datavorous/entrance-exam-dataset huggingface dataset:
The total energy of particle, executing simple harmonic motion is

Options:
A
$\propto \mathbf{X}$

B
$\propto x^2$

C
independent of $x$

D
$\propto x^{1 / 2}$

The correct answer is:
independent of $x$

Solution:
In simple harmonic motion, as a particle is displaced from its mean position, its kinetic energy is converted to potential energy and vice versa and total energy remains constant. The total energy of simple harmonic motion is independent of $x$.

Topic: Physics, Oscillations, JEE Main, JEE Main 2004
The total energy of the body executing S. H. M. is \( T . E \). Then the kinetic energy when the displacement is half of the
amplitude, is

Options:
A \( \frac{T \cdot E .}{2} \)
B \( \frac{T \cdot E .}{4} \)
C \( \frac{3 T \cdot E .}{4} \)
D \( \frac{\sqrt{3}}{4} T . E \)

The correct answer is:
\( \frac{3 T \cdot E .}{4} \)

Solution:
Given,
Displacement from mean position x=A2

Suppose ω is angular velocity and A is amplitude of SHM.
Total energy in SHM is given by,
T.E.=12mω2A2

Velocity at a distance x from mean position in SHM is given by-
v=ωA2-x2

∴ Kinetic energyKE=12mv2=12mω2A2-x2
⇒KE=12mω2A2-A22=34×12mω2A2

⇒KE=34T.E.

Topic: Physics, Oscillations, JEE Main
The total internal energy change for a reversible isothermal cycles is

Options:
A Always 100 calories per degree
B Always negative
C 0
D Always positive

The correct answer is:
0

Solution:
\(\Delta \mathrm{E}=\mathrm{nC} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{T}=0\) for isothermal reversible cycle. This is because in an isothermal process, the temperature remains constant.

Topic: Chemistry, Thermodynamics (C), JEE Main
The total internal energy of two mole monoatomic ideal gas at temperature T=300 K will be _____ J. (Given R=8.31 J mol-1·K)

Options:

The correct answer is:
7479

Solution:
The internal energy of an ideal gas is a state function and is given by, U=nCvT.

For monatomic gas, Cv=32R.

Therefore, U=2×32×8.31×300=7479 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The total kinetic energy of 1 mole of oxygen at 27°C is :
[Use universal gas constant (R)=8.31 J mol-1 K-1 ]

Options:
A 6845.5 J
B 5942.0 J
C 6232.5 J
D 5670.5 J

The correct answer is:
6232.5 J

Solution:
The formula to calculate the kinetic energy can be written as

K=f2nRT ...1

Since oxygen is a diatomic gas, the number of degrees of freedom is 5.

Thus, from equation (1), it follows that

K=52×1×8.31×300 J=6232.5 J

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The total length of a sonometer wire fixed between two bridges is 110 cm. Now, two more bridges are placed to divide the length of the wire in the ratio 6:3:2. If the tension in the wire is 400 N and the mass per unit length of the wire is 0.01 kg m-1, then the minimum common frequency with which all the three parts can vibrate, is

Options:
A 1000Hz
B 1100Hz
C 100Hz
D 110Hz

The correct answer is:
1000Hz

Solution:
l 1 : l 2 : l 3    = 6:3:2

The frequancy in any nth mode for a segment is f=nv2l=constant

no of loops ∝ length of the segment

so, no of loops are in the ratio 6:3:2

hence total loops =11

The string is divided in 60 cm, 30 cm, & 20 cm part such that for minimum frequency, the wavelength is maximum

λ2=111x110=10cm
f = V λ = 1 λ · F μ = 1 0.2 4 0 0 0.01 = 1 0 0 0 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2014 (19 Apr Online)
The total length of a sonometer wire fixed between two bridges is 110 cm. Now, two more bridges are placed to divide the length of the wire in the ratio 6:3:2. If the tension in the wire is 400 N and the mass per unit length of the wire is 0.01 kg m-1, then the minimum common frequency with which all the three parts can vibrate, is

Options:
A 1000Hz
B 1100Hz
C 100Hz
D 110Hz

The correct answer is:
1000Hz

Solution:
l 1 : l 2 : l 3    = 6:3:2

The frequancy in any nth mode for a segment is f=nv2l=constant

no of loops ∝ length of the segment

so, no of loops are in the ratio 6:3:2

hence total loops =11

The string is divided in 60 cm, 30 cm, & 20 cm part such that for minimum frequency, the wavelength is maximum

λ2=111x110=10cm
f = V λ = 1 λ · F μ = 1 0.2 4 0 0 0.01 = 1 0 0 0 Hz

Topic: Physics, Waves and Sound, JEE Main
The total mechanical energy of a planet of mass m moving around the Sun along an ellipse depends only on its semi-major axis a. This energy as a function of a is given by

Options:
A - G Mâ¡ sâ¡ mâ¡ 2 a
B GM⁡s⁡m⁡2a
C GM⁡s⁡m⁡3a
D - G Mâ¡ sâ¡ mâ¡ 3 a

The correct answer is:
- G M⁡ s⁡ m⁡ 2 a

Solution:
- G m⁡ M⁡ s⁡ r⁡ 1 + 1 2 mv⁡ 1 2 = - G m⁡ M⁡ s⁡ r⁡ 2 + 1 2 mv⁡ 2 2
At point A,
Centripetal force mv⁡ 1 2 r⁡ 1 = G m⁡ M⁡ s⁡ r⁡ 1 2
or 12mv12=Gm⁡M⁡s⁡r⁡1 .... (i)
∴ Total energy, E at A⁡ = 1 2 mv 1 2 - G m⁡ M⁡ s r⁡ 1
∴ E⁡ = - G m⁡ M⁡ s⁡ 2 r⁡ 2 ....(ii)
From the property of ellipse,
r1 + r2 = 2a
From eqn. (i) and (ii), we get
E(r1 + r2) = – GMsm
or E⁡ × 2 a = - G M⁡ s⁡ m⁡
∴ E⁡ = - G M⁡ s⁡ m⁡ 2 a

Topic: Physics, Gravitation, JEE Main
The total mole of valence \( \mathrm{e}^{-} \)in \( 2.6 \mathrm{~g} \) of \( \mathrm{N}_{3}^{-} \)

Options:

The correct answer is:
1

Solution:
Given weight of N3- ion =2.6 g

Number of moles present in 2.6 g of azide ion.

=Given WeightMolar Weight=2.6 g(3×14 g)=2.6 g42 g=0.061 moles

Number of valence electrons in one atom of N=5

Therefore, the number of valence electrons in three atoms of N=15

Since the valency on the azide ion is -1, it means it has one extra electron on it.

So, the total number of valence electrons in azide ion is 15+1=16.

Mole of valence electrons in given mass of azide ion is=0.061×16 (valence e- of N3-) =0.99 ≈1

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The total number of 3×3 matrices A having enteries from the set 0,1,2,3 such that the sum of all the diagonal entries of AAT is 9, is equal to

Options:

The correct answer is:
766

Solution:
Let A=abcdefghi, AT=adgbehcfi

Diagonal elements of AAT is a2+b2+c2, d2+e2+f2, g2+b2+c2.

Sum =a2+b2+c2+d2+e2+f2+g2+h2+i2=9

a,b,c,d,e,f,g,h,i∈0,1,2,3

Case	No. of Matrices

(1)	All – 1's	9!9!=1

(2)	One →3's, remaining-0's	9!1!×8!=9

(3)

One-2's

Five-1's

Three-0's

9!1!×5!×3!=8×63

(4)	Two -2's
One-1's
Six-0's	9!2!×6!=63×4

Then, the total no. of ways =1+9+8×63+63×4

=766

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The total number of 3-digit numbers whose sum of digits is 10, is ..........

Options:

The correct answer is:
54

Solution:
Let xyz be the three-digit number such that

x+y+z=10, x≥1, y≥0, z≥0

Let x-1=t⇒x=1+t

So, x≥1⇒t+1≥1⇒t≥0

Also, x+y+z=10⇒t+y+z=10-1=9

So, t+y+z=9, 0≤t, y, z≤9

Thus, total number of non-negative integral solution =C3-19+3-1=C211=11.102=55

But for t=9, x=10, so required number of integers =55-1=54.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The total number of 5-digit numbers, formed by using the digits 1,2,3,5,6,7 without repetition, which are multiple of 6, is

Options:
A 72
B 48
C 24
D 60

The correct answer is:
72

Solution:
Given set 1,2,3,5,6,7

Here sum is 1+2+3+5+6+7 =24

So there will be two cases

Case (i) When numbers are 1,2,5,6,7

Arrangement will be _ _ _ _ 2 & _ _ _ _ 6

In unit place 2 & 6 can take place because for multiple of 6 number should be even & sum of digits should be multiple of 3.

So total way in which remaining number can be taken will be 4!×2 =48 ways

Case (ii) When numbers are 1,2,3,5,7

Arrangement will be _ _ _ _ 2

So total way in which remaining number can be taken will be 4!=24 ways

Adding case (i) & case (ii), we get

required 5-digit numbers =24+48=72

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The total number of 5 -digit numbers that can be composed of distinct digits from 0 to 9 is

Options:
A 45360
B 30240
C 27216
D 15120

The correct answer is:
27216

Solution:
Number of 5 digits numbers with all distinct digit is same as filling of 5 vacent placed out of 10 boxes. First digit of any number can be choosen in 9 ways. Remaining 4 digits can be choosen in remaining 9 digits in ${ }^{9} \mathrm{P}_{4}$ ways. Total number of such number
$=9 \times 9 \times 8 \times 7 \times 6=27216$

Topic: Mathematics, Permutation Combination, NDA, NDA 2018 (Phase 2)
The total number of acidic oxides from the following list is:

NO, N2O, B2O3, N2O5, CO, SO3, P4O10

Options:
A 5
B 4
C 3
D 6

The correct answer is:
4

Solution:
Based on their acid-base characteristics oxides are classified as acidic or basic. An oxide that combines with water to give an acid is termed as an acidic oxide. The oxide that gives a base in water is known as a basic oxide.

P4O10, B2O3, N2O5, SO3 are acidic oxides.

NO, N2O, CO are neutral oxides.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (25 Jul Shift 2)
The total number of amines among the following which can be synthesized by Gabriel synthesis is ________.

Options:

The correct answer is:
3

Solution:
Gabriel phthalimide synthesis is used to prepare 1° aliphatic/alicyclic amine in common.

Hence, amine which can synthesised by Gabriel phthalimide synthesis method is:

(A) Me2CH-CH2-NH2

(B) CH3CH2NH2

(C) Ph-CH2-NH2

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The total number of $\sigma$ and $\pi$-bonds in pyrophosphoric acid are respectively

Options:
A 8, 2
B 10, 2
C 12, 2
D 8, 4

The correct answer is:
12, 2

Solution:
The structure of pyrophosphoric acid $\left(\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7\right)$ is as
Hence, it contains $12 \sigma$ and $2 \pi$-bonds.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The total number of σ and π-bonds present in the following compound are

Options:
A σ=18, π=3
B σ=21, π=4
C σ=23, π=5
D σ=16, π=4

The correct answer is:
σ=21, π=4

Solution:
Count the single bonds and double bonds.

Single bonds are called σ-bonds and double bond contain one σ-bond and a π-bond.

Benzene ring contain three pi bonds in the ring and one more pi bond is carbonyl group of ester. Remaining all are sigma bonds.

Topic: Chemistry, Hydrocarbons, AP EAMCET, AP EAMCET 2020 (23 Sep Shift 1)
The total number of anti bonding molecular orbitals, formed from 2s and 2p atomic orbitals in a diatomic molecule is _____________.

Options:

The correct answer is:
4

Solution:
The atomic orbitals combine to form a new set of equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, hybrid orbitals are used in bond formation. The phenomenon is known as hybridisation, which can be defined as the process of intermixing of the orbitals of slightly different energies to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.

The number of hybrid orbitals is equal to the number of atomic orbitals that get hybridised.

Antibonding molecular orbital from 2s=1 that is σ*2s

Antibonding molecular orbital from 2p=3 that is σ*2px,π*2py and π*2pz.

Total =4

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The total number of boron-oxygen bonds in borax is ′x′ and boron-oxygen-boron bonds are ′y′. Then the value of x-y is:

Options:

The correct answer is:
9

Solution:
B-O bonds =14
B-O-B bonds =5
∴x-y=14-5=9

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The total number of C-C sigma bond/s in mesityl oxide C6H10O is (Round off to the Nearest Integer).

Options:

The correct answer is:
5

Solution:
sigma bonds are the strongest type of covalent chemical bond. They are formed by head-on overlapping between atomic orbitals. Sigma bonding is most simply defined for diatomic molecules using the language and tools of symmetry groups.

Mesityle oxide

∴Cσ¯C=5

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The total number of carboxylic acid groups in the product P is

Options:

The correct answer is:
2

Solution:
Hence in the final product number of -COOH groups is 2.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The total number of carboxylic acid groups in the product P is

Options:

The correct answer is:
2

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The total number of chiral compound/s from the following is

Options:

The correct answer is:
2

Solution:
Following points are to be remembered when checking for chirality in a compound:
1.Compound should have a stereogenic centre.
2. Compound shouldn't have any plane of symmetry as it can convert a chiral compound to a meso compound.
Based on the above two facts, out of the following, only 2 are chiral compounds:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The total number of compounds having at least one bridging oxo group among the molecules given below is
N2O3, N2O5, P4O6, P4O7, H4P2O5, H5P3O10, H2S2O3, H2S2O5

Options:

The correct answer is:
5.00

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The total number of contributing structure showing hyperconjugation (involving \( \mathrm{C}-\mathrm{H} \) bond) for the following carbocation

Options:

The correct answer is:
6

Solution:
The alpha carbon in organic molecules refers to the first carbon atom that attaches to a functional group, such as a carbonyl or sp2 carbon.

Number of α-hydrogen=6

The number of hyperconjugation structures = Number of alpha hydrogens .

Topic: Chemistry, General Organic Chemistry, JEE Main
The total number of correct statements, regarding the nucleic acids is _________.
A. RNA is regarded as the reserve of genetic information.
B. DNA molecule self-duplicates during cell division
C. DNA synthesizes proteins in the cell.
D. The message for the synthesis of particular proteins is present in DNA
E. Identical DNA strands are transferred to daughter cells.

Options:

The correct answer is:
3

Solution:
A. RNA (ribonucleic acid) is not typically regarded as the reserve of genetic information in the same way that DNA (deoxyribonucleic acid) is. While RNA does play important roles in gene expression, protein synthesis, and regulation of cellular processes, it is DNA that is traditionally considered the primary repository of genetic information in cells.

B. DNA molecule self-duplicates during cell division. (True)

DNA (deoxyribonucleic acid) self-duplicates during cell division through a process called DNA replication. DNA replication is a fundamental process that ensures each daughter cell receives an identical copy of the genetic information from the parent cell.

C. While DNA contains the genetic instructions necessary for synthesizing proteins, it does not directly synthesize proteins itself. Instead, the process by which proteins are synthesized from the genetic information encoded in DNA involves several steps and intermediates, collectively known as the central dogma of molecular biology.

D. The message for the synthesis of particular proteins is present in DNA. (True)

The message for the synthesis of particular proteins is present in DNA in the form of a specific sequence of nucleotide bases. This sequence is transcribed into mRNA, which carries the genetic information to the ribosomes for translation into a specific sequence of amino acids, ultimately leading to the synthesis of a particular protein.

E. Identical DNA strands are transferred to daughter cells. (True)

Identical DNA strands are transferred to daughter cells through the processes of DNA replication, chromosome segregation, and cell division. These processes ensure that each daughter cell receives a complete and identical set of genetic information from the parent cell.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The total number of cyclic alcohols with molecular formula C4H7OH are

Options:

The correct answer is:
5

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The total number of cyclic isomers of formula C6H12 is/are

Options:

The correct answer is:
11

Solution:
Total 11 cyclic isomer are shown below here.

Topic: Chemistry, General Organic Chemistry, JEE Main
The total number of diprotic acids among the following is
$\begin{array}{lll}\mathrm{H}_3 \mathrm{PO}_4 & \mathrm{H}_2 \mathrm{SO}_4 & \mathrm{H}_3 \mathrm{PO}_3 \\ \mathrm{H}_2 \mathrm{CO}_3 & \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 & \mathrm{H}_3 \mathrm{BO}_3 \\ \mathrm{H}_3 \mathrm{PO}_2 & \mathrm{H}_2 \mathrm{CrO}_4 & \mathrm{H}_2 \mathrm{SO}_3\end{array}$

Options:

The correct answer is:
6

Solution:
$\mathrm{H}_2 \mathrm{SO}_4, \mathrm{H}_2 \mathrm{CO}_3, \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7, \mathrm{H}_2 \mathrm{CrO}_4, \mathrm{H}_3 \mathrm{PO}_3, \mathrm{H}_2 \mathrm{SO}_3$
Acid-base-salt
Basic structural understanding II

Topic: Chemistry, Ionic Equilibrium, JEE Main
The total number of divisors of the number N=25⋅34⋅510⋅76 that are of the form 4K+2, ∀K∈N is equal to

Options:
A 385
B 384
C 96
D 77

The correct answer is:
384

Solution:
The required number is of the form 22K+1, i.e. it will contain exactly one power of 2.

Also, we have to exclude 2 as K∈N.
Number of required divisors
=1⋅5⋅11⋅7-1=385-1=384

Topic: Mathematics, Permutation Combination, JEE Main
The total number of electrons in all bonding molecular orbitals of O22- is ......... (Round off to the nearest integer)

Options:

The correct answer is:
10

Solution:
M. O. Configuration of O22- ( (18e¯)

σ1 s2σ*1 s2σ2 s2σ*2 s2σ2pz2π2px2=π2py2π*2px2=π*2py2

Total B.M.O electrons =10

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The total number of four digit numbers such that each of the first three digits is divisible by the last digit, is equal to ______.

Options:

The correct answer is:
1086

Solution:
Let the last be d, so total following cases will be there for first three numbers,

No. of such numbers

d=1,	9×10×10=900

d=2	4×5×5=100

d=3	3×4×4=48

d=4	2×3×3=18

d=5	1×2×2=4

Similarly for

d=6, 7, 8, 9

4×4=16

1086

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The total number of hydrogen atoms in product A and product B is__________.

Options:

The correct answer is:
10

Solution:
When bromoethane reacts with NaOH in the presence of ethyl alcohol gives Ethene.

Similarly When bromoethane reacts with NaOH in the aqueous medium gives Ethyl alcohol. This reaction is an example of a nucleophilic substitution reaction.

The reaction is as follows,

Total number of hydrogen atom in A that is Ethene and B that is Ethyl alcohol is 10

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The total number of injections (one-one into mappings) from $\quad\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$
$\left(b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\right)$ is

Options:
A 400
B 420
C 800
D 840

The correct answer is:
840

Solution:
Let $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$
$\begin{aligned} B &=\left\{b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\right\} \\ n(A) &=4, n(B)=7 \end{aligned}$
The total number of infections $={ }^{7} P_{4}$ $=\frac{71}{3 !}=7 \cdot 6 \cdot 5 \cdot 4=840$

Topic: Mathematics, Functions, JEE Main
The total number of intensive properties from the following is...........
Volume, Molar heat capacity, molarity, Ecello,Gibbs free energy change, Molar mass, Mole

Options:

The correct answer is:
4

Solution:
An intensive property is a physical quantity whose value does not depend on the amount of substance which was measured.

Molar heat capacity, molarity, Ecello and molar mass are intensive properties.

An extensive property of a system depends on the system size or the amount of matter in the system.

Volume, Gibbs free energy change, Mole are extensive properties.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (11 Apr Shift 2)
The total number of irrational terms in the binomial expansion of 715-311060 is

Options:
A 48
B 55
C 54
D 49

The correct answer is:
54

Solution:
General term is 60Cr71560-r-3110r

We will get rational term when r is multiple of 10 and 60-r is multiple of 5.

∴r=0, 10, 20, 30, 40 ,50, 60

∴Number of irrational term = Total terms - Number of rational terms

=60+1-7=54.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The total number of isoelectronic species from the given set is ________.

O2-, F–, Al, Mg2+, Na+, O+, Mg, Al3+, F

Options:

The correct answer is:
5

Solution:
Isoelectronic are those which have the same number of electrons. Let us write the number of electrons from the given species.

Species	No. of electrons

O–2	10

F-	10

Al	13

Mg2+	10

Na+	10

O+	7

Mg	12

Al3+	10

F	9

Hence five species among the given are isoelectronic species.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The total number of isomers for a square planar complex: MClFNO2SCN is:

Options:
A 12
B 16
C 4
D 8

The correct answer is:
12

Solution:
Possible Structural isomers	No of geometrical isomer by each

MClFNO2SCN	3

MClFONOSCN	3

MClFNO2NCS	3

MClFONONCS	3

Total number of isomers will be 12 .

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The total number of isomers(including stereo) for the compound with molecular formula C2BrClFI is

Options:

The correct answer is:
6

Solution:
The compound with molecular formula C2BrClFI has following Six geometrical (E, Z type) isomers

Topic: Chemistry, General Organic Chemistry, JEE Main
The total number of isotopes of hydrogen and number of radioactive isotopes among them, respectively, are

Options:
A 2 and 1
B 2 and 0
C 3 and 1
D 3 and 2

The correct answer is:
3 and 1

Solution:
Hydrogen is the first element in the periodic table and has the atomic number one. Those elements which have the same atomic number but a different mass number are called isotopes. There are three isotopes of hydrogen namely, protium H11, deuterium H12 or D and lastly tritium H13 or T. The isotopes are different because of the different number of neutrons present in them.

Isotopes of Hydrogen:
Among all three isotopes only Tritium is radioactive.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The total number of lone pairs of electrons in N2O3 are

Options:

The correct answer is:
8

Solution:
The total number of lone pairs of electrons in N2O3 is 8 as per as its Lewis structure given below

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The total number of lone pairs of electrons on oxygen atoms of ozone is

Options:

The correct answer is:
6

Solution:
From the above Lewis structure of ozone we can see,

There are three oxygen atoms having two and three lone pairs on terminal oxygen atoms and one lone pair central oxygen atom respectively.

Total number of lone pairs on oxygen atoms =6

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The total number of matrices A=02y12xy-12x-y1, x, y∈R, x≠y for which ATA=3I3 is:

Options:
A 6
B 3
C 4
D 2

The correct answer is:
4

Solution:
Given ATA=3I3

02x2x2yy-y1-1102y12xy-12x-y1=300030003

8x20006y20003=300030003

On comparing the two matrices, we get

8x2=3 and 6y2=3

⇒x2=38 and y2=12

⇒x=±38 and y=±12

For each of x & y we have two possible values, hence there are total 2×2=4 combinations of values are possible.

Hence, total 4 matrices are possible.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The total number of Mn=O bonds in Mn2O7 is

Options:
A 4
B 5
C 6
D 3

The correct answer is:
6

Solution:
The manganese heptoxide solubility properties indicate that it is a nonpolar molecular species, which is confirmed by its structure. The molecules consist of a pair of tetrahedra that share a common vertex. The vertices are occupied by oxygen atoms and at the centres of the tetrahedra are the Mn centres. The connectivity is indicated by the formula O3Mn−O−MnO3.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (27 Jul Shift 1)