datavorous/entrance-exam-dataset huggingface dataset:
Air is streaming past a horizontal air plane wing such that its speed in $120 \mathrm{~m} / \mathrm{s}$ over the upper surface and $90 \mathrm{~m} / \mathrm{s}$ at the lower surface. If the density of air is $1.3 \mathrm{~kg}$ per metre ${ }^3$ and the wing is $10 \mathrm{~m}$ long and has an average width of 2 $m$, then the difference of the pressure on the two sides of the wing of

Options:
A 4095.0 Pascal
B 409.50 Pascal
C 40.950 Pascal
D 4.0950 Pascal

The correct answer is:
4095.0 Pascal

Solution:
From the Bernoulli's theorem
$P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)=\frac{1}{2} \times 1.3 \times\left[(120)^2-(90)^2\right]$
$=4095 \mathrm{~N} / \mathrm{m}^2$ or Pascal

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Air of density $1.2 \mathrm{~kg} \mathrm{~m}^{-3}$ is blowing across the horizontal wings of an aeroplane in such a way that its speeds above and below the wings are $150 \mathrm{~ms}^{-1}$ and $100 \mathrm{~ms}^{-1}$, respectively. The pressure difference between the upper and lower sides of the wings, is :

Options:
A
$60 \mathrm{Nm}^{-2}$

B
$180 \mathrm{Nm}^{-2}$

C
$7500 \mathrm{Nm}^{-2}$

D
$12500 \mathrm{Nm}^{-2}$

The correct answer is:
$7500 \mathrm{Nm}^{-2}$

Solution:
Pressure difference
$$
\begin{aligned}
& \mathrm{P}_2-\mathrm{P}_1=\frac{1}{2} \rho\left(\mathrm{v}_2^2-\mathrm{v}_1^2\right) \\
& =\frac{1}{2} \times 1.2\left((150)^2-(100)^2\right) \\
& =\frac{1}{2} \times 1.2(22500-10000) \\
& =7500 \mathrm{Nm}^{-2}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2013 (22 Apr Online)
Air pollution that occurs in sunlight is

Options:
A Acid rain
B Reducing smog
C Fog
D Oxidising smog

The correct answer is:
Oxidising smog

Solution:
Photochemical smog, also known as 'Oxidising smog' is an atmospheric condition that produces various unwanted effects such as eye irritation and poor visibility.
The main ingredients of photochemical smog include energy from sunlight, hydrocarbons and oxides of nitrogen.
The name 'Oxidising smog' is given because it contains high concentration of oxidising agents.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Air pressure in a car tyre increases during driving. Explain.

Options:

None

Solution:
During driving, reaction force due to force on tyres, temperature of the gas increases while its volume remains constant.
So, according to Charle's law, at constant volume $(V)$, temperature of car tyre increases during driving.
Pressure $(\mathrm{P}) \propto$ Temperature $(T)$
Hence, pressure of gas increases.

Topic: Physics, Thermodynamics
$\mathrm{Al}_2 \mathrm{O}_3$ can be converted to anhydrous $\mathrm{AlCl}_3$ by heating:

Options:
A $\mathrm{Al}_2 \mathrm{O}_3$ with $\mathrm{Cl}_2$ gas
B $\mathrm{Al}_2 \mathrm{O}_3$ with $\mathrm{HCl}$ gas
C $\mathrm{Al}_2 \mathrm{O}_3$ with $\mathrm{NaCl}$ in solid state
D A mixture of $\mathrm{Al}_2 \mathrm{O}_3$ and carbon in dry $\mathrm{Cl}_2$ gas

The correct answer is:
A mixture of $\mathrm{Al}_2 \mathrm{O}_3$ and carbon in dry $\mathrm{Cl}_2$ gas

Solution:
$\mathrm{Al}_2 \mathrm{O}_3$ may be converted to anhydrous $\mathrm{AlCl}_3$ by heating a mixture of $\mathrm{Al}_2 \mathrm{O}_3$ and carbon in dry chlorine.
$\mathrm{Al}_2 \mathrm{O}_3+3 \mathrm{C}+3 \mathrm{Cl}_2 \rightarrow \underset{\text {(anhydrous } \mathrm{AlCl}_3)}{\mathrm{Al}_2 \mathrm{Cl}_6}$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
$\mathrm{Al}_2 \mathrm{O}_3$ is reduced by electrolysis at low potentials and high currents. If $4.0 \times 10^4$ amperes of current is passed through molten $\mathrm{Al}_2 \mathrm{O}_3$ for 6 hours, what mass of aluminium is produced ? (Assume 100\% current efficiency, at. mass of $\mathrm{Al}=27 \mathrm{~g} \mathrm{~mol}^{-1}$ ) -

Options:
A $1.3 \times 10^4 \mathrm{~g}$
B $9.0 \times 10^3 \mathrm{~g}$
C $8.1 \times 10^4 \mathrm{~g}$
D $2.4 \times 10^5 \mathrm{~g}$

The correct answer is:
$8.1 \times 10^4 \mathrm{~g}$

Solution:
$$
\begin{aligned}
\mathrm{W} & =\frac{\mathrm{E}}{96500} \times \mathrm{I} \times \mathrm{t} \\
\mathrm{W} & =\frac{9}{96500} \times 4.0 \times 10^4 \times 6 \times 3600 \\
& =8.1 \times 10^4 \mathrm{~g}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, NEET, NEET 2009 (Mains)
$\mathrm{Al}_{2} \mathrm{O}_{3}$ reacts with

Options:
A only water
B only acids
C only alkalis
D both acids and alkalis

The correct answer is:
both acids and alkalis

Solution:
$\mathrm{Al}_{2} \mathrm{O}_{3}$ is amphoteric so it dissolve in acid as well in alkalis

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Alakali metals (IA) are generally soluble with liquid ammonia (Liq. \( \mathrm{NH}_{3} \) ) and provides blue colour solution and
shows paramagnetic behavior, so what could be the reason to provide blue colour and paramagnetic behavior?

Options:
A \( \left[\mathrm{M}\left(\mathrm{NH}_{3}\right)_{\mathrm{x}}\right]^{+} \)and \( \left[\mathrm{e}\left(\mathrm{NH}_{3}\right)_{\mathrm{y}}\right]^{-} \)
B \( \left[\mathrm{e}\left(\mathrm{NH}_{3}\right)_{y}\right]^{-} \)
C \( \left[\mathrm{e}\left(\mathrm{NH}_{3}\right)_{\mathrm{y}}\right]^{-} \)and \( \left[\mathrm{M}\left(\mathrm{NH}_{3}\right)_{\mathrm{x}}\right]^{+} \)
D \( \left[\mathrm{M}\left(\mathrm{NH}_{3}\right)_{\mathrm{x}}\right]^{+} \)

The correct answer is:
\( \left[\mathrm{e}\left(\mathrm{NH}_{3}\right)_{y}\right]^{-} \)

Solution:
The colour is due to the presence of solvated electrons in ammonia. It also imparts paramagnetic character to the solution.

Topic: Chemistry, s Block Elements, JEE Main
Albuminous seeds store their reserve food mainly in:

Options:
A endosperm
B cotyledons
C hypocotyl
D perisperm.

The correct answer is:
endosperm

Solution:
The seed which store food materials in endosperm are called endospermic or albuminous seed. Cotyledon is the seed leaf within the embryo of a seed. These stored nutrients are supplied to the seedling during germination. Upon germination, the hypocotyl pushes the cotyledons above the ground to develop. It eventually becomes part of the plant stem, while radicle forms roots. Perisperm is the remnant of nucellus which are persistent. It surrounds, stores food and provide nourishment to the embryo in an angiosperm seed. It is diploid in nature.
Related Theory
The cotyledon is described as a seed leaf that stores food in the form of starch and protein for use by the embryo. An embryo of a monocotyledon (monocot) plant has one cotyledon, while that of a dicotyledon (dicot) plant has two cotyledons. A monocot stores the bulk of its energy in the endosperm.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2013 (Karnataka)
Albuminous seeds store their reserve food mainly in :

Options:
A Endosperm
B Cotyledons
C Hypocotyl
D Perisperm

The correct answer is:
Endosperm

Solution:
Albuminious seeds retains a part of endosperm as it is not completely used up during embryo development, such seeds are also known as endospermic seeds. Food is stored in endosperm which provides nutrition to embryo during germination. Example : wheat , maize, barley, castor.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Alcohol (X) → aq. NaOH +  I 2 CHI 3 +(Y) → H 3 O + PhCH 2 COOH . the alcohol (X) is

Options:
A CH 3 CHCH OH CH 2        |       Ph
B PhCH2CHOHCH3
C PhCHOHCH2CH3
D PhCHCH3OH

The correct answer is:
PhCH2CHOHCH3

Solution:
I2+2NaOH→NaOI+NaI+H2O
PhCH2CHOHCH3→NaOIPhCH2-C||O-CH3→NaOIPhCH2-C||O-CI3
→NaOHCHI3+PhCH2COONa→H3O+PhCH2COOH .

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Aldehyde turns pink with

Options:
A Benedict solution
B Schiff reagent
C Fehling solution
D Tollen's reagent

The correct answer is:
Schiff reagent

Solution:
Schiff s reagent $\xrightarrow{\mathrm{SO}_2}$ Colourless (Pink) $\xrightarrow{\text { Aldehyde }}$ Pink
colour

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Aldehyde with NH2-NH2 forms

Options:
A Hydrazone
B Aniline
C Nitrobenzene
D None of these

The correct answer is:
Hydrazone

Solution:
Aldehydes and ketones with NH2-NH2 forms hydrazones. This is Nucleophilic additon followed by Elimination reaction
RCHO+ H 2 N− NH 2 →RCH=N− NH 2 + H 2 O

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Aldehydes and ketones give addition reaction with

Options:
A Hydrazine
B Phenyl hydrazine
C Semicarbazide
D Hydrogen cyanide

The correct answer is:
Hydrogen cyanide

Solution:
With hydrogen cyanide, carbonyl compounds give nucleophilic addition reaction while with hydrazine, phenyl hydrazine and semicarbazide, they give addition/ elimination reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Aldol condensation involving $\mathrm{CH}_3 \mathrm{CHO}+\mathrm{CH}_3 \mathrm{CHO}$ gives the product

Options:
A $\mathrm{CH}_3 \mathrm{CHOHCH}{ }_2 \mathrm{CHO}$
B $\mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{CH}_3$
C $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2$
D None of these

The correct answer is:
$\mathrm{CH}_3 \mathrm{CHOHCH}{ }_2 \mathrm{CHO}$

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Aldosterone helps in the

Options:
A elimination of sodium, potassium and water.
B conservation of sodium, potassium and water.
C conservation of potassium and water and elimination of sodium.
D conservation of sodium and water and elimination of potassium.

The correct answer is:
conservation of sodium and water and elimination of potassium.

Solution:
Adrenal cortex secretes many hormones collectively known as corticoids. The corticoids which help to maintain water and electrolyte balance in our body are known as mineralocorticoids. Aldosterone is the most important mineralocorticoid in our body. This hormone acts on kidneys and enhances reabsorption of sodium and water in order to decrease their excretion. It also causes elimination of potassium in the urine.

Topic: Biology, Chemical Coordination and Integration, NEET
Algae are known to reproduce asexually by variety of spores under different environmental conditions. Name these spores and the conditions under which they are produced.

Options:

None

Solution:
Asexual reproduction in algae is very common mean of reproduction. Algae and their spores exhibit significant diversity and vary greatly in their level of specialisation. Asexual reproduction by spores and their types include:
(i) Zoospores are mobile flagellated spores. In this protoplasm of each vegetative cell undergoes repeated longitudinal division either into 2 or 4 , rarely 8 or 16 daughter protoplasts. Before the onset of division the parent cell loses its fiagella.
Each daughter protoplast after the last series of division secretes a cell wall and a neuromotor apparatus that develops two flagella, eyespots and contractile vacuoles. Each of the daughter cell thus formed resembles the parent cell in all aspects except the small size. Under favourable conditions formation of zoospores is very common.
(ii) Aplanospores are the non-motile spores. They are formed asexually within a cell, in which protoplast withdraws itself from the parent wall, rounds up and develops into aplanospores which germinate either directly or may divide to produce zoospores.
(iii) Hypnospores- In this, the protoplasm withdraws from the cell wall, rounds up and develops a thick wall under unfavourable conditions. These resting spores are called as hypnospores. Due to presence of haematochroma they are red in colour e.g., Vaucheria, Ulothrix.
(iv) Akinetes are special vegetative thick walled cells present in the filaments which remain under dormant state and resume germination under favourable conditions. They can also withstand unfavourable condition as Spirogyra.
(v) Statospores are thick walled spores produced in diatoms.
(vi) Neutral spores are the protoplast, of vegetative cells directly functioning as spores (e.g., Ectocarpus).
Following are the characteristic pigments of different classes.

Topic: Biology, Biological Classification
Aliphatic compound is the other name for

Options:
A Acyclic compounds
B Alicyclic compounds
C Ring compounds
D Closed chain compounds

The correct answer is:
Acyclic compounds

Solution:
Explanation: open chain compounds or acyclic compounds are otherwise called as aliphatic compounds.

Topic: Chemistry, General Organic Chemistry, JEE Main
Alkali metal from the following with least melting point is

Options:
A Cs
B Rb
C Na
D K

The correct answer is:
Cs

Solution:
On moving down the group in alkali metals, the melting point generally decreases.

As you move down the alkali metal group (Group 1 of the periodic table), the atomic size or atomic radius increases. This increase in atomic size leads to a decrease in the strength of metallic bonds between the alkali metal atoms. Weaker metallic bonding results in lower melting points because less energy is required to overcome these weaker bonds and convert the solid metal into a liquid state.

Element M.P. (K) Na 371K 336Rb 312Cs 302

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Alkali metals act as

Options:
A good dehydrating agent
B good reducing agent
C good oxidising agent
D none of these

The correct answer is:
good reducing agent

Solution:
Alkali metals have only one electron in their ultimate shell, hence they can easily donate electron and act as reductant.

Topic: Chemistry, s Block Elements, JEE Main
Alkali metals in each period have

Options:
A Smallest size
B Lowest ionization potential
C Highest ionization potential
D Highest electronegativity

The correct answer is:
Lowest ionization potential

Solution:
Lowest ionisation energy due to largest size.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Alkaline earth metal nitrates on heating decompose to give

Options:
A MNO22Â andÂ O2Â only
B MO,Â N2Â andÂ O2
C MO, NO2 and O2
D MOÂ andÂ NO2Â only

The correct answer is:
MO, NO2 and O2

Solution:
For alkaline earth metals
MNO32→MOs+2NO2+12O2

Topic: Chemistry, s Block Elements, JEE Main
Alkaline earth metals are

Options:
A Li, Be, K, Mg, $\mathrm{Ca}$
B $\mathrm{Be}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}$
C $\mathrm{Be}, \mathrm{K}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}$
D $\mathrm{Be}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{K}, \mathrm{Rb}$

The correct answer is:
$\mathrm{Be}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}$

Solution:
$\mathrm{2} \mathrm{Be}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}$ Alkaline earth metals comprises of $\mathrm{Be}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}$. Hence, the correct answer is option (2).

Topic: Chemistry, s Block Elements, JEE Main
Alkaline earth metals come under

Options:
A Halogens
B Representative elements
C Transition elements
D Inner transition elements

The correct answer is:
Representative elements

Solution:
Correct option is $\mathrm{2}$. Representative elements
Alkaline earth metals come under the Representative elements.
Hence, Option "2" is the correct answer

Topic: Chemistry, s Block Elements, JEE Main
Alkaline hydrolysis of esters is known as

Options:
A Saponification
B Hydration
C Esterification
D Alkalisation

The correct answer is:
Saponification

Solution:
Alkaline hydrolysis of esters is known as saponification.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Alkaline oxidative fusion of MnO2 gives "A " which on electrolytic oxidation in alkaline solution produces B. A and B respectively are:

Options:
A Mn2O7 and MnO4-
B MnO42- and MnO4-
C Mn2O3 and MnO42-
D MnO42- and Mn2O7

The correct answer is:
MnO42- and MnO4-

Solution:
The alkaline oxidative fusion is a redox reaction (undergoes both oxidation and reduction simultaneously) in which a compound undergoes fusion with itself only, in the presence of a base or an alkaline solution.

Electrolysis is a chemical process where an electric current is passed through an electrolyte, causing a non-spontaneous chemical reaction to occur, typically resulting in the decomposition of the electrolyte into its constituent elements or ions.

Alkaline oxidative fusion of MnO2 :
2MnO2+4OH-+O2→2MnO42-+2H2O

Electrolytic oxidation of MnO42- in alkaline medium.
MnO42-→MnO4-+e-

Hence, the answer is option B.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Alkanamines have the general formula -

Options:
A CnH2n–1N
B CnH2n+3N
C CnH2n+1N
D CnH2n-3N

The correct answer is:
CnH2n+3N

Solution:
C n H 2n+3 N
e.g. C 2 H 5 NH 2 (Ethylamine)

Topic: Chemistry, Amines, JEE Main
Alkenes convert into alcohols by

Options:
A Hydrolysis by dil. $\mathrm{H}_2 \mathrm{SO}_4$
B Hydration of alkene by alkaline $\mathrm{KMnO}_4$
C Hydrolysis by water yapours and conc. $\mathrm{H}_2 \mathrm{SO}_4$
D Hydration of alkene by aqueous $\mathrm{KOH}$

The correct answers are:
Hydration of alkene by alkaline $\mathrm{KMnO}_4$, Hydrolysis by water yapours and conc. $\mathrm{H}_2 \mathrm{SO}_4$

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Alkyl halides can be converted into Grignard reagents by

Options:
A Boiling them with $M g$ ribbon in alcoholic solution
B Warming them with magnesium powder in dry ether
C Refluxing them with $\mathrm{MgCl}_2$ solution
D Warming them with $\mathrm{MgCl}_2$

The correct answer is:
Warming them with magnesium powder in dry ether

Solution:
$R X+M g \xrightarrow{\text { Dry ether }} R-M g-X$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Alkyl halides react with dialkyl copper reagents to give

Options:
A
alkenes

B
alkyl copper halides

C
alkanes

D
alkenyl halides

The correct answer is:
alkanes

Solution:
$\mathrm{R}_2 \mathrm{CuLi}+\mathrm{R}^{\prime} \mathrm{X} \longrightarrow \mathrm{R}-\mathrm{R}^{\prime}+\mathrm{R}-\mathrm{Cu}+\mathrm{LiX}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2005
Alkyl halides react with $M g$ in dry ether to form

Options:
A Magnesium halide
B Grignard’s reagent
C Alkene
D Alkyne

The correct answer is:
Grignard’s reagent

Solution:
Alkyl halide reacts with Mg in presence of dry ether to give alkyl magnesium halide which is also called as Grignard reagent. This reaction is also called as Grignard’s reaction.
$R-X+M g \xrightarrow[\text { ether }]{\text { dry }} \underset{\text { Grignard's reagent }}{R-M g}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Alkyl halides undergoing $\mathrm{S}_{\mathrm{N}} 2$ reaction do inverse

Options:
A racemic mixture
B retention of configuration
C formation of carbocation
D inversion of configuration

The correct answer is:
inversion of configuration

Solution:
$$
\text { } \mathrm{S}_{\mathrm{N}} 2 \text { mechanism }
$$
Inversion of configuration takes place. There is no carbocation formation.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Alkylation of amines can be carried out by use of

Options:
A alcohol
B ammonia
C alkane
D alkyl halide

The correct answer is:
alkyl halide

Solution:
The alkylation of amines can be carried out by using the alkyl halides. The reactions for the same is given below:

It is a bimolecular nucleophilic substitution reaction.

Topic: Chemistry, Amines, JEE Main
All bonds in benzene are equal due to

Options:
A Tautomerism
B Inductive effect
C Resonance
D Isomerism

The correct answer is:
Resonance

Solution:
All the bonds $(C-C)$ are equal in benzene. The $C-C$ bond length is $1.39 Å$ which is in between $C-C$ bond ( $1.54 Å)$ and $C=C \quad(1.34 Å)$.

Topic: Chemistry, General Organic Chemistry, JEE Main
All but one of the following compounds reacts with aniline to give acetanilide. Which one does not?

Options:
A
B
C
D

The correct answer is:

Solution:
Hence CH3–CH=O does not give acetanilide with aniline.

Topic: Chemistry, Amines, JEE Main
All electromagnetic wave is transporting energy in the negative z direction. At a certain point and certain time the direction of electric field of the wave is along positive y direction. What will be the direction of the magnetic field of the wave at that point and instant?

Options:
A Positive direction of x
B Positive direction of z
C Negative direction of x
D Negative direction of y

The correct answer is:
Positive direction of x

Solution:
In an electromagnetic wave, the direction of energy flow is along the direction of the wave propagation. Therefore, the wave propagates along negative z direction.

Electric field is along the positive y direction and the wave propagates along negative z direction. Thus, magnetic field is in direction perpendicular to both z and y axes.

Using vector algebra,

(-k^)=(+j^)×i^

Hence, magnetic field is in positive x direction.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (25 Jan Shift 1)
All enzymes of the TCA cycle are located in the mitochondrial matrix except one which is located in inner mitochondrial membranes in eukaryotes and in the cytosol in prokaryotes. This enzyme is

Options:
A	Isocitrate dehydrogenase
B Malate dehydrogenase
C Succinate dehydrogenase
D Lactate dehydrogenase

The correct answer is:
Succinate dehydrogenase

Solution:
Succinate:ubiquinone oxidoreductase, also known in mitochondria as Complex II, provides a link between the citric acid cycle and the membrane-bound electron­ transport system. The membrane extrinsic, water-soluble domain, known as succinate dehydrogenase (SDH), contains the fumarate/succinate active site with a covalently bound FAD group and three iron-sulphur clusters: [2Fe-2S]2 +I 1+, [4Fe-4S]2+ 11+, and [3Fe-4S) 1 +10. The enzyme catalyzes the interconversion of fumarate and succinate; and is closely related to fumarate reductase.

Topic: Biology, Respiration in Plants, NEET
All form ideal solutions except

Options:
A $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Br}$
C $\mathrm{C}_6 \mathrm{H}_6$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3$
D $\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

Solution:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ do not form ideal solution.

Topic: Chemistry, Solutions, JEE Main
All letters of the word 'CEASE' are arranged randomly in a row, then the probability that two $\mathrm{E}$ are found together is

Options:
A $\frac{7}{5}$
B $\frac{3}{5}$
C $\frac{2}{5}$
D $\frac{1}{5}$

The correct answer is:
$\frac{1}{5}$

Solution:
Sample space arrangement for the word 'CEASE' $=5$ !
$$
\text { Here, } \mathrm{E} \rightarrow 2, \mathrm{C} \rightarrow 1, \mathrm{~A} \rightarrow 1, \mathrm{~S} \rightarrow 1
$$
Now, consider $2 \mathrm{E}$ as one character, so that arranging for four letters $=4 !$ ways
$\therefore$ Required probability $=\frac{4 !}{5 !}=\frac{1}{5}$

Topic: Mathematics, Permutation Combination, JEE Main
All of the following share the same crystal structure except.

Options:
A
$\mathrm{RbCl}$

B
$\mathrm{NaCl}$

C
$\mathrm{CsCl}$

D
$\mathrm{LiCl}$

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

Solution:
$\mathrm{RbCl}, \mathrm{NaCl}$ and $\mathrm{CsCl}$ share the same crystal structure except $\mathrm{LiCl}$.
$\mathrm{LiCl}$ is deliquescent. It crystalises as a hydrated LiCl. $2 \mathrm{H}_2 \mathrm{O}$. Other alkali metal chlorides do not form hydrates.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
All of the following statements apply to proteins except

Options:
A
Proteins generally have no definite melting point

B
Proteins contain the grouping - $\mathrm{CONH}-$

C
Proteins have high molecular weight

D
Proteins can only contain the elements $\mathrm{C}$, H, O and N.

The correct answer is:
Proteins can only contain the elements $\mathrm{C}$, H, O and N.

Solution:
Statement (d) is not correct. Some proteins also contain $\mathrm{S}$, with $\mathrm{C}, \mathrm{H}, \mathrm{O}$ and $\mathrm{N}$.

Topic: Chemistry, Biomolecules, JEE Main