datavorous/entrance-exam-dataset huggingface dataset:
The compound (s) that exhibit (s) geometrical isomerism is (are)

Options:
A
$\left[\mathrm{Pt}(\mathrm{en}) \mathrm{Cl}_2\right]$

B
$\left[\mathrm{Pt}(\mathrm{en})_2 \mathrm{Cl}_2\right]$

C
$\left[\mathrm{Pt}(\text { en })_2 \mathrm{Cl}_2\right] \mathrm{Cl}_2$

D
$\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$

The correct answers are:
$\left[\mathrm{Pt}(\text { en })_2 \mathrm{Cl}_2\right] \mathrm{Cl}_2$
,
$\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$

Solution:
Only $C$ and $D$ can show geometrical isomerism as

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2009 (Paper 1)
The compound that undergoes decarboxylation most readily under mild condition is

Options:
A
B
C
D

The correct answer is:

Solution:
$\beta-$ Ketoacids undergo decarboxylation easily. It is due to the stability of transition state formed after decarboxylation of $\beta$-keto acids. This is not possible in case of other acids.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Advanced, JEE Advanced 2012 (Paper 2)
The compound that undergoes decarboxylation most readily under mild condition is

Options:
A
B
C
D

The correct answer is:

Solution:
$\beta-$ Ketoacids undergo decarboxylation easily. It is due to the stability of transition state formed after decarboxylation of $\beta$-keto acids. This is not possible in case of other acids.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The compound(s) formed upon combustion of sodium metal in excess air is (are)

Options:
A
$\mathrm{Na}_2 \mathrm{O}_2$

B
$\mathrm{Na}_2 \mathrm{O}$

C
$\mathrm{NaO}_2$

D
$\mathrm{NaOH}$

The correct answers are:
$\mathrm{Na}_2 \mathrm{O}_2$
,
$\mathrm{Na}_2 \mathrm{O}$

Solution:
Sodium on heating in air forms both normal oxide and peroxides.

Topic: Chemistry, s Block Elements, JEE Advanced, JEE Advanced 2009 (Paper 1)
The compounds $P, Q$ and $S$
where separately subjected to nitration using $\mathrm{HNO}_3 / \mathrm{H}_2 \mathrm{SO}_4$ mixture. The major product formed in each case respectively, is

Options:
A

B

C

D

The correct answer is:

Solution:
General organic chemistry (GOC)
Conceptual
II

Topic: Chemistry, Hydrocarbons, JEE Main
The compounds $P, Q$ and $S$
where separately subjected to nitration using $\mathrm{HNO}_3 / \mathrm{H}_2 \mathrm{SO}_4$ mixture. The major product formed in each case respectively, is

Options:
A

B

C

D

The correct answer is:

Solution:
General organic chemistry (GOC)
Conceptual
II

Topic: Chemistry, Hydrocarbons, JEE Advanced, JEE Advanced 2010 (Paper 2)
The concentration of $R$ in the reaction $R \longrightarrow P$ was measured as a function of time and the following data is obtained :
$\begin{array}{lllll}{[R] \text { (molar) }} & 1.0 & 0.75 & 0.40 & 0.10 \\ t \text { (min) } & 0.0 & 0.05 & 0.12 & 0.18\end{array}$
The order of the reaction is

Options:

The correct answer is:
0

Solution:
$k_{\mathrm{I}}=\frac{\Delta[R]}{\Delta t}=\frac{0.25}{0.05}=5$
$$
k_{\text {II }}=\frac{\Delta[R]}{\Delta t}=\frac{0.60}{0.12}=5
$$
So, reaction must be of zero order.
Chemical kinetics
Straight conceptual
II

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2010 (Paper 1)
The coordination number of $\mathrm{Al}$ in the crystalline state of $\mathrm{AlCl}_3$ is

Options:

The correct answer is:
6

Solution:
Coordination number of $\mathrm{Al}=6$.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Advanced, JEE Advanced 2009 (Paper 2)
The correct acidity order of the following is

Options:
A
III $>$ IV $>$ II $>$ I

B
IV $>$ III $>$ I $>$ II

C
III $>$ II $>$ I $>$ IV

D
II $>$ III $>$ IV $>$ I

The correct answer is:
III $>$ IV $>$ II $>$ I

Solution:
A carboxylic acid is stronger acid than phenol, hence both III and IV are stronger acids than both I and II. Also IV has a methyl group that gives electron donating inductive effect and decreases the acid strength. Therefore, III is stronger acid that IV. Between I and II, the dominant electron withdrawing inductive effect of chlorine increases acid strength of phenol slightly, hence II is stronger acid than I. Thus, the overall order is :
(a) III $>$ IV $>$ II $>$ I

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Advanced, JEE Advanced 2009 (Paper 1)
The correct functional group $X$ and the reagent/reaction conditions $Y$ in the following schemes are $\mathrm{X}-\left(\mathrm{CH}_2\right)_4-X$

Options:
A
$X=\mathrm{COOCH}_3, Y=\mathrm{H}_2 / \mathrm{Ni} /$ heat

B
$X=\mathrm{CONH}_2, Y=\mathrm{H}_2 / \mathrm{Ni} /$ heat

C
$X=\mathrm{CONH}_2, \quad Y=\mathrm{Br}_2 / \mathrm{NaOH}$

D
$X=\mathrm{CN}, Y=\mathrm{H}_2 / \mathrm{Ni} /$ heat

The correct answers are:
$X=\mathrm{CONH}_2, \quad Y=\mathrm{Br}_2 / \mathrm{NaOH}$
,
$X=\mathrm{CN}, Y=\mathrm{H}_2 / \mathrm{Ni} /$ heat

Solution:
(a)
(b)
(c)
(d)
$$
\text { In author's opinion (a) and (b) should also be the answer. }
$$
Polyester, a condensation polymer

Topic: Chemistry, Polymers, JEE Advanced, JEE Advanced 2011 (Paper 2)
The correct stability order for the following species is

Options:
A
(II) $>$ (IV) $>$ (I) $>$ (III)

B
(I) $>$ (II) $>$ (III) $>$ (IV)

C
(II) $>$ (I) $>$ (IV) $>$ (III)

D
(I) $>$ (III) $>$ (II) $>$ (IV)

The correct answer is:
(I) $>$ (III) $>$ (II) $>$ (IV)

Solution:
I $>$ III $>$ II $>$ IV
$+M$-effect, hyperconjugation $(6 \mathrm{Hs})$ and $+$ I-effect (2 Me groups)
$+M$-effect, hyperconjugation ( $3 \mathrm{Hs}$ ) and $+I$-effect (1 Me group)
$\overbrace{2^{\circ} \text { carbocation) }}^{+}$Hyperconjugation (5Hs) and +I-effect
Hyperconjugation (2Hs) and $+I$ effect

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2008 (Paper 2)
The correct stability order of the following resonance structure is

Options:
A
I $>$ II $>$ IV $>$ III

B
I $>$ III $>$ II $>$ IV

C
II $>$ I $>$ III $>$ IV

D
III $>$ I $>$ IV $>$ II

The correct answer is:
I $>$ III $>$ II $>$ IV

Solution:
(I) has maximum covalent bonds and negative charge on electronegative nitrogen, most stable. III has more covalent bond than both II and IV, III is second most stable. Between II and IV, II is more stable since it has negative charge on nitrogen while IV has negative charge on carbon.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2009 (Paper 2)
The correct statement about the following disaccharide is

Options:
A
Ring (a) is pyranose with $\alpha$-glycosidic link

B
Ring (a) is furanose with $\alpha$-glycosidic link

C
Ring (b) is furanose with $\alpha$-glycosidic link

D
Ring (b) is pyranose with $\beta$-glycosidic link

The correct answer is:
Ring (a) is pyranose with $\alpha$-glycosidic link

Solution:
Ring $A$ is pyranose ( 6 membered ring containing one O-atom) with $\alpha$-glycosidic linkage and ring $B$ is furanose with $\beta$-glycosidic linkage.
Biomolecules Conceptual structural interpretation III

Topic: Chemistry, Biomolecules, JEE Advanced, JEE Advanced 2010 (Paper 1)
The correct statement(s) about the compound given below is are

Options:
A
the compound is optically active

B
the compound possesses centre of symmetry

C
the compound possesses plane of symmetry

D
the compound possesses axis of symmetry

The correct answers are:
the compound is optically active
,
the compound possesses centre of symmetry

Solution:
The compound does not possess centre of symmetry and plane of symmetry but possesses the proper axis of symmetry $\left(C_n\right)$. A $C_n$ axis of symmetry is an axis about which the molecule can be rotated by $360^{\circ} / n$ and thereby produce a molecule indistinguishable from the original molecule.
From this face $\mathrm{H}$ (on right) and $\mathrm{Cl}$ (on left) are facing towards us and again on rotating at $180^{\circ} \mathrm{C}-2$ comes at $\mathrm{C}-1$ place and again $\mathrm{H}$ (on right) and $\mathrm{Cl}$ (on left) are facing. We get an indistinguishable form on $180^{\circ}$ rotation, so it has a proper axis of symmetry $\left(C_2\right)$.
If a molecule is not superimposable on its mirror image, that molecule can exhibit optical activity, if these have proper axis but no improper axis or if a $C_n$ axis be the only symmetry element present in a molecule, that molecule is optically active.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2008 (Paper 1)
The correct statement(s) about the compound, $\mathrm{H}_3 \mathrm{C}(\mathrm{HO}) \mathrm{HC}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}$ (OH) $\mathrm{CH}_3(\mathrm{X})$ is (are)

Options:
A
the total number of stereoisomers possible for $X$ is 6

B
the total number of diastereomers possible for $X$ is 3

C
if the stereochemistry about the double bond in $X$ is trans, the number of enantiomers possible for $X$ is 4

D
if the stereochemistry about the double bond in $X$ is cis, the number of enantiomers possible for $X$ is 2

The correct answers are:
the total number of stereoisomers possible for $X$ is 6
,
if the stereochemistry about the double bond in $X$ is cis, the number of enantiomers possible for $X$ is 2

Solution:
$$
\text { The various isomers of } X \text { are, }
$$
Xhas six stereoisomers (I-VI). The total number of diasteromers of $X$ are more than three e.g, I + II, I + III, I + IV, I + V, I + VI etc. When stereochemistry at double bond is trans it has only two enantiomers (V and VI). When stereochemistry at doubld bond is cis, it has two enantiomers II and III. Hence, only $A$ and $D$ are correct answers.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2009 (Paper 1)
The correct statement(s) about the following sugars $X$ and $Y$ is (are)

Options:
A
$X$ is a reducing sugar and $Y$ is a non-reducing sugar

B
$X$ is a non-reducing sugar and $Y$ is a reducing sugar

C
the glucosidic linkages in $X$ and $Y$ are $\alpha$ and $\beta$, respectively

D
the glucosidic linkages in $X$ and $Y$ are $\beta$ and $\alpha$, respectively

The correct answers are:
$X$ is a non-reducing sugar and $Y$ is a reducing sugar
,
the glucosidic linkages in $X$ and $Y$ are $\alpha$ and $\beta$, respectively

Solution:
$X$ has acetal linkage and $Y$ has hemiacetal linkage. Carbohydrate with hemiacetal linkage are reducing sugars and carbohydrate with acetal linkage are non reducing sugars.
$X$ is $\alpha$ anomer and $Y$ is $\beta$-anomer of $D(+)$ glucose.

Topic: Chemistry, Biomolecules, JEE Advanced, JEE Advanced 2009 (Paper 2)
$$
\text { The correct statement(s) concerning the structures } E, F \text { and } G \text { is are }
$$

Options:
A
$E, F$ and $G$ are resonance structures

B
$E, F$ and $E, G$ are tautomers

C
$F$ and $G$ are geometrical isomers

D
$F$ and $G$ are diastereomers

The correct answers are:
$E, F$ and $E, G$ are tautomers
,
$F$ and $G$ are geometrical isomers
,
$F$ and $G$ are diastereomers

Solution:
$E$ and $F$ possess different place of atoms, in addition to electron pair, can be isomers only and not the resonance structures. Geometrical isomers are stereoisomers but not enantiomers called diastereomers

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Advanced, JEE Advanced 2008 (Paper 1)
The correct statement(s) pertaining to the adsorption of a gas on a solid surface is(are)

Options:
A
Adsorption is always exothermic

B
Physisorption may transform into chemisorption at high temperature

C
Physisorption increases with increasing temperature but chemisorption decreases with increasing temperature

D
Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy of activation

The correct answers are:
Adsorption is always exothermic
,
Physisorption may transform into chemisorption at high temperature
,
Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy of activation

Solution:
(a) In the process of adsorption, a bond is formed between adsorbate and adsorbent, hence always exothermic.
(b) Physisorption require very low activation energy while chemisorption require high activation energy. Therefore a physisorption may transform into chemisorption but only at high temperature.
(c) It is wrong statement as at higher temperature, physically adsorbed substance starts desorbing.
(d) In physical adsorption, van der Waals' force hold the adsorbate and adsorbent together which is a weak electrostatic attraction. In chemisorption, strong chemical bond binds the adsorbate to the adsorbent. Therefore, chemisorption is more exothermic than physical adsorption.

Topic: Chemistry, Surface Chemistry, JEE Advanced, JEE Advanced 2011 (Paper 1)
The correct statement(s) regarding defects in solids is (are)

Options:
A
Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion

B
Frenkel defect is a dislocation defect

C
Trapping of an electron in the lattice leads to the formation of F-centre

D
Schottky defects have no effect on the physical properties of solids

The correct answers are:
Frenkel defect is a dislocation defect
,
Trapping of an electron in the lattice leads to the formation of F-centre

Solution:
(a) Wrong Frenkel defect is favoured when the difference in ionic radii is very large. In the given case Schottky defect is favoured.
(b) Correct Frenkel defect arises due to dislocation of smaller ion from its normal lattice point to the interstitial space.
(c) Correct When alkali metal in the gaseous state is passed over salts like $\mathrm{NaCl}$, electrons from metal are trapped in the interstitial space giving rise to F-centre defect which is a type of Frenkel defect.
(d) Wrong Schottky defect affects physical properties of solids.

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2009 (Paper 1)
The correct structure of ethylenediaminetetraacetic acid (EDTA) is

Options:
A

B

C

D

The correct answer is:

Solution:
EDTA structure is
Coordination chemistry
Factual
I

Topic: Chemistry, Coordination Compounds, JEE Main
The correct structure of ethylenediaminetetraacetic acid (EDTA) is

Options:
A

B

C

D

The correct answer is:

Solution:
EDTA structure is
Coordination chemistry
Factual
I

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2010 (Paper 1)
The $\beta$-decay process, discovered around 1900 , is basically the decay of a neutron $(n)$. In the laboratory, a proton $(p)$ and an electron $\left(e^{-}\right)$are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, i.e.

$n \rightarrow p+e^{-}+\bar{v}_{e}$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left(\bar{v}_{e}\right)$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $0.8 \times 10^{6} \mathrm{eV}$. The kinetic energy carried by the proton is only the recoil energy.

Question: If the anti-neutrino had a mass of $3 \mathrm{eV} / \mathrm{c}^{2}$ (where $\mathrm{c}$ is the speed of light) instead of zero mass, what should be the range of the kinetic energy, $K$, of the electron?

Options:
A $0 \leq K \leq 0.8 \times 10^{6} \mathrm{eV}$
B $3.0 \mathrm{eV} \leq K \leq 0.8 \times 10^{6} \mathrm{eV}$
C $3.0 \mathrm{eV} \leq K < 0.8 \times 10^{6} \mathrm{eV}$
D $0 \leq K < 0.8 \times 10^{6} \mathrm{eV}$

The correct answer is:
$0 \leq K < 0.8 \times 10^{6} \mathrm{eV}$

Solution:
Kinetic energy $(K)$ of electron will be minimum or zero when total energy is shared by proton and anti-neutrino $\therefore 0 \leq \mathrm{K} < 0.8 \times 10^{6} \mathrm{eV}$

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2012 (Paper 2)
The $\beta$-decay process, discovered around 1900 , is basically the decay of a neutron $(n)$. In the laboratory, a proton $(p)$ and an electron $\left(e^{-}\right)$are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, i.e.

$n \rightarrow p+e^{-}+\bar{v}_{e}$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left(\bar{v}_{e}\right)$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $0.8 \times 10^{6} \mathrm{eV}$. The kinetic energy carried by the proton is only the recoil energy.

Question: What is the maximum energy of the anti-neutrino?

Options:
A Zero
B Much less than $0.8 \times 10^{6} \mathrm{eV}$.
C Nearly $0.8 \times 10^{6} \mathrm{eV}$
D Much larger than $0.8 \times 10^{6} \mathrm{eV}$

The correct answer is:
Nearly $0.8 \times 10^{6} \mathrm{eV}$

Solution:
$\because K_{p}^{-}+K_{e}^{-}+K_{v}^{-}=0.8 \times 10^{6} \mathrm{eV}$

When $K_{e}^{-}=0$ then $K_{p}^{-}+K_{v}^{-}=0.8 \times 10^{6} \mathrm{eV}$

Mass of $\bar{v}< < P \therefore$ maximum energy of anti-neutrino is nearly $0.8 \times 10^{6} \mathrm{eV}$.

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2012 (Paper 2)
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is $0.5 \mathrm{~mm}$ and there are 50 divisions on the circular scale. The reading on the main scale is $2.5 \mathrm{~mm}$ and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of $2 \%$, the relative percentage error in the density is

Options:
A
$0.9 \%$

B
$2.4 \%$

C
$3.1 \%$

D
$4.2 \%$

The correct answer is:
$3.1 \%$

Solution:
Least count of screw gauge $=\frac{0.5}{50}$
$$
=0.01 \mathrm{~mm}=\Delta r
$$
Diameter, $r=2.5 \mathrm{~mm}+20 \times \frac{0.5}{50}$
$$
=270 \mathrm{~mm}
$$
$$
\frac{\Delta r}{r}=\frac{0.01}{2.70}
$$
or $\frac{\Delta r}{r} \times 100=\frac{1}{2.7}$
Now, density $d=\frac{m}{V}=\frac{m}{\frac{4}{3} \pi\left(\frac{r}{2}\right)^3}$
Here, $r$ is the diameter.
$$
\begin{aligned}
\therefore \quad \frac{\Delta d}{d} & \times 100=\left\{\frac{\Delta m}{m}+3\left(\frac{\Delta r}{r}\right)\right\} \times 100 \\
= & \frac{\Delta m}{m} \times 100+3 \times\left(\frac{\Delta r}{r}\right) \times 100
\end{aligned}
$$
$$
\begin{aligned}
& =2 \%+3 \times \frac{1}{27} \\
& =3.11 \%
\end{aligned}
$$
Analysis of Question
(i) Question is moderately difficult.
(ii) In practical part, questions in JEE are asked from vernier callipers and screw gauge.

Topic: Physics, Mathematics in Physics, JEE Advanced, JEE Advanced 2011 (Paper 2)
The difference in the oxidation numbers of the two types of sulphur atoms in $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$ is

Options:

The correct answer is:
5

Solution:
$$
\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6 \text { is a salt of } \mathrm{H}_2 \mathrm{~S}_4 \mathrm{O}_6 \text { which has the following structure }
$$
$$
\Rightarrow \text { Difference in oxidation number of two types of sulphur }=5
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The difference in the oxidation numbers of the two types of sulphur atoms in $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$ is

Options:

The correct answer is:
5

Solution:
$$
\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6 \text { is a salt of } \mathrm{H}_2 \mathrm{~S}_4 \mathrm{O}_6 \text { which has the following structure }
$$
$$
\Rightarrow \text { Difference in oxidation number of two types of sulphur }=5
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2011 (Paper 1)
The differential equation $\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y}$ determines a family of circles with

Options:
A
variable radii and a fixed centre at $(0,1)$

B
variable radii and a fixed centre at $(0,-1)$

C
fixed radius 1 and variable centres along the $\mathrm{X}$-axis

D
fixed radius 1 and variable centres along the $\mathrm{Y}$-axis

The correct answer is:
fixed radius 1 and variable centres along the $\mathrm{X}$-axis

Solution:
$$
\begin{aligned}
& \frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y} \\
& \Rightarrow \quad \int \frac{y}{\sqrt{1-y^2}} d y=\int d x \\
& \Rightarrow \quad-\sqrt{1-y^2}=x+C \\
& \Rightarrow \quad(x+c)^2+y^2=1 \\
&
\end{aligned}
$$
Centre $(-C, 0)$ and radius $=1$.

Topic: Mathematics, Differential Equations, JEE Advanced, JEE Advanced 2007 (Paper 2)
The direct conversion of $A$ to $B$ is difficult, hence it is carried out by the following shown path.
Given $\Delta S_{(A \rightarrow C)}=50 \mathrm{eu}, \quad \Delta S_{(C \rightarrow D)}=30 \mathrm{eu}, \quad \Delta S_{(B \rightarrow D)}=20 \mathrm{eu}$ where eu is entropy unit, then $\Delta S_{(A \rightarrow B)}$ is

Options:
A
$+100 \mathrm{eu}$

B
$+60 \mathrm{eu}$

C
$-100 \mathrm{eu}$

D
$-60$ eu

The correct answer is:
$+60 \mathrm{eu}$

Solution:
Given: $\quad \Delta S_{(A \rightarrow C)}=50 \mathrm{eu}, \Delta S_{(C \rightarrow D)}=30 \mathrm{eu}, \Delta S_{(B \rightarrow D)}=20 \mathrm{eu}$ where eu is entropy unit.
$$
\text { Therefore, } \begin{aligned}
\Delta S_{(A \rightarrow B)} & =\Delta S_{(A \rightarrow C)}+\Delta S_{(C \rightarrow D)}+\Delta S_{(D \rightarrow B)} \\
& =\Delta S_{(A \rightarrow C)}+\Delta S_{(C \rightarrow D)}-\Delta S_{(B \rightarrow D)} \\
& =50+30-20=60 \mathrm{eu}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2006
The dissociation constant of a substituted benzoic acid at $25^{\circ} \mathrm{C}$ is $1.0 \times 10^{-4}$. The $\mathrm{pH}$ of $0.01 \mathrm{M}$ solution of its sodium salt is

Options:

The correct answer is:
8

Solution:
$$
\text { The hydrolysis reaction of conjugate base of acid is }
$$
$$
\begin{gathered}
A^{-}(a q)+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HO}^{-}+\mathrm{HA} \\
K_h=\frac{K_w}{K_a}=\frac{10^{-14}}{10^{-4}}=10^{-10}
\end{gathered}
$$
Since, degree of hydrolysis is negligible;
$$
\left[\mathrm{OH}^{-}\right]=\sqrt{K_h C}=10^{-6} \cdot p[\mathrm{OH}]=6 \text { and } \mathrm{pH}=14-6=8
$$

Topic: Chemistry, Ionic Equilibrium, JEE Advanced, JEE Advanced 2009 (Paper 2)
The distance advanced by the screw of a screw gauge is 2 mm in 4 rotations. Its cap is divided into 50 divisions. There is no zero error. If the screw reads 3 divisions on the main scale and 32 divisions on the cap, then the diameter (in mm) of the wire is

Options:

The correct answer is:
1.82

Solution:
pitch =2/4
LC =2/450= .01 mm
diameter = 3 Ã 0.5 + 32 Ã .01
=1.5 + .32
= 1.82 mm

Topic: Physics, Experimental Physics, JEE Main
The edge length of unit cell of a metal having molecular weight $75 \mathrm{~g} / \mathrm{mol}$ is 5 Å which crystallizes in cubic lattice. If the density is $2 \mathrm{~g} / \mathrm{cc}$ then find the radius of metal atom, $\left(N_A=6 \times 10^{23}\right)$. Give the answer in pm.

Options:

The correct answer is:
216.5

Solution:
For cubic lattice, $\rho=\frac{Z \times A}{N_A \times V}$ or $Z=\frac{\rho \times N_A \times V}{A}$
Given that,
$$
\begin{aligned}
& A=75 \mathrm{~g} / \mathrm{mol} \\
& V=(5 Å)^3=\left(5 \times 10^{-8}\right)^3 \mathrm{~cm}^3 \\
& =125 \times 10^{-24} \mathrm{~cm}^3 \\
& N_{\mathrm{A}}=6.02 \times 10^{23} \\
& \rho=2 \mathrm{gm} / \mathrm{cc} \\
& \therefore \quad Z=\frac{2 \times 6.02 \times 10^{23} \times 125 \times 10^{-24}}{75} \approx 2 \\
&
\end{aligned}
$$
The value of $Z$ represents that element must have body centred cubic (bcc) structure.
For bcc structure atomic radius $(r)=\frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{4} \times 5=2.165 Å$
$$
=216.5 \mathrm{pm}
$$

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2006