datavorous/entrance-exam-dataset huggingface dataset:
The graph between object distance $u$ and image distance $v$ for a lens is given below. The focal length of the lens is

Options:
A
$5 \pm 0.1$

B
$5 \pm 0.05$

C
$0.5 \pm 0.1$

D
$0.5 \pm 0.05$

The correct answer is:
$5 \pm 0.05$

Solution:
From the lens formula:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ we have, $\frac{1}{f}=\frac{1}{10}-\frac{1}{-10}$ or $f=+5$
Further, $\Delta u=0.1$ and $\Delta v=0.1$
(from the graph)
Now, differentiating the lens formula we have,
$$
\frac{\Delta f}{f^2}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2} \text { or } \Delta f=\left(\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}\right) f^2
$$
Substituting the values we have,
$$
\Delta f=\left(\frac{0.1}{10^2}+\frac{0.1}{10^2}\right)(5)^2 \Rightarrow f \pm \Delta f=5 \pm 0.05
$$

Topic: Physics, Ray Optics, JEE Main
The graph between object distance $u$ and image distance $v$ for a lens is given below. The focal length of the lens is

Options:
A
$5 \pm 0.1$

B
$5 \pm 0.05$

C
$0.5 \pm 0.1$

D
$0.5 \pm 0.05$

The correct answer is:
$5 \pm 0.05$

Solution:
From the lens formula:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ we have, $\frac{1}{f}=\frac{1}{10}-\frac{1}{-10}$ or $f=+5$
Further, $\Delta u=0.1$ and $\Delta v=0.1$
(from the graph)
Now, differentiating the lens formula we have,
$$
\frac{\Delta f}{f^2}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2} \text { or } \Delta f=\left(\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}\right) f^2
$$
Substituting the values we have,
$$
\Delta f=\left(\frac{0.1}{10^2}+\frac{0.1}{10^2}\right)(5)^2 \Rightarrow f \pm \Delta f=5 \pm 0.05
$$

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2006
The Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 105 \mathrm{~atm}$. The mole fraction of $\mathrm{N}_2$ in air is $0.8$. The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water of $298 \mathrm{~K}$ and $5 \mathrm{~atm}$ pressure is

Options:
A
$4 \times 10^{-4}$

B
$4.0 \times 10^{-5}$

C
$5.0 \times 10^{-4}$

D
$4.0 \times 10^{-6}$

The correct answer is:
$4 \times 10^{-4}$

Solution:
$P_{\mathrm{N}_2}=K_{\mathrm{H}} \times$ mole-fraction $\left(\mathrm{N}_2\right)$ mole-fraction
$$
\mathrm{N}_2 \frac{1}{10^5} \times 0.8 \times 5=4 \times 10^{-5} \mathrm{~mol}^{-1}
$$
In 10 mole solubility in $4 \times 10^{-4}$.

Topic: Chemistry, Solutions, JEE Advanced, JEE Advanced 2009 (Paper 1)
The integral $\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{\frac{9}{2}}} d x$ equals (for some arbitrary constant $\mathrm{K}$ )

Options:
A $-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
B $\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}-\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
C $-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$
D $\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$

The correct answer is:
$-\frac{1}{(\sec x+\tan x)^{\frac{11}{2}}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K$

Solution:
$\begin{array}{l}

I=\int \frac{\sec ^{2} x}{(\sec x+\tan x)^{9 / 2}} d x \\

\text { Let } \sec x+\tan x=t \Rightarrow \sec x-\tan x=\frac{1}{t} \\

\Rightarrow \sec x=\frac{1}{2}\left(t+\frac{1}{t}\right) \text { and } \sec x(\sec x+\tan x) d x=d t \\

\Rightarrow \sec x d x=\frac{d t}{t} \\

\therefore \quad I=\frac{1}{2} \int \frac{\left(t+\frac{1}{t}\right) d t}{t^{9 / 2} \cdot t}=\frac{1}{2} \int\left(t^{-9 / 2}+t^{-13 / 2}\right) d t \\

=\frac{-1}{7} t^{-7 / 2}-\frac{1}{11} t^{-1 / 2}+K \\

=-\frac{1}{7 t^{7 / 2}}-\frac{1}{11 t^{11 / 2}}+K=-\frac{1}{t^{11 / 2}}\left(\frac{1}{11}+\frac{t^{2}}{7}\right)+K \\

=\frac{-1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^{2}\right\}+K

\end{array}$

Topic: Mathematics, Indefinite Integration, JEE Advanced, JEE Advanced 2012 (Paper 1)
The ionization isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}$ is

Options:
A
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4\left(\mathrm{O}_2 \mathrm{~N}\right)\right] \mathrm{Cl}_2$

B
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]\left(\mathrm{NO}_2\right)$

C
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}(\mathrm{ONO})\right] \mathrm{Cl}$

D
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\left(\mathrm{NO}_2\right)\right] \cdot \mathrm{H}_2 \mathrm{O}$

The correct answer is:
$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]\left(\mathrm{NO}_2\right)$

Solution:
The ionization isomer of $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}\left(\mathrm{NO}_2\right)\right] \mathrm{Cl}$ is $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_2$ because of exchanging of ligand and counter ions
Coordination chemistry
Straight conceptual
II

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2010 (Paper 1)
The IUPAC name of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl}$ is

Options:
A
Benzoyl chloride

B
Benzene chloro ketone

C
Benzene carbonyl chloride

D
Chloro phenyl ketone

The correct answer is:
Benzene carbonyl chloride

Solution:
is known as benzene carbonyl chloride. Although its trivial name but it is also accepted in IUPAC System

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2006
The IUPAC name of $\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_4\right]\left[\mathrm{NiCl}_4\right]$ is

Options:
A
tetrachloronickel (II) - tetraamminenickel (II)

B
tetraamminenickel (II) - tetrachloronickel (II)

C
tetraamminenickel (II) - tetrachloronickelate (II)

D
tetrachloronickel (II) - tetraamminenickelate (0)

The correct answer is:
tetraamminenickel (II) - tetrachloronickelate (II)

Solution:
IUPAC name is tetraamminenickel (II) tetrachloronickelate (II).

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2008 (Paper 2)
The IUPAC name of the following compound, is

Options:
A
4-bromo-3-cyanophenol

B
2-bromo-5-hydroxybenzonitrile

C
2-cyano-4-hydroxybromobenzene

D
6-bromo-3-hydroxybenzonitrile

The correct answer is:
2-bromo-5-hydroxybenzonitrile

Solution:
Cyano group has the highest priority therefore, parent name must be benzonitrile. $\mathrm{Br}$ occurs at 2-position, and hydroxyl at 3-position, hence clockwise numbering will be the correct one.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2009 (Paper 1)
The IUPAC name of the following compound, is

Options:
A
4-bromo-3-cyanophenol

B
2-bromo-5-hydroxybenzonitrile

C
2-cyano-4-hydroxybromobenzene

D
6-bromo-3-hydroxybenzonitrile

The correct answer is:
2-bromo-5-hydroxybenzonitrile

Solution:
Cyano group has the highest priority therefore, parent name must be benzonitrile. $\mathrm{Br}$ occurs at 2-position, and hydroxyl at 3-position, hence clockwise numbering will be the correct one.

Topic: Chemistry, General Organic Chemistry, JEE Main
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is $\left[a_{0}\right.$ is Bohr radius $]$ :

Options:
A $\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}$
B $\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}$
C $\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}$
D $\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}$

The correct answer is:
$\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}$

Solution:
As per Bohr's postulate,

$\begin{array}{l}

m v r=\frac{n h}{2 \pi} \quad \text { So, } v=\frac{n h}{2 \pi m r} \\

\mathrm{KE}=\frac{1}{2} m v^{2} \quad \text { So, } \mathrm{KE}=\frac{1}{2} m\left(\frac{n h}{2 \pi m r}\right)^{2}

\end{array}$

$\text { Since, } r=\frac{a_{\mathrm{o}} \times n^{2}}{z}$

So, for $2^{\text {nd }}$ Bohr orbit

$\begin{array}{l}

r=\frac{a_{\mathrm{o}} \times 2^{2}}{1}=4 a_{\mathrm{o}} \\

\mathrm{KE}=\frac{1}{2} m\left(\frac{2^{2} h^{2}}{4 \pi^{2} m^{2} \times\left(4 a_{\mathrm{o}}\right)^{2}}\right)=\frac{h^{2}}{32 \pi^{2} m a_{\mathrm{o}}^{2}}

\end{array}$

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2012 (Paper 1)
The largest wavelength in the ultraviolet region of the hydrogen spectrum is $122 \mathrm{~nm}$. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is

Options:
A
$802 \mathrm{~nm}$

B
$823 \mathrm{~nm}$

C
$1882 \mathrm{~nm}$

D
$1648 \mathrm{~nm}$

The correct answer is:
$823 \mathrm{~nm}$

Solution:
The series in $U-V$ region is Lymen series. Longest wavelength corresponds to minimum energy which occurs in transition from $n=2$ to $n=1$.
$$
\therefore \quad 122=\frac{\frac{1}{R}}{\left(\frac{1}{1^2}-\frac{1}{2^2}\right)}
$$
The smallest wavelength in the infrared region corresponds to maximum energy of Paschen series.
$$
\lambda=\frac{\frac{1}{R}}{\left(\frac{1}{3^2}-\frac{1}{\infty}\right)}
$$
Solving Eqs. (i) and (ii), we get
$\therefore$ Correct option is (b).
$$
\lambda=823.5 \mathrm{~nm}
$$

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2007 (Paper 1)
The letters of the word $\mathrm{COCHIN}$ are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word $\mathrm{COCHIN}$ is

Options:
A
360

B
192

C
97

D
48

The correct answer is:
97

Solution:
In a word $\mathrm{COCHIN}$, the second place can be filled in ${ }^4 C_1$ ways and the remaining four alphabets can be arranged in $4 !$ ways in four different places. The next $97^{\text {th }}$ word will be COCHIN.

Topic: Mathematics, Permutation Combination, JEE Advanced, JEE Advanced 2007 (Paper 2)
The line $2 x+y=1$ is tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If this line passes through the point of intersection of the nearest directrix and the $x$-axis, then the eccentricity of the hyperbola is

Options:

The correct answer is:
2

Solution:
On substituting $\left(\frac{a}{e}, 0\right)$ in $y=-2 x+1$, we get
$$
\begin{aligned}
0 & =-\frac{2 a}{e}+1 \\
\Rightarrow \quad \frac{a}{e} & =\frac{1}{2}
\end{aligned}
$$
Also, $y=-2 x+1$ is tangent to hyperbola
$$
\begin{array}{ll}
\therefore & 1=4 a^2-b^2 \\
\Rightarrow & \frac{1}{a^2}=4-\left(e^2-1\right)
\end{array}
$$
$$
\begin{array}{ll}
\Rightarrow & \frac{4}{e^2}=5-e^2 \\
\Rightarrow & e^4-5 e^2+4=0 \\
\Rightarrow & \left(e^2-4\right)\left(e^2-1\right)=0 \\
\Rightarrow & e=2, e=1
\end{array}
$$
$e=1$ gives the conic as parabola. But conic is given as hyperbola, hence $e=2$

Topic: Mathematics, Hyperbola, JEE Main
The line $2 x+y=1$ is tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If this line passes through the point of intersection of the nearest directrix and the $x$-axis, then the eccentricity of the hyperbola is

Options:

The correct answer is:
2

Solution:
On substituting $\left(\frac{a}{e}, 0\right)$ in $y=-2 x+1$, we get
$$
\begin{aligned}
0 & =-\frac{2 a}{e}+1 \\
\Rightarrow \quad \frac{a}{e} & =\frac{1}{2}
\end{aligned}
$$
Also, $y=-2 x+1$ is tangent to hyperbola
$$
\begin{array}{ll}
\therefore & 1=4 a^2-b^2 \\
\Rightarrow & \frac{1}{a^2}=4-\left(e^2-1\right)
\end{array}
$$
$$
\begin{array}{ll}
\Rightarrow & \frac{4}{e^2}=5-e^2 \\
\Rightarrow & e^4-5 e^2+4=0 \\
\Rightarrow & \left(e^2-4\right)\left(e^2-1\right)=0 \\
\Rightarrow & e=2, e=1
\end{array}
$$
$e=1$ gives the conic as parabola. But conic is given as hyperbola, hence $e=2$

Topic: Mathematics, Hyperbola, JEE Advanced, JEE Advanced 2010 (Paper 1)
The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^2+9 y^2=9$ meets its auxiliary circle at the point $M$. Then, the area of the triangle with vertices at $A, M$ and the origin $O$ is

Options:
A
$\frac{31}{10}$

B
$\frac{29}{10}$

C
$\frac{21}{10}$

D
$\frac{27}{10}$

The correct answer is:
$\frac{27}{10}$

Solution:
Equation of auxiliary circle is
$$
x^2+y^2=9
$$
Equation of $A M$ is $\frac{x}{3}+\frac{y}{1}=1$
On solving Eqs. (i) and (ii), we get Coordinates of $M=\left(-\frac{12}{5}, \frac{9}{5}\right)$ Now, area of $\triangle A O M=\frac{1}{2} \times O A \times M N$ $=\frac{27}{10}$ sq units

Topic: Mathematics, Ellipse, JEE Main
The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^2+9 y^2=9$ meets its auxiliary circle at the point $M$. Then, the area of the triangle with vertices at $A, M$ and the origin $O$ is

Options:
A
$\frac{31}{10}$

B
$\frac{29}{10}$

C
$\frac{21}{10}$

D
$\frac{27}{10}$

The correct answer is:
$\frac{27}{10}$

Solution:
Equation of auxiliary circle is
$$
x^2+y^2=9
$$
Equation of $A M$ is $\frac{x}{3}+\frac{y}{1}=1$
On solving Eqs. (i) and (ii), we get Coordinates of $M=\left(-\frac{12}{5}, \frac{9}{5}\right)$ Now, area of $\triangle A O M=\frac{1}{2} \times O A \times M N$ $=\frac{27}{10}$ sq units

Topic: Mathematics, Ellipse, JEE Advanced, JEE Advanced 2009 (Paper 1)
The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4 x-$ $5 y=20$ to the circle $x^{2}+y^{2}=9$ is

Options:
A $20\left(x^{2}+y^{2}\right)-36 x+45 y=0$
B $20\left(x^{2}+y^{2}\right)+36 x-45 y=0$
C $36\left(x^{2}+y^{2}\right)-20 x+45 y=0$
D $36\left(x^{2}+y^{2}\right)+20 x-45 y=0$

The correct answer is:
$20\left(x^{2}+y^{2}\right)-36 x+45 y=0$

Solution:
Any point $P$ on line $4 x-5 y=20$ can be considered as $\left(\alpha, \frac{4 \alpha-20}{5}\right)$.

Equation of chord of contact $A B$ to the circle $x^{2}+y^{2}=9$

drawn from point $\mathrm{P}\left(\alpha, \frac{4 \alpha-20}{5}\right)$ is

x. $\alpha+$ y. $\left(\frac{4 \alpha-20}{5}\right)=9...(i)$

Also the equation of chord $A B$ whose mid point is $(h, k)$ is $h x+k y=h^{2}+k^{2}....(ii)$

$\because \quad$ Equations (i) and (ii) represent the same line,

$\therefore \quad \frac{h}{\alpha}=\frac{k}{\frac{4 \alpha-20}{5}}=\frac{h^{2}+k^{2}}{9}$

$\Rightarrow 5 k \alpha=4 h \alpha-20 h$ and $9 h=\alpha\left(h^{2}+k^{2}\right)$

$\Rightarrow \alpha=\frac{20 h}{4 h-5 k} \quad$ and $\alpha=\frac{9 h}{h^{2}+k^{2}}$

$\Rightarrow \quad \frac{20 h}{4 h-5 k}=\frac{9 h}{h^{2}+k^{2}} \Rightarrow 20\left(h^{2}+k^{2}\right)=9(4 h-5 k)$

$\therefore \quad$ Locus of $(h, k)$ is $20\left(x^{2}+y^{2}\right)-36 x+45 y=0$

Topic: Mathematics, Circle, JEE Main
The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4 x-$ $5 y=20$ to the circle $x^{2}+y^{2}=9$ is

Options:
A $20\left(x^{2}+y^{2}\right)-36 x+45 y=0$
B $20\left(x^{2}+y^{2}\right)+36 x-45 y=0$
C $36\left(x^{2}+y^{2}\right)-20 x+45 y=0$
D $36\left(x^{2}+y^{2}\right)+20 x-45 y=0$

The correct answer is:
$20\left(x^{2}+y^{2}\right)-36 x+45 y=0$

Solution:
Any point $P$ on line $4 x-5 y=20$ can be considered as $\left(\alpha, \frac{4 \alpha-20}{5}\right)$.

Equation of chord of contact $A B$ to the circle $x^{2}+y^{2}=9$

drawn from point $\mathrm{P}\left(\alpha, \frac{4 \alpha-20}{5}\right)$ is

x. $\alpha+$ y. $\left(\frac{4 \alpha-20}{5}\right)=9...(i)$

Also the equation of chord $A B$ whose mid point is $(h, k)$ is $h x+k y=h^{2}+k^{2}....(ii)$

$\because \quad$ Equations (i) and (ii) represent the same line,

$\therefore \quad \frac{h}{\alpha}=\frac{k}{\frac{4 \alpha-20}{5}}=\frac{h^{2}+k^{2}}{9}$

$\Rightarrow 5 k \alpha=4 h \alpha-20 h$ and $9 h=\alpha\left(h^{2}+k^{2}\right)$

$\Rightarrow \alpha=\frac{20 h}{4 h-5 k} \quad$ and $\alpha=\frac{9 h}{h^{2}+k^{2}}$

$\Rightarrow \quad \frac{20 h}{4 h-5 k}=\frac{9 h}{h^{2}+k^{2}} \Rightarrow 20\left(h^{2}+k^{2}\right)=9(4 h-5 k)$

$\therefore \quad$ Locus of $(h, k)$ is $20\left(x^{2}+y^{2}\right)-36 x+45 y=0$

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2012 (Paper 1)
The locus of the orthocentre of the triangle formed by the lines $(1+p) x-p y+p(1+p)=0$, $(1+q) x-q y+q(1+q)=0$ and $y=0$, where $p \neq q$, is

Options:
A
a hyperbola

B
a parabola

C
an ellipse

D
a straight line

The correct answer is:
a straight line

Solution:
Given, lines are
$$
(1+p) x-p y+p(1+p)=0
$$
and $(1+q) x-q y+q(1+q)=0$
On solving Eqs. (i) and (ii), we get
$$
C\{p q,(1+p)(1+q)\}
$$
$\therefore$ Equation of altitude $C M$ passing through $C$ and perpendicular to $A B$ is
$$
x=p q
$$
$\because$ Slope of line (ii) is $\left(\frac{1+q}{q}\right)$.
$\therefore$ Slope of altitude $B N$ (as shown in figure) is $\frac{-q}{1+q}$.
$\therefore$ Equation of $B N$ is $y-0=\frac{-q}{1+q}(x+p)$
$$
\Rightarrow \quad y=\frac{-q}{(1+q)}(x+p)
$$
Let orthocentre of triangle be $H(h, k)$, which is the point of intersection of Eqs. (iii) and (iv)
On solving Eqs. (iii) and (iv), we get $x=p q$ and $y=-p q \Rightarrow h=p q$ and $k=-p q \Rightarrow h+k=0$
$\therefore$ Locus of $H(h, k)$ is $x+y=0$.

Topic: Mathematics, Straight Lines, JEE Main
The locus of the orthocentre of the triangle formed by the lines $(1+p) x-p y+p(1+p)=0$, $(1+q) x-q y+q(1+q)=0$ and $y=0$, where $p \neq q$, is

Options:
A
a hyperbola

B
a parabola

C
an ellipse

D
a straight line

The correct answer is:
a straight line

Solution:
Given, lines are
$$
(1+p) x-p y+p(1+p)=0
$$
and $(1+q) x-q y+q(1+q)=0$
On solving Eqs. (i) and (ii), we get
$$
C\{p q,(1+p)(1+q)\}
$$
$\therefore$ Equation of altitude $C M$ passing through $C$ and perpendicular to $A B$ is
$$
x=p q
$$
$\because$ Slope of line (ii) is $\left(\frac{1+q}{q}\right)$.
$\therefore$ Slope of altitude $B N$ (as shown in figure) is $\frac{-q}{1+q}$.
$\therefore$ Equation of $B N$ is $y-0=\frac{-q}{1+q}(x+p)$
$$
\Rightarrow \quad y=\frac{-q}{(1+q)}(x+p)
$$
Let orthocentre of triangle be $H(h, k)$, which is the point of intersection of Eqs. (iii) and (iv)
On solving Eqs. (iii) and (iv), we get $x=p q$ and $y=-p q \Rightarrow h=p q$ and $k=-p q \Rightarrow h+k=0$
$\therefore$ Locus of $H(h, k)$ is $x+y=0$.

Topic: Mathematics, Straight Lines, JEE Advanced, JEE Advanced 2009 (Paper 2)
$\text { The major product } \mathbf{H} \text { of the given reaction sequence is }$

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
$\text { The major product } \mathbf{H} \text { of the given reaction sequence is }$

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2012 (Paper 2)
The major product of the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{PhS}^{-}$is a strong nucleophile and dimethylformamide (DMF) is a highly polar aprotic solvent. Condition indicates that nucleophilic substitution $\left(\mathrm{S}_{\mathrm{N}} 2\right)$ takes place at $2^{\circ}$ benzylic place, stereochemically it involves inversion of configuration.
Aryl group can exhibit neighbouring group or can show anchimeric assistance but not possible here due to deactivity imparted by $-\mathrm{NO}_2$ and $-\mathrm{F}$. It can be if $-\mathrm{OCH}_3$ would have been instead of $-\mathrm{NO}_2$.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Advanced, JEE Advanced 2008 (Paper 1)
The major product of the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
It is the first step of Gabriel's phthalimide synthesis. The hydrogen bonded to nitrogen is sufficiently acidic due to two $\alpha$-carbonyls.
The conjugate base formed above act as nucleophile in the subsequent step of reaction. As shown above, the nucleophile exist in three resonating form, one may think of oxygen being the donor atom in the nucleophilic attack. However, nitrogen act as donor as it is better donor than oxygen.
Bromine is not substituted in the above reaction as it is in resonance with benzene ring giving partial double bond character to $\mathrm{C}-\mathrm{Br}$ bond, hence difficult to break.

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2011 (Paper 1)
The major product of the following reaction is

Options:
A
a hemiacetal

B
an acetal

C
an ether

D
an ester

The correct answer is:
an acetal

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2011 (Paper 2)
The major product of the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{PhS}^{-}$is a strong nucleophile and dimethylformamide (DMF) is a highly polar aprotic solvent. Condition indicates that nucleophilic substitution $\left(\mathrm{S}_{\mathrm{N}} 2\right)$ takes place at $2^{\circ}$ benzylic place, stereochemically it involves inversion of configuration.
Aryl group can exhibit neighbouring group or can show anchimeric assistance but not possible here due to deactivity imparted by $-\mathrm{NO}_2$ and $-\mathrm{F}$. It can be if $-\mathrm{OCH}_3$ would have been instead of $-\mathrm{NO}_2$.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product of the following reaction is

Options:
A
a hemiacetal

B
an acetal

C
an ether

D
an ester

The correct answer is:
an acetal

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The major product of the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
It is the first step of Gabriel's phthalimide synthesis. The hydrogen bonded to nitrogen is sufficiently acidic due to two $\alpha$-carbonyls.
The conjugate base formed above act as nucleophile in the subsequent step of reaction. As shown above, the nucleophile exist in three resonating form, one may think of oxygen being the donor atom in the nucleophilic attack. However, nitrogen act as donor as it is better donor than oxygen.
Bromine is not substituted in the above reaction as it is in resonance with benzene ring giving partial double bond character to $\mathrm{C}-\mathrm{Br}$ bond, hence difficult to break.

Topic: Chemistry, Amines, JEE Main
$$
\text { The major products } P \text { and } Q \text { are }
$$

Options:
A

B

C

D

The correct answers are:

Solution:
$$
\text { It is cumene hydroperoxide rearrangement. }
$$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Advanced, JEE Advanced 2006
The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point $P$ is

Options:
A
$\frac{k_a A}{k_2}$

B
$\frac{k_2 A}{k_1}$

C
$\frac{k_1 A}{k_1+k_2}$

D
$\frac{k_2 A}{k_1+k_2}$

The correct answer is:
$\frac{k_2 A}{k_1+k_2}$

Solution:
$x_1+x_2=A$ and $k_1 x_1=k_2 x_2$ or $\frac{x_1}{x_2}=\frac{k_2}{k_1}$
Solving these equations, we get
$$
x_1=\left(\frac{k_2}{k_1+k_2}\right) A
$$

Topic: Physics, Oscillations, JEE Main
The mass $M$ shown in the figure oscillates in simple harmonic motion with amplitude $A$. The amplitude of the point $P$ is

Options:
A
$\frac{k_a A}{k_2}$

B
$\frac{k_2 A}{k_1}$

C
$\frac{k_1 A}{k_1+k_2}$

D
$\frac{k_2 A}{k_1+k_2}$

The correct answer is:
$\frac{k_2 A}{k_1+k_2}$

Solution:
$x_1+x_2=A$ and $k_1 x_1=k_2 x_2$ or $\frac{x_1}{x_2}=\frac{k_2}{k_1}$
Solving these equations, we get
$$
x_1=\left(\frac{k_2}{k_1+k_2}\right) A
$$

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2009 (Paper 2)
The maximum number of electrons that can have principal quantum number, $n=3$ and spin quantum number, $m_s=-\frac{1}{2}$, is

Options:

The correct answer is:
9

Solution:
When $n=3, l=0,1,2$ i.e., there are $3 s, 3 p$ and $3 d$ orbitals. If all these orbitals are completely occupied as
$$
\text { Total } 18 \text { electrons, } 9 \text { electrons with } s=+\frac{1}{2} \text { and } 9 \text { with } s=-\frac{1}{2} .
$$
Alternatively In any $n$th orbit, there can be a maximum of $2 n^2$ electrons. Hence, when $n=3$, number of maximum electrons $=18$. Out of these 18 electrons, 9 can have spin $-\frac{1}{2}$ and remaining nine with spin $=+\frac{1}{2}$.

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2011 (Paper 1)
The maximum number of isomers (including stereoisomers) that are possible on monochlorination of the following compounds, is

Options:

The correct answer is:
8

Solution:
I has one chiral carbon $=$ two isomers II has two chiral carbons and no symmetry $=$ four isomers.
III and IV have no chiral carbon, no stereoisomers.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Advanced, JEE Advanced 2011 (Paper 2)