datavorous/entrance-exam-dataset huggingface dataset:
Electromagnetic waves are transverse in nature is evident by

Options:
A
polarization

B
interference

C
reflection

D
diffraction

The correct answer is:
polarization

Solution:
No Solution Available

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2002
Electromagnetic waves travel in a medium with a speed of 2× 10 8  m  s −1 . The relative permeability of the medium is 1. Find the relative permittivity.

Options:
A 2.25
B 3.25
C 3.70
D 4.20

The correct answer is:
2.25

Solution:
Here, v=2× 10 8  m  s −1 and μ r =1
The speed of electromagnetic waves in a medium is given by
v = 1 μ ε , ....(i)
where μ and ε are absolute permeability and absolute permittivity of the medium.
Now, μ= μ 0    μ r and ε= ε 0   ε r
Therefore, the equation (i) becomes
v = 1 μ 0 μ r × ε 0 ε r = 1 μ 0 ε 0 × 1 μ r ε r = c μ r ε r
or ε r = c 2 v 2 μ r = 3 × 10 8 2 2 × 10 8 2 × 1 = 2.25

Topic: Physics, Electromagnetic Waves, JEE Main
Electrometallurgical process is used to extract

Options:
A $\mathrm{Fe}$
B $\mathrm{Pb}$
C $\mathrm{Na}$
D $A g$

The correct answer is:
$\mathrm{Na}$

Solution:
Because Na is very reactive and can not be extracted by means of the reduction by $\mathrm{C}, \mathrm{CO}$ etc. So extracted by electrolysis

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Electrometallurgical process is used to extract

Options:
A Fe
B Pb
C Na
D Ni

The correct answer is:
Na

Solution:
Because Na is very reactive and can not be extracted by means of the reduction by C, CO. etc. So, it is extracted by electrolysis of molten NaCl Solution.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Electron affinity is the

Options:
A Energy absorbed when an electron is added to an isolated atom in the gaseous state
B Energy released when an electron is added to an isolated atom in the gaseous state
C Energy required to take out an electron from an isolated gaseous atom
D Power of an atom to attract an electron to itself

The correct answer is:
Energy released when an electron is added to an isolated atom in the gaseous state

Solution:
Energy released when an electron is added to an isolated atom in gaseous state.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV has a de-Broglie wavelength of λ0. If the voltage is increased to 40 kV then the de-Broglie wavelength associated with the electron beam would be:

Options:
A 3λ0
B 9λ0
C λ02
D λ02

The correct answer is:
λ02

Solution:
The de-Broglie wavelength of an electron accelerated through potential V is

λ=hmv=h2 m(KE)=h2 meV

where, h is Planck's constant, e is charge of electron and m is mass of electron.

From above relation, λ∝1V

∴ λλ0=2040

⇒ λ=λ02.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Electron gain enthalpy and ionisation energy of an atom are –a and +b eV respectively. The electronegativity of that atom on Mulliken scale is given by :

Options:
A a – b
B b-a2
C a + b
D a+b2

The correct answer is:
a+b2

Solution:
EN=IE+E⁡A⁡2andE⁡A⁡=-E⁡ga

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Electron gain enthalpy with negative sign of fluorine is less than that of chlorine due to :

Options:
A
High ionization enthalpy of fluorine

B
Smaller size of chlorine atom

C
Smaller size of fluorine atom

D
Bigger size of $2 p$ orbital of fluorine

The correct answer is:
Smaller size of fluorine atom

Solution:
The electron gain enthalpy order for halogens is
$\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I}$
Due to small size of fluorine the extra electron to be added feels more electronelectron repulsion. Therefore fluorine has less value for electron affinity than chlorine.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2013 (09 Apr Online)
Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength $\lambda_{1}: \lambda_{2}$ emitted in the two cases is

Options:
A $7 / 5$
B $27 / 20$
C $27 / 5$
D $20 / 7$

The correct answer is:
$27 / 5$

Solution:
The wave number $(\bar{v})$ of the radiation
$$
\begin{array}{l}
=\frac{1}{\lambda} \\
=R_{\infty}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]
\end{array}
$$
Now for case (I) $n_{1}=3, n_{2}=2$
$$
\begin{array}{l}
\frac{1}{\lambda_{1}}=R_{\infty}\left[\frac{1}{9} \frac{-1}{4}\right], R_{\infty}=\text { Rydberg constant } \\
\frac{1}{\lambda_{1}}=R_{\infty}\left[\frac{4-9}{36}\right]=\frac{-5 R_{\infty}}{36} \\
\Rightarrow \quad \lambda_{1}=\frac{-36}{5 R_{\infty}} \\
\frac{1}{\lambda_{2}}=R_{\infty}\left[\frac{1}{4}-\frac{1}{1}\right]=\frac{-3 R_{\infty}}{4} \Rightarrow \lambda_{2}=\frac{-4}{3 R_{\infty}} \\
\Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=\frac{-36}{5 R_{\infty}} \times \frac{3 R_{\infty}}{-4} \Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=\frac{27}{5}
\end{array}
$$

Topic: Physics, Atomic Physics, JEE Main
Electronic configurations of few elements are given below. Mark the incorrect match.

Options:
A 1s22s22p5; Most electronegative element
B 1s22s22p3; An element belonging to 3 period and 5 group
C 1s22s22p63s23p63d84s2; A d-block element
D 1s22s22p63s23p6; An element from 18 group

The correct answer is:
1s22s22p3; An element belonging to 3 period and 5 group

Solution:
Since n=2 so it belongs to 2 period and 5 group. It is nitrogen.

Topic: Chemistry, Structure of Atom, JEE Main
Electrons are accelerated through a potential difference $\mathrm{V}$ and protons are accelerated through a potential difference $4 \mathrm{~V}$. The de - Broglie wavelengths are $\lambda_{e}$ and $\lambda_{p}$ for electrons and protons respectively. The ratio of $\frac{\lambda_{e}}{\lambda_{p}}$ is given by: (given $m_{e}$ is mass of electron and $m_{p}$ is mass of proton).

Options:
A $\sqrt{\frac{m_{p}}{m_{e}}}$
B $\sqrt{\frac{m_{e}}{m_{p}}}$
C $\frac{1}{2} \sqrt{\frac{m_{e}}{m_{p}}}$
D $2 \sqrt{\frac{m_{p}}{m_{e}}}$

The correct answer is:
$2 \sqrt{\frac{m_{p}}{m_{e}}}$

Solution:
Energy $=$ charge $\times$ potential diff.

$\mathrm{E}_{\text {electron }}=\mathrm{q}_{\mathrm{e}} \mathrm{V}$ and $\mathrm{E}_{\text {proton }}=\mathrm{q}_{\mathrm{p}} 4 \mathrm{~V}$

de $-$ Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$

$\lambda_{e}=\frac{h}{\sqrt{2 m_{e} \mathrm{eV}}}$ and $\lambda_{\mathrm{P}}=\frac{h}{\sqrt{2 m_{\mathrm{p}} e 4 \mathrm{~V}}} \because\left(q_{e}=q_{p}\right)$

$\therefore \frac{\lambda_{e}}{\lambda_{\mathrm{P}}}=\frac{\frac{h}{\sqrt{2 m_{e} \mathrm{eV}}}}{\frac{h}{\sqrt{2 m_{\mathrm{p}} e 4 \mathrm{~V}}}}=2 \sqrt{\frac{m_{\mathrm{P}}}{m_{e}}}$

Topic: Physics, Dual Nature of Matter, JEE Main
Electrons are accelerated through a potential difference $\mathrm{V}$ and protons are accelerated through a potential difference $4 \mathrm{~V}$. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by: (given $m_e$ is mass of electron and $m_p$ is mass of proton).

Options:
A
$\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_p}{m_e}}$

B
$\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_e}{m_p}}$

C
$\frac{\lambda_e}{\lambda_p}=\frac{1}{2} \sqrt{\frac{m_e}{m_p}}$

D
$\frac{\lambda_e}{\lambda_p}=2 \sqrt{\frac{m_p}{m_e}}$

The correct answer is:
$\frac{\lambda_e}{\lambda_p}=2 \sqrt{\frac{m_p}{m_e}}$

Solution:
Energy in joule (E)
$=$ charge $\times$ potential diff. in volt $\mathrm{E}_{\text {electron }}=\mathrm{q}_{\mathrm{c}} \mathrm{V}$ and $\mathrm{E}_{\text {proton }}=\mathrm{q}_{\mathrm{p}} 4 \mathrm{~V}$
de-Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$
$$
\begin{array}{r}
\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}} \text { and } \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{e} 4 \mathrm{~V}}} \\
\left(\because \mathrm{q}_{\mathrm{e}}=\mathrm{q}_{\mathrm{P}}\right)
\end{array}
$$
$\begin{aligned} \therefore \frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{P}}} & =\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}}}{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{P}} \mathrm{e} 4 \mathrm{~V}}}}=\sqrt{\frac{2 \mathrm{~m}_{\mathrm{P}} \mathrm{e} 4 \mathrm{~V}}{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}} \\ & =2 \sqrt{\frac{\mathrm{m}_{\mathrm{P}}}{\mathrm{m}_{\mathrm{e}}}}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2013 (23 Apr Online)
Electrons behave like gas in J.J. Thompson experiment because they

Options:
A ionise the gas
B are effected by electric field
C are deflected by magnetic field
D diffracted by a crystal

The correct answer is:
diffracted by a crystal

Solution:
The English physicists named J. J. Thomson discovered electron using the cathode ray experiment and was awarded a Nobel Prize in physics for this discovery.

In this experiment the electrons behave like gas because they are diffracted by a crystal.

Topic: Physics, Atomic Physics, JEE Main
Electrons from excited chlorophyll molecule of photosystem II are accepted first by

Options:
A Ferredoxin
B Cytochrome - b
C Cytochrome - f
D Quinone

The correct answer is:
Quinone

Solution:
Plastoquinone is used as mobile carrier and primary electron acceptor from PS II.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2008 (Mains)
Electrons having kinetic energy 30 eV are made to collide with atomic hydrogen gas (in ground state) and 42.5% of electron energy is used to excite the hydrogen atoms. The number of emitted wavelengths in emission spectra are

Options:
A 3
B 6
C 12
D 18

The correct answer is:
6

Solution:
42.5% of 30 eV=12.75 eV
â´E4-E1=-0.85--13.6=12.75 eV
So, it means electron has been transmitted from ground state to 3rd excited state n=4, so total number of unique wavelengths emitted are
=4Ã4-12=6

Topic: Physics, Atomic Physics, JEE Main
Electrons in a cathode ray tube have been emitted with a velocity of 1000 ms-1. The number of following statements which is/are true about the emitted radiation is

Given : h=6×10-34Js,me=9×10-31 kg

(A ) The deBroglie wavelength of the electron emitted is 666.67 nm

(B ) The characteristic of electrons emitted depend upon the material of the electrodes of the cathode ray tube.

(C) The cathode rays start from cathode and move towards anode.

(D ) The nature of the emitted electrons depends on the nature of the gas present in cathode ray tube..

Options:

The correct answer is:
2

Solution:
(A) Ve=1000 m/s;h=6×10-34Js

me=9×10-31 kg
Debroglie wavelength:

⇒λ=hmv=6×10-349×10-31×1000=666.67×10-9 m

=666.67 nm

(B) The characteristic of electrons emitted is independent of the material of the electrodes of the cathode ray tube. Electrons are fundamental to all traditional forms of matter.

(C) The cathode rays start from cathode and move towards anode. It was discovered by Sir J.J. Thompson.

(D) The nature of the emitted electrons is independent on the nature of the gas present in cathode ray tube. They are fundamental particles.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Electrons in a certain energy level n⁡ = n⁡ 1 , can emit 3 spectral lines. When they are in another energy level, n⁡ = n⁡ 2 , they can emit 6 spectral lines. The ratio of orbital speeds of the electrons is

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

The correct answer is:
4:3

Solution:
Number of emitted spectral lines, N=n(n-1)2.

Case I
N=3
â3=n1(n1-1)2
ân12-n1-6=0
â(n1 -3)(n1 +2)=0
ân1=3
Case II
N=6
â6=n2(n2-1)2
ân22-n2-12=0
â(n2-4)(n2+3)=0
ân2=4

Velocity of electron v= Ze22Îµ0hn
â´v1v2=n2n1
âv1v2=43

Topic: Physics, Atomic Physics, JEE Main
Electrons move at right angles to a magnetic field of $1.5 \times 10^{-2}$ Tes/a with a speed of $6 \times 10^7 \mathrm{~m} / \mathrm{s}$. If the specific charge of the electron is $1.7 \times 10^{11} \mathrm{C} / \mathrm{kg}$. The radius of the circular path will be

Options:
A $2.9 \mathrm{~cm}$
B $3.9 \mathrm{~cm}$
C $2.35 \mathrm{~cm}$
D $3 \mathrm{~cm}$

The correct answer is:
$2.35 \mathrm{~cm}$

Solution:
Radius of circular path
$\begin{aligned}
r & =\frac{\mathrm{mV}}{\mathrm{eB}}=\frac{\mathrm{V}}{\left(\frac{\mathrm{e}}{\mathrm{m}}\right) \mathrm{B}} \\
\mathrm{r} & =\frac{6 \times 10^7}{1.7 \times 10^{11} \times 15 \times 10^{-2}} \\
& =2.35 \times 10^{-2} \mathrm{~m}=2.35 \mathrm{~m}
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
Electrons with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

Options:
A
$\lambda_0=\frac{2 m c \lambda^2}{h}$

B
$\lambda_0=\frac{2 h}{m c}$

C
$\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2}$

D
$\lambda_0=\lambda$

The correct answer is:
$\lambda_0=\frac{2 m c \lambda^2}{h}$

Solution:
Momentum of striking electrons, $p=\frac{h}{\lambda}$
$\therefore$ Kinetic energy of striking electrons, $K=\frac{p^2}{2 m}=\frac{h^2}{2 m \lambda^2}$
This is also, maximum energy of $X$-ray photons.
Therefore, $\quad \frac{h c}{\lambda_0}=\frac{h^2}{2 m \lambda^2}$ or $\lambda_0=\frac{2 m \lambda^2 c}{h}$
$\therefore$ Correct option is (a).

Topic: Physics, Dual Nature of Matter, JEE Main
Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have 72.5 keV energy. X-rays emitted by the tube contain only

Options:
A A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of 0.155 Å
B A continuous X-ray spectrum (Bremsstrahlung) with all wavelengths
C The characteristic X-ray spectrum of tungsten
D A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of0.155 Å and the characteristic X-ray spectrum of tungsten

The correct answer is:
A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of0.155 Å and the characteristic X-ray spectrum of tungsten

Solution:
Minimum wavelength of continuous X - ray spectrum is given
by λ min (in Å) = 12375E
Here E = energy of incident electrons (in eV) = energy corresponding to minimum wavelength
E = 80 keV = 80 x 103 eV
Therefore, λmin (ni Å)= 1237580×103=0.155
Also, the energy of the incident electrons (80 keV) is more than the ionization energy of the K - shell electrons (i. e., 72. 5 keV), Therefore, characteristic X-ray spectrum will also be obtained because the energy of the incident electron is high enough to knock out the electron from K or L- shells.

Topic: Physics, Atomic Physics, JEE Main
Electrovalent bond formation depends on

Options:
A Ionization energy
B Electron affinity
C Lattice energy
D All the three above

The correct answer is:
All the three above

Solution:
Electrovalent bond formation depends on ionization energy of cation, electron affinity of anion and on lattice energy.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Electrovalent bond-formation depends on :

Options:
A ionization energy
B lattice energy
C electron affinity
D all of these

The correct answer is:
all of these

Solution:
The electrovalent bond is favoured with a metal having low ionization energy and non-metal with high electron affinity.
The amount of energy released when one mole of an ionic compound is formed is called lattice energy. Higher the lattice energy more the stability of the ionic compound.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Element B forms ccp structure and A occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is:

Options:
A A2BO4
B AB2O4
C A2B2O
D A4B2O

The correct answer is:
AB2O4

Solution:
For ccp structure:
Number of octahedral voids are =4
Number of tetrahedral voids are =8
ZB=4 (number of atoms present in ccp structure)
ZA=12×4=2
ZO=8
So formula is A2B4O8=AB2O4.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Element not present in Nessler’s reagent is

Options:
A N
B Hg
C I
D K

The correct answer is:
N

Solution:
Nessler's reagent is an aqueous solution of potassium iodide, mercuric chloride, and potassium hydroxide. Its chemical formula is K2HgI4. It is used to detect the presence of ammonia. It turns into pale brown colour in presence of ammonia.

So, N is not present in Nessler's reagent.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Element not showing variable oxidation state is :

Options:
A Bromine
B Iodine
C Chlorine
D Fluorine

The correct answer is:
Fluorine

Solution:
Because of the absence of d-orbitals, it is not possible for Fluorine to show variable oxidation state. Fluorine does not show variable oxidation states because it is the most electronegative element and has seven valence electrons. Its high electronegativity and strong attraction for electrons prevent it from losing electrons to exhibit different oxidation states. Fluorine always show -1oxidation state in compounds. Fluorine does not show variable oxidation state.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2024 (27 Jan Shift 1)
Element showing the phenomenon of allotropy is

Options:
A Aluminium
B Tin
C Lead
D Copper

The correct answer is:
Tin

Solution:
Among the given elements tin $\left(\mathrm{Sn}\right)$ shows the allotropy.
$\beta$-tin (the metallic form, or white tin), which is stable at and above room temperature, is malleable. In contrast, $\alpha$-tin (nonmetallic form, or gray tin), which is stable below 13.2 ${ }^{\circ} \mathrm{C}\left(55.8^{\circ} \mathrm{F}\right)$, is brittle.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Element X crystallizes in 12 co-ordination fcc lattice. On applying high temperature it changes to 8 co-ordination bcc lattice. Find the ratio of the density of the crystal lattice before and and after applying high temperature

Options:
A 1:1
B 3:2
C 2:3
D 223:33

The correct answer is:
223:33

Solution:
Co. number is 12 means it is fcc lattice
Co. number is 8 means it is bcc lattice
fcc bcc
Ï1=4Ãat.massNaÃa13 Ï2=2Ãat.massNaÃa23
ρ 1 ρ 2 = 4 2 × a 2 3 a 1 3 = 2× 4r 3 3 (2 2r ) 3
=2Ã4Ã4Ã42Ã2Ã2Ã2Ã2(33)
=4233=2(2)3(3)3

Topic: Chemistry, Solid State, JEE Main
Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4 K. The ratio of molal elevation constant to molal depression constant Kb/Kf is

Options:

The correct answer is:
3

Solution:
Elevation in boiling point ∆Tb=ikbm1

Depression in freezing point ∆Tf=ikfm2

4=ikb1.5

4=ikf4.5

Dividing both:

kbkf=31

Hence, The ratio of molal elevation constant to molal depression constant is 3:1

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Elevation in boiling point was \( 0.52^{\circ} \mathrm{C} \) when \( 6 \mathrm{~g} \) of a compound \( \mathrm{X} \) was dissolved in \( 100 \mathrm{~g} \) of water. Molecular weight of \( \mathrm{X} \) is
\( \left(\mathrm{K}_{\mathrm{b}}\right. \) of water is \( 5.2^{\circ} \mathrm{C} \) per \( 100 \mathrm{~g} \) of water \( ): \)

Options:
A \( 120 \)
B \( 60 \)
C \( 600 \)
D \( 180 \)

The correct answer is:
\( 60 \)

Solution:
When a non-volatile solute is added in a pure solvent, then boiling point of solution increases. This increase in boiling point is known as Elevation in boiling point.

We know that,
m=100.Kb.wΔTb. W

where,

Kb= molal elevation constant, m = molecular weight of solute.

m=100×5.2×60.52×100=60

Topic: Chemistry, Solutions, JEE Main
Eleven apples are distributed among a girl and a boy, Then, which one of the following statements is true?

Options:
A atleast one of them will receive 7 apples
B the girl receives atleast 4 apples or the boy receives atleast 9 apples
C the girl receives atleast 5 apples or the boy receives atleast 8 apples
D the girl receives atleast 4 apples or the boy receives atleast 8 apples

The correct answer is:
the girl receives atleast 4 apples or the boy receives atleast 8 apples

Solution:
When eleven apples are distributed among a gil and a boy, then the girl receives atieast
14 apples or the boys receives atieast 8 apples. (by hypothesis)

Topic: Mathematics, Permutation Combination, JEE Main
Eleven books consisting of 5 Mathematics, 4 Physics and 2 Chemistry are placed on a shelf. The number of possible ways of arranging them on the assumption that the books of the same subject are all together is

Options:
A $4!2!$
B $11 !$
C $5!4!3!2$ !
D None of these

The correct answer is:
$5!4!3!2$ !

Solution:
Possible ways $=5!, 4!, 3!\cdot 2$ !

Topic: Mathematics, Permutation Combination, JEE Main
Eleven objects A, B, C, D, E, F, Î±, Î±, Î±, Î² and Î² are arranged in a row, then the probability that every Î² has two Î± as neighbors is

Options:
A 11320
B 17920
C 1110
D 1660

The correct answer is:
1660

Solution:
Total cases =11!3!.2!
The number of favourable cases when Î² is placed in gap of Î±
=1Ã2C2Ã1Ã7!
=7!
Required probability =7!Ã3!Ã2!11!=1660

Topic: Mathematics, Probability, JEE Main
Elimination of bromine from 2-bromobutane results in the formation of-

Options:
A
equimolar mixture of 1 and 2-butene

B
predominantly 2-butene

C
predominantly 1-butene

D
predominantly 2-butyne

The correct answer is:
predominantly 2-butene

Solution:
Saytzeffs product

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2005
Ellingham diagram is a graphical representation of :

Options:
A ΔG vs P
B ΔG-TΔS vs T
C ΔH vs T
D ΔG vs T

The correct answer is:
ΔG vs T

Solution:
Ellingham diagram is a graphical representation of ΔG vs T when metal heated with oxygen to form metal oxide.

An Ellingham diagram is a graph showing the temperature dependence of the stability for compounds. This analysis is usually used to evaluate the ease of reduction of metal oxides and sulphides.

Ellingham diagram represents change in free energy ΔG with temperature

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Emergent land plants that can tolerate the salinities of the sea are called.

Options:

None

Solution:
Emergent land plants that can withstand the salinities of sea and are able of maintain their water supply from the same are known as halophytes.

Topic: Biology, Organisms and Populations
E.M.F. diagram for some ions is given as
FeO42-→Eo=+2.20VFe3+→Eo=+0.77VFe2+→Eo=-0.445VFeo
Determine the value of EFeO42-/Fe2+o

Options:
A 1.84 V
B 1.42 V
C 1.3 V
D 2.0 V

The correct answer is:
1.84 V

Solution:
Fe6++3e→Fe3+; -ΔG1o=3×2.20×F …(i)
Fe3++e→Fe2+; -ΔG2o=1×0.77×F …(ii)
Fe2++2e→Feo; -ΔG3o=2×(-0.445)×F …(iii)
By equation (i) and (ii)
Fe6++4e→Fe2+; -ΔG4o=-ΔG1o-ΔG2o
Hence 4×E4o×F=3×2.20×F+1×0.77×F
So E4o=1.84V
or Fe6++4e→Fe2+; Eo=1.84V

Topic: Chemistry, Electrochemistry, JEE Main
Emf is most closely related to

Options:
A Mechanical force
B Potential difference
C Electric field
D Magnetic field

The correct answer is:
Potential difference

Solution:
Correct option is B. Potential difference
EMF is most closely related to Potential difference.

Topic: Physics, Current Electricity, JEE Main
EMF of a cell in terms of reduction potential of its left and right electrodes is

Options:
A
$E=E_{\text {left }}-E_{\text {right }}$

B
$E=E_{\text {left }}+E_{\text {right }}$

C
$E=E_{\text {right }}-E_{\text {left }}$

D
$E=-\left(E_{\text {right }}+E_{\text {left }}\right)$

The correct answer is:
$E=E_{\text {right }}-E_{\text {left }}$

Solution:
$\begin{aligned} \mathrm{E}_{\text {cell }} & =\text { Reduction potential of cathode (right) - reduction potential of anode (left) } \\ & =\mathrm{E}_{\text {right }}-\mathrm{E}_{\text {left }}\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2002