datavorous/entrance-exam-dataset huggingface dataset:
Match the following:

(i) Foam	(a) smoke

(ii) Gel	(b) cell fluid

(iii) Aerosol	(c) jellies

(iv) Emulsion	(d) rubber

(e) froth

(f) milk

Options:
A (i) – (b), (ii) – (c), (iii) – (e), (iv) – (d)
B (i) – (d), (ii) – (b), (iii) – (a), (iv) – (e)
C (i) – (e), (ii) – (c), (iii) – (a), (iv) – (f)
D (i) – (d), (ii) – (b), (iii) – (e), (iv) – (f)

The correct answer is:
(i) – (e), (ii) – (c), (iii) – (a), (iv) – (f)

Solution:
Dispersed Phase

Dispersion medium

Type of colloid	Examples

Solid	Solid	Solid sol	Some coloured glasses and gem stones

Solid	Liquid	Sol	Paints, cell fluids

Solid	Gas	Aerosol	Smoke, dust

Liquid	Solid	Gel	Cheese, butter, jellies

Liquid	Liquid	Emulsion	Milk, hai cream

Liquid	Gas	Aerosol	Fog, mist cloud, insecticide sprays

Gas	Solid	Solid sol	Pumice stone, foam rubber

Gas	Liquid	Foam	Froth, whipped cream, soap lather

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Match the following :

i	Riboflavin	a	Beriberi

ii	Thiamine	b	Scurvy

iii	Pyridoxine	c	Cheilosis

iv	Ascorbic acid	d	Convulsions

Options:
A i-a,ii-d,iii-c,iv-b
B i-c,ii-d,iii-a,iv-b
C i-c,ii-a,iii-d,iv-b
D i-d,ii-b,iii-a,iv-c

The correct answer is:
i-c,ii-a,iii-d,iv-b

Solution:
Vitamin	Deficiency Disease

Vitamin B 1	Beri Beri

Vitamin B 2	Cheilosis

Vitamin B 6	convulsions

Vitamin C	Scurvy

Topic: Chemistry, Biomolecules, JEE Main
Match the following :

i	Riboflavin	a	Beriberi

ii	Thiamine	b	Scurvy

iii	Pyridoxine	c	Cheilosis

iv	Ascorbic acid	d	Convulsions

Options:
A i-a,ii-d,iii-c,iv-b
B i-c,ii-d,iii-a,iv-b
C i-c,ii-a,iii-d,iv-b
D i-d,ii-b,iii-a,iv-c

The correct answer is:
i-c,ii-a,iii-d,iv-b

Solution:
Vitamin	Deficiency Disease

Vitamin B 1	Beri Beri

Vitamin B 2	Cheilosis

Vitamin B 6	convulsions

Vitamin C	Scurvy

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Match the following items in column I with the corresponding items in column II.

Options:
A (i) $\rightarrow$ (B); (ii) $\rightarrow$ (C); (iii) $\rightarrow$ (A); (iv) $\rightarrow$ (D)
B (i) $\rightarrow(\mathrm{C}) ;$ (ii) $\rightarrow(\mathrm{B}) ;$ (iii) $\rightarrow(\mathrm{D}) ;$ (iv) $\rightarrow(\mathrm{A})$
C (i) $\rightarrow$ (D); (ii) $\rightarrow$ (A); (iii) $\rightarrow$ (B); (iv) $\rightarrow$ (C)
D (i) $\rightarrow$ (C); (ii) $\rightarrow$ (D); (iii) $\rightarrow$ (B); (iv) $\rightarrow$ (A)

The correct answer is:
(i) $\rightarrow$ (C); (ii) $\rightarrow$ (D); (iii) $\rightarrow$ (B); (iv) $\rightarrow$ (A)

Solution:
(i) $\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O} \rightarrow$ Solvay process

(ii) $\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2} \rightarrow$ Temporary hardness

(iii) $\mathrm{NaOH} \rightarrow$ Castner-Kellner process

(iv) $\mathrm{Ca}_{3} \mathrm{Al}_{2} \mathrm{O}_{6} \rightarrow$ Portland cement

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Match the following List-I with List-II.
\(\begin{array}{cccc}
\hline & \text { List-I } & & \text { List-II } \\
\hline \text { A. } & \oint E d A & \text { (i) } & 0 \\
\hline \text { B. } & \oint B d A & \text { (ii) } & -\frac{d \phi_B}{d t} \\
\hline \text { C. } & \oint E d l & \text { (iii) } & \frac{Q}{E_0} \\
\hline \text { D. } & \oint B d l & \text { (iv) } & \mu_0\left(i_c+i_d\right)
\end{array}\)

Options:
A \(\begin{array}{cccc} \text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iii) } & \text { (ii) } & \text { (i) } & \text { (iv) }\end{array}\)
B \(\begin{array}{cccc} \text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iv) } & \text { (i) } & \text { (iii) } & \text { (ii) } \end{array}\)
C \(\begin{array}{cccc} \text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iii) } & \text { (i) } & \text { (ii) } & \text { (iv) }\end{array}\)
D \(\begin{array}{cccc} \text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iii) } & \text { (i) } & \text { (iv) } & \text { (ii) } \end{array}\)

The correct answer is:
\(\begin{array}{cccc} \text { A } & \text { B } & \text { C } & \text { D } \\ \text { (iii) } & \text { (i) } & \text { (ii) } & \text { (iv) }\end{array}\)

Solution:
\(\mathrm{A} \rightarrow\) III According to Gauss's law, total electric flux passing through a closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the total charge enclosed within the surface. i.e., \(\quad \phi=\oint \mathbf{E}. d \mathbf{A}=\frac{Q}{\varepsilon_0}\)
\(\mathrm{B} \rightarrow \mathrm{I}\), According to gauss' law of magnetism, total magnetic flux passing through a bar magnet around its enclosed surface is zero because the magnetic lines of forces emerging from North-pole is equal to magnetic lines of force entering into South direction.
i.e., \(\oint B. d A=0\)
\(\mathrm{C} \rightarrow\) II, The electromotive force in a wire is the line integral,
\(E=\int \mathbf{E} \cdot d \mathbf{l}\)...(i)
But, by Faraday's law of electromagnetic induction, (EMI)
\(\varepsilon=-\frac{d \phi B}{d t}\)
From Eqs. (i) and (ii), we get
\(\int \mathbf{E} \cdot d \mathbf{l}=-\frac{d \phi B}{d t}\)
\(\mathrm{D} \rightarrow \mathrm{IV}\), According to the Maxwell, when electric field and hence electric flux charges with time, then an additional current comes into existence to the conduction current called displacement current. Therefore, the total current across the loop,
\(I=I_c+I_d\)
Hence, the modified form of the Ampere's law is
\(\oint \mathbf{B} \cdot d \mathbf{l}=\mu_0\left[I_c+I_d\right]\)

Topic: Physics, Electromagnetic Waves, JEE Main
Match the following lists :

Options:
A

B

C

D

The correct answer is:

Solution:
Grignard reagent $=\mathrm{CH}_3 \mathrm{Mg} X$
Clemmensen reduction $=\mathrm{Zn}-\mathrm{Hg}$ / Conc. $\mathrm{HCl}$
Rosenmund reduction $=\mathrm{H}_2 / \mathrm{Pd}-\mathrm{BaSO}_4$
Wolff-Kishner reduction $=$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Match the following organisms with their respective structures used for reproduction and choose the correct option.

Organism	Reproductive structures

i. Green thallophyte	a. Gemmules

ii. Organism used to produce the first antibiotic	b. Conidia

iii. Animals having a cellular level of organization	c. Buds

iv. A freshwater cnidarian	d. Zoospore

Options:
A i - c , ii - d, iii - a, iv - b
B i - d, ii - a, iii - b, iv - c
C i - b, ii - c, iii - d, iv - a
D i - d, ii -b, iii - a, iv - c

The correct answer is:
i - d, ii -b, iii - a, iv - c

Solution:
Algae are green thallophytes. They reproduce by forming zoospores. Penicillium is the organism used to produce the first antibiotic. It reproduces by forming conidia. Hydra is the freshwater Cnidarian. It reproduces by buds. Sponges are animals with the cellular level of organization. They reproduce by gemmules. These are asexual reproductive structures. The organisms form these reproductive structures which grow into new organisms and detach from the parents or sometimes remain attached to the parents. They are genetically similar to their parents.

Topic: Biology, Reproduction in Organisms, NEET
Match the following processes of metallurgy with their corresponding ore for which they are used::

(i) Froth floatation method (a) Germanium

(ii) Electrolytic refining of metals (b) ZnS

(iii) Zone refining of metals (c) Copper

Options:
A

(i)-(c),(ii)-(a),(iii)-(b)

B

(i)-(b),(ii)-(c),(iii)-(a)

C

(i)-(a),(ii)-(c),(iii)-(b)

D

(i)-(a),(ii)-(b),(iii)-(c)

The correct answer is:

(i)-(b),(ii)-(c),(iii)-(a)

Solution:
(i) Froth floatation method :

The method has been in use for removing gangue from sulphide ores. In this process, a suspension of the powered ore is made with water. To it, collectors and froth stabilisers are added. Collectors (e.g., pine oils, fatty acid, xanthates, etc.) enhance non-wettability of the mineral particles and froth stabilisers (e.g., Cresols, aniline) stabilise the froth.

The mineral particles become wet by oils while the gangue particles by water. A rotating paddle agitates the mixture and draws air in it. As a result, froth is formed which carries the mineral particles. The froth is light and is skimmed off. It is then dried for recovery of the ore particles.

(ii) Electrolytic refining of metals: In this method, the impure metal in pure form is used as Cathode. They are put in a suitable electrolyte both containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud. The reactions are

Anode : M→ M n+ +n e −

Cathode : M n+ +n e − →M

Copper and zinc are refined by this method.

(iii) Zone refining of metals: This method is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.

A circular mobile heater is fixed at one end of the rod of the impure metal. The molten zone moves along with the heater which is moved forward. As the heater moves forward, the pure metal crystalises out of melt and the impurities pass on into the adjacent molten zone. This process is repeated several times and the heater is moved in the same direction. At one end, impurities get concentrated.

This method is very useful for producing semiconductor and other metals of very high purity like germanium, silicon, boron etc.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
Match the following species with the correct number of electrons present in them:

Species	Number of Electrons

(i) Be2+	(a) 0

(ii) H+	(b) 10

(iii) Na+	(c) 2

(iv) Mg+	(d) 11

(e) 4

Options:
A (i-d),(i i-c),(i i-b),(i v-a)
B (i-a),(i i-b),(i i i-c),(i v-d)
C (i-e),(i i-d),(i i-a),(i v-c)
D (i-c),(i i-a),(i i i-b),(i v-d)

The correct answer is:
(i-c),(i i-a),(i i i-b),(i v-d)

Solution:
We know that the number of electrons is equal to the atomic number of element in neutral state.

Atomic number of beryllium is 4 but Be2+ can be formed by losing the two electrons so remaining electrons in Be2+ will be 4-2 = 2.

Similarly, the number of electrons in H+ is 1-1=0.

Similarly, the number of electrons in Na+ is 11-1=10.

Similarly, the number of electrons in Mg+ is 12-1=11.

Hence, the correct match will be (i-c), (ii-a), (iii-b), (iv-d).

Topic: Chemistry, Structure of Atom, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 1)
Match the following (Take the relative strength of the strongest fundamental forces in nature as one)
\(\begin{array}{|c|c|}\hline I & II \\\hline \text{Fundamental forces in nature} & \text{Relative strength} \\\hline \text{(A) Strong nuclear force} & (e) 10^{-2} \\\hline \text{(B) Weak nuclear force} & (f) 1 \\\hline \text{(C) Electromagnetic force} & (g) 10^{-13} \\\hline \text{(D) Gravitational force} & (h) 10^{-39} \\\hline\end{array}\)
The correct match is:

Options:
A \( \mathrm{A}-\mathrm{f}, \mathrm{B}-\mathrm{h}, \mathrm{C}-\mathrm{e}, \mathrm{D}-\mathrm{g} \)
B \( \mathrm{A}-\mathrm{h}, \mathrm{B}-\mathrm{f}, \mathrm{C}-\mathrm{e}, \mathrm{D}-\mathrm{g} \)
C \( A-f, B-g, C-e, D-h \)
D \( \mathrm{A}-\mathrm{f}, \mathrm{B}-\mathrm{e}, \mathrm{C}-\mathrm{h}, \mathrm{D}-\mathrm{g} \)

The correct answer is:
\( A-f, B-g, C-e, D-h \)

Solution:
I	II

Fundamental forces in nature	Relative strength

(A) Strong nuclear force	(f) 1

(B) Weak nuclear force	(g) 10-13

(C) Electromagnetic force	(e) 10-2

(D) Gravitational force	(h) 10-39

So, A-f, B-g, C-e, D-h

Topic: Physics, Laws of Motion, JEE Main
Match the following :

Test/Method	Reagent

(i)	Lucas Test	(a)	C6H5SO2Cl/aq. KOH

(ii)	Dumas method	(b)	HNO3/AgNO3

(iii)	Kjeldahl's method	(c)	CuO/CO3

(iv)	Hinsberg Test	(d)	Conc. HCl and ZnCl2

(e)	H2SO4

Options:
A (i)-(d), (ii)-(c), (iii)-(b), (iv)-(e)
B (i)-(b), (ii)-(d), (iii)-(e), (iv)-(a)
C (i)-(d), (ii)-(c), (iii)-(e), (iv)-(a)
D (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d)

The correct answer is:
(i)-(d), (ii)-(c), (iii)-(e), (iv)-(a)

Solution:
(I) Lucas reagent →Only ZnCl2/Conc.HCI

(II) Dumas method → CuO/Δ

(III) Kjeldahl's method → Conc. H2SO4/Δ

(IV) Heinsberg reagent → C6H5SO2CI/aq.NaOH

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Match the following
The correct answer is

Options:
A A-III, B-I, C-II, D-IV
B A-IV, B-III, C-I, D-II
C A-III, B-I, C-IV, D-II
D A-III, B-IV, C-II, D-I

The correct answer is:
A-III, B-IV, C-II, D-I

Solution:
Sn belongs to group 14 and has four valence electrons out of which two are used for forming two
$\mathrm{Sn}-\mathrm{Cl}$ bonds and the remaining two form a lone pair of electrons.
Thus, $\mathrm{SnCl}_2$ will be angular or bent in shape. $\mathrm{NH}_3$ has three bond pairs and one lone pair so it acquires a trigonal pyramidal shape.
$\mathrm{I}_3{ }^{-}$has a linear geometry as the two bond pairs lie along a straight line while the three lone pairs form the equatorial geometry. $\mathrm{SO}_3$ has three bond pairs only so it forms a trigonal planar shape.
Thus, $\mathrm{A} \rightarrow \mathrm{III}, \mathrm{B} \rightarrow \mathrm{IV}, \mathrm{C} \rightarrow \mathrm{II}, \mathrm{D} \rightarrow \mathrm{I}$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, TS EAMCET, TS EAMCET 2023 (12 May Shift 1)
Match the following
The correct answer is

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

The correct answer is:
A-(iv), B-(iii), C-(i), D-(ii)

Solution:
$\rightarrow$ We need high retentivity for permanent magnet so that it does not loses its magnetic property even if the external magnetic field is removed.
$\rightarrow$ Dimagnetic substance is feebly induced in the direction opposite of magnetic field. So it has negative susceptibility.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Match the following.
The correct answer is

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(a)

Solution:
(a)
(B) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{H}_2 \mathrm{SO}_4 \stackrel{443 \mathrm{~K}}{\rightarrow} \rightarrow \mathrm{CH}_2=\mathrm{CH}_2$
(C) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HBr} \underset{\text { Anti-Markownikof }}{\stackrel{\text { Peroxide }}{\rightarrow}}$ $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br}$
(D)

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Match the following.
The correct match is
$\text { A } \quad B \quad C$

Options:
A II I IV
B III IV IV
C III I I
D II IV I

The correct answer is:
III IV IV

Solution:
(A) Addition of HI takes place by the formation of $1^{\circ}$ and $2^{\circ}$ carbonium ion. $2^{\circ}$ carbonium ion being more stable will form major product, i.e. 2-iodopropane.
(B) The reaction of bromine with 1-phenylethane proceed with formation of free radical. $2^{\circ}$ free radical being more stable than $1^{\circ}$ free radical, will form the product as 1-bromo-1-phenylethane.
(C) Reaction of $\mathrm{HBr}$ to an alkene follows Markownikoff rule. So, the product will be 1-bromo-1-phenylethane.
Note : Markownikoff's rule states that, electron rich component adds to carbon atom with less number of hydrogen atoms.
Hence, the correct match is A-III, B-IV, C-IV. The reactions are :

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Match the following.
The correct match is
A B C D

Options:
A I V IV III
B III IV V I
C I V II IV
D IV V I III

The correct answer is:
III IV V I

Solution:
The correct match is : A-III, B-IV, C-V, D-I.
Dipole in different direction add up to give larger value.
Molecules with their geometry and dipole moments are given below
$(\mathrm{Cl} \rightarrow 2$ chlorine atoms are present opposite to each other, hence cancel dipole).

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Match the following with the types of leucocytes and their function.

P	Neutrophils	A	Heparin and Histamine

Q	Basophils	B	Antibody formation

R	Acidophils	C	Anti-allergic

S	Lymphocytes	D	Phagocytes

Options:
A P-D, Q-A, R-C, S-B
B P-A,Â Q-D,Â R-C,Â S-B
C P-C,Â Q-D,Â R-B,Â S-A
D P-B,Â Q-A,Â R-C,Â S-D

The correct answer is:
P-D, Q-A, R-C, S-B

Solution:
White blood cells are granulocytes and agranulocytes. Granulocytes are neutrophils, eosinophils, and basophils, depending on their staining properties. Neutrophils are the main phagocytic cells of the body. Eosinophils are acidophils and are anti-allergic in function. Basophils are least in number but play an essential role as these release heparin, histamine and serotonin. Agranulocytes are lymphocytes and monocytes. Monocytes are phagocytic, while lymphocytes are immune cells and help in antibody formation.

Topic: Biology, Human Health and Diseases, NEET
Match the followings and choose the correct option from below.

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

B
A-(iii), B-(i), C-(ii), D-(v), E-(iv) (c) A-(ii), B-(iv), C-(v), D-(i), E-(iii)

C
A-(ii), B-(iv), C-(v), D-(i), E-(iii) (d) A-(v), B-(iv), C-(iii), D-(ii), E-(i)

D
A-(v), B-(1v), C-(111), D-(11), E-(1)

The correct answer is:
A-(iii), B-(i), C-(ii), D-(v), E-(iv) (c) A-(ii), B-(iv), C-(v), D-(i), E-(iii)

Solution:
A meristem is a simple tissue made of a group of similar and immature cells (meristematic cells) which can divide and form new cells.
Parenchyma cells having chloroplasts are termed as chlorenchyma. It helps in the manufacture of food (photosynthesis). Storage parenchyma is made of large sized vacuolated cells which are used to store water, mucilage and food,
e.g., Aloe, Opuntia, potato tuber.
Collenchyma gives mechanical strength to young dicot stems, petioles and leaves.
Sclerenchyma is of two types, sclerenchyma fibres and sclereids.
Epidermal tissue system make the outermost covering of plant body. It mainly consists of epidermis and epidermal appendages. Epidermis is made of epidermal cells and stomata.

Topic: Biology, Anatomy of Flowering Plants
Match the functions of List-I with their nature in List-II and choose the correct option.
Then the correct match is

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(d)

Solution:
(A) $f: R \rightarrow R$
$$
f(x)=\cos (112 x-37)
$$
$f(x)$ is periodic function.
$\therefore$ It is not injective.
Range of $f(x)$ is $[-1,1]$
$\therefore$ It is also not surjective.
$\therefore \quad \mathrm{A} \rightarrow \mathrm{IV}$
(B) $f: A \rightarrow B, A \in[-2,2]$ and $B \in[-4,4]$
$$
\begin{aligned}
& f(x)=x|x| \\
& f(x)=\left\{\begin{array}{cc}
-x^2 & -2 \leq x < 0 \\
x^2 & 0 \leq x \leq 2
\end{array}\right.
\end{aligned}
$$
Clearly $f(x)$ is bijective function.
$$
\therefore \mathrm{B} \rightarrow \mathrm{III}
$$
(C) $f: R \rightarrow R$
$$
\begin{aligned}
& f(x)=(x-2)(x-3)(x-5) \\
& f(2)=f(3)=f(5)=0
\end{aligned}
$$
$\therefore$ It is not injective.
Range of $f(x)=R$
$\therefore$ It is surjective.
$\therefore \mathrm{C} \rightarrow \mathrm{II}$
(D) $f: N \rightarrow N$
$$
f(x)=n+1
$$
Clearly it is injective.
Range of $f(x)=\{2,3,4, \ldots\}$
Range $\neq$ Codomain
$\therefore$ It is not surjective.
$\therefore \mathrm{D} \rightarrow \mathrm{I}$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2019 (03 May Shift 1)
Match the given figures of corolla aestivation with the related plants.

(i) Cassia
(ii) Pea
(iii) Calotropis
(iv) Gossypium

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

The correct answer is:
A - (iii), B - (iv), C - (i), D - (ii)

Solution:
The mode of arrangement of sepals or petals in floral bud with respect to the other members of the same whorl is known as aestivation. The main types of aestivation are valvate, twisted, imbricate and vexillary .When sepals or petals in a whorl just touch one another at the margin, without overlapping, as in Calotropis, it is said to be valvate. If one margin of the appendage overlaps that of the next one and so on as in china rose, lady’s finger and cotton, it is called twisted. If the margins of sepals or petals overlap one another but not in any particular direction as in Cassia and gulmohur, the aestivation is called imbricate. 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
Match the hormone with its functions.

P	Melatonin	I	It plays an important role in the regulation of the basal metabolic rate.

Q	Aldosterone	II	It stimulates the reabsorption of Na+ and water and excretion of K+ and phosphate ions.

R	Thyroxine	III	It stimulates reabsorption of Ca+2 by the renal tubules and increases Ca +2 absorption from the digested food.

IV	It helps in maintaining the normal rhythms of the sleep-wake cycle and body temperature.

V	It stimulates the breakdown of glycogen resulting in an increased concentration of glucose in the blood.

Options:
A P - IV, Q - I, R - II
B P - II, Q - IV, R - V
C P - IV, Q - II , R - I
D P -II, Q - III, R - V

The correct answer is:
P - IV, Q - II , R - I

Solution:
Functions of melatonin: Melatonin is very important for the regulation of a 24-hour (diurnal) rhythm of our body, for example, It helps in maintaining the normal rhythms of the sleep-wake cycle, body temperature, etc. It also affects metabolism, pigmentation, menstrual cycle and our immune system.
Functions of aldosterone: Aldosterone mainly acts at the renal tubules and increases the reabsorption of Na+ and water and excretion of K+ and phosphate ions. Thus it helps in maintaining electrolyte levels, body fluid volume, osmotic pressure and blood pressure.
Functions of Thyroxine:
a. Regulation of the basal metabolic rate (BMR).
b. Regulate the process of red blood cell formation and the menstrual cycle.
c. Controlling the metabolism of carbohydrates, proteins and fats.
d. Maintenance of water and electrolyte balance.

Topic: Biology, Chemical Coordination and Integration, NEET
Match the hormones listed under column I with their functions listed under column II. Choose the answer which gives the CORRECT combination of the alphabets of the two columns.

Column I	Column II

A . Oxytocin	p. Stimulates ovulation

B. Prolactin	q. Implantation and maintenance of pregnancy

C. Luteinizing hormone	r. Lactation after childbirth

D. Progesterone	s. Release of milk from the mammary gland

t. Reabsorption of water by nephrons

Options:
A A = s, B = q, C = r, D = t
B A = t, B = r, C = p, D = s
C A = s, B = r, C = p, D = q
D A = t, B = p, C = s, D = r

The correct answer is:
A = s, B = r, C = p, D = q

Solution:
Oxytocin is known as a birth hormone as it brings about uterine contraction and also ejection of milk. Prolactin is responsible for lactation after childbirth. This hormone is secreted by anterior pituitary. Luteinizing hormone is a gonadotropin secreted by anterior pituitary and it is the main hormone responsible for ovulation. Progesterone and estrogen are responsible for maintaining the endometrium and hence prepares the uterus for implantation.

Topic: Biology, Chemical Coordination and Integration, NEET
Match the hormones with its source of secretion:

Column-I	Column- II

(A)	Somatostatin	(1)	Pineal gland

(B)	Melatonin	(2)	Pancreas

(C)	Aldosterone	(3)	Placenta

(D)	Erythropoietin	(4)	Adrenal cortex

(E)	hCG	(5)	Kidney

(6)	Liver

Options:
A A-4, B-1, C-2, D-3, E-5
B A-2, B-1, C-4, D-3, E-5
C A-4, B-1, C-2, D-3, E-6
D A-2, B-1, C-4, D-5, E-3

The correct answer is:
A-2, B-1, C-4, D-5, E-3

Solution:
1. Somatostatin is secreted by delta cells at several locations in the digestive system, namely, the pyloric antrum, the duodenum and the pancreatic islets.
2. Melatonin is mainly produced by the pineal gland.
3. Aldosterone, the main mineralocorticoid hormone, is a steroid hormone produced by the zona glomerulosa of the adrenal cortex in the adrenal gland. It is essential for sodium conservation in the kidney, salivary glands, sweat glands and colon.
4. Erythropoietin, also known as hematopoietic or hemopoietin, is a glycoprotein cytokine secreted by the kidney in response to cellular hypoxia; it stimulates red blood cell production in the bone marrow.
5. hCG (Human chorionic gonadotropin) is a hormone produced by the placenta after implantation. The presence of hCG is detected in some pregnancy tests.

Topic: Biology, Chemical Coordination and Integration, NEET