datavorous/entrance-exam-dataset huggingface dataset:
The figure shows two isotherms at temperatures T1 and T2. A gas is taken from one isotherm to another isotherm through different processes. Then change in internal energy ÎU has a relation â

Options:
A Δ U ab > Δ U ac > Δ U ad > Δ U ae
B Δ U ab = Δ U ac > Δ U ad > Δ U ae
C Δ U ab = Δ U ac = Δ U ad = Δ U ae
D Δ U ab < Δ U ac < Δ U ad < Δ U ae

The correct answer is:
Δ U ab = Δ U ac = Δ U ad = Δ U ae

Solution:
Internal energy depends only on state but not on the process. Hence its change is same for all the processes between two same states.

Topic: Physics, Thermodynamics, JEE Main
The figure shows two regions of uniform magnetic fields of strengths B and 2B. A charged particle of mass m and charge q enters the region of the magnetic field with a velocity v=qBwm, where w is the width of each region of the magnetic field. The time taken by the particle to come out of the region of the magnetic field is

Options:
A 4πmqB
B 2πmqB
C πm2qB
D ÏmqB

The correct answer is:
πmqB

Solution:
Time = π m 2 qB + π m 2 qB = π m qB

Topic: Physics, Magnetic Effects of Current, JEE Main
The figure shows two solid discs with radius R and r respectively. If mass per unit area is the same for both, what is the ratio of MI of bigger disc around axis AB (Which is ⊥ to the plane of the disc and passing through its centre) of MI of smaller disc around one of its diameters lying on its plane? Given M is the mass of the larger disc.

(MI stands for a moment of inertia)

Options:
A R2: r2
B 2r4: R4
C 2R2: r2
D 2R4: r4

The correct answer is:
2R4: r4

Solution:
Ratio of the moment of inertia =12MR214mr2

=2σπR2R2σπr2r2=2R4r4

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The figure that is not a direct manifestation of the quantum nature of atom is :

Options:
A
B
C
D

The correct answer is:

Solution:
1, 2 and 3 are according to quantum theory but (4) is statement of kinetic theory of gases.

Figure 1 is associated with theory of black body, figure 2 is associated with photoelectric effect and figure 3 shows absorption spectra, so we can say that these 3 figures are associated with quantum nature of atom while figure 4 shows variation of internal energy with temperature .

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The final acid obtained during the manufacture of $\mathrm{H}_2 \mathrm{SO}_4$ by contact process is

Options:
A $\mathrm{H}_2 \mathrm{SO}_4$ (conc.)
B $\mathrm{H}_2 \mathrm{SO}_4$ (dil.)
C $\mathrm{H}_2 \mathrm{SO}_4$
D $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$

The correct answer is:
$\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$

Solution:
\(\mathrm{H}_2 \mathrm{SO}_4+\mathrm{SO}_3 \rightarrow \underset{\text{(oleum)}}{\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7}\)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The final product A, formed in the following multistep reaction sequence is:

Options:
A
B
C
D

The correct answer is:

Solution:
Bromobenzene reacts with magnesium in dry ether (diethyl ether) to form phenylmagnesium bromide, often abbreviated as PhMgBr.

The resulting phenylmagnesium bromide reacts with carbon dioxide in the presence of an acidic medium to form benzoic acid.

Heating benzoic acid with ammonia can lead to a reaction known as the Schotten-Baumann type reaction, which is a method for the amidation of carboxylic acids. In this reaction, the carboxylic acid group of benzoic acid reacts with ammonia to form an amide, benzamide, under heating conditions.

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine.

In this case, it is an hoffmann bromamide reaction, aniline is formed.

Topic: Chemistry, Amines, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The final product A formed in the following multistep reaction sequence is

Options:
A
B
C
D

The correct answer is:

Solution:
During the reaction, first addition of water takes place according to Markownikoff's rule, alcohol is formed, CrO3 acts as an oxidising agents which converts alcohol into ketone, then Wolf Kishner reduction occurs.

Hence option A is the answer.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The final product A, formed in the following reaction sequence is:

Options:
A Ph-CH2-CH2-CH3
B
C
D Ph-CH2-CH2-CH2-OH

The correct answer is:
Ph-CH2-CH2-CH2-OH

Solution:
The Hydroboration Oxidation reaction is an organic chemical reaction which is employed for the conversion of alkenes into alcohols that are neutral.

The hydroboration oxidation reaction mechanism can be considered as an anti-markownikov's reaction where a hydroxyl group attaches itself to the carbon which is less substituted.

Hence, the answer is option D.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The final product 'A' in the following reaction sequence

Options:
A

B

C

D

The correct answer is:

Solution:
Butan-2-one undergoes nucleophilic addition reaction with HCN to form a cyanohydrin which on hydrolysis forms 2-hydroxy-2methyl butanoic acid which undergoes dehydration to form 2-methyl but-2-enoic acid

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The final product C in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
This conversion occurs as follows

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The final product ' $\mathrm{c}$ ', obtained in this reaction would be

Options:
A

B

C

D

The correct answer is:

Solution:
After bromination, bromoacetanilide is acid hydrolysed to yield the desired halogenated amine.

Topic: Chemistry, Amines, JEE Main
The final product C, obtained in this reaction would be

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
The final product formed when methylamine is treated with NaNO 2 and HCl followed by hydrolysis is:

Options:
A Nitromethane
B Methylcyanide
C Methyl alcohol
D Diazomethane

The correct answer is:
Methyl alcohol

Solution:
CH 3 NH 2 ⟶ HCl NaNO 2 CH 3 N 2 + Cl - ⟶ H 2 O CH 3 OH Methyl alcohol

Topic: Chemistry, Amines, JEE Main, JEE Main 2014 (19 Apr Online)
The final product formed when methylamine is treated with NaNO 2 and HCl followed by hydrolysis is:

Options:
A Nitromethane
B Methylcyanide
C Methyl alcohol
D Diazomethane

The correct answer is:
Methyl alcohol

Solution:
CH 3 NH 2 ⟶ HCl NaNO 2 CH 3 N 2 + Cl - ⟶ H 2 O CH 3 OH Methyl alcohol

Topic: Chemistry, Amines, JEE Main
The final product (IV) in the sequence of reactions

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The final product obtained in the reaction

Options:
A
B
C
D

The correct answer is:

Solution:
As Hydration (addition of H+/OH- ) takes place according to Markownikoff rule.

Topic: Chemistry, Hydrocarbons, JEE Main
The final product of the following reaction is/are

Options:
A
B
C
D

The correct answer is:

Solution:
Phenol on reaction with chloroform and KOH gives salicylaldehyde, which with $50 \%$ KOH solution undergoes Cannizaro's reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The final product (T) in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The final step for the extraction of copper from copper pyrite in Bessemer converter involves the reaction-

Options:
A 4Cu2O+FeS→8Cu+FeSO4
B Cu2S+2Cu2O→6Cu+SO2
C 2 Cu 2 O+FeS→4Cu+Fe+ SO 2
D Cu2S+2FeO→2Cu+2FeCO+SO2

The correct answer is:
Cu2S+2Cu2O→6Cu+SO2

Solution:
In Bessemer converter copper sulphide is partially oxidised to cuprous oxide which further reacts with remaining copper sulphide to form copper and sulphur dioxide. Cu2S+2Cu2O→6Cu+SO2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The finch species of Galapagos islands are grouped according to their food sources. Which of the following is not a finch food?

Options:
A Carrion
B Insects
C Tree buds
D Seeds

The correct answer is:
Carrion

Solution:
Darwin's Finches shows adaptive radiation, which occurred in response to various food resources available. They feed on insects, tree buds, seeds, cactus, etc. Carrion are dead bodies and no finches feed on them.
Related Theory
Darwin studied the climatic conditions, fauna and flora of Galapagos islands. He noticed varieties of small birds in these islands which differed mainly in the shape and size of their beak and in the colour of their plumage. These birds are now called Darwin's finches. Darwin concluded that the American mainland species was the original one from which different forms migrated to the different islands of Galapagos and adapted to the different environmental conditions of these islands. These adapted forms are considered to be the new species. Thus, Darwin's finches provide a beautiful example of adaptive radiation supporting the concept of evolution.

Topic: Biology, Evolution, NEET, NEET 2013 (Karnataka)
The finction $f: R \rightarrow R$ defined by $f(x)=e^x$ is

Options:
A Onto
B Many-one
C One-one and into
D Many one and onto

The correct answer is:
One-one and into

Solution:
Function $f: R \rightarrow R$ is defined by $f(x)=e^x$. Let $x_1, x_2 \in R$ and $f\left(x_1\right)=f\left(x_2\right)$ or $e^{x_1}=e^{x_2}$ or $x_1=x_2$. Therefore $f$ is one-one. Let $f(x)=e^x=y$. Taking log on both sides, we get $x=$ logy. We know that negative real numbers have no pre-image or the function is not onto and zero is not the image of any real number. Therefore function $f$ is into.

Topic: Mathematics, Functions, JEE Main
The first and second dissociation constants of an acid $\mathrm{H}_2 \mathrm{A}$ are $1.0 \times 10^{-5}$ and $5.0 \times 10^{-10}$ respectively. The overall dissociation constant of the acid will be

Options:
A
$5.0 \times 10^{-5}$

B
$5.0 \times 10^{15}$

C
$5.0 \times 10^{-15}$

D
$0.0 \times 10^5$

The correct answer is:
$5.0 \times 10^{-15}$

Solution:
$\begin{array}{ll}\mathrm{H}_2 \mathrm{A} \rightleftharpoons \mathrm{HA}^{-}+\mathrm{H}^{+} & \mathrm{K}_1=\frac{\left[\mathrm{HA}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{H}_2 \mathrm{A}\right]} \quad \dots (1)\\ \mathrm{HA}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{2-} & \mathrm{K}_2=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{2-}\right]}{\left[\mathrm{HA}^{-}\right]}\quad \dots(2)\end{array}$
For the reaction
$\begin{aligned}
& \mathrm{H}_2 \mathrm{A} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{A}^{2-} \\
& \mathrm{K}=\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~A}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{A}\right]}=\mathrm{K}_1 \times \mathrm{K}_2 \\
& =1 \times 10^{-5} \times 5 \times 10^{-10} \\
& =5 \times 10^{-15}
\end{aligned}$
Hence, (C) is correct.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The first and second dissociation constants of an acid $\mathrm{H}_2 \mathrm{A}$ are $1.0 \times 10^{-5}$ and $5.0 \times 10^{-10}$ respectively. The overall dissociation constant of the acid will be

Options:
A
$5.0 \times 10^{-5}$

B
$5.0 \times 10^{15}$

C
$5.0 \times 10^{-15}$

D
$0.0 \times 10^5$

The correct answer is:
$5.0 \times 10^{-15}$

Solution:
$\begin{array}{ll}\mathrm{H}_2 \mathrm{A} \rightleftharpoons \mathrm{HA}^{-}+\mathrm{H}^{+} & \mathrm{K}_1=\frac{\left[\mathrm{HA}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{H}_2 \mathrm{A}\right]} \quad \dots (1)\\ \mathrm{HA}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{2-} & \mathrm{K}_2=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{2-}\right]}{\left[\mathrm{HA}^{-}\right]}\quad \dots(2)\end{array}$
For the reaction
$\begin{aligned}
& \mathrm{H}_2 \mathrm{A} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{A}^{2-} \\
& \mathrm{K}=\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~A}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{A}\right]}=\mathrm{K}_1 \times \mathrm{K}_2 \\
& =1 \times 10^{-5} \times 5 \times 10^{-10} \\
& =5 \times 10^{-15}
\end{aligned}$
Hence, (C) is correct.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2007
The first clinical gene therapy was given for treating

Options:
A diabetes mellitus
B chicken pox
C rheumatoid arthritis
D adenosine deaminase deficiency

The correct answer is:
adenosine deaminase deficiency

Solution:
Gene therapy is the treatment of disease by the transfer and extension of genetic material in the patient's cell in order to restore normal cellular functions. The first clinical gene therapy was given for treating adenosine deaminase deficiency.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2012 (Mains)
The first diffraction minimum due to the single slit diffraction is seen at $\theta=30^{\circ}$ for a light of wavelength $5000 Å$ falling perpendicularly on the slit. The width of the slit is

Options:
A
$2.5 \times 10^{-5} \mathrm{~cm}$

B
$1.25 \times 10^{-5} \mathrm{~cm}$

C
$10 \times 10^{-5} \mathrm{~cm}$

D
$5 \times 10^{-5} \mathrm{~cm}$

The correct answer is:
$10 \times 10^{-5} \mathrm{~cm}$

Solution:
For first minimum, $d \sin \theta=\lambda$
$$
\begin{aligned}
\Rightarrow \quad d & =\frac{\lambda}{\sin \theta}=\frac{5000 \times 10^{-8} \mathrm{~cm}}{\sin 30^{\circ}} \\
& =\frac{5000 \times 10^{-8} \mathrm{~cm}}{1 / 2}=10 \times 10^{-5} \mathrm{~cm}
\end{aligned}
$$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2012 (12 May Online)
The first element of rare–earth metals is

Options:
A Cerium
B Actinium
C Uranium
D Lanthanum

The correct answer is:
Cerium

Solution:
Cerium

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The first four terms in the expansion of $(1-x)^{3 / 2}$ are

Options:
A $1-\frac{3}{2} x+\frac{3}{8} x^2-\frac{1}{16} x^3$
B $1-\frac{3}{2} x-\frac{3}{8} x^2-\frac{x^3}{16}$
C $1-\frac{3}{2} x+\frac{3}{8} x^2+\frac{x^3}{16}$
D None of these

The correct answer is:
$1-\frac{3}{2} x+\frac{3}{8} x^2+\frac{x^3}{16}$

Solution:
$\begin{aligned} & (1-x)^{3 / 2} \\ & =\left[1+\frac{3}{2}(-x)+\frac{3}{2} \cdot \frac{1}{2} \cdot \frac{1}{2!}(-x)^2+\frac{3}{2} \cdot \frac{1}{2}\left(-\frac{1}{2}\right) \frac{1}{3!}(-x)^3+\ldots\right] \\ & =1-\frac{3}{2} x+\frac{3}{8} x^2+\frac{x^3}{16} \text { (only four terms). }\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
The first IE1 and second IE2 ionization energies (kJ/mol) of a few elements designated by Roman numerals are given below. Which of these would be an alkali metal?

Options:
A

IE1	IE2

I	2372	5251

B

IE1	IE2

II	520	7300

C

IE1	IE2

III	900	1760

D

IE1	IE2

IV	1680	3380

The correct answer is:

IE1	IE2

II	520	7300

Solution:
As alkali metals have the lowest IE1 values in their respectively periods and also there is a large jump between IE1 and IE2 due to the removal of second electron from the noble gas core, therefore, out of the given IE1 and IE2 values (II) belongs to alkali metals.

Topic: Chemistry, s Block Elements, JEE Main
The first ΔiH1 and second ΔiH2 ionization enthalpies (in kJ mol-1 ) and the electron gain enthalpy ΔegH (in kJ mol-1 ) of the elements I, II, III, IV and V are given below

Element	ΔiH1	ΔiH2	ΔegH

I	520	7300	− 60

II	419	3051	−48

III	1681	3374	−328

IV	1008	1846	−295

V	2372	5251	+48

the least reactive non- metal and the most reactive metal of these are respectively

Options:
A IV and V
B II and V
C V and III
D V and II

The correct answer is:
V and II

Solution:
Generally metals having low I.E.are more reactive and ÎegH=+ve for noble gas.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The first integral term in the expansion of 3+239 , is the

Options:
A 2ndÂ term
B 3rdÂ term
C 4th term
D 5thÂ term

The correct answer is:
4th term

Solution:
3+239=312+2139
∴ Tr+1=9Cr3129-r213r
For first integral term,

r=3
⇒T3+1= 9C333.21
i.e., T3+1=T4

Topic: Mathematics, Binomial Theorem, JEE Main
The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na + will be :

Options:
A -10.2 eV
B +2.55 eV
C -2.55 eV
D -5.1 eV

The correct answer is:
-5.1 eV

Solution:
Electron affinity is defined as the change in energy (in kJ/mole) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion.

In other words, the neutral atom's likelihood of gaining an electron.
Na(g)      →  Na+(g)+e−;
1 st ionisation energy ; ΔHI E(1)=+5.1eV
Electron gaining is the reversed reaction of ionization. Hence, electron gain enthalpy is ;

IENa = EANa+
Na+(g)+e−    →  Na
So, ΔHeg=−5.1eV

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2013 (07 Apr)
The first ionization energy (in kJ/mol ) of Na, Mg, Al and Si respectively are:

Options:
A 496,737,577,786
B 496,577,737,786
C 786,737,577,496
D 496,577,786,737

The correct answer is:
496,737,577,786

Solution:
Correct order of ionisation energy will be : Na < Al < Mg < Si.

On going from left to right in periodic table effective nuclear charge increases due to which ionisation energy increases. But there is exception in order of Al and Mg, in case of Mg electron is removed from fulfilled 's' subshell due to which it has higher I.E.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The first ionization energy of magnesium is smaller, as compared to that of elements X and Y, but higher than that of Z. The elements X, Y and Z, respectively, are :

Options:
A chlorine, lithium and sodium
B argon, lithium and sodium
C argon, chlorine and sodium
D neon, sodium and chlorine

The correct answer is:
argon, chlorine and sodium

Solution:
The 1st IE order of 3rd period is

Na<Al<Mg<Si<S<P<Cl<Ar

X & Y are Ar & Cl

Z is sodium (Na).

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The first ionization enthalpies of Be, B, N and O follow the order

Options:
A B<Be<O<N
B O<N<B<Be
C Be<B<N<O
D B<Be<N<O

The correct answer is:
B<Be<O<N

Solution:
The ionisation enthalpy increases as we go from left to right in a period, while it decreases as we come down a group. So according to this, the order will be Be<B<C<N<O. But there are two exceptions, in the case of Be and N because of the highly stable 2s2 (completely filled orbital) and 2s22p3 (half-filled orbital 2p3) valence configuration. Due to this, the correct order is:
B<Be<C<O<N.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The first ionization enthalpy values (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) of group 13 elements are :
\(\begin{array}{lllll}\text { B } & \text { Al } & \text { Ga } & \text { In } & \text { Tl } \\ 801 & 577 & 579 & 558 & 589\end{array}\)
How would you explain this deviation from the general trend?

Options:

None

Solution:
On moving down the group 13 from \(\mathrm{B}\) to \(\mathrm{Al}\), the ionization enthalpy decreases as expected due to an increase in atomic size and screening effect which outweigh the effect of increased nuclear charge.
However \(\Delta_{\mathrm{i}} \mathrm{H}\) of \(\mathrm{Ga}\) is only slightly higher \(\left(2 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) than that of \(\mathrm{Al}\) while that of \(\mathrm{Tl}\) is much higher than those of \(\mathrm{Al}\), Ga and In. These deviations can be explained as follows:
\(\mathrm{Al}\) follows immediately after \(s\)-block elements while Ga and In follow after \(d\)-block elements and Tl after \(d\) - and \(f\)-block elements. These extra \(d\)-and \(f\)-electrons do not shield(or screen) the outer shell-electrons from the nucleus very effectively. As a result, the valence electrons remain more tightly held by the nucleus and hence larger amount of energy is needed for their removal. This explains why Ga has higher ionization enthalpy than \(\mathrm{Al}\). Further on moving down the group from Ga to In, the increased shielding effect (due to the presence of additional \(4 d\)-electrons) outweighs the effect of increased nuclear charge ( \(49-31=18\) units) and hence the \(\Delta_1 \mathrm{H}_1\) of In is lower than that of Ga. Thereafter, the effect of increased nuclear charge (81-49=32 units) outweighs the shielding effect due to the presence of additional \(4 f\) and \(5 d\) electrons and hence the \(\Delta_{\mathrm{i}} \mathrm{H}_1\) of \(\mathrm{Tl}\) is much higher than that of \(\mathrm{Al}, \mathrm{Ga}\) and \(\mathrm{In}\).

Topic: Chemistry, Classification of Elements and Periodicity in Properties
The first law of thermodynamic is expressed as:

Options:
A \( \Delta \mathrm{E}=\mathrm{q}-\mathrm{W} \)
B \( \mathrm{q}=\Delta \mathrm{E}-\mathrm{W} \)
C \( \mathrm{W}=\mathrm{q}+\Delta \mathrm{E} \)
D None of the above

The correct answer is:
\( \mathrm{q}=\Delta \mathrm{E}-\mathrm{W} \)

Solution:
The first law of thermodynamics can be expressed as:
∆E=q+WHeat q=+Ve endothermic q=-Ve exothermicWork =+Ve on the systemWork =-Ve by the system
q=∆E-W

Topic: Physics, Thermodynamics, JEE Main
The first \( \left(\Delta_{\mathrm{i}} \mathrm{H}_{1}\right) \) and the second \( \left(\Delta_{\mathrm{i}} \mathrm{H}_{2}\right) \) ionization enthalpies ( in \( \mathrm{kJ} \mathrm{mol}^{-1} \) ) and the \( \Delta \mathrm{e}_{\mathrm{g}} \mathrm{H} \) electron gain enthalpy (in \( \mathrm{kJ} \) mol
\( -1 \) ) of a few elements are given below :
\(\begin{array}{|c|c|c|c|}\hline \text { Elements } & \Delta \mathrm{H}_{1} & \Delta \mathrm{H}_{2} & \Delta_{\mathrm{eg}} \mathrm{H} \\\hline \mathrm{I} & 52 \mathrm{O} & 7300 & -60 \\\hline \mathrm{II} & 419 & 3051 & -48 \\\hline \mathrm{III} & 1681 & 3374 & -328 \\\hline \mathrm{IV} & 1008 & 1846 & -295 \\\hline \mathrm{V} & 2372 & 5251 & +48 \\\hline \mathrm{VI} & 738 & 1451 & -40 \\\hline\end{array}\)
Which of the above likely to be the least reactive non-metal?

Options:

The correct answer is:
4

Solution:
The least reactive non-metal is element IV. The element IV has not so high ΔiH1 but it has high negative electron gain enthalpy (ΔegH). The given values for element IV match with I, (iodine).

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The first meiotic division during oogenesis is completed at the stage of

Options:
A primary oocyte within primary follicle
B secondary oocyte within secondary follicle
C primary oocyte within the tertiary follicle
D secondary oocyte within tertiary follicle

The correct answer is:
primary oocyte within the tertiary follicle

Solution:
Oogenesis is the process of formation of oocyte involving the process of folliculogenesis. Oogenesis begins before birth and the ovaries before birth consist of many primary oocytes. At birth these oocytes are surrounded by follicular cells and this is primary follicle. These remain in the ovary till puberty. As puberty is attained one oocyte completes its cycle. At the beginning primary oocyte is within primary follicle. Oocyte begins its Meiosis I and arrests at Prophase-I. Primary follicle then develops into secondary and further into tertiary follicle but the follicle remains in the arrested prophase-I phase. When tertiary follicle is formed the primary oocyte finally completes its meiosis I and forms secondary oocyte and first polar body after which the tertiary follicle forms the Graffian follicle.

Topic: Biology, Human Reproduction, NEET
The first orbital of H is represented by
ψ=1π1a0e-r/a0 , where a0 is Bohr's radius. The probability of finding the electron at a distance r, from the nucleus in the region dV is

Options:
A ψ2dr
B ∫ψ24πr2dv
C ψ24πr2dr
D ∫ψdv

The correct answer is:
ψ24πr2dr

Solution:
This can be obtained by the solution of Schrodinger wave equation as follows
Probability =ψ2dV
As orbital is spherically symmetrical
So V=43πr3
i.e., dVdr=4πr2
Hence, probability =ψ24πr2dr.

Topic: Chemistry, Structure of Atom, JEE Main
The first orbital of H is represented by: Ï=1r1a03/2e-r/a0,
where a0 is Bohrâs radius. The probability of finding the electron at a distance r, from the nucleus in the region dV is:

Options:
A ψ2dr
B ∫ψ24πr2dv
C ψ24πr2dr
D ∫ψdv

The correct answer is:
ψ24πr2dr

Solution:
This can be obtained by the solution of Schrodinger wave equation as follows probability =Ï2dV
As orbital is spherically symmetrical
So, V =  4 3 πr 3 , i.e.  dV dr  =  4πr 2
Hence, Probability =Ï24Ïr2dr

Topic: Chemistry, Structure of Atom, JEE Main