datavorous/entrance-exam-dataset huggingface dataset:
The centres of a set of circles, each of radius 3 , lie on the circle $x^2+y^2=25$. The locus of any point in the set is

Options:
A
$4 \leq x^2+y^2 \leq 64$

B
$x^2+y^2 \leq 25$

C
$x^2+y^2 \geq 25$

D
$3 \leq x^2+y^2 \leq 9$

The correct answer is:
$4 \leq x^2+y^2 \leq 64$

Solution:
Let $(h, k)$ be the centre of any such circle. Equation of such circle is $(x-h)^2+(y-k)^2=3^2$. Since $(\mathrm{h}, \mathrm{k})$ lies on $\mathrm{x}^2+\mathrm{y}^2=25 \quad \therefore \mathrm{h}^2+\mathrm{k}^2=25$.
$x^2+y^2-(2 x h+2 y k)+25=9$; Locus of $(h, k)$ is $x^2+y^2=16$, which clearly satisfies (a).

Topic: Mathematics, Circle, JEE Main, JEE Main 2002
The centres of those circles which touch the circle, x2+y2-8x-8y-4=0, externally and also touch the x - axis, lie on

Options:
A A hyperbola
B A parabola
C A circle
D An ellipse which is not a circle

The correct answer is:
A parabola

Solution:
Given circle x2+y2-8x-8y-4=0
has centre 4, 4 and radius 6.
Let h, k be the centre of the circle which is touching the given circle externally
Then
h-42+k-42=6+k
⇒h2-8h+16+k2-8k+16=36+12k+k2
⇒h2-8h-20k-4=0 ,
Replacing h by x and k by y, locus is
x2-8x-20y-4=0
Which is equation of parabola.

Topic: Mathematics, Circle, JEE Main, JEE Main 2016 (03 Apr)
The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from each other. Let P be the midpoint of the line segment joining the centres of C1 and C2 and C be a circle touching circles C1 and C2 externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 and C, then the radius (in units) of the circle C is

Options:

The correct answer is:
8

Solution:
In Δ CPC 2 , r + 1 2 = α 2 + 3 2 ...(i)
In Δ PTC ,
α 2 = r 2 + 3 2 - 1 2
α 2 = r 2 + 8
Putting the value of α 2 in equation (i), we get,
1 + r 2 + 2 r = 1 r 2 + 8 + 9
⇒2r=16⇒r=8

Topic: Mathematics, Circle, JEE Main
The centroid of the triangle formed by joining the feet of the normals drawn from any point to the parabola $y^2=4 a x$, lies on

Options:
A Axis
B Directrix
C Latus rectum
D Tangent at vertex

The correct answer is:
Axis

Solution:
Normal to parabola will be, $y=m x-2 a m-a m^3$
For three values of $m$. Three normal can be drawn on parabola $y^2=4 a x$. So three feet of normals can be obtained, hence centroid of triangle lies on axis of parabola.

Topic: Mathematics, Parabola, JEE Main
The centromere is that part of the chromosome where _________.

Options:
A nicking occurs
B chromatids are attached
C Nucleoli are formed
D Crossing-over takes place

The correct answer is:
chromatids are attached

Solution:
A chromosome consists of two chromatids that remain attached at a point called centromere or primary constriction. At centromere two disc-shaped bodies are present called kinetochore to which spindle fibers attach during cell division.

Topic: Biology, Principles of Inheritance and Variation, NEET
The cerebral cortex is greyish in appearance due to

Options:
A the large number of cell bodies of neurons.
B the cell body of neurons lacks myelin sheath.
C the neuron cell bodies are concentrated here.
D all of the above.

The correct answer is:
the large number of cell bodies of neurons.

Solution:
The central nervous system is made up of two types of tissues: the grey matter and the white matter. In the grey matter, the axons are mainly unmyelinated, i.e., they are not covered by a whitish-coloured fatty protein called myelin.

Topic: Biology, Neural Control and Coordination, NEET
The certain endodermal cell which is present opposite to the xylem bundle remain thin-walled, these cells are

Options:
A bulliform cells
B passage cells
C companion cells
D subsidiary cells

The correct answer is:
passage cells

Solution:
The cell wall of endodermis is deposited with suberin in the form of strips known as Casparin strips, which makes the layer impermeable. Few cells of endodermis lying opposite to the protoxylem lack suberin on their tangential wall, these are called passage cells which maintains the unidirectional flow of water from outsite to inside.

Topic: Biology, Anatomy of Flowering Plants, NEET
The CFSE for [CoCl6]4- complex is 18000 cm-1 . The Δ for [CoCl4]2- will be

Options:
A 9000 cm-1
B 4000 cm-1
C 8000 cm-1
D 2000 cm-1

The correct answer is:
8000 cm-1

Solution:
The relation of CFSE for tetrahedral and octahedral complex is given as
Ît=49Î0
So Ît for [CoCl4]2-=49Ã18000cm-1
=8000 cm-1 .

Topic: Chemistry, Coordination Compounds, JEE Main
The change in energy when a big dropis split in sma $\| n$ droplets is

Options:
A $4 R^{2}\left(n^{2 / 3}-1\right) I$
B $4 \pi R^{2}\left(n^{1 / 3}-1\right) T$
C $4 \pi R^{2}\left(n^{-1 / 3}-1\right) T$
D $4 \pi R^{2}\left[n^{-2 / 3}-1\right] T$

The correct answer is:
$4 \pi R^{2}\left(n^{1 / 3}-1\right) T$

Solution:
When a drop of radius $R$ split into $n$ smaller drops (each of radius $r$ ), then surface area of liquid increases. Hence, the work is to be done against surface tension. Since, the volume of liquid remains constant
$$
\begin{array}{l}
\therefore \quad \frac{4}{3} \pi R^{3}=n \frac{4}{3} \pi r^{3} \quad \therefore R^{3}=n r^{3} \\
\text { Work done }=T \times \Delta A \\
=T\left[n 4 \pi r^{2}-4 \pi R^{2}\right] \\
=4 \pi R^{2} T\left(n^{1 / 3}-1\right)
\end{array}
$$

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2009
The change in entropy when the pressure of perfect gas is changed isothermally from P1 to P2 is

Options:
A ΔS=nRln(P1+P2)
B ΔS=nRln(P2/P1)
C ΔS=nRln(P1/P2)
D ΔS=nRlnP1+P2P2

The correct answer is:
ΔS=nRln(P1/P2)

Solution:
the change in entropy when the pressure is changed isothermally from P1 to P2 is given by relation
ÎS=nRln(P1/P2)

Topic: Chemistry, Thermodynamics (C), JEE Main
The change in optical rotation of a freshly prepared solution of sugar is known as

Options:
A Mutarotation
B Hydrolysis
C Rotatory motion
D Specific angle rotation

The correct answer is:
Mutarotation

Solution:
Glucose has two forms α and β . When either of these two is dissolved in water and allowed to stand, it gets converted to an equilibrium mixture of α and β forms.

Topic: Chemistry, Biomolecules, JEE Main
The change in optical rotation, with time, of freshly prepared solution of sugar is known as

Options:
A Rotatory motion
B Inversion
C Specific rotation
D Mutarotation

The correct answer is:
Mutarotation

Solution:
$\begin{aligned} & \alpha-D-\text { Glucose Equilibriu m mixture } \beta-D-\text { Glucose } \\ & {[a]-+112^{\circ} \quad[a]_D-+52^{\circ} \quad[a]_D-+19^{\circ}} \\ & (36 \%) \quad \Rightarrow \quad(0.02 \%) \quad \rightleftharpoons \quad(64 \%) \end{aligned}$

Topic: Chemistry, Biomolecules, JEE Main
The change in the internal energy of a substance

Options:
A Increases with increase in temperature
B Decreases with increase in temperature
C Can be calculated by the relation E=mc2
D Remains unaffected with change in temperature

The correct answer is:
Increases with increase in temperature

Solution:
ΔE increases with temperature.

Topic: Chemistry, Thermodynamics (C), NEET
The change in the magnitude of the volume of an ideal gas when a small additional pressure ∆P is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity ∆T at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm respectively. If |ΔT|=C|ΔP| then value of C in (K/atm) is __________

Options:

The correct answer is:
150

Solution:
PV=nRT

PΔV+VΔP=0

ΔV=-ΔPPV ....(i)

In second case

PΔV=-nRΔT

ΔV=-nRΔTP ...(ii)

equating (i) and (ii)

nRΔTP=-ΔPPV

ΔT=ΔP VnR

c=vnR

Putting the value of V, n and R , C = 150

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The change in the oxidation state of iodine when excess chlorine water is added to an iodide salt is

Options:

The correct answer is:
6

Solution:
2 KI +  Cl 2 →2 KCl +  I 2
5  Cl 2  +  I 2 + 6  H 2 O→10 HCl + 2  HIO 3
Change in oxidation state of iodine =5−(−1)
=6

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The change in the value of acceleration of earth towards sun, when the moon comes from the position of solar eclipse to the position on the other side of earth in line with sun is:
(mass of the moon $=7.36 \times 10^{22} \mathrm{~kg}$, radius of the moon's orbit $=3.8 \times 10^8 \mathrm{~m}$ ).

Options:
A
$6.73 \times 10^{-5} \mathrm{~m} / \mathrm{s}^2$

B
$6.73 \times 10^{-3} \mathrm{~m} / \mathrm{s}^2$

C
$6.73 \times 10^{-2} \mathrm{~m} / \mathrm{s}^2$

D
$6.73 \times 10^{-4} \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$6.73 \times 10^{-5} \mathrm{~m} / \mathrm{s}^2$

Solution:
No Solution Available

Topic: Physics, Gravitation, JEE Main, JEE Main 2013 (22 Apr Online)
The change in the value of $g$ at a height ' $h$ ' above the surface of the earth is the same as at a depth 'd' below the surface of earth. When both ' $d$ ' and ' $h$ ' are much smaller than the radius of earth, then which one of the following is correct?

Options:
A
$d=\frac{h}{2}$

B
$\mathrm{d}=\frac{3 \mathrm{~h}}{2}$

C
$\mathrm{d}=2 \mathrm{~h}$

D
$\mathrm{d}= \mathrm{~h}$

The correct answer is:
$\mathrm{d}=2 \mathrm{~h}$

Solution:
$\frac{G M}{(R+h)^2}=\frac{G M}{R^3}(R-d)$
$\Rightarrow d=2 h$

Topic: Physics, Gravitation, JEE Main, JEE Main 2005
The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants $\mathrm{G}, \mathrm{h}$ and $\mathrm{c}$. Which of the following correctly gives the Planck length?

Options:
A
$\mathrm{G}^2 \mathrm{hc}$

B
$\left(\frac{\mathrm{Gh}}{\mathrm{c}^3}\right)^{\frac{1}{2}}$

C
$\mathrm{G}^{\frac{1}{2}} \mathrm{~h}^2 \mathrm{c}$

D
$G\left(h^2 \mathrm{c}^3\right.$

The correct answer is:
$\left(\frac{\mathrm{Gh}}{\mathrm{c}^3}\right)^{\frac{1}{2}}$

Solution:
Plank length is a unit of length,
$$
\begin{aligned}
&l_p=1.616229 \times 10^{-35} \mathrm{~m} \\
&l_p=\sqrt{\frac{h G}{c^3}}
\end{aligned}
$$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The characteristic equation of a matrix $\mathrm{A}$ is $\lambda^{3}-5 \lambda^{2}-3 \lambda+2=0$, then $\mid$ adj (A) $\mid$ is equal to

Options:
A 9
B 25
C $\frac{1}{2}$
D 4

The correct answer is:
4

Solution:
We know that, $\left[\begin{array}{ccc}0 & 0 & -\mathrm{c} \\ 1 & 0 & -\mathrm{b} \\ 0 & 1 & -\mathrm{a}\end{array}\right]$ iff
$$
\lambda^{3}+\mathrm{a} \hat{\lambda}^{2}+\mathrm{b} \lambda+\mathrm{c}=0
$$
But given that,
$\therefore$
$\lambda^{3}-5 \lambda^{2}-3 \lambda+2=0$
$A=\left[\begin{array}{rrr}0 & 0 & -2 \\ 1 & 0 & 3 \\ 0 & 1 & 5\end{array}\right]$
$|A|=\left[\begin{array}{rrr}0 & 0 & -2 \\ 1 & 0 & 3 \\ 0 & 1 & 5\end{array}\right]$
$=0+0-2(1-0)=-2$
$\therefore \mid$ adj $\left.\mathrm{A}|=| \mathrm{A}\right|^{2}=(-2)^{2}=4$

Topic: Mathematics, Matrices, JEE Main
The characteristic grouping of secondary alcohols is

Options:
A $-\mathrm{CH}_2 \mathrm{OH}$
B $\gt\mathrm{CHOH}$
C $-\stackrel{1}{\mathrm{C}}-\mathrm{OH}$
D

The correct answer is:
$\gt\mathrm{CHOH}$

Solution:
In a secondary alcohol, the carbon atom bonded to the hydroxyl group is bonded to two other carbon atoms. The $\gt\mathrm{C}-\mathrm{OH}$ represents this functional group.
Primary alcohols have the hydroxyl group bonded to one carbon atom, and tertiary alcohols have the hydroxyl group bonded to three carbon atoms.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The characteristic pigment of Cyanobacteria is

Options:
A chlorophyll a
B fucoxanthin
C phycocyanin
D chlorophyll d

The correct answer is:
chlorophyll a

Solution:
Cyanobacteria are a group of free-living photosynthetic bacteria that perform oxygenic photosynthetic process which are included in the kingdom Monera.
It contains the main green pigment chlorophyll a for photosynthesis as this group have both the photosystems -PS I and PS II.
Phycobilins are also present that gives the blue colour.

Topic: Biology, Biological Classification, NEET
The characteristic X-rays wavelength is related to atomic number by the relation

Î½=aZ-b

When Z is the atomic number, a and b are Mosley's constants. If Î»1= 2.886 â« and Î»2= 2.365 â« corresponding to Z1=55 and Z2=60 respectively, the value of Z corresponding to Î»=2.660 â« is

Options:
A 63
B 67
C 74
D 57

The correct answer is:
57

Solution:
v =a(Z−b)
c λ =a(Z−b)
∴ c 2.886 =a(55−b)
c 2.365 =a(60−b)
a=67.5
b=7.235
Now,  c 2.660 =67.5(Z−7.235)
Z=57

Topic: Chemistry, Structure of Atom, JEE Main
The characteristics of elements X, Y and Z with atomic numbers, respectively, 33, 53 and 83 are:

Options:
A X and Y are metalloids and Z is a metal
B X is a metalloid, Y is a non-metal and Z is a metal.
C X,Y and Z are metals.
D Xand Z are non-metals and Y is a metalloid

The correct answer is:
X is a metalloid, Y is a non-metal and Z is a metal.

Solution:
X=As33→ Metalloid

Y=I53→ Non-metal

Z=Bi83→ Metal

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The characteristics of elements X, Y and Z with atomic numbers, respectively, 33, 53 and 83 are:

Options:
A X and Y are metalloids and Z is a metal
B X is a metalloid, Y is a non-metal and Z is a metal.
C X,Y and Z are metals.
D Xand Z are non-metals and Y is a metalloid

The correct answer is:
X is a metalloid, Y is a non-metal and Z is a metal.

Solution:
X=As33→ Metalloid

Y=I53→ Non-metal

Z=Bi83→ Metal

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The charge accumulated on the capacitor connected in the following circuit is ______μC.

(Given C=150 μF)

Options:

The correct answer is:
400

Solution:
Under the balanced condition, the capacitor will act as an open circuit.

The current i1 through the 1 Ω resistor is given by

i1=10 V1 Ω+2 Ω=103 A

The current i2 through the 6 Ω resistor is given by

i2=10 V6 Ω+4 Ω=1 A

From the above diagram, it can be written that

VA+i1×1-i2×6=VB⇒VA-VB=1×6-103×1=83 V

Hence, the charge on the capacitor is given by

Q=CVA-VB=150 μF×83 V=400 μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The charge flowing in a conductor changes with time as Qt=αt-βt2+γt3, where α, β and γ are constants. Minimum value of current is:

Options:
A α-3β2γ
B α-γ23β
C β-α23γ
D α-β23γ

The correct answer is:
α-β23γ

Solution:
Given here, Q=αt-βt2+γt3

Current flowing through conductor is given by

i=dQdt=α-2βt+3γt2 ...1

For current to be minimum, didt=0

So, didt=6γt-2β=0

⇒t=β3γ

Substituting in equation (1),

i=α-2βt+3γt2=α-2ββ3γ+3γβ3γ2=α-2β23γ+3β29γ=α-β23γ

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (30 Jan Shift 1)
The charge on \(4 \mu \mathrm{F}\) capacitor, in the given circuit is

Options:
A \(24 \mu \mathrm{C}\)
B \(100 \mu \mathrm{C}\)
C \(2.4 \mu \mathrm{C}\)
D \(30 \mu \mathrm{C}\)

The correct answer is:
\(24 \mu \mathrm{C}\)

Solution:
The circuit can be redrawn as,
\(C_{\mathrm{cq}}=\frac{4 \times 6}{4+6}=2.4 \mu \mathrm{F}\)
As we know, potential drop across parallel branch \(A B\) is \(V_{A B}=10 \mathrm{~V}\)
So, charge \(Q=C_{\mathrm{eq}} V=2.4 \times 10 \times 10^{-6} \mathrm{C}\)
\(Q=24 \mu \mathrm{C}\)
Since, in a series capacitor branch, there is equal storage of charge in each capacitor.
So, \(4 \mu \mathrm{F}\) capacitor store a charge of \(24 \mu \mathrm{C}\). Hence, the correct option is (a).

Topic: Physics, Capacitance, JEE Main
The charge on a capacitor of capacitance $10 \mu \mathrm{F}$ connected as shown in the figure is

Options:
A $20 \mu \mathrm{C}$
B $15 \mu \mathrm{C}$
C $10 \mu \mathrm{C}$
D Zero

The correct answer is:
$20 \mu \mathrm{C}$

Solution:
In steady state potential difference a cross capacitor $=2 \mathrm{~V}$.
So charge on capacitor $Q=10.2=20 u C$

Topic: Physics, Capacitance, JEE Main
The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure:
What is the value of current at t=4 s?

Options:
A 1.5 Î¼A
B 2 Î¼A
C Zero
D 3 Î¼A

The correct answer is:
Zero

Solution:
As we know that the dqdt at any point t gives as current at that point

and in this case the curve is constant at the point of 4 so that dqdt is zero

Since dqdtt=4s=0
Therefore current =0

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The charge on $\mathrm{As}_2 \mathrm{~S}_3$ sol is due to the adsorbed

Options:
A $\mathrm{H}^{+}$
B $\mathrm{OH}^{-}$
C $\mathrm{O}^{2-}$
D $\mathrm{S}^{2-}$

The correct answer is:
$\mathrm{S}^{2-}$

Solution:
Charge on $A s_2 S_3$ sol is due to the adsorbed sulphide ion.

Topic: Chemistry, Surface Chemistry, JEE Main
The charge on colloidal particles is due to

Options:
A Presence of electrolyte
B Very small size of particles
C Adsorption of ions from the solution
D None of these

The correct answer is:
Adsorption of ions from the solution

Solution:
Some colloidal particles develop electrical charges due to the dissociation/ionization of the surface molecules. The charge on the colloidal particles is balanced by the oppositely charged ions in the sol.

Topic: Chemistry, Surface Chemistry, JEE Main
The charge on electron was discovered by

Options:
A J.J. Thomson
B Neil Bohr
C Millikan
D Chadwick

The correct answer is:
Millikan

Solution:
The charge on electrons was discovered by Millikan.

Topic: Physics, Atomic Physics, JEE Main
The charge on the 4μF capacitor in the steady state is

Options:
A 10/3 μC
B 32/3 μC
C 4/3 μC
D 8/3 μC

The correct answer is:
32/3 μC

Solution:
In steady state, the capacitors are fully charged and act as open circuit, so the equivalent circuit in steady state would be as shown:-

∴ Steady State current is I=122+4=2 A
∴ Potential difference across AB is V=2×4=8 V
∴ Sum of potential difference across 2 μF and 4 μF capacitors is 8 V. As these are in series, so charges on them would be same. Let q be the charge on them, then:-
From KVL,

q2+q4=8⇒q=323 μC

Topic: Physics, Current Electricity, JEE Main
The charges on the colloidal CdS sol and TiO2 sol are, respectively :

Options:
A positive and positive
B positive and negative
C negative and negative
D negative and positive

The correct answer is:
negative and positive

Solution:
CdS sol→-ve sol

TiO2 sol→+ve sol

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizinig order of the polarizing power of the cationic species, $\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}$ ?

Options:
A
$\mathrm{Mg}^{2+}, \mathrm{Be}^{2+}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}$

B
$\mathrm{Be}^{2+}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}$

C
$\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}$

D
$\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}, \mathrm{K}^{+}$

The correct answer is:
$\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}$

Solution:
Higher the charge/size ratio, more is the polarizing power.
$\therefore \mathrm{K}^{+} < \mathrm{Ca}^{2+} < \mathrm{Mg}^{2+} < \mathrm{Be}^{2+}$
Hence, (C) is correct.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizinig order of the polarizing power of the cationic species, $\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}$ ?

Options:
A
$\mathrm{Mg}^{2+}, \mathrm{Be}^{2+}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}$

B
$\mathrm{Be}^{2+}, \mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}$

C
$\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}$

D
$\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}, \mathrm{K}^{+}$

The correct answer is:
$\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}$

Solution:
Higher the charge/size ratio, more is the polarizing power.
$\therefore \mathrm{K}^{+} < \mathrm{Ca}^{2+} < \mathrm{Mg}^{2+} < \mathrm{Be}^{2+}$
Hence, (C) is correct.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2007
The charring of sugar takes place when treated with concentrated $\mathrm{H}_2 \mathrm{SO}_4$. What is the type of reaction involved in it?

Options:
A Dehydration reaction
B Hydrolysis reaction
C Addition Reaction
D Disproportionation reaction

The correct answer is:
Dehydration reaction

Solution:
The charring of sugar is its dehydration process. When sugar is treated with conc. $\mathrm{H}_2 \mathrm{SO}_4$, water molecules gets escaped and sugar charcoal is formed. Due to its further reaction with acid, smell of $\mathrm{SO}_2$ is obtained.
$$
\begin{array}{rl}
\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\left[\mathrm{H}_2 \mathrm{SO}_4\right] & \longrightarrow \\
\text { Sugar } & 12 \mathrm{C}+\left[\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{SO}_4\right] \\
& \text { Sugar charcoal } \\
\mathrm{C}+2 \mathrm{H}_2 \mathrm{SO}_4 & 2 \mathrm{SO}_2+\mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}
\end{array}
$$

Topic: Chemistry, Biomolecules, JEE Main
The chemical composition of slag formed during the smelting process in the extraction of copper is

Options:
A Cu 2 O + FeS
B FeSiO 3
C CuFeS 2
D Cu 2 S + FeO

The correct answer is:
FeSiO 3

Solution:
In the smelting process of copper from copper pyrites the following reactions takes place
Cu 2 O + FeS→ Cu 2 S + FeO
2FeS +  3O 2 →2FeO +  2SO 2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The chemical formula of Hinsberg's reagent is

Options:
A $\mathrm{HNO}_{2}$
B $\mathrm{NaOH}+\mathrm{CaO}$
C $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}$
D $\mathrm{CH}_{3} \mathrm{CONH}_{2}$

The correct answer is:
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}$

Solution:
The chemical formula of Hinsberg's reagent is $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{CI}$

Topic: Chemistry, Amines, JEE Main
The chemical nature of hydrogen peroxide is:

Options:
A Oxidizing and reducing agent in both acidic and basic medium.
B Oxidizing agent in acidic medium, but not in basic medium
C Oxidizing and reducing agent in acidic medium, but not in basic medium
D Reducing agent in basic medium, but not in acidic medium.

The correct answer is:
Oxidizing and reducing agent in both acidic and basic medium.

Solution:
H2O2 acts as an oxidizing agent, as well as a reducing agent in both acidic as well as basic mediums.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The chemical that is added to reduce the melting point of the reaction mixture during the extraction of aluminium is:

Options:
A Cryolite
B Bauxite
C Calamine
D Kaolite

The correct answer is:
Cryolite

Solution:
To reduce the melting point of reaction mixture, cryolite is added.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The chemicals and the reaction conditions required for the preparation of ethane are

Options:
A $\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}, \mathrm{Zn}-\mathrm{Cu}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
B $\mathrm{CH}_3 \mathrm{Cl}, \mathrm{Na}, \mathrm{H}_2 \mathrm{O}$
C $\mathrm{KOOC}-\mathrm{CH}=\mathrm{CH}-\mathrm{COOK}$, electrolysis
D $\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{Na}, \mathrm{NaOH}, \mathrm{CaO}, \Delta$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}, \mathrm{Zn}-\mathrm{Cu}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

Solution:
Alkyl halides can be reduced to hydrocarbons by means of $\mathrm{Zn}-\mathrm{Cu}$ couple in presence of alcohol.

Topic: Chemistry, Hydrocarbons, JEE Main
The chemicals used for preparing acetophenone are
(A) $\mathrm{C}_6 \mathrm{H}_6$
(B) $\mathrm{CH}_3 \mathrm{COCH}_3$
(C) $\mathrm{CH}_3 \mathrm{COCl}$
(D) Anhydrous $\mathrm{AlCl}_3$

Options:
A $A, B, C$
B $B, C, D$
C $A, B, D$
D $A, C, D$

The correct answer is:
$A, B, D$

Solution:
Acetophenone can be prepared as
$\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{COCl} \underset{\text { Friedel-Crafts reaction }}{\stackrel{\text { anhyd }\cdot \mathrm{AlCl}_3}{\longrightarrow}}$$\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3+\mathrm{HCl}$

Topic: Chemistry, Hydrocarbons, JEE Main