datavorous/entrance-exam-dataset huggingface dataset:
One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the centre of the loop BL to that at the centre of the coil BC, i.e. BLBC will be

Options:
A 1N2
B 1N
C N
D N2

The correct answer is:
1N2

Solution:
For a single loop, radius
r1=L2Ï
BL=Î¼0I2r1=Î¼0I2Â·LÃ2Ï ...(1)
For a circular coil, radius
r2=L2ÏN
Bc=Î¼0Â·IÂ·N2r2=Î¼0Â·IÂ·N2Â·Lâ2ÏN ...(2)
BLBC=1N2 (from eqns (1) and (2))

Topic: Physics, Magnetic Effects of Current, JEE Main
One of the two small circular coils, (none of them having any self - inductance) is suspended with a V-shaped copper wire, with plane horizontal. The other coil is placed just below the first one with plane horizontal. Both the coils are connected in series with a dc supply. The coils are found to attract each other with a force.
Which one of the following statements is incorrect?

Options:
A
Both the coils carry currents in the same direction.

B
Coils will attract each other, even if the supply is an ac source.

C
Force is proportional to $\mathrm{d}^{-1}$

D
Force is proportional to $\mathrm{d}^{-2}$

The correct answer is:
Force is proportional to $\mathrm{d}^{-2}$

Solution:
As the force between the two coils is attractive hence currents in the coil must be in the same direction and the force, $\mathrm{F} \propto \frac{\mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{d}}$
where $I_1$ and $I_2$ are the currents flowing through the coils.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2013 (25 Apr Online)
One plano-convex and one plano-concave lens of the same radius of curvature R but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is μ1 and that of 2 is μ2, then the focal length of the combination is:

Options:
A R2-μ1-μ2
B R2μ1-μ2
C 2Rμ1-μ2
D Rμ1-μ2

The correct answer is:
Rμ1-μ2

Solution:
From lens maker’s formula,
1f=μ-11R1-1R2
1f1=μ1-11∞-1-R=μ1-1R
1f2=μ2-11-R-1∞=-μ2-1R
1feq=1f1+1f2=μ1-1R-μ2-1R
=μ1-μ2R
feq=Rμ1-μ2

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (10 Apr Shift 1)
One plano-convex and one plano-concave lens of the same radius of curvature R but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is Î¼1 and that of 2 is Î¼2, then the focal length of the combination is:

Options:
A R2-μ1-μ2
B R2μ1-μ2
C 2Rμ1-μ2
D Rμ1-μ2

The correct answer is:
Rμ1-μ2

Solution:
From lens maker’s formula,
1f=μ-11R1-1R2
1f1=μ1-11∞-1-R=μ1-1R
1f2=μ2-11-R-1∞=-μ2-1R
1feq=1f1+1f2=μ1-1R-μ2-1R
=μ1-μ2R
feq=Rμ1-μ2

Topic: Physics, Ray Optics, JEE Main
One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each plate of the capacitor is $A$ and the distance between the plates is $d$, when the battery is not connected and the spring is unstretched. After connecting the battery, in the steady state the distance between the plates is $0.75 d$, then the force constant of the spring is

Options:
A $\frac{3}{8} \frac{\varepsilon_0 V^2 A}{d^3}$
B $\frac{8}{3} \frac{\varepsilon_0 V^2 A}{d^3}$
C $\frac{9}{32} \frac{\varepsilon_0 V^2 A}{d^3}$
D $\frac{32}{9} \frac{\varepsilon_0 V^2 A}{d^3}$

The correct answer is:
$\frac{32}{9} \frac{\varepsilon_0 V^2 A}{d^3}$

Solution:
In equilibrium, force between plates of capacitor $=$ spring force $\Rightarrow \frac{q^2}{2 \varepsilon_0 A}=k x$ where, $x=$ extension in spring. Now before charging, when spring is unstretched $(x=0)$ distance of plates is $d$ and after charging it is $\frac{3}{4} d$.
So, $\quad x=d-\frac{3}{4} d=\frac{1}{4} d$
Hence, $\begin{aligned} k & =\frac{q^2}{2 \varepsilon_0 A \cdot x}=\frac{C^2 V^2}{2 \varepsilon_0 A x} \\ & =\left\{\frac{\frac{\varepsilon_0^2 A^2}{\left(\frac{3}{4} d\right)} \cdot V^2}{2 \varepsilon_0 A \cdot \frac{1}{4} d}\right\}=\frac{32}{9}\left(\frac{\varepsilon_0 V^2 A}{d^3}\right)\end{aligned}$

Topic: Physics, Capacitance, JEE Main
One scientist cultured Cladophora in a suspension of Azotobacter and illuminated the culture by splitting light through a prism. He observed that bacteria accumulated mainly in the region of

Options:
A violet and green light
B indigo and green light
C orange and yellow light
D blue and red light

The correct answer is:
blue and red light

Solution:
Engelmann used a prism to split light into its spectral components and then illuminated a green alga, Cladophora placed in a suspension of aerobic bacteria (Azotobacter). The bacteria were used to detect the sites of oxygen evolution.
He observed that bacteria mainly accumulated in the region of blue and red light of the split spectrum, thus giving the first action spectrum of photosynthesis.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2019 (Odisha)
One slit of a double-slit experiment is covered by a thin glass plate of refractive index 1.4 , and the other by a thin glass plate of the refractive index 1.7 . The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by the fifth bright fringe. Assume the plate have the same thickness t and wavelength of light 480 nm , then the value of t is

Options:
A 2.4 μ m
B 4.8 μ m
C 8 μ m
D 16 μ m

The correct answer is:
8 μ m

Solution:
Due to the induction of the glass plate, the change in path difference is μ2-μ1t .
∵ Before inserting the glass plate, the path difference for central maxima is zero.
After introducing glass plate, the change in path difference is equal to 5λ .
μ2-μ1t=5λ
t=5λμ2-μ1=5×480×10-91.7-1.4
=8 micrometre.

Topic: Physics, Wave Optics, JEE Main
One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moment of inertia about their diameters are respectively $\mathrm{I}_{\mathrm{A}}$ and $\mathrm{I}_{\mathrm{B}}$ such that

Options:
A
$I_A=I_B$

B
$I_A>I_B$

C
$I_A < I_B$
D
yes, $45^{\circ}$
$I_A / I_B=d_A / d_B$
Where $d_A$ and $d_B$ are their densities.

The correct answer is:
$I_A < I_B$

Solution:
Moment of inertia of a uniform density solid sphere, $A=\frac{2}{5} M R^2$
Since $M$ and $R$ are same, $I_A < I_B$.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2004
One solid sphere $A$ and another hollow sphere B are of same mass and same outer radii. Their moments of inertia about their diameters are respectively $I_A$ and $I_B \mid$ such that Here $\rho_A$ and $\rho_B$ represent their densities.

Options:
A $I_A=I_B$
B $I_A \gt I_B$
C $I_A \lt I_B$
D $\quad I_A / I_B=\rho_A=\rho_B$

The correct answer is:
$I_A \lt I_B$

Solution:
In a hollow sphere, the mass is distributed away from the axis of rotation. So, its moment of inertia is greater than that of a solid sphere.

Topic: Physics, Rotational Motion, JEE Main
One stone is dropped from a tower from rest and simultaneously another stone is projected vertically upwards from the tower with some initial velocity. The graph of the distance(s) between the two stones varies with time t as (before either stone hits the ground)

Options:
A
B
C
D

The correct answer is:

Solution:
s⁡1t=12gt2downwards and s⁡2t=u⁡t-12gt2upwards
∴ Distance between the two stones will be
s=s1(t)+s2(t)=ut
Therefore, s-t graph will be a straight line passing through origin.

Topic: Physics, Motion In One Dimension, JEE Main
One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to total initial surface energy is

Options:
A $1000: 1$
B $1: 1000$
C $10: 1$
D $1: 10$

The correct answer is:
$1: 10$

Solution:
Let radius of big drop $=R$
Radius of small drop $=r$
Number of small drops, $n=1000$
In this process, the volume remains same
$V_f=V_i$
$\frac{4}{3} \pi R^3=n\left(\frac{4}{3} \pi r^3\right)$
$R^3=1000 r^3 \Rightarrow R=10 r$ ...(i)
Initial surface area,
$A_1=n 4 \pi r^2$
Final surface area, $A_2=4 \pi R^2$
Now, increase in surface area,
$\Delta A=A_2-A_1=4 \pi\left[R^2-n r^2\right]$
We know that, surface energy,
$\mathrm{SE}=\mathrm{TA}$ [where, $T=$ surface tension of liquid]
According to question,
$\frac{\mathrm{SE}_f}{\mathrm{SE}_i}=\frac{\mathrm{TA}_2}{\mathrm{TA}_1}=\frac{4 \pi R^2}{n\left(4 \pi r^2\right)}=\frac{R^2}{1000 r^2}$ [From Eq. (i)]
$=\frac{(10 r)^2}{1000 r^2}=\frac{100 r^2}{1000 r^2}=\frac{1}{10}$ or $1: 10$

Topic: Physics, Mechanical Properties of Fluids, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 2)
One ticket is selected at random from 50 tickets numbered $00,01,02, \ldots, 49$. Then the probability that the sum of the digits on the selected ticket is 8 , given that the product of these digits is zero, equals

Options:
A
$\frac{1}{14}$

B
$\frac{1}{7}$

C
$\frac{5}{14}$

D
$\frac{1}{50}$

The correct answer is:
$\frac{1}{14}$

Solution:
$$
S=\{00,01,02, \ldots .49\}
$$
Let $A$ be the even that sum of the digits on the selected ticket is 8 then
$$
A=\{08,17,26,35,44\}
$$
Let $B$ be the event that the product of the digits is zero
$$
B=\{00,01,02,03, \ldots .09,10,20,30,40\}
$$
$A \cap B=\{8\}$
$$
\text { Required probability }=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{50}}{\frac{14}{50}}=\frac{1}{14}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2009
One vertex of a rectangular parallelopiped is at the origin O and the lengths of its edges along x, y and z axes are 3, 4 and 5 units respectively. Let P be the vertex (3, 4, 5). Then the shortest distance between the diagonal OP and an edge parallel to z axis, not passing through O or P is

Options:
A 125
B 125
C 1255
D 125

The correct answer is:
125

Solution:
Given,

One vertex of a rectangular parallelopiped is at the origin O and the lengths of its edges along x, y and z axes are 3, 4 and 5 units respectively,

And P be the vertex (3, 4, 5),

Now plotting the diagram we get,

Now equation of line OP:x3=y4=z5

And equation of line AB:x-30=y0=z1

Now finding, n1→×n2→=i^j^k^345001

⇒n1→×n2→=i^4-j^3+k^0

⇒n1→×n2→=4i^-3j^

Now we know that,

Shortest Distance, S.D=a2→-a1→·n1→×n2→n1→×n2→

⇒S.D=3i^+0j^+0k^-0i^+0j^+0k^·4i^-3j^5

⇒S.D=3i^·4i^-3j^5

⇒S.D=125

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (06 Apr Shift 1)
One word answer is given for the following definitions. Mark the one which is incorrect.

Options:
A The process in which temperature remains constant: Isobaric
B The process in which volume remains constant: Isochoric
C The relation between ΔH and ΔE when all the reactants and products are solid: ΔH=ΔE
D The relation between ΔG,ΔH and ΔS : ΔG=ΔH-TΔS

The correct answer is:
The process in which temperature remains constant: Isobaric

Solution:
As the process in which temperature remains constant is called isothermic process.

Topic: Chemistry, Thermodynamics (C), NEET
One-fourth length of a uniform rod is placed on a rough horizontal surface and it starts rotating about the edge as soon as we release it. The rod starts slipping on the edge when it has turned through an angle Î¸. If the coefficient of friction between rod and surface is Î¼, and it satisfies the relation xtanÎ¸=4Î¼, then what is the value of x? [Take g=10 m/s2]

Options:

The correct answer is:
13

Solution:
mgl4sin⁡θ=12Iω2
α=mg l4cos⁡θ I

μN-mgsin⁡θ=mω2l4
mgcos⁡θ-N=ml4α
Solving,
13tan⁡θ=4μ .

Topic: Physics, Rotational Motion, JEE Main
Only \(2 \mathrm{~mL}\) of \(\mathrm{KMnO}_4\) solution of unknown molarity is required to reach the end point of a titration of \(20 \mathrm{~mL}\) of oxalic acid \((2 \mathrm{M})\) in acidic medium. The molarity of \(\mathrm{KMnO}_4\) solution should be \(\qquad\) M.

Options:

The correct answer is:
8

Solution:
$\begin{aligned} & \text { eq. }\left(\mathrm{KMnO}_4\right)=\text { eq. }\left(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4\right) \\ & \mathrm{M} \times 2 \times 5=2 \times 20 \times 2 \\ & \mathrm{M}=8 \mathrm{M}\end{aligned}$

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Only an aldehyde having..... can undergo the aldol condensation

Options:
A At least one beta $H$ atom
B At least one alpha $H$ atom
C An aromatic ring
D No alpha $H$ atom

The correct answer is:
At least one alpha $H$ atom

Solution:
presence of At least one alpha $\mathrm{H}$ atom in aldehyde is the necessary condition for aldol condensation.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
\( \operatorname{Lim}_{x \rightarrow \infty} \frac{x(\log x)^{3}}{1+x+x^{2}} \) equals

Options:
A \( 0 \)
B \( -1 \)
C \( 1 \)
D Does not exist

The correct answer is:
\( 0 \)

Solution:
Limx→∞log⁡x3+x.3log⁡x2×1x1+2x
(By D.L. Hospital rule)
⇒ Limx→∞3log⁡x2×1x+6log⁡x×1x2
(By D.L. Hospital rule)
⇒ Limx→∞3log⁡x2+6log⁡x2x
(By D.L. Hospital rule)
⇒Limx→∞6log⁡x×1x+6x2
(By D.L. Hospital rule)
⇒ Limx→∞6log⁡x+62x
(By D.L. Hospital rule)
⇒Limx→∞61x+02
(By D.L. Hospital rule)
=6∞2=0

Topic: Mathematics, Limits, JEE Main
Optical activity is shown by

Options:
A Glucose
B Fructose
C Sucrose
D All of these

The correct answer is:
All of these

Solution:
Glucose and sucrose are dextrorotatory
Fructose is leavorotatory

Topic: Chemistry, Biomolecules, JEE Main
Optical activity of an enantiomeric mixture is +12.6° and the specific rotation of + isomer is +30°. The optical purity is____%

Options:

The correct answer is:
42

Solution:
The "optical purity" is a comparison of the optical rotation of a pure sample of unknown stereochemistry versus the optical rotation of a sample of pure enantiomer. It is expressed as a percentage. If the sample only rotates plane-polarized light half as much as expected, the optical purity is 50%.

% optical purity =observed rotation of mixture×100rotation of pure enantiomer

=+12.6°+30°×100=42

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Optical isomers

Options:
A 2
B 4
C Zero
D 3

The correct answer is:
Zero

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Optically active compound is

Options:
A 3-chloropentane
B 2-chlorobutane
C 2-chloropropane
D None of these

The correct answer is:
2-chlorobutane

Solution:
2-chloro butane will be optically active.

Topic: Chemistry, General Organic Chemistry, JEE Main
Optically active
was found to have lost its optical activity after standing in water containing a few drops of acids, mainly due to the formation of

Options:
A $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2$
B $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
C

D $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}$

The correct answer is:
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$

Solution:
The above optically active compound undergo dehydration with few drops of acid and form stable alkene through the formation of stable carbocation.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Optically active
was found to have lost its optical activity after standing in water containing a few drops of acids, mainly due to the formation of

Options:
A $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2$
B $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$
C

D $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}$

The correct answer is:
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$

Solution:
The above optically active compound undergo dehydration with few drops of acid and form stable alkene through the formation of stable carbocation.

Topic: Chemistry, Alcohols Phenols and Ethers, TS EAMCET, TS EAMCET 2017
Orange light of wavelength 6000×10–10 m illuminates a single slit of width 0.6×10–4 m. The maximum possible number of diffraction minima produced on both sides of the central maximum is __________

Options:

The correct answer is:
200

Solution:
Light of wavelength 6000 × 10–10 m passes through a single slit of width 0.6 × 10–4 m. Find height of highest order of minima on both side central maxima

for minima

dsinθ=nλ

sinθ=nλd<1

n≤dλ

n<0.6×10-46000×10-10

n<100

The total number of maxima of both side at central maxima =100+100=200

Topic: Physics, Wave Optics, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Orbit of a planet around a star is

Options:
A A circle
B An ellipse
C A parabola
D A straight line

The correct answer is:
An ellipse

Solution:
According to the Kepler's Law of Orbits, the planets revolve around the star in elliptical orbits.

Topic: Physics, Gravitation, JEE Main
Orbital interaction between the sigma bonds of a substituent group and a neighboring pi orbital is known as

Options:
A Hyperconjugation
B Inductive effect
C Steric effect
D Dipole-dipole interactions

The correct answer is:
Hyperconjugation

Solution:
It is hyperconjugation process.

Topic: Chemistry, General Organic Chemistry, JEE Main
Orbital velocity of an artificial satellite does not depend upon

Options:
A Mass of the earth
B Mass of the satellite
C Radius of the earth
D Acceleration due to gravity

The correct answer is:
Mass of the satellite

Solution:
Orbital velocity is independent of mass of satellite.
$v=\sqrt{\frac{G M}{r}}$

Topic: Physics, Gravitation, JEE Main
Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the $n^{\text {th }}$ orbital will therefore be proportional to :

Options:
A
$n^2$

B
$n$

C
$n^{1 / 2}$

D
$n^{1 / 4}$

The correct answer is:
$n^{1 / 2}$

Solution:
According to the question,
$$
\begin{aligned}
& 2 \pi \mathrm{r}=\mathrm{n} \lambda=\frac{\mathrm{nh}}{\mathrm{p}}=\frac{\mathrm{nh}}{\mathrm{mv}} \\
& \text { or } \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \text { or } \mathrm{mv}=\frac{\mathrm{nh}}{2 \pi \mathrm{r}} \\
& \mathrm{F}=\mathrm{qv}_{\mathrm{B}}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\
& \text { or, } \mathrm{q}_{\mathrm{B}}=\frac{\mathrm{mv}}{\mathrm{r}}=\frac{\mathrm{nh}}{2 \pi \mathrm{r} \cdot \mathrm{r}} \\
& \text { or, } \mathrm{r}^2=\frac{\mathrm{nh}}{2 \pi \mathrm{qB}} \\
& \text { or, } \mathrm{r}=\sqrt{\frac{\mathrm{nh}}{2 \pi \mathrm{qB}}} \\
& \text { i.e., } \mathrm{r} \propto \mathrm{n}^{1 / 2}
\end{aligned}
$$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2013 (22 Apr Online)
order and degree of differential equation $\frac{d^2 y}{d x^2}=\left\{y+\left(\frac{d y}{d x}\right)^2\right\}^{1 / 4}$

Options:
A 4 and 2
B 1 and 2
C 1 and 4
D 2 and 4

The correct answer is:
2 and 4

Solution:
Making fourth power both the sides, we get the differential equation$\left(\frac{d^2 y}{d x^2}\right)^4=y+\left(\frac{d y}{d x}\right)^2$Obviously, order is 2 and degree is 4 .

Topic: Mathematics, Differential Equations, JEE Main
Order of Covalent bond;

A. KF>KI; LiF>KF

B. KF<KI; LiF>KF

C. SnCl4>SnCl2; CuCl>NaCl

D. LiF>KF; CuCl>NaCl

E. KF<KI; CuCl>NaCl

Options:
A C, E Only
B

B, C Only

C

B, C, E Only

D A, B Only

The correct answer is:

B, C, E Only

Solution:
According to Fajan's Rule, small cation has more polarising power and larger anion has more polarisability. These small cations and larger anions favours covalent character.

A. KF>KI-False (Iodide is larger anion than fluoride)

LiF>KF-True (Lithium-ion is smaller cation than potassium ion)

B. KF<KI-True; LiF>KF-True

C. SnCl4 >SnCl2-True(Sn4+ is smaller than Sn2+);

The ions with pseudo inert gas configuration prefers the covalent character.

CuCl>NaCl-True(Cu+(3s23p63d10) ion has pseudo inert gas configuration)

D. LiF>KF-True; CuCl<NaCl-False

E. KF<KI-True; CuCl>NaCl-True

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Order of $\mathrm{g} / \mathrm{m}$ ratio of proton, $\alpha$-particle and electron is

Options:
A $\mathrm{e} \gt p \gt \alpha$
B $p \gt \alpha \gt $ e
C $\mathrm{e} \gt \alpha \gt p$
D None of these

The correct answer is:
$\mathrm{e} \gt p \gt \alpha$

Solution:
From the given question,
$\because m_e \lt m_p \lt m_\alpha \Rightarrow\left(\frac{q}{m}\right)_e \gt \left(\frac{q}{m}\right)_p \gt \left(\frac{q}{m}\right)_\alpha$

Topic: Physics, Atomic Physics, JEE Main
Order of Ka will be :

Options:
A I > II > III
B II > I > III
C I > III > II
D III > I > II

The correct answer is:
I > III > II

Solution:
Acidic strength K a α - m
α 1 + m
Order : I > III > II

Topic: Chemistry, General Organic Chemistry, JEE Main
Order of radioactive disintegration reaction is

Options:
A Zero
B First
C Second
D Third

The correct answer is:
First

Solution:
The radioactive disintegration reactions are of first order because this rate of disintegration depends on the concentration term of radioactive material only.

Topic: Chemistry, Chemical Kinetics, JEE Main
Order of rate of reaction of following compounds with AgNO3 (following SN1 mechanism).

Options:
A II > III > I
B III > II > I
C I > II > III
D III > I > II

The correct answer is:
I > II > III

Solution:
Here the reactivity order is
(I) becomes aromatic after removal of Br-
(II) becomes non aromatic after removal of Br-
(III) becomes anti aromatic after removal of Br-

Topic: Chemistry, General Organic Chemistry, JEE Main
Order of reactivity of $\mathrm{C}_2 \mathrm{H}_6, \mathrm{C}_2 \mathrm{H}_4$ and $\mathrm{C}_2 \mathrm{H}_2$ is

Options:
A $\mathrm{C}_2 \mathrm{H}_6\gt\mathrm{C}_2 \mathrm{H}_4\gt\mathrm{C}_2 \mathrm{H}_2$
B $\mathrm{C}_2 \mathrm{H}_2\gt\mathrm{C}_2 \mathrm{H}_6\gt\mathrm{C}_2 \mathrm{H}_4$
C $\mathrm{C}_2 \mathrm{H}_2\gt\mathrm{C}_2 \mathrm{H}_4\gt\mathrm{C}_2 \mathrm{H}_6$
D All are equally reactive

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_2\gt\mathrm{C}_2 \mathrm{H}_4\gt\mathrm{C}_2 \mathrm{H}_6$

Solution:
Unsaturated hydrocarbons are more reactive than saturated hydrocarbons. Among ethyne $\left(C_2 \mathrm{H}_2\right)$ and ethene $\left(\mathrm{C}_2 \mathrm{H}_4\right)$ the later is more reactive as $C \equiv C$ triple bond is quite strong bond and therefore ethyne generally require catalysts (like $\mathrm{Hg}^{2+}$ etc) in its reactions.

Topic: Chemistry, Hydrocarbons, JEE Main
Ordinarily the barrier to rotation about a carbon-carbon double bond is quite high but in compound P double bond between two rings was observed by NMR to have a rotational energy barrier of only about 20 cal/mol., showing that it has lot of single bond character.

The reason for this is

Options:
A Double bond having partial triple bond character because of resonance
B Double bond undergoes flipping
C Double bond having very high single bond character because of aromaticity gained in both three and five membered rings.
D +I effect of nC3H7 groups makes double bond having partial single bond character.

The correct answer is:
Double bond having very high single bond character because of aromaticity gained in both three and five membered rings.

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Ordinary glass is:

Options:
A Sodium silicate
B Calcium silicate
C Sodium and calcium silicate
D Mixed salt of $\mathrm{Na}$ and $\mathrm{Ca}$

The correct answer is:
Sodium silicate

Solution:
Normal glass is calcium alkali silicate glass made by fusing the alkali metal carbonate, $\mathrm{CaCO}_{3}$ and $\mathrm{SiO}_{2}$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Organic compounds are broadly classified as

Options:
A Open chain compounds and acyclic compounds
B Open chain compounds and linear chain compounds
C Cyclic compounds and alicyclic compounds
D alicyclic compounds and acyclic compounds

The correct answer is:
alicyclic compounds and acyclic compounds

Solution:
Explanation: Organic compounds are broadly classified into open chain and closed chain compounds.

Topic: Chemistry, General Organic Chemistry, JEE Main
Organic compounds can be classified even based upon the function groups. Identify the one which is not a functional group?

Options:
A Isocyanide
B Isocyano
C Carboxyl
D Carbonyl

The correct answer is:
Isocyanide

Solution:
Explanation: Isocyanide is a compound and it is not a functional group.

Topic: Chemistry, General Organic Chemistry, JEE Main