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

Choose the correct answer from the options given below :-

List I
Isomeric pairs	List II
Type of isomers

A	Propanamine and N-
Methylethanamine	I	Metamers

B	Hexan-2-one
Hexan-3-one	II	Positional
isomers

C	Ethanamide and
Hydroxyethanimine	III	Functional isomers

D	o-nitrophenol and pnitrophenol	IV	Tautomers

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

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

Solution:
Metamers are the isomers having the same molecular formula but different alkyl groups on two sides of functional groups. Positional isomers are constitutional isomers with the same carbon skeleton and will have the same functional group. Functional isomers are structural isomers that have the same molecular formula (that is, the same number of atoms of the same elements), but the atoms are connected in different ways so that the groupings are dissimilar. Tautomerism is a phenomenon where a single chemical compound tends to exist in two or more interconvertible structures that are different in terms of the relative position of one atomic nucleus which is generally the hydrogen.

A. Propanamine and N- Methylethanamine (CH3CH2CH2NH2 and CH3CH2NHCH3)

B. Hexan-2-one Hexan-3-one

C. Ethanamide Hydroxyethanimine

D. o-Nitrophenol p-nitrophenol

So, the correct option is D.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Match List I with List II :
Choose the correct answer from the options given below :

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

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

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Match List I with List II

Chose the correct answer from the options given below

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

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

Solution:
Ampere – Maxwell law

→∮B→·dl→=μ0ic+μ0ε0dϕEdt

Faraday law

→∮E→·dl→=dϕBdt

Gauss’ law for electricity →∮E→·dA→=Qε0

Gauss' law for magnetism →∮B→·dA→=0

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Match list I with list II for the compositions of substances and select the correct answer using the code given below the lists.

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

The correct answer is:
(b)

Solution:
(A) Plaster of Paris $=\mathrm{CaSO}_4 \cdot \frac{1}{2} \mathrm{H}_2 \mathrm{O}$
(B) Epsomite $=\mathrm{MgSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}$
(C) Kieserite $=\mathrm{MgSO}_4 \cdot \mathrm{H}_2 \mathrm{O}$
(D) Gypsum $=\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, s Block Elements, NEET, NEET 2011 (Mains)
Match List I with List II
$\begin{array}{|l|l|r|l|}
\hline \text{ LIST I } & \text{ LIST II } \\
\hline \text{A.} & \text{Torque} & \text{I.} & {\left[M^1 L^1 T^{-2} A^{-2}\right]} \\
\hline \text{B.} & \text{Magnetic field} & \text{II.} & {\left[L^2 A^1\right]} \\
\hline \text{C.} & \text{Magnetic moment} & \text{III.} & {\left[M^1 T^{-2} A^{-1}\right]} \\
\hline \text{D.} & \text{Permeability of free space} & \text{IV.} & {\left[M^1 L^2 T^{-2}\right]} \\
\hline
\end{array}$
Choose the correct answer from the options given below:

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

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

Solution:
$\begin{aligned} & {[\vec{\tau}]=[\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]} \\ & {[\mathrm{F}]=[\mathrm{qVB}]} \\ & \Rightarrow \mathrm{B}=\left(\frac{\mathrm{F}}{\mathrm{qV}}\right)=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{ATLT}^{-1}}\right]=\left[\mathrm{MA}^{-1} \mathrm{~T}^{-2}\right] \\ & {[\mathrm{M}]=[\mathrm{I} \times \mathrm{A}]=\left[\mathrm{AL}^2\right]} \\ & \mathrm{B}=\frac{\mu_0}{4 \pi} \frac{\mathrm{Idl} \sin \theta}{\mathrm{r}^2} \\ & \Rightarrow[\mu]=\left[\frac{\mathrm{Br}^2}{\mathrm{Idl}}\right]=\left[\frac{\mathrm{MT}^{-2} \mathrm{~A}^{-1} \times \mathrm{L}^2}{\mathrm{AL}}\right] \\ & =\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right]\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Match List I with List II

List I (Amines)	List II pKb

A.	Aniline	I.	3.25

B.	Ethanamine	II.	3.00

C.	N-Ethylethanamine	III.	9.38

D.	N, N-Diethylethanamine	IV.	3.29

Choose the correct answer from the options given below:-

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

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

Solution:
Among given aliphatic amines, 2° amine is most basic, followed by 3° amine and 1° amine.

Therefore, the correct basic strength (Kb) order of the given amines is

(C) > (D) > (B) > (A)

The pKb order of the given amines will be just the opposite of their basic strength order.

The order is (A) > (B) > (D) > (C)

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

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

List-I
(Chemicals)	List-II
(Use / Preparation / Constituent)

(a)	Alcoholic potassium hydroxide	(i)	Electrodes in batteries

(b)	Pd/BaSO4	(ii)	Obtained by addition reaction

(c)	BHC (Benzene hexachloride)	(iii)	Used for β-elimination reaction

(d)	Polyacetylene	(iv)	Lindlar's catalyst

Choose the most appropriate match

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

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

Solution:
(a) Alcoholic potassium hydroxide→used for β-elimination
(b) Pd/BaSO4→ Lindlar's catalyst
(c) BHC (Benzene hexachloride)→Obtained by addition reactions
(d) Polyacetylene→Electrodes in batteries

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Match List - I with List - II.

List-I
(Colloid Preparation Method)	List-II
(Chemical Reaction)

(a)	Hydrolysis	(i)	2AuCl3+3HCHO+3H2O→2Au(sol)+3HCOOH+6HCl

(b)	Reduction	(ii)	As2O3+3H2 S→As2 S3(sol)+3H2O

(c)	Oxidation	(iii)	SO2+2H2 S→3 S(sol)+2H2O

(d)	Double Decomposition	(iv)	FeCl3+3H2O→Fe(OH)3(sol)+3HCl

Choose the most appropriate answer from the options given below :

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

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

Solution:
Colloid preparation by Hydrolysis reaction: (no change in oxidation number.)

FeCl3+3H2O→Fe(OH)3(sol)+3HCl

Gold sol is prepared by reduction reaction:

2AuCl3+3HCHO+3H2O→2Au(sol)+3HCOOH+6HCl

Preparation of sulphur sol:(oxidation)

SO2+2H2 S→3 S(sol)+2H2O

Preparation of colloid by double decomposition reaction:

As2O3+3H2 S→As2 S3(sol)+3H2O

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Match List I with List II:

List I
(Complexes)	List II
(Hybridisation)

(A)	Ni(CO)4	I	sp3

(B)	CuNH342+	II	dsp2

(C)	FeNH362+	III	sp3d2

(D)	FeH2O62+	IV	d2sp3

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

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

Solution:
Ni is in zero oxidation state in [Ni(CO)4]. So, the electronic configuration of Ni is 3d8 4s2. As CO is a strong ligand, it pushes all the electrons in the 3d orbital, therefore the hybridisation of [Ni(CO)4] is sp3 and it has tetrahedral geometry. It is diamagnetic due to the absence of unpaired electrons.

CuNH342+ ion, oxidation state of Cu is +2 and its valence shell electronic configuration is 3d9. So, there would be a rearrangement of electrons in Cu2+ because of the NH3 strong field ligand . And the last electron in the d-orbital would be out waiting for the N's electrons to fill up first. So, the 4 electron pairs from N would be in one 3d, one 4s, & two 4p orbitals and in the third place of 4p orbital, the e- from 3d would take place. So, you can say the hybridisation here would be dsp2.

FeNH362+ ion, oxidation state of Fe is +2 and its valence shell electronic configuration is 3d6. NH3 is strong field ligand with complex. So, no pairing will be there. So, you can say the hybridisation here would be d2sp3.

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.

So, correct order of matching is given in the below table:

List I
(Complexes)	List II
(Hybridisation)

Ni(CO)4	sp3

CuNH342+	dsp2

FeNH362+	d2sp3

FeH2O62+	sp3d2

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Match List - I with List - II :

List - I (compound)	List - II
(effect/affected species)

a	Carbon monoxide	i	Carcinogenic

b	Sulphur dioxide	ii	Metabolized by pyrus plants

c	Polychlorinated biphenyls	iii	Haemoglobin

d	Oxides of Nitrogen	iv	Stiffness of flower buds

Choose the correct answer from the options given below:

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

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

Solution:
Carbon monoxide combines with haemoglobin to form carboxyhaemoglobin. High concentration of SO2 leads to stiffness of flower buds. Polychlorinated biphenyls, are highly toxic industrial compounds. Pyrus and Vitis can metabolise nitrogen oxide.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Match List I with List II

List – I
(Current configuration)	List – II
(Magnetic field at point O)

A	I.	B0=μ0I4πr[π+2]

B	II.	B0=μ04Ir

C	III.	B0=μ0I2πr[π-1]

D	IV.	B0=μ0I4πr[π+1]

Choose the correct answer from the option given below:

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

Here net magnetic field will be sum of magnetic field due to straight wires ab, de and due to loop bcd.

So, Bab=μ04πIr=Bde (directed out of the plane in both cases)

Bbcd=μ04πIr(2π) (in the plane)

Thus, magnetic field at O is

BO=-μ04πIr+μ04πIr(2π)-μ04πIr

BO=μ02πIr(π-1) (III)

(B)

Here net magnetic field will be sum of magnetic field due to straight wires ab, de and semicircular arc bcd.

Bab=μ04πIr=Bde (directed out of the plane in both cases)

Bbcd=μ04πIr(π) (out of the plane)

Thus, net magnetic field at O is

BO=μ04πIr+μ04πIr(π)+μ04πIr

B0=μ04πIr(π+2) (I)

(C)

Here net magnetic field will be sum of magnetic field due to straight wires ab, de and due to loop bcd.

Bab=μ04πIr (directed in the plane)

Bbcd=μ04πIr(π) (in the plane)

Bde=0 (passing through the axis)

Thus, magnetic field at O is

BO=μ04πIr(π+1) (IV)

(D)

Here net magnetic field will be sum of magnetic field due to straight wires ab, de and due to loop bcd.

Bab=0=Bde (passing through the axis)

Bbcd=μ04πIr(π) (out of the plane)

Thus, magnetic field at O is

BO=μ0I4r (II).

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

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Match List - I with List - II:

List - I
(Electromagnetic waves)	List - II
(Wavelength)

(a)	AM radio waves	(i)	10-10 m

(b)	Microwaves	(ii)	102 m

(c)	Infrared radiations	(iii)	10-2 m

(d)	X-rays	(iv)	10-4 m

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

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

Solution:
List - I
(Electromagnetic waves)	List - II
(Wavelength)

(a)	AM radio waves	(ii)	102 m

(b)	Microwaves	(iii)	10-2 m

(c)	Infrared radiations	(iv)	10-4 m

(d)	X-rays	(i)	10-10 m

Topic: Physics, Electromagnetic Waves, JEE Main
Match List I with List II

LIST I – Enzymatic reaction	LIST II - Enzyme

A	Sucrose→ Glucose and Fructose	I	Zymase

B	Glucose → ethyl alcohol and CO2	II	Pepsin

C	Starch → Maltose	III	Invertase

D	Proteins → Amino acids	IV	Diastase

Choose the correct answer from the options given below.

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

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

Solution:
The Zymase enzyme converts glucose and fructose into ethanol and carbon dioxide. And the Invertase enzyme is used to convert sucrose or cane sugar into glucose and fructose. The enzyme diastase converts starch to maltose. The enzyme pepsin converts proteins to amino acids.

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

List I
(Example)	List II
(Type)

A	2-chloro-1, 3-butadiene	I	Biodegradable polymer

B	Nylon 2-nylon 6	II	Synthetic Rubber

C	Polyacrylonitrile	III	Polyester

D	Dacron	IV	Addition Polymer

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

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

Solution:
2-chloro-1,3-butadiene can undergo polymerization to form a polymer known as polychloroprene. Polychloroprene is a synthetic rubber commonly known as neoprene.

Polyacrylonitrile (PAN) is an addition polymer. Addition polymers are formed through a polymerization process called addition polymerization, where monomers add to each other without the loss of any small molecules. Nylon-2-nylon-6 is Biodegradable polymer. Dacron is a type of polyester. Polyester is a polymer that belongs to the polyester family of polymers.

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

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

List I Industry	List II Waste Generated

(A)	Steel plants	(I)	Gypsum

(B)	Thermal power plants	(II)	Fly ash

(C)	Fertilizer Industries	(III)	Slag

(D)	Paper mills	(IV)	Bio-degradable wastes

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

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

Solution:
The various solid wastes in the form of slags and sludges that are emerged from steel plants are blast furnace slag. Fly ash is a fine by-product recovered from gases of burning coal in thermal power plants. Gypsum is the waste of fertiliser industry. Non-cellulosic part of plant is wastes of paper industry and it is biodegradable.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Match List I with List II:

List I

(Interacting species)

List II

(Name of Interaction)

A. A Leopard and a Lion in a forest/grassland	I. Competition

B. A Cuckoo laying egg in a Crow's nest	II. Brood parasitism

C. Fungi and root of a higher plant in Mycorrhizae	III. Mutualism

D. A cattle egret and a Cattle in a field	IV. Commensalism

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

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

Solution:
A Leopard and a Lion in a forest/grassland

This is an example of interspecific Competition. The two species fight for similar resources.

A Cuckoo laying egg in a Crow's nest

This is an example of Brood parasitism. The bird lays egg in the nest of crow.

Fungi and root of a higher plant in Mycorrhizae

Mutualism is a positive interaction where both the interacting species are benefitted.

Mycorrhizae help in better absorption of phosphate and minerals and in turn get shelter in the roots of higher plants.

A cattle egret and a cattle in a field

Commensalism is a type of interaction where one species is benefitted but the other is neutral. The canttle in this case remain neutral but the cattle erget feeds on ticks from the body of the grazing cattle.

Topic: Biology, Organisms and Populations, NEET, NEET 2023 (All India)
Match List I with List II:

List I	List II

(A)	3 Translational degrees of freedom	(I)	Monoatomic gases

(B)	3 Translational, 2 rotational degrees of freedoms	(II)	Polyatomic gases

(C)	3 Translational, 2 rotational and 1 vibrational degrees of freedom	(III)	Rigid diatomic gases

(D)	3 Translational, 3 rotational and more than one vibrational degrees of freedom	(IV)	Nonrigid diatomic gases

Choose the correct answer from the options given below:

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

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

Solution:
The formula for degree of freedom is given by

f=3N-k

where N is the number of particles, k is the individual relationship in the system.

For a monoatomic gas particle, N=1, k=0. So, f=3

Hence, monoatomic gas has 3 translational degree of freedom.

For a rigid diatomic gas molecule, there are 5 degrees of freedom, 3 translational and 2 rotational. A diatomic gas can have two extra degrees of freedom due to rotation along two independent axes.

A non rigid diatomic molecule has one extra degree of freedom than a rigid one. This is the vibrational degree of freedom. Hence, it has 6 degrees of freedom.

So, a non rigid diatomic gas has 3 translational, 2 rotational and 1 vibrational degrees of freedom.

A polyatomic gas has more than two atoms per molecule. Hence, the total number of degrees of freedom is 3 translational, 3 rotational and more than one vibrational degree of freedom.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Match List–I with List–II.

List–I	List–II

(a)	10 km height over earth's surface	(i)	Thermosphere

(b)	70 km height over earth's surface	(ii)	Mesosphere

(c)	180 km height over earth's surface	(iii)	Stratosphere

(d)	270 km height over earth's surface	(iv)	Troposphere

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

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

Solution:
Order of atmosphere stratification from bottom Troposphere, stralospherre, Mesosphre, Thermosphere
(a)→(iv)
(b)→(iii)
(c)→(ii)
(d)→(i)

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Match List I with List II

LIST I	LIST II

A	AM Broadcast	I	88-108 MHz

B	FM Broadcast	II	540-1600 kHz

C	Television	III	3.7-4.2 GHz

D	Satellite Communication	IV	54 MHz- 590 MHz

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

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

Solution:
AM broadcasting is radio broadcasting using amplitude modulation (AM) transmissions.

AM Broadcast range → 540–1600 kHz

FM broadcasting is the method of radio broadcasting that uses frequency modulation (FM).

FM Broadcast range →88 – 108 MHz

Television →54 – 890 MHz

Satellite communication →3.7–4.2 GHz

∴A-II, B-I, C-IV, D-III

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Match List I with List II

List I	List II

A	Angular momentum	I	ML2T-2

B	Torque	II	ML-2T-2

C	Stress	III	ML2T-1

D	Pressure gradient	IV	ML-1T-2

Choose the correct answer from the options given below :

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

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

Solution:
As we know, angular momentum is given by, L→=r→×p→⇒L=M0 L1 T0M1 L1 T-1

=M1 L2 T-1

Torque is given by,

τ→=r→×F→⇒τ=L1MLT-2

=ML2 T-2

Also, Stress ≡ Pressure =FA

Hence, [Stress] =ML-1 T-2

Pressure Gradient =dPdx

Therefore, [Pressure Gradient]=ML-2 T-2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Match List I with List II

List I	List II

A	Attenuation	I	Combination of a receiver and transmitter

B	Transducer	II	Process of retrieval of information from the carrier wave at receiver

C	Demodulation	III	Converts one form of energy into another

D	Repeater	IV	Loss of strength of a signal while propagating through a medium

Choose the correct answer from the options given below :

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

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

Solution:
Attenuation is the loss of signal strength in networking cables or connections.

Transducers are used in electronic communications systems to convert signals of various physical forms to electronic signals, and vice versa.

Demodulation is defined as extracting the original information-carrying signal from a modulated carrier wave.

Repeaters are used to extend transmissions so that the signal can cover longer distances or be received on the other side of an obstruction.

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Match List I with List II.

List I	List II

(a)	C→-A→-B→=0	(i)

(b)	A→-C→-B→=0	(ii)

(c)	B→-A→-C→=0	(iii)

(d)	A→+B→=-C→	(iv)

Choose the correct answer from the options given below:

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

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

Solution:
(a) C→=A→+B→

Option (iv)

(b) A→=B→+C→=C→+B→

Option (iii)

(c) B→=A→+C→

Option (i)

(d) A→+B→+C→=0

Option (ii)

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Match List-I with List-II

List-I	List-II

(a)	Ca(OCl)2	(i)	Antacid

(b)	CaSO4·12H2O	(ii)	Cement

(c)	CaO	(iii)	Bleach

(d)	CaCO3	(iv)	Plaster of paris

Choose the most appropriate answer from the

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

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

Solution:
Ca(OCl)2 is Bleach.
CaSO4·12H2O is plaster of paris.
CaCO3 is used as an antacid.

CaO is major component of cement.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Match List-I with List-II

List-I	List-II

(a)	Chlorophyll	(i)	Ruthenium

(b)	Vitamin-B12	(ii)	Platinum

(c)	Anticancer drug	(iii)	Cobalt

(d)	Grubbs catalyst	(iv)	Magnesium

Choose the most appropriate answer from the options given below:

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

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

Solution:
Chlorophyll is a coordination compound of magnesium. Vitamin B-12, cyanocobalamine is a coordination compound of cobalt. Cisplatin is used as an anti-cancer drug and is a coordination compound of platinum. Grubbs catalyst is a compound of Ruthenium.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (18 Mar Shift 1)