datavorous/entrance-exam-dataset huggingface dataset:
The rupture and fractionation do not usually occur in the water column in vessel/tracheids during the ascent of sap because of

Options:
A lignified thick walls
B cohesion and adhesion
C weak gravitational pull
D transpiration pull

The correct answer is:
cohesion and adhesion

Solution:
The vertical conduction of water from root to aerial parts of plant is called ascent of sap. The water molecules remain joined to each other due to a force of attraction called cohesion force. This attraction is due to the presence of hydrogen bonds between them. The magnitude of this force is very high therefore, the continuous water column in the xylem cannot be broken easily due to the force of gravity or other obstructions offered by internal tissues in the upward movement of water. This adhesive property of water, i.e., attraction between the water molecules and the walls of xylem ensures the continuity of water column in the xylem.
Due to transpiration water evaporates to outer atmosphere through stomata. As a result leaf cells develop low water potential and water from leaf veins moves into leaf cells. The xylem vessel in turn draw water from xylem of main stem. A pressure (pull) is thus, exerted by all the leaves on the stems, which is called transpiration pull. This is strong enough to pull up the column of water to a great height.

Topic: Biology, Transport in Plants, JIPMER, JIPMER 2012
The rupture and fractionation do not usually occur in the water column in vessel/tracheids during the ascent of sap because of

Options:
A Transpiration pull
B Lignified thick walls
C Cohesion and adhesion
D Weak gravitational pull

The correct answer is:
Cohesion and adhesion

Solution:
Continuity of water column is maintained by cohesion and adhesion.

Topic: Biology, Transport in Plants, NEET, NEET 2008 (Mains)
The SN2 reaction involves back-side attack and therefore results in a "Walden Inversion." For which one of the substrates shown would you be able to demonstrate that such back-side attack with "Walden Inversion" has infact occurred?

Options:
A 1-bromopropane
B 2-bromobutane
C 3-bromopentane
D Methyl bromide

The correct answer is:
2-bromobutane

Solution:
To observe inversion, the substitution must occur at a chiral center.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The $\left(\mathrm{S}^{\circ}\right)$ of the following substances are: $\left.\mathrm{CH}_4 \mathrm{~g}\right) 186.2 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{O}_2$ (g) $205.2 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{CO}_2(\mathrm{~g}) 213.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
$\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ 69.9.JK ${ }^{-1} \mathrm{~mol}^{-1}$
The entropy change $\left(\Delta \mathrm{S}^{\circ}\right)$ for the reaction $\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}$ (l) is:

Options:
A
$-312.5 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

B
$-242.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

C
$-108.1 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

D
$-37.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

The correct answer is:
$-242.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

Solution:
$\Delta \mathrm{S}^{\circ}=\mathrm{S}^{\circ}{ }_{\mathrm{CO}_2}+2 \times \mathrm{S}^{\circ}{ }_{\mathrm{H}_2 \mathrm{O}}-\left(\mathrm{S}^{\circ} \mathrm{CH}_4+2 \times \mathrm{S}_{\mathrm{O}_2}^{\circ}\right)$
$=(213.6+2 \times 69.9)-(186.2+2 \times 205.2)$
$=-242.8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2014 (12 Apr Online)
The S-S bond energy is if ΔHfoEt-S-Et=-147 kJ/mol; ΔHfoEt-S-S-Et=-202 kJ/mol and ΔHfoSg=+223 kJ/mol

Options:
A 168 kJ
B 126 kJ
C 278 kJ
D 572 kJ

The correct answer is:
278 kJ

Solution:
C2H5-S-C2H5+Ss→C2H5-S-S-C2H5
ΔHreaction=ΣΔHfproductso-ΣΔHfreactantso
=-202--147=-55 kJ
ΔHreaction=Σ(BE)reaction-Σ(BE)products
-55= Heat of sublimation or enthalpy of atomisation of sulphur -BES-S
-55=223-BES-S
BE(S-S)=223+55
=278 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main
The saliva in the mouth and the tears from the eyes will be included under which category of innate immunity?

Options:
A Physiological barrier
B Physical barrier
C Cytokine barrier
D Cellular barrier

The correct answer is:
Physiological barrier

Solution:
Innate immunity is non-specific type of defence, that is present at the time of birth. This is accomplished by providing different types of barriers to the entry of the foreign agents into our body. Innate immunity consist of four types of barriers. These are (i) Physical barriers : Skin on our body is the main barrier which prevents entry of the micro-organisms. Mucus coating of the epithelium lining the respiratory, gastrointestinal and urogenital tracts also help in trapping microbes entering our body. (ii) Physiological barriers: Acid in the stomach, saliva in the mouth, tears from eyes-all prevent microbial growth. (iii) Cellular barriers: Certain types of leukocytes (WBC) of our body like polymorpho-nuclear leukocytes (PMNL-neutrophils) and monocytes and natural killer (type of lymphocytes) in the blood as well as macrophages in tissues can phagocytose and destroy microbes. (iv) Cytokine barriers : Virus-infected cells secrete proteins called interferons which protect non-infected cells from further viral infection.

Topic: Biology, Human Health and Diseases, NEET
The salt which is least likely to be found in minerals is

Options:
A Chloride
B Sulphate
C Sulphide
D Nitrate

The correct answer is:
Nitrate

Solution:
Correct Option is 4 Nitrate
Nitrate salts are highly soluble in water, making them less likely to be found in minerals.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The salts of ___________ like sodium benzoate are used as food preservatives.

Options:
A acetic acid
B ethanoic acid
C benzoic acid
D none of the above

The correct answer is:
benzoic acid

Solution:
Benzoic acid is prepared by oxidation of toluene. In the form of esters, benzoic acid is used in perfume industry. The salts of benzoic acid like sodium benzoate are used as food preservatives.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is

Options:

The correct answer is:
15

Solution:
m=fu+f

+m=f-10+f ....1

-m=f-20+f .....2

1/2

-1=f-20f-10

10-f=f-20

30=2f

f=15 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius $r=2.5 \times 10^{-5} \mathrm{~m}$. The surface tension of sap is $T=7.28 \times 10^{-2} \mathrm{Nm}^{-1}$ and the angle of contact is $0^{\circ}$. Does surface tension alone account for the supply of water to the top of all trees?

Options:

None

Solution:
As given that, radius $(r)=2.5 \times 10^{-5} \mathrm{~m}$
Surface tension $(\sigma)=7.28 \times 10^{-2} \mathrm{~N} / \mathrm{m}$
$$
g=9.8 \mathrm{~m} / \mathrm{s}^2
$$
Angle of contact $(\theta)=0^{\circ}, \rho=10^3 \mathrm{Kg} / \mathrm{m}^3$
As we know that the maximum height to which sap can rise in trees through capillarity action is
$$
\begin{aligned}
&h=\frac{2 \sigma \cos \theta}{r \rho g} \text { where } \sigma=\text { surface tension, } \\
&\rho=\text { Density, } r=\text { Radius } \\
&=\frac{2 \times 7.28 \times 10^{-2} \times \cos 0^{\circ}}{2.5 \times 10^{-5} \times 1 \times 10^{-3} \times 9.8} \\
&=\frac{104}{175}=0.594 \mathrm{~m} \cong 0.6 \mathrm{~m} \\
&
\end{aligned}
$$
This is the maximum height to which the sap can rise due to surface tension. Hence most of tree are of more than $0.6$ $\mathrm{m}$ height. Since, many trees have heights much more than this, capillary action alone cannot account for the rise of water in all other trees.

Topic: Physics, Mechanical Properties of Fluids
The satements that are TRUE:
(A) Methane leads to both global warming and photochemical smog
(B) Methane is generated from paddy fields
(C) Methane is a stronger global warming gas than CO2
(D) Methane is a part of reducing smog

Choose the most appropriate answer from the options given below:

Options:
A (A),(B),(C) only
B (A) and (B) only
C (B),(C),(D) only
D (A),(B),(D) only

The correct answer is:
(A),(B),(C) only

Solution:
Methane leads to both global warming & photochemical smog. Methane is generated in large amounts from paddy fields. CO2 can be absorbed by photosynthesis, or by formation of acid rain etc., while no such activities are there for methane. Hence methane is stronger global warming gas than CO2, Methane is not a component of reducing smog.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The saturated heterocyclic compound is :

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The schematic symbol of light emitting diode (LED) is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Physics, Semiconductors, JEE Main
The screening effect of $d$ -electrons is

Options:
A Equal to $p$ -electrons
B Much more than $p$ -electrons
C Same as $f$ -electrons
D Less than $p$ -electrons.

The correct answer is:
Less than $p$ -electrons.

Solution:
The screening effect of inner electron of the nucleus causes the decrease in ionization potential, therefore the order of the screening effect is

$$

f
$$

Hence, the screening effect of $d$ -electron is less than $p$ -electron.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The screening effect of d-electons is

Options:
A Equal to that of p-electrons
B More than that of p-electrons
C Same as f-electrons
D Less than p-electrons

The correct answer is:
Same as f-electrons

Solution:
Screening effect of d and f block elements is nearly same

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The scrotum helps in maintaining the low temperature of the testes, necessary for __________________.

Options:
A spermatogenesis.
B spermiogenesis.
C insemination.
D spermiation.

The correct answer is:
spermatogenesis.

Solution:
The testes are situated outside the abdominal cavity within a pouch called scrotum. The scrotum helps in maintaining the low temperature of the testes (2â2.5Â°C lower than the normal internal body temperature) necessary for spermatogenesis.

Topic: Biology, Human Reproduction, NEET
The scrotum helps in maintaining the low temperature of the testes

Options:
A 4-4.5 oC temperature higher than body temperature to facilitate spermatogenesis.
B 3- 3.5 oC temperature higher than body temperature to facilitate spermiogenesis.
C 2- 2.5 oC temperature lower than body temperature to facilitate spermatogenesis.
D 5-5.5 oC temperature lower than body temperature to facilitate spermiogenesis.

The correct answer is:
2- 2.5 oC temperature lower than body temperature to facilitate spermatogenesis.

Solution:
The testes are situated outside the abdominal cavity within a pouch called the scrotum. The scrotum helps in maintaining the low temperature of the testes (2–2.5oC lower than the normal internal body temperature) which is necessary for spermatogenesis.

Topic: Biology, Human Reproduction, NEET
The S.D of 15 items is 6 and if each item is decreased or increased by 1 , then standard deviation will be

Options:
A $5$
B $7$
C $91 / 15$
D $6$

The correct answer is:
$6$

Solution:
If each item of a data is increased or decreased by the same constant, the standard deviation of the data remains unchanged.

Topic: Mathematics, Statistics, JEE Main
The second order differential equation is

Options:
A $y^{\prime 2}+x=y^2$
B $y^{\prime} y^{\prime \prime}+y=\sin x$
C $y^{\prime \prime \prime}+y^{\prime \prime}+y=0$
D $y^{\prime}=y$

The correct answer is:
$y^{\prime} y^{\prime \prime}+y=\sin x$

Solution:
The order of the differential equation is order of highest derivatives in the differential equation.

Topic: Mathematics, Differential Equations, JEE Main
The second stage of hydrosere is occupied by plants like

Options:
A Azolla
B Typha
C Salix
D Vallisneria

The correct answer is:
Vallisneria

Solution:
Hydrosere originating in a pond starts with colonization of some phytoplanktons which forms the pioneer plant community. The stages are
Ist-Bacteria, Blue-green algae, Algae
IInd-Hydrilla Potamogeton, Vallisneria
IIIrd-Nelumbo, Nymphaea, Trapa, Azolla, Wolffia
IVth-Typha, Sagitaria
V-Juncus, Cyperus
VI-Salix, Populus, Almus

Topic: Biology, Ecosystem, NEET, NEET 2012 (Mains)
The second term of an infinite geometric progression is 2 and its sum to infinity is 8. The first term is

Options:

The correct answer is:
4

Solution:
Second term is 2
⇒ar=2
Sum to infinity is 8
⇒a1-r=8
Eliminating a we get,
2/r1-r=8⇒r1-r8=2
⇒8r-8r2=2⇒4r2-4r+1=0
⇒r=12⇒a=212=4.

Topic: Mathematics, Sequences and Series, JEE Main
The secondary structure of a protein refers to

Options:
A
$\alpha$-helical backbone

B
hydrophobic interactions

C
sequence of $\alpha$-amino acids

D
fixed configuration of the polypeptide backbone

The correct answer is:
$\alpha$-helical backbone

Solution:
Secondary structure of proteins involves $\alpha$-helical back bond and $\beta$-sheet structures. These structures are formed as a result of $\mathrm{H}$-bonding between different peptide groups.
Hence, (A) is correct

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2007
The secondary structure of protein is stabilised by:

Options:
A Peptide bond
B glycosidic bond
C Hydrogen bonding
D van der Waals forces

The correct answer is:
Hydrogen bonding

Solution:
The secondary structure of protein includes two type :

(a) α-Helix

(b) β-pleated sheet

In α-Helix structure, the poly peptide chain is coil around due to presence of Intramolecular H-Bonding.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The secondary valency and the number of hydrogen bonded water molecule(s) in CuSO4·5H2O, respectively, are

Options:
A 6 and 4
B 4 and 1
C 6 and 5
D

5 and 1

The correct answer is:
4 and 1

Solution:
In CuSO4.5H2O, Cu is coordinated with 4 water molecules and two more oxygen atoms from Sulphate ion.

Now fifth water molecule is hydrogen-bonded and is deeply embedded in a crystal. It is not coordinated.

Only 4 water molecules are coordinated and the fifth is the only hydrogen bonded.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The secretory phase in the human menstrual cycle is also called

Options:
A luteal phase and lasts for about 6 days
B follicular phase lasting for about 6 days
C luteal phase and lasts for about 13 days
D follicular phase and lasts for about 13 days

The correct answer is:
luteal phase and lasts for about 13 days

Solution:
The menstrual cycle of human is generally divided into three phases namely proliferative (follicular), ovulatory and secretory (luteal) phase. The secretory phase occurs between ovulation and onset of menses, i.e., the phase lasts about 14 days. The luteinizing hormone secreted from anterior lobe of pituitary causes ovulation.

Topic: Biology, Human Reproduction, NEET, NEET 2012 (Mains)
The segment of DNA which acts as the instrument manual for the synthesis of the protein is :

Options:
A Nucleoside
B Nucleotide
C Ribose
D Gene

The correct answer is:
Gene

Solution:
Gene

Topic: Chemistry, Chemistry in Everyday Life, NEET, NEET 2009 (Mains)
The selective reabsorption of glucose and amino acids into the blood takes place mainly in:

Options:
A Proximal convoluted tubule
B Henle's loop
C Distal convoluted tubules
D Collecting duct

The correct answer is:
Proximal convoluted tubule

Solution:
The complete absorption nutrients takes place in proximal convoluted tubule.
PCT is lined by microvilli contain large number of Mitochondria in their cells as the absorption of nutrients is an active process.

Topic: Biology, Excretory Products and their Elimination, NEET
The selective secretion of hydrogen ion, ammonia and potassium into the filtrate takes place mainly in

Options:
A Proximal convoluted tubule
B Henle's loop
C Distal convoluted tubules
D Collecting duct

The correct answer is:
Proximal convoluted tubule

Solution:
Functions of PCT:
1. PCT absorbs almost all essential nutrients.
2. PCT absorbs about 70%â80% electrolytes and water.
3. As PCT selectively secretes H+ ions, ammonia and potassium ions into the filtrate and also absorbs HCO3- from it, it maintains the pH and ionic balance of the body fluids.

Topic: Biology, Excretory Products and their Elimination, NEET
The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10A to 25A in 1s , the change in the energy of the inductance is:

Options:
A 637.5 J
B 740 J
C 437.5 J
D 540 J

The correct answer is:
437.5 J

Solution:
Induced emf / back emf in an inductor coil is given by

ε=Ldidt
⇒25=L×25-101

⇒L=2515=53H
Energy of inductor is given by

E=12Li2

Energy change △E=12LI22-I12
⇒∆E=12×53625-100 J
∆E=437.5 J

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The self inductance of a solenoid of length $L$, area of cross-section $A$ and having $N$ turns is

Options:
A $\frac{\mu_0 N^2 A}{L}$
B $\frac{\mu_0 N A}{L}$
C $\mu_0 N^2 L A$
D $\mu_0 N A L$

The correct answer is:
$\frac{\mu_0 N^2 A}{L}$

Solution:
Given,
length of Solenoid $=2$
Ara of cross - section $=A$
No. of tums $\quad=\mathrm{N}$
We know that total number of tums in solenoid, $N=n \&$ Magnetic field inside tong solenoid, $B=\mu_0 n_i$
So, flux, through $N$ thurs, $\phi_t=N \phi_1 \quad\left\{\theta_1=B A=\mu_0 n i A\right\}$
$\begin{aligned} & =n l \times \mu_0 \cdot n i A \\ & =\mu_0 n^2 l A i\end{aligned}$
$u \sin g_1 \quad \psi_1=2 i$
Inductanes of coil, $\because \mu \cdot n^2 A l i=l i$
$\Rightarrow \quad L \quad=\mu_0 n^2 A l=\mu_0 \frac{N^2}{l^2} A l=\frac{\mu_0 N^2 A}{l}$
Hence, option (1) is correct

Topic: Physics, Electromagnetic Induction, JEE Main
The self inductance of the motor of an electric fan is $10 \mathrm{H}$. In order to impart maximum power at $50 \mathrm{~Hz}$, it should be connected to a capacitance of

Options:
A
$C=4 \mu F$

B
$C=8 \mu F$

C
$C=1 \mu F$

D
$C=2 \mu F$

The correct answer is:
$C=1 \mu F$

Solution:
$f=\frac{1}{2 \pi \sqrt{L C}}$
$C=\frac{1}{4 \times \pi^2 f^2 \times 10}$
$C=1 \mu F$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2005
The self-inductance of two coils of a transformer is $20 \mathrm{mH}$ and $30 \mathrm{mH}$. What is the resistance between them?

Options:
A 0
B $1.5 \Omega$
C $600 \Omega$
D Infinite

The correct answer is:
Infinite

Solution:
Transformers are built in such a way that both primary and secondary coils are well insulated from each other. Hence, resistance between them is considered to be infinite.

Topic: Physics, Electromagnetic Induction, JEE Main
The sense organs like eyes, statocysts, and osphradia are present in

Options:
A Porifera
B Arthropods
C Mollusca
D Chordata

The correct answer is:
Mollusca

Solution:
Head- anterior part of the body, which contains mouth, eyes and tentacles and other sense organs. Foot-is ventral in position usually thick and muscular and generally forms the main locomotory organs. Visceral mass- on dorsal side contains digestive, circulatory and reproductive organ of the body and it forms a hump or dome.
Sense organs include eyes, statocysts and osphradia. Statocyst are organs for balancing. Osphradium is used to test chemical nature of water.

Topic: Biology, Animal Kingdom, NEET
The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance A and substance B from the base line are 3.25 cm.2.08 cm and 1.05 cm. respectively. The ratio of Rf values of A to B is (Answer the nearest integer)

Options:

The correct answer is:
2

Solution:
Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography. In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase.

The relative partition of each component of the mixture is expressed in terms of its retardation factor i.e. Rf value

Rf = distance travel by the solute from the base line distance travel by the solvent from the base line

RfARfH=2.083.251.053.25=2.081.05≃2

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The sequence of an mRNA molecule produced from a DNA template strand with the composition
5'-AGCTACACT-3' is

Options:
A 5'-AGUGUAGCU-3'
B 5'-UCGAUGUGA-3'
C 5'-AGTGTAGCT-3'
D 5'-TCGATGTGA-3'

The correct answer is:
5'-UCGAUGUGA-3'

Solution:
The two stands of DNA are complementary to each other and the bases of one stand are paired with the bases of other strand by means of hydrogen bonding, i.e., A = T and G≡C . But uracil (U) instead of thymine (T) is present in case of RNA. Thus, the sequence of base on m-RNA molecule is
DNA strand 5'-AGCTACACT-3'
m-RNA 5'-UCGAUGUCA-3'

Topic: Chemistry, Biomolecules, JEE Main
The sequence of ionic mobility in aqueous solution is

Options:
A $\mathrm{Na}^{+}>\mathrm{K}^{+}>\mathrm{Rb}^{+}>\mathrm{Cs}^{+}$
B $\mathrm{K}^{+}>\mathrm{Na}^{+}>\mathrm{Rb}^{+}>\mathrm{Cs}^{+}$
C $\mathrm{Cs}^{+}>\mathrm{Rb}^{+}>\mathrm{K}^{+}>\mathrm{Na}^{+}$
D $\mathrm{Rb}^{+}>\mathrm{K}^{+}>\mathrm{Cs}^{+}>\mathrm{Na}^{+}$

The correct answer is:
$\mathrm{Cs}^{+}>\mathrm{Rb}^{+}>\mathrm{K}^{+}>\mathrm{Na}^{+}$

Solution:
More the charge density of ion, more will be the ion-dipole interaction, so more will be hydration of ion and hence less will be the ionic mobility.

Topic: Chemistry, s Block Elements, NEET, NEET 2008 (Mains)
The series combination of two batteries, both of the same emf 10V, but different internal resistance of 20 Ω and 5 Ω, is connected to the parallel combination of two resistors 30 Ω and x Ω. The voltage difference across the battery of internal resistance 20 Ω is zero, the value of x (in Ω) is____

Options:

The correct answer is:
30

Solution:
V1=ϵ1-i.r1
0=10-i×20
i=0.5A
V2=7.530+7.5x
0.5=0.25+7.5x
7.5x=0.25x=7.50.25=30 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The serum has all of the following, except

Options:
A Ions
B Glucose
C Clotting factors
D Amino acids

The correct answer is:
Clotting factors

Solution:
Plasma: Straw coloured viscous fluid forming nearly 55% of the blood.
Plasma = 90%â92% water + 6%â8% Proteins + Minerals + Simple molecules (glucose, amino acids, lipids, etc).
Major proteins in the blood are:
a. Albumins: Mainly concerned with maintaining the osmotic balance.
b. Globulins: Play an important role in the defence mechanisms of the body.
c. Fibrinogen: Required for clotting or coagulation of blood.
Minerals like Na+, Ca++, Mg++, HCO3-, Cl-, etc are present in small quantities.
Clotting factors are also present in the plasma in an inactive form.
Serum = Plasma â clotting factors.

Topic: Biology, Body Fluids and Circulation, NEET
The set A∪B∪C ∩ A∩ B ′ ∩ C ′ ′ ∩C is equal to

Options:
A Aâ©B
B A
C B∩ C ′
D C

The correct answer is:
C

Solution:
A∪B∪C ∩ A∩ B ′ ∩ C ′ ′ ∩C
= A∪B∪C ∩ A ′ ∪B∪C ∩C
= A∩ A ′ ∪ B∪C ∩C
=ϕ∪B∪C∩C=B∪C∩C=C

Topic: Mathematics, Sets and Relations, JEE Main
The set A∪B∪C ∩ A∩ B ′ ∩ C ′ ′ is equal to

Options:
A Aâ©B
B A∩ C ′
C BâªC
D Bâ©C

The correct answer is:
B∪C

Solution:
A∪B∪C ∩ A∩ B ′ ∩ C ′ ′
= A∪B∪C ∩ A ′ ∪B∪C
=A∩A'∪B∪C
=ϕ∪B∪C=(B∪C)

Topic: Mathematics, Sets and Relations, JEE Main