datavorous/entrance-exam-dataset huggingface dataset:
The increasing order of basicity of the following compounds is :

Options:
A (A)<(B)<(C)<(D)
B (B)<(A)<(D)<(C)
C (D)<(A)<(B)<(C)
D (B)<(A)<(C)<(D)

The correct answer is:
(B)<(A)<(D)<(C)

Solution:
The basic nature of amines can be explained on the basis hybridisation. The basic nature order is sp3>sp2>sp. If the lone pair on nitrogen involved in resonance, then it is less basic than others. In the molecules A, B and D, the nitrogen is sp2 hybridised. In the pyrrole molecule the lone pair on nitrogen involving in conjugation to get aromaticity. Hence, C is most basic and B is least basic.

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The increasing order of basicity of the following compounds is :

Options:
A (iv) < (ii) < (i) <(iii)
B (i) < (ii) < (iii) < (iv)
C (ii) < (i) < (iii) < (iv)
D (ii) < (i) < (iv) < (iii)

The correct answer is:
(ii) < (i) < (iv) < (iii)

Solution:
Correct order should be : ii < i < iv < iii
(i) N is sp3 hybridised, 1°Amine without delocalization.
(ii) N is sp2 hybridised.
(iii) Guanidine is a stronger base than amidine (resonance stabilised)
(iv) 2° Amine, N is sp3 hybridised.

Topic: Chemistry, Amines, JEE Main, JEE Main 2018 (08 Apr)
The increasing order of diazotisation of the following compounds is:
(a)
(b)
(c)
(d)

Options:
A
(D) $ < $ (C) $ < $ (B) $ < $ (A)
B
(A) $ < $ (D) $ < $ (B) $ < $ (C)
C
(A) $ < $ (B) $ < $ (C) $ < $ (D)
D
(A) $ < $ (D) $ < $ (C) $ < $ (B)

The correct answer is:
(A) $ < $ (D) $ < $ (B) $ < $ (C)

Solution:
Aromatic diazonium salts are more stable than aliphatic diazonium salts. The higher stability of aryl diazonium salts is due to resonance. Electron donating substituents increase electron density on benzene ring. Hereby they increase the stability of diazonium salts. Electron withdrawing substituents decrease electron density on benzene ring. Hereby they decrease the stability of diazonium salts. $-\mathrm{COCH}_3$ group is electron withdrawing and hence, diazonium salts from (D) is less stable than that from (B). Although-O- $\mathrm{COCH}_3$ is electron donating substituent, but it is present in meta position. Hence, it will not have significant effect on stability. The increasing order of diazotisation is (A) $ < $ (D) $ < $ (B) $ < $ (C).

Topic: Chemistry, Amines, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The increasing order of nitration in the acidic medium of the following compounds is:

Options:
A a<b<d<c
B a<b<c<d
C b<a<c<d
D b<a<d<c

The correct answer is:
a<b<d<c

Solution:
–OCH3=+M effect so, ring activating group and ortho para directing.
–CH3=+ hyperconjugation & +I so, ring activating group (ortho para directing).
–NH2= Here the aniline is the least reactive due to formation of anilinium ion in acidic medium. It will now act as strong -I group.
–Cl=–I>+M so, ring deactivating & meta directing.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2018 (15 Apr)
The increasing order of nitration of the following compounds is :

Options:
A
$ < $ (B) $ < $ (D) $ < $ (C)
B
$ < $ (B) $ < $ (C) $ < $ (D)
C
$ < $ (A) $ < $ (C) $ < $ (D)
D
$ < $ (A) $ < $ (D) $ < $ (C)

The correct answer is:
$ < $ (B) $ < $ (D) $ < $ (C)

Solution:
In the given substituted benzene rings, the substitutents methoxy $\left(-\mathrm{OCH}_3\right)$ and amino $\left(-\mathrm{NH}_2\right)$ are strongly activating groups while methyl $\left(-\mathrm{CH}_3\right)$ is weakly activating and chloro $(-\mathrm{Cl})$ is a deactivating group towards electrophilic aromatic substituation reaction. Since among methyl and methoxy group, methoxy group is more reactive than methyl group, (c) is more reactive than (d).
Although amino group is strongly activating group, it gets protonated in presence of acid to form anilinium ion
which is strongly deactivating. Hence, (a) is less reactive than (c) and (d). Chloro group is also deactivating group but less deactivating than
.
Thus order is (a) $ < $ (b) $ < $ (d) $ < $ (c).
Note:
The activating groups increases the electron density on benzene ring and increases the rate of electrophilic aromatic substitution reaction. The deactivating groups decreases the electron density on benzene ring and decreases the rate of electrophilic aromatic substitution reaction.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The increasing order of nucleophilicity of the following nucleophiles is;

Options:
A (i)<(iv)<(iii)<(ii)
B (ii)<(iii)< (iv)<(i)
C iv<i<iii<ii
D (ii)<(iii)<(i)<(iv)

The correct answer is:
(ii)<(iii)<(i)<(iv)

Solution:
Anions are more nucleophilic than neutral. So, H2O is least nucleophilic.
nucleophilicity∝e- density
Greater the delocalization of charge, lesser will be nucleophilicity.
OH->CH3COO->CH3SO3->H2O

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The increasing order of pKb for the following compounds will be :

Options:
A B<C<A
B A<B<C
C C<A<B
D B<A<C

The correct answer is:
B<A<C

Solution:
Compound ‘’B’’ represents Guanidine, the conjugate acid of which is resonance stabilised. Compound "C" is aliphatic amine, here the ‘N’ is sp3 whereas in compound ''A" the ''N'' is sp2 .

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The increasing order of pKa for the following phenols is

(A) 2,4-Dinitrophenol

(B) 4-Nitrophenol

(C) 2,4,5-Trimethylphenol

(D) Phenol

(E) 3-Chlorophenol

Choose the correct answer from the option given below :

Options:
A (A), (E), (B), (D), (C)
B (A), (B), (E), (D), (C)
C (C), (D), (E), (B), (A)
D (C), (E), (D), (B), (A)

The correct answer is:
(A), (B), (E), (D), (C)

Solution:
Order of acidity for following phenol is

Increasing order of pKa means most acidic will be placed first.

- M groups at ortho and para position increases acidic strength in phenol. Hence, (A) is most acidic.

+M, Hyperconjugation, +I decreases Ka Value of 2,4,5-Trimethylphenol. Hence (C), is least acidity.

So, the order of pKa value is (A) < (B )< (E)< (D)< (C).

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The increasing order of pKb values of the following compounds is :

Options:
A II<IV<III<I
B I<II<IV<III
C II<I<III<IV
D I<II<III<IV

The correct answer is:
I<II<IV<III

Solution:
PKb∝1 Basic strength Kb
Kb∝+1,+M effect e- density)
Kb∝1-L-M effect

Kb=1>2>4>3

PKb=1<2<4<3

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The increasing order of pKa of the following amino acids in aqueous solution is

Glycine, Aspartate, Lysine, Arginine.

Options:
A Arginine<Lysine<Glycine<Aspartate
B Aspartate<Glycine<Arginine<Lysine
C Glycine<Aspartate<Arginine<Lysine
D Aspartate<Glycine<Lysine<Arginine

The correct answer is:
Aspartate<Glycine<Lysine<Arginine

Solution:
Hence, the correct order is
Aspartate<Glycine<Lysine<Arginine.

Topic: Chemistry, Biomolecules, JEE Main
The increasing order of pKa of the following amino acids in aqueous solution is

Glycine, Aspartate, Lysine, Arginine.

Options:
A Arginine<Lysine<Glycine<Aspartate
B Aspartate<Glycine<Arginine<Lysine
C Glycine<Aspartate<Arginine<Lysine
D Aspartate<Glycine<Lysine<Arginine

The correct answer is:
Aspartate<Glycine<Lysine<Arginine

Solution:
Hence, the correct order is
Aspartate<Glycine<Lysine<Arginine.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The increasing order of reactivity of the following compounds towards aromatic electrophilic substitution reaction is:

Options:
A A<B<C<D
B B<C<A<D
C D<B<A<C
D D<A<C<B

The correct answer is:
D<A<C<B

Solution:
Reactivity of E+ substitution∝πe- can be present on the aromatic compound.
Electrophilic substitution ∝+M ∝1-M
∝+HC ∝1-HC
∝+I ∝1-I
(order of e- density α electrophilic substitution)
Ring e- density is highest.
So B>C>A>D

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The increasing order of reactivity of the following compounds towards reaction with alkyl halides directly is:
(i)
(ii)
(iii)
(iv)

Options:
A i<ii<iii<iv
B ii<i<iii<iv
C ii<i<iv<iii
D i<iii<iv<ii

The correct answer is:
ii<i<iii<iv

Solution:
Alkyl halide undergoes nucleophilic substitution reaction. More nucleophilic nitrogen of amine group, more reactive will it be with alkyl halides.

Presence of electron-withdrawing group will decrease the nucleophilic character of amino group.

Hence the order of reactivity;

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The increasing order of stability of the following free radicals is

Options:
A
$\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}}$
B
$\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H}$
C
$\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H}$
D
$\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H}$

The correct answer is:
$\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}}$

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2006
The increasing order of the acidity of the following carboxylic acids is:

Options:
A
III $ < $ II $ < $ IV $ < $ I
B
I $ < $ III $ < $ II $ < $ IV
C
IV $ < $ II $ < $ III $ < $ I
D
II $ < $ IV $ < $ III $ < $ I

The correct answer is:
III $ < $ II $ < $ IV $ < $ I

Solution:
The increasing order of the acidity of the carboxylic acids is III $ < $ II $ < $ IV $ < $ I. In aromatic acids, electron withdrawing groups like $\mathrm{Cl},-\mathrm{CN},-\mathrm{NO}_2$ increases the acidity, whereas electron releasing groups like $\mathrm{CH}_3,-\mathrm{OH},-\mathrm{OCH}_3,-\mathrm{NH}_2$ decreases the acidity.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The increasing order of the acidity of the α-hydrogen of the following compounds is:

A B C D

Options:
A (D)<(C)<(A)<(B)
B (B)<(C)<(A)<(D)
C (A)<(C)<(D)<(B)
D (C)<(A)<(B)<(D)

The correct answer is:
(D)<(C)<(A)<(B)

Solution:
Acidity α stability of conjugate base

(Stability of conjugate bases)

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The increasing order of the atomic radii of the following elements is:
aC
bO
cF
dCl
eBr

Options:
A b<c<d<a<e
B d<c<b<a<e
C c<b<a<d<e
D a<b<c<d<e

The correct answer is:
c<b<a<d<e

Solution:

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The increasing order of the boiling points for the following compounds is:
IC2H5OH IIC2H5Cl IIIC2H5CH3 IV C2H5OCH3

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

The correct answer is:
III<IV<II<I

Solution:
Boiling Point∝ Dipole moment∝H-bonding in the molecules
C2H5CH3<C2H5OCH3<C2H5Cl<C2H5OH,
C2H5OH-O-H bond present.
Boiling point also increases with the increasing molecular mass.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2017 (09 Apr Online)
The increasing order of the boiling points of the major products A, B and C of the following reactions will be :

(a)

(b)

(c)

Options:
A B<C<A
B C<A<B
C A<B<C
D A<C<B

The correct answer is:
B<C<A

Solution:
(a)

(b)

(c)

The boiling points of isomeric halo alkanes decrease with increase in branching.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is

Options:
A
F < S < P < B
B
P < S < B < F
C
B < P < S < F
D
B < S < P < F

The correct answer is:
B < S < P < F

Solution:
No Solution Available

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2006
The increasing order of the following compounds towards HCN addition is :

(i) (ii) (iii)(iv)

Options:
A (i) < (iii) < (iv) < (ii)
B (iii) < (iv) < (i) < (ii)
C (iii) < (i) < (iv) < (ii)
D (iii) < (iv) < (ii) < (i)

The correct answer is:
(iii) < (i) < (iv) < (ii)

Solution:
–I, – M effect of NO2 increase reactivity towards nucleophilic addition reaction with HCN.

Nucleophile attacks on carbon atom of carbonyl group which will be more reactive if any wlectron withdrawing group is attached in ring.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The increasing order of the ionic radii of the given isoelectronic species is :

Options:
A
$\mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{S}^{2-}$

B
$\mathrm{S}^{2-}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{K}^{+}$

C
$\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-}$

D
$\mathrm{K}^{+}, \mathrm{S}^{2-}, \mathrm{Ca}^{2+}, \mathrm{Cl}^{-}$

The correct answer is:
$\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-}$

Solution:
For isoelectronic species, as the z/e decreases, ionic radius increases

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2012 (Offline)
The increasing order of the pKa values of the following compounds is:

Options:
A B<C<A<D
B D<A<C<B
C B<C<D<A
D C<B<A<D

The correct answer is:
B<C<A<D

Solution:
pKa=-logKa
Ka↑pKa↓ & vice versa
Acidic character α Ka α1pKa
Acidic character α stability of anion (Conjugate base)
Conjugate base are
Stability of conjugate base α-M or -I
α1+M or1+I
-M is more dominating than -I
( B ) −NO2 at para so exerts -M or -I & -M > -I
So it is highly stable anion in
( C ) -NO2 at meta exerts only -I effect
( A ) No e- withdrawing or electron releasing group.
( D ) -OCH3 has +M & -I but +M dominates
So acidic strength will be
B>C>A>D
Therefore pKa values will be in the order
B<C<A<D

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The increasing order of the pKa values of the following compounds is

Options:
A C < B < A < D
B B < C < A < D
C D < A < C < B
D B < C < D < A

The correct answer is:
B < C < A < D

Solution:
Ka∝ stability of phenoxide ion ∝ dispersal of –ve charge
pKa∝1Ka
Stability of phenoxide ion = b > c > a > d
Ka= B > C > A > D
Hence values of pKa in increasing order are B < C < A < D.

Topic: Chemistry, General Organic Chemistry, JEE Main
The increasing order of the pKb of the following compund is:
(A)
(B)
(C)
(D)

Options:
A (B)<(D)<(C)<(A)
B (B)<(D)<(A)<(C)
C (C)<(A)<(D)<(B)
D (A)<(C)<(D)<(B)

The correct answer is:
(B)<(D)<(A)<(C)

Solution:
pKb=-log⁡kb kb↑pkb↓
Basic character in amine and its derivative
α -ve charge density
α availability of lone pair
α+M α 1-M
α+HC α 1-HC
α+I α 1-I
Kb basic character B>D>A>(C)
pKb=B<D<A<C

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The increasing order of the rate of $\mathrm{HCN}$ addition to compounds $\mathrm{A}-\mathrm{D}$ is (A) $\mathrm{HCHO}$
(B) $\mathrm{CH}_3 \mathrm{COCH}_3$
(C) $\mathrm{PhCOCH}_3$
(D) PhCOPh

Options:
A
$A < B < C < D$
B
$D < B < C < A$
C
$D < C < B < A$
D
$C < D < B < A$

The correct answer is:
$D < C < B < A$

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2006
The increasing order of the reactivity of the following compounds in nucleophile addition reaction is:

Propanal, Benzaldehyde, Propanone, Butanone

Options:
A Benzaldehyde > Butanone < Propanone < Propanal
B Butanone < Propanone < Benzaldehyde < Propanal
C Propanal < Propanone < Butanone < Benzaldehyde
D Benzaldehyde < Propanal < Propanone < Butanone

The correct answer is:
Butanone < Propanone < Benzaldehyde < Propanal

Solution:
Rate of NAR α -I,-M group attache to carbonyl carbonMoreover aldehydes are more reactive than ketones.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The increasing order of the reactivity of the following compounds towards electrophilic aromatic substitution reactions is:

Options:
A II<I<III
B III<II<I
C III<I<II
D I<III<II

The correct answer is:
III<I<II

Solution:
Rate of E+ substitution reaction ∝-ve charge density on benzene ring
Electron releasing group; increasing the rate of ESR ; halogen has -I>+M so deactivating group & but show +M but show -M>-I so less deactivating group only
Reactivity order is:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The increasing order of the reactivity of the following halides for the SN1 reaction is

Options:
A (II)<(I)<(III)
B (I)<(III)<(II)
C (II)<(III)<(I)
D (III)<(II)<(I)

The correct answer is:
(II)<(I)<(III)

Solution:
Rate of SN1 reaction is directly dependent on the stability of carbocation intermediate.
(II) is primary alkyl halide.
(I) is secondary alkyl halide.
Primary alkyl halide is less reactive towards SN1 reaction than secondary alkyl halide.
In (III), the carbocation generated is stabilised due to resonance. Hence, (III) is most reactive towards SN1 reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2017 (02 Apr)
The increasing order of the reactivity of the following with LiAlH4 is:
(A)
(B)
(C)
(D)

Options:
A B<A<D<C
B A<B<C<D
C A<B<D<C
D B<A<C<D

The correct answer is:
A<B<D<C

Solution:
In the mentioned structures H- will get added to the carbon having slight positive charge.
Presence of Cl increases the positive charge whereas an electron donating group like NH2 decreases the tendency.
Rate of reaction ∝ -M or -I group
Rate of reaction ∝ 1+M and 1+I
Hence, the correct order will be:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The indefinite integral â«eexxexâlnâ¡x+1xdx simplifies to (where, c is the constant of integration)

Options:
A xln⁡ln⁡x+c
B eex+c
C eexx+c
D eex⋅ln⁡x+c

The correct answer is:
eex⋅ln⁡x+c

Solution:
Given integral is I=∫eex⋅ex⋅ln⁡x+eexxdx
Let, eex=fx and ln⁡x=gx
∴I=∫f'xgx+fxg'xdx
=fx·gx+c
=eex⋅ln⁡x+c

Topic: Mathematics, Indefinite Integration, JEE Main
The indefinite integral I=∫sec2⁡xtan⁡xsec⁡x+tan⁡xdxsec5⁡x+sec2⁡xtan3⁡x-sec3⁡xtan2⁡x-tan5⁡x simplifies to 13ln⁡fx+c, where fπ4=22+1 and c is the constant of integration. If the value of fπ3 is a+b, then the value of b-3a is equal to

Options:

The correct answer is:
3

Solution:
I=13∫3sec3⁡xtan⁡x+3sec2⁡xtan2⁡xdxsec2⁡x-tan2⁡xsec3⁡x+tan3⁡x
Let sec3⁡x+tan3⁡x=t
Then, I=13∫dtt
=13ln⁡t+c
=13ln⁡sec3⁡x+tan3⁡x+c
∴fx=sec3⁡x+tan3⁡x
⇒fπ3=8+27

⇒a=8, b=27
⇒b-3a=3

Topic: Mathematics, Indefinite Integration, JEE Main
The indefinite integral I=â«sin2â¡x-cos2â¡x2019sinâ¡x2021cosâ¡x2021dx simplifies to (where c is an integration constant)

Options:
A sin2⁡x-cos2⁡x20202020+c
B tan⁡x-cot⁡x20202020+c
C sin⁡x-cos⁡x20202020+c
D tan2⁡x+cot2⁡x20202020+c

The correct answer is:
tan⁡x-cot⁡x20202020+c

Solution:
I=∫sin2⁡x-cos2⁡xsin⁡xcos⁡x20191sin2⁡xcos2⁡xdx
=∫tan⁡x-cot⁡x2019sin2⁡x+cos2⁡xsin2⁡xcos2⁡xdx
=∫tan⁡x-cot⁡x2019sec2⁡x+cosec2xdx
Let tan⁡x-cot⁡x=t
⇒sec2⁡x+cosec2xdx=dt
∴I=∫t2019dt
=t20202020+c
=tan⁡x-cot⁡x20202020+c

Topic: Mathematics, Indefinite Integration, JEE Main
The Indian rhinoceros is a natural inhabitant of which one of the Indian states?

Options:
A Uttarakhand
B Uttar Pradesh
C Himachal Pradesh
D Assam

The correct answer is:
Assam

Solution:
Indian rhinoceros (Rhinoceros unicornis) is a natural inhabitant of Asom is India. It is now protected in Kaziranga National Park situated at Sibsagar in Jorhat district of Asom.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2010 (Mains)
The indicated atom is not a nucleophilic site in

Options:
A \( \underset{\uparrow}{\mathrm{B}} \mathrm{H}_{4}^{-} \)
B \( \underset{\uparrow}{\mathrm{C}} \mathrm{H}_{3} \mathrm{Mgl} \)
C \( \mathrm{CH}_{3} \underset{\uparrow}{\mathrm{OH}} \)
D \( \mathrm{CH}_{3} \underset{\uparrow}{\mathrm{NH}_{2}} \)

The correct answer is:
\( \underset{\uparrow}{\mathrm{B}} \mathrm{H}_{4}^{-} \)

Solution:
In indicated atom is not a nucleophilic site because there is no lone-pair on 'B' atom.

Hence, the option (a) is correct answr.

Topic: Chemistry, General Organic Chemistry, JEE Main
The induced emf cannot be produced by:

Options:
A Moving a magnet near a circuit
B Moving a circuit near a magnet
C Changing the current in one circuit placed near the other
D Maintaining large but constant current in a circuit

The correct answer is:
Maintaining large but constant current in a circuit

Solution:
Faraday's law states that changing magnetic field induces a changing emf in a nearby conductor and changing current produces a changing magnetic field around the conductor.

Therefore, induced emf cannot be produced maintaining large but constant current in a circuit.

Topic: Physics, Electromagnetic Induction, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 1)
The inductance between A and D is

Options:
A
3.66 H

B
9 H

C
0.66 H

D
1 H

The correct answer is:
1 H

Solution:
The given circuit clearly shows that the inductors are in parallel we have, $\frac{1}{L}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$ or $L=1$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2002
The inductive effect does not operate in

Options:
A \( \mathrm{HF} \)
B \( \mathrm{F}_{2} \)
C \( \mathrm{CH}_{3} \mathrm{Cl} \)
D \( \mathrm{CH}_{3} \mathrm{COOH} \)

The correct answer is:
\( \mathrm{F}_{2} \)

Solution:
An effect shown by atoms or molecules due to the difference in electronegativity is known as an inductive effect.

As we can see in F2, both atoms are the same. So, there is no electronegativity difference and hence, it cannot show an inductive effect.

Topic: Chemistry, General Organic Chemistry, JEE Main
The industrial activity held least responsible for global warming is :

Options:
A manufacturing of cement

B steel manufacturing

C Electricity generation in thermal power plants.

D Industrial production of urea

The correct answer is:
Industrial production of urea

Solution:
In urea industry, N2, CO2 and H2 are consumed and no greenhouse gases are emitted. So, it is least resposible for global warming.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The inequality C n+1 6 + C n 4 > C n+2 5 − C n 5 holds true for all n greater than

Options:
A 8
B 9
C 7
D 1

The correct answer is:
9

Solution:
C n+1 6 + C n 4 + C n 5 > C n+2 5
⇒ C n+1 6 + C n+1 5 > C n+2 5
⇒ C n+2 6 > C n+2 5
⇒n+2!6!n-4!>n+2!5!n-3!
⇒16>1n-3
⇒n-3>6
⇒n>9

Topic: Mathematics, Permutation Combination, JEE Main
The initial concentration of X and Y were 2 and 4âmole/L respectively. For the following equilibrium X+2Yâ Z which of the following relationship among equilibrium concentrations of X,âY and Z is not feasible?

Options:
A [X] < [Z]
B [X] < [Y]
C [X] > [Y]
D [Y] > [Z]

The correct answer is:
[X] > [Y]

Solution:
Now X  >  Y
2-a>4-2a
a>2
which is not feasible.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The initial configuration of the system is as shown in the figure.

The string and the pulley are light. The bob is released and the string starts wrapping around the pulley. (The pulley is held in place by a force applied at the centre)

Find the rate r at which the length of the string wrapped around the pulley increases (as a function of θ).

Options:
A r=Rl-Rθ2gl-Rθsinθ
B r=Rl-Rθ2gl-Rθsinθ+Rcosθ
C r=Rl-Rθ2gl-Rθsinθ+R1-sinθ
D r=Rl-Rθ2gl-Rθsinθ+R1-cosθ

The correct answer is:
r=Rl-Rθ2gl-Rθsinθ+R1-cosθ

Solution:
At any instant of time, the length of the string wrapped on the pulley is Rθ and the remaining length L=l-Rθ

The rate r=Rdθdt=Rω=Rvl-Rθ, where v is the speed of the particle

The height through which the particle has come down

h=R1-cosθ+l-Rθsinθ

⇒v=2gR1-cosθ+l-Rθsinθ

So,

r=Rl-Rθ2gl-Rθsinθ+R1-cosθ

Topic: Physics, Work Power Energy, JEE Main