datavorous/entrance-exam-dataset huggingface dataset:
Match List-I with List-II
\begin{array}{|l|l|l|l|}\hline & List-I & & List-II \\\hline A. & Cells are metabolically active and proliferate & I. & \mathrm{G}_2 phase \\\hline B. & DNA replication takes place & II. & \mathrm{G}_1 phase \\\hline C. & Proteins are synthesised & III. & \mathrm{G}_0 phase \\\hline D. & Quiescent stage with metabolically active cells & IV. & S phase \\\hline\end{array}
Choose the correct answer from the options given below:

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

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

Solution:
(1) During $\mathrm{G}_1$ phase the cell is metabolically active and continuously grows but does not replicate its DNA.
(2) S or synthesis phase marks the period during which DNA synthesis or replication takes place.
(3) During the $\mathrm{G}_2$ phase, proteins are synthesized in preparation for mitosis while cell growth continues.
(4) Cells that do not divide further exit $\mathrm{G}_1$ phase to enter an inactive stage called quiescent stage $\left(\mathrm{G}_0\right)$ of the cell cycle. Cells in this stage remain metabolically active but no longer proliferate unless called on to do so.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2024 (Re-NEET)
Match List-I with List-II

List - I

(Elements)

List - II

(Properties)

(a)	Ba	(i)	Organic solvent soluble compounds

(b)	Ca	(ii)	Outer electronic configuration 6 s2

(c)	Li	(iii)	Oxalate insoluble in water

(d)	Na	(iv)	Formation of very strong monoacidic base

Choose the correct answer from the options given below:

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

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

Solution:
(a) 'Ba' having outer electronic configuration 6 s2.
(b) CaC2O4 is water insoluble.
(c) 'Li' compounds are soluble in organic solvents.
(d) NaOH is strong Monoacidic base.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Match List-I with List-II

List - I

(Elements)

List - II

(Properties)

(a)	SF4	(i)	sp3 d2

(b)	IF5	(ii)	d2sp3

(c)	NO2+	(iii)	sp3 d

(d)	NH4+	(iv)	sp3

(v)	sp

Choose the correct answer from the options given below:

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

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

Solution:
(a) SF4-sp3d hybridisation
(b) IF5-sp3d2 hybridisation
(c)NO2+-sp hybridisation
(d) NH4+-sp3 hybridisation

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Match List-I with List-II :

List - I	List - II

(a)	Chloroprene	(i)

(b)	Neoprene	(ii)

(c)	Acrylonitrile	(iii)

(d)	Isoprene	(iv)

Choose the correct answer from the options given
below:

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

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

Solution:
List - I	List - II

(a)	Chloroprene	(i)

(b)	Neoprene	(ii)

(c)	Acrylonitrile	(iii)

(d)	Isoprene	(iv)

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Match List-I with List-II :

List-I Chemical Compound	List-II Used as

a	Sucralose	i	Synthetic detergent

b	Glyceryl ester of stearic acid	ii	Artificial sweetener

c	Sodium benzoate	iii	Antiseptic

d	Bithional	iv	Food preservative

Choose the correct match :

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

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

Solution:
Artificial sweetner: Sucralose

Antiseptic: Bithional

Food preservative: Sodium Benzoate
Glyceryl ester of stearic acid: Sodium benzoate.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Match List-I with List-II

LIST-I
Coordination
Complex	LIST-II
Number of unpaired electrons

A.	Cr(CN)63-	I.	0

B.	FeH2O62+	II.	3

C.	CoNH363+	III.	2

D.	NiNH362+	IV.	4

Choose the correct answer from the options given below:

Options:
A

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

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

The correct answer is:

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

Solution:
Cr(CN)63- ion, oxidation state of Cr is +3 and its valence shell electronic configuration is 3d3. There are 3 unpaired electrons in 3d orbital.

(A). Cr(CN)63-
No. of unpaired electrons =3

FeH2O62+ ion, oxidation state of Fe is +2 and its valence shell electronic configuration is 3d6. There are 4 unpaired electrons in 3d orbital. So, you can say the hybridisation here would be sp3d2.

(B). FeH2O62+
No. of unpaired electrons =4

CoNH363+ in this oxidation state of central metal atom is +3 and it has no unpaired electrons.
(C). CoNH363+
No. of unpaired electrons =0

NiNH362+ the oxidation state of Ni is Ni2+ and it has two unpaired electrons.
(D) NiNH362+
No. of unpaired electrons =2
So the correct option among the given is A.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Match List-I with List-II :
$\begin{array}{|c|c|}
\hline \begin{array}{l}
\text{List-I} \\
\text{(Defects)}
\end{array} & \begin{array}{l}
\begin{array}{l}
\text{List-II} \\
\text{(Shown by)}
\end{array}
\end{array} \\
\hline \begin{array}{l}
\text{(a) Frenkel defect}
\end{array} & \begin{array}{l}
\text{(i) non-ionic solids and} \\
\text{density of the solid} \\
\text{decreases}
\end{array} \\
\hline \begin{array}{l}
\text{(b) Schottky defect}
\end{array} & \begin{array}{l}
\text{(ii) non-ionic solids and} \\
\text{density of the solid} \\
\text{increases}
\end{array} \\
\hline \text{(c) Vacancy defect} & \begin{array}{l}
\text{(iii) ionic solids and} \\
\text{density of the solid} \\
\text{decreases}
\end{array} \\
\hline \begin{array}{l}
\text{(d) Interstitial defect}
\end{array} & \begin{array}{l}
\text{(iv) ionic solids and} \\
\text{density of the solid} \\
\text{remains constant.}
\end{array} \\
\hline
\end{array}$
Choose the correct answer from the options given below :

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

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

Solution:
(a) Frenkel Defect-formed in ionic compounds where an ion is missing from its lattice site and occupies interstitial site. Compounds suffering from such defects does not have any change in their density.
(b) Schottky defect-Observed inionic compounds where equal number of ions are found missing from their lattice site. Such compounds have lower density.
(c) Vacancy defect-Observed in non-ionic solids where either an anion or cation is missing from their sites. Density of such solids are lower.
(d) Interstitial defects-Observed in nonionic solids where an extra cation may get into interstitial site, increasing the density of solid.

Topic: Chemistry, Solid State, NEET, NEET 2022 (Phase 2)
Match List-I with List-II.

List-I

Element detected

List-II

Reagent used/Product formed

A	Nitrogen	I	Na2FeCN5NO

B	Sulphur	II	AgNO3

C	Phosphorus	III	Fe4FeCN63

D	Halogen	IV	NH42MoO4

Choose the correct answer from the options given below:

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

The correct answer is:
A→III;B→I;C→IV;D→II

Solution:
The nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide (NaCN) soluble in water. This is converted into sodium ferrocyanide by the addition of sufficient quantity of ferrous sulphate.

6NaCN + FeSO4 → Na4[Fe(CN)6]Sodium ferrocyanide + Na2SO4Ferric ions generated during the process react with ferrocyanide to form Prussian blue precipitate of ferric ferrocyanide.

Na4[Fe(CN)6] + Fe3+ → Fe4[Fe(CN)6]3Ferric ferrocyanideprussian blue

Sulphur present in organic compounds are detected by Lassaigne’s test. Here, a small piece of Na metal is heated in a fusion tube with the organic compound. The principle is that, in doing so, Na converts all the elements present into ionic form.

2Na+S → Na2S

Test of sulphur :

The extract is treated with sodium nitroprusside. The appearance of violet colour indicates the presence of sulphur.The following reaction occurs:

S2-+ [Fe(CN)5NO]2- → [Fe(CN)5NOS]4-

The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.

The halogen present in the compound converted into sodium halide, and it will be precipitated with silver nitrate.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Match List-I with List-II

List-I

(Layer of atmosphere)

List-II

(Approximate height over earth's surface)

(A)	F1-Layer	(I)	10 km

B	D-Layer	II	170-190 km

C	Troposphere	III	100 km

D	E-Layer	V	65-75 km

Choose the correct answer from the options given below.

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

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

Solution:
The F1 layer is above the E layer and the ionization is less compared to E layer. It ranges to 170-190 km.

The D layer is located above the stratosphere. It ranges from 60-90 km above the earth's surface.

The troposphere starts from the ground and ranges to 100 km.

The E layer is located above the D layer. The intensity of ionization is less than the D layer. It ranges from 100-125 km above the earth's surface.

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Match List-I with List-II.

List-I	List-II

A	4NH3g+5O2g→4NOg+6H2Og	I	NOg

B	N2g+3H2g→2NH3g	II	H2SO4l

C	C12H22O11aq+H2Ol→C6H12O6Glucose+C6H12O6Fructose	III	Pts

D	2SO2g+O2g→2SO3g	IV	Fes

Choose the correct answer from the options given below

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

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

Solution:
(a)
4NH3g+5O2g⟶Pts4NOg+6H2Og
Ostwald process 500 K

(b) N2+3H2⟶Fes2NH3g
Haber's process

(c)C12H22O11aq+H2Ol⟶H+C6H12O6glucose+C6H12O6fructose
Inversion of sugar cane

(d) 2SO2g+O2g⟶NOg2SO3g

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (28 Jul Shift 1)
Match List-I with List-II

List-I	List-II

A	16 g of CH4 (g)	I	Weighs 28g

B	1 g of H2(g)	II	60.2 × 1023 electrons

C	1 mole of N2 (g)	III	Weighs 32g

D	0.5 mol of SO2(g)	IV	Occupies11.4 L volume at STP

Choose the correct answer from the options given below:

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

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

Solution:
The number of moles =MassMolar mass

The number of molecules = moles×NA

One mole of any gas occupies 22.7 L at STP.

(A) Moles of CH4=1616=1
Molecules =6.02×1023
Electrons =10×6.02×1023
=60.2×1023

(B) Moles of H2=12=0.5 mole
Volume at STP =12×22.7=11.35 lit .

(C) 1 mole N2=28gm

(D) 0.5 mole SO2=12×64=32gm

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Match List-I with List-II

List-I	List-II

(A)	Antipyretic	(I)	Reduces pain

(B)	Analgesic	(II)	Reduces stress

(C)	Tranquilizer	(III)	Reduces fever

(D)	Antacid	(IV)	Reduces acidity(stomach)

Choose the correct answer from the options given below

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

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

Solution:
Analgesics reduce or abolish pain without causing impairment of consciousness, mental confusion, incoordination, paralysis or some other disturbances of nervous system.

Tranquillizers are a class of chemical compounds used for the treatment of stress, and mild or even severe mental diseases. These relieve anxiety, stress, irritability or excitement by inducing a sense of well-being. They form an essential component of sleeping pills.

Treatment for acidity was administration of antacids, such as sodium hydrogen carbonate or a mixture of aluminium and magnesium hydroxide. However, excessive hydrogen carbonate can make the stomach alkaline and trigger the production of even more acid.

An antipyretic is a substance that reduces fever. Antipyretics cause the hypothalamus to override a prostaglandin-induced increase in temperature.

Hence,

(A)	Antipyretic Reduces fever

(B)	Analgesic Reduces pain

(C)	Tranquilizer Reduces stress

(D)	Antacid Reduces acidity (Stomach)

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Match List-I with List-II

List-I	List-II

(A)	Cl2O7	(I)	Amphoteric

(B)	Na2O	(II)	Basic

(C)	Al2O3	(III)	Neutral

(D)	N2O	(IV)	Acidic

Choose the correct answer from the options given below

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

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

Solution:
Non-metal oxides are generally acidic in nature.

Cl2O7- Acidic

Metal oxides are generally basic in nature

Na2O- Basic

Amphoteric oxides reacts with both acids and bases.

Al2O3- Amphoteric.

Some non-metal oxides don't react with both acids and bases they are neutral in nature.

N2O- Netural

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Match List-I with List-II

List-I	List-II

A	Glucose +HI	I	Gluconic acid

B	Glucose +Br2 water	II	Glucose pentacetate

C	Glucose + acetic anhydride	III	Saccharic acid

D	Glucose +HNO3	IV	Hexane

Choose the correct answer from the options given below

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

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

Solution:
A) On prolonged heating with HI, Glucose forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.

B) Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.

C) Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.

D) On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Match List-I with List-II

List-I	List-II

A	I	Br2 in CS2

B	II	Na2Cr2O7/H2SO4

C	III	Zn

D	IV	CHCl3/NaOH

Question: Choose the correct answer from the options given below

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

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

Solution:
A) Phenol reacts with chloroform in presence of aqueous sodium hydroxide to form salicylaldehyde, which is named as Riemer-Tiemann reaction.

B)Phenol when heated with zinc dust forms benzene.

C)Oxidation of phenol with acidified potassium dichromate gives benzoquinone.

D) Bromination of phenol in presence of a non-polar solvent like CS2 gives para-bromo phenol as the major product.

(A)

(B)

(C)

(D)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Match List-I with List-II.

List-I	List-II

A

I	Gatterman Koch reaction

B

II	Etard reaction

C

III	Stephen reaction

D

IV	Rosenmund reaction

Choose the correct answer from the options given below

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

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

Solution:
Benzoyl chloride is hydrogenated over catalyst, palladium on barium sulphate. This reaction is called Rosenmund reduction.

Nitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid, which on hydrolysis give corresponding aldehyde.

This is named as Stephen reaction

Use of Chromyl chloride oxidises methyl group to a chromium complex, which on hydrolysis gives corresponding benzaldehyde.This is called etard reaction

When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, it gives benzaldehyde or substituted benzaldehyde.

This is called Gattermann-koch reaction.The mechanism of this reaction is electrophilic substitution.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Match List-I with List-II.

List-I	List-II

A

I	Spiro compound

B

II	Aromatic compound

C

III	Non-planar Heterocyclic compound

D

IV	Bicyclo compound

Choose the correct answer from the options given below

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

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

Solution:
A spiro compound, or spirane, is a chemical compound, typically an organic compound, that presents a twisted structure of two or more rings (a ring system), in which 2 or 3 rings are linked together by one common atom.

Bicyclo compounds are a class of saturated compounds consisting of two fused rings, having two or more atoms in common, and that take the name of an open chain hydrocarbon containing the same total number of atoms.

Aromatic compounds are chemical compounds that consist of conjugated planar ring systems accompanied by delocalised pi-electron clouds in place of individual alternating double and single bonds.

Heterocyclic compounds are cyclic compounds with the ring containing carbon and other element, the component being oxygen, nitrogen and sulphur.

The name of is tetrahydro furan. The name of is furan. The name of is 2,2,4,4-tetramethylbicyclo[1.1.0]butane. The name of is 1,2,4,5-tetramethylspiro[2.2]pentane

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Match List-I with List-II:

List-I	List-II

A.	K	I.	Thermonuclear reactions

B.	KCI	II.	Fertilizer

C.	KOH	III.	Sodium potassium pump

D.	Li	IV.	Absorbent of CO2

Options:
A A(III), B(II), C(IV), D(I)
B A(III), B(IV), C(II), D(I)
C A(IV), B(I), C(III), D(II)
D A(IV), B(III), C(I), D(II)

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

Solution:
Lithium can indeed participate in thermonuclear reactions, specifically in certain fusion reactions.

The movement of K+ ions (potassium ions) is indeed involved in the functioning of the sodium-potassium pump.

KOH (potassium hydroxide) can act as an absorbent of carbon dioxide. This property is primarily due to the alkaline nature of KOH.

KCl (potassium chloride) is commonly used as a fertilizer in agriculture due to its high potassium content.

So option A is correct.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Match list-I with list-II :

List-I	List-II

(a)	Mercury	(i)	Vapour phase refining

(b)	Copper	(ii)	Distillation refining

(c)	Silicon	(iii)	Electrolytic refining

(d)	Nickel	(iv)	Zone refining

Choose the most appropriate answer from the option given below:

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

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

Solution:
(a) Mercury → Distillation refining
(b) Copper → Electrolytic refining
(c) Silicon → Zone refining
(d) Nickel → Vapour phase refining

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Match List-I with List-II

List-I	List-II

(A)	negatively charged sol	(I)	Fe2O3·xH2O

(B)	macromolecular colloid	(II)	CdS sol

(C)	positively charged sol	(III)	Starch

(D)	Cheese	(IV)	a gel

Choose the correct answer from the options given below

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

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

Solution:
Colloidal particles always carry an electric charge. The nature of this charge is the same on all the particles in a given colloidal solution and may be either positive or negative.

Negatively charged sol ⇒CdS

Positively charged sol ⇒Fe2O3.xH2O

Macromolecular colloids are individual particles that are large enough to be considered colloids (diameter around 100 nm). These particles are made up of chemicals with a high molecular weight.

Macromolecular colloid ⇒Starch

Gel ⇒ Cheese

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Match List-I with List-II

List-I	List-II

A	Nylon 6,6	I	Buckets

B	Low density
polythene	II	Non-stick
utensils

C	High density
polythene	III	Bristles of
brushes

D	Teflon	IV	Toys

Choose the correct answer from the options given below

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

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

Solution:
Column I	Column II

A.	Nylon-6, 6	III	Bristles of brush

B.	Low density polythene	IV	Toys

C.	High density polythene	I	Buckets

D.	Teflon	II	Non-stick utensils

Teflon is a brand name for a synthetic chemical called polytetrafluoroethylene (PTFE). Teflon is used to coat a variety of products because it's waterproof, cuts down on friction, and creates a nonstick surface.

Nylon 66 is a type of polyamide or nylon. It, and nylon 6, are the two most common for textile and plastic industries. Nylon 66 is made of two monomers each containing 6 carbon atoms, hexamethylenediamine and adipic acid, which give nylon 66 its name.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Match List-I with List-II

List-i	List-II

a	Phase difference between current and voltage in a purely resistive AC circuit	i	π2; current leads
voltage

b	Phase difference between current and
voltage in a pure inductive AC circuit	ii	zero

c	Phase difference between current and
voltage in a pure capacitive AC circuit	iii	π2; current lags
voltage

d	Phase difference between current and
voltage in an LCR series circuit	iv	tan-1XC-XLR

Choose the most appropriate answer from the options given below :

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

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

Solution:
(d) tanϕ=VL-VCVR=XL-XCR

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Match List-I with List-II

List-I	List-II

(A)	Television signal	I.	03KHz

(B)	Radio signal	II.	20KHz

(C)	High Quality Music	III.	02MHz

(D)	Human speech	IV.	06MHz

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

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

Solution:
Television signal →6 MHz

Radio signal →2 MHz

High quality music →20 kHz

Human speech →3 kHz

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Match List-I with List-II :

List-I	List-II

Industrial process	Application

(a)	Haber's process	(i)	HNO3 synthesis

(b)	Ostwald's process	(ii)	Aluminium extraction

(c)	Contact process	(iii)	NH3 synthesis

(d)	Hall-Heroult process	(iv)	H2SO4 synthesis

Choose the correct answer from the options given below:

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

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

Solution:
(a) Haber's process is used for NH3 synthesis.

(b) Ostwald's process is used for HNO3 synthesis.

(c) Contact process is used for H2SO4 synthesis.

(d) In Hall-Heroult process, electrolytic reduction of impure alumina can be done. (Aluminium extraction)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (16 Mar Shift 1)