datavorous/entrance-exam-dataset huggingface dataset:
Using the digits $1,2,3,4$ and 5 only once, how many numbers $\begin{array}{ll}\text { greater than } 41000 can be formed?

Options:
A 41
B 48
C 50
D 55

The correct answer is:
48

Solution:
We have to construct 5 digit numbers which are greater than 41000 . So, we have only 2 ways to choose $5^{\text {th }}$ digit.
$\left[\because\right.$ Only 4 or 5 can come at $5^{\text {th }}$ place $]$ Thus, for $4^{\text {th }}$ place we have 4 ways to choose digits. For $3^{\text {rd }}$ place we have 3 ways. For $2^{\text {nd }}$ place we have 2 ways. and for unit place we have only 1 way. Required number of ways $=2 \times 4 \times 3 \times 2 \times 1=48$

Topic: Mathematics, Permutation Combination, NDA, NDA 2011 (Phase 1)
Using the standard electrode potentials given below identify the correct statements from the following.
$$
\begin{aligned}
\mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe} ; E^{\circ} & =-0.44 \mathrm{~V} \\
\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu} ; E^{\circ} & =+0.34 \mathrm{~V} \\
\mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag} ; E^{\circ} & =+0.80 \mathrm{~V}
\end{aligned}
$$
(i) Copper can displace iron from $\mathrm{FeSO}_4$ solution.
(ii) Iron can displace copper from $\mathrm{CuSO}_4$ solution.
(iii) Silver can displace copper from $\mathrm{CuSO}_4$ solution.
(iv) Iron can displace silver from $\mathrm{AgNO}_3$ solution.

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

The correct answer is:
(ii), (iv)

Solution:
$$
\begin{gathered}
\mathrm{Fe}^{2+}+2 e^{-} \longrightarrow \mathrm{Fe} ; E^{\circ}=-0.44 \\
\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu} ; E^{\circ}=+0.34 \\
\mathrm{Ag}+1 e^{-} \longrightarrow \mathrm{Ag} ; E^{\circ}=+0.80
\end{gathered}
$$
If a metal have negative electrode potential than metal easily displace, other metal, i.e. Fe can displace $\mathrm{Cu}$ and $\mathrm{Ag}$.

Topic: Chemistry, Electrochemistry, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 2)
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible Fe3+(aq) + I-(aq) →

Options:
A Fe2+(aq) + I2(s)
B No Reaction
C Fe(s) + I3-(aq)
D FeI3(s)

The correct answer is:
Fe2+(aq) + I2(s)

Solution:
Correct Option is : (A)
Fe2+(aq) + I2(s)

Topic: Chemistry, Electrochemistry, NEET
Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:
Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible: Fe³⁺(aq) and I⁻(aq)

Options:
A Fe³⁺ will reduce I⁻
B I⁻ will reduce Fe³⁺
C The reaction is feasible
D Both will not react

The correct answer is:
The reaction is feasible

Solution:
Correct Option is : (C)
The reaction is feasible

Topic: Chemistry, Redox Reactions, NEET
Using the standard reduction potentials of the electrodes $\mathrm{Li}, \mathrm{Zn}, \mathrm{Mg}, \mathrm{Ni}$ as $-3.05,-0.76$, $-2.36,-0.25 \mathrm{~V}$ respectively, identify the correct statement.

Options:
A $\mathrm{Mg}$, displaces $\mathrm{Zn}$ from its solution.
B $\mathrm{Ni}$, reduces $\mathrm{Zn}^{2+}$ to $\mathrm{Zn}$
C $\mathrm{Mg}$, reduces $\mathrm{Li}^{+}$to $\mathrm{Li}$
D $\mathrm{Zn}$, reduces $\mathrm{Mg}^{2+}$ to $\mathrm{Mg}$

The correct answer is:
$\mathrm{Mg}$, displaces $\mathrm{Zn}$ from its solution.

Solution:
No solution. Refer to answer key.

Topic: Chemistry, s Block Elements, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 1)
Value of standard electrode potential for the oxidation of Cl–ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl–oxidized at anode instead of water?

Options:
A The concentration of Cl–ions in the solution is higher than that of water molecules.
B Under the conditions of electrolysis of aqueous sodium chloride, oxidation of water at anode requires overpotential hence Cl–is oxidized instead of water
C The standard electrode potential alone determines the species oxidized at the anode.
D The concentration of Na⁺ ions in the solution promotes the oxidation of Cl–ions.

The correct answer is:
Under the conditions of electrolysis of aqueous sodium chloride, oxidation of water at anode requires overpotential hence Cl–is oxidized instead of water

Solution:
Correct Option is : (B)
Under the conditions of electrolysis of aqueous sodium chloride, oxidation of water at anode requires overpotential hence Cl–is oxidized instead of water

Topic: Chemistry, Electrochemistry, NEET
Value of standard electrode potential for the oxidation of $\mathrm{Cl}^{-}$ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is $\mathrm{Cl}^{-}$ oxidised at anode instead of water?

Options:

None

Solution:
Under the condition of electrolysis of aqueous sodium chloride, oxidation of water at anode requires over potential. So, $\mathrm{Cl}^{-}$is oxidized at anode instead of water. Possible oxidation half cell reactions occurring at anode are
$$
\begin{aligned}
&\mathrm{Cl}^{-}(a q) \longrightarrow \frac{1}{2} \mathrm{Cl}_2(g)+e^{-} ; \mathrm{E}_{\text {Cell }}^{\circ}=1.36 \mathrm{~V} \\
&2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q)+4 e^{-} ; \mathrm{E}_{\text {Cell }}^{\circ}=1.23 \mathrm{~V}
\end{aligned}
$$
Species having lower $\mathrm{E}_{\text {Cell }}^{\circ}$ undergo oxidation first than the higher value but oxidation of $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$ is kinetically so slow that it needs some overvoltage.

Topic: Chemistry, Electrochemistry
Variance is always independent of the change of

Options:
A origin but not scale
B scale only
C both origin and scale
D None of the above

The correct answer is:
origin but not scale

Solution:
Variance is always independent of change of the origin but not scale

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 1)
Vascular bundles in a dicot stem are

Options:
A Closed, collateral, endarch
B Open, collateral, endarch
C Closed, collateral, exarch
D Open, collateral, exarch

The correct answer is:
Open, collateral, endarch

Solution:
In dicot stem, vascular bundles are conjoint, collateral in which xylem is present towards inner side and phloem towards the center. Xylem is endarch as protoxylem is present towards center and metaxylem towards the periphery.

Topic: Biology, Anatomy of Flowering Plants, NEET
Vascular tissues in flowering plants develop from

Options:
A phellogen
B plerome
C periblem
D dermatogen

The correct answer is:
plerome

Solution:
Histogen theory for shoot apical meristem has been proposed by Hanstein (1870). It advocates that there are three distinct meristematic zones (layers) called dermatogen periblem and plerome. The dermatogen is the outermost histogen giving rise to epidermis, periblem is the middle one producing the cortex and plerome is the innermost resulting in central cylinder (ie, vascular tissue).
Cork cambium (phellogen) is the secondary lateral meristem found in outer cortical region. Its cells divide periclinally cutting off cells towards the outside (forming cork or phellem) and inside (forming secondary cortex or phelloderm).

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2008 (Screening)
Vascular tissues in flowering plants develop from

Options:
A phellogen
B plerome
C periblem
D dermatogen

The correct answer is:
plerome

Solution:
Histogen theory for shoot apical meristem has been proposed by Hanstein (1870). It advocates that there are three distinct meristematic zones (layers) called dermatogen periblem and plerome. The dermatogen is the outermost histogen giving rise to epidermis, periblem is the middle on producing the cortex and plerome is the innermost resulting in central cylinder (i.e., vascular tissue).
Cork cambium (phellogen) is the secondary lateral meristem found in outer cortical region. Its cells divide periclinally cutting off cells towards the outside (forming cork or phellem) and inside (forming secondary cortex or phelloderm).

Topic: Biology, Anatomy of Flowering Plants, JIPMER, JIPMER 2012
Vertical distribution of different species occupying different levels in a biotic community is known as:

Options:
A Divergence
B Stratification
C Zonation
D Pyramid

The correct answer is:
Stratification

Solution:
By the definition of stratification, it is the distribution of a community in different levels by various socioeconomic means. The different growth form (trees, shrubs, under shrubs, herbs) determines the structure of a plant community. Stratification is based on mode of arrangement of various growth forms.

In the deep lake, zonation or stratification may be according to the need of light. There are three types of zones differentiated is a deep lake -

Littoral Zone

Limnetic zone

Profundal zone

The clear stratification (vertical arrangement) in various growths from plants according to the need for light is any dense forest.

Surface dwellers → Herbs → under shrubs → Shrubs → Trees

Topic: Biology, Ecosystem, NEET, NEET 2015 (Phase 1)
Vessels in xylem are found in

Options:
A all angiosperms and gymnosperms
B most of angiosperms and few gymnosperms
C few angiosperms, all gymnosperms and some pteridophytes
D all pteridophytes.

The correct answer is:
most of angiosperms and few gymnosperms

Solution:
Vessels occur in the wood (both primary and secondary xylem) of almost all angiosperms. But in some angiosperms like Dracaena, Yucca etc., they are absent. Vessels are also absent in the wood of most gymnosperms (except Gnetales) and pteridophytes.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2018 (27 May)
Vexillary aestivation is seen in

Options:
A China rose
B Bean
C Gulmohur
D Cotton

The correct answer is:
Bean

Solution:
In pea and bean flowers, there are five petals, the largest (standard) overlaps the two lateral petals (wings) which in turn overlap the two smallest anterior petals (keel); this type of aestivation is known as vexillary or papilionaceous.

Topic: Biology, Morphology of Flowering Plants, NEET
Viruses that infect bacteria multiply and cause their lysis, are called:

Options:
A lysozymes
B lipolytic
C lytic
D lysogenic.

The correct answer is:
lytic

Solution:
Bacteriophage is a kind of virus that can infect bacteria and start replicating inside the bacterial cell because the lysis represents the lytic cycle. Lysogenic cycle allows the host cell to continue to survive and reproduce, the virus is reproduced in all of the cell's offspring. Any hormone (such as epinephrine, glucagon, and cortisol) that promotes release of free fatty acids from fat tissue is known as lypolytic. Lysozyme is a naturally occurring enzyme found in bodily secretions such as tears, saliva, and milk and act as an antimicrobial agent
Related Theory
Human lysozyme act as physiological barrier in immune system against some bacteria. It is found in abundance in tears, saliva and mucous.

Topic: Biology, Biological Classification, NEET, NEET 2004
Voltage in the secondary coil of a transformer does not depend upon

Options:
A frequency of the source
B voltage in the primary coil
C ratio of number of turns in the two coils
D Both (b) and (c)

The correct answer is:
frequency of the source

Solution:
The voltage in primary and secondary coil is related as:
$\Rightarrow \frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\text { no. of turns in secondary coil }\left(\mathrm{N}_{\mathrm{s}}\right)}{\text { no. of turns in primary coil }\left(\mathrm{N}_{\mathrm{p}}\right)}$
Clearly, voltage in secondary coil depends on voltage in primary coil and number of turns in primary and secondary coil. It is independent of frequency of the source.
Hence, option 1 is correct.

Topic: Physics, Alternating Current, JIPMER, JIPMER 2011
Water in plants is transported by :

Options:
A cambium
B phloem
C xylem
D epidermis

The correct answer is:
xylem

Solution:
The water absorbed by root hairs is transported to other parts of the plant through xylem. Phloem transfers food prepared in leaves to the other parts. Cambium is responsible for secondary growth while epidermis is the outermost protective layer of cells.

Topic: Biology, Transport in Plants, JIPMER, JIPMER 2006
Water potential is equal to

Options:
A ψs + O.P.
B ψs + T.P.
C ψp + ψw
D Ïs + Ïp

The correct answer is:
ψs + ψp

Solution:
Water potential is defined as the difference between the chemical potential of water at any point in a system and that of pure water under standard conditions. The unit of measurement of water potential is the pascal, Pa (1 megapascal, Mpa = 10 bars). It is represented by greek letter PSi (ψ). Water potential has two components solute potential which is always negative and pressure potential which is usually positive. Therefore, water potential ψw is expressed as ψw = ψs + ψp.

Topic: Biology, Transport in Plants, NEET
Water potential of pure water at standard temperature is equal to

Options:
A
10

B
20

C
zero

D
None of these

The correct answer is:
zero

Solution:
Water potential of pure water at standard temperature is zero. It is the highest value of water potential.

Topic: Biology, Transport in Plants
$ What is $\sin A \cos A \tan A+$ cos $A$ sin AcotA equal to?

Options:
A $\sin \mathrm{A}$
B $\cos \mathrm{A}$
C $\tan \mathrm{A}$
D 1

The correct answer is:
1

Solution:
Given expression
$\begin{aligned}=& \sin A \cos A \cdot \frac{\sin A}{\cos A}+\cos A \sin A \frac{\cos A}{\sin A} \\=& \sin ^{2} A+\cos ^{2} A=1 \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
What $\int_{0}^{2} \frac{\mathrm{dx}}{\mathrm{x}^{2}+1}$ equal to ?

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{4}$
C $\frac{\pi}{8}$
D None of the above

The correct answer is:
$\frac{\pi}{8}$

Solution:
Let $\mathrm{I}=\int_{0}^{2} \frac{\mathrm{dx}}{\mathrm{x}^{2}+4}=\int_{0}^{2} \frac{\mathrm{dx}}{\mathrm{x}^{2}+(2)^{2}}$
$=\frac{1}{2}\left[\tan ^{-1} \frac{\mathrm{x}}{2}\right]_{0}^{-2}=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right]$
$=\frac{1}{2}\left[\frac{\pi}{4}-0\right]=\frac{\pi}{8}$

Topic: Mathematics, Definite Integration, NDA, NDA 2012 (Phase 2)
What advantage do the fuel cells have over primary and secondary batteries?

Options:
A Fuel cells have higher energy density.
B Fuel cells produce fewer environmental pollutants.
C Fuel cells are smaller and more portable.
D Primary batteries contain a limited amount of reactants and are discharged when the reactants have been consumed. Secondary batteries can be recharged but take a long time to recharge. Fuel cell runs continuously as long as the reactants are supplied to it and products are removed continuously.

The correct answer is:
Primary batteries contain a limited amount of reactants and are discharged when the reactants have been consumed. Secondary batteries can be recharged but take a long time to recharge. Fuel cell runs continuously as long as the reactants are supplied to it and products are removed continuously.

Solution:
Correct Option is : (D)
Primary batteries contain a limited amount of reactants and are discharged when the reactants have been consumed. Secondary batteries can be recharged but take a long time to recharge. Fuel cell runs continuously as long as the reactants are supplied to it and products are removed continuously.

Topic: Chemistry, Electrochemistry, NEET
What advantage do the fuel cells have over primary and secondary batteries?

Options:

None

Solution:
Reaction occuring in Daniel cell is as following:
$$
\mathrm{Zn}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}
$$
Nernst equation for Daniel cell is given as following:
$$
\begin{aligned}
&\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \left[\frac{\mathrm{Zn}^{2+}}{\mathrm{Cu}^{2+}}\right] \\
&\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}+\frac{0.0591}{2} \log \left[\frac{\mathrm{Cu}^{2+}}{\mathrm{Zn}^{2+}}\right]
\end{aligned}
$$
According to this equation $E_{\text {cell }}$ is directly dependent on concentration of $\mathrm{Cu}^{2+}$ and inversely dependent upon concentration of $\mathrm{Zn}^{2+}$ ions.
Therefore, $E_{\text {cell }}$ decreases when concentration of $\mathrm{Zn}^{2+}$ ions is increased.

Topic: Chemistry, Electrochemistry
What angle does the line segment joining $(5,2)$ and $(6,-15)$ subtend at $(0,0) ?

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{2}$
D $\frac{3 \pi}{4}$

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

Solution:
Slope of line joining $(5,2)$ and $(0,0)$
$\tan \mathrm{A}=\mathrm{m}_{1}=\frac{2-0}{5-0}=\frac{2}{5}$
Slope of line joining $(6,-15)$ and $(0,0)$
$\tan \mathrm{B}=\mathrm{m}_{2}=\frac{-15}{6}=\frac{-5}{2}$
Now, $\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{2}{5}\left(-\frac{5}{2}\right)=-1$
Hence, both lines are perpendicular. and than angle
between them $=\frac{\pi}{2}$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 2)
What are animals that feed on herbivores called?

Options:
A Primary carnivores
B Secondary carnivores
C Tertiary carnivores
D Herbivores

The correct answer is:
Primary carnivores

Solution:
The consumers that feed on herbivores are carnivores, or more correctly, primary carnivores.

Topic: Biology, Ecosystem, NEET
What are coordinates of the point equidistant from the points $(a, 0,0),(0, a, 0),(0,0, a)$ and $(0,0,0) ?$

Options:
A $\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{a}}{3}, \frac{\mathrm{a}}{3}\right)$
B $\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right)$
C $(a, a, a)$
D $(2 a, 2 a, 2 a) \quad$

The correct answer is:
$\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right)$

Solution:
Let the point $\mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is equidistant from the points $\begin{aligned} & \mathrm{B}(\mathrm{a}, 0,0), \mathrm{C}(0, \mathrm{a}, 0), \mathrm{D}(0,0, \mathrm{a}) \text { and } \mathrm{E}(0,0,0) . \\ & \text { Hence, }(\mathrm{x}-\mathrm{a})^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=\mathrm{x}^{2}+(\mathrm{y}-\mathrm{a})^{2}+\mathrm{z}^{2} \\ &=\mathrm{x}^{2}+\mathrm{y}^{2}+(\mathrm{z}-\mathrm{a})^{2} \\ &=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2} \\ \Rightarrow &(\mathrm{x}-\mathrm{a})^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=\mathrm{x}^{2}+(\mathrm{y}-\mathrm{a})^{2}+\mathrm{z}^{2} \\ \Rightarrow & \mathrm{x}^{2}+\mathrm{a}^{2}-2 \mathrm{ax}+\mathrm{y}^{2}+\mathrm{z}^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{a}^{2}-2 \mathrm{ay}+\mathrm{z}^{2} \\ \Rightarrow &-2 \mathrm{ax}=-2 \mathrm{ay} \\ \Rightarrow & \mathrm{ax}=\mathrm{ay} \Rightarrow \mathrm{x}=\mathrm{y} \\ & \text { Similarly, ay }=\mathrm{az} \Rightarrow \mathrm{y}=\mathrm{z} \end{aligned}$
$\Rightarrow x=y=z$
$(x-a)^{2}+x^{2}+x^{2}=x^{2}+x^{2}+x^{2}$
$\Rightarrow x^{2}+a^{2}-2 a x+x^{2}+x^{2}=3 x^{2}$
$\Rightarrow \mathrm{a}^{2}=2 \mathrm{ax} \Rightarrow \mathrm{x}=\frac{\mathrm{a}}{2}$
$\therefore$ Point is $\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}, \frac{\mathrm{a}}{2}\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 1)
What are sacred groves and their role in biodiversity conservation?

Options:

None

Solution:
Sacred tracts which are held in high esteem by local communities are sacred groves. The sacred groves are dedicated to local deities or ancestral spirits. They are protected by local communities through social traditions and taboos that incorporate spiritual as well as ecological values.
A native vegetation in a natural or near natural state is represented by sacred groves and are thus, rich in biodiversity and harbour many rare species of plants and animals.
sacred groves as such are found in Meghalaya (Khasi and Jaintia hills), Aravalli hills of Rajasthan, Western Ghats, regions of Karnataka and Maharashtra and Madhya Pradesh (Bastar. Chanda and Sarguia region).

Topic: Biology, Biodiversity and its Conservation
What are secondary medullary rays?

Options:
A Narrow band of parenchyma
B Arranged transversely
C Dead and lignified
D Formed by primary cambium

The correct answer is:
Narrow band of parenchyma

Solution:
During secondary growth, at some places, the cambium forms a narrow band of parenchyma, which passes through the secondary xylem and the secondary phloem in the radial directions. These are the secondary medullary rays.

Topic: Biology, Anatomy of Flowering Plants, NEET
What are some examples of secondary metabolites that are useful to human welfare?

Options:
A Amino acids
B Sugars
C Rubber, drugs, spices, scents and pigments
D Alkaloids

The correct answer is:
Rubber, drugs, spices, scents and pigments

Solution:
Many secondary metabolites are useful to human welfare, including rubber, drugs, spices, scents and pigments.

Topic: Biology, Biomolecules (B), NEET
What are the actions of estrogens in female individuals?

Options:
A Stimulation of male secondary sex organs
B Development of male secondary sex characters
C Stimulation of growth and activities of female secondary sex organs
D Stimulation of alveoli formation

The correct answer is:
Stimulation of growth and activities of female secondary sex organs

Solution:
Estrogens produce wide ranging actions such as stimulation of growth and activities of female secondary sex organs, development of growing ovarian follicles, appearance of female secondary sex characters, mammary gland development, and regulation of female sexual behaviour.

Topic: Biology, Chemical Coordination and Integration, NEET
What are the changes in the oogonia during the transition of a primary follicle to Graafian follicle?

Options:

None

Solution:
After the repeated division of the germinal epithelial cells many diploid oogonia are formed. The oogonia grow further to form primary oocytes, which then gets surrounded by a layer of granulosa cells called the primary follicles.
When more layers of granulosa cells surrounds primary follicles. It forms secondary follicles. The secondary follicle soon transforms into a tertiary follicle which is characterised by a fluid filled cavity called antrum.
In primary oocyte within the tertiary follicle meiotic division occurs to become a secondary oocyte and a first polar body (haploid). The tertiary follicle further changes into the mature follicle or Graafian follicle that ruptures and releases the secondary oocyte (ovum) from the ovary by the process called ovulation.

Topic: Biology, Human Reproduction
What are the characteristics of the leafy stage in mosses?

Options:
A Upright, slender axes bearing spirally arranged leaves
B Creeping, green, branched and filamentous
C Multicellular and branched rhizoids
D All of the above

The correct answer is:
Upright, slender axes bearing spirally arranged leaves

Solution:
The leafy stage in mosses develops from the secondary protonema as a lateral bud. They consist of upright, slender axes bearing spirally arranged leaves.

Topic: Biology, Plant Kingdom, NEET
What are the coefficients 'a' and 'b' in the equation for calculating the enthalpy change using standard enthalpies of formation?

Options:
A 'a' represents the coefficients of the products, and 'b' represents the coefficients of the reactants.
B 'a' represents the coefficients of the reactants, and 'b' represents the coefficients of the products.
C 'a' represents the coefficients of the products, and 'b' represents the coefficients of the products.
D 'a' represents the coefficients of the reactants, and 'b' represents the coefficients of the reactants.

The correct answer is:
'a' represents the coefficients of the products, and 'b' represents the coefficients of the reactants.

Solution:
In the equation for calculating the enthalpy change using standard enthalpies of formation, 'a' represents the coefficients of the products, and 'b' represents the coefficients of the reactants in the balanced chemical equation.

Topic: Chemistry, Thermodynamics (C), NEET