datavorous/entrance-exam-dataset huggingface dataset:
The major product of the following reaction sequence is

Options:
A \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CdCH}_{3} \)
B \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOH}\left(\mathrm{CH}_{3}\right)_{2} \)
C \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3} \)
D \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{3} \)

The correct answer is:
\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3} \)

Solution:
Acyl chlorides upon treatment with Grignard reagent and a metal halide, yield ketones.For example, when cadmium chloride is reacted with the Grignard reagent, dialkylcadmium is formed. Dialkylcadmium thus formed is further reacted with acyl chlorides to form ketones.

3CH3-CH2-CO-Cl + (CH3)2Cd→ 2CH3-CH2-CO-CH3 + CdCl2

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The major product of the following reaction

Options:
A
B
C
D

The correct answer is:

Solution:
Chromium oxide is an oxidising agent and oxidizes alcohol to carboxylic acid, which on reaction with thionyl chloride convert into acyl halide followed by hydrolysis.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The major product of the following reactions

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHBrCO}_2 \mathrm{H}$
B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHBrCO}_2 \mathrm{H}$
C $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHBrCO}_2 \mathrm{H}$
D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHBrCOBr}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHBrCO}_2 \mathrm{H}$

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The major product of the following reactions is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product of the reaction is

Options:
A

​

B

​

C

​

D

​

The correct answer is:

​

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2014 (09 Apr Online)
The major product of the reaction is

Options:
A

​

B

​

C

​

D

​

The correct answer is:

​

Solution:

Topic: Chemistry, Amines, JEE Main
The major product of the reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
The major product 'P' for the following sequence of reactions is:

Options:
A
B
C
D

The correct answer is:

Solution:
In the Clemmesen reduction Zn-Hg/HCl reduces aldehydes and ketone to alkane but doesn't reduce amide to amine, lithium aluminium hydride subsequently can reduce amide to amine. Chain of reactions are:

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The major product P formed in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
Chlorobenzene tends to readily react with Chloral in the presence of some sulfuric acid to produce a compound called DDT. This is condensation reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product P of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
Here addition reaction takes place to give 2,4-dibromo pentane

Topic: Chemistry, Hydrocarbons, JEE Main
The major product P of the given reaction is (where, Me is -CH3)

Options:
A
B
C
D

The correct answer is:

Solution:
Dehydration of alcohols involves the formation of carbocation intermediate and this carbocation rearrange to form more stable carbocation which looses proton to form a more stable alkene.

In the following dehydration reaction also initially a secondary carbocation is formed which undergoes methyl shift to form a more stable tertiary carbocation which looses proton to form a stable alkene.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (28 Jun Shift 2)
The major product Q of the following schemes is

Options:
A

B

C

D

The correct answer is:

Solution:
We know that the reagent of KMnO4/KOH is act as strong oxidising agent when it is react with toluene then it cause of toluene into benzoic acid

While we also know that about the Grignard reagent that it changes the aldehyde group into the alcoholic group i.e.

So the correct product is

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product [R] in the following sequence of reaction is :

Options:
A
B
C
D

The correct answer is:

Solution:
Alcohol upon reaction with protic acids tends to lose a molecule of water to form alkenes. These reactions are known as dehydrogenation or dehydration of alcohols. It is an example of an elimination reaction.The order of the ease of dehydration of alcohols is: tertiary > secondary > primary.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The major product (s) obtained in the following reaction is/are:

Options:
A
B
C
D

The correct answer is:

Solution:
First step of the reaction is dehydrohalogenation and then ozonolysis.

The dehydrohalogenation of alkyl halides, another ß elimination reaction, involves the loss of a hydrogen and a halide from an alkyl halide (RX). Dehydrohalogenation is normally accomplished by reacting the alkyl halide with a strong base, such as sodium ethoxide.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The major product U in the following reactions is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The major product (X) of the monobromination reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
In the mono bromination of cycloalkene with NBS the bromination takes place at the allylic position.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product Y in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The major product ′Y′ in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
In the first step, elimination reaction takes place as sodium ethoxide is a strong base
in the second step, hydrobromination takes place in accordance with Markovnikov's rule

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The major product Y in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The major product Y in the following reactions is:

Options:
A
B
C
D

The correct answer is:

Solution:
Grignard reagents add to carbonyl compounds to give primary, secondary, and tertiary alcohols. Reacting a Grignard reagent with an aldehyde gives a secondary alcohol. Reacting a Grignard reagent with a ketone gives a tertiary alcohol.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The major product Z obtained in the following reaction scheme is:

Options:
A
B
C
D

The correct answer is:

Solution:
When a diazonium salt is treated with copper (I) chloride (Cu2Cl2Cu2Cl2) or copper (I) bromide (Cu2Br2), the corresponding haloarene is formed. This reaction is known as Sandmeyer reaction. It is used for introducing a chloro or bromo group in the benzene ring.

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The major product $(Z)$ of the following chemical reaction is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The major product(A) of the reaction given below is

Options:
A
B
C
D

The correct answer is:

Solution:
It is a case of free radical addition to C=C bond.

Topic: Chemistry, Hydrocarbons, JEE Main
The major product(P) in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
In this reaction, the given ether is converted to phenolic derivative and ethyl bromide.

Treatment of alkenes with hydrobromic acid will result in the formation of alkyl bromides according to Markonvikov's rule. Markonvikov's rule is that when an unsymmetrical reagent adds to an unsymmetrical alkene, the more positive part of the agent goes to the carbon that has more hydrogen atoms.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The major products A and B for the following reactions are, respectively:

Options:
A
B
C
D

The correct answer is:

Solution:
1. In polar aprotic solvent, primary alkyl halides undergo S N 2 reaction with KCN, to give alkyl nitrile.
2. Alkyl nitriles, reduce in presence of H 2 /Pd, give 1 o − amines.
3. Cyclic ketones, on reduction with H 2 /Pd, form 2 o − alcohol.
H2/Pd reduces -CN into -CH2-NH2

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The major products A and B from the following reactions are :

Options:
A
B
C
D

The correct answer is:

Solution:
Acetanilide undergo electrophilic bromination with bromine in acetic acid. p-bromoacetaniide is the major product in this reaction.

Acetanilide undergo reduction reaction with lithium aluminium hydride.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The major products A and B in the following reactions are

Options:
A

and

B

and

C

and

D

and

The correct answer is:

and

Solution:
A is a radical, which undergoes addition to give a stable radical, and results like the Markovnikov product B.

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The major products A and B in the following set of reactions are:

Options:
A
B
C
D

The correct answer is:

Solution:
Alcohol undergoes elimination reaction inpresence of H2SO4/H3O+

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The major products ‘A’ and ‘B’, respectively, are

Options:
A
B
C
D

The correct answer is:

Solution:
Alkenes react with concentrated sulfuric acid in the cold to produce alkyl hydrogensulphates. At 80°C, the carbocation intermediate formed reacts with another molecule of alkene.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The major products formed in the following reaction sequence A and B are:

Options:
A
B
C
D

The correct answer is:

Solution:
It is Haloform reaction. The haloform reaction is the reaction of a methyl ketone with chlorine, bromine, or iodine in the presence of hydroxide ions to give a carboxylate ion and a haloform.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The major products from the following reaction sequence are product \(A\) and product B.
The total sum of \(\pi\) electrons in product \(A\) and product \(B\) are _______ (nearest integer)

Options:

The correct answer is:
8

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2024 (06 Apr Shift 1)
The major products of the following reaction are:

Options:
A
B
C
D

The correct answer is:

Solution:
Initial reaction is Reimer -Tiemann reaction, which gives salicylaldehyde. Which will undergo cross Cannizzaro reaction followed by hydrolysis.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The major products $P$ and $Q$ formed in the following reactions of benzonitrile are

Options:
A

B

C

D

The correct answer is:

Solution:
When $\mathrm{C}_d \mathrm{H}_5 \mathrm{CN}$ reacts with $\mathrm{Na} / \mathrm{E}$ toH, $\mathrm{C}_e \mathrm{H}_5 \mathrm{CN}$
undergo for reduction and $-\mathrm{C}=\mathrm{N}$, changes to $-\mathrm{CH}_2-\mathrm{NH}_2$.
Thus, the product $(P)$ is
The reaction occurs as follows :
When $\mathrm{C}_s \mathrm{H}_5 \mathrm{CN}$ reacts with (i) $\mathrm{HCl}$ (conc. cold) and (ii) $\mathrm{Br}_2 / \mathrm{KOH}$, it gives corresponding primary amine with one $\mathrm{C}$-atom less, as in parent formula (using Hoffmann-Bromamide degradation reaction). Thus, the final product is
The whole reaction is as follows:
Hence, option (c) is the correct answer.

Topic: Chemistry, Amines, JEE Main
The major products P and Q in the following reaction sequence are

Options:
A
B
C
D

The correct answer is:

Solution:
KMnO4 is a strong oxidising agent. It can oxidise the alkyl group to carboxylic acid and ketone also oxidise into carboxylic acid.LiAlH4 is strong reducing reagent. It can reduce the carboxylic acid to alcohols.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The major products $P$ and $Q$ in the following reactions are
$\begin{array}{ll}P & Q\end{array}$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
The major products Q and R from the following reactions, respectively are

Options:
A

Q	R

B

Q	R

C

Q	R

D

Q	R

The correct answer is:

Q	R

Solution:
We know that when Br2/FeBr react with benzoic acid then Br attack at meta position because carboxylic acid act as meta directing group and on further reaction of meta bromine benzoic acid with B2H6 carboxylic acid is reduce to the methanol

But on reaction of meta bromine benzoic acid with MeOH, water molecule is eliminated from the molecule and we get the final product

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The major products X and Y in the following sequence of transformations

are

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The major role of minor elements inside living organisms is to act as

Options:
A binding factor of cell structure
B co-factors of enzymes
C building blocks of important amino acids
D constituent of hormones

The correct answer is:
co-factors of enzymes

Solution:
The micronutrients or the minor elements are the ones which are not required in large quantities. The minerals required can be either macronutrients or micronutrients. Macronutrients are the elements which are required in large amounts like carbohydrates, proteins, fats whereas micronutrients are those which mainly act as a co-factor of enzymes. Eg: Copper acts as a co-factor of the enzyme cytochrome oxidase, iron is needed by the enzyme catalase etc.

Topic: Chemistry, Biomolecules, NEET
The manifestation of band structure in solids is due to

Options:
A
Heisenberg's uncertainty principle

B
Pauli's exclusion principle

C
Bohr's correspondence principle

D
Boltzmann's law

The correct answer is:
Pauli's exclusion principle

Solution:
No Solution Available

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2004
The marks obtained by 9 students in a chemistry test are 50,69,20,33,53,39,40,65 and 59. If the mean deviation about the median of this data is Î», then the value of 9Î»10 is equal to

Options:

The correct answer is:
11.4

Solution:
20,33,39,40,50,53,59,65,69
Median, M =9+12th observation =5th observation =50
The absolute values of respective deviations from the median is
30,17,11,10,0,3,9,15,19
∴∑i=19xi-M=114
Mean deviation about M=19∑i=19xi-M=19×114

i.e λ=1149⇒9λ10=11.4

Topic: Mathematics, Statistics, JEE Main
The marks of some students were listed out of \( 75 \). The SD of marks was found to be \( 9 \). Subsequently the marks were raised to a maximum of \( 100 \) and variance of new marks was calculated. The new variance is,

Options:
A \( 144 \)
B \( 122 \)
C \( 81 \)
D None of these

The correct answer is:
\( 144 \)

Solution:
Given, σ=9

Let a student obtains x out of 75.

Then his marks out of 100 are 4x3.

⇒ Each observation is multiplied by 4 3

⇒ New σ=43×9=12,

⇒ New Variance =σ2=144

Topic: Mathematics, Statistics, JEE Main
The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

Options:
A $9.8 \mathrm{~m} / \mathrm{sec}^2$
B $4.9 \mathrm{~m} / \mathrm{sec}^2$
C $980 \mathrm{~m} / \mathrm{sec}^2$
D $19.6 \mathrm{~m} / \mathrm{sec}^2$

The correct answer is:
$4.9 \mathrm{~m} / \mathrm{sec}^2$

Solution:
$\frac{g^{\prime}}{g}=\frac{M^{\prime}}{M}\left(\frac{R}{R^{\prime}}\right)^2=\left(\frac{2 M}{M}\right)\left(\frac{R}{2 R}\right)^2=\frac{1}{2}$
$\Rightarrow g^{\prime}=\frac{g}{2}=\frac{9.8}{2}=4.9 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Gravitation, JEE Main
“The mass and energy both are conserved in an isolated system”, is the statement of

Options:
A Second law of thermodynamics
B Third law of thermodynamics
C Modified first law of thermodynamics
D First law of thermodynamics

The correct answer is:
Modified first law of thermodynamics

Solution:
“The mass and energy both are conserved in an isolated system” is the statement of modified first law of thermodynamics. According to this law, if a certain amount of one kind of energy is consumed to that the total energy in the universe remains constant.

Topic: Chemistry, Thermodynamics (C), MHT CET, MHT CET 2019 (Shift 1)
The mass and the diameter of a planet are three times of the respective values for the earth. If the time period of oscillation of a simple pendulum on earth is 2 s, then the time period of oscillation of the same pendulum on the surface of the given planet would be

Options:
A 32 s
B 23 s
C 32 s
D 23 s

The correct answer is:
23 s

Solution:
We know that,

g=GMR2

⇒gpge=MpMe×ReRp2=3×19=13

Also,
T=2π1g

⇒TpTe=gegp=3
⇒Tp=3Te

Time period at earth for seconds pendulum=2 s

∴Tp=23 s

Topic: Physics, Gravitation, JEE Main
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is $2 \mathrm{~s}$. The period of oscillation of the same pendulum on the planet would be:

Options:
A $\frac{\sqrt{3}}{2} \mathrm{~s}$
B $\frac{2}{\sqrt{3}} \mathrm{~s}$
C $\frac{3}{2} s$
D $2 \sqrt{3}$ s

The correct answer is:
$2 \sqrt{3}$ s

Solution:
Acceleration due to gravity $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$

$\frac{g_{p}}{g_{e}}=\frac{M_{p}}{M_{e}}\left(\frac{R_{e}}{R_{p}}\right)^{2}=3\left(\frac{1}{3}\right)^{2}=\frac{1}{3}$

Also $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}} \Rightarrow \frac{\mathrm{T}_{\mathrm{p}}}{\mathrm{T}_{\mathrm{e}}}=\sqrt{\frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{p}}}}=\sqrt{3}$

$\Rightarrow \mathrm{T}_{\mathrm{p}}=2 \sqrt{3} \mathrm{~s}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The mass and volume of a body are found to be \( 5.00 \pm 0.05 \mathrm{~kg} \) and \( 1.00 \pm 0.05 \mathrm{~m}^{3} \) respectively. Then the maximum possible percentage error in its density is

Options:
A \( 6 \% \)
B \( 3 \% \)
C \( 5 \% \)
D \( 7 \% \)

The correct answer is:
\( 6 \% \)

Solution:
∆pp=∆mm+∆VV
=0.055×100+0.051×100
=6
∆pp=6%

Topic: Physics, Mathematics in Physics, JEE Main
The mass defect in a particular reaction is 0.4 g. The amount of energy liberated is n×107 kW h, where n= _____. (speed of light =3×108 m s-1)

Options:

The correct answer is:
1

Solution:
Energy liberated is given by, E=∆mc2

=0.4×10-3×3×1082

=3600×107 kW s

=3600×1073600 kW h=1×107 kW h

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The mass density of a planet of radius \( \mathrm{R} \) varies with the distance \( \mathrm{r} \) from its centre as \( \rho(r)=\rho_{0}\left(1-\frac{r^{2}}{R^{2}}\right) \) Then the gravitational field is maximum at:

Options:
A \( r=\sqrt{\frac{3}{4}} R \)
B \( r=R \)
C \( r=\frac{1}{\sqrt{3}} R \)
D \( r=\sqrt{\frac{5}{9}} R \)

The correct answer is:
\( r=\sqrt{\frac{5}{9}} R \)

Solution:
dm=ρ×4πx2dx=ρ01-x2r2×4πx2dx

m=4πρ∫0rx2-x4R2 dx s

m=4πρ0r33-r55R2

E=Gmr2=Gr2×4πρ0r33-r55R2

E=4πGρ0r3-r35R2

E is maximum when dEdt=0 ⇒ dEdr=4πGρ013-3r25R2=0

⇒ r=53R

Emax=4πGρ0×5R313-15×59

Emax=8527πGρ0R

Topic: Physics, Gravitation, JEE Main
The mass density of a planet of radius R varies with the distance r from its centre as ρ(r)=ρ01-r2R2 Then the gravitational field is maximum at:

Options:
A r=34R
B r=R
C r=13R
D r=59R

The correct answer is:
r=59R

Solution:
dm=ρ×4πx2dx=ρ01-x2r2×4πx2dx

m=4πρ∫0rx2-x4R2 dx s

m=4πρ0r33-r55R2

E=Gmr2=Gr2×4πρ0r33-r55R2

E=4πGρ0r3-r35R2

E is maximum when dEdt=0 ⇒ dEdr=4πGρ013-3r25R2=0

⇒ r=53R

Emax=4πGρ0×5R313-15×59

Emax=8527πGρ0R

Topic: Physics, Gravitation, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The mass density of a spherical body is given by ρr=kr for r≤R and ρr=0 for r>R, where r is the distance from the center. The correct graph that describes qualitatively the acceleration, a of a test particle as a function of r is:

Options:
A
B
C
D

The correct answer is:

Solution:
For Inside point r≤R
g=E=Gr2∫0r4πr2drKr=GKr2×4πr22=2πGK=constant
For outside point r>R
E=gE=GMr2E=Gr2∫0R(4πr2dr')(Kr')E=4πKGr2[r'22]0R
E=2πKGr2R2
E∝1r2.

Topic: Physics, Gravitation, JEE Main
The mass density of a spherical body is given by ρr=kr for r≤R and ρr=0 for r>R, where r is the distance from the center. The correct graph that describes qualitatively the acceleration, a of a test particle as a function of r is:

Options:
A
B
C
D

The correct answer is:

Solution:
For Inside point r≤R
g=E=Gr2∫0r4πr2drKr=GKr2×4πr22=2πGK=constant
For outside point r>R
E=gE=GMr2E=Gr2∫0R(4πr2dr')(Kr')E=4πKGr2[r'22]0R
E=2πKGr2R2
E∝1r2.

Topic: Physics, Gravitation, JEE Main, JEE Main 2017 (09 Apr Online)
The mass density of a spherical body is given by ρr=kr for r≤R and ρr=0 for r>R, where r is the distance from the centre. The correct graph that describes qualitatively the acceleration, a, of a test particle as a function of r is :

Options:
A
B
C
D

The correct answer is:

Solution:
For Inside point râ¤R
g=E=Gr2â«0r4Ïr2drKr GKr2Ã4Ïr22=2ÏGK=constant
For outside point r>R
E=g= GM r 2 = G r 2 â« 0 R ( 4Ï r 2 d r â² )( K r â² )= 4ÏKG r 2 [ r â² 2 2 ] 0 R
E=g=2ÏKGr2R2
Eâ1r2

Topic: Physics, Gravitation, JEE Main
The mass (in grams) of glucose that should be dissolved in 100 g of water in order to produce same lowering of vapour pressure as is produced by dissolving 1 g of urea (molecular mass = 60) in 50 g of water is

Options:

The correct answer is:
6.00

Solution:
ΔPP0=XB
≈nBnA=wB×mAmB×wA
wB×mAmB×wAglucose=wB×mAmB×wAurea
wB×18180×100=1×1860×50
wB=6 g.

Topic: Chemistry, Solutions, JEE Main
The mass M shown in figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is

Options:
A k1Ak2
B k2Ak2-k1
C k2Ak1+k2
D k2Ak1

The correct answer is:
k2Ak1+k2

Solution:
If a force F is applied to M, say to the right, let A be the distance moved by M. If the system is released, it executes simple harmonic motion of amplitude A. If A1 and A2 ae the extensions in springs k1 and k2 then A=(A1 + A2) and
F=k1A1=k2A2
⇒ A1=Fk1 and A2=Fk2
∴ A=A1+A2=F1k1+1k2
=Fk1+k2k1k2
⇒ F=k1k2Ak1+k2
The amplitude of point P = amplitude of oscillation of spring k1 which is
A1=Fk1=k2Ak1+k2

Topic: Physics, Oscillations, JEE Main
The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is:

Options:
A 24
B 32
C 40
D 20

The correct answer is:
24

Solution:
Given: R1=R22

As we know, size of the nucleus is given by R=R0A1/3

Therefore, we can write R0A11/3=R02A21/3

⇒A1=18A2

⇒A1=1928=24

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The mass of $_{7} \mathrm{~N}^{15}$ is $15.00011 \mathrm{amu}$, mass of $_{8} \mathrm{O}^{16}$ is 15.99492 amu and $m_{p}=1.00783$ amu. Determine binding energy of last proton of ${ }_{8} \mathrm{O}^{16}$.

Options:
A $2.13 \mathrm{MeV}$
B $0.13 \mathrm{MeV}$
C $10 \mathrm{MeV}$
D $12.13 \mathrm{MeV}$

The correct answer is:
$12.13 \mathrm{MeV}$

Solution:
$\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)=\mathrm{M}\left({ }_{7} \mathrm{~N}^{15}\right)+1 \mathrm{~m}_{\mathrm{p}}$

binding energy of last proton $=M\left(N^{15}\right)+m_{P}-M\left({ }_{1} O^{16}\right)$

$=15.00011+1.00783-15.99492$

$=0.01302 \mathrm{amu}=12.13 \mathrm{MeV}$

Topic: Physics, Nuclear Physics, JEE Main
The mass of a hydrogen molecule is 3.32×10-27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of the area 2 cm2 at an angle of 45o to the normal, and rebound elastically with a speed of 103 m s-1, then the pressure on the wall is nearly:

Options:
A 4.70×102 N m-2
B 2.35×103 N m-2
C 4.70×103 N m-2
D 2.35×102 N m-2

The correct answer is:
2.35×103 N m-2

Solution:
F=2mnVcosθ⇒P=2mnVcosθA
⇒P=2×3.32×10-27×1023×1032×2×10-4
⇒P=2.35×103 N m-2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2018 (08 Apr)