datavorous/entrance-exam-dataset huggingface dataset:
Match the reactions given in column (I) with process/products in column (II) and select the proper code of choice from the choices given at the end.

(I)	(II)

(A) NaCl(aq) â Na + +Cl â Na + +â e â â Na(s) 2Cl â â Cl 2 g +2e â	p)	Castner Kellner Cell

(B) NaCl aq â¶ Na aq + + Cl aq - Na + + e - â¶ Hg Na - Hg 2 Cl - â¶ Cl 2 + 2 e -
Na - Hg + H 2 O â¶ NaOH + Hg + 1 2 H 2	q)	Bleaching Powder

(C) 2 Ca OH 2 + 2 Cl 2 â¶ CaCl 2 + Ca OCl 2 + 2 H 2 O Slaked lime	r)	Solvay's Process

(D) NaCl + NH 3 + CO 2 + H 2 O â¶ NaHCO 3 + NH 4 Cl 2 NaHCO 3 â¶ Na 2 CO 3 + H 2 O + CO 2	s)	Down's Cell

Options:
A Aâ(s); Bâ(p); Câ(q); Dâ(r)
B A→(r); B→(p); C→(s); D→(q)
C A→(p); B→(s); C→(r); D→(q)
D A→(r); B→(q); C→(p); D→(s)

The correct answer is:
Aâ(s); Bâ(p); Câ(q); Dâ(r)

Solution:
Sodium metal It is manufactured by electrolysis of a fused mixture of NaCl (40%) and CaCl 2  (60%) in Down's cell at 873 K. Na liberated at the cathode is collected, in kerosene oil while Cl 2 is evolved at the anode.
NaCl l⁡ → Electrolysis Na + + Cl - At cathode Na + + e - ⟶ Na l⁡ At anode 2 Cl - ⟶ Cl 2 g + 2 e -
Sodium hydroxide It is manufactured by electrolysis of an aqueous solution of NaCl (brine) in Castner - Kellner cell using mercury cathode and carbon anode. Sodium metal which is discharged at the cathode combines with mercury to form sodium amalgam. Cl 2 .
At cathode Na + + e - ⟶ Hg Na - amalgam At anode 2 Cl - ⟶ Cl 2 + 2 e -
The sodium amalgam thus obtained is treated with water to form sodium hydroxide and hydrogen gas.
2 Na - amalgam + 2 H 2 O ⟶ 2 NaOH + 2 Hg + H 2
Sodium carbonate It is obtained by Solvay ammonia process. When CO 2 is passed through a concentrated solution of brine (aqueous solution of NaCl ) saturated with ammonia, NaHCO 3 gets precipitated. NaHCO 3 on subsequent heating gives Na 2 CO 3 .
NaCl + NH 3 + CO 2 + H 2 O ⟶ NaHCO 3 ↓ + NH 4 Cl 2 NaHCO 3 ⟶ Δ Na 2 CO 3 + CO 2 + H 2 O
It reacts with Cl 2 to form calcium hypochlorite, Ca(OCl) 2
2CaOH2+2Cl2→ CaCl2+CaOCl2+2H2OSlaked limeBleaching power

Topic: Chemistry, s Block Elements, JEE Main
Match the refining methods (Column I) with metals (Column II).

Column I	Column II

(Refining methods)	(Metals)

(I)	Liquation	(a) Zr

(II)	Zone Refining	(b) Ni

(III)	Mond Process	(c) Sn

(IV)	Van Arkel Method	(d) Ga

Options:
A (I)-(b); (II)-(c); (III)-(d); (IV)-(a)
B (I)-(b); (II)-(d); (III)-(a); (IV)-(c)
C I-c; II-d; III-b; IV-a
D I-(c); II-a; (III)-b; IV-(d)

The correct answer is:
I-c; II-d; III-b; IV-a

Solution:
Liquation is used to purify the metal whose melting point is lower than that of impurities like Sn, Bi,Pb etc.
Mond’s process The impure nickle is heated in presence of carbon monoxide.

Nis + COg →NiCO4g → Nis + 4COg
Zone refining is based on the principle that impurities are more soluble in melt that in the pure solid metal. The metals like Ga, Ge etc. are purified by this method.
Van Arkell method is used for those metals which are very difficult to obtain in pure state like Zr.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Match the refining methods (Column I) with metals (Column II)

Column-I
(Refining methods)
Column-II
(Metals)

(I) Liquiation
(p) Zr

(II) Zone Refining
(q) Ni

(III) Mond process
(r) Sn

(IV) Van Arkel Method
(s) Ga

Options:
A (I) – (q); (II) – (r); (III) – (s); (IV) – (p)
B (I) – (q); (II) – (s); (III) – (p); (IV) – (r)
C (I) – (r); (II) – (p); (III) – (q); (IV) – (s)
D (I) – (r); (II) – (s); (III) – (q); (IV) – (p)

The correct answer is:
(I) – (r); (II) – (s); (III) – (q); (IV) – (p)

Solution:
Liquation is used for Sn.
Zone refining is used for Ga.
Mond's process is used for Ni.
Van Arkel process is used for Zr.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Match the scientists present in columns I with their contribution in column II, and choose the correct combination from the options given.

Column I	Column II

(Scientists)	(Contribution)

a.	Ivanowsky	1.	Contagium vivum fluidum

b.	Pasteur	2.	Crystallization of viruses

c.	Beijerinck	3.	Discovery of viruses

d.	Stanley	4.	Term virus

Options:
A aâ3,Â bâ4,Â câ2,Â dâ1
B aâ4,Â bâ3,Â câ2,Â dâ1
C aâ3, bâ4, câ 1, dâ2
D aâ4,Â bâ3,Â câ1,dâ2

The correct answer is:
aâ3, bâ4, câ 1, dâ2

Solution:
The term virus means venom or poisonous fluid, this term was given by Pasteur. D.J. Ivanowsky discovered certain microbes as a causal organism of the mosaic disease of tobacco, which is smaller than bacteria and can pass through bacteria-proof filters. M.W. Beijerinck told that the extract of the infected tobacco plants could cause infection in healthy plants and called the fluid as "contagium vivum fluidum" (infectious living fluid). W.M. Stanley crystallized them and these crystals were mainly consisting of proteins.

Topic: Biology, Biological Classification, NEET
Match the source gland with its respective hormone as well as the function.

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(a)

Solution:
The pituitary gland is located in a bony cavity called sella tursica and is attached to hypothalamus by a stalk. It is divided anatomically into an adenohypophysis and a neurohypophysis. The latter is also called pars nervosa or posterior pituitary. It stores and releases two hormone called oxytocin and vasopressin, which are actually synthesized by the hypothalamus and are transported axonally to neurohypophysis. Vasopressin acts mainly at the kidney and stimulates resorption of water and electrolytes by the distal tubules in the nephron and therby reduces loss of water through urine (divereses). Hence, it is also called as anti-diuretic hormone (ADH) .

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2011 (Screening)
Match the terms given in Column I with their physiological processes given in Column II and choose the correct answer

Column I

Column II

A. PCT

i. Formation of concentrated urine

B. DCT

ii. Filtration of blood

C. Henle’s loop

iii. Reabsorption of 70-80% of electrolytes

D. Counter-current

iv. Ionic balance

E. Renal corpuscle

v. maintenance of concentration gradient in medulla

Options:
A A-iii, B-v, C-iii, D-ii, E-i
B A-iii, B-iv, C-i, D-v, E-ii
C A-i, B-iii, C-ii, D-v, E-iv
D A-iii, B-i, C-iv, D-v, E-ii

The correct answer is:
A-iii, B-iv, C-i, D-v, E-ii

Solution:
PCT reabsorbs 70-80% of electrolytes. DCT helps in ionic balance. Henle’s loop helps in formation of concentrated urine Counter-current mechanism helps in maintenance of concentration gradient in medulla. Renal corpuscle helps in filtration of blood.

Topic: Biology, Excretory Products and their Elimination, NEET
Match the thermodynamics processes taking place in a system with the correct conditions. In the table : ∆Q is the heat supplied, ∆W is the work done and ∆U is change in internal energy of the system.

Process	Condition

(I)	Adiabatic	(A)	∆W=0

(II)	Isothermal	(B)	∆Q=0

(III)	Isochoric	(C)	∆U≠0,∆W≠0,∆Q≠0

(IV)

Isobaric

(D)	∆U=0

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

The correct answer is:
(I) - (B), (II) - (D), (III) - (A), (IV) - (C)

Solution:
In Adiabatic ΔQ=0,i.e., Exchange of heat does not take place.

In Isothermal ΔU=0,i.e., Temperature does not change therefore internal energy also remains constant.

In Isochoric ΔW=0,i.e., Volume remains constant therefore work done is zero.

In Isobaric ∆U≠0, ∆W≠0, ∆Q≠0
W=P.∆V, Volume changes so work done is not equal to zero.
∆U=f2P∆V, Volume changes therefore internal energy changes.
∆Q=nCp∆T, Temperature changes therefore heat also changes.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Options:
A
1

B
-1

C
zero

D
does not exist

The correct answer is:
1

Solution:
$$
\operatorname{Lim}_{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{\sqrt{2} x} \Rightarrow \operatorname{Lim}_{x \rightarrow 0} \frac{\sqrt{1-\left(1-2 \sin ^2 x\right)}}{\sqrt{2} x} ; \operatorname{Lim}_{x \rightarrow 0} \frac{\sqrt{2 \sin ^2 x}}{\sqrt{2} x} \Rightarrow \operatorname{Lim}_{x \rightarrow 0} \frac{|\sin x|}{x}
$$
the function does not exist or $\mathrm{LHS} \neq \mathrm{RHS}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2002
Options:
A
2925

B
1469

C
1728

D
1456

The correct answer is:
2925

Solution:
Consider $1^2+3^2+5^2+\ldots \ldots+25^2$ $n^{\text {th }}$ term $\mathrm{T}_n=(2 n-1)^2, n=1, \ldots \ldots 13$
Now, $\mathrm{S}_n=\sum_{n=1}^{13} \mathrm{~T}_n=\sum_{n=1}^{13}(2 n-1)^2$
$$
\begin{aligned}
&=\sum_{n=1}^{13} 4 n^2+\sum_{n=1}^{13} 1-\sum_{n=1}^{13} 4 n \\
&=4 \sum n^2+13-4 \sum n
\end{aligned}
$$
$$
=4\left[\frac{n(n+1)(2 n+1)}{6}\right]+13-4 \frac{n(n+1)}{2}
$$
Put $n=13$, we get
$$
\begin{aligned}
&\mathrm{S}_n=26 \times 14 \times 9+13-26 \times 14 \\
&=3276+13-364 \\
&=2925 .
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2013 (25 Apr Online)
\( \mathrm{A} \) (Light pink colour complex) \( \frac{\mathrm{Pb}_{3} \mathrm{O}_{4} / \mathrm{dil} . \mathrm{HNO}_{3}}{\Delta} \mathrm{HMnO}_{4} \stackrel{\mathrm{H}_{2} \mathrm{~S} / \mathrm{H}^{+}}{\longrightarrow} \mathrm{A} \) (Light pink colour complex)
Calculate the sum total of CFSE value in light pink colour complex for both sets of orbitals as compared to the average
energy of \( \mathrm{d} \)-orbital.

Options:

The correct answer is:
0

Solution:
⇒ Pink colour complex: [Mn(H2O)6]2+
Mn2+ (3d5) in [Mn(H2O)6]++, Δ 0 < P
CFSE of the pink colour complex = 3 × - 2 5 Δ 0 + 2 × 3 5 Δ 0 = 0

Topic: Chemistry, Coordination Compounds, JEE Main
\( \mathrm{A} \rightarrow \mathrm{A}^{-}+\mathrm{e}, \mathrm{E}_{1} \) and \( \mathrm{A}^{+} \rightarrow \mathrm{A}^{2+}+\mathrm{e}, \mathrm{E}_{2} \). The energy required to pull out the two electrons are \( \mathrm{E}_{1} \) and \( \mathrm{E}_{2} \) respectively. The correct relationship between two energy would be :

Options:
A \( \mathrm{E}_{1} < \mathrm{E}_{2} \)
B \( \mathrm{E}_{1}>\mathrm{E}_{2} \)
C \( \mathrm{E}_{1}=\mathrm{E}_{2} \)
D \( \mathrm{E}_{1} \neq \mathrm{E}_{2} \)

The correct answer is:
\( \mathrm{E}_{1} < \mathrm{E}_{2} \)

Solution:
The second ionization energy is always higher than the first ionization energy as an electron is to be removed from unipositive cation. There is a strong electrostatic attractive force between the unipositive cation and the electron. This makes the removal of electron difficult and requires a large amount of energy.

Always IP2>IP1
So E2>E1

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
\( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \underset{373 \mathrm{~K}}{\stackrel{\mathrm{HI} / \operatorname{Red}}{\longrightarrow}} \mathrm{A}+\mathrm{B} \)
The products \( \mathrm{A} \) and \( \mathrm{B} \) respectively are

Options:
A \( \mathrm{n}- \) heptane and \( 2- \) iodo hexane.
B \( \mathrm{n}-\mathrm{octane} \) and \( 2 \)-iodo hexane.
C \( \mathrm{n}- \) hexane and \( 2- \) iodo heptane.
D \( \mathrm{n}- \) hexane and \( 2 \)-iodo hexane.

The correct answer is:
\( \mathrm{n}- \) hexane and \( 2 \)-iodo hexane.

Solution:
HI/Red P is a reducing agent that reduces the carbonyl group and alcoholic group to hydrocarbons.

Glucose on reduction forms alkane (major) along with 2-iodohexane (minor).

This reaction confirms the presence of straight chain of carbon in glucose.

C6H12O6→373kHI/Red PCH3CH2CH2CH2CH2n-hexane CH3+CH3CH(I)CH2CH2CH2CH32-iodo hexane

Topic: Chemistry, Biomolecules, JEE Main
\( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3} \stackrel{\mathrm{Cl}_{2}, 500^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{A} . \) The product \( \mathrm{A} \) is

Options:
A
B
C
D

The correct answer is:

Solution:
It is free radical substitution at allylic position due to high temperature.

Topic: Chemistry, Hydrocarbons, JEE Main
\( \mathrm{CH}_{3} \mathrm{COOH} \stackrel{\mathrm{NH}_{3}}{\longrightarrow} \stackrel{\Delta}{\rightarrow} ? \)
The product of the reaction is isomeric with

Options:
A
B \( \mathrm{CH}_{3} \mathrm{CH}=\mathrm{NHO} \)
C \( \mathrm{HCONH}-\mathrm{CH}_{3} \)
D All of these

The correct answer is:
All of these

Solution:
CH3COOH →NH3CH3COONH4 → â -H2O CH3CONH2
acetic acid ammonium acetate acetamide
The isomers of CH3CONH2 is
1. NH2CH2CHO
2. CH3-CH=NOH
3. H-CONH-CH3

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
\( -\mathrm{CONH}_{2} \stackrel{\text { reduction }}{\longrightarrow}-\mathrm{CH}_{2} \mathrm{NH}_{2} \). In this reaction, the hybridisation state of carbon changes from

Options:
A \( \mathrm{sp} \rightarrow \mathrm{sp}^{2} \)
B \( \mathrm{sp} \rightarrow \mathrm{sp}^{3} \)
C \( \mathrm{sp}^{2} \rightarrow \mathrm{sp}^{3} \)
D \( \mathrm{sp}^{2} \rightarrow \mathrm{sp} \)

The correct answer is:
\( \mathrm{sp}^{2} \rightarrow \mathrm{sp}^{3} \)

Solution:
In acid amide group, the carbon atom forms a double bond with the oxygen atom due to which it shows sp2 hybridisation. In contrast, in alkyl amines, the carbon atom is attached with a hydrogen atom, and it shows sp3 hybridisation, so the hybridisation changes from sp2 to sp3.

Topic: Chemistry, Amines, JEE Main
\( [\mathrm{EDTA}]^{4-} \) is a

Options:
A nodentate ligand.
B bidentate ligand.
C quadridentate ligand
D hexadentate ligand.

The correct answer is:
hexadentate ligand.

Solution:
[EDTA]-4 is ethylenediaminetetracetate. It is a member of the aminopolycarboxylic acid family of ligands. It usually binds to a metal cation through its two amines and four carboxylates. Therefore, it can act as a hexadentate ligand since it has six donor atoms.

Topic: Chemistry, Coordination Compounds, JEE Main
\(\mathrm{Fe}+\mathrm{CuSO}_4 \rightarrow \mathrm{FeSO}_4+\mathrm{Cu}\)
This is an example of _____ reaction.

Options:
A displacement
B redox
C neutralisation
D More than one of the above

The correct answer is:
More than one of the above

Solution:
- The reaction \(\mathrm{Fe}+\mathrm{CuSO}_4 \rightarrow \mathrm{FeSO}_4+\mathrm{Cu}\) involves the displacement of copper from copper sulfate by iron.
- This is because iron is more reactive than copper and can displace it from its compound.
- This is an example of a displacement reaction, which is a type of redox (reduction-oxidation) reaction where there is a transfer of electrons between the reactants.
- Iron (Fe) is oxidized to iron(II) sulfate \(\left(\mathrm{FeSO}_4\right)\), and copper \(\left(\mathrm{Cu}^{2+}\right)\) is reduced to copper \((\mathrm{Cu})\). This means there is a transfer of electrons, making it a redox reaction.

Topic: Chemistry, Redox Reactions, JEE Main
\( \mathrm{NH}_{3} \) is basic while \( \mathrm{PH}_{3} \) is

Options:
A acidic
B neutral
C amphoteric
D basic

The correct answer is:
basic

Solution:
PH3 is also basic in nature, but PH3 is less basic in comparison to NH3. The stability of hydrides decreases down the group.

The order of basicity hydrides of Group 15 elements is shown below:

NH3 > PH3 > AsH3 > SbH3 ≥ BiH3.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
\( \mathrm{RCOOH}+\mathrm{NaHCO}_{3} \rightarrow \mathrm{RCOONa}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \) The above reaction is showing the property of carboxylic acid.

Options:
A acidic
B basic
C oxidizing
D reducing

The correct answer is:
acidic

Solution:
RCOOH+NaHCO3→RCOONa+H2O+CO2

In the reaction, it can be clearly seen that carboxylic group gives proton which indicates its acidic character and sodium bicarbonate is taking proton, so it is behaving as base.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
\( \mathrm{RNH}_{2} \) reacts with \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl} \) in aqueous \( \mathrm{KOH} \) to give a clear solution. On acidification, a precipitate is obtained which is due to the formation of

Options:
A
B
C
D \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{NH}_{2} \)

The correct answer is:

Solution:
Benezene sulphonyl chloride is known as Hinsberg reagent and in presence of this reagent primary amines form precipitate.

N-alkyl benzene sulphonamide is formed as precipitate.

Tertiary amines do not react with hinsberg reagent and do not give hinsberg test.

Topic: Chemistry, Amines, JEE Main
\( \mathrm{R}-\underset{|| \atop \huge\mathrm{O}}{\mathrm{C}}-\mathrm{Cl} \frac{\mathrm{H}_{2}}{\mathrm{Pd}-\mathrm{BaSO}_{4}} \mathrm{P} \)

Options:
A \( \mathrm{RCH}_{2} \mathrm{OH} \)
B \( \mathrm{RCOOH} \)
C \( \mathrm{RCHO} \)
D \( \mathrm{RCH}_{3} \)

The correct answer is:
\( \mathrm{RCHO} \)

Solution:
It is Rosenmund reaction. In this reaction, acid halide in the presence of H2/Pd-BaSO4 converts into aldehydes. It is also a reduction process.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
\(\mathrm{X} \mathrm{g}\) of ethanamine was subjected to reaction with \(\mathrm{NaNO}_2 / \mathrm{HCl}\) followed by hydrolysis to liberate \(\mathrm{N}_2\) and \(\mathrm{HCl}\). The \(\mathrm{HCl}\) generated was completely neutralised by 0.2 moles of \(\mathrm{NaOH} . X\) is ______ g.

Options:

The correct answer is:
9

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2024 (05 Apr Shift 2)
\(\mathrm{X} \mathrm{g}\) of ethylamine is subjected to reaction with \(\mathrm{NaNO}_2 / \mathrm{HCl}\) followed by water; evolved dinitrogen gas which occupied \(2.24 \mathrm{~L}\) volume at STP. \(\mathrm{X}\) is \(\qquad\) \(\times 10^{-1} \mathrm{~g}\).

Options:

The correct answer is:
45

Solution:
$\underset{\substack{\text { Mol.wt.45g }}}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2} \xrightarrow{\mathrm{NaNO}_2+\mathrm{HCl}} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{CH}_2-\mathrm{OH}+\underset{14 \mathrm{~g}}{\mathrm{~N}_2}$
given : $\mathrm{N}_2$ evolved is 2.24 L i.e. 0.1 mole.
i.e. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2$ (ethyl amine) will be $4.5 \mathrm{~g}$
$(=0.1 \mathrm{~mole})$
Hence the answer $=45 \times 10^{-1} \mathrm{~g}$

Topic: Chemistry, Amines, JEE Main, JEE Main 2024 (04 Apr Shift 1)
\( \mathrm{XeF}_{6} \) dissolves in anhydrous \( \mathrm{HF} \) to give a conducting solution which contains

Options:
A \( \mathrm{H}^{+} \)and \( \mathrm{XeF}_{7}^{-} \)ions.
B \( \mathrm{HF}_{2}^{-} \)and \( \mathrm{XeF}_{5}^{+} \)ions.
C \( \mathrm{HXeF}_{6}^{+} \)and \( \mathrm{F}^{-} \)ions.
D None of these.

The correct answer is:
\( \mathrm{HF}_{2}^{-} \)and \( \mathrm{XeF}_{5}^{+} \)ions.

Solution:
HF + XeF 6 → XeF 5 + + HF 2 -

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Matrix $\left[\begin{array}{ccc}0 & -4 & 1 \\ 4 & 0 & -5 \\ -1 & 5 & 0\end{array}\right]$ is

Options:
A Orthogonal
B Idempotent
C Skew-symmetric
D Symmetric

The correct answer is:
Skew-symmetric

Solution:
It is skew-symmetric.

Topic: Mathematics, Matrices, JEE Main
Mature Graafian follicle is generally present in the ovary of a healthy human female around

Options:
A
$5-8$ day of menstrual cycle

B
11-17 day of menstrual cycle

C
18-23 day of menstrual cycle

D
$24-28$ day of menstrual cycle

The correct answer is:
11-17 day of menstrual cycle

Solution:
v. In humans (female), the menstrual cycle lasts for about $28 / 29$ days. In the follilcular phase in which the primary follicles in the ovary grows to become a fully mature graafian follicle due to stimulation of FSH. This phase (follicular) lasts for about 14 days.
During this phase, the secretion of gonadotropins (LH and FSH) increases gradually and stimulates secretion of estrogen by the growing follicles both LH and FSH attain a peak level in the middle of cycle (about 14th day).
This rapid secretion of LH called LH surge, induces rupture of Graafian follicle and thereby the release of ovum. This ovulatory phase is followed by the luteal phase during which the remaining follicular cells enlarge to become the corpus luteum.

Topic: Biology, Human Reproduction
max0≤x≤πx-2sinxcosx+13sin3x=

Options:
A π+2-336
B π
C 0
D 5π+2+336

The correct answer is:
5π+2+336

Solution:
Let fx=x-2sinxcosx+13sin3x

⇒f'x=1-2 cos2x+cos3x

⇒f"x=4sin2x-3sin3x

For maxima/minima f'x=0

⇒1-22cos2x-1+4cos3x-3cosx=0

⇒(2 cosx+3)(2 cosx-3)(cosx-1)=0

cosx=-32,32,1

x=5π6,π6,0

f"5π6=-23-3<0

f"π6=23-3>0

f"(0)=0

So x=5π6is local maxima point

Maximum value of fx=f5π6=5π6+32+13

=5π+2+336

Hence this is the correct option.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Maximize $z=4 x+6 y$
subject to
$3 x+2 y \leq 12, x+y \geq 2, x, y \geq 0$

Options:
A 24 at $(6,0)$
B 36 at $(0,6)$
C 36 at $(0,6)$
D More than one of the above

The correct answer is:
36 at $(0,6)$

Solution:
The objective function $z=4 x+6 y$
The constraints are -
$\begin{aligned}
& 3 x+2 y \leq 12 \\
& x+y \geq 2
\end{aligned}$
and $\mathrm{x}, \mathrm{y} \geq 0$
Concept:
The maximum or minimum value of a function occurs at one of the corner points of the feasible
Solution:
Making the graph of the constraints -
We see that the feasible region has corner points $(0,2),(2,0),(4,0)$ and $(0,6)$
The values of the objective function $z=4 x+6 y$ at these points respectively are $12,8,16$ and $\mathbf{3 6}$
Hence maximum value occurs at $(0,6)$ and is equal to 36

Topic: Mathematics, Linear Programming, JEE Main
Maximum acceleration of the train in which a 50 kg box lying on its floor will remain stationary (Given, coefficient of static friction between the box and the train's floor is 0.3 and g=10 m s-2)

Options:
A 5.0Â m s-2
B 3.0 m s-2
C 1.5Â m s-2
D 15Â m s-2

The correct answer is:
3.0 m s-2

Solution:
box remain stationary when maâ¤Î¼mg
â aâ¤Î¼sg
Here, Î¼s=0.3,g=10 m s-2
aâ¤0.3Ã10
â aâ¤3 m s-2

Topic: Physics, Laws of Motion, NEET
Maximum acceleration of the train in which a \( 50 \mathrm{~kg} \) box lying on its floor will remain stationary
(Given : Co-efficient of static friction between the box and the train's floor is \( 0.3 \) and
\( g=10 m s^{-2} \) )

Options:
A \( 5.0 \mathrm{~ms}^{-2} \)
B \( 3.0 \mathrm{~ms}^{-2} \)
C \( 1.5 \mathrm{~ms}^{-2} \)
D \( 15 m s^{-2} \)

The correct answer is:
\( 3.0 \mathrm{~ms}^{-2} \)

Solution:
Given, mass of box $=50 \mathrm{~kg}$; coefficient of static friction $=0.3 ; g=10 \mathrm{~ms}^{-2}$
Train moving along this direction
In order that the box remains stationary, we must have $m a=\mu N$
$\Rightarrow m a=\mu m g$
$\Rightarrow a=\mu g=0.3 \times 10=3$
Therefore, maximum acceleration of the train $=3 \mathrm{~ms}^{-2}$

Topic: Physics, Laws of Motion, KCET, KCET 2016
Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon

Options:
A
temperature

B
nature of solute

C
pressure

D
nature of solvent

The correct answer is:
pressure

Solution:
Maximum amount of solid that can be dissolved in a specified amount of a given solvent does not depend upon pressure. This is because solid and liquid are highly incompressible and practically remain unaffected by change in pressure.

Topic: Chemistry, Solutions
Maximum energy is evolved during which of the following transitions?

Options:
A n=1 to n=2
B n=2 to n=6
C n=2 to n=1
D n=6 to n=2

The correct answer is:
n=2 to n=1

Solution:
When transition is from upper state to lower state, the energy is evolved. Therefore, options (C) and (d) are practicable.
In option (C)
E=-Rhc 112-122=-34Rhc
In option (d)
E=-Rhc 122-162
=-29Rhc
Thus, maximum energy is evolved in option (c).

Topic: Physics, Atomic Physics, JEE Main
Maximum no. of moles of grignard reagent that can be used is x. Find x.

Options:

The correct answer is:
6

Solution:
4 mole with acyl chloride and two mole with -OH,-SH.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Maximum number of bromine atoms present in the final product $P$ upon complete bromination are

Options:
A 4
B 6
C 10
D 8

The correct answer is:
8

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Maximum number of electrons in a subshell of an atom is determined by the following

Options:
A $4 l+2$
B $2 l+1$
C $4 l-2$
D $2 n^2$

The correct answer is:
$4 l+2$

Solution:
Total number of subshells $=(2 l+1)$
$\therefore$ Maximum number of electrons in the subshell $=2(2 \mathrm{l}+1)=4 \mathrm{l}+2$

Topic: Chemistry, Structure of Atom, JEE Main