datavorous/entrance-exam-dataset huggingface dataset:
Native silver metal forms a water soluble complex with a dilute aqueous solution of $\mathrm{NaCN}$ in the presence of

Options:
A
nitrogen

B
oxygen

C
carbon dioxide

D
argon

The correct answer is:
oxygen

Solution:
A water soluble complex of silver with a dilute aqueous solution of $\mathrm{NaCN}$ is sodium argentocyanide. In the cyanide process, the native form is crushed and treated with $0.1-0.2 \%$ solution of $\mathrm{NaCN}$ and aerated.
$$
4 \mathrm{Ag}+8 \mathrm{NaCN}+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \longrightarrow 4 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+4 \mathrm{NaOH}
$$
Argentocyanide is soluble. Further, metals are recovered from the complex by reduction with zinc

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2008 (Paper 1)
$\mathrm{NiCl}_{2}\left\{\mathrm{P}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)\right\}_{2}$ exhibits temperature depend-ent magnetic behaviour (paramagnetic/diamagnetic). The coordination geometries of $\mathrm{Ni}^{2+}$ in the paramagnetic and diamagnetic states are respectively

Options:
A tetrahedral and tetrahedral
B square planar and square planar
C tetrahedral and square planar
D square planar and tetrahedral

The correct answer is:
tetrahedral and square planar

Solution:
In both states (paramagnetic and diamagnetic) of the given complex, $\mathrm{Ni}$ exists as $\mathrm{Ni}^{2+}$ whose electronic configuration is $[\mathrm{Ar}] 3 d^{8} 4 s^{0}$.

In the above paramagnetic state the geometry of the complex is $s p^{3}$ giving tetrahedral geometry.

The diamagnetic state is achieved by pairing of electrons in $3 d$ orbital.

Thus, the geometry of the complex will be $d s p^{2}$ giving square planar geometry.

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2012 (Paper 2)
One Indian and four American men and their wives are to be seated randomly around a circular table. Then, the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife, is

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

B
$\frac{1}{3}$

C
$\frac{2}{5}$

D
$\frac{1}{5}$

The correct answer is:
$\frac{2}{5}$

Solution:
Let $E=$ event when each American man is seated adjacent to his wife and $A=$ event when Indian man is seated adjacent to his wife.
Now, $n(A \cap E)=(4 !) \times(2 !)^5$
Even when each American man is seated adjacent to his wife.
Again
$$
\begin{aligned}
n(E) & =(5 !) \times(2 !)^4 \\
P\left(\frac{A}{E}\right) & =\frac{n(A \cap E)}{n(E)} \\
& =\frac{(4 !) \times(2 !)^5}{(5 !) \times(2 !)^4}=\frac{2}{5}
\end{aligned}
$$
$$
\Rightarrow \quad P\left(\frac{A}{E}\right)=\frac{n(A \cap E)}{n(E)}
$$
ALITER
Fixing four American couples and one Indian man in between any two couples, we have 5 different ways in which his wife can be seated, of which 2 cases are favourable. $\therefore$ Required probability $=\frac{2}{5}$.

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2007 (Paper 1)
One mole of a monatomic ideal gas is taken through a cycle $A B C D A$ as shown in the $p$-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.

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

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

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

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

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

Solution:
Internal energy $\propto T \propto p V$
This is because
$$
U=\frac{n f}{2} R T=\frac{f}{2} p V
$$
Here, $n=$ number of moles
$f=$ degree of freedom
If the product $p V$ increases, then internal energy will increase and if product decreases, the internal energy will decrease.
Further, work is done on the gas, if volume of gas decreases. For heat exchange.
$$
Q=W+\Delta U
$$
Work done is area under $p-V$ graph. If volume increases work done by gas is positive and if volume decreases work done by gas is negative. Further $\Delta U$ is positive if product of $p V$ is increasing and $N U$ is negative, if product of $p V$ is decreasing. If heat is taken by the gas $Q$ is positive and if heat is lost by the gas $Q$ is negative.
Keeping the above points in mind the answer to this question is as under.
(A) $\rightarrow(\mathrm{p}, \mathrm{r}, \mathrm{t})$
(B) $\rightarrow(\mathrm{p}, \mathrm{r})$
(C) $\rightarrow$ (q, s)
(D) $\rightarrow(\mathrm{r}, \mathrm{t})$
Analysis of Question
(i) Calculation wise, question is slightly lengthy. Otherwise question is theory based and simple.
(ii) In process $D A$,
$\begin{array}{rlrl} & & p_A V_A & =p_D V_D \\ \therefore & & T_A & =T_D \\ \text { or } & \Delta U & =0\end{array}$
Further, volume of gas is decreasing. Therefore, work is done on the gas or work done by gas is negative. Therefore, $Q$ is negative or heat is lost.
(iii) This question covers almost all the concepts of first law of thermodynamics.

Topic: Physics, Thermodynamics, JEE Main
One mole of a monatomic ideal gas is taken through a cycle $A B C D A$ as shown in the $p$-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.

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

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

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

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

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

Solution:
Internal energy $\propto T \propto p V$
This is because
$$
U=\frac{n f}{2} R T=\frac{f}{2} p V
$$
Here, $n=$ number of moles
$f=$ degree of freedom
If the product $p V$ increases, then internal energy will increase and if product decreases, the internal energy will decrease.
Further, work is done on the gas, if volume of gas decreases. For heat exchange.
$$
Q=W+\Delta U
$$
Work done is area under $p-V$ graph. If volume increases work done by gas is positive and if volume decreases work done by gas is negative. Further $\Delta U$ is positive if product of $p V$ is increasing and $N U$ is negative, if product of $p V$ is decreasing. If heat is taken by the gas $Q$ is positive and if heat is lost by the gas $Q$ is negative.
Keeping the above points in mind the answer to this question is as under.
(A) $\rightarrow(\mathrm{p}, \mathrm{r}, \mathrm{t})$
(B) $\rightarrow(\mathrm{p}, \mathrm{r})$
(C) $\rightarrow$ (q, s)
(D) $\rightarrow(\mathrm{r}, \mathrm{t})$
Analysis of Question
(i) Calculation wise, question is slightly lengthy. Otherwise question is theory based and simple.
(ii) In process $D A$,
$\begin{array}{rlrl} & & p_A V_A & =p_D V_D \\ \therefore & & T_A & =T_D \\ \text { or } & \Delta U & =0\end{array}$
Further, volume of gas is decreasing. Therefore, work is done on the gas or work done by gas is negative. Therefore, $Q$ is negative or heat is lost.
(iii) This question covers almost all the concepts of first law of thermodynamics.

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2011 (Paper 2)
One mole of an ideal gas in initial state A undergoes a cyclic process $A B C A$, as shown in the figure. Its pressure at $A$ is $p_0$. Choose the correct option(s) from the following

Options:
A
Internal energies at $A$ and $B$ are the same

B
Work done by the gas in process $A B$ is $p_0 V_0 \ln 4$

C
Pressure at $C$ is $\frac{p_0}{4}$

D
Temperature at $C$ is $\frac{T_0}{4}$

The correct answers are:
Internal energies at $A$ and $B$ are the same
,
Work done by the gas in process $A B$ is $p_0 V_0 \ln 4$

Solution:
$$
\begin{aligned}
& T_A=T_B \quad \therefore U_A=U_B \\
& W_{A B}=(1)(R) T_0 \ln \left(\frac{V_f}{V_i}\right) \\
& =R T_0 \ln \left(\frac{4 V_0}{V_0}\right) \\
& =p_0 V_0 \ln (4)
\end{aligned}
$$
Information regarding $p$ and $T$ at $C$ can not be obtained from the given graph. Unless it is mentioned that line $B C$ passes through origin or not.
Hence, the correct options are (a) and (b).

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2010 (Paper 1)
One mole of an ideal gas is taken from $a$ to $b$ along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is $W_s$ and that along the dotted line path is $W_d$, then the integer closest to the ratio $\frac{W_d}{W_s}$ is

Options:

The correct answer is:
2

Solution:
Solid line represents reversible isothermal process.
So, work $W_s=-4 \times 0.5 \ln \left(\frac{5.5}{0.5}\right)=-2 \ln 11 \mathrm{~L}-\mathrm{atm}$
Dotted line represents irreversible process
So, work
$$
W_d=-\left\{4 \times 1.5+1 \times 1+\frac{2}{3} \times 2.5\right\}=-\left\{6+1+\frac{5}{3}\right\} \mathrm{L}-\mathrm{atm}=-\frac{26}{3} \mathrm{~L}-\mathrm{atm}
$$
So $\quad \frac{W_d}{W_s}=\frac{26}{3 \times 2 \ln 11} \approx 2$
Energetics
Conceptual
III

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2010 (Paper 2)
Oogamous type of sexual reproduction is found in

Options:
A Chlamydomonas
B Volvox
C Spirogyra
D Both (A) and (B)

The correct answer is:
Volvox

Solution:
Oogamy is a form of anisogamy (heterogamy) in which the female gamete (e.g. egg cell) is significantly larger than the male gamete and is non-motile. Oogamy is the most advanced form of reproduction, with the egg cell or oogonium retained and fertilized on the parent plant.

Topic: Biology, Plant Kingdom, NEET
Oxidation states of the metal in the minerals haematite and magnetite, respectively, are

Options:
A
II, III in haematite and III in magnetite

B
II, III in haematite and II in magnetite

C
II in haematite and II, III in magnetite

D
III in haematite and II, III in magnetite

The correct answer is:
III in haematite and II, III in magnetite

Solution:
Haematite is $\mathrm{Fe}_2 \mathrm{O}_3$, in which oxidation number of iron is III. Magnetite is $\mathrm{Fe}_3 \mathrm{O}_4$ which is infact a mixed oxide (FeO $\cdot \mathrm{Fe}_2 \mathrm{O}_3$ ) hence, iron is present in both II and III oxidation state.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2011 (Paper 2)
Paragraph:
A carbonyl compound $P$, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin $Q$. Ozonolysis of $Q$ leads to a dicarbonyl compound $R$, which undergoes intramolecular aldol reaction to give predominantly $S$.
$$
P \underset{\text { (ii) } \mathrm{H}^{+}, \mathrm{H}_2 \mathrm{O}}{\stackrel{\text { (i) } \mathrm{MeMgBr}}{\longrightarrow}} \mathrm{Q} \underset{\text { Heat }}{\stackrel{\mathrm{O}_3 / \mathrm{Zn}-\mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \mathrm{P}
$$
(iii) $\mathrm{H}_2 \mathrm{SO}_4$ /HeatQuestion:
$$
\text { The structure of the carbonyl compound } P \text {, is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
A carbonyl compound $P$, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin $Q$. Ozonolysis of $Q$ leads to a dicarbonyl compound $R$, which undergoes intramolecular aldol reaction to give predominantly $S$.
$$
P \underset{\text { (ii) } \mathrm{H}^{+}, \mathrm{H}_2 \mathrm{O}}{\stackrel{\text { (i) } \mathrm{MeMgBr}}{\longrightarrow}} \mathrm{Q} \underset{\text { Heat }}{\stackrel{\mathrm{O}_3 / \mathrm{Zn}-\mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \mathrm{P}
$$
(iii) $\mathrm{H}_2 \mathrm{SO}_4$ /HeatQuestion:
$$
\text { The structures of the products } Q \text { and } R \text {, respectively, are }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
A carbonyl compound $P$, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin $Q$. Ozonolysis of $Q$ leads to a dicarbonyl compound $R$, which undergoes intramolecular aldol reaction to give predominantly $S$.
$$
P \underset{\text { (ii) } \mathrm{H}^{+}, \mathrm{H}_2 \mathrm{O}}{\stackrel{\text { (i) } \mathrm{MeMgBr}}{\longrightarrow}} \mathrm{Q} \underset{\text { Heat }}{\stackrel{\mathrm{O}_3 / \mathrm{Zn}-\mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \mathrm{P}
$$
(iii) $\mathrm{H}_2 \mathrm{SO}_4$ /HeatQuestion:
$$
\text { The structures of the products } Q \text { and } R \text {, respectively, are }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Paragraph:
A carbonyl compound $P$, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin $Q$. Ozonolysis of $Q$ leads to a dicarbonyl compound $R$, which undergoes intramolecular aldol reaction to give predominantly $S$.
$$
P \underset{\text { (ii) } \mathrm{H}^{+}, \mathrm{H}_2 \mathrm{O}}{\stackrel{\text { (i) } \mathrm{MeMgBr}}{\longrightarrow}} \mathrm{Q} \underset{\text { Heat }}{\stackrel{\mathrm{O}_3 / \mathrm{Zn}-\mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \mathrm{P}
$$
(iii) $\mathrm{H}_2 \mathrm{SO}_4$ /HeatQuestion:
$$
\text { The structure of the product } S \text {, is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$ respectively. The line $P Q$ is given by the equation
$\sqrt{3} x+y-6=0$ and the point $D$ is $\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$.Question:
Equations of the sides $Q R, R P$ are

Options:
A
$y=\frac{2}{\sqrt{3}} x+1, y=-\frac{2}{\sqrt{3}} x-1$

B
$y=\frac{1}{\sqrt{3}} x, y=0$

C
$y=\frac{\sqrt{3}}{2} x+1, y=-\frac{\sqrt{3}}{2} x-1$

D
$y=\sqrt{3} x, y=0$

The correct answer is:
$y=\sqrt{3} x, y=0$

Solution:
Clearly, point $E$ and $F$ satisfy the equations given in option (d).

Topic: Mathematics, Straight Lines, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$ respectively. The line $P Q$ is given by the equation
$\sqrt{3} x+y-6=0$ and the point $D$ is $\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$.Question:
Points $E$ and $F$ are given by

Options:
A
$\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$

B
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right),(\sqrt{3}, 0)$

C
$\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

D
$\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

The correct answer is:
$\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$

Solution:
Slope of line joining centre of circle to point $D$
$$
=\frac{\frac{3}{2}-1}{\frac{3 \sqrt{3}}{2}-\sqrt{3}}=\frac{1}{\sqrt{3}}
$$
[makes an angle $30^{\circ}$ with $X$-axis]
$\therefore$ Points $E$ and $F$ will make angle $150^{\circ}$ and $-90^{\circ}$ with $X$-axis.
$\therefore E$ and $F$ are given by and
$\therefore$
$$
\begin{aligned}
\frac{x-\sqrt{3}}{\cos 150^{\circ}} & =\frac{y-1}{\sin 150^{\circ}}=1 \\
\frac{x-\sqrt{3}}{\cos \left(-90^{\circ}\right)} & =\frac{y-1}{\sin \left(-90^{\circ}\right)}=1 \\
E & =\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right) \\
and
F & =(\sqrt{3}, 0)
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$ respectively. The line $P Q$ is given by the equation
$\sqrt{3} x+y-6=0$ and the point $D$ is $\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$.Question:
Points $E$ and $F$ are given by

Options:
A
$\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$

B
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right),(\sqrt{3}, 0)$

C
$\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

D
$\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

The correct answer is:
$\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)$

Solution:
Slope of line joining centre of circle to point $D$
$$
=\frac{\frac{3}{2}-1}{\frac{3 \sqrt{3}}{2}-\sqrt{3}}=\frac{1}{\sqrt{3}}
$$
[makes an angle $30^{\circ}$ with $X$-axis]
$\therefore$ Points $E$ and $F$ will make angle $150^{\circ}$ and $-90^{\circ}$ with $X$-axis.
$\therefore E$ and $F$ are given by and
$\therefore$
$$
\begin{aligned}
\frac{x-\sqrt{3}}{\cos 150^{\circ}} & =\frac{y-1}{\sin 150^{\circ}}=1 \\
\frac{x-\sqrt{3}}{\cos \left(-90^{\circ}\right)} & =\frac{y-1}{\sin \left(-90^{\circ}\right)}=1 \\
E & =\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right) \\
and
F & =(\sqrt{3}, 0)
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
Paragraph:
A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$ respectively. The line $P Q$ is given by the equation
$\sqrt{3} x+y-6=0$ and the point $D$ is $\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$.Question:
The equation of circle $C$ is

Options:
A
$(x-2 \sqrt{3})^2+(y-1)^2=1$

B
$(x-2 \sqrt{3})^2+\left(y+\frac{1}{2}\right)^2=1$

C
$(x-\sqrt{3})^2+(y+1)^2=1$

D
$(x-\sqrt{3})^2+(y-1)^2=1$

The correct answer is:
$(x-\sqrt{3})^2+(y-1)^2=1$

Solution:
Let centre of circle $C$ be $(h, k)$.
Then, $\left|\frac{\sqrt{3} h+k-6}{\sqrt{3+1}}\right|=1$
$$
\Rightarrow \quad \sqrt{3} h+k-6=2,-2
$$
$$
\Rightarrow \quad \sqrt{3} h+k=4
$$
(Rejecting 2 because origin and centre of $C$ are on the same side of $P Q$ ) The point $(\sqrt{3}, 1)$ satisfies Eq. (i).
$\therefore$ Equation of circle $C$ is $(x-\sqrt{3})^2+(y-1)^2=1$.

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let $N$ be the number density of free electrons, each of mass $m$. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions.If the electric field becomes zero, the electrons being to oscillate about the positive ions with a natural angular frequency $\omega_p$, which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency $\omega$, where a part of the energy is absorbed and a part of it is reflected. As $\omega$ approaches $\omega_p$, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals.Question:
Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons $N \approx 4 \times 10^{27} \mathrm{~m}^{-1}$. Take $\varepsilon_0=10^{-11}$ and $m=10^{-30}$, where these quantities are in proper SI units.

Options:
A
$800 \mathrm{~nm}$

B
$600 \mathrm{~nm}$

C
$300 \mathrm{~nm}$

D
$200 \mathrm{~nm}$

The correct answer is:
$600 \mathrm{~nm}$

Solution:
$$
\begin{gathered}
\omega=2 \pi f=\frac{2 \pi c}{\lambda} \\
\therefore \quad \lambda=\frac{2 \pi c}{\omega}=\frac{2 \pi c}{\sqrt{N e^2 / m \varepsilon_0}}
\end{gathered}
$$
Substituting the values, we get
$$
\lambda=600 \mathrm{~nm}
$$
$\therefore$ Correct option is (b).
Analysis of Question
(i) Paragraph does not make the solution very clear.
(ii) Solution is simple but based on error and trial method.

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let $N$ be the number density of free electrons, each of mass $m$. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions.If the electric field becomes zero, the electrons being to oscillate about the positive ions with a natural angular frequency $\omega_p$, which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency $\omega$, where a part of the energy is absorbed and a part of it is reflected. As $\omega$ approaches $\omega_p$, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals.Question:
Taking the electronic charge as $e$ and the permittivity as $\varepsilon_0$, use dimensional analysis to determine the correct expression for $\omega_p$.

Options:
A
$\sqrt{\frac{N e}{m \varepsilon_0}}$

B
$\sqrt{\frac{m \varepsilon_0}{N e}}$

C
$\sqrt{\frac{N e^2}{m \varepsilon_0}}$

D
$\sqrt{\frac{m \varepsilon_0}{N e^2}}$

The correct answer is:
$\sqrt{\frac{N e^2}{m \varepsilon_0}}$

Solution:
$N=$ Number of electrons per unit volume
$$
\begin{aligned}
\therefore \quad[N] & =\left[\mathrm{L}^{-3}\right],[e]=[q] \\
& =[I t]=[\mathrm{AT}] \\
{\left[\varepsilon_0\right] } & =\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]
\end{aligned}
$$
Substituting the dimensions, we can see that,
$$
\left.\sqrt{\frac{N e^2}{m \varepsilon_0}}\right]=\left[\mathrm{T}^{-1}\right]
$$
Angular frequency has also the dimension $\left[\mathrm{T}^{-1}\right]$.
$\therefore$ Correct option is (c).
Analysis of Question
(i) From calculation point of view, question is moderately difficult. Otherwise it is simple.
(ii) Students may commit a mistake in the dimensions of $N$. It is not dimensionless.

Topic: Physics, Units and Dimensions, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
A fair die is tossed repeatedly until a six is obtained. Let $X$ denotes the number of tosses required.Question:
The conditional probability that $X \geq 6$ given $X>3$ equals

Options:
A
$\frac{125}{216}$

B
$\frac{25}{216}$

C
$\frac{5}{36}$

D
$\frac{25}{36}$

The correct answer is:
$\frac{25}{36}$

Solution:
$P((X \geq 6) /(X>3))$
$$
\begin{aligned}
& =\frac{P((X>3) /(X \geq 6)) \cdot P(X \geq 6)}{P(X>3)} \\
& =\frac{1 \cdot\left[\left(\frac{5}{6}\right)^5 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^6 \cdot \frac{1}{6}+\ldots \infty\right]}{\left[\left(\frac{5}{6}\right)^3 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}+\ldots \infty\right]}=\frac{25}{36} \\
&
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
A fair die is tossed repeatedly until a six is obtained. Let $X$ denotes the number of tosses required.Question:
The probability that $X=3$ equals

Options:
A
$\frac{25}{216}$

B
$\frac{25}{36}$

C
$\frac{5}{36}$

D
$\frac{125}{216}$

The correct answer is:
$\frac{25}{216}$

Solution:
$P(X=3)=\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}=\frac{25}{216}$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
A fair die is tossed repeatedly until a six is obtained. Let $X$ denotes the number of tosses required.Question:
The probability that $X \geq 3$ equals

Options:
A
$\frac{125}{216}$

B
$\frac{25}{36}$

C
$\frac{5}{36}$

D
$\frac{25}{216}$

The correct answer is:
$\frac{25}{36}$

Solution:
$P(X \geq 3)=\frac{5}{6} \cdot \frac{5}{6} \cdot 1=\frac{25}{36}$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2009 (Paper 1)