crystals is observed when

Options:
A Density of crystal is increased
B Unequal number of cations and anions are missing from the lattice
C An ion leaves its normal site and occupies an interstitial site
D Equal number of cations and anions are missing from the lattice

The correct answer is:
Equal number of cations and anions are missing from the lattice

Solution:
Schottky defect is due to missing of equal number of cations and anions.

Topic: Chemistry, Solid State, JEE Main
Schottky defect in crystals is observed when

Options:
A an ion leaves its normal site and occupies an interstitial site.
B unequal number of cations and anions are missing from the crystal lattice.
C equal number of cations and anions are missing from the crystal lattice.
D there is large difference in size of positive and negative ions.

The correct answer is:
equal number of cations and anions are missing from the crystal lattice.

Solution:
The number of missing cations and anions are equal in Schottky defect to maintain electrical neutrality.

Topic: Chemistry, Solid State, COMEDK, COMEDK 2014
Schottky defect is

Options:
A vacancy of ions
B delocalization of ions
C interstitial vacancy of ions
D vacancy of only cations

The correct answer is:
vacancy of ions

Solution:
In Schottky defect, some of the lattice points are unoccupied (vacancies or holes). The number of missing cations and anions is the same, thus crystal remains neutral.

Topic: Chemistry, Solid State, AIIMS, AIIMS 2011
Schottky defects occurs mainly in electrovalent compounds where

Options:
A positive ions and negative ions are of different size
B positive ions and negative ions are of same size
C positive ions are small and negative ions are big
D positive ions are big and negative ions are small

The correct answer is:
positive ions and negative ions are of same size

Solution:
Schottky defect is generally shown by compounds which have ionic nature and small difference in the size of cations \& anions.
In this defect equal no. of cations \& anions is found missing from their lattice sites.

Topic: Chemistry, Solid State, VITEEE, VITEEE 2008
Scrotum in human males contains:

Options:
A Testis and epididymis
B Testis, vasa efferentia and epididymis
C Testis, vasa efferentia, epididymis and a small part of vas deferens
D Testis, vasa efferentia, epididymis, a small part of vas deferens and ejaculatory duct

The correct answer is:
Testis, vasa efferentia, epididymis and a small part of vas deferens

Solution:
Scrotum is a pouch of loose wrinkled skin present outside the abdominal cavity in males. It maintains the temperature of testis 2â2.5oC less than body temperature. It encloses a testis, vasa efferentia, epididymis and a small part of vas deferens

Topic: Biology, Human Reproduction, NEET
Sea is the greatest source of some halogens. Comment.

Options:

None

Solution:
Sea water contains \(\mathrm{Cl}^{-}, \mathrm{Br}^{-}\)and \(\mathrm{I}^{-}\)of \(\mathrm{Na}, \mathrm{K}, \mathrm{Mg}\) and \(\mathrm{Ca}\), but mainly \(\mathrm{NaCl}\). Dried up sea beds contain \(\mathrm{NaCl}\) and carnallite,
\(\mathrm{KCl} . \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O}\). Certain seaweeds contain upto \(0.5 \%\) of iodine as \(\mathrm{NaI}\) and chile saltpetre \(\left(\mathrm{NaNO}_3\right)\) contains upto \(0.2 \%\) of \(\mathrm{NaIO}_3\). Thus, sea is a great source of halogens.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18)
Sea water is 3.5% by mass of common salt and has a density 1.04 g cm-3 at 293 K. Assuming the salt to be sodium chloride, then osmotic pressure of sea water will be (assume complete ionisation of the salt)

Options:
A 25.45 atm
B 11.56 atm
C 29.98 atm
D 30.20 atm

The correct answer is:
29.98 atm

Solution:
Let mass of solution be 100 g
Volume of solution =MassDensity=1001.04
=96.154 mL=0.096 L
wB=3.5 g
MB=58.5 g mol-1 (molar mass of NaCl)
For complete ionisation of NaCl, van't Hoff factor will be '2'.
i = 2
we know,
π=iCRT=inVRT
=2×3.5×0.0821×29358.5×0.096=29.98 atm

Topic: Chemistry, Solutions, JEE Main
Seawater at a frequency f=9×102 Hz, has permittivity ε=80ε0 and resistivity ρ=0.25 Ω m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t)=V0sin(2πft). Then the conduction current density becomes 10x times the displacement current density after time t=1800 s. The value of x is _______

(Given :14πε0=9×109 N m2 C-2

Options:

The correct answer is:
6

Solution:
Jc=Eρ=Vρd

Jd=1 Adqdt=CAdVcdt=ϵddVcdt

⇒V0sin2πftρd=10x×80ε0 d V0(2πf)cos2πft

tan2π×900800=10x×409×109×900

=x=6

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (17 Mar Shift 2)
$\sec ^{-1}\left[\sec \left(-30^{\circ}\right)\right]=$

Options:
A $-60^{\circ}$
B $-30^{\circ}$
C $30^{\circ}$
D $150^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
$\sec ^{-1}\left[\sec \left(-30^{\circ}\right)\right]=\sec ^{-1}\left(\sec 30^{\circ}\right)=30^{\circ}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\int \sec ^{-1} x d x=$

Options:
A $x \sec ^{-1} x+\log \left|x+\sqrt{x^2-1}\right|+c$
B $x \sec ^{-1} x-\log \left|x+\sqrt{x^2-1}\right|+c$
C $x \sec ^{-1} x-\log \left|x+\sqrt{x^2+1}\right|+c$
D $x \sec ^{-1} x+\log \left|x+\sqrt{x^2+1}\right|+c$

The correct answer is:
$x \sec ^{-1} x-\log \left|x+\sqrt{x^2-1}\right|+c$

Solution:
$\begin{aligned} & \text { Let } I=\int \sec ^{-1} x d x=\int \sec ^{-1} x \cdot d x \\ & =\left(x \sec ^{-1} x\right)-\int \frac{x}{x \sqrt{x^2-1}} d x \\ & =\left(x \sec ^{-1} x\right)-\log \left|x+\sqrt{x^2-1}\right|+C\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$(\sec 2 A+1) \sec ^2 A=$

Options:
A $\sec A$
B $2 \sec A$
C $\sec 2 A$
D $2 \sec 2 A$

The correct answer is:
$2 \sec 2 A$

Solution:
$(\sec 2 A+1) \sec ^2 A=\left(\frac{1+\tan ^2 A}{1-\tan ^2 A}+1\right)\left(1+\tan ^2 A\right)$
$=\frac{2\left(1+\tan ^2 A\right)}{1-\tan ^2 A}=2 \sec 2 A \text {. }$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\sec ^2\left(\tan ^{-1} 2\right)+\operatorname{cosec}^2\left(\cot ^{-1} 3\right)=$

Options:
A $5$
B $13$
C $15$
D $6$

The correct answer is:
$15$

Solution:
$\begin{aligned} & \text { Let } \tan ^{-1} 2=\alpha \Rightarrow \tan \alpha=2 \\ & \text { and } \cot ^{-1} 3=\beta \Rightarrow \cot \beta=3 \\ & \sec ^2\left(\tan ^{-1} 2\right)+\operatorname{cosec}^2\left(\cot ^{-1} 3\right) \\ & =\sec ^2 \alpha+\operatorname{cosec}^2 \alpha=1+\tan ^2 \alpha+1+\cot ^2 \alpha \\ & =2+(2)^2+(3)^2=15 .\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sec 50^{\circ}+\tan 50^{\circ}$ is equal to

Options:
A $\tan 20^{\circ}+\tan 50^{\circ}$
B $2 \tan 20^{\circ}+\tan 50^{\circ}$
C $\tan 20^{\circ}+2 \tan 50^{\circ}$
D $2 \tan 20^{\circ}+2 \tan 50^{\circ}$

The correct answer is:
$\tan 20^{\circ}+2 \tan 50^{\circ}$

Solution:
$\sec 50^{\circ}+\tan 50^{\circ}$
$\begin{aligned}
& \Rightarrow \tan \left(70^{\circ}-20^{\circ}\right)=\frac{\tan 70^{\circ}-\tan 20^{\circ}}{1+\tan 70^{\circ} \tan 20^{\circ}} \\
& \Rightarrow \tan 50^{\circ}+\tan 70^{\circ} \tan 20^{\circ} \tan 50^{\circ}=\tan 70^{\circ}-\tan 20^{\circ} \\
& \Rightarrow \tan 50^{\circ}+\tan 50^{\circ}=\tan 70^{\circ}-\tan 20^{\circ} \\
& \left[ \tan 70^{\circ}=\cot 20^{\circ}\right] \\
& \Rightarrow 2 \tan 50^{\circ}+\tan 20^{\circ}=\tan 70^{\circ} \\
& \Rightarrow 2 \tan 50^{\circ}+\tan 20^{\circ}=\tan 50^{\circ}+\sec 50^{\circ} .
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
∫sec8⁡xcosec⁡xdx=

Options:
A sec8â¡x8+c
B sec7â¡x7+c
C sec6â¡x6+c
D sec9â¡x9+c

The correct answer is:
sec7â¡x7+c

Solution:
I= ∫ sec 8 x cosec x dx
Let cosâ¡x=t
- sinx dx=dt
I= â«-dtt8
=171t7+c
=sec7â¡x7+c

Topic: Mathematics, Indefinite Integration, JEE Main
$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ equals to

Options:
A $\log _e(3(2+\sqrt{3}))$
B $\log _e\left(\frac{1+\sqrt{3}}{3}\right)$
C $\log _e\left(\frac{2+\sqrt{3}}{3}\right)$
D $\log _e\left(\frac{2-\sqrt{3}}{3}\right)$

The correct answer is:
$\log _e\left(\frac{2+\sqrt{3}}{3}\right)$

Solution:
$\begin{aligned} & \because \sec h^{-1} x=\log _e\left(\frac{1+\sqrt{1-x^2}}{x}\right) \\ & \text { and cosec } h^{-1} x=\log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right) \\ & \therefore \sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h\left(\frac{3}{4}\right) \\ & =\log _e\left(\frac{1+\sqrt{1-\left(\frac{1}{2}\right)^2}}{\frac{1}{2}}\right)-\log _e\left(\frac{1+\sqrt{1+\frac{9}{16}}}{\frac{3}{4}}\right) \\ & =\log _e\left(\frac{1+\sqrt{\frac{3}{2}}}{\frac{1}{2}}\right)-\log _e\left(\frac{1+\frac{5}{4}}{\frac{3}{4}}\right) \\ & =\log _e(2+\sqrt{3})-\log _e\left(\frac{9}{3}\right) \\ & =\log _e(2+\sqrt{3})-\log _e(3) \\ & =\log _e \frac{(2+\sqrt{3})}{3}\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
$\int \sec x \tan ^3 x d x=$

Options:
A $\frac{1}{3} \sec ^3 x-\sec x+c$
B $\sec ^3 x-\sec x+c$
C $\frac{1}{3} \sec ^3 x+\sec x+c$
D None of these

The correct answer is:
$\frac{1}{3} \sec ^3 x-\sec x+c$

Solution:
$\begin{aligned} & \int \sec x \tan ^3 x d x=\int \sec x\left(\sec ^2 x-1\right) \tan x d x \\ & =\int \sec x \tan x \sec ^2 x d x-\int \sec x \tan x d x \\ & =\frac{\sec ^3 x}{3}-\sec x+c, \quad \text { (Putting } \sec x=t \text { in first part). }\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
Second ionization potential value is lowest for -

Options:
A Sodium
B Magnesium
C Fluorine
D Oxygen

The correct answer is:
Magnesium

Solution:
2nd ionization energy of Mg is lower because it gets the inert gas configuration after losing 2nd electron.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Secondary rainbow in the atmosphere is

Options:
A the result of polarization and dispersion of light
B brighter than the primary rainbow
C due to the phenomenon of double refraction
D formed due to two reflections in the rain drop

The correct answer is:
formed due to two reflections in the rain drop

Solution:
Secondary rainbow in the atmosphere is formed due to two or double reflection in rain drops, suspended in clouds.

Topic: Physics, Ray Optics, JEE Main
Secondary sewage treatment is mainly a

Options:
A mechanical process
B chemical process
C biological process
D physical process

The correct answer is:
biological process

Solution:
In secondary or biological treatment of municipal waste rich in sewage, the organic matter is decomposed with the help of microbes. Decomposition of organic matter occurs by one of the three methods-water hyacinth pod, trickling filter method and activated sludge method. After decomposition the treated water is sterilized through chlorination.

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2011 (Screening)
Secondary sewage treatment is mainly a

Options:
A Physical process
B Mechanical process
C Chemical process
D Biological process

The correct answer is:
Biological process

Solution:
Secondary treatment of effluent is a biological process where the primary effluent is passed into large aeration tanks and allowed for the vigorous growth of useful aerobic microbes like bacteria and fungi to form flocs (mesh-like structure). A constant mechanical agitation with the pumping of air enables the growth of microbes. It consumes the major part of the organic matter and significantly reduces the BOD (biochemical oxygen demand) of the effluent.

Topic: Biology, Microbes in Human Welfare, NEET
Secondary structure of a protein refers to

Options:
A Mainly denatured proteins and structures of prosthetic groups
B Three dimensional structure, specially the bond between amino acid residues that are distant from each other in the polypeptide chain
C Linear sequence of amino acid residues in the polypeptide chain
D Regular folding patterns of continuous portions of the polypeptide chain

The correct answer is:
Regular folding patterns of continuous portions of the polypeptide chain

Solution:
Secondary structure of a protein refers to regular folding patterns of continuous portions of the polypeptide chain

Topic: Chemistry, Biomolecules, JEE Main
Secondary structure of protein refers to

Options:
A mainly denatured proteins and structure of prosthetic groups
B three-dimensional structure, especially the bond between amino acid residues that are distinct from each other in the polypeptide chain
C linear sequence of amino acid residues in the polypetide chain
D regular folding patterns of continous portions of the polypeptide chain

The correct answer is:
regular folding patterns of continous portions of the polypeptide chain

Solution:
No Solution Available

Topic: Chemistry, Biomolecules, VITEEE, VITEEE 2021
Secretin and cholecystokinin are digestive hormones. They are secreted in

Options:
A oesophagus
B ileum
C duodenum
D pyloric stomach

The correct answer is:
duodenum

Solution:
Secretin and Cholecystokinin (CСK) are two main Gastrointestinal (GI) hormones secretes in duodenum of alimentary canal. CCK stimulates gall bladder contraction and thus increases the flow of bile salts into the intestine. Secretin stimulates the release of an alkaline pancreatic fluid that neutralizes stomach acid as it enters the intestine.

Topic: Biology, Chemical Coordination and Integration, JIPMER, JIPMER 2007
Secretin and cholecystokinin are digestive hormones. They are secreted in:

Options:
A pyloric stomach
B duodenum
C ileum
D oesophagus.

The correct answer is:
duodenum

Solution:
Secretin and cholecystokinin (CCK) are two main gastrointestinal (GI) hormones secreted in duodenum of alimentary canal. CCK stimulates gall bladder contraction and thus increases the flow of bile salts into the intestine. While, secretin stimulates the release of an alkaline pancreatic fluid that neutralizes stomach acid as it enters the intestine.
Related Theory

Topic: Biology, Digestion and Absorption, NEET, NEET 2005
Secretions from which one of the following are rich in fructose, calcium and some enzymes?

Options:
A Male accessory glands
B Liver
C Pancreas
D Salivary glands

The correct answer is:
Male accessory glands

Solution:
Seminal plasma is the fluid in which the sperms are bathed. It contains secretions from male accessory glands (ie, seminal vesicles, prostate gland and bulbourethral glands). Seminal fluid contains fructose, citrate, inositol, prostaglandins and several proteins. Prostate secretes a thin, milky fluid that contains calcium, citrate ions, phosphate ions, a clotting enzyme and a profibrinolysin.

Topic: Biology, Human Reproduction, NEET, NEET 2010 (Mains)
See the electrical circuit shown in this figure. Which of the following equations is a correct equation for it?

Options:
A $\in_1-\left(i_1+i_2\right) R+i_1 r_1=0$
B $\in_1-\left(i_1+i_2\right) R-i_1 r_1=0$
C $\in_2-i_1 \mathrm{r}_2-\varepsilon_1-\mathrm{i}_1 \mathrm{r}_1=0$
D $-\epsilon_2-\left(i_1+i_2\right) R+i_2 r_2=0$

The correct answer is:
$\in_1-\left(i_1+i_2\right) R-i_1 r_1=0$

Solution:
For loop (3)
$E_1-\left(i_1+i_2\right) R-i_1 r_1=0$
For loop (4)
$-\mathrm{E}_1+\mathrm{i}_1 \mathbf{r}_1-\mathrm{i}_2 \mathbf{r}_2+\mathrm{E}_2=0$
For loop (1)
$E_2-\left(i_1+i_2\right) R-i_2 r_2=0$

Topic: Physics, Current Electricity, NEET, NEET 2009 (Mains)
See the following chemical reaction:

Cr2O72-+XH++6Fe2+→YCr3++6Fe3++ZH2O

The sum of X, Y and Z is

Options:

The correct answer is:
23

Solution:
The given equation can be balanced using ion-electron method as shown below.

Fe2+→Fe3++e- -oxidation Half reaction

6e-+14H++Cr2O72-→2Cr3++7H2O -Reduction Half reaction

The overall reaction is

6Fe2++14H++Cr2O72-→6Fe3++2Cr3++7H2O

The values X=14,Y=2 and Z = 7

(X+ Y +Z) = 14 + 2 +7

= 23

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Seed and bud dormancy can be induced by

Options:
A IAA
B GA
C ABA
D Ethylene

The correct answer is:
ABA

Solution:
Abscisic acid (ABA) acts as a growth inhibitor and induces dormancy of buds towards the approach of winter. Dormancy of seeds is mainly caused by abscisic acid. Because of its role in inducing dormancy, abscisic acid is also called as dormin. The buds as well as seeds sprout only when the concentration of abscisic acid is overcome by gibberellins.

Topic: Biology, Plant - Growth and Development, NEET
Seed dormancy may be due to

Options:
A immature embryo
B hard seed coat
C presence of germination inhibitors
D all of these

The correct answer is:
all of these

Solution:
The presence of dormancy in a seed is the most important characteristic feature, because of this character, seeds remain viable for many years. Most of the seeds are unable to germinate just after dispersal. They germinate after sometime. The time between the maturation and germination of seed is known as the dormancy period. The state of inhibited germination as a result of internal causes is usually called dormancy. This seed dormancy is of considerable advantage to the plant which helps in adverse environmental conditions. The embryo remains inactive in this period and all the growth processes suspended temporarily. There are three main basic reasons of dormancy of seed, they are:

Impermeability of seed coat

Dormant Embryo

Germination Inhibitors

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Seeds offer several advantages to angiosperms. One of them is that seed formation is more dependable. This is because:

Options:
A Being products of sexual reproduction they generate new genetic combinations leading to variations.
B They have sufficient food reserves to nourish young seedlings.
C They have better strategies for dispersal to new habitats help the species colonize new areas.
D Reproductive processes such as pollination and fertilization are independent of water.

The correct answer is:
Reproductive processes such as pollination and fertilization are independent of water.

Solution:
The fertilized mature ovule containing the embryo surrounded by layer of hard coat is generally called as seed. Seed has hard covering layers called seed coat that protect the embryo from unfavorable conditions. The seed provides lot of advantage to the angiosperm as it shows adaptive strategies for being dispersed to new region. Seed can sustain in harsh environment and maintain dormancy of the embryo. Also, seed formation is dependable as fertilization and pollination process are not dependent on water anymore.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Selaginella and Salvinia are considered to represent a significant step towards evolution of seed habit because

Options:
A female gametophyte is free and gets dispersed like seeds
B female gametophyte lacks archegonia
C magaspores possess endosperm and embryo surrounded by seed coat
D embryo develops in female gametophyte which is retained on parent sporophyte

The correct answer is:
embryo develops in female gametophyte which is retained on parent sporophyte

Solution:
Selaginella and Salvinia are heterosporous as they produce two kinds of spores-mirco (small) spores and macro (large) spores. The microspores and megaspores germinate and give rise to male and female gametophyte, respectively. The female gametophyte in these plants are retained on the parent sporophytes for variable periods. The development of the zygotes into young embryos takes place within the female gametophytes. This event is the precursor to the seed habit considered to an important step in evolution.

Topic: Biology, Plant Kingdom, NEET, NEET 2011 (Mains)
Select an incorrect statement w.r.t. capitulum inflorescence of sunflower

Options:
A Main axis becomes a flat receptacle
B Florets are sessile and many
C Ray florets are bisexual
D Disc florets are actinomorphic

The correct answer is:
Ray florets are bisexual

Solution:
The common inflorescence of Asteraceae is capitulum. The flowers of this inflorescence are called florets. There are two types of florets in heterogamous heads, the central florets are tubular, bisexual and also known as disc florets. The marginal florets are ligulate, pistillate and also known as ray florets. Ray florets are unisexual.

Topic: Biology, Morphology of Flowering Plants, NEET
Select correct adsorption isobars for physisorption and chemisorption respectively
(where xm= extent of adsorption, T = temperature)

Options:
A
B
C
D

The correct answer is:

Solution:
The plots for physical and chemical adsorption are given below respectively
Physical adsorption Chemical adsorption.

Topic: Chemistry, Surface Chemistry, JEE Main
Select correct statement :

Options:
A Geometrical isomers of complexes may differ in dipole moment and visible / UV spectra
B Complexes of the type Ma 3 b 3 can also have facial (fac) and meridional (mer) isomer
C No optical isomer exists for the complex trans - Co en 2 Cl 2 +
D All are correct

The correct answer is:
All are correct

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main