datavorous/entrance-exam-dataset huggingface dataset:
Number of molecules having bond order 2 from the following molecules is ________ \(\mathrm{C}_2, \mathrm{O}_2, \mathrm{Be}_2, \mathrm{Li}_2, \mathrm{Ne}_2, \mathrm{~N}_2, \mathrm{He}_2\)

Options:

The correct answer is:
2

Solution:
$\begin{aligned} & \mathrm{C}_2 \\ & \left(12 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma^* 2 \mathrm{~s}^2\left[\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2\right] \\ & \text { B.O. }=\frac{8-4}{2}=2 \\ & \mathrm{O}_2 \\ & \left(16 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma^* 2 \mathrm{~s}^2, \sigma 2 \mathrm{pz}^2 \\ & {\left[\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2\right]\left[\pi * 2 \mathrm{p}_{\mathrm{x}}^1=\pi * 2 \mathrm{p}_{\mathrm{y}}^1\right]} \\ & \text { B.O. }=\frac{10-6}{2}=2 \\ & \text { Be }{ }_2 \\ & \left(8 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma^* 2 \mathrm{~s}^2 \\ & \text { B.O. }=\frac{4-4}{2}=0 \\ & \mathrm{Li}_2 \\ & \left(6 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2\end{aligned}$
$\begin{aligned} & \text { B.O. }=\frac{4-2}{2}=1 \\ & \mathrm{Ne}_2 \\ & \left(20 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma^* 2 \mathrm{~s}^2, \sigma 2 \mathrm{pz}^2 \\ & {\left[\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2\right]\left[\pi * 2 \mathrm{p}_{\mathrm{x}}^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}^2\right] \sigma^* 2 \mathrm{p}_{\mathrm{z}}^2} \\ & \text { B.O. }=\frac{10-10}{2}=0 \\ & \mathrm{~N}_2 \\ & \left(14 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2, \sigma 2 \mathrm{~s}^2, \sigma^* 2 \mathrm{~s}^2\left[\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2\right] \sigma 2 \mathrm{p}_{\mathrm{z}}^2 \\ & \text { B.O. }=\frac{10-4}{2}=6 \\ & \mathrm{He}_2 \\ & \left(4 \mathrm{e}^{-}\right): \sigma 1 \mathrm{~s}^2, \sigma^* 1 \mathrm{~s}^2 \\ & \text { B.O. }=\frac{2-2}{2}=0\end{aligned}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Number of molecules in a volume of 4 cm3 of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to? (Given, mean kinetic energy of a molecule (at T) is 4×10−14 erg, g=980 cm s-2 density of mercury =13.6 g cm-3)

Options:
A 4.8×1018
B 4.0×1016
C 5.8×1016
D 5.8×1018

The correct answer is:
4.8×1018

Solution:
N=PVKT

U=32KT

∴ N=3PV2U=3.99×1018

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (05 Sep Shift 1)
Number of molecules/species from the following having one unpaired electron is \(\qquad\) \(\mathrm{O}_2, \mathrm{O}_2^{-1}, \mathrm{NO}, \mathrm{CN}^{-1}, \mathrm{O}_2^{2-}\)

Options:

The correct answer is:
2

Solution:
According to M.O.T.
$\mathrm{O}_2 \rightarrow$ no. of unpaired electrons $=2$
$\mathrm{O}_2^{-} \rightarrow$ no. of unpaired electron $=1$
$\mathrm{NO} \rightarrow$ no. of unpaired electron $=1$
$\mathrm{CN}^{-} \rightarrow$ no. of unpaired electron $=0$
$\mathrm{O}_2^{2-} \rightarrow$ no. of unpaired electron $=0$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Number of moles of H+ ions required by 1 mole of MnO4− to oxidise oxalate ion to CO2 is ______.

Options:

The correct answer is:
8

Solution:
The reaction between MnO4- ​ with C2O42- ​ is as given below:

2MnO4-+16H++5C2O42- →2Mn+2+8H2O+10CO2

From this, we can see that 16 moles of H+react with 2 moles of MnO4-​ . Hence, one mole of permanganate requires eight moles of H+ ions

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Number of moles of methane required to produce 22g CO2(g) after combustion is x×10–2 moles. The value of x is

Options:

The correct answer is:
50

Solution:
CH4+2O2→CO2+2H2O

Gram molecular mass of CO2=12+2(16)=44 g/mol

From the reaction, it is clear that 1 mole of methane on complete combustion produces 44 g (1 mole) of carbon dioxide.

Therefore moles of CH4 required to produce 22 g of CO2 are:

=144×22=0.5 mol =50×10-2 mol

Hence, m=50

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Number of moles of $\mathrm{MnO}_4^{-}$required to oxidize one mole of ferrous oxalate completely in acidic medium will be

Options:
A 0.2 moles
B 0.6 moles
C 0.4 moles
D $7.5 \mathrm{moles}$

The correct answer is:
0.6 moles

Solution:
$$
\begin{aligned}
& \left.\mathrm{FeC}_2 \mathrm{O}_4 \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{CO}_2+3 \mathrm{e}^{-}\right] \times 5 \\
& \left.\mathrm{MnO}_4{ }^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 3 \\
& 5 \mathrm{FeC}_2 \mathrm{O}_4+3 \mathrm{MnO}_4^{-}+24 \mathrm{H}^{+} \longrightarrow 5 \mathrm{Fe}^{3+}+10 \mathrm{CO}_2+3 \mathrm{Mn}^{2+}+12 \mathrm{H}_2 \mathrm{O} \\
& 5 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv 3 \mathrm{~mol} \mathrm{MnO}_4^{-} \\
& \therefore 1 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv \frac{3}{5} \mathrm{~mol} \mathrm{MnO}_4^{-} \\
& =0.6 \mathrm{~mol} \mathrm{MnO}_4^{-} \\
&
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, JEE Main
Number of moles of $\mathrm{MnO}_4^{-}$required to oxidize one mole of ferrous oxalate completely in acidic medium will be

Options:
A 0.2 moles
B 0.6 moles
C 0.4 moles
D $7.5 \mathrm{moles}$

The correct answer is:
0.6 moles

Solution:
$$
\begin{aligned}
& \left.\mathrm{FeC}_2 \mathrm{O}_4 \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{CO}_2+3 \mathrm{e}^{-}\right] \times 5 \\
& \left.\mathrm{MnO}_4{ }^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 3 \\
& 5 \mathrm{FeC}_2 \mathrm{O}_4+3 \mathrm{MnO}_4^{-}+24 \mathrm{H}^{+} \longrightarrow 5 \mathrm{Fe}^{3+}+10 \mathrm{CO}_2+3 \mathrm{Mn}^{2+}+12 \mathrm{H}_2 \mathrm{O} \\
& 5 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv 3 \mathrm{~mol} \mathrm{MnO}_4^{-} \\
& \therefore 1 \mathrm{~mol} \mathrm{FeC}_2 \mathrm{O}_4 \equiv \frac{3}{5} \mathrm{~mol} \mathrm{MnO}_4^{-} \\
& =0.6 \mathrm{~mol} \mathrm{MnO}_4^{-} \\
&
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, NEET, NEET 2008 (Mains)
Number of nodal centres for $2 s$ orbital

Options:
A 1
B 0
C 4
D 3

The correct answer is:
1

Solution:
Number of nodal centre for $2 s$ orbitals $(n-1)=2-1=1$.

Topic: Chemistry, Structure of Atom, JEE Main
Number of optical isomers possible for 2 – chlorobutane ........

Options:

The correct answer is:
2

Solution:
The number of optical isomers possible for a compound is 2n where n= number of asymmetric carbon atoms.

As n=1 for 2-chlorobutane, it is an optically active molecule with one chiral carbon atom and it forms dextro and laevo isomers.

2n=21=2.

Hence, it has two optical isomers.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Number of oxygen atoms present in chemical formula of fuming sulphuric acid is
______.

Options:

The correct answer is:
7

Solution:
Fuming sulphuric acid is a mixture of
conc. $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{SO}_3$ Or $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$
So, Number of Oxygen atoms = 7

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2024 (09 Apr Shift 2)
Number of oxygen atoms shared per SiO 4 4 - tetrahedron in single chain silicates are ______

Options:

The correct answer is:
2.00

Solution:
In case of simple chain silicate, two corners of each tetrahedron are shared and they form a long chain of tetrahedron. Their general formula is also same as the cyclic silicates i.e., (SiO 3 ) n 2n−

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Number of paramagnetic oxides among the following given oxides is ___________.

Li2O, CaO, Na2O2, KO2, MgO and K2O

Options:
A 1
B 3
C 0
D 2

The correct answer is:
1

Solution:
Compound	Nature

K2O	diamagnetic

Li2O	diamagnetic

MgO	diamagnetic

CaO	diamagnetic

Na2O2	diamagnetic

KO2	paramagnetic

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Number of particles, given by n=-Dn2-n1x2-x1, is crossing a unit area perpendicular to x-axis in unit time, where n1 and n2 are number of particles per unit volume for the value of x meant to x2 and x1 . Find dimensions of D, called as diffusion constant.

Options:
A M0 LT2
B M0 L2 T-4
C M0 LT-3
D M0 L2 T-1

The correct answer is:
M0 L2 T-1

Solution:
Given, number of particle passing from unit area in unit time =n
=Number of particleA×t=M0 L0 T0L2 T=L-2 T-1
n1=n2=Number of particle in unit volume=L-3
Now, from the given formula D=n x2-x1n2-n1
=L-2 T-1 LL-3
=L2 T-1

Topic: Physics, Units and Dimensions, JEE Main
Number of photons having wavelength 632.8 nm, emitted by 5 mW laser source in 1 second is

Options:
A 1.6×1019
B 1.6×1016
C 1.6×1025
D 1.6×1013

The correct answer is:
1.6×1016

Solution:
E=nhcÎ»
5Ã10-3 J/s=nÃ1240632.8Ã1.6Ã10-19 J
n=1.6Ã1016

Topic: Chemistry, Structure of Atom, JEE Main
Number of P-N junctions that are there in a bipolar junction transistor is

Options:
A \( 0 \)
B \( 1 \)
C \( 2 \)
D \( 3 \)

The correct answer is:
\( 2 \)

Solution:
A Bipolar Junction Transistor (BJT) is a three terminal circuit or device that amplifies flow of current. There are two types of transistors: NPN and PNP junction. In each type, there will be 2 P-N junctions present (one type of semiconductor is sandwiched between two of other type).

∴Number of P-N junctions in a dipolar junction transistor is 2.

Topic: Physics, Semiconductors, JEE Main
Number of points from where perpendicular tangents to the curve x216-y225=1 can be drawn, is/are

Options:
A 1
B 2
C 0
D infinite

The correct answer is:
0

Solution:
Director circle is the set of points from where drawn tangents are perpendicular, in this case x2+y2=a2-b2 (equation of director circle) i.e., x2+y2=-9 is not a real circle, so there is no point from where perpendicular tangents can be drawn.

Topic: Mathematics, Hyperbola, JEE Main
Number of possible isomers (stereoisomers) of D-glucose are

Options:
A 10
B 14
C 8
D 20

The correct answer is:
8

Solution:
Chiral centers are four so 24 = 16 stereoisomers. Hence, 16 isomers are for glucose, and half of them will be for D-glucose and other half for L-glucose.

Topic: Chemistry, General Organic Chemistry, JEE Main
Number of products in the given reaction
C6H5CHO+CH3–CHO→O⊕HPr⁡oducts
Will be

Options:

The correct answer is:
2

Solution:
Here 2 Aldol products are possible as follows as only CH3CHO has α -H-atoms.
(i) CH3–CHO→O⊕HC⊕H2–CHO→CH3CHOCH3CHOH–CH2–CHO
(ii) CH3–CHO→O⊕HC⊕H–CHO→C6H5CHOC6H5CHOH–CH2–CHO

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Number of protons, neutrons and electrons in the element ${ }^{239} Y$ is

Options:
A $89,231,89$
B $89,89,242$
C $89,142,89$
D $89,71,89$

The correct answer is:
$89,142,89$

Solution:
In number of protons and electrons is 89 and No. of neutrons $=A-Z=$ $231-89=142$

Topic: Chemistry, Structure of Atom, JEE Main
Number of roots of the equation cos2⁡x+3+12sin⁡x-34-1=0 which lie in the interval -π,π is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
Given equation is
1-sin2â¡x+3+12sinâ¡x-34-1=0
â sin2â¡x-3+12sinâ¡x+34=0;4sin2â¡x-23sinâ¡x-2sinâ¡x+3=0
2sinx( 2sinxâ 3 )â( 2sinxâ 3 )=0
â( 2sinxâ1 )( 2sinxâ 3 )=0
On solving we get sinâ¡x=12 ;32
x=Ï6,5Ï6 ;Ï3,2Ï3

Topic: Mathematics, Trigonometric Equations, JEE Main
Number of \(\sigma\) and \(\pi\) bonds present in ethylene molecule is respectively :

Options:
A 4 and 1
B 5 and 2
C 3 and 1
D 5 and 1

The correct answer is:
5 and 1

Solution:
it has $5 \sigma$ bonds and $1 \pi$ bond.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Number of sigma bonds in $\mathrm{P}_4 \mathrm{O}_{10}$ is

Options:
A
6

B
7

C
17

D
16

The correct answer is:
16

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2002
Number of sigma bonds in $\mathrm{P}_4 \mathrm{O}_{10}$ is

Options:
A
6

B
7

C
17

D
16

The correct answer is:
16

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Number of sigma bonds in P4O10 is

Options:
A 6
B 7
C 17
D 16

The correct answer is:
16

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Number of solutions of 2sin(|x|)=3|cosx| in [ −π,π ], is equal to

Options:

The correct answer is:
4

Solution:
In the interval âÏ, Ï all the values of sin|x| are positive as well as |cosx|
Hence in âÏ, Ï, equation gets reduced to,

2sinâ¡x=3cosâ¡x
Taking log on both sides,
sinâ¡xlogâ¡2=cosâ¡xlogâ¡3
â tanâ¡x=logâ¡3/logâ¡2 , value of tanx are positive in 1st and 3rd quadrant and negative in 2nd and 4th quadrant but positive for tanâ¡x in all the values of x in âÏ, Ï.

Hence, tanx will repeat its value in all 4 quadrants, so the number of solutions are 4 .

Topic: Mathematics, Trigonometric Equations, JEE Main
Number of spectral lines obtained in He+ spectra, when an electron makes transition from fifth excited state to first excited state will be

Options:

The correct answer is:
10

Solution:
The formula used here is:

Number of spectral lines =∑n2-n1

n2= Highest energy level=6(fifth excited state)

n1=Lowest energy level=2(first excited state)

Number of spectral lines

=∑6-2=10

Number of spectral lines in He+ for transition from n=6 to n=2 are 10

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Number of stereo centers present in linear and cyclic structures of glucose are respectively:

Options:
A 4 and 4
B 5 and 5
C 4 and 5
D 5 and 4

The correct answer is:
4 and 5

Solution:
The respective structures are (with for stereocentres),
Thus, no. of stereocentres respectively are 4 and 5.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Number of stereoisomers possible for the following compound is

Options:

The correct answer is:
4

Solution:
1, 2 have plane of symmetry and are optically inactive. 3 is optically active as there is no plane of symmetry.

Topic: Chemistry, General Organic Chemistry, JEE Main
Number of the following statements is/are correct?
A Greater the stability constant of a complex ion, greater is its stability
B Greater the oxidation state of the central metal ion, greater is the stability of the complex
C CO stabilizes complex because of its synergic Bonding.
D Chelate complexes have low stability constants

Options:

The correct answer is:
3

Solution:
A Greater the stability constant of a complex ion, greater is its stability
B Greater the oxidation state of the central metal ion, greater is the stability of the complex
C CO stabilizes complex because of its synergic Bonding.
Above three are correct statements.

Topic: Chemistry, Coordination Compounds, JEE Main
Number of unpaired electrons in $1 s^2 2 s^2 2 p^3$ is

Options:
A 2
B 0
C 3
D 1

The correct answer is:
3

Solution:
In this type of electronic configuration the number of unpaired electrons are 3.
= 3

Topic: Chemistry, Structure of Atom, JEE Main
Number of values of \( x \in(-\pi, 3 \pi)-\left\{\frac{-\pi}{2}, 0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}, 2 \pi, \frac{5 \pi}{2}\right\} \) satisfying the equation \( \tan x \cot \frac{x}{2}=2-\cot x \tan \frac{x}{2} \) is/are

Options:

The correct answer is:
0

Solution:
tanx cotx2+tanx2cotx=2

Let tanx cotx2=t

⇒t+1t=2

⇒t-12=0

⇒t=1

⇒tanx cotxx=1

⇒sinx cosx2=cosx sinx2

⇒sinx2=0

⇒x2=0, π as x2∈-π2, 3π2

⇒x=0, 2π both are not possible.

Topic: Mathematics, Trigonometric Equations, JEE Main
Number of water molecules in washing soda and soda ash respectively are :

Options:
A 1 and 10
B 10 and 1
C 10 and 0
D 1 and 0

The correct answer is:
10 and 0

Solution:
The chemical formula for washing soda is Na2CO3.10H2O, which means that each molecule of washing soda contains 10 water molecules.

The chemical formula for soda ash is Na2CO3, which does not contain any water molecules.

Therefore, there are 10 water molecules in washing soda and 0 water molecules in soda ash.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Number of ways in which \( 5 \) boys and \( 4 \) girls can be arranged on a circular table such that no two girls sit together and two particular boys are always together.

Options:
A \( 276 \)
B \( 288 \)
C \( 296 \)
D \( 304 \)

The correct answer is:
\( 288 \)

Solution:
Total number of ways = (3!) (2!) (4!) = 288

Topic: Mathematics, Permutation Combination, JEE Main
Number of wildlife is continuously decreasing. What is the main reason of this?

Options:
A Predation
B Cutting down of forest
C Destruction of habitat
D Hunting

The correct answer is:
Destruction of habitat

Solution:
Habitat loss and fragmentation is a very important cause of species extinction. The deforestation of tropical forests is expected to be the greatest cause of mass extinctions caused by human activity. Cutting down forests and hedgerows to provide agricultural land has increased. The loss of tropical forests, which holds $50 \%$ of the world's biodiversity proving to be the most significant threat to species in the modern age. This leads to fragmentation of habitats and species which require large habitats such as bears cannot survive if the area is too small.
Related Theory
The accelerated rate of species extinction that the world is facing now is largely due to human activities. This is due to Habitat loss and fragmentation, overexploitation, introduction or invasion of alien species, pollution, climate change and co-extinction.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2002
Number of words from the letters of the words BHARAT in which $B$ and $\mathrm{H}$ will never come together is

Options:
A 210
B 240
C 422
D 400

The correct answer is:
240

Solution:
There are 6 letters in the word BHARAT, 2 of them are identical. Hence total number of words $=6 ! / 2 !=360$
Number of words in which $B$ and $\mathrm{H}$ come together
$=\frac{5 ! 2 !}{2 !}=120$
final ans $=360-120=240$

Topic: Mathematics, Permutation Combination, JEE Main
Number of words that can be formed with the letters of the word ALGEBRA so that all the vowels are separated (or no two vowels come together) is

Options:
A 720
B 2160
C 1440
D 1200

The correct answer is:
720

Solution:
Vowels are A,E,A
Consonants are L,G,B,R
XLXGXBXRX
Since there is no restriction on consonants, we must place them first and they can be arranged in 4!=24 days.
Now to arrange vowels so that no two of them are together we have 5 places X. Out of these 5 places, 3 can be selected in 5C3 ways and then one arrangement can be made in 3!2! ways.
Thus, required ways =4!×5C3×3!2!=720

Topic: Mathematics, Permutation Combination, JEE Main
Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1,2,3,4,5 and 6 without repetition of digits. Then the total number of such numbers is _______.

Options:

The correct answer is:
30

Solution:
To find total numbers between 1000 and 3000 divisible by 4 using the digits 1,2,3,4,5,6,

We will solve in two cases.

Case I : When first digit is 1.

Then last two digits can be 24,32,36,52,56 or 64

So, total ways of choosing last two digit is 6 and second digit will be chosen in 3 ways

So, number of such numbers =6×3=18

Case II: When first digit is 2

Then last two digits can be 16,36,56 or 64

So, total ways of choosing last two digit is 4 and second digit will be chosen in 3

So, number of such numbers =4×3=12

Total numbers of numbers =18+12=30

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Nylon threads are made of

Options:
A
polyester polymer

B
polyamide polymer

C
polyethylene polymer

D
polyvinyl polymer

The correct answer is:
polyamide polymer

Solution:
Nylon is a polyamide polyer

Topic: Chemistry, Polymers, JEE Main, JEE Main 2003
Nylon-66 is an example of

Options:
A Poly propylene
B Polyester
C Polyamide
D Polystyrene

The correct answer is:
Polyamide

Solution:
Nylon-66 is a polyamide fibre.

Topic: Chemistry, Polymers, JEE Main
O2F2 is an unstable yellow orange solid and H2O2 is colourless liquid. O-O bond length in H2O2 and O2F2 respectively are

Options:
A 1.22 Å, 1.48 Å
B 1.48 Å, 1.22 Å
C 1.22 Å, 1.22 Å
D 1.48 Å, 1.48 Å

The correct answer is:
1.48 Å, 1.22 Å

Solution:
x1>x2
According to Bent's rule in O2F2 , there is more p-character in O-F bond in comparison to O-H bond in H2O2; hence s-character in O-O bond is greater in O2F2 .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
OA is the chord of the parabola y2=4x (where O is the origin). AB is a chord of y2=4x and perpendicular to OA which cuts the axis of the parabola at C. If the foot of A on the axis of the parabola is D, then the length CD is equal to

Options:
A 2 units
B 3 units
C 4 units
D 6 units

The correct answer is:
4 units

Solution:
Let, A=t2,2t
The slope of OA is equal to 2t
⇒ The slope of AB is equal to -t2
⇒The equation of AB is y-2tx-t2=-t2
⇒C=4+t2,0 & D=t2,0
⇒CD=t2+4-t2=4 units

Topic: Mathematics, Parabola, JEE Main
OHCCH23COCH3âOH-â

The major product is ?

Options:
A
B
C
D

The correct answer is:

Solution:
In the presence of base catalyst, intramolecular aldol condensation and ring closure takes place

Topic: Chemistry, Aldehydes and Ketones, JEE Main
→ OH-/Heat  Product

Options:
A
B
C
D Ph - CH = C = CH - Ph

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
OPQR is a square and M,N are the mid points of the sides PQ and QR respectively, then the ratio of the areas of the square and the triangle OMN is

Options:
A 4:1
B 2:1
C 8:3
D 7:3

The correct answer is:
8:3

Solution:
Let the coordinates of the vertices O, P, Q, R be 0, 0 , a, 0 , a,âa ,â 0, a , respectively. Then, we get the coordinates of M as a,a2 and those of N as a2, a
Therefore, area of âOMN is
12 001aa21a2a1=3a28
Area of the square is a2.
Hence, the required ratio is 8:3

Topic: Mathematics, Straight Lines, JEE Main
Observation of "Rhumann's purple" is a confirmatory test for the presence of:

Options:
A Starch
B Reducing sugar
C Protein
D Cupric ion

The correct answer is:
Protein

Solution:
Observation of "Rhumann's purple" is a confirmatory test for the presence of proteins. Rhumann's purple is ninhydrin. When the proteins reacts with Rhumann's purple then violet precipitates are formed. This confirms the presence of proteins in the compounds.

The reaction is given below.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2016 (10 Apr Online)
Observe structures of the following compounds

The total number of structures/compounds which possess asymmetric carbon atoms is

Options:

The correct answer is:
3

Solution:
An asymmetric carbon atom is a carbon atom that is attached to four different types of atoms or groups of atoms.

Number of compounds containing asymmetric carbons are three.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (29 Jun Shift 1)
Observe the following conversion.

Which of following is best correct sequence of reaction for following conversion?

Options:
A i Br2/FeBr31.eq iiKMnO4/Î iiiConc.HNO3+H2SO4
B i Conc. HNO3+H2SO4 iiBr2/FeBr31.eq iiiKMNO4/Δ
C i KMNO4/Δ ii Conc.HNO3+H2SO4 iii Br2/FeBr3(1 eq.)
D i Br2/FeBr31 eq. ii Conc.HNO3+H2SO4 iii KMnO4/Δ

The correct answer is:
i Br2/FeBr31.eq iiKMnO4/Î iiiConc.HNO3+H2SO4

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Observe the following lists
Then the correct match for List I from List II is
A B C D

Options:
A 1 2 3 4
B 3 5 2 1
C 3 5 5 1
D 3 5 5 1

The correct answer is:
3 5 2 1

Solution:
(A) $[\mathbf{a} \mathbf{b} \mathbf{c}]=\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}$
(B) $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}=(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}$
(C) $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}$
(D) $\mathbf{a} \cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \cos (\mathbf{a} \cdot \mathbf{b})$
$\therefore$ Option (b) is correct.

Topic: Mathematics, Vector Algebra, JEE Main