datavorous/entrance-exam-dataset huggingface dataset:
In the reaction, $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2 \xrightarrow{\substack{\text { coldalkaline } \\ \mathrm{KMnO}_4}}(A)$ :
Product A is :

Options:
A Ethylene glycol
B Acetic acid
C Ethane
D Butyric acid

The correct answer is:
Ethylene glycol

Solution:
When ethylene is treated with cold alkaline $\mathrm{KMnO} 4$, ethylene glycol is formed.

Topic: Chemistry, Hydrocarbons, JEE Main
In the reaction $\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-}$

Options:
A $\mathrm{H}_2 \mathrm{O}$ is the conjugate base of $\mathrm{HCl}$ acid
B $\mathrm{Cl}^{-}$is the conjugate base of $\mathrm{HCl}$ acid
C $\mathrm{Cl}^{-}$is the conjugate acid of $\mathrm{H}_2 \mathrm{O}$ base
D $\mathrm{H}_3 \mathrm{O}^{+}$is the conjugate base of $\mathrm{HCl}$

The correct answer is:
$\mathrm{Cl}^{-}$is the conjugate base of $\mathrm{HCl}$ acid

Solution:
$\mathrm{HCl}$ is a strong acid and their conjugate base is a very weak base.

Topic: Chemistry, Ionic Equilibrium, JEE Main
In the reaction $\mathrm{HNO}_3+\mathrm{P}_4 \mathrm{O}_{10} \rightarrow 4 \mathrm{HPO}_3+x$, the product $x$ is

Options:
A $\mathrm{N}_2 \mathrm{O}_3$
B $\mathrm{N}_2 \mathrm{O}_5$
C $\mathrm{NO}_2$
D $\mathrm{H}_2 \mathrm{O}$

The correct answer is:
$\mathrm{N}_2 \mathrm{O}_5$

Solution:
We know that, $4 \mathrm{HNO}_3+\mathrm{P}_4 \mathrm{O}_{10} \rightarrow 4 \mathrm{HPO}_3+2 \mathrm{~N}_2 \mathrm{O}_5$
The product is dinitrogen pentaoxide $\left(\mathrm{N}_2 \mathrm{O}_5\right)$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
In the reaction $\mathrm{I}_2+2 \mathrm{~S}_2 \mathrm{O}_3^{--} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_4 \mathrm{O}_6^{--}$ equivalent weight of iodine will be equal to

Options:
A $1 / 2$ of molecular weight
B Molecular weight
C $1 / 4$ of molecular weight
D None

The correct answer is:
$1 / 2$ of molecular weight

Solution:
Molecular weight/2 = Equivalent weight of lodine.

Topic: Chemistry, Redox Reactions, JEE Main
In the reaction; \( \mathrm{F} e(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{F} e^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \), if the concentration of \( \mathrm{OH}^{-} \)ions is
decreased by \( \frac{1}{4} \) times, then the equilibrium concentration of \( F e^{3+} \) will increase by,

Options:
A \( 8 \) times
B \( 16 \) times
C \( 64 \) times
D \( 4 \) times

The correct answer is:
\( 64 \) times

Solution:
\[
\begin{array}{l}
\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \\
K_{\text {eq }}=\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3} \\
K_{\text {eq }}=\left[\mathrm{Fe}^{3+}\right]\left[\frac{1}{4} \mathrm{OH}^{-}\right]^{3}=\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}}{64}
\end{array}
\]
Therefore, to maintain the equilibrium, the equilibrium concentration of \( F e^{3+} \) will increase by \( 64 \) times

Topic: Chemistry, Chemical Equilibrium, KCET, KCET 2016
In the reaction \(\mathrm{NH}_3+\mathrm{BF}_3 \rightarrow \mathrm{F}_3 \mathrm{BNH}_3, \mathrm{BF}_3\) is

Options:
A Lewis acid
B Lewis base
C Neither Lewis acid nor Lewis base
D Both Lewis acid and Lewis base

The correct answer is:
Lewis acid

Solution:
\(\mathrm{BF}_3\) accepts an electron pair from \(\mathrm{NH}_3\), so it is a Lewis acid.
The correct answer is option 1: Lewis acid.

Topic: Chemistry, Ionic Equilibrium, JEE Main
In the reaction of 1-bromo-3-Chlorocyclobutane with two equivalents of sodium inether, the major product is

Options:
A
B
C
D

The correct answer is:

Solution:
Wurtz Reaction

Topic: Chemistry, Hydrocarbons, JEE Main
In the reaction of formation of sulphur trioxide by contact process $2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$ the rate of reaction was measured as
$\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}=-2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of reaction is terms of $\left[\mathrm{SO}_2\right]$ in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$ will be:

Options:
A
$-1.25 \times 10^{-4}$

B
$-2.50 \times 10^{-4}$

C
$-3.75 \times 10^{-4}$

D
$-5.00 \times 10^{-4}$

The correct answer is:
$-5.00 \times 10^{-4}$

Solution:
From rate law
$$
\begin{aligned}
&-\frac{1}{2} \frac{\mathrm{dSO}_2}{\mathrm{dt}}=-\frac{\mathrm{dO}_2}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{dSO}_3}{\mathrm{dt}} \\
&\therefore-\frac{\mathrm{dSO}_2}{\mathrm{dt}}=-2 \times \frac{\mathrm{dO}_2}{\mathrm{dt}} \\
&=-2 \times 2.5 \times 10^{-4} \\
&=-5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2014 (11 Apr Online)
In the reaction of gold with aquaregia, oxidation state of Nitrogen changes from

Options:
A \( +5 \) to \( +2 \)
B \( +3 \) to \( +1 \)
C \(+4 \) to \( +2 \)
D \(+6 \) to \( +4 \)

The correct answer is:
\( +5 \) to \( +2 \)

Solution:
(A)
\( \mathrm{Au}+4 \mathrm{H}^{+}+\mathrm{NO}_{3}^{+5}{ }^{-}+4 \mathrm{Cl}^{-} \rightarrow \mathrm{AuCl}{ }^{-}{ }^{-}+\mathrm{NO}^{+2} \mathrm{O}+2 \mathrm{H}_{2} \mathrm{O} \)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
In the reaction of hypobromite with amide, the carbonyl carbon is lost as:

Options:
A CO32-
B HCO3-
C CO2
D CO

The correct answer is:
CO32-

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (18 Mar Shift 2)
In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is:

Options:
A 2
B 10
C 1
D 5

The correct answer is:
1

Solution:
2 MnO4-+5C2O42-+16H+→10 CO2+2 Mn2++8H2O
C goes from +3 oxidation state to +4 oxidation state.
Total 10 e- are transferred for 10 molecules of CO2.
Hence, for 1 molecule of CO2, only one e- is transferred.
So, number of electrons involved for 1 molecule of CO2 is 1.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2019 (10 Jan Shift 2)
In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is + _______.

Options:

The correct answer is:
6

Solution:
The oxidation state or oxidation number is defined as the loss or gain of electrons from an atom, molecule, or compound.

The reaction of potassium dichromate, potassium chloride and sulphuric acid (conc.)is as follows,

K2Cr2O7s+4KCls+6H2SO4 (conc.)

→2CrO2Cl2g+6KHSO4+3H2O

This reaction takes place in chromyl chloride test.

The Chromyl chloride is a neutral molecule as it does not have any charge on it.

Here, oxidation state of Cr in CrO2Cl2 is +6.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (31 Jan Shift 2)
In the reaction of sodium thiosulphate with $\mathrm{I}_2$ in aqueous medium the equivalent weight of sodium thiosulphate is equal to

Options:
A molar mass of sodium thiosulphate
B the average of molr masses of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{I}_2$
C half the molar mass of sodium thiosulphate
D molar mass of sodium thiosulphate $\times 2$

The correct answer is:
molar mass of sodium thiosulphate

Solution:
$\begin{gathered}
\text {Hints : } 2 \mathrm{Na}_2 \stackrel{+2}{\mathrm{~S}_2} \mathrm{O}_3+\mathrm{I}_2 \longrightarrow \mathrm{Na}_2 \stackrel{+25}{\mathrm{~S}_4} \mathrm{O}_6+2 \mathrm{NaI} \\
\text {n-factor }=1 \\
\mathrm{E}=\frac{\mathrm{M}}{1}=\mathrm{M}
\end{gathered}$

Topic: Chemistry, Redox Reactions, JEE Main
In the reaction $Ph-O H+R x \xrightarrow{\text { alkali }} A, A$ is

Options:
A An aldehyde
B An aryl chloride
C An ether
D A ketone

The correct answer is:
An ether

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the reaction, P+Q→R+S
The time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is

Options:
A 2
B 3
C 0
D 1

The correct answer is:
1

Solution:
For P, t75%=2t50% Ist order w.r.t 'P'
For Q, Q=Q0-kt. i.e. zero order.
i.e. overall order of a reaction =1

Topic: Chemistry, Chemical Kinetics, JEE Main
In the reaction, product $P$ is

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{MgBr} \xrightarrow[\text { (i) } \mathrm{H}_3^{+} \mathrm{O}]{(\mathrm{z}) \mathrm{CO}_2} \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In the reaction sequence
2CH3CHO →OH- A →∆ B ; the product B is:

Options:
A CH3-CH=CH-CHO
B
C CH3-CH2-CH2-CH3
D CH3-CH2-CH2-CH2-OH

The correct answer is:
CH3-CH=CH-CHO

Solution:
This reaction is aldol condensation.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2015 (11 Apr Online)
In the reaction sequence
C 6 H 5 CHO → Δ NH 2 OH /  H ⊕ (X) → P 2 O 5  / Δ (Y) → H 2 O /  H ⊕ (Z)
(X), (Y) and (Z) respectively be

Options:
A C6H5CH=N-OH,C6H5CN,C6H5COOH
B C6H5CH=NOH,C6H5CONH2,C6H5COOH
C C6H5-CH=NOH,C6H5COOH,C6H5CONH2
D C6H5-CH=NOH,C6H5COOH,C6H5CN

The correct answer is:
C6H5CH=N-OH,C6H5CN,C6H5COOH

Solution:
Here X is benzladehyde oxime, Y is cyanobenzene and Z is benzoic acid
C6H5-CHO+NH2OH→C6H5-CH=N-OH(X)
→P2O5/ΔC6H5-C≡N(Y)→H3O⊕/ΔC6H5-COOH(Z)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the reaction sequence.

. Correct statement regarding B is:

Options:
A It is less acidic than
B Degree of unsaturation in B is 6
C It reacts with Br2 /acetic acid to form tetra bromo product
D B give violet colour with neutral FeCl3 .

The correct answer is:
B give violet colour with neutral FeCl3 .

Solution:
is more acidic than phenol
Degree of unsaturation =5

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the reaction sequence,
the major products $\mathrm{X}$ and $\mathrm{Y}$, respectively, are-

Options:
A i
B ii
C iii
D iv

The correct answer is:
iii

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the reaction sequence

[X] will be

Options:
A
B CH3COOH
C C6H5–NH2
D CH3–CO–NH–C6H5

The correct answer is:
CH3–CO–NH–C6H5

Solution:
Here X is CH3–CO–NH–C6H5 and the reaction is as follows

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the reaction sequence [X will be]

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the reaction shown below, identify the correct combination of te intermediate P and the product Q.

Options:
A
B
C
D

The correct answer is:

Solution:
Here the reaction sequence is as follows

Topic: Chemistry, General Organic Chemistry, JEE Main
In the reaction,

The compound (C) is?

Options:
A
B
C
D

The correct answer is:

Solution:
(C - Br) bond is weaker than (C - Cl) bond. Grignard's reagent is formed more predominantly with (C - Br).

Topic: Chemistry, Hydrocarbons, JEE Main
In the reaction,
The number of possible isomers for the organic compound $X$ is

Options:
A 4
B 5
C 3
D 2

The correct answer is:
2

Solution:
In the given reaction,
So, for $X$,
we can have another isomer as
$\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3} \text {. }$
So, total two isomers are possible.

Topic: Chemistry, Amines, JEE Main
In the reaction,
The product E is

Options:
A
B
C
D

The correct answer is:

Solution:
Firstly, the given compound is treated with dilute hydrochloric acid with an aqueous solution of sodium nitrite at low temperature to give benzene diazonium chloride product, this reaction is known as diazotisation reaction. Now, the obtained product is heated with anhydrous cuprous cyanide, the halogen atom is replaced by cyano group.

Topic: Chemistry, Amines, JEE Main, JEE Main 2015 (04 Apr)
In the reaction $\underset{\substack{\text { (Vapour) }}}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \underset{300^{\circ} \mathrm{C}}{\stackrel{\mathrm{Cu}}{\longrightarrow}} X$. The molecular formula of $X$ is

Options:
A $\mathrm{C}_4 \mathrm{H}_6 \mathrm{O}$
B $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$
C $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}$
D $\mathrm{C}_2 \mathrm{H}_6$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}$

Solution:
Alcohols are oxidised by hot copper to give aldehydes.
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \frac{\mathrm{Cu}}{300^{\circ} \mathrm{C}} \underset{\text { Acetaldehyde }}{\mathrm{CH}_3 \mathrm{CHO}} \text { or } \mathrm{C}_2 \mathrm{H}_4 \mathrm{O}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the reaction xA→yB ,
log⁡-d[A]dt=log⁡+d[B]dt+0.3
Then, x : y is

Options:
A 2 : 1
B 1 : 2
C 3 : 1
D 3 : 10

The correct answer is:
2 : 1

Solution:
For the reaction
xA→yB
-1xd[A]dt=+1yd[B]dt
-d[A]dt=+d[B]dt×xy
log⁡-d[A]dt=log⁡d[B]dt+log⁡xy …(i)
Given log⁡-d[A]dt=log⁡d[B]dt+0.3 …(ii)
Comparing (i) and (ii), we get
log⁡xy=0.3 ⇒log xy= log 2
So xy=21 .

Topic: Chemistry, Chemical Kinetics, JEE Main
In the reaction:

Znâ+â NaNO 3 â+âNaOH â Na 2 ZnO 2 â+âAâ+â H 2 O

The sum of stoichiometric coefficients of Zn and A in the balanced reaction with simplest integer coefficient is

Options:

The correct answer is:
5

Solution:
Balanced reaction is
4Zn+NaNO3+7NaOH→4Na2ZnO2 +NH3+ 2H2O [A]
The sum of stoichiometric coefficients of Zn and A=4+1=5

Topic: Chemistry, s Block Elements, JEE Main
In the reaction

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In the reactions $\mathrm{I}$ and $\mathrm{II}$ the covalencies of $\mathrm{Be}$ and $\mathrm{Al}$ in $\mathrm{X}$ and $\mathrm{Y}$ are respectively
I $\mathrm{Be}(\mathrm{OH})_2+\underset{\text { (excess) }}{\mathrm{NaOH}} \rightarrow \mathrm{X}$ (excess)
II $\mathrm{Al}(\mathrm{OH})_3+\underset{\text { (excess) }}{\mathrm{NaOH}} \rightarrow \mathrm{Y}$

Options:
A 4,6
B 4,4
C 6,4
D 3,6

The correct answer is:
4,4

Solution:
The covalencies of $\mathrm{Be}$ and $\mathrm{Al}$ in $\mathrm{X}$ and $\mathrm{Y}$ respectively are 4,4 .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In the reactions sequence
CH 3 −CO− CH 3 → C 2 H 5 OH/OH − (A) → C 2 H 5 OH/H + [X] will be

Options:
A
B
C
D

The correct answer is:

Solution:
Here formation of Hemiacetal and ketal takes place as follows

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the real number system, the equation $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ has $-$

Options:
A No solution
B Exactly two distinct solutions
C Exactly four distinct solutions
D Infinitely may solutions

The correct answer is:
Infinitely may solutions

Solution:
$\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1 ; x \geq 1$
$\sqrt{(x-1)-2 \times 2 \sqrt{x-1}+4}+\sqrt{(x-1)-6 \sqrt{x-1}+9}=1$
$\begin{array}{l}
2 \sqrt{x-1}=6 \\
x=10 \\
\text { Case -II } \\
\sqrt{x-1}-2-\sqrt{x-1}+3=1 \quad 5 \leq x \leq 10
\end{array}$
Case -III
$\begin{array}{l}
-\sqrt{x-1}+2-\cdot \sqrt{x-1}+3=1 \\
2 \cdot \sqrt{x-1}=4 \\
x=5
\end{array}$

Topic: Mathematics, Quadratic Equation, JEE Main
In the reduction of nitriles to primary amines, what is the main purpose of using a catalyst such as hydrogen gas (H2)?

Options:
A To increase the reaction temperature
B To increase the reaction rate
C To reduce the need for a reducing agent
D To increase the yield of the reaction

The correct answer is:
To increase the reaction rate

Solution:
Catalysts like hydrogen gas with a catalyst are used to increase the reaction rate in the reduction of nitriles to primary amines.

Topic: Chemistry, Amines, NEET
In the replacement reaction
The reaction will be most favourable if $M$ happens to be

Options:
A $\mathrm{Na}$
B $\mathrm{K}$
C $\mathrm{Rb}$
D $\mathrm{Li}$

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

Solution:
For the reaction,
The reaction will be faster with $\mathrm{Rb}$ because lattice energy of RbF is less than LiF, NaF, KF.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2012 (Mains)
In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be: (Given the area of the plate =A )

Options:
A 1534Kε0 A d
B 156Kε0 A d
C 256Kε0 A d
D 96Kε0 A d

The correct answer is:
1534Kε0 A d

Solution:
1Ceff=dK∈0A+2d3K∈0A+3d5∈0A

Ceff=15K∈0A34d

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (27 Jul Shift 1)
In the reported figure, heat energy absorbed by a system in going through a cyclic process is ______πJ.

Options:

The correct answer is:
100

Solution:
For complete cyclic process
ΔU=0
∴ from ΔQ=ΔU+W
=0+W
ΔQ=W
=Area
=πr1·r2
=π×10×103×10×10-3
ΔQ=100π
∴ Answer =100

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (20 Jul Shift 1)
In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA:AB will be x:5. The value of x is

Options:

The correct answer is:
4

Solution:
gA=GMrR3

gC=GMR+R22
gA=gC

rR3=194R2⇒r=4R9

so OA=4R9; AB=R-r=5R9

OA:AB=4:5

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (26 Feb Shift 2)
In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be _______ rad s-1 when k=20 N m-1.

Options:

The correct answer is:
10

Solution:
ω=keqμ

μ= reduced mass

Springs are in series connection,

keq=k1k2k1+k2

keq=k×4k5k=4k5

keq=4×205 N m-1=16 N m-1

μ=m1 m2 m1+m2=0.2×0.80.2+0.8=0.16 kg

ω=160.16=100=10

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (25 Jul Shift 1)
In the resting state, which protein masks the active binding sites of actin filaments?

Options:
A troponin
B tropomyosin
C myosin
D dynein

The correct answer is:
troponin

Solution:
Actin and Myosin are contractile proteins of muscle while Troponin and Tropomyosin are regulatory proteins. Myosin is a polymer of Meromysoin units and this is the main protein of Dark bands. Light bands consist of Actin, Troponin and Tropomyosin. Actin protein consists of two double stranded polypetides while troponin is attached on these strands at intervals. Actin shows presence of Myosin binding site and when myosin attaches to this site the muscle shows contraction due to formation of cross bridges. However this cross bridge formation is not seen when muscle relaxes as this site is masked by Troponin molecule. At intervals on actin, trimeric molecule Tropomyosin is also seen. This includes TpC, TpT and TpI.

Topic: Biology, Locomotion and Movement, NEET
In the rock-salt arrangement, the shortest distance between two \( \mathrm{Na}^{+} \)ions having edge length equal to a \( \sqrt{2} \) picometers is

Options:
A 2a
B \( 2 \mathrm{a} \sqrt{2} \)
C \( \frac{3}{\sqrt{2}} \)
D a

The correct answer is:
a

Solution:
Distance between two Na+ is 12 times of edge length at the edge centres.Let the edge length =a2, and distance between two Na+ ions at the edges=d. Hence, d=2a22=a.

Topic: Chemistry, Solid State, JEE Main
In the Rutherford experiment, $\alpha$-particles are scattered from a nucleus as shown. Out of the four paths, which path is not possible?

Options:
A
$D$

B
$B$

C
$C$

D
$A$

The correct answer is:
$C$

Solution:
As $\alpha$-particles are doubly ionised helium $\mathrm{He}^{++}$i.e. Positively charged and nucleus is also positively charged and we know that like charges repel each other.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2012 (07 May Online)
In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle.

Options:
A remains the same
B is always $90^{\circ}$
C decreases
D increases

The correct answer is:
decreases

Solution:
In the Rutherford's $\alpha$-scattering experiment,
impact parameter $(b)$ and scattering angle $\theta$ are related as,
$\begin{aligned} b & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{z e^2 \cot \frac{\theta}{2}}{\left(\frac{1}{2} m v^2\right)} \\ \Rightarrow \quad b & \propto \cot \frac{\theta}{2}\end{aligned}$
Thus, with the increase of impact parameter $(b)$, $\cot \frac{\theta}{2}$ increases. It is possible only when $\theta$ will decrease.

Topic: Physics, Atomic Physics, KCET, KCET 2023
In the 'S' phase of the cell cycle:

Options:
A Amount of DNA doubles in each cell
B Amount of DNA remains same in each cell
C Chromosome number is increased
D Amount of DNA is reduced to half in each cell

The correct answer is:
Amount of DNA doubles in each cell

Solution:
S-phase/Synthetic phase

(a) DNA replicates and its amount becomes double (2C→ 4C).
(b) Synthesis of histone proteins.
(c) Euchromatin replicates earlier than heterochromatin.
(d) Synthesis of NHC (non-histone chromosomal proteins).
(e) Each chromosome has 2 chromatids. Amount of DNA doubles but the chromosome number remains constant.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2014
In the scheme given below, the total number of intramolecular aldol condensation products formed from 'Y' is

Options:

The correct answer is:
1

Solution:
The number of intra molecular aldol condensation products (α,β -unsaturated carbonyl compound formed from Y is 1.)

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the sequence, $(1,2,3),(4,5,6),(7,8,9,10)$ $\ldots$ of sets, the sum of elements in the 50th set is

Options:
A $62525$
B $65225$
C $56255$
D $557625$

The correct answer is:
$62525$

Solution:
First term of each sets are $1,2,4,7, \ldots$
Let
$$
\begin{aligned}
& S=1+2+4+7+\ldots+T_n \\
& S=1+2+4+\ldots+T_n
\end{aligned}
$$
On subtracting, we get
$$
\begin{aligned}
& 0=1+1+2+3+\ldots-T_n \\
& T_n=1+(1+2+3+\ldots(n-1) \text { terms }) \\
& T_n=1+\frac{(n-1) n}{2} \\
& T_{50}=1+\frac{49 \times 50}{2} \\
& \therefore \quad T_{50}=1226 \\
&
\end{aligned}
$$
$\Rightarrow$ First term of 50th set is 1226 , therefore series is $1226,1227, \ldots 50$ terms
$$
S=\frac{50}{2}[2 \cdot 1226+49]=62525
$$

Topic: Mathematics, Sequences and Series, JEE Main
In the sequence of reactions

the product (Y) is

Options:
A biphenyl
B m-bromobenzoic acid
C benzoic acid
D benzyl alcohol

The correct answer is:
benzoic acid

Solution:
Here the product Y is benzoic acid
C6H5Br→MgetherC6H5MgBrX→CO2C6H5CO2MgBr→H3O+C6H5COOHY+MgBrOH

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In the series \( L C R \) circuit as shown in the figure, the voltmeter and ammeter readings are

Options:
A \( V=100 \mathrm{~V}, I=2 \mathrm{~A} \)
B \( V=100 \mathrm{~V}, I=5 \mathrm{~A} \)
C \( V=1000 \mathrm{~V}, I=2 \mathrm{~A} \)
D \( V=300 \mathrm{~V}, I=1 \mathrm{~A} \)

The correct answer is:
\( V=100 \mathrm{~V}, I=2 \mathrm{~A} \)

Solution:
V→=V→R+V→L+V→C (considering phasor).
Here, VL and VC should be out of phase for a given current and as their magnitudes are also equal, they will cancel each other.
⇒V→=V→R=100 V
⇒I=VR=10050=2 A

Topic: Physics, Alternating Current, JEE Main
In the shown wireframe, each side of a square (the smallest square) has a resistance 7 Î©. What is the equivalent resistance value (in ohm) of the circuit between the points A and B?

Options:

The correct answer is:
14

Solution:
The circuit can be folded about B and redrawn as
Hence equivalent resistance between A and B is 2R=2×7 Ω=14 Ω.

Topic: Physics, Current Electricity, JEE Main
In the sieve elements, which one of the following is the most likely function of P-proteins?

Options:
A Deposition of callose on sieve plates
B Providing energy for active translocation
C Autolytic enzymes
D Sealing off mechanism on wounding

The correct answer is:
Sealing off mechanism on wounding

Solution:
In most angiospermic plants, sieve tube elements of phloem, are P-proteins. Later is found in all dictos
and in many monocots but it is absent in gymnosperms and cryptogams.
The main function of P -proteins is in sealing off damaged sieve elements by plugging up sieve plate pores.

Topic: Biology, Anatomy of Flowering Plants, JIPMER, JIPMER 2011
In the sixth period, the orbitals that are filled are :

Options:
A 6s, 4f, 5d, 6p
B 6s, 5d, 5f, 6p
C 6s, 5f, 6d, 6p
D 6s, 6p, 6d, 6f

The correct answer is:
6s, 4f, 5d, 6p

Solution:
In 6th period 6s, 4f, 5d and 6p orbitals are gradually filled.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (05 Sep Shift 1)