datavorous/entrance-exam-dataset huggingface dataset:
A complex of molecular formula CrCl3.6H2O has green colour 0.1 mole of the complex when treated with excess of AgNO3 gave 28.7 g of white precipitate of AgCl (Molar mass = 143.5 g). The formula of the complex would be

Options:
A [Cr(H2O)6]Cl3
B [CrCl(H2O)5]Cl2.H2O
C [CrCl2(H2O)4]Cl.2H2O
D [Cr(H2O)3Cl3]3H2O

The correct answer is:
[CrCl(H2O)5]Cl2.H2O

Solution:
CrCl3.6H2O0.1mol+AgNO3excess→AgCl28.7g
No. of moles of AgCl=28.7143.5≈0.2mol
Hence it required 2 mole free Cl atom.

Topic: Chemistry, Coordination Compounds, JEE Main
A composite slab consists of two slabs A and B of different materials but of the same thickness placed one on top of the other. The thermal conductivities of A and B are k1 and k2 respectively. A steady temperature difference of 12 Â°C is maintained across the composite slab. If k1=k22, the temperature difference across the slab A will be

Options:
A 4°C
B 8°C
C 12°C
D 16°C

The correct answer is:
8°C

Solution:
For slab A ,
ΔQΔt=k1Al2ΔT1
For slab B ,
ΔQΔt=k2Al2ΔT2
∴ k1Al2ΔT1=k2Al2ΔT2
k22ΔT1=k212-ΔT2
ΔT1=24-2ΔT1
ΔT1=8 °C

Topic: Physics, Thermal Properties of Matter, NEET
A composite solid propellant is -

Options:
A N2O4+acrylic rubber
B N2O4+monomethyl hydrazine
C Polyurethane+Ammonium perchloride
D Nitrocellulose+nitroglycerine

The correct answer is:
Polyurethane+Ammonium perchloride

Solution:
Composite propellant are the mixture of polymeric binder like polyurethane or polybutadiene as fuel and ammonium perchlorate as an oxidizing agent along with on additive like finely divided Al or Mg to modify the working of propellants.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
A composite spherical shell is made up of two materials having thermal conductivities K and 2K respectively as shown in the diagram. The temperature at the innermost surface is maintained at T whereas the temperature at the outermost surface is maintained at 10T. A, B, C and D are four points in the outer material such that AB = BC = CD.

The effective thermal resistance between the inner surface of the shell and the outer surface of the shell for the radial heat flow is

Options:
A 1 8 Ï KR
B 1 7 π KR
C 7 4 8 π KR
D 6 4 9 π KR

The correct answer is:
7 4 8 π KR

Solution:
R eff = â« R 2 R dx K 4 Ï x 2 + â« 2 R 3 R dx 2 K 4 Ï x 2
= 1 4 Ï K - 1 x R 2 R + 1 8 Ï K - 1 x 2 R 3 R
â R eff = 7 4 8 Ï KR

Topic: Physics, Thermal Properties of Matter, NEET
A composite wire of length 2L is made by joining two different wires A and B having the same length, made of the same material but of different radii r and 2r respectively. The composite wire is vibrating at such a frequency, that the junction of the two wires form a node. If the number of antinodes in the wire A is p and that in the wire B is q , then the ratio p:q is

Options:
A 1:2
B 3:5
C 1:4
D 4:9

The correct answer is:
1:2

Solution:
f1=n2âTÏA; f2=m2âTÏ4A
Given f1=f2
f1f2=nmÃ2

nm=12

Topic: Physics, Waves and Sound, JEE Main
A compound A dissociate by two parallel first order paths at certain temperature

A g â k 1 ( min -1 ) 2 B g k 1 = 6 . 9 3 Ã 1 0 - 3 min -1

A g â k 2 ( min -1 ) C g k 2 = 6 . 9 3 Ã 1 0 - 3 min -1

If reaction started with pure 'A' with 1 mole of A in 1 litre closed container with initial pressure 2 atm. What is the pressure (in atm) developed in container after 50 minutes from start of experiment ?

Options:
A 1.25
B 0.75
C 1.50
D 2.50

The correct answer is:
2.50

Solution:
A g â¶ 2 B g ; k 1 = 6 . 9 3 Ã 1 0 - 3 min -1 A g â¶ C g ; k 2 = 6 . 9 3 Ã 1 0 - 3 min -1
k = ( k 1 + k 2 ) overall velocity constant = 6.93 x 2 x 10-3 min-1
t 1 / 2 = 0 . 6 9 3 Ã 1 0 0 0 6 . 9 3 Ã 2 = 5 0 min
i.e., after = 50 min
P A = 1 atm.
Since k1 = k2
P B = 0.5 x 2 = 1 atm (due to stoichio metric coefficient)
P C = 0.5 x 1 = 0.5 atm.
Total pressure = 2.5 atm.

Topic: Chemistry, Chemical Kinetics, JEE Main
A compound A has a molecular formula C2Cl3OH. It reduces Fehling solution and on oxidation gives a monocarboxylic acid B . A is obtained by action of chlorine on ethyl alcohol. A is

Options:
A Chloral
B CHCl3
C CH3Cl
D Chloroacetic acid

The correct answer is:
Chloral

Solution:
C 2 Cl 3 OH A â+âFehlingâsol.âââ Cu 2 O Redâppt
It means -CHO group is present.
C 2 Cl 3 OH A → Oxidation CCl 3 −COOH Monocarboxylic acid
It means only one âCHO group is present.
C2H5OH+Cl2âCH3CHO+2HCl
CH3CHO+3Cl2âCCl3CHOChloral+3HCl

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A compound $\mathrm{A}$ has a molecular formula $\mathrm{C}_2 \mathrm{Cl}_3 \mathrm{OH}$. It reduces Fehling solution and on oxidation gives a monocarboxylic acid ( $B$ ). A is obtained by action of chlorine on ethyl alcohol. $A$ is

Options:
A Chloral
B $\mathrm{CHCl}_3$
C $\mathrm{CH}_3 \mathrm{Cl}$
D Chloroacetic acid

The correct answer is:
Chloral

Solution:
It means $-\mathrm{CHO}$ group is present.
It means only one $-\mathrm{CHO}$ group is present.
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{Cl}_2 \rightarrow \mathrm{CH}_3 \mathrm{CHO}+2 \mathrm{HCl}$
$\mathrm{CH}_3 \mathrm{CHO}+3 \mathrm{Cl}_2 \rightarrow \underset{\text { Chloral }}{\mathrm{CCl}_3 \mathrm{CHO}}+3 \mathrm{HCl}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
A compound A is used in preparation of washing soda to recover ammonia in Solvay's process. When CO2 is bubbled through an aqueous solution of A , the solution turns milky. It is used in white washing due to disinfectant nature. What is the chemical formula of A?

Options:
A Ca(HCO3)2
B CaO
C Ca(OH)2
D CaCO3

The correct answer is:
Ca(OH)2

Solution:
Here Compound A is calcium hydroxide and the milky solution is of calcium carbonate.

Topic: Chemistry, s Block Elements, JEE Main
A compound A with molecular formula C10H13Cl, gives a white precipitate on adding silver nitrate solution. A on reacting with alcoholic KOH gives compound B as the main product. B on ozonolysis, gives C and D. C gives Cannizaro reaction, but not aldol condensation. D gives aldol condensation, but not Cannizaro reaction. A is

Options:
A
B
C
D

The correct answer is:

Solution:
Chlorine attached to tertiary carbon will give alkene (Saytzeff Rule).

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2015 (10 Apr Online)
A compound absorbs light in the wavelength region $490-500 \mathrm{~nm}$. Its complementary colour is

Options:
A red
B blue
C orange
D blue-green

The correct answer is:
red

Solution:
When a compound absorbs a certain wavelength (i.e., $490 \mathrm{~nm}-500 \mathrm{~nm}$ ) from the visible light which corresponds to the blue-green light in the visible spectrum, the colour or light transmitted by it does not contain the colour of absorbed radiation and thus shows the complementary colour, i.e., red.

Topic: Chemistry, Coordination Compounds, JEE Main
A compound C6H14O2 has two tertiary alcoholic groups. The IUPAC name of this compound is

Options:
A 2, 3-dimethyl-1, 2-butanediol
B 3, 3-dimethyl-1, 2-butanediol
C 2, 3-dimethyl-1, 2-butanediol
D 2-methyl-2, 3-pentanediol

The correct answer is:
2, 3-dimethyl-1, 2-butanediol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
A compound contains three elements X, Y and Z. The oxidation number of X, Y and Z are +3, +5 and -2 respectively. The possible formula of the compound is

Options:
A X2YZ
B XYZ3
C X3(YZ4)3
D X2Y2Z

The correct answer is:
X3(YZ4)3

Solution:
The sum of all oxidation values must be equal to zero. Let, number of atoms of X, Y and Z are respectively a, b and c.
Then, (+3)a+(+5)b+(-2)c=0
Or
3a+5b-2c=0
Now, putting values of a, b and c for all the given options, which will give sum equal to zero, will be our formula.
(a) For, X2YZ
a=2
b=1
c=1
Thus, 3a+5b-2c=0
3×2+5-2≠0
(b) For, XYZ3
a=1,b=1,c=3
3a+5b-2c=0
3+5-6≠0
(c) X3(YZ4)3
a=3,b=3,c=12
3a+5b-2c=0
(3×3+5×3)-(2×12)=0
9+15-24
24-24=0
(d) X2Y2Z
a=2,b=2,c=1
3a+5b-2c=0
(3×2)+(5×2)-(2×1)=0
6+10-2≠0
Hence, option (c) is the correct answer.

Topic: Chemistry, Redox Reactions, JEE Main
A compound formed by elements X and Y crystallises in in a cubic structure in which atoms X are at the corners of the cube and atoms Y are at two non-adjacent face-centres. The formula of the compound is

Options:
A X3Y
B XY
C XY2
D XY3

The correct answer is:
XY

Solution:
•→X
→Y
Number of X atoms =18×8=1
Number of Y atoms =12×2=1
formula of the solid compound =XY

Topic: Chemistry, Solid State, JEE Main
A compound has the following composition by weight ; $\mathrm{Na}=18.60 \%, \mathrm{~S}=25.80 \%, \mathrm{H}=4.02 \%$ and $\mathrm{O}=51.58 \%$ Assuming that all the hydrogen atoms in the compound are part of water of crystallization, the correct molecular formula of the compound is

Options:
A $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}$
B $\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$
C $\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}$
D $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}$

The correct answer is:
$\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}$

Solution:
\begin{array}{|c|c|c|c|}
\hline Elements & \% & moles & Simplest ratio \\
\hline \mathrm{Na} & 18.6 & \frac{18.6}{23}=0.8 & 1 \times 2 \\
\hline \mathrm{S} & 25.8 & \frac{25.8}{32}=0.8 & 1 \times 2 \\
\hline \mathrm{O} & 51.58 & \frac{51.58}{16}=3.22 & 4 \times 2 \\
\hline \mathrm{H} & 4.02 & \frac{4.02}{1}=4.02 & 5 \times 2 \\
\hline
\end{array}
$\Rightarrow$ Formula is $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A compound is formed by two elements X and Y. The element Y forms cubic close packed arrangement and those of element X occupy one third of the tetrahedral voids. What is the formula of the compound?

Options:
A X2Y3
B X3Y2
C X3Y
D XY3

The correct answer is:
X2Y3

Solution:
Atoms of Element Y, are present in cubic close packing. In cubic close packing the number of atoms per unit cell is 4.

So, the effective number of element Y, per unit cell is 4

Element X, are present at one third of the tetrahedral voids.

Number of tetrahedral void =Twice the number of atoms in unit cell

Number of atoms of Element X are present unit cell =13×8=83

X : Y83 : 42 : 3

Formula of the compound =X2Y3

Topic: Chemistry, Solid State, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A compound is treated with NaNH2 to give sodium salt. Identify the compound

Options:
A C2H2
B C6H6
C C2H6
D C2H4

The correct answer is:
C2H2

Solution:
Only C2H2 (acetylene) has acidic H-atoms and hence reacts with NaNH2 to form sodium salt, i.e.,
HC≡CH+ NaNH 2 →HC≡ C Θ Na ⊕ + NH 3

Topic: Chemistry, General Organic Chemistry, JEE Main
A compound \( \mathrm{X} \) has the following characteristics:
a. It is used as a disinfectant.
b. It is used to prepare bleaching powder.
c. Its suspension in water is called milk of lime.
Then compound \( \mathrm{X} \) is

Options:
A \( \mathrm{Ca}(\mathrm{OH})_{2} \)
B \( \mathrm{CaO} \)
C \( \mathrm{CaCO}_{3} \)
D \( \mathrm{CaSO}_{4} 2 \mathrm{H}_{2} \mathrm{O} \)

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

Solution:
Bleaching powder is prepared by passing chlorine gas (Cl2) through a solution of calcium hydroxide Ca(OH)2.

The reaction that occurs is, Ca(OH)2+Cl2 (g)→CaOCl2+H2O.

When an excess of calcium oxide is added to water, Ca(OH)2 forms,

CaO+H2O→Ca(OH)2

It partially dissolves in water to produce a solution called lime water, which is a moderate base.

Ca(OH)2 is highly basic and antibacterial due to which it is used as a disinfectant.

Topic: Chemistry, s Block Elements, JEE Main
A compound microscope consists of an objective lens of focal length 1 cm and an eye piece of focal length 5 cm with a separation of 10 cm. The distance between an object and the objective lens, at which the strain on the eye is minimum is n40cm. The value of n is.......

Options:

The correct answer is:
50

Solution:
L=10
Ve=∞

ue=fe=5
V0=10-5=5

1v0-1u0=1f0
15-1u0=11
u0=-54cm
54=n40
n=50

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A compound of formula A2B3 has the HCP lattice. Which atom forms the HCP lattice and what fraction of the tetrahedral voids are occupied by the other atoms?

Options:
A

HCP lattice - B

23tetrahedral voids - A

B

HCP lattice - A

13 tetrahedral voids - B

C

HCP lattice - B

13 tetrahedral voids - A

D

HCP lattice - A

23 tetrahedral voids - B

The correct answer is:

HCP lattice - B

13 tetrahedral voids - A

Solution:
Atom B forms HCP as anions are larger in size.

From the formula A2B3,

Number of B atoms=3,

Then, number of tetrahedral voids =2×Tetrahedral voids=2×3=6.

As 'A' is occupying tetrahedral voids,

1 3 of tetrahedral voids=13×6=2
Therefore, formula of the compound=A2B3

Topic: Chemistry, Solid State, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A compound of molecular formula C8H8O2 reacts with acetophenone to form a single cross-aldol product in the presence of base. The same compound on reaction with concentrated NaOH forms benzyl alcohol as one of the products. The structure of the compound is:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2017 (09 Apr Online)
A compound of vanadium chloride has spin only magnetic moment of 1.73 BM. Its formula is

Options:
A VCl2
B VCl5
C VCl4
D VCl3

The correct answer is:
VCl4

Solution:
Spin only Magnetic Moment
μ = n n + 2 BM = 1 . 7 3 = 3
∴ n = 1 (unpaired e-)
Electronic configuration of V= 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2
to get 1 unpaired electron V has to lose 4 electrons to form V 4+.
So, formula of chloride is VCl 4

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2014 (09 Apr Online)
A compound of Xe and $\mathrm{F}$ is found to have $53.5 \%$ of Xe. What is oxidation number of Xe in this compound?

Options:
A -4
B 0
C 4
D 6

The correct answer is:
6

Solution:
$\mathrm{Xe}=53.5 \% \therefore \mathrm{F}=46.5 \%$

Relative number of atoms Xe

$$

=\frac{53.5}{131.2}=0.4 \text { and } \mathrm{F}=\frac{46.5}{19}=2.4

$$

Simple ratio $\mathrm{Xe}=1$ and $\mathrm{F}=6 ;$ Molecular formula is $\mathrm{XeF}_{6}$

O. $\mathrm{N}$.of $\mathrm{Xe}$ is +6

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A compound that shows positive iodoform test is

Options:
A 2-pentanone
B 3 -pentanone
C 3- pentanol
D 1-pentanol

The correct answer is:
2-pentanone

Solution:
(A)
(B)
(C)
(D) $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A compound with molecular formula C4H10O3 in converted by the action of acetyl chloride to a compound of molecular mass 190. The original compound C4H10O3 has how many -OH groups?

Options:

The correct answer is:
2

Solution:
In acetylation, replacement of H of O-H occurs by CH3CO- group. -O-H+ClC||O-CH3→-OCOCH3+HCl
i.e., H atom of mass 1 amu is lost and an acetyl group of mass 43 amu is added. Thus, there is a net gain of 43 - 1 = 42 amu for every acetyl group introduced. Mass of difference of final product and original compound is 190 - 106 = 84.
Hence, number of -OH groups =8442=2 .

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
A compound with molecular formula \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}_{4} \) on acylation with acetic anhydride gives a compound with molecular formula \( \mathrm{C}_{12} \mathrm{H}_{18} \mathrm{O}_{8} \). How many hydroxyl groups are present in the compound?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Acylation is used to determine the number of -OH group in the compound.
R - OH ⟶ R - OCOCH 3
âAcylation of one hydroxyl group results in the increment of -COCH2 (or C2H2O). In the question, there is an increment of (C8H8O4). Hence, there should be four -OH groups.
−OH+ ( CH 3 CO ) 2 O→− OCOCH 3 + CH 3 COOH

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is

Options:

The correct answer is:
5

Solution:
R-NH2M+CH3COCl→RNHCOCH3M+42+HCl
If the increase in molecular mass is 42, the reactant has one NH2 group.
Actual increase in molecular mass =390-180=210
Number of NH2 groups =21042=5

Topic: Chemistry, Amines, JEE Main
A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups presents per molecule of the former compound is

Options:
A 4
B 6
C 2
D 5

The correct answer is:
5

Solution:
After acylation the formula will be R-(NHCOCH3)x. Hence, the change in the molecular mass is calculated as below

For 1one NH2→14+2=16 unitsWhen It turns into NH−COCH3 ​ →14+1+12+16+12+3=58 Change in mass⇒58−16=42units. So for every NH2 , there will be an increase of 42g after acylation. ⇒ The total increase in mass = 390−180 = 210 ⇒ No. of amino groups=21042= 5 Hence number of amino groups are 5.
(12+16+12+3-1)x=390-180
or 42x=210; or x=5

Topic: Chemistry, Amines, JEE Main
A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups presents per molecule of the former compound is

Options:
A 4
B 6
C 2
D 5

The correct answer is:
5

Solution:
After acylation the formula will be R-(NHCOCH3)x. Hence, the change in the molecular mass is calculated as below

For 1one NH2→14+2=16 unitsWhen It turns into NH−COCH3 ​ →14+1+12+16+12+3=58 Change in mass⇒58−16=42units. So for every NH2 , there will be an increase of 42g after acylation. ⇒ The total increase in mass = 390−180 = 210 ⇒ No. of amino groups=21042= 5 Hence number of amino groups are 5.
(12+16+12+3-1)x=390-180
or 42x=210; or x=5

Topic: Chemistry, Amines, JEE Main, JEE Main 2013 (07 Apr)
A compound $X$ contains $32 \%$ of $A, 20 \%$ of $B$ and remaining percentage of $C$. Then, the empirical formula of X is :
(Given atomic masses of $A=64 ; B=40 ; C=32 u$)

Options:
A $\mathrm{ABC}_3$
B $\mathrm{AB}_2 \mathrm{C}_2$
C $\mathrm{ABC}_4$
D $\mathrm{A}_2 \mathrm{BC}_2$

The correct answer is:
$\mathrm{ABC}_3$

Solution:
$\begin{array}{|c|c|c|c|c|}\hline \text{Element} & \begin{array}{c} \text{Mass} \\ \text{percentage} \%\end{array} & \begin{array}{c} \text{No. of} \\ \text{moles} \end{array} & \begin{array}{c} \text{No. of moles} / \\ \text{Smallest number} \end{array} & \begin{array}{c} \text{Simplest whole} \\ \text{number} \end{array} \\\hline A & 32 \% & \frac{32}{64}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\\hline B & 20 \% & \frac{20}{40}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\\hline C & 48 \% & \frac{48}{32}=\frac{3}{2} & \frac{3}{2} \times 2 & =3 \\\hline\end{array}$
So, empirical formula of
$X=\begin{array}{clll}A & : & B & \text { C } \\ 1 & : & 1 & :\end{array}$
$\therefore$ The correct empirical formula of compound X is $\mathrm{ABC}_3$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2024
A compound 'X' is a weak acid and it exhibits colour change at pH close to the equivalence point during neutralization of NaOH with CH3COOH. Compound 'X' exists in ionized form in basic medium. The compound 'X' is

Options:
A methyl orange
B methyl red
C phenolphthalein
D erichrome Black T

The correct answer is:
phenolphthalein

Solution:
phenolphthalein, (C20H14O4), an organic compound of the phthalein family that is widely employed as an acid-base indicator. As an indicator of a solution's pH, phenolphthalein is colourless below pH 8.5 and attains a pink to deep red hue above pH 9.0.

It is used for the titration of weak acid and strong base. Phenolphthalein is weak acid give colour in basic medium. It exist in ionized form in basic medium.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A compound 'X' is acidic and it is soluble in NaOH solution, but insoluble in NaHCO3 solution. Compound 'X' also gives violet colour with neutral FeCl3 solution. The compound 'X' is

Options:
A

B

C

D

The correct answer is:

Solution:
Generally alcohols are acidic in nature but they does not react with NaOH solution. Among the given options only phenol is acidic enough to react with NaOH solution.

So, compound 'X' is phenol which is acidic and it is soluble in NaOH solution, but insoluble in NaHCO3 solution. Phenol also gives violet colour with neutral FeCl3 solution according to the following reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (29 Jul Shift 1)
A compound (X) of molecular formula C3H6O forms bisulphate complex, gives iodoform test but does not reduce Tollens reagent. (X) on reaction with CH3MgBr/H3O+ gives a compound (Y) that cannot

Options:
A give red colour with Cerium Ammonium Nitrate (CAN)
B give white turbidity immediately with Lucas reagent
C give iodoform test
D be dehydrated to alkene on reaction with heated Cu

The correct answer is:
give iodoform test

Solution:
In given conditions X C3H6O should be a methyl ketone.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
A compound 'X' on treatment with Br $/ \mathrm{NaOH}$, provided $\mathrm{C}_{3} \mathrm{H}_{9} \mathrm{~N},$ which gives positive carbylamine test. Compound $\mathrm{X}^{\prime}$ is:

Options:
A $\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{NHCH}_{3}$
B $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{NH}_{2}$
C $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CONH}_{2}$
D $\mathrm{CH}_{3} \mathrm{CON}\left(\mathrm{CH}_{3}\right)_{2}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CONH}_{2}$

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by ΔT. The net change in its length is zero. Let l be the length of the rod, A its area of cross-section, Y its Young's modulus, and α its coefficient of linear expansion. Then, F is equal to:

Options:
A lAYα ΔT
B A YÎ± ÎT
C AYÎ±ÎT
D l2 YÎ± ÎT

The correct answer is:
A YÎ± ÎT

Solution:
Net change in length =0
Thermal expansion. Îl =lÎ± Ît
Y=FAÎll
Îll=FAY
Îl=FlAY
FlAY=l Î± Ît
F=AY Î± Ît

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2017 (08 Apr Online)
A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by $1 \%$. If the Poisson's ratio for the material of the rod be $0.2,$ which of the following statements is correct? "The volume approximately ...........

Options:
A decreases by $1 \%$
B decreases by $0.8 \%$
C decreases by $0.6 \%$
D increases by $0.2 \%$

The correct answer is:
decreases by $0.6 \%$

Solution:
Given, Decrement in the length $=1 \%$ Poisson's ratio for material of the rod, $\sigma=0.2$
As we know that, Volume, $v=\pi r^{2} l,[$ where $, l$ is the length of the rod $]$
$$
\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}...(i)
$$
since, Poisson's ratio, $\sigma=\frac{-\Delta D / D}{\Delta L / L}$
$$
=\frac{-\Delta r / r}{\Delta L / L}
$$
So, Eq. (i) can be written as,
$$
\begin{aligned}
\frac{\Delta V}{V}=\frac{\Delta l}{l}(1-2 \sigma) \\
\frac{\Delta V}{V} \times 100 &=\left(\frac{\Delta l}{l} \times 100\right) \times(1-2 \sigma) \\
&=-1 \times[1-2 \times(0.2)] \\
&=-1 \times[1-0.4] \\
&=-0.6 \%
\end{aligned}
$$
Here, negative sign shows the decrement in the volume.

Topic: Physics, Mechanical Properties of Solids, JEE Main