datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers $+2,+4$ and $+6 . \mathrm{XeF}_4$ reacts violently with water to give $\mathrm{XeO}_3$. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.Question:
Argon is used in are welding because of its

Options:
A
low reactivity with metal

B
ability to lower the melting point of metal

C
flammability

D
high calorific value

The correct answer is:
low reactivity with metal

Solution:
Argon is used in arc welding to cause the inert atmosphere around without interaction with metal.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers $+2,+4$ and $+6 . \mathrm{XeF}_4$ reacts violently with water to give $\mathrm{XeO}_3$. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.Question:
The structure of $\mathrm{XeO}_3$ is

Options:
A
linear

B
planar

C
pyramidal

D
$T$-shaped

The correct answer is:
pyramidal

Solution:
$\mathrm{XeO}_3$ total $e^{-}$pairs $=4(3 p b+1 l p)$ $s p^3$-hybridisation $\Rightarrow$ pyramidal shape.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers $+2,+4$ and $+6 . \mathrm{XeF}_4$ reacts violently with water to give $\mathrm{XeO}_3$. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.Question:
$\mathrm{XeF}_4$ and $\mathrm{XeF}_6$ are expected to be

Options:
A
oxidising

B
reducing

C
unreactive

D
strongly basic

The correct answer is:
oxidising

Solution:
$\mathrm{XeF}_4$ oxidises potassium iodide
$$
\mathrm{XeF}_4+4 \mathrm{I}^{-} \longrightarrow 2 \mathrm{I}_2+4 \mathrm{~F}^{-1}+\mathrm{Xe}
$$
$\mathrm{XeF}_6$ oxidises hydrogen like other xenon fluorides.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
The nuclear charge $(\mathrm{Ze})$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho(r)$ [charge per unit volume] is dependent only on the radial distance $r$ from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.
Question:
The electric field within the nucleus is generally observed to be linearly dependent on $r$. This implies

Options:
A
$a=0$

B
$a=\frac{R}{2}$

C
$a=R$

D
$a=\frac{2 R}{3}$

The correct answer is:
$a=R$

Solution:
In case of solid sphere of charge of uniform volume density.
$$
E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^3} \cdot r \text { or } E \propto r
$$
Thus, for $E$ to be linearly dependent on $r$, volume charge density should be constant.
or $\quad a=R$
$\therefore$ correct option is (c).

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
The nuclear charge $(\mathrm{Ze})$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho(r)$ [charge per unit volume] is dependent only on the radial distance $r$ from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.
Question:
The electric field at $r=R$ is

Options:
A
independent of $a$

B
directly proportional to $a$

C
directly proportional to $a^2$

D
inversely proportional to $a$

The correct answer is:
independent of $a$

Solution:
At $r=R$, from Gauss's law
or
$$
\begin{aligned}
E\left(4 \pi R^2\right) & =\frac{q_{\text {net }}}{\varepsilon_0}=\frac{Z e}{\varepsilon_0} \\
E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e}{R^2}
\end{aligned}
$$
$E$ is independent of ' $a$ '
$\therefore$ correct option is (a).

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
The nuclear charge $(\mathrm{Ze})$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho(r)$ [charge per unit volume] is dependent only on the radial distance $r$ from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.
Question:
The electric field at $r=R$ is

Options:
A
independent of $a$

B
directly proportional to $a$

C
directly proportional to $a^2$

D
inversely proportional to $a$

The correct answer is:
independent of $a$

Solution:
At $r=R$, from Gauss's law
or
$$
\begin{aligned}
E\left(4 \pi R^2\right) & =\frac{q_{\text {net }}}{\varepsilon_0}=\frac{Z e}{\varepsilon_0} \\
E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e}{R^2}
\end{aligned}
$$
$E$ is independent of ' $a$ '
$\therefore$ correct option is (a).

Topic: Physics, Electrostatics, JEE Main
Paragraph:
The nuclear charge $(\mathrm{Ze})$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho(r)$ [charge per unit volume] is dependent only on the radial distance $r$ from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.
Question:
For $a=0$, the value of $d$ (maximum value of $\rho$ as shown in the figure) is

Options:
A
$\frac{3 Z e}{4 \pi R^3}$

B
$\frac{3 Z e}{\pi R^3}$

C
$\frac{4 Z e}{3 \pi R^3}$

D
$\frac{Z e}{3 \pi R^3}$

The correct answer is:
$\frac{3 Z e}{\pi R^3}$

Solution:
For $a=0$
$$
\begin{aligned}
& \rho(r)=\left(-\frac{d}{R} \cdot r+d\right) \\
& \text { Now } \int_0^R\left(4 \pi r^2\right)\left(d-\frac{d}{R} r\right) d r=\text { net charge }=Z e
\end{aligned}
$$
Solving this equation, we get $d=\frac{3 Z e}{\pi R^3}$

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
There are $n$ urns each containing $(n+1)$ balls such that the ith urn contains $i$ white balls and $(n+1-i)$ red balls. Let $u_i$ be the event of selecting ith urn, $i=1,2,3, \ldots, n$ and $W$ denotes the event of getting a white balls.Question:
If $P\left(u_i\right)=c$, where $c$ is a constant, then $P\left(u_i / W\right)$ is equal to

Options:
A
$\frac{2}{n+1}$

B
$\frac{1}{n+1}$

C
$\frac{n}{n+1}$

D
$\frac{1}{2}$

The correct answer is:
$\frac{2}{n+1}$

Solution:
$P\left(\frac{u_i}{W}\right)=\frac{\frac{n}{n+1}}{\frac{\Sigma i}{n+1}}=\frac{2}{n+1}$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2006
Paragraph:
There are $n$ urns each containing $(n+1)$ balls such that the ith urn contains $i$ white balls and $(n+1-i)$ red balls. Let $u_i$ be the event of selecting ith urn, $i=1,2,3, \ldots, n$ and $W$ denotes the event of getting a white balls.Question:
If $n$ is even and $E$ denotes the event of choosing even numbered urn $\left(P\left(u_i\right)=\frac{1}{n}\right)$, then the value of $P(W / E)$ is

Options:
A
$\frac{n+2}{2 n+1}$

B
$\frac{n+2}{2(n+1)}$

C
$\frac{n}{n+1}$

D
$\frac{1}{n+1}$

The correct answer is:
$\frac{n+2}{2(n+1)}$

Solution:
$P(W / E)=\frac{2+4+6+\ldots}{\frac{n(n+1)}{2}}=\frac{n+2}{2(n+1)}$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2006
Paragraph:
There are $n$ urns each containing $(n+1)$ balls such that the ith urn contains $i$ white balls and $(n+1-i)$ red balls. Let $u_i$ be the event of selecting ith urn, $i=1,2,3, \ldots, n$ and $W$ denotes the event of getting a white balls.Question:
If $P\left(u_i\right) \propto i$, where $i=1,2,3, \ldots, n$, then $\lim _{n \rightarrow \infty} P(W)$ is equal to

Options:
A
1

B
$\frac{2}{3}$

C
$\frac{1}{4}$

D
$\frac{3}{4}$

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

Solution:

Topic: Mathematics, Probability, JEE Main
Paragraph:
There are $n$ urns each containing $(n+1)$ balls such that the ith urn contains $i$ white balls and $(n+1-i)$ red balls. Let $u_i$ be the event of selecting ith urn, $i=1,2,3, \ldots, n$ and $W$ denotes the event of getting a white balls.Question:
If $P\left(u_i\right) \propto i$, where $i=1,2,3, \ldots, n$, then $\lim _{n \rightarrow \infty} P(W)$ is equal to

Options:
A
1

B
$\frac{2}{3}$

C
$\frac{1}{4}$

D
$\frac{3}{4}$

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

Solution:

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2006
Paragraph:
There are some deposits of nitrates and phosphates in earth's crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridisation easily explains the ease of sigma donation capability of $\mathrm{NH}_3$ and $\mathrm{PH}_3$. Phosphine is a flammable gas and is prepared from white phosphorus.Question:
Among the following, the correct statement is

Options:
A
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{NH}_3$ is a better electron donor because the lone pair of electrons occupies spherical 's' orbital and is less directional

B
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{PH}_3$ is a better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$ orbital and is more directional

C
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{NH}_3$ is a better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$ orbital and is more directional

D
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{PH}_3$ is a better electron donor because the lone pair of electrons occupies spherical ' $s$ ' orbital and is less directional

The correct answer is:
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{NH}_3$ is a better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$ orbital and is more directional

Solution:
$\mathrm{NH}_3$ is better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$-orbital and is more directional.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
There are some deposits of nitrates and phosphates in earth's crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridisation easily explains the ease of sigma donation capability of $\mathrm{NH}_3$ and $\mathrm{PH}_3$. Phosphine is a flammable gas and is prepared from white phosphorus.Question:
White phosphorus on reaction with $\mathrm{NaOH}$ gives $\mathrm{PH}_3$ as one of the products. This is a

Options:
A
dimerisation reaction

B
disproportionation reaction

C
condensation reaction

D
precipitation reaction

The correct answer is:
disproportionation reaction

Solution:
White phosphorus on reaction with $\mathrm{NaOH}$ gives $\mathrm{PH}_3$ as one of the product in disproportionation reaction
$$
\mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{NaH}_2 \mathrm{PO}_2+\mathrm{PH}_3
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
There are some deposits of nitrates and phosphates in earth's crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridisation easily explains the ease of sigma donation capability of $\mathrm{NH}_3$ and $\mathrm{PH}_3$. Phosphine is a flammable gas and is prepared from white phosphorus.Question:
Among the following, the correct statement is

Options:
A
phosphates have no biological significance in humans

B
between nitrates and phosphates, phosphates are less abundant in earth's crust

C
between nitrates and phosphates, nitrates are less abundant in earth's crust

D
oxidation of nitrates is possible in soil

The correct answer is:
between nitrates and phosphates, nitrates are less abundant in earth's crust

Solution:
Due to greater solubility and nature to be prone to microbial action, nitrates are less abundant in earth's crust.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Tollen's reagent is used for the detection of aldehyde when a solution of $\mathrm{AgNO}_3$ is added to glucose with $\mathrm{NH}_4 \mathrm{OH}$ then gluconic acid is formed
$\begin{array}{rr}\mathrm{Ag}^{+}+e^{-} \rightarrow \mathrm{Ag} & E_{\text {red }}^{\circ}=0.8 \mathrm{~V} \\ \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow \text { Gluconic acid }\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7\right)+2 \mathrm{H}^{+}+2 e^{-} ; & -E_{\mathrm{red}}^{\circ}=-0.05 \mathrm{~V} \\ \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \rightarrow \mathrm{Ag}(s)+2 \mathrm{NH}_3 ; & E_{\mathrm{red}}^{\circ}=0.337 \mathrm{~V}\end{array}$
[Use $2.303 \times \frac{R T}{F}=0.0592$ and $\frac{F}{R T}=38.92$ at $298 \mathrm{~K}$ ]Question:
When ammonia is added to the solution, $\mathrm{pH}$ is raised to 11 . Which half-cell reaction is affected by $\mathrm{pH}$ and by how much?

Options:
A
$E_{\text {oxd }}$ will increase by a factor of $0.65$ from $E_{\text {oxd }}^{\circ}$

B
$E_{\text {oxd }}$ will decrease by a factor of $0.65$ from $E_{\text {oxd }}^{\circ}$

C
$E_{\text {red }}$ will increase by a factor of $0.65$ from $E_{\text {red }}^{\circ}$

D
$E_{\text {red }}$ will decrease by a factor of $0.65$ from $E_{\text {red }}^{\circ}$

The correct answer is:
$E_{\text {red }}$ will increase by a factor of $0.65$ from $E_{\text {red }}^{\circ}$

Solution:
On increasing concentration of $\mathrm{NH}_3$, the concentration of $\mathrm{H}^{+}$ion decreases, therefore
$$
\begin{aligned}
E_{\mathrm{red}} & =E_{\mathrm{red}}^{\circ}-\frac{0.0592}{n} \log _{10}\left[\mathrm{H}^{+}\right]^2 \\
& =0-\frac{0.0592}{2} \times 2 \log _{10} 10^{-11} \quad\left(\because \mathrm{pH}=11 ; \therefore\left[\mathrm{H}^{+}\right]=10^{-11} \mathrm{M}\right) \\
& =-0.0592 \times 11=0.6512 \mathrm{~V}=0.65 \mathrm{~V}
\end{aligned}
$$
$E_{\text {red }}$ increased by a factor of $0.65$ from $E_{\text {red }}^{\circ}$.

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2006
Paragraph:
Tollen's reagent is used for the detection of aldehyde when a solution of $\mathrm{AgNO}_3$ is added to glucose with $\mathrm{NH}_4 \mathrm{OH}$ then gluconic acid is formed
$\begin{array}{rr}\mathrm{Ag}^{+}+e^{-} \rightarrow \mathrm{Ag} & E_{\text {red }}^{\circ}=0.8 \mathrm{~V} \\ \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow \text { Gluconic acid }\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7\right)+2 \mathrm{H}^{+}+2 e^{-} ; & -E_{\mathrm{red}}^{\circ}=-0.05 \mathrm{~V} \\ \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \rightarrow \mathrm{Ag}(s)+2 \mathrm{NH}_3 ; & E_{\mathrm{red}}^{\circ}=0.337 \mathrm{~V}\end{array}$
[Use $2.303 \times \frac{R T}{F}=0.0592$ and $\frac{F}{R T}=38.92$ at $298 \mathrm{~K}$ ]Question:
Ammonia is always added in this reaction. Which of the following must be incorrect?

Options:
A
$\mathrm{NH}_3$ combines with $\mathrm{Ag}^{+}$to form a complex

B
$\mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}$is a stronger oxidising reagent than $\mathrm{Ag}^{+}$.

C
In absence of $\mathrm{NH}_3$ silver salt of gluconic acid is formed.

D
$\mathrm{NH}_3$ has affected the standard reduction potential of glucose/gluconic acid electrode.

The correct answer is:
$\mathrm{NH}_3$ has affected the standard reduction potential of glucose/gluconic acid electrode.

Solution:
Ammonia has no effect on the standard reduction potential.

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2006
Paragraph:
Tollen's reagent is used for the detection of aldehyde when a solution of $\mathrm{AgNO}_3$ is added to glucose with $\mathrm{NH}_4 \mathrm{OH}$ then gluconic acid is formed
$\begin{array}{rr}\mathrm{Ag}^{+}+e^{-} \rightarrow \mathrm{Ag} & E_{\text {red }}^{\circ}=0.8 \mathrm{~V} \\ \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow \text { Gluconic acid }\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7\right)+2 \mathrm{H}^{+}+2 e^{-} ; & -E_{\mathrm{red}}^{\circ}=-0.05 \mathrm{~V} \\ \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \rightarrow \mathrm{Ag}(s)+2 \mathrm{NH}_3 ; & E_{\mathrm{red}}^{\circ}=0.337 \mathrm{~V}\end{array}$
[Use $2.303 \times \frac{R T}{F}=0.0592$ and $\frac{F}{R T}=38.92$ at $298 \mathrm{~K}$ ]Question:
$2 \mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_4+2 \mathrm{H}^{+}$. Find $\ln \mathrm{K}$ of this reaction.

Options:
A
$66.13$

B
$58.38$

C
$28.30$

D
$46.29$

The correct answer is:
$58.38$

Solution:
Cell reaction,
(i) $\mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag}(\mathrm{s}), E_{\mathrm{red}}^{\circ}=0.8 \mathrm{~V}$
(ii) $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7+2 \mathrm{H}^{+}+2 e^{-}, E_{\text {oxid }}=-0.05 \mathrm{~V}$
Hence, for reaction
$$
\begin{aligned}
2 \mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} & \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O} 7+2 \mathrm{H}^{+} \\
E_{\text {cell }}^{\circ} & =(0.8-0.05)=0.75 \mathrm{~V} \\
E_{\text {cell }}^{\circ} & =E_{\text {anode }}^{\circ}-E_{\text {cathode }}^{\circ}=[-0.05-(-0.8)]_{\mathrm{V}}
\end{aligned}
$$
On the basis of oxidised electrode potential $=0.75 \mathrm{~V}$
By Nernst equation, $E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log _{10} K$
At equilibrium, $E_{\text {cell }}=0$
$$
\begin{aligned}
E_{\text {cell }}^{\circ} & =\frac{2.303 R T}{n F} \log _{10} K \\
0.75 & =\frac{0.0592}{2} \log _{10} K \\
\log _{10} K & =\frac{2 \times 0.75}{0.0592}=25.33
\end{aligned}
$$
or In $K=2.303 \times \log _{10} K=2.303 \times 25.33=58.35$

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2006
Paragraph:
Two aliphatic aldehydes $\mathrm{P}$ and $\mathrm{Q}$ react in the presence of aqueous $\mathrm{K}_2 \mathrm{CO}_3$ to give compound R, which upon treatment with HCN provides compound $\mathrm{S}$. On acidification and heating, S gives the product shown below :
Question:
$$
\text { The compound } R \text { is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
Some inferences from above data
- $a q \mathrm{~K}_2 \mathrm{CO}_3$ can facilitate aldol condensation to aliphatic aldehydes.
- HCN can lead to cyanohydrin formation
- $\mathrm{H}^{+}, \Delta$ can lead to hydrolysis of cyanide to acid and following esterification. So $S$ is cyanohydrin.
Carbonyl compounds (Aldol condensation) Conceptual reaction and mechanistic understanding IV

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
Two aliphatic aldehydes $\mathrm{P}$ and $\mathrm{Q}$ react in the presence of aqueous $\mathrm{K}_2 \mathrm{CO}_3$ to give compound R, which upon treatment with HCN provides compound $\mathrm{S}$. On acidification and heating, S gives the product shown below :
Question:
The compounds $P$ and $Q$ respectively are

Options:
A

B

C

D

The correct answer is:

Solution:
Some inferences from above data
- $a q \mathrm{~K}_2 \mathrm{CO}_3$ can facilitate aldol condensation to aliphatic aldehydes.
- HCN can lead to cyanohydrin formation
- $\mathrm{H}^{+}, \Delta$ can lead to hydrolysis of cyanide to acid and following esterification. So $S$ is cyanohydrin.
Carbonyl compounds (Aldol condensation) Conceptual reaction and mechanistic understanding IV

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
Two aliphatic aldehydes $\mathrm{P}$ and $\mathrm{Q}$ react in the presence of aqueous $\mathrm{K}_2 \mathrm{CO}_3$ to give compound R, which upon treatment with HCN provides compound $\mathrm{S}$. On acidification and heating, S gives the product shown below :
Question:
The compounds $P$ and $Q$ respectively are

Options:
A

B

C

D

The correct answer is:

Solution:
Some inferences from above data
- $a q \mathrm{~K}_2 \mathrm{CO}_3$ can facilitate aldol condensation to aliphatic aldehydes.
- HCN can lead to cyanohydrin formation
- $\mathrm{H}^{+}, \Delta$ can lead to hydrolysis of cyanide to acid and following esterification. So $S$ is cyanohydrin.
Carbonyl compounds (Aldol condensation) Conceptual reaction and mechanistic understanding IV

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Paragraph:
Two aliphatic aldehydes $\mathrm{P}$ and $\mathrm{Q}$ react in the presence of aqueous $\mathrm{K}_2 \mathrm{CO}_3$ to give compound R, which upon treatment with HCN provides compound $\mathrm{S}$. On acidification and heating, S gives the product shown below :
Question:
$$
\text { The compound } S \text { is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
Some inferences from above data
- $a q \mathrm{~K}_2 \mathrm{CO}_3$ can facilitate aldol condensation to aliphatic aldehydes.
- HCN can lead to cyanohydrin formation
- $\mathrm{H}^{+}, \Delta$ can lead to hydrolysis of cyanide to acid and following esterification. So $S$ is cyanohydrin.
Carbonyl compounds (Aldol condensation) Conceptual reaction and mechanistic understanding IV

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
Two plane harmonic sound waves are expressed by the equations.
$$
\begin{aligned}
& y_1(x, t)=A \cos (0.5 \pi x-100 \pi t) \\
& y_2(x, t)=A \cos (0.46 \pi x-92 \pi t)
\end{aligned}
$$
(All parameters are in MKS)Question:
At $x=0$ how many times the amplitude of $y_1+y_2$ is zero in one second at $x=0$ ?

Options:
A
192

B
48

C
100

D
96

The correct answer is:
96

Solution:
At $x=0, y=y_1+y_2=2 A \cos 96 \pi t \cos 4 \pi t$
Frequency of $\cos (96 \pi t)$ function is $48 \mathrm{~Hz}$ and that of $\cos (4 \pi t)$ function is $2 \mathrm{~Hz}$. In one second cos function becomes zero at $2 f$ times, where $f$ is the frequency. Therefore, first function will become zero at 96 times and the second at 4 times. But second overlaps with first. Hence, net $y$ will become zero 96 times in 1 second.

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2006
Paragraph:
Two plane harmonic sound waves are expressed by the equations.
$$
\begin{aligned}
& y_1(x, t)=A \cos (0.5 \pi x-100 \pi t) \\
& y_2(x, t)=A \cos (0.46 \pi x-92 \pi t)
\end{aligned}
$$
(All parameters are in MKS)Question:
What is the speed of the sound?

Options:
A
$200 \mathrm{~m} / \mathrm{s}$

B
$180 \mathrm{~m} / \mathrm{s}$

C
$192 \mathrm{~m} / \mathrm{s}$

D
$96 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$200 \mathrm{~m} / \mathrm{s}$

Solution:
Speed of wave $v=\frac{\omega}{R}$ or $\quad v=\frac{100 \pi}{0.5 \pi}$ or $\frac{92 \pi}{0.46 \pi}=200 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2006
Paragraph:
Two plane harmonic sound waves are expressed by the equations.
$$
\begin{aligned}
& y_1(x, t)=A \cos (0.5 \pi x-100 \pi t) \\
& y_2(x, t)=A \cos (0.46 \pi x-92 \pi t)
\end{aligned}
$$
(All parameters are in MKS)Question:
How many times does an observer hear maximum intensity in one second?

Options:
A
4

B
10

C
6

D
8

The correct answer is:
4

Solution:
In one second number of maximas is called the beat frequency. Hence,
$$
f_b=f_1-f_2=\frac{100 \pi}{2 \pi}-\frac{92 \pi}{2 \pi}=4 \mathrm{~Hz}
$$

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2006