datavorous/entrance-exam-dataset huggingface dataset:
Integrating factor of the differential equation $\frac{d y}{d x}+y \tan x-\sec x=0$ is

Options:
A $e^{\sin x}$
B $\frac{1}{\sin x}$
C $\frac{1}{\cos x}$
D $\mathrm{e}^{\cos x}$

The correct answer is:
$\frac{1}{\cos x}$

Solution:
$\frac{d y}{d x}+y \tan x=\sec x$, where $p=\tan x, Q=\sec x$
This is the linear differential equation in $y$, hence
I.F. $=e^{\int \operatorname{ddx}}=e^{\int \tan x d x} ;$ I.F. $=e^{\log \sec x}=\sec x$.

Topic: Mathematics, Differential Equations, JEE Main
Integrating factor of \( x \frac{d y}{d x}-y=x^{4}-3 x \) is

Options:
A \( x \)
B \( \log x \)
C \( \frac{1}{x} \)
D \( -x \)

The correct answer is:
\( \frac{1}{x} \)

Solution:
Given differential equation is,
\[
\begin{array}{l}
x \frac{d y}{d x}-y=x^{4}-3 x \\
\Rightarrow \frac{d y}{d x}-\frac{1}{x} y=x^{3}-3
\end{array}
\]
General differential equation of this type is given by
\( \frac{d y}{d x}+P(x) y=Q(x) \)
Here \( P=-\frac{1}{x} \) and \( Q=x^{3}-3 \) So, I.F. factor is given by,
\[
\begin{array}{l}
\text { I.F. }=e^{\int P d x}=e^{\int-\frac{1}{x} d x} \\
=e^{-\log x}=e^{\log \left(\frac{1}{x}\right)}=\frac{1}{x}
\end{array}
\]

Topic: Mathematics, Differential Equations, JEE Main
Intensities of the two waves of light are $I$ and $4 I$. The maximum intensity of the resultant wave after superposition is

Options:
A $51$
B $91$
C 16 /
D $251$

The correct answer is:
$91$

Solution:
$I_{\max }=I_1+I_2+2 \sqrt{I_1 I_2}$
So, $I_{\max }=I+4 I+2 \sqrt{I .4 I}=9 I$

Topic: Physics, Wave Optics, JEE Main
Intensity of an electric field $E$ due to a dipole, depends on distance $r$ as

Options:
A $E \propto \frac{1}{r^4}$
B $E \propto \frac{1}{r^3}$
C $E \propto \frac{1}{r^2}$
D $E \propto \frac{1}{r}$

The correct answer is:
$E \propto \frac{1}{r^2}$

Solution:
Electric field intensity at a far point from the centre of the dipole varies with distance as $\frac{1}{r^3}$

Topic: Physics, Electrostatics, JEE Main
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source λ=632.8 nm. The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :

Options:
A 1.27μm
B 2 .87nm
C 2 nm
D 2.05 μm

The correct answer is:
1.27μm

Solution:
ΔP=dsinθ

=dθ

=dyD=10-3×1.270 mm1 m=1.27μm

1.27μm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Interference fringes are produced on a screen by using two light sources of intensities I and 9I. The phase difference between the beams is Ï2 at the point P and Ï at the point Q on the screen. The difference between the resultant intensities at point P and Q is

Options:
A 2I
B 4I
C 6I
D 8I

The correct answer is:
6I

Solution:
At point P, resultant intensity of interfering wave is
IPres=I1+I2+2I1I2cosâ¡ Î¸
For Î¸=Ï2,IPres=I+9I=10I
At point Q, resultant intensity of interfering wave is
IQres=10I-2.9I2=10I-6I=4I for Î¸=Ï
â´ âI=IPres-Igres=10I-10I+6I=6I

Topic: Physics, Wave Optics, JEE Main
Interference pattern is observed at ' $\mathrm{P}$ ' due to superimposition of two rays coming out from a source ' $S$ ' as shown in the figure. The value of ' $\mathrm{l}$ ' for which maxima is obtained at ' $\mathrm{P}$ ' is:
( $\mathrm{R}$ is perfect reflecting surface)

Options:
A
$1=\frac{2 n \lambda}{\sqrt{3}-1}$

B
$\left.1=\frac{(2 \mathrm{n}-1) \lambda}{2(\sqrt{3}-1}\right)$

C
$\left.1=\frac{(2 n-1 \lambda) \sqrt{3}}{4(2-\sqrt{3}}\right)$

D
$1=\frac{(2 \mathrm{n}-1) \lambda}{\sqrt{3}-1}$

The correct answer is:
$\left.1=\frac{(2 n-1 \lambda) \sqrt{3}}{4(2-\sqrt{3}}\right)$

Solution:
No Solution Available

Topic: Physics, Wave Optics, JEE Main, JEE Main 2014 (12 Apr Online)
Internal energy and pressure of a gas per unit volume are related as

Options:
A $P=\frac{2}{3} E$
B $P=\frac{3}{2} E$
C $P=\frac{1}{2} E$
D $P=2 E$

The correct answer is:
$P=\frac{2}{3} E$

Solution:
$K E=\frac{3}{2} R T=\frac{3}{2} P V$
$\therefore P=\frac{2}{3} \frac{E}{V}$ for unit volume $(V=1) \Rightarrow P=\frac{2}{3} E$

Topic: Chemistry, States of Matter, JEE Main
Internodal elongation is stimulated by:

Options:
A Auxin
B Ethylene
C Cytokinin
D Gibberellin

The correct answer is:
Gibberellin

Solution:
Gibberellic acid (GA3) was one of the first gibberellins to be discovered and remains the most intensively studied form. All GAs are acidic. They produce a wide range of physiological responses in the plants. Their ability to cause an increase in length of axis is used to increase the length of grapes stalks. Gibberellins, cause fruits like apple to elongate and improve its shape. They also delay senescence. Thus, the fruits can be left on the tree longer so as to extend the market period. GA3 is used to speed up the malting process in brewing industry. Sugarcane stores carbohydrate as sugar in their stems. Spraying sugarcane crop with gibberellins increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre. Spraying juvenile conifers with GAs hastens the maturity period, thus leading to early seed production. Gibberellins also promotes bolting (internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.

Topic: Biology, Plant - Growth and Development, NEET
Intersection of two perpendicular tangents to the hyperbola \(\frac{x^2}{4}-\frac{y^2}{2}=1\) lies on the circle \(x^2+y^2=\ldots \ldots \ldots\)

Options:
A 2
B 12
C \(\sqrt{2}\)
D \(2 \sqrt{3}\)

The correct answer is:
2

Solution:
Locus of point of intersection of perpendicular tangents to hyperbola \(\frac{x^2}{4}-\frac{y^2}{2}=1\) is the director circle \(x^2+y^2=4-2=2\)
Hence, option (a) is correct.

Topic: Mathematics, Hyperbola, JEE Main
\( \int\left(\cos ^{-3 / 7} x\right)\left(\sin ^{-11 / 7} x\right) d x \) is equal to

Options:
A \( \log \left|\sin ^{4 / 7} x\right|+c \)
B \( \frac{4}{7} \tan ^{4 / 7} x+c \)
C \( -\frac{7}{4} \tan ^{-4 / 7} x+c \)
D \( \log \left|\cos ^{3 / 7} x\right|+c \)

The correct answer is:
\( -\frac{7}{4} \tan ^{-4 / 7} x+c \)

Solution:
∫cos-3/7xsin-11/7xdx=∫sin-11/7xcos-11/7x.sec2x dx
=∫tan-11/7xsec2x dx
Put tanx=t ⇒ sec2x dx=dt
∴ I=∫ t-11/7dt=-74tan-4/7x+c

Topic: Mathematics, Indefinite Integration, JEE Main
\(\int_\pi^{16 \pi}|\sin x| d x=\)

Options:
A 0
B 32
C 30
D 28

The correct answer is:
30

Solution:
Hints: \(15 \int_0^\pi \sin x d x=15(-\cos x)_0^\pi=30\)

Topic: Mathematics, Definite Integration, JEE Main
Intramolecular forces between n-hexane and n-heptane are nearly same as between hexane and heptane individually. When these two are mixed, which of the following is not true about the solution formed?

Options:
A It obeys Raoult's law, i.e. pA=xApAo and pB=xBpBo
B ΔHmixing is zero
C ΔVmixing is zero
D Its forms minimum boiling azeotrope

The correct answer is:
Its forms minimum boiling azeotrope

Solution:
n-hexane and n-heptane form an ideal solution so here azeotropic mixture is not possible.

Topic: Chemistry, Solutions, JEE Main
Introduction of inert gas (at the same temperature) will affect the equilibrium if

Options:
A Volume is constant and Δng≠0
B Pressure is constant and Δng≠0
C Volume is constant and Δng=0
D Pressure is constant and Δng=0

The correct answer is:
Pressure is constant and Δng≠0

Solution:
Introduction of inert gas (at the same temperature) will affect the equilibrium if Pressure is constant and Δng≠0
For constant volume, reaction quotient (Q) will remain constant.
For constant pressure, reaction quotient (Q) will remain constant when Δng=0 .

Topic: Chemistry, Chemical Equilibrium, NEET
\( \int_{\sin x}^{1} t^{2} f(t) d t=1-\sin x \forall x \in(0, \pi / 2) \), then \( f\left(\frac{1}{\sqrt{3}}\right) \) is :

Options:

The correct answer is:
3

Solution:
∫sin x1t2f(t)dt=1−sin x

Applying Leibnitz Rule,

−sin2x f(sinx)cosx=−cosx

⇒f(sinx)=1sin2x

⇒f(t)=1t2

f(13)=1(13)2=3

Topic: Mathematics, Definite Integration, JEE Main
Inverse of the function $y=2 x-3$ is

Options:
A $\frac{x+3}{2}$
B $\frac{x-3}{2}$
C $\frac{1}{2 x-3}$
D None of these

The correct answer is:
$\frac{x+3}{2}$

Solution:
$\begin{aligned} & y=2 x-3 \Rightarrow x=\frac{y+3}{2} \\ & \Rightarrow f^{-1}(y)=\frac{y+3}{2} \Rightarrow f^{-1}(x)=\frac{x+3}{2} .\end{aligned}$

Topic: Mathematics, Functions, JEE Main
Inverse of the matrix $\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]$ is

Options:
A $\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]$
B $\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ \sin 2 \theta & -\cos 2 \theta\end{array}\right]$
C $\left[\begin{array}{ll}\cos 2 \theta & \sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]$
D $\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$

Solution:
Let $A=\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right],|A|=1$
$\operatorname{adj}(A)=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
$A^{-1}=\frac{\operatorname{adj}(A)}{|A|}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$.

Topic: Mathematics, Matrices, JEE Main
Invertase brings about the conversion of

Options:
A Starch to glucose
B Sucrose to glucose and fructose
C Maltose to glucose
D Glucose to $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{CO}_2$

The correct answer is:
Sucrose to glucose and fructose

Solution:
Invertase catalyzes the hydrolysis of the disaccharide sucrose (table sugar) into glucose and fructose and is a major enzyme present in plants and microorganism

Topic: Chemistry, Biomolecules, JEE Main
Iodine is a solid and sublimes at ordinary temperature. This is because of:

Options:
A weak I-I bonds
B strong I-I bonds
C lone pair-bond pair repulsions
D weak van der Waals forces between I2 molecules

The correct answer is:
weak van der Waals forces between I2 molecules

Solution:
I2 is solid and sublimes at ordinary temperature because of weak vanderwaal's force between I2 molecules.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Iodine is obtained from

Options:
A Laminaria
B Chlorella
C Polysiphonia
D Porphyra

The correct answer is:
Laminaria

Solution:
Brown algae such as Laminaria, Macrocystis, Fucus etc are the main source of iodine.

Topic: Biology, Plant Kingdom, JIPMER, JIPMER 2008
Iodine reacts with concentrated HNO3 to yield Y along with other products. The oxidation state of iodine in Y , is:

Options:
A 1
B 5
C 3
D 7

The correct answer is:
5

Solution:
I2+10 HNO3→2 HIO3+10 NO2+4 H2O
Iodine gets oxidized to IO3- when it reacts with an oxidizing agent HNO3. The oxidation number of I in product will be
Oxidation: I 2 −I O 3 −
-1=x+3×-2
-1=x-6
x=-1+6=+5
Reduction: concentrated HNO3 gives NO2.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (12 Jan Shift 1)
Iodoform test is not given by

Options:
A 2 -pentanone
B ethanol
C ethanal
D 3-pentanone

The correct answer is:
3-pentanone

Solution:
Iodoform test is given by compounds which have
group or
group, Hence, $\quad$ 2-pentanone, $\mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
give this test. But 3-pentanone $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\right)$ does not give iodoform test.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Ion having highest hydration enthalpy among the given alkaline earth metal ions is:

Options:
A Be2+
B Sr2+
C Ba2+
D Ca2+

The correct answer is:
Be2+

Solution:
Smaller the size of the cation, more the hydration of ion, hence the more values of the hydration enthalpies. Now, as the down the group the size increases so the hydration enthalpies decreases, hence as the increasing order of the size of the ions is as follows:
Be2+<Mg2+<Ca2+<Sr2+<Ba2+
and reverse of this order is true for the hydration enthalpies.
Be2+>Mg2+>Ca2+>Sr2+>Ba2+

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Ionic and covalent bonds are present in

Options:
A $\mathrm{CCl}_4$
B $\mathrm{CaCl}_2$
C $\mathrm{NH}_4 \mathrm{Cl}$
D $\mathrm{H}_2 \mathrm{O}$

The correct answer is:
$\mathrm{NH}_4 \mathrm{Cl}$

Solution:
$\mathrm{NH}_4 \mathrm{Cl}$ has covalent as well as ionic bond.
$\left[\begin{array}{c}\mathrm{H} \\ \mathrm{H}-\mathrm{N} \rightarrow \mathrm{H}^{+}\end{array}\right] \mathrm{Cl}^{-}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Ionic bonds are usually formed by combination of elements with

Options:
A High ionisation potential and low electron affinity
B Low ionisation potential and high electron affinity
C High ionisation potential and high electron affinity
D Low ionisation potential and low electron affinity

The correct answer is:
Low ionisation potential and high electron affinity

Solution:
Correct option is 2. Low ionization potential and high electron affinity

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Ionic compounds do not have

Options:
A Hard and brittle nature
B High melting and boiling point
C Directional properties
D Soluble in polar solvents

The correct answer is:
Directional properties

Solution:
Direction properties

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Ionic hydrides are usually

Options:
A Good electrically conductors when solid
B Easily reduced
C Good reducing agents
D Liquid at room temperature

The correct answer is:
Good reducing agents

Solution:
lonic hydrides are good reducing agents.

Topic: Chemistry, Hydrogen, JEE Main
Ionic product of water increases, if

Options:
A Pressure is reduced
B $H^{+}$is added
C $\mathrm{OH}^{-}$is added
D Temperature increases

The correct answer is:
Temperature increases

Solution:
$K_w$ increases with increase in temperature

Topic: Chemistry, Ionic Equilibrium, JEE Main
Ionic radii are

Options:
A Directly proportional to effective nuclear charge
B Directly proportional to square of effective nuclear charge
C Inversely proportional to effective nuclear charge
D Inversely proportional to square of effective nuclear charge.

The correct answer is:
Inversely proportional to effective nuclear charge

Solution:
lonic radius in the $\mathrm{n}^{\text {th }}$ orbit is given by $r_n=\frac{n^2 a}{z}$ or $r_n \propto \frac{1}{Z}$ where $n$ is principal equation no., ao. bohr's radius of hydrogen atom and $Z$ is the effective nuclear energy.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Ionic reactions with organic compounds proceed through:

(A) Homolytic bond cleavage (B) Heterolytic bond cleavage (C) Free radical formation

(D) Primary free radical (E) Secondary free radical

Choose the correct answer from the options given below:

Options:
A (A) only
B (C) only
C (B) only
D (D) and (E) only

The correct answer is:
(B) only

Solution:
Heterolytic fission, also known as heterolysis, is a type of bond fission in which a covalent bond between two chemical species is broken in an unequal manner, resulting in the bond pair of electrons being retained by one of the chemical species (while the other species does not retain any of the electrons from the bond pair). When a neutrally charged molecule undergoes heterolytic fission, one of the products will have a positive charge whereas the other product will have a negative charge.
Polar reactions occur when two bonded atoms come apart, one taking more of the shared electrons than the other. They involve heterolytic cleavage. The result is two charged species—one cation and one anion.
Hence, organic reactions which proceed through heterolytic bond cleavage are called polar reactions.

In homolytic reactions, free radicals are formed and these reaction takes place in gaseous phase.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Ionic solids, with Schottky defects, contain in their structure

Options:
A Equal number of cation and anion vacancies
B Anion vacancies and interstitial anions
C Cation vacancies only
D Cation vacancies and interstitial cations

The correct answer is:
Equal number of cation and anion vacancies

Solution:
Schottky defect is due to missing of equal number of cations and anions.

Topic: Chemistry, Solid State, JEE Main
Ionisation energy of a hydrogen like species is 54.4 eV. Calculate radius of second bohr orbit of this species.

Options:
A 0.529   A o
B 1.058   A o
C 2.116   A o
D 2.116   A o

The correct answer is:
1.058   A o

Solution:
Ionisation energy 13.6Ãz2=54.4
∴ Z = 2
r = 0.529  n 2 z A o  = 0.529 ×  2 2 2  = 1.058  A o

Topic: Chemistry, Structure of Atom, JEE Main
Ionisation energy of He + is 19.6× 10 −18  J per atom. The energy of the first stationary state (n = 1) of Li 2+ per atom is

Options:
A 4.41Ã 10 â16 âJ
B 4.41× 10 −17  J
C 2.2× 10 −15  J
D 8.82Ã 10 â17 âJ

The correct answer is:
4.41× 10 −17  J

Solution:
IEâ=â Z 2 n 2 âÃâ13.6âeV â¦(i)
IE 1 IE 2 = Z 1 2 n 1 2 Ã n 2 2 Z 2 2 â¦(ii)
Given IE 1 â=â19.6âÃâ 10 â18 ,â Z 1 â=â2,
n 1 â=â1,â Z 2 â=â3âandâ n 2 =1 .
Substituting these values in equation (ii)
19.6âÃâ 10 â18 IE 2 â=â 4 1 âÃâ 1 9
Or IE 2  = 19.6 ×  10 −18  ×  9 4
= 4.41 ×  10 −17  J/atom

Topic: Chemistry, Structure of Atom, JEE Main
Ionisation energy of $\mathrm{He}^{+}$is $19.6 \times 10^{-18} \mathrm{~J} \mathrm{atom}^{-1}$. The energy of the first stationary state $(\mathrm{n}=1)$ of $\mathrm{Li}^{2+}$ is

Options:
A
$4.41 \times 10^{-16} \mathrm{~J} \mathrm{atom}^{-1}$

B
$-4.41 \times 10^{-17} \mathrm{~J}_{\text {atom }}{ }^{-1}$

C
$-2.2 \times 10^{-15} \mathrm{~J} \mathrm{atom}^{-1}$

D
$8.82 \times 10^{-17} \mathrm{~J}_{\text {atom }^{-1}}$

The correct answer is:
$-4.41 \times 10^{-17} \mathrm{~J}_{\text {atom }}{ }^{-1}$

Solution:
$$
\mathrm{IE}_{\mathrm{He}^{+}}=13.6 \mathrm{Z}_{\mathrm{He}^{+}}^2\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right]=13.6 \mathrm{Z}_{\mathrm{He}^{+}}^2 \text { where }\left(\mathrm{Z}_{\mathrm{He}^{+}}=2\right)
$$
Hence $13.6 \times Z_{\mathrm{He}^{+}}^2=19.6 \times 10^{-18} \mathrm{~J} \mathrm{atom}^{-1}$.
$$
\left(\mathrm{E}_1\right)_{\mathrm{L}^{+2}}=-13.6 \mathrm{Z}_{\mathrm{Uj}^{+2}}^2 \times \frac{1}{1^2}=-13.6 \mathrm{Z}_{\mathrm{He}^{+}}^2 \times\left[\frac{\mathrm{Z}_{\mathrm{L}^{+2}}^2}{\mathrm{Z}_{\mathrm{He}^{+}}^2}\right]=-19.6 \times 10^{-18} \times \frac{9}{4}=-4.41 \times 10^{-17} \mathrm{~J} / \text { atom }
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2010
Ionisation energy of Li (Lithium) atom in ground state is $5.4 \mathrm{eV}$. Binding energy of an electron in $\mathrm{Li}^{+}$ion in ground state is $75.6 \mathrm{eV}$. Energy required to remove all three electrons of Lithium (Li) atom is

Options:
A
$81.0 \mathrm{eV}$

B
$135.4 \mathrm{eV}$

C
$203.4 \mathrm{eV}$

D
$156.6 \mathrm{eV}$

The correct answer is:
$156.6 \mathrm{eV}$

Solution:
No Solution Available

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2012 (19 May Online)
Ionization energy is highest for

Options:
A Noble gases
B Platinum metals
C Transition elements
D Inner–transition elements

The correct answer is:
Noble gases

Solution:
As the $e^{-}$is to be removed from stable configuration

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Iron crystallizes in body centered cubic system with edge length 2.86 Å . The density of iron is nearly 'X' g/ml. What is the value of 'X' here?
Report your answer by rounding it upto nearest whole number.

Options:

The correct answer is:
8

Solution:
As d=Zma3NA=2×562.86×10-8×6.02×1023
≈7.92 g cm3≃8 g cm3
≃8 g/ml

Topic: Chemistry, Solid State, JEE Main
Iron displaces copper from its salt solution, because

Options:
A Atomic number of iron is less than that of copper
B The standard reduction potential of iron is less than that of copper
C The standard reduction potential of iron is more than that of copper
D The iron salt is more soluble in water than the copper salt

The correct answer is:
The standard reduction potential of iron is less than that of copper

Solution:
The answer to the question is (2) The standard reduction potential of iron is less than that of copper.
Here's why this is the correct answer:
- The standard reduction potential is a measure of a metal's tendency to gain electrons and be reduced from a cation to a metal atom in an electrochemical reaction. Metals with more negative reduction potentials are better reducing agents (more reactive) than metals with less negative or positive reduction potentials.
- In the reaction between iron and copper sulfate, iron ( $\mathrm{Fe}$ ) is displacing copper $(\mathrm{Cu})$ from its salt solution $\left(\mathrm{CuSO}_4\right)$. This means iron is more reactive than copper.
- Iron has a more negative standard reduction potential than copper, which means it is easier for iron to be reduced (gain electrons) than copper. This greater ease of reduction allows iron to take electrons away from the copper ions in copper sulfate, reducing the copper ions to copper metal and oxidizing itself to ferrous iron $\left(\mathrm{Fe}^{2+}\right)$.

Topic: Chemistry, Electrochemistry, JEE Main
Iron exhibits $+2$ and $+3$ oxidation states. Which of the following statements about iron is incorrect?

Options:
A
Ferrous oxide is more basic in nature than the ferric oxide.

B
Ferrous compounds are relatively more ionic than the corresponding ferric compounds

C
Ferrous compounds are less volatile than the corresponding ferric compounds.

D
Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds.

The correct answer is:
Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds.

Solution:
$\mathrm{FeO} \rightarrow$ More basic, more ionic, less volatile

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2012 (Offline)
Iron exhibits bcc structure at room temperature. Above $500^{\circ} \mathrm{C}$, it transforms to fcc structure. Find the ratio of the density of iron at room temperature to that at $500^{\circ} \mathrm{C}$. (Assume the atomic radii and the molar mass of iron remain constant even with variation in temperature)

Options:
A $3 \sqrt{3}: 4 \sqrt{2}$
B $\sqrt{3}: \sqrt{2}$
C $\sqrt{2}: \sqrt{3}$
D $10: 92$

The correct answer is:
$3 \sqrt{3}: 4 \sqrt{2}$

Solution:
We know,
Density, $d=Z M / a^3 N_0$
For bcc unit cell, $Z=2, a=\frac{4}{\sqrt{3}} r$
$$
d_{\mathrm{bcc}}=\frac{2 M}{\left(\frac{4}{\sqrt{3}} r\right)^3 N_0}
$$
For fcc unit cell, $\quad Z=4, a=\frac{4}{\sqrt{2}} r$
$$
d_{\mathrm{fcc}}=\frac{4 M}{\left(\frac{4}{\sqrt{2}} r\right)^3 N_0}
$$
Ratio of density of iron bcc : fcc can be calculated by dividing Eq. (i) by Eq. (ii)
$$
\frac{d_{\mathrm{bcc}}}{d_{\mathrm{fcc}}}=\frac{\frac{2 M}{\left(\frac{4}{\sqrt{3}} r\right)^3 N_0}}{\frac{4 M}{\left(\frac{4}{\sqrt{2}} r\right)^3 N_0}}=\frac{1}{2} \times\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^3
$$
$$
=\frac{1}{2} \times \frac{3}{2} \frac{\sqrt{3}}{\sqrt{2}}=\frac{3 \sqrt{3}}{4 \sqrt{2}}
$$
The ratio is $3 \sqrt{3}: 4 \sqrt{2}$.

Topic: Chemistry, Solid State, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
Iron exhibits BCC structure at room temperature. Above 900°C, it transforms to FCC structure. The ratio of the density of iron at room temperature to that at 900°C is (assuming molar mass and atomic radii of iron remains constant with temperature)

Options:
A 3342
B 4332
C 32
D 12

The correct answer is:
3342

Solution:
BCC=4r=3a⇒a=4r3
FCC=4r=2a⇒a=4r2
dBCCdFCC=ZBCC×MNA a3ZFCC×MNA a3=2×MNA4r334×MNA×4r23=3432 .

Topic: Chemistry, Solid State, NEET, NEET 2018
Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is

Options:
A 3342
B 4332
C 32
D 12

The correct answer is:
3342

Solution:
For BCC lattice: Z=2,a=4r3
For FCC lattice: Z=4,a=22r
Hence d25°Cd900°C=ZMNAa3BCCZMNAa3FCC=2422r4r33=3342

Topic: Chemistry, Solid State, NEET