datavorous/entrance-exam-dataset huggingface dataset:
Among the following, the coloured compound is

Options:
A
$\mathrm{CuCl}$

B
$\mathrm{K}_3\left[\mathrm{Cu}(\mathrm{CN})_4\right]$

C
$\mathrm{CuF}_2$

D
$\left[\mathrm{Cu}\left(\mathrm{CH}_3 \mathrm{CN}_4\right)\right] \mathrm{BF}_4$

The correct answer is:
$\mathrm{CuF}_2$

Solution:
In $\mathrm{CuF}_2, \mathrm{Cu}^{2+}$ ions exist, having $d^9$-configuration. Unpaired electron causes colour ( $d-d$ transition)
In the crystalline form, $\mathrm{CuF}_2$ is blue coloured

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2008 (Paper 2)
Among the following, the least stable resonance structure is

Options:
A

B

C

D

The correct answer is:

Solution:
Tow positive charges are present at the adjacent place, elevates the energy lowers the stability most.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 2)
Among the following, the number of compounds than can react with $\mathrm{PCl}_5$ to give $\mathrm{POCl}_3$ is $\mathrm{O}_2, \mathrm{CO}_2, \mathrm{SO}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{P}_4 \mathrm{O}_{10}$.

Options:

The correct answer is:
4

Solution:
$$
\text { } \mathrm{PCl}_5 \text { produces } \mathrm{POCl}_3 \text { with the following reagents }
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2011 (Paper 2)
Among the following, the number of elements showing only one non-zero oxidation state is O, Cl, F, N, P, Sn, Tl, $\mathrm{Na}, \mathrm{Ti}$

Options:

The correct answer is:
8

Solution:
F is the most electronegative element which cannot loose electron to other so only $-1$ Na is alkali metal which can loose only one electron so $+1$ Basics of general chemistry Basic understanding I

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2010 (Paper 2)
Among the following, the paramagnetic compound is

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

B
$\mathrm{O}_3$

C
$\mathrm{N}_2 \mathrm{O}$

D
$\mathrm{KO}_2$

The correct answer is:
$\mathrm{KO}_2$

Solution:
For $\mathrm{KO}_2, \mathrm{O}_2^{-}$has unpaired $e^{-} \Rightarrow$ paramagnetic
$$
\begin{gathered}
\mathrm{O}_2^{-} \text {: Total } e^{-} s=16+1=17 \\
\sigma 1 s^2, \sigma^* 1 s^2 \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2 \pi 2 p_x^{1+1} \\
=\pi 2 p_y^{1+1}, \pi^* 2 p_x^{1+1}=\pi 2 p_y^1
\end{gathered}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Advanced, JEE Advanced 2007 (Paper 1)
Among the following, the state function(s) is (are)

Options:
A
internal energy

B
irreversible expansion work

C
reversible expansion work

D
molar enthalpy

The correct answers are:
internal energy
,
reversible expansion work
,
molar enthalpy

Solution:
State function are internal energy and molar enthalpy. Work is a path dependent function, however in reversible process the path is fixed i.e. between a given initial and final state there is just on reversible path and one work. Hence, reversible work is also a state function.

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2009 (Paper 2)
Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient conditions is

Options:
A
$\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{15} \mathrm{~N}^{+}\left(\mathrm{CH}_3\right)_3 \mathrm{Br}^{-}$

B
$\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{11} \mathrm{OSO}_3^{-} \mathrm{Na}^{+}$

C
$\mathrm{CH}_3\left(\mathrm{CH}_2\right)_6 \mathrm{COO}^{-} \mathrm{Na}^{+}$

D
$\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{11} \mathrm{~N}^{+}\left(\mathrm{CH}_3\right)_3 \mathrm{Br}^{-}$

The correct answer is:
$\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{15} \mathrm{~N}^{+}\left(\mathrm{CH}_3\right)_3 \mathrm{Br}^{-}$

Solution:
Sodium Hexadecyl trimethylammonium bromide (CTAB) dodecylsulphate
(SDS)
CMC $(\mathrm{mm})>-10$
1

Topic: Chemistry, Surface Chemistry, JEE Advanced, JEE Advanced 2008 (Paper 2)
Amongst the compounds given, the one that would form a brilliant coloured dye on treatment with $\mathrm{NaNO}_2$ in dil. $\mathrm{HCl}$ followed by addition to an alkaline solution of $\beta$-naphthol is

Options:
A

B

C

D

The correct answer is:

Solution:
As we know, benzenediazonium salt forms brilliant coloured dye with $\beta$-naphthol, the compound under consideration must be $p$-toludine (c) as it is a primary aromatic amine. Primary aromatic amine, on treatment with $\mathrm{NaNO}_2$ in dil. $\mathrm{HCl}$ forms the corresponding diazonium chloride salt.

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2011 (Paper 2)
Amongst the following, the total number of compounds soluble in aqueous $\mathrm{NaOH}$ is

Options:

The correct answer is:
4

Solution:
are soluble in aqueous $\mathrm{NaOH}$. Benzylic alcohol is less acidic than water so not soluble in aq. $\mathrm{NaOH}$, General organic chemistry Conceptual (structural interpretation) III

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2010 (Paper 1)
Amongst the following, the total number of compounds soluble in aqueous $\mathrm{NaOH}$ is

Options:

The correct answer is:
4

Solution:
are soluble in aqueous $\mathrm{NaOH}$. Benzylic alcohol is less acidic than water so not soluble in aq. $\mathrm{NaOH}$, General organic chemistry Conceptual (structural interpretation) III

Topic: Chemistry, General Organic Chemistry, JEE Main
Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is
$$
\begin{array}{llll}
\mathrm{KCN} & \mathrm{K}_2 \mathrm{SO}_4 & \left(\mathrm{NH}_4\right)_2 \mathrm{C}_2 \mathrm{O}_4 & \mathrm{NaCl} \\
\mathrm{Zn}\left(\mathrm{NO}_3\right)_2 & \mathrm{FeCl}_3 & \mathrm{~K}_2 \mathrm{CO}_3 & \mathrm{NH}_4 \mathrm{NO}_3 \\
\mathrm{LiCN} & & &
\end{array}
$$

Options:

The correct answer is:
3

Solution:
The aqueous solution of $\mathrm{KCN}, \mathrm{K}_2 \mathrm{CO}_3$ and LiCN turn red litmus blue because of alkaline nature.
The anionic hydrolysis turns the solution alkaline.
Ionic equilibrium
Conceptual
III

Topic: Chemistry, Ionic Equilibrium, JEE Advanced, JEE Advanced 2010 (Paper 1)
Amongst the given options, the compound (s) in which all the atoms are in one plane in all the possible conformations (if any), is (are)

Options:
A

B

C

D

The correct answers are:
,

Solution:
$$
\text { In both (b) and (c), all the atoms are present in one single plane }
$$
In (a) 1, 3-butadiene, conformational change is possible between $\mathrm{C}_2-\mathrm{C}_3$ bond in which atoms will be present in more than one single plane.
In (d) allene, the terminals $\mathrm{H}-\mathrm{C}-\mathrm{H}$ planes are perpendicular to one another.
$$
\text { Paragraph (12 and 13) The final ozonolysis product indicates that the alkene before }
$$
Ozonolysis is
Also $\mathrm{P}\left(\mathrm{C}_6 \mathrm{H}_{10}\right)$ has two degree of unsaturation and oxymercuration demercuration hydration indicates that it is an alkyne. As alkyne, on hydration, gives a carbonyl compound which on reduction with $\mathrm{NaBH}_4$ gives a $2^{\circ}$ alcohol.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2011 (Paper 1)
An AC voltage source of variable angular frequency $\omega$ and fixed amplitude $V_0$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega$ is increased

Options:
A
the bulb glows dimmer

B
the bulb glows brighter

C
total impedance of the circuit is unchanged

D
total impedance of the circuit increases

The correct answer is:
the bulb glows brighter

Solution:
$Z=\sqrt{R^2+X_C^2}: I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}: P=I_{\mathrm{rms}}^2 R$ where, $\quad X_C=\frac{1}{\omega C}$
As $\omega$ is increased, $X_C$ will decrease or $Z$ will decrease. Hence $I_{\mathrm{mms}}$ or $P$ will increase. Therefore, bulb glows brighter.
Hence the correct option is (b).

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2010 (Paper 1)
An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true?

Options:
A
They will never come out of the magnetic field region

B
They will come out travelling along parallel paths

C
They will come out at the same time

D
They will come out at different times

The correct answers are:
They will come out travelling along parallel paths
,
They will come out at different times

Solution:
$\quad r=\frac{m v}{B q}$
or $\quad r \propto m$
$\therefore r_e < r_p$ as $m_e < m_p$
Further, $\quad T=\frac{2 \pi m}{B q}$
or $\quad T \propto m$
$$
\begin{aligned}
\therefore \quad T_e & < T_p \\
t_e & =\frac{T_e}{2}
\end{aligned}
$$
and
or $\quad t_e < t_p$
$\therefore$ Correct options are (b) and (d).
Note In JEE 2011, official answer key, correct options were given bc, bd, bcd, here student should realise that both the electron and proton enter the magnetic field at the same time or at different times.
Analysis of Question
(i) Question is simple.
(ii) In the situation given above particle will never complete the circle unless magnetic field is all around.

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2011 (Paper 1)
An ellipse intersects the hyperbola $2 x^2-2 y^2=1$ orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

Options:
A
equation of ellipse is $x^2+2 y^2=2$

B
the foci of ellipse are $(\pm 1,0)$

C
equation of ellipse is $x^2+2 y^2=4$

D
the foci of ellipse are $(\pm \sqrt{2}, 0)$

The correct answers are:
equation of ellipse is $x^2+2 y^2=2$
,
the foci of ellipse are $(\pm 1,0)$

Solution:
Given, $2 x^2-2 y^2=1$
$$
\Rightarrow \quad \frac{x^2}{\left(\frac{1}{2}\right)}-\frac{y^2}{\left(\frac{1}{2}\right)}=1
$$
Eccentricity of hyperbola $=\sqrt{2}$
So, eccentricity of ellipse $=1 / \sqrt{2}$
Let equation of ellipse be
$$
\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) \Rightarrow \frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}} \\
& \Rightarrow \quad \frac{b^2}{a^2}=\frac{1}{2} \Rightarrow a^2=2 b^2 \\
& \therefore \quad x^2+2 y^2=2 b^2 \\
&
\end{aligned}
$$
Let ellipse and hyperbola intersect at
$$
A\left(\frac{1}{\sqrt{2}} \sec \theta, \frac{1}{\sqrt{2}} \tan \theta\right)
$$
On differentiating Eq. (i),
$$
4 x-4 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{x}{y}
$$
$$
\therefore \quad\left(\frac{d y}{d x}\right)_{\text {at } A}=\frac{\sec \theta}{\tan \theta}=\operatorname{cosec} \theta
$$
On differentiating Eq. (ii),
$$
2 x+4 y \frac{d y}{d x}=0
$$
$$
\therefore \quad\left(\frac{d y}{d x}\right)_{\text {at } A}=-\frac{x}{2 y}=-\frac{1}{2} \operatorname{cosec} \theta
$$
Since, ellipse and hyperbola are orthogonal.
$$
\begin{aligned}
& \therefore \quad-\frac{1}{2} \operatorname{cosec}^2 \theta=-1 \\
& \Rightarrow \quad \operatorname{cosec}^2 \theta=2 \Rightarrow \theta=\pm \frac{\pi}{4} \\
& \therefore A\left(1, \frac{1}{\sqrt{2}}\right) \text { or }\left(1,-\frac{1}{\sqrt{2}}\right)
\end{aligned}
$$
From Eq. (i), $1+2\left(\frac{1}{\sqrt{2}}\right)^2=2 b^2$ $\Rightarrow \quad b^2=1$
Equation of ellipse is $x^2+2 y^2=2$
Coordinate of foci $(\pm a e, 0)$
$$
=\left(\pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0\right)=(\pm 1,0)
$$
Hence, options (a) and (b) are correct. If major axis is along $Y$-axis, then
$$
\begin{aligned}
& \frac{1}{\sqrt{2}}=\sqrt{1-\frac{a^2}{b^2}} \Rightarrow b^2=2 a^2 \\
& \therefore 2 x^2+y^2=2 a^2 \Rightarrow y^{\prime}=-\frac{2 x}{y} \\
& \Rightarrow \quad y^{\prime}\left(\frac{1}{\sqrt{2}} \sec \theta \frac{1}{\sqrt{2}} \tan \theta\right)=\frac{-2}{\sin \theta} \\
&
\end{aligned}
$$
As ellipse and hyperbola are orthogonal.
$$
\begin{array}{ll}
\therefore & -\frac{2}{\sin \theta} \cdot \operatorname{cosec} \theta=-1 \\
\Rightarrow & \operatorname{cosec}^2 \theta=1 \Rightarrow \theta=\pm \frac{\pi}{4} \\
\therefore & 2 x^2+y^2=2 a^2 \Rightarrow 2+\frac{1}{2}=2 a^2 \\
\Rightarrow & a^2=\frac{5}{4} \Rightarrow 2 x^2+y^2=\frac{5}{2}
\end{array}
$$
Corresponding foci are $(0, \pm 1)$.

Topic: Mathematics, Ellipse, JEE Advanced, JEE Advanced 2009 (Paper 2)
An experiment has 10 equally likely outcomes. Let $A$ and $B$ be two non-empty events of the experiment. If $A$ consists of 4 outcomes, the number of outcomes that $B$ must have so that $A$ and $B$ are independent, is

Options:
A
2,4 or 8

B
3,6 or 9

C
4 or 8

D
5 or 10

The correct answer is:
5 or 10

Solution:
$P(A)=\frac{2}{5}$
For independent events,
$$
\begin{aligned}
& P(A \cap B)=P(A) P(B) \Rightarrow P(A \cap B) \leq \frac{2}{5} \\
& \Rightarrow P(A \cap B)=\frac{1}{10}, \frac{2}{10}, \frac{3}{10}, \frac{4}{10} \\
& \text { (i) } P(A \cap B)=\frac{1}{10} \\
& \Rightarrow P(A) \cdot P(B)=\frac{1}{10} \\
& \Rightarrow \quad P(B)=\frac{1}{10} \times \frac{5}{2}=\frac{1}{4}, \text { not possible. } \\
& \text { (ii) } P(A \cap B)=\frac{2}{10} \Rightarrow \frac{2}{5} \times P(B)=\frac{2}{10} \\
& \Rightarrow \quad P(B)=\frac{5}{10}, \text { outcomes of } B=5
\end{aligned}
$$
$$
\text { (iii) } \begin{aligned}
& P(A \cap B)=\frac{3}{10} \Rightarrow P(A) \cdot P(B)=\frac{3}{10} \\
& \Rightarrow \quad \frac{2}{5} \times P(B)=\frac{3}{10} \\
& P(B)=\frac{3}{4}, \text { not possible }
\end{aligned}
$$
$$
\text { (iv) } \begin{aligned}
& P(A \cap B)=\frac{4}{10} \\
\Rightarrow & P(A) \cdot P(B)=\frac{4}{10} \\
\Rightarrow & P(B)=1, \text { outcomes of } B=10 .
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2008 (Paper 2)
An ideal gas is expanding such that $p T^2=$ constant. The coefficient of volume expansion of the gas is

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

B
$\frac{2}{T}$

C
$\frac{3}{T}$

D
$\frac{4}{T}$

The correct answer is:
$\frac{3}{T}$

Solution:
$p T^2=$ constant
$\begin{array}{ll}\therefore \quad & \left(\frac{n R T}{V}\right) T^2=\text { constant } \\ \text { or } \quad T^3 V^{-1}=\text { constant }\end{array}$
Differentiating the equation, we get
$$
\frac{3 T^2}{V} \cdot d T-\frac{T^3}{V^2} \cdot d V=0
$$
or
$$
3 \cdot d T=\frac{T}{V} \cdot d V
$$
From the equation, $d V=V \gamma \cdot d T$
$\gamma$ = coefficient of volume expansion of gas $=\frac{d V}{V \cdot d T}$
From Eq. (i) $\gamma=\frac{d V}{V \cdot d T}=\frac{3}{T}$
$\therefore$ correct answer is (c).

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2008 (Paper 1)
An infinitely long hollow conducting cylinder with inner radius $R / 2$ and outer radius $R$ carries a uniform current density along its length. The magnitude of the magnetic field, $|\vec{B}|$ as a function of the radial distance $r$ from the axis is best represented by

Options:
A
B
C
D

The correct answer is:

Solution:
For $r < \frac{R}{2}, \quad B=0$

$\begin{array}{l}

\text { For } \frac{R}{2} \leq r < R, \\

B=\frac{\mu_{0}}{2}\left[r-\frac{R^{2}}{2 r}\right] J

\end{array}$

$\begin{array}{l}

\text { For } r>R, \quad B=\frac{\mu_{0} i}{2 \pi r} \\

\text { i.e., } B \propto \frac{1}{r}

\end{array}$

Hence graph $(d)$ correctly depicts $|\vec{B}|$ versus $r$ graph.

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2012 (Paper 2)
An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho$. It has a spherical cavity of radius $R / 2$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$, which is at a distance $2 R$ from the axis of the cylinder, is given by the expression $\frac{23 \rho R}{16 K \varepsilon_{0}}$. The value of $k$ is

Options:

The correct answer is:
6

Solution:
The magnitude of the electric field at the point $P$ which is at a distance $2 R$ from the axis of the cylinder

$\begin{array}{l}

\mathrm{E}=\mathrm{E}_{\text {total }}-\mathrm{E}_{\text {cavity }} \\

=\frac{\gamma}{2 \pi \varepsilon_{0}(2 R)}-\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{(2 R)^{2}} \\

\mathrm{Q}_{\text {sphere }}=\frac{4}{3} \pi\left(\frac{R}{2}\right)^{3} \rho=\frac{\pi R^{3} \rho}{6} \\

\begin{array}{l}

\lambda_{\text {cylinder }}=\pi R^{2} \rho \\

\therefore E=\frac{\pi R^{2} \rho}{4 \pi \varepsilon_{0} R}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\pi R^{3} \rho / 6}{4 R^{2}} \\

=\frac{23 \rho R}{96 \varepsilon_{0}}=\frac{23 \rho R}{16 \times 6 \times \varepsilon_{0}} \\

\therefore k=6

\end{array}

\end{array}$

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2012 (Paper 1)
An optical component and an object $S$ placed along its optic axis are given in Column I. The distance between the object and the component can be varied. The properties of images are given in Column II. Match all the properties of images from Column II with the appropriate components given in Column I. Indicate your answer by darkening the appropriate bubble of the $4 \times 4$ matrix given in the ORS.

Options:
A
(A) p,q,r,s, (B) q, (C) p,q,r,s, (D) p,q,s

B
(A) p,q,r,s, (B) q, (C) p,q,r,s, (D) p,q,r,s

C
(A) p,s, (B) q, (C) p,q,r,s, (D) p,q,s

D
(A) p,q,r,s, (B) p,q,r, (C) p,q,r,s, (D) p,q,s

The correct answer is:
(A) p,q,r,s, (B) q, (C) p,q,r,s, (D) p,q,r,s

Solution:
(A) : (C) and (D) : In case of concave mirror or convex lens image can be real, virtual, diminished magnified or of same size. (B) : In case of convex mirror image is always virtual (for real object)

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2008 (Paper 2)
An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1 /$ 8 and $1 / 10$ of its initial concentration are $t_{1 / 8}$ and $t_{1 / 10}$ respectively. What is the value of $\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10$ ? $\left(\log _{10} 2=0.3\right)$

Options:

The correct answer is:
9

Solution:
$\begin{array}{l}

t_{1 / 8}=\frac{2.303 \log 8}{k}=\frac{2.303 \times 3 \log 2}{k} \\

t_{1 / 10}=\frac{2.303}{k} \log 10=\frac{2.303}{k} \\

{\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10=\frac{\left(\frac{2.303 \times 3 \log 2}{k}\right)}{\left(\frac{2.303}{k}\right)} \times 10=9}

\end{array}$

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2012 (Paper 1)
An $\alpha$-particle and a proton are accelerated from rest by a potential difference of $100 \mathrm{~V}$. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\frac{\lambda_p}{\lambda_\alpha}$, to the nearest integer, is

Options:

The correct answer is:
3

Solution:
$$
\begin{aligned}
\lambda & =\frac{h}{P}=\frac{h}{\sqrt{2 q q V m}} \text { or } \lambda \propto \frac{1}{\sqrt{q m}} \\
\frac{\lambda_p}{\lambda_\alpha} & =\sqrt{\frac{q_\alpha}{q_p} \cdot \frac{m_{\propto}}{m_p}}=\sqrt{\frac{(2)(4)}{(1)(1)}}=2.828
\end{aligned}
$$
The nearest integer is 3 .
$\therefore$ answer is 3 .

Topic: Physics, Dual Nature of Matter, JEE Advanced, JEE Advanced 2010 (Paper 1)
$$
\text { Aqueous solution of } \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \text { on reaction with } \mathrm{Cl}_2 \text { gives }
$$

Options:
A
$\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$

B
$\mathrm{NaHSO}_4$

C
$\mathrm{NaCl}$

D
$\mathrm{NaOH}$

The correct answer is:
$\mathrm{NaHSO}_4$

Solution:
Sodium thiosulphate, $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ gets oxidised by chlorine water.
$$
\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+4 \mathrm{Cl}_2+5 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaHSO}_4+8 \mathrm{HCl}
$$
$\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ gets oxidised by $\mathrm{FeCl}_3$ into $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$
$$
2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+2 \mathrm{FeCl}_3 \longrightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{FeCl}_2+2 \mathrm{NaCl}
$$
Further, $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ decomposes on treatment with dilute acids
$$
\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+2 \mathrm{HCl} \longrightarrow 2 \mathrm{NaCl}+\mathrm{SO}_2+\mathrm{S}+\mathrm{H}_2 \mathrm{O}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2008 (Paper 1)
Aqueous solutions of $\mathrm{HNO}_3, \mathrm{KOH}$, $\mathrm{CH}_3 \mathrm{COOH}$, and $\mathrm{CH}_3 \mathrm{COONa}$ of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is(are)

Options:
A
$\mathrm{HNO}_3$ and $\mathrm{CH}_3 \mathrm{COOH}$

B
$\mathrm{KOH}$ and $\mathrm{CH}_3 \mathrm{COONa}$

C
$\mathrm{HNO}_3$ and $\mathrm{CH}_3 \mathrm{COONa}$

D
$\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COONa}$

The correct answers are:
$\mathrm{HNO}_3$ and $\mathrm{CH}_3 \mathrm{COONa}$
,
$\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{CH}_3 \mathrm{COONa}$

Solution:
$\mathrm{HNO}_3$ and $\mathrm{CH}_3 \mathrm{COONa}$ can form buffer if the amount of $\mathrm{HNO}_3$ is limited. However concentrations are same but the amount can differ with manipulation of volume.
Ionic equilibrium
Conceptual
II

Topic: Chemistry, Ionic Equilibrium, JEE Advanced, JEE Advanced 2010 (Paper 1)
Area of the region bounded by the curve $y=e^x$ and lines $x=0$ and $y=e$ is

Options:
A
$e-1$

B
$\int_1^e \ln (e+1-y) d y$

C
$e-\int_0^1 e^x d x$

D
$\int_1^e \ln y d y$

The correct answers are:
$\int_1^e \ln (e+1-y) d y$
,
$e-\int_0^1 e^x d x$
,
$\int_1^e \ln y d y$

Solution:
Shaded area $=e-\left(\int_0^1 e^x d x\right)=1$
Also, $\quad \int_1^e \ln (e+1-y) d y$
$$
\begin{aligned}
& =\int_e^q \ln t(-d t) \\
& \text { [put } e+1-y=t \Rightarrow-d y=d t \text { ] } \\
& =\int_1^e \ln t d t=\int_1^e \ln y d y=1
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Advanced, JEE Advanced 2009 (Paper 1)
As per IUPAC nomenclature, the name of the complex $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}_{3}$ is :

Options:
A Tetraaquadiaminecobalt (III) chloride
B Tetraaquadiamminecobalt (III) chloride
C Diaminetetraaquacoblat (II) chloride
D Diamminetetraaquacobalt (III) chloride

The correct answer is:
Diamminetetraaquacobalt (III) chloride

Solution:
(achiral)

$\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}_{3}$

Diamminetetraaquacobalt (III) chloride.

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2012 (Paper 1)
Assertion: Consciousness is considered as the defining property of living organisms.
Reason : All organisms, from the prokaryotes to the most complex eukaryotes can sense and respond to environmental stimuli.

Options:
A If both assertion and reason are true and reason is the correct explanation of assertion.
B If both assertion and reason are true but reason is not the correct explanation of assertion.
C If assertion is true but reason is false.
D If both assertion and reason are false.

The correct answer is:
If both assertion and reason are true and reason is the correct explanation of assertion.

Solution:
Consciousness is awareness of the surroundings and responding to external stimuli. The external stimuli can be physical, chemical or biological. All organisms, from primitive prokaryotes to most advanced and complex eukaryotes, are able to sense and respond to environmental factors. Consciousness is thus said to be the defining property of living organisms.

Topic: Biology, The Living World, AIIMS, AIIMS 2016
Assertion : Protostele is the simplest stele.
Reason : Protostele is the most advanced type of stele.

Options:
A If both assertion and reason are true and reason is the correct explanation of assertion.
B If both assertion and reason are true but reason is not the correct explanation of assertion.
C If assertion is true but reason is false
D If both assertion and reason are false

The correct answer is:
If assertion is true but reason is false

Solution:
Protostele is the simplest and considered to be the most primitive type of stele. It consists of a solid core of xylem surrounded by the cylinder of phloem, enclosing no pith. All other types of steles have evolved from it in the course of evolutionary specialization. Protosteles may be found in Selaginella, Lycopodium, Gleichenia and Lygodium among present day forms.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2013