datavorous/entrance-exam-dataset huggingface dataset:
Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional change in density 1ρdρdt is constant. The velocity v of any point on the surface of the expanding sphere is proportional to

Options:
A R3
B 1R
C R
D R23

The correct answer is:
R

Solution:
Density of sphere is ρ=mv=3m4πR3

⇒ 1ρdρdt= -3R dRdt

Since ⇒ 1ρdρdt is constant

∴ dRdt ∝R

Topic: Physics, Thermal Properties of Matter, NEET
Consider an experiment of a single throw of a pair of unbiased normal dice. Let three events ε1,ε2 and ε3 are defined as follows
ε1: getting prime numbered face on each dice
ε2: getting the same number on each dice
ε3: getting the sum of 4 on two dice
Which of the following is not true?

Options:
A The probabilities PÎµ1,PÎµ2,PÎµ3 are in arithmetic progression.
B The events Îµ1 and Îµ2 are dependent
C PÎµ3Îµ1=29
D PÎµ3Îµ1=19

The correct answer is:
Pε3ε1=29

Solution:
ε1=2,22,32,5,3,2,3,3,3,5,5,2,5,3,5,5
ε2=1,12,23,3,4,4,5,5,6,6
ε3=1,3,3,1,2,2
Pε1=936=14,Pε2=636=16,Pε3=336=112
Pε1,Pε2,Pε3 are in A.P.
Pε1∩ε2=336≠Pε1⋅Pε2Dependent
Pε3ε1=Pε3∩ε1Pε1=136936=19

Topic: Mathematics, Probability, JEE Main
Consider an ideal gas at pressure $p$, volume $V$ and temperature $T$. The mean free path for molecules of the gas is $L$. If the radius of gas molecules, as well as pressure, volume and temperature of the gas are doubled, then the mean free path will be

Options:
A $\frac{5 L}{2}$
B $\frac{L}{4}$
C $\frac{L}{8}$
D $2 L$

The correct answer is:
$\frac{L}{4}$

Solution:
Given, initial values of pressure, volume,
temperature, radius of molecule of gas and mean free path are $P p, V, T, r$ and $L$, respectively.
$\therefore$ Mean free path,
$\begin{aligned} \lambda & =\frac{2 \sqrt{2} R T}{\pi r^2 N p} \\ \therefore \quad L & =\frac{2 \sqrt{2} R T}{\pi r^2 N p}\end{aligned}$
According to question if the radius of gas molecules, as well as pressure, volume and temperature of gas are doubled then
$\begin{aligned} \lambda^{\prime} & =\frac{2 \sqrt{2} R \cdot 2 T}{\pi \times 4 r^2 \times N \times 2 p} \\ & =\frac{1}{4} \cdot \frac{2 \sqrt{2} R T}{\pi r^2 N p}=\frac{L}{4}\end{aligned}$
(From Eq. (i))
Hence, the mean free path will be $\frac{L}{4}$.

Topic: Physics, Kinetic Theory of Gases, JEE Main
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq , where V is the volume of the gas. The value of q is:

γ=CPCv

Options:
A γ-12
B 3 γ + 5 6
C 3γ-56
D Î³+12

The correct answer is:
γ+12

Solution:
For an adiabatic process TVγ−1= constant.
We know that average time of collision between molecules

τ=1nπ2vrmsd2

where, n= number of molecules per unit volume Vrms = rms velocity of molecules

As n∝1V and vrms∝T

τ∝VT

Thus, we can write

n=K1V-1 and Vrms=K2T12.

where, K1 and K2 are constants.

For adiabatic process, TVγ-1= constant. Thus, we can write

τ∝VT-12∝VV1-γ-12

or τ∝Vγ+12

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2015 (04 Apr)
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq , where V is the volume of the gas. The value of q is:

γ=CPCv

Options:
A γ-12
B 3 γ + 5 6
C 3γ-56
D Î³+12

The correct answer is:
γ+12

Solution:
For an adiabatic process TVγ−1= constant.
We know that average time of collision between molecules

τ=1nπ2vrmsd2

where, n= number of molecules per unit volume Vrms = rms velocity of molecules

As n∝1V and vrms∝T

τ∝VT

Thus, we can write

n=K1V-1 and Vrms=K2T12.

where, K1 and K2 are constants.

For adiabatic process, TVγ-1= constant. Thus, we can write

τ∝VT-12∝VV1-γ-12

or τ∝Vγ+12

Topic: Physics, Thermodynamics, JEE Main
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq , where V is the volume of the gas. The value of q is:

Î³=CPCv

Options:
A γ-12
B 3γ+16
C 3γ-56
D γ+12

The correct answer is:
γ+12

Solution:
Average time of collision
t=mean free path(λ)average speed (v)
t∝λv
∵λ∝1no. of molecules per unit volume
λ∝1Nv
⇒λ∝V
And v¯ ∝vrv
So average time
∴tavg∝VV1-r2
tavg∝V1-1-r2
tavg∝V1+r2
∴q=1+r2

Topic: Physics, Thermodynamics, JEE Main
Consider an imaginary ion X3-2248. The nucleus contains 'a'% more neutrons than the number of electrons in the ion. The value of 'a' is

Options:

The correct answer is:
4

Solution:
X22483-

In an atomic representation XZA

Z is atomic number = 22

A is atomic mass number = 48.

No. of protons = Z = 22

No. of neutrons =A-Z = 48-22 = 26

No. of electrons =P-charge = 22+3 = 25

% of extra neutrons than electrons =26-2525×100=4 % = a%

So, the value of a=4.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Consider circles C1 and C2 touching both the axes and passing through 4,4, then the x-intercept of the common chord of the circles is

Options:

The correct answer is:
8

Solution:
Since the circle touches both the axes and passes through 4,4, it lies in the first quadrant, so its equation is
x-r2+y-r2=r2
4,4 lies on it, hence
r2-16r+32âr1+r2=16
C1 is x-r12+y-r12=r12
C2 is x-r22+y-r22=r22
Equation of the common chord is
2x-r1+r2+2y-r1+r2+r1+r2=0
âx+y=8

Hence, the x-intercept is 8

Topic: Mathematics, Circle, JEE Main
Consider circles C1 & C2 touching both the axes and passing through 4,4, then the product of radii of these circles is

Options:

The correct answer is:
32

Solution:
Since circle touches both the axes and passes through 4,4, it lies in the first quadrant, so its equation is
x-r2+y-r2=r2
4,4 lies on it hence
r2-16r+32=0
Hence, the product of roots (radii) =32

Topic: Mathematics, Circle, JEE Main
Consider $f: R \rightarrow R$ given by $f(x)=4 x+3$. Show that $f$ is invertible. Find the inverse of $f$.

Options:

None

Solution:
$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ given by $\mathrm{f}(\mathrm{x})=4 \mathrm{x}+3$
$f\left(x_1\right)=4 x_1+3, f\left(x_2\right)=4 x_2+3$
If $\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)$, then $4 \mathrm{x}_1+3=4 \mathrm{x}_2+3$
or $4 x_1=4 x_2$ or $x_1=x_2 \therefore$ fis one-one
Also let $y=4 x+3$, or $4 x=y-3 \therefore x=\frac{y-3}{4}$
For each value of $\mathrm{y} \in \mathrm{R}$ and belonging to co-domain of $\mathrm{y}$ has a pre-image in its domain.
$\therefore f$ is onto i.e. $f$ is one-one and onto
$\therefore$ fis invertible and $\mathrm{f}^{-1}(\mathrm{y})=\mathrm{g}(\mathrm{y})=\frac{\mathrm{y}-3}{4}$.

Topic: Mathematics, Relations and Functions (Class 12)
Consider fx=-2,-2≤x<0x2-2,0≤x≤2 and gx=fx+fx. Then, in the interval -2,2, gx is

Options:
A not differentiable at one point
B differentiable at all points
C not continuous
D not differentiable at two points

The correct answer is:
not differentiable at one point

Solution:
Given, fx=-2,-2≤x<0x2-2,0≤x≤2
∴fx=2,-2≤x<02-x2, 0≤x<2x2-2,2≤x≤2
And fx=x2-2, x∈-2,2
∴gx=x2,-2≤x<00,0≤x<22x2-2,2≤x≤2
g ′ x = 2x −2<x<0 0 0<x< 2 4x 2 <x<2
g'0−=0=g'0+g'2−=0,g'2+=42
So, gx is not differentiable at x=2

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider fx=minimumx+2,4-x, âxâ¤4. If the area bounded by y=fx and the x-axis is 22k square units, then the value of k is

Options:

The correct answer is:
3

Solution:
Required area =∫024-y2-y-2dy
=∫026-y2-ydy
=6y-y33-y2202=12-83-2
=223 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
Consider fx=tan-1⁡1+sin⁡x1-sin⁡x, x ∈0,π2. A normal to y=fx at x=π6 also passes through the point

Options:
A Ï6,Â 0
B Ï4,Â 0
C 0, 0
D 0,2Ï3

The correct answer is:
0,2π3

Solution:
fx=tan-1⁡1+sin⁡x1-sin⁡x, x ∈0,π2
f′(x)= 11+1+sin⁡x1-sin⁡x .12 1+sin⁡x1-sin⁡x-12 . cos⁡x1-sin⁡x+cos⁡1+sin⁡x1-sin⁡x2
at x=π6
f′x=122 .12 13 314=12
Slope of tangent =12
So slope of normal = -2
Also at x=π6, y=tan-1⁡3=π3 .
So equation of the Normal would be y-π3= -2 x-π6
It passes through 0,2π3

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2016 (03 Apr)
Consider fx=x+-x x≠2λx=2, where ·denotes the greatest integer function. If fx is continuous at x=2, then the value of λ is equal to

Options:
A -1
B 0
C 1
D No value is possible

The correct answer is:
-1

Solution:
limx→2-⁡fx=limx→2-⁡x+-x
=limx→2-⁡x+limx→2-⁡-x
=1-2=-1
limx→2+⁡fx=limx→2+⁡x+-x
=limx→2+⁡x+limx→2+⁡-x
=2+-3=-1
∴λ=-1 wil make f continuous at x=2

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider figure for photoemission. How would you reconcile with momentum - conservation? Note light (photons) have momentum in a different direction than the emitted electrons.

Options:

None

Solution:
During photoelectric emission, when photons strike to metal surface, photons transfer their momentum to atoms of metal by decreasing its own speed upto zero. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.
The excited electron is emitted. So, the conservation of momentum is to be considered as the momentum of incident photon transferred to the nucleus and electrons.

Topic: Physics, Dual Nature of Matter
Consider following reactions:

The sum of molecular masses of gas A and B is ____________ u.

Options:

The correct answer is:
46

Solution:
Gas 'A' is hydrogen gas H2 Molecular mass of H2=2u
Gas 'B' is carbon dioxide CO2 .
Molecular mass of CO2=44 u
â´âSum=2+44=46 u

Topic: Chemistry, Practical Chemistry, JEE Main
Consider four conducing materials copper, tungsten, mercury and aluminum with resistivity ρC, ρT, ρM and ρA respectively. Then :

Options:
A ρC>ρA>ρT
B ρM>ρA>ρC
C ρA>ρT>ρC
D ρA>ρM>ρC

The correct answer is:
ρM>ρA>ρC

Solution:
ρM=98×10-8Ω m

ρA=2.65×10-8Ω m

ρC=1.724×10-8 Ω m

ρT=5.65×10-8 Ω m

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Consider four steel wires of dimensions given below ( \( \mathrm{d}= \) dimeter and \( \mathrm{l}= \) length ):
I. \( \mathrm{l}=1 \mathrm{~m}, \mathrm{~d}=1 \mathrm{~mm} \)
II. \( \mathrm{l}=2 \mathrm{~m}, \mathrm{~d}=2 \mathrm{~mm} \)
III. \( \mathrm{l}=2 \mathrm{~m}, \mathrm{~d}=1 \mathrm{~mm} \)
IV. \( \mathrm{l}=1 \mathrm{~m}, \mathrm{~d}=2 \mathrm{~mm} \)
If same force is applied to all the wires then the elastic potential energy stored will be maximum in wire.

Options:
A I
B II
C III
D IV

The correct answer is:
III

Solution:
We have, U=F22k
Where k=YlA=Yl14πd2
∴U∝d2l

Topic: Physics, Mechanical Properties of Solids, JEE Main
Consider four triangles having sides $(5,12,9),(5,12,11),(5,12,13)$ and $(5,12,15)$. Among these, the triangle having maximum areahas sides

Options:
A $(5,12,9)$
B $(5,12,11)$
C $(5,12,13)$
D $(5,12,15)$

The correct answer is:
$(5,12,13)$

Solution:
Clearly area of $\Delta$ having sides $(5,12,13)$
is greatest (use $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$

Topic: Mathematics, Straight Lines, JEE Main
Consider function f:A→B and g:B→CA,B,C⊆R such that gof-1 exists, then:

Options:
A f and g both are one-one
B f and g both are onto
C f is one-one and g is onto
D f is onto and g is one-one

The correct answer is:
f is one-one and g is onto

Solution:
∴gof-1 exist ⇒ gof=gfx is bijective

⇒f must be one-one and g must be onto.

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Consider I=â«01dx1+x5. Then, I satisfies

Options:
A I>1
B I=1
C I<1
D I+1<0

The correct answer is:
I<1

Solution:
As, xâ0,1
x5â0,1
â´1+x5â1,2
â1+x5>1
â´â«01dx1+x5<â«011dx
âI<1

Topic: Mathematics, Definite Integration, JEE Main
Consider I1=∫1020ln⁡xln⁡x+ln⁡30-xdx and I2=∫2030ln⁡xln⁡x+ln⁡50-xdx. Then, the value of I1I2 is

Options:
A 10
B 2
C 1
D 12

The correct answer is:
1

Solution:
Applying a+b-x in I1, and adding, we get,

2I1=∫1020ln⁡x+ln⁡30-xln⁡x+ln⁡30-xdx
2I1=∫10201dx=x1020=20-10
⇒I1=5
Similarly, 2I2=x2030=30-20
⇒I2=5
∴I1I2=1

Topic: Mathematics, Definite Integration, JEE Main
Consider \( \mathrm{I}(\alpha)=\int_{\alpha}^{\alpha^{2}} \frac{d x}{x} \) (where \( \alpha>0 \) ), then the value of \( \sum_{r=2}^{5} \mathrm{I}(r)+\sum_{k=2}^{5} \mathrm{I}\left(\frac{1}{k}\right) \) is

Options:
A \( 0 \)
B \( 1 \)
C \( \ln 2 \)
D \( \ln 4 \)

The correct answer is:
\( 0 \)

Solution:
Ια=ln⁡xαα2=ln⁡α2-ln⁡α
=ln⁡α2α
=ln⁡α
Thus, ∑r=25Ιr=ln⁡2+ln⁡3+ln⁡4+ln⁡5 and

∑r=25Ι1k=ln⁡12+ln⁡13+ln⁡14+ln⁡15=-ln⁡2+ln⁡3+ln⁡4+ln⁡5

Hence, their sum equals to zero

Topic: Mathematics, Definite Integration, JEE Main
Consider one dimensional motion of a particle of mass m . It has potential energy U=a+bx2, where a and b are positive constants. At origin (x=0) it has initial velocity V 0 . It performs simple harmonic oscillations. The frequency of the simple harmonic motion depends on

Options:
A b and m alone
B b, a and m alone
C b alone
D b and a alone

The correct answer is:
b and m alone

Solution:
Given,
Potential energy, U=a+bx2
Force, F=-dUdx
=-d.(a+bx2)dx
=-ddxa+ddx(bx2)
â F=0-b.ddx(x2)
â F=0-b.2x=-2bx
â a=Fm=-2bmx .......(i)
Compare equation. (i) with the standard relation of displacement in case of simple harmonic motion.
i.e., a=F2m=-Ï2x
i.e., Ï2=2bm
Ï=2bm
i.e., the frequency of the simple harmonic motion depends on b and m alone.

Topic: Physics, Oscillations, JEE Main
Consider points A, B, C and D with position vectors $7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}, \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}},-\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and $5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ respectively. Then $\mathrm{ABCD}$ is a

Options:
A
parallelogram but not a rhombus

B
square

C
rhombus

D
rectangle

The correct answer is:
rhombus

Solution:
No option satisfied wrong.
$\mathrm{A}=(7,-4,7), \mathrm{B}=(1,-6,10), \mathrm{C}=(-1,-3,4)$ and $\mathrm{D}=(5,-1,5)$
$A B=\sqrt{(7-1)^2+(-4+6)^2+(7-10)^2}=\sqrt{36+4+9}=7$
Similarly $\mathrm{BC}=7, \mathrm{CD}=\sqrt{41}, \mathrm{DA}=\sqrt{17}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003
Consider reaction,

CaCO 3 s â CaO s + CO 2 g

In a closed container at equilibrium. What would be the effect of addition of CaCO 3 on the equilibrium concentration of CO 2  ?

Options:
A Increases
B Decreases
C Data is not sufficient to predict it
D Remains unaffected

The correct answer is:
Remains unaffected

Solution:
This is a heterogeneous equilibrium. The equilibrium constant for it can be written as:
Kâ¡ c = CO 2
The concentration of CaCO3 and CaO are assumed to be constant as they are present as solid.
Hence, CaCO3 has no effect on equilibrium concentration.

Topic: Chemistry, Chemical Equilibrium, JEE Main
Consider:

S1:p⇒q∨p∧~q is a tautology.

S2:~p⇒~q∧~p∨q is a contradiction.

Then

Options:
A only S2 is correct
B both S1 and S2 are correct
C both S1 and S2 are wrong
D only S1 is correct

The correct answer is:
only S1 is correct

Solution:
p	q	p⇒q	~q	p∧~q	p⇒q∨p∧~q

T	T	T	F	F	T

T	F	F	T	T	T

F	T	T	F	F	T

F	F	T	T	F	T

So, p⇒q∨p∧~q is a tautology.

~p	~q	~p⇒~q	~p∨q	~p⇒~q∧~p∨q

F	F	T	T	T

F	T	T	F	F

T	F	F	T	F

T	T	T	T	T

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Consider separate solutions of 0.500 M C2H5OH aq, 0.100 M Mg3PO42 aq, 0.250 M KBr aq and 0.125 M Na3PO4 aq at 25°C. Which statement is true about these solutions, assuming all salts to be strong electrolytes?

Options:
A They all have the same osmotic pressure.
B 0.100 M Mg3(PO4)2 aq has the highest osmotic pressure.
C 0.125 M Na3PO4 aq has the highest osmotic pressure.
D 0.500 M C2H5OH aq has the highest osmotic pressure.

The correct answer is:
They all have the same osmotic pressure.

Solution:
Osmotic pressure π=iCRT(i is Van't Hoff factor, C is concentration)
π=iC at same temperature.
C2H5OH=1x0.5=0.5Mg3(PO4)2=5x0.1=0.5KBr=2x0.25=0.5Na3PO4=4x0.125=0.5
The solutions are isotonic as they have the same osmotic pressure.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2014 (06 Apr)
Consider single slit experiment of diffraction of light. If the light of wavelength 5000 Aâ fall on a slit of width 1 Î¼m then the angular width of central maximum is

Options:
A 30°
B 15°
C 60°
D None of these

The correct answer is:
60°

Solution:
dsin⁡θ=λ
sin⁡θ=λd
For first minima
=5 × 10-710-6
sin⁡θ=510
θ=30° Angular width of first minima
2θ=60°

Topic: Physics, Wave Optics, JEE Main
Consider :
Statement - I : p∧~q∧~p∧q is a fallacy.
Statement - II : p→q↔~q→~p is a tautology.

Options:
A Statement - I is true; statement - II is false.
B Statement - I is false; Statement -II is true.
C Statement - I true; Statement -II is true; Statement -II is a correct explanation for Statement - I.
D Statement - I is true; Statement - II is true; Statement - II is not a correct explanation for Statement - I.

The correct answer is:
Statement - I is true; Statement - II is true; Statement - II is not a correct explanation for Statement - I.

Solution:
The truth table for Statement I is

p	q	~p	~q	p∧~q	~p∧q	p∧~q∧~p∧q

TTFF

TFTF	FFTT	FTFT	FTFF	FFTF

FFFF

Hence, it is a fallacy.

The truth table for statement II is

p	q	~p	~q	p→q	~q→~p	p→q↔~q→~p

TTFF	TFTF	FFTT	FTFT	TFTT	TFTT	TTTT

Hence, it is a tautology.

But, the second statement is not a correct explanation for the first.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2013 (07 Apr)
Consider telecommunication through optical fibres. Which of the following statements is not true?

Options:
A
Optical fibres can be of graded refractive index

B
Optical fibres are subjective to electromagnetic interference from outside

C
Optical fibres have extremely low transmission loss

D
Optical fibres may have homogeneous core with a suitable cladding.

The correct answer is:
Optical fibres are subjective to electromagnetic interference from outside

Solution:
No Solution Available

Topic: Physics, Communication System, JEE Main, JEE Main 2003
Consider telecommunication through optical fibres. Which of the following statements is not true?

Options:
A optical fibres can be of graded refractive index
B optical fibres are subjected to electro-magnetic interference from outside
C optical fibres have extremely low transmission loss
D optical fibres may have homogeneous core with a suitable cladding

The correct answer is:
optical fibres are subjected to electro-magnetic interference from outside

Solution:
Some of the characteristics of an optical fibre are as follows.
(i) This works on the principle of total internal reflection.

(ii) It consists of core made up of glass/silica/plastic with refractive index n1, which is surrounded by a glass or plastic cladding with refractive index n2n2>n1. The refractive index of cladding can be either changing abruptly or gradually changing (Graded index fibre).

(iii) There is a very little transmission loss through optical fibres.

(iv) There is no interference from stray electric and magnetic fields to the signals through optical fibres.

Topic: Physics, Communication System, JEE Main
Consider that d6 metal ion M2+ forms a complex with aqua ligands, and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilization energy of the complex is :

Options:
A octahedral and -2.4Δ0+2P
B tetrahedral and −0.6Δt
C octahedral and -1.6Δ0
D tetrahedral and -1.6Δt+1P

The correct answer is:
tetrahedral and −0.6Δt

Solution:
Since spin only magnetic moment is 4.90 BM so number of unpaired electrons must be 4. There are two possibilities:

(1) If the complex is octahedral, then it has to be high spin complex with configuration t2g2,1,1eg1,1, in that case CFSE=4×-0.4Δ0+2×0.6Δ0=-0.4Δ0
(2) If the complex is tetrahedral then its electronic configuration will be =egt2g2,11,1,1 and CFSE will be=3X -0.6Δt+3X0.4Δt=-0.6Δt

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Consider the above chemical reaction and identify product A

Options:
A
B
C
D

The correct answer is:

Solution:
It is a hydrolysis reaction. In this reaction first cyclo hexane cyanide forms cyclohexyl methanamide which on further hydrolysis forms cyclohexyl methanoic acid. In this reaction water behaves as nucleophile and attacks on electrophilic site of carbonyl group.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Consider the above chemical reaction. Product " \(\mathrm{A}\) " is :

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Consider the above chemical reaction. The total number of stereoisomers possible for Product "P" is _________

Options:

The correct answer is:
2

Solution:
ENT denote enantiomer of product.

The total number of products possible =2

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Consider the above chemical reaction. The total number of stereoisomers possible for Product "P" is

Options:

The correct answer is:
2

Solution:
ENT denote enantiomer of product.

The total number of products possible =2

Topic: Chemistry, Hydrocarbons, JEE Main
consider the above reaction, and choose the correct statement:

Options:
A The reaction is not possible in acidic medium
B Both compounds A and B are formed equally
C Compound A will be the major product
D Compound B will be the major product

The correct answer is:
Compound A will be the major product

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Consider the above reaction and identify the missing reagent/chemical.

Options:
A B2H6
B Red Phosphorus
C CaO
D DIBAL-H

The correct answer is:
CaO

Solution:
Given reaction is soda lime decarboxylation which takes place with NaOH and CaO. In this reaction carbanion is formed as intermediate.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Consider the above reaction and identify the Product P :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Consider the above reaction and predict the major product.

Options:
A

B

C
D

The correct answer is:

Solution:
DiBAL-H is Diisobutylaluminium hydride, which is a very useful reducing agent. It is used for selectively converting ester or nitriles into respective aldehydes. It does not affect other groups like aldehyde or double bond.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Consider the above reaction, compound B is:

Options:
A
B
C
D

The correct answer is:

Solution:
In first step diazonium salt is formed which will react with given compound to show coupling reaction. In coupling reaction diazonium salt will act as electrophile and activating group will direct electrophile in para position.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Consider the above reaction sequence and identify the product B.

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Consider the above reaction sequence, Product "A" and Product "B" formed respectively are:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Consider the above reaction, the compound 'A' is

Options:
A

B

C

D

The correct answer is:

Solution:
The first step of the reaction is nucleophilic substitution reaction in which ammonia being a strong nucleophile will substitute chloride ion and amide will be formed in place of acid chloride. In 2nd step Hoffmann degradation reaction will take place and will give aniline in place of amide.

Topic: Chemistry, Amines, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Consider the above reaction, the major product P formed is :-

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Consider the above reaction, the major product 'P' is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (25 Jul Shift 1)