datavorous/entrance-exam-dataset huggingface dataset:
Match List I with List II

List I	List II

A.	Cobalt catalyst	I.	H2+Cl2 production

B.	Syngas	II.	Water gas production

C.	Nickel catalyst	III.	Coal gasification

D.	Brine solution	IV.	Methanol production

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

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

Solution:
Hydrogen reacts with carbon monoxide in presence of cobalt catalyst to give methanol.

2H2g + COg →Cobalt catalyst CH3OHl

As the mixture of CO and H2 is used for the synthesis of methanol and a number of hydrocarbons, it is called synthesis gas or syngas. The process of producing syngas from coal is called coal gasification.

Cs + H2Og →1270K COg + H2g

The reaction of steam with hydrocarbons or coke at high temperature in presence of nickel catalyst gives a mixture of CO and H2, called water gas.

CH4g + H2Og →Ni1270K COg + 3H2g

Electrolysis of brine solution produces H2 gas at cathode and Cl2 gas at anode.

Cathode: 2H2Ol+ 2e- → H2g + 2OH-aq

Anode: 2Cl- (aq) → Cl2 (g) + 2e-

The overall reaction is 2Na+ (aq) + 2Cl- (aq) + 2H2O (l) → Cl2 (g) + H2 (g) + 2Na+ (aq) + 2OH- (aq)

So, the correct answer is A-IV, B-III, C-II, D-I

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Match List I with List II

List I	List II

A.	Gauss’s law of magnetostatics	i.	∮E→·da→=1ε0∫ρdV

B.	Faraday’s law of electro magnetic induction	ii.	∮B→·da→=0

C.	Ampere’s law	iii.	∮E→·dl→=-ddt∫B→·da→

D.	Gauss’s law of electrostatics	iv.	∮B→·dl→=μ0I

Choose the correct answer from the options given below:

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

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

Solution:
Gauss’s law of magnetostatics:

∮B→·da→=0

Faraday’s law of electro magnetic induction:

∮E→·dl→=-ddt∫B→·da→

Ampere’s law:

∮B→·dl→=μ0I

Gauss’s law of electrostatics:

∮E→·da→=1ε0∫ρdV

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Match List - I with List - II :

List - I	List - II

(a)	Gravitational constant (G)	(i)	L2 T-2

(b)	Gravitational potential energy	(ii)	M-1 L3 T-2

(c)	Gravitational potential	(iii)	LT-2

(d)	Gravitational intensity	(iv)	ML2 T-2

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) - (iv), (b) - (ii), (c) - (i), (d) - (iii)
D (a)-(ii), (b)-(i), (c) - (iv), (d) - (iii)

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

Solution:
List - I	List - II

(a)	Gravitational constant (G)	(ii)	M-1 L3 T-2

(b)	Gravitational potential energy	(iv)	ML2 T-2

(c)	Gravitational potential	(i)	L2 T-2

(d)	Gravitational intensity	(iii)	LT-2

Hence, (a) - (ii), (b) - (iv), (c) - (i), (d) - (iii).

Topic: Physics, Units and Dimensions, JEE Main
Match List - I with List - II :

List-I	List-II

a	h (Planck's constant)	i	MLT-1

b	E (kinetic energy)	ii	ML2 T-1

c	V (electric potential)	iii	ML2 T-2

d	P (linear momentum)	iv	ML2I-1 T-3

Choose the correct answer from the options given below:

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

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

Solution:
By dimensional analysis.

K.E.=ML2 T-2
P (linear momentum) =MLT-1
h ( planck's constant )=ML2 T-1
V (electric potential )=ML2 T-3I-1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Match List I with List II

List I	List II

A	I	Fittig reaction

B	II	Wurtz Fittig reaction

C	III	Finkelstein reaction

D	C2H5Cl+NaI→C2H5I+Nacl	IV	Sandmeyer reaction

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

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

Solution:
(i) Fittig reaction: Aryl halides give analogous compounds when treated with sodium in dry ether, in which two aryl groups are joined together. It is called Fittig reaction.

(ii) Wurtz-Fittig reaction: A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether and is called Wurtz-Fittig reaction.

(iii) Finkelstein reaction: Alkyl iodides are often prepared by the reaction of alkyl chlorides/ bromides with NaI in dry acetone. This reaction is known as Finkelstein reaction.

(iv) Sandmeyer reaction: When a primary aromatic amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed. Mixing the solution of freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by –Cl or –Br

Hence, the correct matching is given in the below table:

List I	List II

A

Wurtz Fitting reaction

B	Fitting reaction

C

Sandmeyer reaction

D	C2H5Cl+NaI→C2H5I+Nacl	Finkelstein reaction

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Match List I with List II.

List I	List II

(a)	Isothermal	(i)	Pressure constant

(b)	Isochoric	(ii)	Temperature constant

(c)	Adiabatic	(iii)	Volume constant

(d)	Isobaric	(iv)	Heat content is constant

Choose the correct answer from the options given below:

Options:
A (a)→(ii),(b)→(iii),(c)→(iv),(d)→(i)
B (a)→(iii),(b)→(ii),(c)→(i),(d)→(iv)
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)→(iii),(c)→(iv),(d)→(i)

Solution:
(a) Isothermal ⇒Temperature constant
(a)→ii

(b) Isochoric ⇒ Volume constant
(b)→iii

(c) Adiabatic ⇒ΔQ=0

⇒Heat content is constant

(c)→iv

(d) Isobaric⇒Pressure constant

(d)→i

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Match List I with List II :

List I	List II

A	Isothermal Process	I	Work done by the gas decreases internal energy

B	Adiabatic Process	II	No change in internal energy

C	Isochoric Process	III	The heat absorbed goes partly to increase internal energy and partly to do work

D	Isobaric Process	IV	No work is done on or by the gas

Choose the correct answer from the options given below :

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

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

Solution:
(A) Change in internal energy is expressed as∆U=nCv∆T, here, ∆T is change in temperature.

Since, in an isothermal process temperature remains constant, thus, ∆U=0

A→II

(B) In an adiabatic process, heat transfer, Q = 0.

So, from first law of thermodynamics,

Q=∆U +W

∆U=-W

Since, work done by gas is positive, thus, ∆U is negative

B→I

(C) In an isochoric process, volume remains constant, so work done by or on the gas W =P∆V=0

C→IV

(D) In an isobaric process, pressure remains constant, so work doneW=P∆V≠0 and change in internal energy ∆U=nCv∆T≠0

Thus, heat absorbed goes partly to increase internal energy and partly do work.

D→III

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Match List - I with List - II :

List -I	List-II

a	NaOH	i	Acidic

b	Be(OH)2	ii	Basic

c	Ca(OH)2	iii	Amphoteric

d	B(OH)3

e	Al(OH)3

Choose the most appropriate answer from the options given below

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

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

Solution:
NaOH→ Basic
Be(OH)2→ Amphoteric
Ca(OH)2→ Basic
B(OH)3→ Acidic
Al(OH)3→ Amphoteric

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Match List I with List II

List I	List II

A	Nitrogen oxides in air	I	Eutrophication

B	Methane in air	II	pH of rain water becomes 5.6

C	Carbon dioxide	III	Global warming

D	Phosphate fertilisers in water	IV	Acid rain

Choose the correct answer from the options given below :

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

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

Solution:
Nitogen oxides are produced by various natural and human activities, including combustion processes, industrial emissions, and vehicle exhaust. These are responsible for acid rains.

Methane (CH4) is a potent greenhouse gas that contributes to global warming. While carbon dioxide (CO2) is the most well-known greenhouse gas. Carbon dioxide present in air, react with the rain water to form carbonic acid.

The excessive use or runoff of phosphate fertilizers into bodies of water can contribute to eutrophication.

A	Nitrogen oxides in air	IV	Acid rain

B	Methane in air	III	Global warming

C	Carbon dioxide	II	pH of rain water becomes 5.6

D	Phosphate fertilisers in water	I	Eutrophication

∴ A-IV; B-III; C-II; D-I

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Match List I with List II

List I	List II

A	Physisorption	I	Single Layer Adsorption

B	Chemisorption	II	20-40 kJ mol-1

C	N2g+3H2g⟶Fe(s)2NH3g	III	Chromatography

D	Analytical Application or Adsorption	IV	Heterogeneous catalysis

Choose the correct answer from the options given below:

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

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

Solution:
(A) In physisorption, the enthalpy heat is low and lies between 20 - 40 kJ mol-1. In physisorption, multilayers of adsorbed particles are formed.

(B) In chemisorption, the enthalpy heat is high and lies within the range of 40 - 200 kJ mol-1. In chemisorption, a monolayer of adsorbed particles is formed.

(C) N2g+3H2g⟶Fe(s)2NH3g

Heterogeneous catalysts occupy a different phase. Generally, heterogeneous catalysts are solid compounds that are added to liquid or gas reaction mixtures. Fe is used as a heterogeneous catalyst in the Haber process.

(D) In the process of adsorption chromatography, different compounds are adsorbed on the adsorbent to different degrees based on the absorptivity of the component.

So, the correct matching is A-IV, B-II, C-III, D-I.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Match List-I with List-II.

List-I	List-II

a	RH (Rydberg constant)	i	kg m-1 s-1

b	h (Planck's constant)	ii	kg m2 s-1

c	μB (Magnetic field energy density)	iii	m-1

d	η (coefficient of viscosity)	iv	kg m-1 s-2

Choose the most appropriate answer from the options given below:

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

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

Solution:
The relation between wavelength and the principal quantum number is, 1λ=R1n12-1n22, for hydrogen-like atom, so the unit to Rydberg constant is, m-1.

Now recall the formula of energy of a photon in terms of frequency, E=hν, it means, h=Eν=kg m2 s-2s-1=kg m2 s-1.

The coefficient of viscosity,
η=FA×dvdz=kg m s-2 mm2×m s-1=kg m-1 s-1.

Energy density is defined as, EV=kg m2 s-2m3=kg m-1 s-2.

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Match List I with List II.

List I	List II

(a)	Rectifier	(i)	Used either for stepping up or stepping down the A.C. voltage

(b)	Stabilizer	(ii)	Used to convert A.C. voltage into D.C. voltage

(c)	Transformer	(iii)	Used to remove any ripple in the rectified output voltage

(d)	Filter	(iv)	Used for constant output voltage even when the input voltage or load current change

Choose the correct answer from the options given below:

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

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

Solution:
(a) Rectifier → A.C. to D.C.

(b) Stabilizer → used for constant output voltage even when input voltage or current change.

(c) Transformer → Step- up or step - down A.C. voltage.

(d) Filter → used to remove any ripple in the rectified output voltage.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Match List I with List II.

List I	List II

A.	Reverberatory furnace	I.	Pig iron

B.	Electrolytic cell	II.	Aluminium

C.	Blast furnace	III.	Silicon

D.	Zone refining furnace	IV.	Copper

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

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

Solution:
Reverberatory furnace, in copper, tin, and nickel production, a furnace used for smelting or refining in which the fuel is not in direct contact with the ore but heats it by a flame blown over it from another chamber.Purification of aluminium by electrolytic refining is known as Hoope's process. By this process 99.9% pure aluminium metal is obtained.

The iron manufactured in Blast Furnaces contain about 3 – 4 % of Carbon and smaller quantities of many other impurities such as sulphur, Silicon, etc. This is called Pig Iron.

Zone refining is used for the purification of metals like Si, Ge, Ga which are used in semiconductors.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Match List I with List II :

List I	List II

A.	Saccharin	I.	High potency sweetener

B.	Aspartame	II.	First artificial sweetening agent

C	Alitame	III.	Stable at cooking temperature

D.	Sucralose	IV.	Unstable at cooking temperature

Choose the correct answer from the options given below :

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

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

Solution:
Saccharin is an artificial sweetener that is commonly used as a sugar substitute. It is about 300-400 times sweeter than sugar and has a slightly bitter aftertaste.

Aspartame use is limited to cold foods and soft drinks because it is unstable at cooking temperature.

Alitame is high potency sweetener, which means that only a small amount is needed to achieve the same level of sweetness as sugar.

Sucralose is stable at cooking temperature.

So the correct option is , A→II, B→IV, C→I, D→III

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Match List-I with List-II.

List-I	List-II

(a)	Siderite	(i)	Cu

(b)	Calamine	(ii)	Ca

(c)	Malachite	(iii)	Fe

(d)	Cryolite	(iv)	Al

(v)	Zn

Choose the correct answer from the options given below:

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

The correct answer is:
a→iii, b→v, c→i, d→iv

Solution:
Siderite is a mineral composed of iron (II) carbonate FeCO3. It is a widespread mineral that is an ore of iron.

Calamine: Zinc ore is most commonly found as zinc carbonate ZnCO3 known as calamine or smithsonite.

Malachite - CuCO3·Cu(OH)2, Malachite is a copper carbonate hydroxide mineral

Cryolite - Na3AlF6, which consists of Aluminium, Sodium & Fluorine.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Match List-I with List-II

List-I	List-II

(a)	Sodium Carbonate	(i)	Deacon

(b)	Titanium	(ii)	Castner-Kellner

(c)	Chlorine	(iii)	van-Arkel

(d)	Sodium hydroxide	(iv)	Solvay

Choose the correct answer from the options given below:

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

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

Solution:
(a) Sodium carbonate Na2CO3 & NaHCO3 are prepared by Solvay process.

(b) Titanium is refined by Van-Arkel process.

Titanium : Van arkel method

Ti+I2⟶T1TiI4g

TiI4g→T2>T1T2TiRefined titanium+2I2g

(c) Chlorine is prepared by Deacon process.

Chlorine : Decon's process

HCl+O2⟶CuCl2H2O+Cl2

(d) Sodium hydroxide is prepared by Castner - Kellner process.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Match List-I with List-II.

List-I	List-II

(a)	Sucrose	(i)	β-D-Galactose and β-D-Glucose

(b)	Lactose	(ii)	α-D-Glucose and β-D-Fructose

(c)	Maltose	(iii)	α-D-Glucose and α-D-Glucose

Choose the correct answer from the options given below:

Options:
A a→iii, b→ii, c→i
B a→iii, b→i, c→ii
C a→ii, b→i, c→iii
D a→i, b→iii, c→ii

The correct answer is:
a→ii, b→i, c→iii

Solution:
(1) Sucrose →α-D-Glucose and β-D-Fructose

(2) Lactose →β-D-Galactose and β-D-Glucose

(3) Maltose →α-D-Glucose and α-D-Glucose

a→II

b→I

c→III

Sucrose is made up of two monomers α-D-Glucose and β-D-Fructose.
Lactose is made up of two monomers β-D-Galactose and β-D-Glucose.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Match List I with List II

List - I	List - II

A	Surface tension	I.	kg m-1 s-1

B	Pressure	II.	kg m s-1

C	Viscosity	III.	kg m-1 s-2

D	Impulse	IV.	kg s-2

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

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

Solution:
(A) Surface tension is related to force and length as S=Fl=MLT-2 L=M T-2

=kg s-2 (IV)

(B) Pressure is related to force and area as P=FA

=MLT-2 L2=kg m-1 s-2 (III)

(C) Viscosity in terms of force, area and velocity gradient is given by η=FAdvdx=MLT-2 L2LT-1L

=ML-1 T-1=kg m-1 s-1 (I)

(D) Impulse is related to force and time of application as J=∫F dt=MLT-2×T

=MLT-1=kg m s-1 (II).

So A-(IV), B-(III), C-(I), D-(II)

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Match List I with List II

List I	List II

A	Torque	I	kg m–1 s–2

B	Energy density	II	kg m s–1

C	Pressure gradient	III	kg m–2 s–2

D	Impulse	IV	kg m2 s–2

Choose the correct answer from the options given below :

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

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

Solution:
Unit of force is written as, kg m s-2.

Now, torque can be written as, τ=Fs therefore its unit can be written as kg m2 s–2.

Pressure is given by, P=FA, therefore unit of pressure will be kg m-1 s-2.

Now, pressure gradient is given by ratio of pressure and distance, so its unit will be kg m-2 s-2.

Impulse is given by product of force and time, so its unit will be kg m s-1.

As we know unit of torque and energy is the same, therefore energy density can written as =energyvolume=kg m-1 s–2.

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Match List I with List II

List I	List II

A.	Torque	I.	M L-2 T-2

B.	Stress	II.	M L2 T-2

C.	Pressure gradient	III.	M L-1 T-1

D.	Coefficient of
viscosity	IV.	M L-1 T-2

Choose the correct answer from the options given below :

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

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

Solution:
The dimensions of the quantities are as follows.

F=M LT-2r=Lv=LT-1A=L2

The dimensional form of stress is same as pressure.

Torque = Force × distance=ML2 T-2
Stress = Force Area =M L-1 T-2
Pressure Gradient =dpdx=ML-2 T-2
η=F6πrv=MLT-2L·LT-1=ML-1 T-1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Match List - I with List - II.

List - I	List - II

a	Torque	i	MLT-1

b	Impulse	ii	MT-2

c	Tension	iii	ML2 T-2

d	Surface Tension	iv	MLT-2

Choose the most appropriate answer from the option given below :

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

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

Solution:
The torque is defined as, τ=⊥rFsinθ=ML2T-2, the formula of impulse, I=∆p=MLT-1 and tension, T=MLT-2 and the formula of surface tension is defined as force per unit length.s=Fl=MLT-2L=MT-2.

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Match List I with List II.

List I	List II

(A)	Torque	(I)	Nms-1

(B)	Stress	(II)	Jkg-1

(C)	Latent Heat	(III)	Nm

(D)	Power	(IV)	Nm-2

Choose the correct answer from the options given below:

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

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

Solution:
Torque =Force×perpendicular distance r⊥	N m

Stress= Force Area	N m-2

Latent heat = Energy Mass	J kg-1

Power = Work Time =Force×displacementTime	N m s-1

Thus, the correct answer is A-III, B-IV, C-II, D-I.

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Match List I with List II

List I	List II

A	Troposphere	I	Approximate 65-75 km over Earth’s surface

B	E-Part of Stratosphere	II	Approximate 300 km over Earth’s surface

C	F2-Part of Thermosphere	III	Approximate 10 km over Earth’s surface

D	D-Part of Stratosphere	IV	Approximate 100 km over Earth’s surface

Choose the correct answer from the options given below :

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

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

Solution:
(A) The troposphere extends upward to about 10 km (6.2 miles or about 33,000 feet) above sea level.

(B) The Electrical-layer or E-part of stratosphere lies approximately 100 km over Earth’s surface

(C) The F2-part of thermosphere lies approximately 300 km over Earth’s surface.

(D) D-part of Stratosphere lies approximate 65-75 km
over Earth’s surface

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

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Match List I with List II.

List I	List II

A.	van't Hoff
factor, i	I.	Cryoscopic constant

B.	kf	II.	Isotonic solutions

C.	Solutions with same osmotic pressure	III.	Normal molar massAbnormal molar mas

D.	Azeotropes	IV.	Solutions with same composition of vapour above it

Choose the correct answer from the options given below:

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

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

Solution:
(A) van't Hoff factor, it is used in measuring colligative properties

i= Normal molar mass Abnormal molar mass

(B) kf= Cryoscopic constant or molal freezing constant

(C) Solutions with same osmotic pressure are known as isotonic solutions.

(D) Solutions with same composition of vapour over them are called Azeotrope or azeotrophic mixtures.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Match List I with List II

List I	List II

A. XeF4	I.See-saw

B. SF4	II. Square planar

C. NH4+	III. Bent T- shaped

D.BrF3	IV. Tetrahedral

Choose the correct answer from the options given below :

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

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

Solution:
(A) XeF4:

Xe has 4 bond pairs along with 2 lone pairs in the XeF4. Thus, the hybridisation of Xe is sp3d2. Now since it contains two lone pairs it will show square planar geometry.

(B) SF4:

The lone pair is an equatorial position, and there are two lone-pair—bond pair repulsions. Hence, it is more stable. So, the shape is described as a distorted tetrahedron, a folded square or a see-saw.

(C) NH4+:

In this case steric number = lone pair + sigma bond = 0 + 4 = 4.

So, it is sp3 hybridisation. There is no lone pair present. Thus, the shape of this ion is tetrahedral.

(D) BrF3:

To decrease repulsion between the lone pairs, the molecule's structure is bent, making it T-shaped.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (31 Jan Shift 1)