datavorous/entrance-exam-dataset huggingface dataset:
A uniform electric field, E→=-4003 j^ N C-1 is applied in a region. A charged particle of mass m carrying positive charge q is projected in this region with an initial speed of 210×106 m s-1. This particle is aimed to hit a target T, which is 5 m away from its entry point into the field as shown schematically in the figure. Take qm=1010 C kg-1. Then

Options:
A the particle will hit T if projected at an angle 45° from the horizontal
B the particle will hit T if projected either at an angle 30° or 60° from the horizontal
C time taken by the particle to hit T could be 56μs as well as 52μs
D time taken by the particle to hit T is 53μs

The correct answers are:
the particle will hit T if projected either at an angle 30° or 60° from the horizontal, time taken by the particle to hit T could be 56μs as well as 52μs

Solution:
ay=EM=43×1012 m s2

R=u2sin2θay

sin2θ=Rayu2=32

θ=30° or 60°

T=2usin 30°ay or 2usin 60°ay

=56μs or 52μs

Topic: Physics, Motion In Two Dimensions, JEE Advanced, JEE Advanced 2020 (Paper 1)
A uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure, and rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is

Options:
A
$\frac{1}{2 \pi} \sqrt{\frac{2 k}{M}}$

B
$\frac{1}{2 \pi} \sqrt{\frac{k}{M}}$

C
$\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}$

D
$\frac{1}{2 \pi} \sqrt{\frac{24 k}{M}}$

The correct answer is:
$\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}$

Solution:
$$
\begin{aligned}
& \text { Restoring torque }=-(2 k x) \cdot \frac{L}{2} \\
& \qquad \alpha=-\frac{k L(L / 2 \theta)}{I}=-\left[\frac{k L^2 / 2}{M L^2 / 12}\right] \cdot \theta
\end{aligned}
$$
$$
\begin{aligned}
& =-\left(\frac{6 k}{M}\right) \theta \\
\therefore \quad f & =\frac{1}{2 \pi} \sqrt{\left|\frac{\alpha}{\theta}\right|}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main
A uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure, and rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is

Options:
A
$\frac{1}{2 \pi} \sqrt{\frac{2 k}{M}}$

B
$\frac{1}{2 \pi} \sqrt{\frac{k}{M}}$

C
$\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}$

D
$\frac{1}{2 \pi} \sqrt{\frac{24 k}{M}}$

The correct answer is:
$\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}$

Solution:
$$
\begin{aligned}
& \text { Restoring torque }=-(2 k x) \cdot \frac{L}{2} \\
& \qquad \alpha=-\frac{k L(L / 2 \theta)}{I}=-\left[\frac{k L^2 / 2}{M L^2 / 12}\right] \cdot \theta
\end{aligned}
$$
$$
\begin{aligned}
& =-\left(\frac{6 k}{M}\right) \theta \\
\therefore \quad f & =\frac{1}{2 \pi} \sqrt{\left|\frac{\alpha}{\theta}\right|}=\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2009 (Paper 2)
A uniformly charged thin spherical shell of radius $R$ carries uniform surface charge density of $\sigma$ per unit area. It is made of two hemispherical shells, held together by pressing them with force $F$ (see figure). $F$ is proportional to

Options:
A
$\frac{1}{\varepsilon_0} \sigma^2 R^2$

B
$\frac{1}{\varepsilon_0} \sigma^2 R$

C
$\frac{1}{\varepsilon_0} \frac{\sigma^2}{R}$

D
$\frac{1}{\varepsilon_0} \frac{\sigma^2}{R^2}$

The correct answer is:
$\frac{1}{\varepsilon_0} \sigma^2 R^2$

Solution:
Electrical force per unit area $=\frac{1}{2} \varepsilon_0 E^2$
$$
=\frac{1}{2} \varepsilon_0\left(\frac{\sigma}{\varepsilon_0}\right)^2=\frac{\sigma^2}{2 \varepsilon_0}
$$
Projected area $=\pi R^2$
$\therefore$ net electrical force $=\left(\frac{\sigma^2}{2 \varepsilon_0}\right)\left(\pi R^2\right)$
In equilibrium, this force should be equal to the applied force.
$\therefore \quad F=\frac{\pi \sigma^2 R^2}{2 \varepsilon_0}$
or $\quad F \propto \frac{\sigma^2 R^2}{\varepsilon_0}$
$\therefore$ The correct option is (a).

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2010 (Paper 2)
A vernier calipers has $1 \mathrm{~mm}$ marks on the main scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is

Options:
A
$0.02 \mathrm{~mm}$

B
$0.05 \mathrm{~mm}$

C
$0.1 \mathrm{~mm}$

D
$0.2 \mathrm{~mm}$

The correct answer is:
$0.2 \mathrm{~mm}$

Solution:
Least count of vernier calipers
$$
\begin{aligned}
& \text { LC }=1 \mathrm{MSD}-1 \text { VSD } \\
& =\frac{\text { Smallest division on main scale }}{\text { Number of divisions on vernier scale }} \\
& 20 \text { divisions of vernier scale } \\
& =16 \text { divisions of main scale } \\
& \therefore \quad 1 \mathrm{VSD}=\frac{16}{20} \mathrm{~mm}=0.8 \mathrm{~mm} \\
& \therefore \quad \text { LC }=1 \mathrm{MSD}-1 \mathrm{VSD} \\
& =1 \mathrm{~mm}-0.8 \mathrm{~mm} \\
& \therefore \quad=0.2 \mathrm{~mm} \\
&
\end{aligned}
$$
$\therefore$ The correct option is (d).

Topic: Physics, Experimental Physics, JEE Advanced, JEE Advanced 2010 (Paper 2)
A vibrating string of certain length $l$ under a tension $T$ resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $75 \mathrm{~cm}$ inside a tube closed at one end. The string also generates 4 beat/s when excited along with a tuning fork of frequency $n$. Now when the tension of the string is slightly increased the number of beats reduces to 2 per second. Assuming the velocity of sound in air to be $340 \mathrm{~m} / \mathrm{s}$, the frequency $n$ of the tuning fork (in $\mathrm{Hz}$ ) is

Options:
A
344

B
336

C
$117.3$

D
$109.3$

The correct answer is:
344

Solution:
With increase in tension, frequency of vibrating string will increase. Since number of beats are decreasing. Therefore, frequency of vibrating string or third harmonic frequency of closed pipe should be less than the frequency of tuning fork by 4 .
$\therefore$ Frequency of tuning fork $=$ Third harmonic frequency of closed pipe $+4$
$$
=3\left(\frac{v}{4 l}\right)+4=3\left(\frac{340}{4 \times 0.75}\right)+4=344 \mathrm{~Hz} .
$$
$\therefore$ correct option is (a).

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2008 (Paper 2)
A wooden block performs SHM on a frictionless surface with frequency $v_0$. The block carries a charge $+Q$ on its surface. If now a uniform electric field $\mathbf{E}$ is switched on as shown, then the SHM of the block will be

Options:
A
of the dame frequency and with shifted mean position

B
of the same frequency and with the same mean position

C
of changed frequency and with shifted mean position

D
of changed frequency and with the same mean position

The correct answer is:
of the dame frequency and with shifted mean position

Solution:
Frequency or time period of SHM depends on variable forces. It does not depend on constant external force. Constant external force can only change the mean position. For example, in the given question mean position is at natural length of spring in the absence of electric field. Whereas in the presence of electric field mean position will be obtained after a compression of $x_0$. Where $x_0$ is given by
or
$$
\begin{aligned}
k x_0 & =Q E \\
x_0 & =\frac{Q E}{k}
\end{aligned}
$$
$\therefore$ Correct answer is (a).
Analysis of Question
(i) Question is simple.
(ii) I think almost all the serious aspirants of JEE can solve this problem easily.

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2011 (Paper 2)
According to Bohr's theory $E_n=$ Total energy, $K_n=$ Kinetic energy, $V_n=$ Potential energy, $r_n=$ Radius of $n$th orbit

Options:
A
(A) R, (B) Q, (C) P, (D) Q

B
(A) S, (B) R, (C) R, (D) S

C
(A) R, (B) Q, (C) P, (D) S

D
(A) S, (B) P, (C) R, (D) Q

The correct answer is:
(A) R, (B) Q, (C) P, (D) S

Solution:
According to Bohr's theory,
Total energy is $E_n=K_n+V_n$
Kinetic energy $=K_n=\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}$
Potential energy $=V_n=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}$
$\therefore \quad E_n=\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}$
Radius of $n$th orbit $\left(r_n\right)=\frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}$ $$ E_n=-\frac{m e^4}{8 \varepsilon_0^2 h^2} \times \frac{1}{n^2} $$ (A) $\frac{V_n}{K_n}=\frac{-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}}{\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}}=-2$
Hence, $(A)$ match with $(R)$.
(B) $E_n \propto \frac{1}{n^2}$ or $E_n \propto \frac{1}{r_n}$
Radius of $n$th orbit $r_n \propto E_n^x$ $\therefore \quad x=-1$
Hence, (B) match with (Q).
(C) Angular momentum $=\frac{h}{2 \pi} \sqrt{l(l+1) l}=0,1,2, \ldots$
For the lower orbit $n=1$
$\therefore \quad l=0$ and $m=0$
Hence, angular momentum of lowest orbit $=\frac{h}{2 \pi} \sqrt{0(0 H)}=0$
(C) match with (P)
(D) $\frac{1}{r^n} \propto Z^y$ as $r_n \propto \frac{1}{Z} \therefore y=1$
Hence, (D) match with (S).

Topic: Chemistry, Structure of Atom, JEE Main
According to Bohr's theory $E_n=$ Total energy, $K_n=$ Kinetic energy, $V_n=$ Potential energy, $r_n=$ Radius of $n$th orbit

Options:
A
(A) R, (B) Q, (C) P, (D) Q

B
(A) S, (B) R, (C) R, (D) S

C
(A) R, (B) Q, (C) P, (D) S

D
(A) S, (B) P, (C) R, (D) Q

The correct answer is:
(A) R, (B) Q, (C) P, (D) S

Solution:
According to Bohr's theory,
Total energy is $E_n=K_n+V_n$
Kinetic energy $=K_n=\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}$
Potential energy $=V_n=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}$
$\therefore \quad E_n=\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}$
Radius of $n$th orbit $\left(r_n\right)=\frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}$ $$ E_n=-\frac{m e^4}{8 \varepsilon_0^2 h^2} \times \frac{1}{n^2} $$ (A) $\frac{V_n}{K_n}=\frac{-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}}{\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Z e^2}{r}}=-2$
Hence, $(A)$ match with $(R)$.
(B) $E_n \propto \frac{1}{n^2}$ or $E_n \propto \frac{1}{r_n}$
Radius of $n$th orbit $r_n \propto E_n^x$ $\therefore \quad x=-1$
Hence, (B) match with (Q).
(C) Angular momentum $=\frac{h}{2 \pi} \sqrt{l(l+1) l}=0,1,2, \ldots$
For the lower orbit $n=1$
$\therefore \quad l=0$ and $m=0$
Hence, angular momentum of lowest orbit $=\frac{h}{2 \pi} \sqrt{0(0 H)}=0$
(C) match with (P)
(D) $\frac{1}{r^n} \propto Z^y$ as $r_n \propto \frac{1}{Z} \therefore y=1$
Hence, (D) match with (S).

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2006
According to kinetic theory of gases

Options:
A
collisions are always elastic

B
heavier molecules transfer more momentum to the wall of the container

C
only a small number of molecules have very high velocity

D
between collisions, the molecules move in straight lines with constant velocities

The correct answers are:
collisions are always elastic
,
heavier molecules transfer more momentum to the wall of the container
,
between collisions, the molecules move in straight lines with constant velocities

Solution:
(a) According to a postulate of kinetic theory of gases, collision between the molecules as well as with the wall of container is perfectly elastic in nature.
(b) If a gas molecule of man ' $m$ ' moving with speed ' $u$ ' collide to the wall of container, the change in momentum is $\Delta p=-2 \mathrm{mu}$. Therefore, heavier molecule will transfer more momentum to the wall as there will be greater change in momentum of the colliding gas malecule. However this was not postulated in kinetic theory of Gases.
(c) According to Maxwell-Boltzmann distribution of molecular speed, very few molecules have either very high or very low speeds. Most of the molecules moves in a specific, intermediate speed range.
(d) According to kinetic theory of gases, a gas molecule moves in straight line unless it collide with another molecular or to the wall of container and change in momentum is observed only after collision.

Topic: Chemistry, States of Matter, JEE Advanced, JEE Advanced 2011 (Paper 1)
Advanced oogamous sexual reproduction is found in:

Options:
A Bryophytes
B Pteridophytes
C Gymnosperms
D More than one option is correct

The correct answer is:
More than one option is correct

Solution:
Oogamy is the form of sexual reproduction and it is a type of heterogamy. Also comes under the part of anisogamy. Oogamy is the fertilization process where the male gamete is received by the female and fusion occurs within the female gamete cell.
In it, the male and female gamete are not of same size and shape.
Such type of reproductive fertilization is observed in Gymnosperm, Bryophytes, and even Pteridophytes

Topic: Biology, Plant Kingdom, NEET
$\mathrm{Ag}^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; k_1=3.5 \times 10^{-3}$ $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} ; k_2=1.7 \times 10^{-3}$ then the formation constant of $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$is

Options:
A
$6.08 \times 10^{-6}$

B
$6.08 \times 10^6$

C
$6.08 \times 10^{-9}$

D
None of these

The correct answer is:
$6.08 \times 10^{-6}$

Solution:
(i) $\mathrm{Ag}^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} ; k_1=3.5 \times 10^{-3}$
(ii) $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}+\mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}, k_2=1.7 \times 10^{-3}$
on the basis of above reaction
$$
\begin{aligned}
& k_1=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_3\right]} \\
& k_2=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}}{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+}\left[\mathrm{NH}_3\right]}
\end{aligned}
$$
For the formation of $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$
$$
\mathrm{Ag}^{+}+2 \mathrm{NH}_3 \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}
$$
Formation constant $(K)=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_3\right]^2}$
From eq. (i) and (ii)
$$
K=k_1 \times k_2=3.5 \times 10^{-3} \times 1.7 \times 10^{-3}=5.95 \times 10^{-6} \approx 6.08 \times 10^{-6}
$$

Topic: Chemistry, Chemical Equilibrium, JEE Advanced, JEE Advanced 2006
$\mathrm{AgNO}_3(a q)$ was added to an aqueous $\mathrm{KCl}$ solution gradually and the conductivity of the solution was measured. The plot of conductance $(\Lambda)$ versus the volume of $\mathrm{AgNO}_3$ is

Options:
A
$P$

B
$Q$

C
$R$

D
$S$

The correct answer is:
$S$

Solution:
As $\mathrm{AgNO}_3$ is added to solution, $\mathrm{KCl}$ will be displaced accrding to following reaction
$$
\mathrm{AgNO}_3(a q)+\mathrm{KCl}(a q) \longrightarrow \operatorname{AgCl}(s)+\mathrm{KNO}_3(a q)
$$
For every mole of $\mathrm{KCl}$ displaced from solution, one mole of $\mathrm{KNO}_3$ comes in solution resulting in almost constant conductivity. As the end point is reached, added $\mathrm{AgNO}_3$ remain in solution increasing ionic concentration, hence conductivity increases.

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2011 (Paper 1)
All the compounds listed in Column I react with water. Match the result of the respective reactions with the appropriate options listed in Column II.

Options:
A
(A) P,Q, (B) P,Q,R,T, (C) P,Q,T, (D) P

B
(A) P,Q, (B) P,Q,T, (C) P,Q,R,T, (D) Q

C
(A) P,S, (B) P,Q,R,T, (C) P,Q,T, (D) P

D
(A) P,S, (B) P,Q,T, (C) P,Q,R,T, (D) Q

The correct answer is:
(A) P,S, (B) P,Q,R,T, (C) P,Q,T, (D) P

Solution:
$$
\begin{aligned}
& \quad 3 \mathrm{XeF}_4+6 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{XeO}_3+2 \mathrm{Xe}+\frac{3}{2} \mathrm{O}_2+12 \mathrm{HF} \\
& \quad \mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{HCl}+\frac{1}{2} \mathrm{O}_2 \\
& \quad \mathrm{VCl}_5+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{VOCl}_3+2 \mathrm{HCl} \\
& \text { Representative elements } \\
& \text { Factual } \\
& \text { III }
\end{aligned}
$$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
All the compounds listed in Column I react with water. Match the result of the respective reactions with the appropriate options listed in Column II.

Options:
A
(A) P,Q, (B) P,Q,R,T, (C) P,Q,T, (D) P

B
(A) P,Q, (B) P,Q,T, (C) P,Q,R,T, (D) Q

C
(A) P,S, (B) P,Q,R,T, (C) P,Q,T, (D) P

D
(A) P,S, (B) P,Q,T, (C) P,Q,R,T, (D) Q

The correct answer is:
(A) P,S, (B) P,Q,R,T, (C) P,Q,T, (D) P

Solution:
$$
\begin{aligned}
& \quad 3 \mathrm{XeF}_4+6 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{XeO}_3+2 \mathrm{Xe}+\frac{3}{2} \mathrm{O}_2+12 \mathrm{HF} \\
& \quad \mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{HCl}+\frac{1}{2} \mathrm{O}_2 \\
& \quad \mathrm{VCl}_5+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{VOCl}_3+2 \mathrm{HCl} \\
& \text { Representative elements } \\
& \text { Factual } \\
& \text { III }
\end{aligned}
$$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Advanced, JEE Advanced 2010 (Paper 2)
Alzhimer disease in humans is associated with the deficiency of

Options:
A dopamine
B glutamic acid
C acetylcholine
D Gamma Amino Butyric Acid (GABA)

The correct answer is:
acetylcholine

Solution:
Alzhimer disease in humans is associated with the deficiency of acetylcholine. Alzhimer's Disease (AD) is an irreversible, progressive disorder, in which brain cells (neurons) deteriorate, resulting in the loss of cognitive functions, primarily memory, judgement and reasoning, movement coordination and pattern recognition. In advanced stages of the disease, all memory and mental functioning may be lost.
Dopamine and glutamic acid are excitatory neurotransmitters, while Gama Amino Butyric Acid (GABA) is inhibitory neurotransmitter.

Topic: Biology, Neural Control and Coordination, NEET, NEET 2009 (Screening)
Among cellulose, polyvinyl chloride, nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest in

Options:
A
nylon

B
polyvinyl chloride

C
cellulose

D
natural rubber

The correct answer is:
natural rubber

Solution:
$$
\text { Nylon has amid linkage capable of forming intermolecular } \mathrm{H} \text {-bonding as }
$$
Due to $\mathrm{H}$-bonding, nylon has strong intermolecular attraction.
Cellulose is a polyhydroxy compound, also capable of forming strong intermolecular $\mathrm{H}$-bonding. Polyvinyl chloride is a polar polymer due to carbon chlorine bond and it possess strong dipole-dipole attraction. Natural rubber is poly (isoprene), a hydrocarbon, possess weak van der Waal's attraction.

Topic: Chemistry, Polymers, JEE Advanced, JEE Advanced 2009 (Paper 1)
Among the electrolytes $\mathrm{Na}_2 \mathrm{SO}_4, \mathrm{CaCl}_2 \quad \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$ and $\mathrm{NH}_4 \mathrm{Cl}$, the most effective coagulating agent for $\mathrm{Sb}_2 \mathrm{~S}_3$ sol is

Options:
A
$\mathrm{Na}_2 \mathrm{SO}_4$

B
$\mathrm{CaCl}_2$

C
$\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$

D
$\mathrm{NH}_4 \mathrm{Cl}$

The correct answer is:
$\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3$

Solution:
$\mathrm{Sb}_2 \mathrm{~S}_3$ is an anionic $\mathrm{sol}_2$, therefore cation of highest valency ( $\mathrm{Al}^{3+}$ in the present case) would be most effective coagulating agent.

Topic: Chemistry, Surface Chemistry, JEE Advanced, JEE Advanced 2009 (Paper 1)
Among the following complexes $(K-P), \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right](K)$,
$\left[\right.$ Co $\left.\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3(L), \mathrm{Na}_3\left[\mathrm{Co}(\mathrm{ox})_3\right](M)$,
$\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2(\mathrm{~N}) ; \mathrm{K}_2\left[\mathrm{Pt}(\mathrm{CN})_4\right](\mathrm{O})$ and $\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]\left(\mathrm{NO}_3\right)_2(P)$
the diamagnetic complexes are

Options:
A
$K, L, M, N$

B
$K, M, O, P$

C
$L, M, O, P$

D
$L, M, N, O$

The correct answer is:
$L, M, O, P$

Solution:
For a diamagnetic complex, there should not be any unpaired electron in the valence shell of central metral. In $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$, Fe (III) has $d^5$-configuration (odd electrons), hence it is paramagnetic. In [Co( $\left.\left.\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$, Co (III) has $d^6$ configuration in a strong ligand field, hence all the electrons are paired and the complex is diamagnetic. In $\mathrm{Na}_3\left[\mathrm{Co}(\mathrm{ox})_3\right]$, $\mathrm{Co}$ (III) has $d^6$-configuration and oxalate being a chelating ligand, very strong ligand and all the six electrons remains paired in lower $t_{2 g}$ level, diamagnetic. In $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_2, \mathrm{Ni}$ (II) has $3 d^8$-configuration and $\mathrm{H}_2 \mathrm{O}$ is a weak ligand hence
In $\mathrm{K}_2\left[\mathrm{Pt}(\mathrm{CN})_4\right]$, Pt (II) has $d^8$-configuration and $\mathrm{CN}^{-}$is a strong ligand, hence all the eight electrons are spin-paired. Therefore, complex is diamagnetic.
In $\left[\mathrm{Zn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]\left(\mathrm{NO}_3\right)_2, \mathrm{Zn}$ (II) has ${ }_3 \mathrm{~d}^{10}$ configuration spin paired, hence diamagnetic.

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2011 (Paper 2)
Among the following compounds, the most acidic is

Options:
A
p-nitrophenol

B
p-hydroxybenzoic acid

C
o-hydroxybenzoic acid

D
p-toluic acid

The correct answer is:
o-hydroxybenzoic acid

Solution:
A monosubstituted benzoic acid is stronger than a monosubstituted phenol as former being a carboxylic acid. Among the given substituted benzoic acid, ortho-hydroxy acid is strongest acid although - $\mathrm{OH}$ cause electron donation by resonance effect which tends to decrease acid strength. It is due to a very high stabilization of conjugate base by intramolecular $\mathrm{H}$-bond which outweight the electron donating resonance effect of $-\mathrm{OH}$.
The overall order of acid-strength of given four acids is ortho-hydro $x y$ benzoic acid $\left(p K_a=298\right)>$ Toluic acid $\left.p K_a=4.37\right)>p$-hydroxybenzoic acid $\left(p K_a=4.58\right)>$ $p$-nitrophenol $\left(p K_a=7.15\right)$.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2011 (Paper 1)
Among the following, intensive property is (properties are)

Options:
A
molar conductivity

B
electromotive force

C
resistance

D
heat capacity

The correct answers are:
molar conductivity
,
electromotive force

Solution:
Intensive property is one whose magnitude does not depend on the mass of the substance. Opposite to that is extensive property.
Energetics
Conceptual, concept analysis
III

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2010 (Paper 1)
Among the following metal carbonyls, the $\mathrm{C}-\mathrm{O}$ bond order is lowest in

Options:
A
$\left[\mathrm{MO}(\mathrm{CO})_6\right]^{+}$

B
$\left[\mathrm{Fe}(\mathrm{CO})_5\right]$

C
$\left[\mathrm{Cr}(\mathrm{CO})_6\right]$

D
$\left[\mathrm{V}(\mathrm{CO})_6\right]^{-}$

The correct answer is:
$\left[\mathrm{Fe}(\mathrm{CO})_5\right]$

Solution:
(a) $\mathrm{Mn}^{+}=3 d^5 4 s^1$ In presence of $\mathrm{CO}$ effective configuration $=3 d^6 4 s^0, 3$ lone pair for back bonding with vacant orbital of $\mathrm{C}$ in $\mathrm{CO}$.
(b) $\mathrm{Fe}^{\circ}=3 d^6 4 s^2$ in presence of $\mathrm{CO}$ effective configuration $=3 d^8, 4$ lone pair for back bonding with $\mathrm{CO}$.
(c) $\mathrm{Cr}^{\circ}=3 d^5 4 s^1$, effective configuration $=3 d^6$
3 lone pair for back bonding with $\mathrm{CO}$.
(d) $\mathrm{V}^{-}=3 d^4 s^2$, effective configuration $=3 d^6, 3$ lone pair for back bonding with $\mathrm{CO}$. Maximum back bonding in $\mathrm{Fe}(\mathrm{CO})_5$, therefore $\mathrm{CO}$ bond order is lowest here.

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2007 (Paper 2)