datavorous/entrance-exam-dataset huggingface dataset:
Phytohormones are

Options:
A hormones regulating growth from seed to adulthood
B growth regulators synthesised by plants and influencing physiological process
C hormones regulating flowering
D hormones regulating secondary growth.

The correct answer is:
growth regulators synthesised by plants and influencing physiological process

Solution:
Growth regulators are organic substances, other than nutrients, which in low concentration regulate growth, differentiation and development by promoting or inhibiting the same. Phytohormones are growth regulators produced naturally in plants and translocated to another region for regulating one or more physiological reactions when present in low concentration. Phytohormone can have a promoting or inhibiting effect on a process.

Topic: Biology, Plant - Growth and Development, AIIMS, AIIMS 2008
Pick out incorrect statement.

Options:
A Colloidal sols are heterogenous
B Colloids carry + ve or â ve charge
C Colloidal sols show Tyndall effect
D The size range of colloidal particles 10−2,000 Å

The correct answer is:
The size range of colloidal particles 10−2,000 Å

Solution:
Here option D is incorrect as the colloidal dimension is 10−1,000 Å (1−100 nm).

Topic: Chemistry, Surface Chemistry, NEET
Pick the wrong answer in the context with rainbow.

Options:
A When the light rays undergo two internal reflections in a water drop, a secondary rainbow is formed.
B The order of colours is reversed in the secondary rainbow.
C An observer can see a rainbow when his front is towards the sun.
D Rainbow is a combined effect of dispersion, refraction and reflection of sunlight.

The correct answer is:
An observer can see a rainbow when his front is towards the sun.

Solution:
An observer can see a rainbow only when his back is towards the sun. The sunlight from his back falls on water droplets suspended in the air in front of her. It is refracted inside it and different colours of white light get separated due to refraction. These different rays strike the inner surface of the droplet and reflected if the angle of incidence is greater than the critical angle(nearly equal to 48∘). This reflected ray is again refracted into the air. Hence, rainbow formation is a combination of dispersion, refraction and reflection of light.

Topic: Physics, Ray Optics, NEET, NEET 2019 (All India)
Planck's constant (h), speed of light in a vacuum ( c ) and Newton's gravitational constant ( G ) are three fundamental constants. Which of the following combinations of these has the dimension of length?

Options:
A hGc32
B hGc52
C hcG
D cG⁡ ⁡ h⁡ 3 2

The correct answer is:
hGc32

Solution:
L ∝hacbGc
L1=M1L2T-1aLT-1bM-1L3T-2c
Solving,
a=12, c=12, b=-32
⇒L=hGc32

Topic: Physics, Units and Dimensions, NEET, NEET 2016 (Phase 2)
Planck's constant (h), speed of light in a vacuum ( c ) and Newton's gravitational constant ( G ) are three fundamental constants. Which of the following combinations of these has the dimension of length?

Options:
A hGc32
B hGc52
C hcG
D cG⁡ ⁡ h⁡ 3 2

The correct answer is:
hGc32

Solution:
L ∝hacbGc
L1=M1L2T-1aLT-1bM-1L3T-2c
Solving,
a=12, c=12, b=-32
⇒L=hGc32

Topic: Physics, Units and Dimensions, NEET
Plants do not get benefit from

Options:
A $\mathrm{N}_2$ in air
B $\mathrm{O}_2$ in air
C $\mathrm{CO}_2$ in air
D $\mathrm{O}_3$ in air.

The correct answer is:
$\mathrm{O}_3$ in air.

Solution:
Ozone is a harmful secondary air pollutant. Ozone, being a strong oxidant destroys chlorenchyma, produces necrosis, hardens rubber, damages textiles, corrodes the surface of marble statues and heritage buildings.

Topic: Biology, Photosynthesis in Higher Plants, AIIMS, AIIMS 2014
Plants having little or no secondary growth are

Options:
A Conifers
B Deciduous angiosperms
C Grasses
D Cycads

The correct answer is:
Grasses

Solution:
The tissues responsible for the increase in girth of plants comprises secondary growth. Such tissues include two types of lateral meristem: vascular cambium and cork cambium. Secondary growth is primarily observed in dicots. Monocots do not undergo secondary growth. In the given options, grasses are monocot and have closed type of vascular bundle, therefore, they do not show secondary growth.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2018
β -pleated sheets of polypeptide chains are present in

Options:
A Secondary structure
B Primary structure
C Tertiary structure
D quaternary structure

The correct answer is:
Secondary structure

Solution:
In the β- sheets structure of proteins, polypeptide chains are present as parallel or anti-parallel to each other.

Topic: Chemistry, Polymers, MHT CET, MHT CET 2019 (02 May Shift 1)
Point $(1,2)$ relative to the circle $x^{2}+y^{2}+4 x-2 y-4=0$ is a/ an

Options:
A exterior point
B interior point, but not centre
C boundary point
D centre

The correct answer is:
exterior point

Solution:
We put the co-ordinates of the given point in the given equation of circle $x^{2}+y^{2}+4 x-2 y-4=0$
$\operatorname{At}(1,2)$
$(1)^{2}+(2)^{2}+4(1)-2(2)-4$
$=1+4+4-4-4=1>0$
$\Rightarrow$ Point $(1,2)$ lies out side the circle i.e, an exterior point.

Topic: Mathematics, Circle, NDA, NDA 2008 (Phase 1)
Point out the false statement.

Options:
A Colloidal sols are homogeneous
B Colloids carry + ve or - ve charges
C Colloids show Tyndall effect
D The size range of colloidal particles is $10-1000 \mathrm{A}$

The correct answer is:
Colloidal sols are homogeneous

Solution:
Option (a) is false as colloidal sols are not homogeneous.
Colloidal sols are heterogeneous mixture of dispersed phase and dispersion medium.

Topic: Chemistry, Surface Chemistry, WBJEE, WBJEE 2016
Polluted water bodies have usually very high abundance of plants like Nostoc and Oscillitoria. Give reasons.

Options:

None

Solution:
Polluted water bodies have high growth of algae due to the presence of nutrient. These nutrients increase the rapid growth of water plants, i.e., algae especially Nostoc and Oscillitoria, etc., and result in colonies. These colonies are generally surrounded by a gelatinous sheath and leads to the formation of blooms in water bodies.

Topic: Biology, Biological Classification
Porous wood has abundance of

Options:
A vessels
B fibres only
C tracheids
D sieve tubes.

The correct answer is:
vessels

Solution:
Porous wood possesses abundant vessels. Due to the presence of vessels, they are called porous woods.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2001
PQRS is a parallelogram, where $\overrightarrow{\mathbf{P Q}}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\mathrm{m} \hat{\mathbf{k}}$,
$\overrightarrow{\mathrm{PS}}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and the area of the parallelogram is $\sqrt{90}$.
What is the value of $m ?$

Options:
A 1
B $-1$
C 2
D $-2$

The correct answer is:
1

Solution:
Let $\overrightarrow{\mathrm{PQ}}=3 \hat{i}+2 \hat{j}-m \hat{k}$ and $\overrightarrow{\mathrm{PS}}=\hat{i}+3 \hat{j}+\hat{k}$
where PQRS is a parallelogram.
Area of parallelogram $=\left\|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -m \\ 1 & 3 & 1\end{array}\right\|$
$=|\hat{i}(2+3 m)-\hat{j}(3+m)+\hat{k}(9-2)|$
$=\sqrt{(2+3 m)^{2}+(3+m)^{2}+7^{2}}$
$\Rightarrow 90=4+9 m^{2}+12 m+9+m^{2}+6 m+49$
$\Rightarrow 10 m^{2}+18 m-28=0$
$\Rightarrow 5 m^{2}+9 m-14=0$
$\Rightarrow 5 m^{2}+14 m-5 m-14=0$
$\Rightarrow m(5 m+14)-1(5 m+14)=0$
$\Rightarrow(5 m+14)(m-1)=0$
$\Rightarrow \mathrm{m}=1$ or $\frac{-14}{5}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 2)
Predict if the reaction between the following is feasible:
(i) \(\mathrm{Fe}^{3+}(a q)\) and \(\mathrm{I}^{-}(a q)\)
(ii) \(\mathrm{Ag}^{+}(a q)\) and \(\mathrm{Cu}(s)\)
(iii) \(\mathrm{Fe}^{3+}(a q)\) and \(\mathrm{Br}^{-}(a q)\)
(iv) \(\mathrm{Ag}(s)\) and \(\mathrm{Fe}^{3+}(a q)\)
(v) \(\mathrm{Br}_2(a q)\) and \(\mathrm{Fe}^{2+}(a q)\).
Given standard electrode potentials:
\(\begin{aligned}
&\mathbf{E}_{1 / 2 \mathrm{I}_2, \mathrm{I}^{-}}^{\mathrm{a}}=+0.541 \mathrm{~V}, \\
&\mathbf{E}_{\mathrm{Cu}^{2+}, \mathrm{Cu}}^{\mathrm{a}}=+0.34 \mathrm{~V}, \\
&\mathbf{E}_{1 / 2^{\mathrm{Br}}, \mathrm{Br}^{-}}^{\mathrm{a}}=+1.09 \mathrm{~V}, \\
&\mathbf{E}_{\mathrm{Ag}^{+}, \mathrm{Ag}}^{\circ}=+0.80 \mathrm{~V}, \mathbf{E}_{\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}}^{\circ}=+0.77 \mathrm{~V} .
\end{aligned}\)

Options:

None

Solution:
The reaction is feasible if the EMF of the cell reaction is positive.
(i) \(\mathrm{Fe}^{3+}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\frac{1}{2} \mathrm{I}_2(g)\)
i.e., \(\mathrm{Pt} / \mathrm{I}_2 / \mathrm{I}^{-}(a q)|| \mathrm{Fe}^{3+}(a q)\left|\mathrm{Fe}^{2+}(a q)\right| \mathrm{Pt}\)
\(\therefore \mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\mathrm{Fe}^{3+}}^{\circ}, \mathrm{Fe}^{2+}-\mathrm{E}_{1 / 2 \mathrm{I}_2, \mathrm{I}^{-}}^{\circ}\)
\(=0.77-0.54=0.23 \mathrm{~V}\) (Feasible).
(ii) \(2 \mathrm{Ag}^{+}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\),
i.e., \(\left.\mathrm{Cu} \mid \mathrm{Cu}^{2+}(a q)\right) \| \mathrm{Ag}^{+}(a q) \mid \mathrm{Ag}\)
\(\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\mathrm{Ag}^{+}, \mathrm{Ag}}^{\circ}-\mathrm{E}_{\mathrm{Cu}^{2+}, \mathrm{Cu}}^{\circ}\)
\(=0.80-0.34=0.46 \mathrm{~V}\) (Feasible).
(iii) \(\mathrm{Fe}^{3+}(a q)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\frac{1}{2} \mathrm{Br}_2(g)\) \(\mathrm{E}_{\text {cell }}^{\circ}=0.77-1.09=-0.32 \mathrm{~V}\) (Not feasible).
(iv) \(\mathrm{Ag}(s)+\mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Fe}^{2+}(a q)\) \(\mathrm{E}_{\text {cell }}^{\circ}=0.77-0.80=-0.03 \mathrm{~V}\) (Not feasible \()\).
(v) \(\frac{1}{2} \mathrm{Br}_2(g)+\mathrm{Fe}^{+2}(a q) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{Fe}^{3+}(a q)\) \(\mathrm{E}_{\text {cell }}^{\circ}=1.09-0.77=0.32 \mathrm{~V}\) (Feasible).

Topic: Chemistry, Electrochemistry
Predict the atomic weight of Eka-aluminium (i) and Eka-silicon(ii)

Options:
A i
B i>ii
C i=ii
D None of above

The correct answer is:
i

Solution:
Correct Option is : 1
i

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Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Predict the atomic weight of Gallium (i) and Germanium (ii)

Options:
A i
B i>ii
C i=ii
D None of above

The correct answer is:
i

Solution:
Correct Option is : 1
i

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NEET with chapter-wise PYQs, revision notes, formula sheets,
custom tests & much more.

Use on iOS or Desktop Download from Google Play

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Predict the melting point of Gallium (i) and Germanium (ii)

Options:
A i
B i>ii
C i=ii
D None of above

The correct answer is:
i

Solution:
Correct Option is : 1
i

Looking for more such questions to practice?

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NEET with chapter-wise PYQs, revision notes, formula sheets,
custom tests & much more.

Use on iOS or Desktop Download from Google Play

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Predict the product,

Options:
A

B

C

D

The correct answer is:

Solution:
Key Idea With nitric acid $\left(\mathrm{NaNO}_2+\mathrm{HCl}\right)$ primary aromatic amines give azo compounds and secondary aromatic amines give nitroso compound

Topic: Chemistry, Amines, NEET, NEET 2009 (Screening)
Preparation of alcohols from alkenes involves the electrophilic attack on alkene carbon atom. Explain its mechanism.

Options:

None

Solution:
Reaction involved is:
This addition reaction takes place in accordance with Markownikoff's rule.
Mechanism Step 1 : Protonation of alkene to form carbocation by electrophilic attack of $\mathrm{H}_3 \mathrm{O}^{+}$
Step 2 : Attack of water molecule to the secondary carbocation.
Step 3: Loss of the hydrogen from the protonated alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give a reason.

Options:
A Secondary and tertiary alcohols are less reactive than primary alcohols.
B Secondary and tertiary alcohols are more prone to oxidation during the reaction.
C Ethers are formed as a result of acid dehydration by SN2 mechanism from primary alcohols.If secondary or tertiary alcohols are used, due to steric hindrance, alkenes are formed and not ethers.
D Secondary and tertiary alcohols form unstable intermediates in this process.

The correct answer is:
Ethers are formed as a result of acid dehydration by SN2 mechanism from primary alcohols.If secondary or tertiary alcohols are used, due to steric hindrance, alkenes are formed and not ethers.

Solution:
Correct Option is : (C)
Ethers are formed as a result of acid dehydration by SN2 mechanism from primary alcohols.If secondary or tertiary alcohols are used, due to steric hindrance, alkenes are formed and not ethers.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Options:

None

Solution:
Acid catalysed dehydration of primary alcohols to ethers occurs by $\mathrm{S}_{\mathrm{N}} 2$ reaction involving nucleophilic attack by the alcohol molecule on the protonated alcohol molecule.
Under these conditions, $2^{\circ}$ and $3^{\circ}$ alcohols, however, give alkenes rather than ethers. The reason being that due to steric hindrance, nucleophilic attack by the alcohol molecule on the protonated alcohol molecule does not occur. Instead protonated $2^{\circ}$ and $3^{\circ}$ alcohols lose a molecule of water to form stable $2^{\circ}$ and $3^{\circ}$ carbocation. These carbocations prefer to lose a proton to form alkenes rather than undergoing nucleophilic attack by alcohol molecules to form ethers.
Similarly, $3^{\circ}$ alcohols give alkenes rather than ethers.

Topic: Chemistry, Alcohols Phenols and Ethers
Primary endosperm nucleus (PEN) is formed by the fusion of

Options:
A 2 polar nuclei +1 synergid cell nucleus
B 1 polar nucleus +1 antipodal cell nucleus +1 synergid cell nucleus
C 2 polar nuclei +1 male gamete nucleus
D 2 antipodal cell nuclei +1 male gamete nucleus.

The correct answer is:
2 polar nuclei +1 male gamete nucleus

Solution:
During the event of double fertilisation in angiosperms, the second male gamete fuses with the two haploid polar nuclei or diploid secondary nucleus of the central cell to form a triploid primary endosperm nucleus (PEN). This second fertilisation is called vegetative fertilisation.

Topic: Biology, Sexual Reproduction in Flowering Plants, AIIMS, AIIMS 2015
Primary exposure to antigenic proteins of pathogen results in the production of all except

Options:
A B cell - antibodies
B T cell - antibodies
C B memory cells
D T memory cells

The correct answer is:
T cell - antibodies

Solution:
The primary and secondary immune responses are carried out with the help of two special types of lymphocytes present in our blood, i.e., B-lymphocytes and T-lymphocytes. The B-lymphocytes produce an army of proteins in response to pathogens into our blood to fight with them. These proteins are
called antibodies. The T-cells themselves do not secrete antibodies but help B cells to produce them.

Topic: Biology, Human Health and Diseases, NEET
Product is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, NEET
Proteins are found to have two different types of secondary structures viz. $\alpha$-helix and $\beta$-pleated sheet structure. $\alpha$-helix structure of protein is stabilised by:

Options:
A
peptide bonds

B
van der Waal's forces

C
hydrogen bonds

D
dipole-dipole interactions

The correct answer is:
hydrogen bonds

Solution:
In $\alpha$-helix structure, - NH group of one amino acid is hydrogen bonded to $\rangle \mathrm{C}=\mathrm{O}$ group of adjacent amino acid, forming a helix.

Topic: Chemistry, Biomolecules
Proteins are needed in diet because

Options:
A all amino acids are not available in body
B during fasting, body utilises proteins
C proteins act as building blocks of our body
D All of the above

The correct answer is:
All of the above

Solution:
Proteins are polymers of amino acids. These are divided into two categories; essential and nonessential amino acids. Essential amino acids cannot be synthesised in the body and therefore must be included in the diet. Proteins are used as structural components of tissues as channels, transporters, regulatory molecules and enzymes. Proteins can also be utilised as energy sources. Normally, this is a secondary function and becomes important only when there is not enough carbohydrate and fat in the body, e.g. during fasting.

Topic: Biology, Biomolecules (B), JIPMER, JIPMER 2018
Provascular tissue gives rise to

Options:
A Secondary phloem only
B Primary xylem only
C Primary xylem and primary phloem
D Primary xylem and secondary phloem

The correct answer is:
Primary xylem and primary phloem

Solution:
Procambium is primary meristematic tissue which undergoes division to form primary vascular tissue such as primary xylem and primary phloem. Secondary xylem and secondary phloem are formed during secondary growth from vascular cambium.

Topic: Biology, Anatomy of Flowering Plants, NEET
Provide appropriate technical term in the space provided.
(a) Blood-filled cavity in arthropods .......
(b) Free-floating form of cnidaria .......
(c) Stinging organ of jelly fishes ........
(d) Lateral appendages in aquatic annelids .........

Options:

None

Solution:
(a) The blood-filled cavity in arthropods containing haemolymph is haemocoel.
(b) A form in cnidarians in which the body is shaped like an umbrella which can float freely in sed water is medusa.
(c) Capsules of specialised cells in cnidarians which act as a paraylysing sting are nematocytes.
(d) The paired unjointed lateral outgrowth in annelids bearing chaetae are parapodia.

Topic: Biology, Animal Kingdom
PT, a tower of height $2^{\mathrm{x}}$ metre, Pbeing the foot, $\mathrm{T}$ being the top of the tower. $A, B$ are points on the same line with $P$. If $\mathrm{AP}=2^{\mathrm{x}+1} \mathrm{~m}, \mathrm{BP}=192 \mathrm{~m}$ and if the angle of elevation of the
tower as seen from $b$ is double the angle of the elevation of the tower as seen from $\mathrm{A}$, then what is the value of $\mathrm{x}$ ?

Options:
A 6
B 7
C 8
D $9 \quad$

The correct answer is:
8

Solution:
Let PT be the tower such that $\mathrm{PT}=2^{\mathrm{x}} \mathrm{m}, \mathrm{AP}=2^{\mathrm{x}+1} \mathrm{~m}$
A and $\mathrm{B}$ are two points joining foot of the tower. T is top of the tower.
Also, given $\mathrm{BP}=192 \mathrm{~m}, \angle \mathrm{TAP}=\theta$ and $\angle \mathrm{TBP}=2 \theta$
In $\Delta \mathrm{PTA}$
$\tan \theta=\frac{\mathrm{PT}}{\mathrm{AP}} \Rightarrow \tan \theta=\frac{2^{\mathrm{x}}}{2^{\mathrm{x}+1}}=\frac{1}{2}$
Now, in $\Delta \mathrm{PTB}$
$\tan 2 \theta=\frac{\mathrm{PT}}{\mathrm{PB}}=\frac{2^{\mathrm{x}}}{192}$
$\Rightarrow \frac{2\left(\frac{1}{2}\right)}{1-\frac{1}{4}}=\frac{2^{\mathrm{x}}}{192} \Rightarrow \frac{1}{3 / 4}=\frac{2^{\mathrm{x}}}{192} \Rightarrow \frac{4}{3} \times 192=2^{\mathrm{x}}$
$2^{\mathrm{x}}=256$
$\Rightarrow 2^{\mathrm{x}}=2^{8} \Rightarrow \mathrm{x}=8$

Topic: Mathematics, Heights and Distances, NDA, NDA 2008 (Phase 1)
Question: Why cannot an electrostatic field line have sudden breaks? Options:

Options:
A Because an electrostatic field is always continuous and cannot have interruptions.
B Because sudden breaks in a field line would violate the principles of electrostatics.
C Because charges in an electrostatic field cannot jump from one point to another abruptly.
D Because it is a fundamental law of nature that field lines must be continuous.

The correct answer is:
Because an electrostatic field is always continuous and cannot have interruptions.

Solution:
Correct Option is : (A)
Because an electrostatic field is always continuous and cannot have interruptions.

Topic: Chemistry, Electrostatics, NEET
Rate of growth of bacteria is proportional to the number of bacteria present at that time. If $x$ is the number of bacteria present at any instant $t$, then which one of the following is correct? (Take proportional constant equal to 1)

Options:
A $x=\log t$
B $x=c e^{t}$
C $e^{x}=t$
D $x=\sqrt{t}$

The correct answer is:
$x=c e^{t}$

Solution:
Rate of growth of bacteria $\propto$ number of bacteria present at that time
$\Rightarrow \quad \frac{d x}{d t} \propto x \Rightarrow \frac{d x}{d t}=x(\because$ Proportional constant $=1)$
$$
\begin{array}{l}
\Rightarrow \quad \int \frac{1}{x} d x=\int d t \\
\Rightarrow \quad \log x=t+\log c \\
\Rightarrow \quad \log x-\log c=t \\
\Rightarrow \quad \log \left(\frac{x}{c}\right)=t \\
\Rightarrow \quad \frac{x}{c}=e^{t} \\
x=c e^{t}
\end{array}
$$

Topic: Mathematics, Differential Equations, NDA, NDA 2008 (Phase 2)
Reaction: 2Fe3+ + 3I- â 2Fe2+ + I3-
The standard reduction potentials in acidic conditions are 0.77 V and 0.54 V respectively for cathodic and anodic reactions. The equilibrium constant for the reaction is approximately.(Given 10 7.79 =6.26Ã 10 7 )

Options:
A 6.26Â ÃÂ 10-7
B 5.33Â ÃÂ 10-4
C 6.26 Ã 107
D 5.33Â ÃÂ 104

The correct answer is:
6.26 Ã 107

Solution:
For the change , 2Fe3++ 3I- â 2Fe2+ + I3-
EcellÂ° = 0.77 - 0.54 = 0.23 V
E cell â = 0.059 2 log K C
0.23 = 0.05912logâ¡Kc
Kc = 6.26 Ã 107

Topic: Chemistry, Electrochemistry, NEET
Read statements (a-c) regarding pollen grains and select the option that has correct statements.
a. The generative cell floats in the cytoplasm of the vegetative cell.
b. Sporopollenin is absent at germ pores.
c. Pollen grain measures about 25-50 micrometer in diameter.

Options:
A a, b and c
B Only b and c
C Only a
D Only a and b

The correct answer is:
a, b and c

Solution:
Pollen grains are generally spherical measuring about 25-50 micrometers in diameter. It has a prominent two-layered wall. The hard outer layer called the exine is made up of sporopollenin, which is one of the most resistant organic material known. It can withstand high temperatures, strong acids and alkali. No enzyme that degrades sporopollenin is so far known. Pollen grain exine has prominent apertures called germ pores where sporopollenin is absent. Pollen grains are well preserved as fossils because of the presence of sporopollenin.
The cytoplasm of the pollen grain is surrounded by a plasma membrane. When the pollen grain is mature, it contains two cells, the vegetative cell and generative cell. The vegetative cell is bigger, has abundant food reserve and a large irregularly shaped nucleus. The generative cell is small and floats in the cytoplasm of the vegetative cell.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Read the below characteristics of organisms belonging to the Plant Kingdom and choose the wrong one.

Options:
A Selaginella is a pteridophytes which is produces two types of spores
B Pinus is a gymnosperms with a branched stem
C Fucus is an algae with haplo-diplontic life cycle
D Funaria is a bryophytes with protonema stage

The correct answer is:
Fucus is an algae with haplo-diplontic life cycle

Solution:
In the majority of the pteridophytes, all the spores are of similar kinds; such plants are called homosporous. Genera like Selaginella and Salvinia which produce two kinds of spores, macro (large) and micro (small) spores, are known as heterosporous. The stems of gymnosperms are unbranched (Cycas) or branched (Pinus, Cedrus). The predominant stage of the life cycle of a moss is the gametophyte which consists of two stages. The first stage is the protonema stage, which develops directly from a spore. It is a creeping, green, branched and frequently filamentous stage. The second stage is the leafy stage, which develops from the secondary protonema as a lateral bud. Common examples of mosses are Funaria, Polytrichum, and Sphagnum. While most algal genera are haplontic, some of them such as Ectocarpus, Polysiphonia, kelps are haplo-diplontic. Fucus, an alga is diplontic.

Topic: Biology, Plant Kingdom, NEET