datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
Read the following passage and answer the questions.
Let $A B C D$ be a square of side length 2 units. $C_2$ is the circle through vertices $A, B, C, D$ and $C_1$ is the circle touching all the sides of square $A B C D$. $L$ is the line through $A$.Question:
If $P$ is a point on $C_1$ and $Q$ is a point on $C_2$, then $\frac{P A^2+P B^2+P C^2+P D^2}{Q A^2+Q B^2+Q C^2+Q D^2}$ is equal to

Options:
A
$0.75$

B
$1.25$

C
1

D
$0.5$

The correct answer is:
$0.75$

Solution:
$$
\text { Here, equation of } C_2:(x-1)^2+(y-1)^2=(\sqrt{2})^2 \text { and } C_1:(x-1)^2+(y-1)^2=(1)^2
$$
$$
\begin{aligned}
& \therefore P(1+\cos \theta, 1+\sin \theta) \\
& \text { and } Q(1+\sqrt{2} \sin \theta, 1+\sqrt{2} \sin \theta) \\
& \therefore P A^2+P B^2+P C^2+P D^2 \\
& =\left\{(1+\cos \theta)^2+(1+\sin \theta)^2\right\} \\
& +\left\{(\cos \theta-1)^2+(1+\sin \theta)^2\right\} \\
& +\left\{(1+\cos \theta)^2+(1-\sin \theta)^2\right\} \\
& +\left\{(1-\cos \theta)^2+(1-\sin \theta)^2\right\} \\
& =16 \\
& \therefore \quad \frac{\Sigma Q A^2}{\Sigma P A^2}=\frac{12}{16}=0.75 \\
&
\end{aligned}
$$
Hence (a) is the correct answer.

Topic: Mathematics, Circle, JEE Main
Paragraph:
Read the following passage and answer the questions.
Let $A B C D$ be a square of side length 2 units. $C_2$ is the circle through vertices $A, B, C, D$ and $C_1$ is the circle touching all the sides of square $A B C D$. $L$ is the line through $A$.Question:
A circle touches the line $L$ and the circle $C_1$ externally such that both the circles are on the same side of the line, then the locus of centre of the circle is

Options:
A
ellipse

B
hyperbola

C
parabola

D
parts of straight line

The correct answer is:
parabola

Solution:
Let $C$ be the centre of the required circle.
Now, draw a line parallel to $L$ at a distance of $r_1$ (radius of $C_1$ ) from it. Now, $C P_1=A C$
$\Rightarrow C$ lies on a parabola.

Topic: Mathematics, Parabola, JEE Main
Paragraph:
Read the following passage and answer the questions.
Let $A B C D$ be a square of side length 2 units. $C_2$ is the circle through vertices $A, B, C, D$ and $C_1$ is the circle touching all the sides of square $A B C D$. $L$ is the line through $A$.Question:
A line $M$ through $A$ is drawn parallel to $B D$. Points $S$ moves such that its distances from the line $B D$ are the vertex $A$ are equal. If locus of $S$ cuts $M$ at $T_2$ and $T_3$ and $A C$ at $T_1$, then area of $\Delta T_1 T_2 T_3$ is

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

B
$\frac{2}{3}$ sq unit

C
1 sq unit

D
2 sq unit

The correct answer is:
1 sq unit

Solution:
$\because \quad A G=\sqrt{2}$
$$
\therefore \quad A T_1=T_1 G=\frac{1}{\sqrt{2}}
$$
\{ as $A$ is the focus, $T_1$ is the vertex and $B D$ is the directrix of parabola.\}
Also, $T_2 T_3$ is latus rectum
$$
\begin{aligned}
& \therefore \quad T_2 T_3=4 \cdot \frac{1}{\sqrt{2}} \\
& \therefore \text { Area of } \Delta T_1 T_2 T_3=\frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=1 \text { sq unit }
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Main
Paragraph:
Read the following passage and answer the questions.
Let $A B C D$ be a square of side length 2 units. $C_2$ is the circle through vertices $A, B, C, D$ and $C_1$ is the circle touching all the sides of square $A B C D$. $L$ is the line through $A$.Question:
If $P$ is a point on $C_1$ and $Q$ is a point on $C_2$, then $\frac{P A^2+P B^2+P C^2+P D^2}{Q A^2+Q B^2+Q C^2+Q D^2}$ is equal to

Options:
A
$0.75$

B
$1.25$

C
1

D
$0.5$

The correct answer is:
$0.75$

Solution:
$$
\text { Here, equation of } C_2:(x-1)^2+(y-1)^2=(\sqrt{2})^2 \text { and } C_1:(x-1)^2+(y-1)^2=(1)^2
$$
$$
\begin{aligned}
& \therefore P(1+\cos \theta, 1+\sin \theta) \\
& \text { and } Q(1+\sqrt{2} \sin \theta, 1+\sqrt{2} \sin \theta) \\
& \therefore P A^2+P B^2+P C^2+P D^2 \\
& =\left\{(1+\cos \theta)^2+(1+\sin \theta)^2\right\} \\
& +\left\{(\cos \theta-1)^2+(1+\sin \theta)^2\right\} \\
& +\left\{(1+\cos \theta)^2+(1-\sin \theta)^2\right\} \\
& +\left\{(1-\cos \theta)^2+(1-\sin \theta)^2\right\} \\
& =16 \\
& \therefore \quad \frac{\Sigma Q A^2}{\Sigma P A^2}=\frac{12}{16}=0.75 \\
&
\end{aligned}
$$
Hence (a) is the correct answer.

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2006
Paragraph:
Read the following passage and answer the questions.
Question:
The value of $\left[\begin{array}{lll}3 & 2 & 0\end{array}\right]\left[\begin{array}{l}3 \\ 2 \\ 0\end{array}\right]$ is

Options:
A
$[5]$

B
$\left[\frac{5}{2}\right]$

C
[4]

D
$\left[\frac{3}{2}\right]$

The correct answer is:
$[5]$

Solution:

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2006
Paragraph:
Read the following passage and answer the questions.
Question:
The value of the elements of $U^{-1}$ is

Options:
A
$-1$

B
0

C
1

D
3

The correct answer is:
$-1$

Solution:

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2006
Paragraph:
Read the following passage and answer the questions.
Question:
The value of $|U|$ is

Options:
A
3

B
$-3$

C
$\frac{3}{2}$

D
2

The correct answer is:
3

Solution:

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2006
Paragraph:
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E^{\circ}\right)$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}(V$ with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 14-19.
$$
\begin{array}{cl}
\mathrm{I}_2+2 e^{-} \longrightarrow 2 \mathrm{I}^{-} & E^{\circ}=0.54 \\
\mathrm{Cl}_2+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-} & E^{\circ}=1.36 \\
\mathrm{Mn}^{3+}+e^{-} \longrightarrow \mathrm{Mn}^{2+} & E^{\circ}=1.50 \\
\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} & E^{\circ}=0.77 \\
\mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} & E^{\circ}=1.23
\end{array}
$$Question:
Among the following, identify the correct statement

Options:
A
Chloride ion is oxidised by $\mathrm{O}_2$

B
$\mathrm{Fe}^{2+}$ is oxidised by iodine

C
Iodide ion is oxidised by chlorine

D
$\mathrm{Mn}^{2+}$ is oxidised by chlorine

The correct answer is:
Iodide ion is oxidised by chlorine

Solution:
$2 \mathrm{I}^{-}+\mathrm{Cl}_2 \longrightarrow \mathrm{I}_2+2 \mathrm{Cl}^{-}$
$$
E_{\text {cell }}^{\circ}=1.36-0.54=0.82 \mathrm{~V}
$$
$E_{\text {cell }}^{\circ}$ is $+v e$ cell reactions is feasible.

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E^{\circ}\right)$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}(V$ with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 14-19.
$$
\begin{array}{cl}
\mathrm{I}_2+2 e^{-} \longrightarrow 2 \mathrm{I}^{-} & E^{\circ}=0.54 \\
\mathrm{Cl}_2+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-} & E^{\circ}=1.36 \\
\mathrm{Mn}^{3+}+e^{-} \longrightarrow \mathrm{Mn}^{2+} & E^{\circ}=1.50 \\
\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} & E^{\circ}=0.77 \\
\mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} & E^{\circ}=1.23
\end{array}
$$Question:
Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and $\mathrm{H}_2 \mathrm{SO}_4$ in presence of air gives a prussian blue precipitate. The blue colour is due to the formation of

Options:
A
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

B
$\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$

C
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$

D
$$
\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3
$$

The correct answer is:
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

Solution:

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E^{\circ}\right)$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}(V$ with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 14-19.
$$
\begin{array}{cl}
\mathrm{I}_2+2 e^{-} \longrightarrow 2 \mathrm{I}^{-} & E^{\circ}=0.54 \\
\mathrm{Cl}_2+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-} & E^{\circ}=1.36 \\
\mathrm{Mn}^{3+}+e^{-} \longrightarrow \mathrm{Mn}^{2+} & E^{\circ}=1.50 \\
\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} & E^{\circ}=0.77 \\
\mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} & E^{\circ}=1.23
\end{array}
$$Question:
Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and $\mathrm{H}_2 \mathrm{SO}_4$ in presence of air gives a prussian blue precipitate. The blue colour is due to the formation of

Options:
A
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

B
$\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$

C
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$

D
$$
\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3
$$

The correct answer is:
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

Solution:

Topic: Chemistry, Redox Reactions, JEE Main
Paragraph:
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E^{\circ}\right)$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}(V$ with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 14-19.
$$
\begin{array}{cl}
\mathrm{I}_2+2 e^{-} \longrightarrow 2 \mathrm{I}^{-} & E^{\circ}=0.54 \\
\mathrm{Cl}_2+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-} & E^{\circ}=1.36 \\
\mathrm{Mn}^{3+}+e^{-} \longrightarrow \mathrm{Mn}^{2+} & E^{\circ}=1.50 \\
\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} & E^{\circ}=0.77 \\
\mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} & E^{\circ}=1.23
\end{array}
$$Question:
While $\mathrm{Fe}^{3+}$ is stable, $\mathrm{Mn}^{3+}$ is not stable in acid solution because

Options:
A
$\mathrm{O}_2$ oxidises $\mathrm{Mn}^{2+}$ to $\mathrm{Mn}^{3+}$

B
$\mathrm{O}_2$ oxidises both $\mathrm{Mn}^{2+}$ and $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$

C
$\mathrm{Fe}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$

D
$\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$

The correct answer is:
$\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$

Solution:
$$
\begin{aligned}
& {\left[\mathrm{Mn}^{3+}+e^{-}\right.}\left.\longrightarrow \mathrm{Mn}^{2+}\right] \times 4 \\
& 2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}^{+}+\mathrm{O}_2+4 e^{-} \\
& 4 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Mn}^{2+}+4 \mathrm{H}^{+}+\mathrm{O}_2 \\
& E_{\text {cell }}^{\circ}= 1.50-1.23=0.27 \mathrm{~V} \\
& E_{\text {cell }}^{\circ} \text { is + ve }
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Reimer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes as depicted below.
Question:
Which one of the following reagents is used the above reaction?

Options:
A
aq. $\mathrm{NaOH}+\mathrm{CH}_3 \mathrm{Cl}$

B
aq. $\mathrm{NaOH}+\mathrm{CH}_2 \mathrm{Cl}_2$

C
aq. $\mathrm{NaOH}+\mathrm{CHCl}_3$

D
aq. $\mathrm{NaOH}+\mathrm{CCl}_4$

The correct answer is:
aq. $\mathrm{NaOH}+\mathrm{CHCl}_3$

Solution:
aq. $\mathrm{NaOH}+\mathrm{CHCl}_3$
$$
\mathrm{Cl}_3 \mathrm{C}-\mathrm{H} \underset{-\mathrm{H}_2 \mathrm{O}}{\stackrel{\mathrm{OH}^{-}}{\longrightarrow}} \mathrm{Cl}_3 \mathrm{C}^{-} \underset{-\mathrm{Cl}^{-}}{\longrightarrow} \mathrm{Cl}_2 \mathrm{C}:
$$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Reimer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes as depicted below.
Question:
The electrophile in this reaction is

Options:
A
: CHCL

B
${ }^{+} \mathrm{CHCl}_2$

C
$: \mathrm{CCl}_2$

D
$\cdot \mathrm{CCl}_3$

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

Solution:
In Reimer-Tiemann reaction

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Reimer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes as depicted below.
Question:
The structure of the intermediate $\mathrm{I}$ is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Paragraph:
Reimer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes as depicted below.
Question:
Which one of the following reagents is used the above reaction?

Options:
A
aq. $\mathrm{NaOH}+\mathrm{CH}_3 \mathrm{Cl}$

B
aq. $\mathrm{NaOH}+\mathrm{CH}_2 \mathrm{Cl}_2$

C
aq. $\mathrm{NaOH}+\mathrm{CHCl}_3$

D
aq. $\mathrm{NaOH}+\mathrm{CCl}_4$

The correct answer is:
aq. $\mathrm{NaOH}+\mathrm{CHCl}_3$

Solution:
aq. $\mathrm{NaOH}+\mathrm{CHCl}_3$
$$
\mathrm{Cl}_3 \mathrm{C}-\mathrm{H} \underset{-\mathrm{H}_2 \mathrm{O}}{\stackrel{\mathrm{OH}^{-}}{\longrightarrow}} \mathrm{Cl}_3 \mathrm{C}^{-} \underset{-\mathrm{Cl}^{-}}{\longrightarrow} \mathrm{Cl}_2 \mathrm{C}:
$$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Paragraph:
Reimer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes as depicted below.
Question:
The structure of the intermediate $\mathrm{I}$ is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, ${ }_1^2 \mathrm{H}$ known as deuteron and denoted by $D$ can be thought of as a candidate for fusion reactor. The $D$ - $D$ reaction is ${ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+n+$ energy. In the core of fusion reactor. A gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ${ }_1^4 \mathrm{H}$ nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time $t_0$ before the particles fly away from the core. If n is the density (number/volume) of deutrons, the product t $_0$ is called Lawson number. In one of the criteria, $a$ reactor is termed successful if Lawson number is greater than $5 \times 10^{14} \mathrm{scm}^{-3}$.
It may be helpful to use the following : Boltzmann constant $k=8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K}$; $\frac{e^2}{4 \pi \varepsilon_0}=1.44 \times 10^{-9} \mathrm{eVm}$Question:
In the core of nuclear fusion reactor, the gas becomes plasma because of

Options:
A
strong nuclear force acting between the deuterons

B
Coulomb fore acting between the deuterons

C
Coulomb force acting between deuteron-electron pairs

D
the high temperature maintained inside the reactor core

The correct answer is:
the high temperature maintained inside the reactor core

Solution:
No Solution Available

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, ${ }_1^2 \mathrm{H}$ known as deuteron and denoted by $D$ can be thought of as a candidate for fusion reactor. The $D$ - $D$ reaction is ${ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+n+$ energy. In the core of fusion reactor. A gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ${ }_1^4 \mathrm{H}$ nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time $t_0$ before the particles fly away from the core. If n is the density (number/volume) of deutrons, the product t $_0$ is called Lawson number. In one of the criteria, $a$ reactor is termed successful if Lawson number is greater than $5 \times 10^{14} \mathrm{scm}^{-3}$.
It may be helpful to use the following : Boltzmann constant $k=8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K}$; $\frac{e^2}{4 \pi \varepsilon_0}=1.44 \times 10^{-9} \mathrm{eVm}$Question:
Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion?

Options:
A
Deuteron density $=20 \times 10^{12} \mathrm{~cm}^{-3}$, confinement time $=5.0 \times 10^{-3} \mathrm{~s}$

B
Deuteron density $=8.0 \times 10^{14} \mathrm{~cm}^{-3}$, confinement time $=9.0 \times 10^{-1} \mathrm{~s}$

C
Deuteron density $=4.0 \times 10^{23} \mathrm{~cm}^{-3}$, confinement time $=1.0 \times 10^{-11} \mathrm{~s}$

D
Deuteron density $=1.0 \times 10^{24} \mathrm{~cm}^{-3}$, confinement time $=4.0 \times 10^{-12} \mathrm{~s}$

The correct answer is:
Deuteron density $=8.0 \times 10^{14} \mathrm{~cm}^{-3}$, confinement time $=9.0 \times 10^{-1} \mathrm{~s}$

Solution:
As given in the paragraph, a reactor is termed successful, if
$$
n t_0>5 \times 10^{14} \mathrm{scm}^{-3}
$$

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, ${ }_1^2 \mathrm{H}$ known as deuteron and denoted by $D$ can be thought of as a candidate for fusion reactor. The $D$ - $D$ reaction is ${ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+n+$ energy. In the core of fusion reactor. A gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ${ }_1^4 \mathrm{H}$ nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time $t_0$ before the particles fly away from the core. If n is the density (number/volume) of deutrons, the product t $_0$ is called Lawson number. In one of the criteria, $a$ reactor is termed successful if Lawson number is greater than $5 \times 10^{14} \mathrm{scm}^{-3}$.
It may be helpful to use the following : Boltzmann constant $k=8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K}$; $\frac{e^2}{4 \pi \varepsilon_0}=1.44 \times 10^{-9} \mathrm{eVm}$Question:
Assume that two deuteron nuclei in the core of fusion reactor at temperature $T$ are moving towards each other, each with kinetic energy $1.5 \mathrm{kT}$, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature $T$ required for them to reach a separation of $4 \times 10^{-15} \mathrm{~m}$ is in the range

Options:
A
$1.0 \times 10^9 \mathrm{~K} < T < 2.0 \times 10^9 \mathrm{~K}$
B
$20 \times 10^9 \mathrm{~K} < T < 3.0 \times 10^9 \mathrm{~K}$
C
$3.0 \times 10^9 \mathrm{~K} < T < 4.0 \times 10^9 \mathrm{~K}$
D
$4.0 \times 10^9 \mathrm{~K} < T < 5.0 \times 10^9 \mathrm{~K}$

The correct answer is:
$1.0 \times 10^9 \mathrm{~K} < T < 2.0 \times 10^9 \mathrm{~K}$

Solution:
From conservation of mechanical energy, we have
$$
\begin{gathered}
U_i+K_i=U_f+U_f \\
0+2(1.5 k T)=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(e)(e)}{d}+0
\end{gathered}
$$
Substituting the values, we get
$$
T=1.4 \times 10^9 \mathrm{~K}
$$

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$.Question:
The coordinates of $A$ and $B$ are

Options:
A
$(3,0)$ and $(0,2)$

B
$\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

C
$\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $(0,2)$

D
$(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

The correct answer is:
$(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$

Solution:
$$
\text { Figure is self explanatory }
$$

Topic: Mathematics, Ellipse, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$.Question:
The equation of the locus of the point whose distance from the point $P$ and the line $A B$ are equal, is

Options:
A
$9 x^2+y^2-6 x y-54 x-62 y$
$$
+241=0
$$

B
$x^2+9 y^2+6 x y-54 x+62 y$
$$
-241=0
$$

C
$9 x^2+9 y^2-6 x y-54 x-62 y$
$$
-241=0
$$

D
$x^2+y^2-2 x y+27 x+31 y$
$$
-120=0
$$

The correct answer is:
$9 x^2+y^2-6 x y-54 x-62 y$
$$
+241=0
$$

Solution:
Equation of $A B$ is $y-0=-\frac{1}{3}(x-3)$
$$
\begin{gathered}
x+3 y-3=0 \\
|x+3 y-3|^2=10\left[(x-3)^2+(y-4)^2\right]
\end{gathered}
$$
(Look at coefficient of $x^2$ and $y^2$ in the
answers)

Topic: Mathematics, Ellipse, JEE Advanced, JEE Advanced 2010 (Paper 2)
Paragraph:
Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$.Question:
The orthocentre of the triangle $P A B$ is

Options:
A
$\left(5, \frac{8}{7}\right)$

B
$\left(\frac{7}{5}, \frac{25}{8}\right)$

C
$\left(\frac{11}{5}, \frac{8}{5}\right)$

D
$\left(\frac{8}{25}, \frac{7}{5}\right)$

The correct answer is:
$\left(\frac{11}{5}, \frac{8}{5}\right)$

Solution:
$$
\text { Equation of } A B \text { is }
$$
$$
\begin{aligned}
y-0 & =\frac{\frac{8}{5}}{-\frac{9}{5}-3}(x-3)=\frac{8}{-24}(x-3) \\
\Rightarrow \quad y & =-\frac{1}{3}(x-3) \\
\Rightarrow \quad x & +3 y=3
\end{aligned}
$$
Equation of the straight line perpendicular to $A B$ through $P$ is $3 x-y=5$.
Equation of $P A$ is $x-3=0$.
The equation of straight line perpendicular to $P A$ through $B\left(\frac{-9}{5}, \frac{8}{5}\right)$. is $y=\frac{8}{5}$.
Hence, the orthocentre is $\left(\frac{11}{5}, \frac{8}{5}\right)$.

Topic: Mathematics, Straight Lines, JEE Main