datavorous/entrance-exam-dataset huggingface dataset:
Let $z_1$ and $z_2$ be two distinct complex numbers and let $z=(1-t) z_1+t z_2$ for some real number $t$ with $0 < t < 1$. If $\arg (w)$ denotes the principal argument of a non-zero complex number $w$, then

Options:
A
$\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right|$

B
$\arg \left(z-z_1\right)=\arg \left(z-z_2\right)$

C
$\left|\begin{array}{cc}z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1\end{array}\right|=0$

D
$\arg \left(z-z_1\right)=\arg \left(z_2-z_1\right)$

The correct answers are:
$\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right|$
,
$\left|\begin{array}{cc}z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1\end{array}\right|=0$
,
$\arg \left(z-z_1\right)=\arg \left(z_2-z_1\right)$

Solution:
Given, $z=\frac{(1-t) z_1+t z_2}{(1-t)+t}$
Clearly, $z$ divides $z_1$ and $z_2$ in the ratio of $t:(1-t), 0 < t < 1$
$$
\begin{aligned}
& \Rightarrow \quad A P+B P=A B \\
& \text { ie, }\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right| \\
& \Rightarrow \text { Option (a) is true. } \\
& \text { and } \begin{aligned}
\arg \left(z-z_1\right) & =\arg \left(z_2-z\right) \\
& =\arg \left(z_2-z_1\right)
\end{aligned}
\end{aligned}
$$
$\Rightarrow$ (b) is false and (d) is true.
Also, $\arg \left(z-z_1\right)=\arg \left(z_2-z_1\right)$
$$
\begin{aligned}
& \Rightarrow \quad \arg \left(\frac{z-z_1}{z_2-z_1}\right)=0 \\
& \therefore \frac{z-z_1}{z_2-z_1} \text { is purely real. } \\
& \Rightarrow \quad \frac{z-z_1}{z_2-z_1}=\frac{\bar{z}-\bar{z}_1}{\bar{z}_2-\bar{z}_1} \\
& \text { or }\left|\begin{array}{cc}
z-z_1 & \bar{z}-\bar{z}_1 \\
z_2-z_1 & \bar{z}_2-\bar{z}_1
\end{array}\right|=0 \\
&
\end{aligned}
$$
$\therefore$ Option (c) is correct.
Hence, (a, c, d) is the correct option.

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a=z^{2}+z+1$ is real. Then a cannot take the value

Options:
A $-1$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{3}{4}$

The correct answer is:
$\frac{3}{4}$

Solution:
$\because \quad \operatorname{Im}(z) \neq 0$ $\Rightarrow \quad z$ is non-real

and equation $z^{2}+z+(1-a)=0$

will have non-real roots, if $D < 0$

$\Rightarrow \quad 1-4(1-a) < 0$ $\Rightarrow \quad 4 a < 3 \Rightarrow a < \frac{3}{4}$

$\therefore \quad a$ can not take the value $\frac{3}{4}$.

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2012 (Paper 1)
Let $z=\cos \theta+i \sin \theta$. Then, the value of $\sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right)$ at $\theta=2^{\circ}$ is

Options:
A
$\frac{1}{\sin 2^{\circ}}$

B
$\frac{1}{3 \sin 2^{\circ}}$

C
$\frac{1}{2 \sin 2^{\circ}}$

D
$\frac{1}{4 \sin 2^{\circ}}$

The correct answer is:
$\frac{1}{4 \sin 2^{\circ}}$

Solution:
Given that $z=\cos \theta+i \sin \theta=e^{i \theta}$
$$
\begin{aligned}
\therefore \sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right) & =\sum_{m=1}^{15} \operatorname{Im}\left(e^{i \theta}\right)^{2 m-1}=\sum_{m=1}^{15} \operatorname{Im} e^{i(2 m-1) \theta} \\
& =\sin \theta+\sin 3 \theta+\sin 5 \theta+\ldots+\sin 29 \theta \\
& =\frac{\sin \left(\frac{\theta+29 \theta}{2}\right) \sin \left(\frac{15 \times 2 \theta}{2}\right)}{\sin \left(\frac{2 \theta}{2}\right)}=\frac{\sin (15 \theta) \sin (15 \theta)}{\sin \theta}=\frac{1}{4 \sin 2^{\circ}}
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2009 (Paper 1)
Let $z=x+i y$ be a complex number, where $x$ and $y$ are integers. Then, the area of the rectangle whose vertices are the roots of the equation $z \bar{z}^3+\bar{z} z^3=350$ is

Options:
A
48

B
32

C
40

D
80

The correct answer is:
48

Solution:
Since, $z \bar{z}\left(z^2+\bar{z}^2\right)=250$
$$
\begin{aligned}
& \Rightarrow \quad 2\left(x^2+y^2\right)\left(x^2-y^2\right)=350 \\
& \Rightarrow \quad\left(x^2+y^2\right)\left(x^2-y^2\right)=175
\end{aligned}
$$
Since, $x, y \in I$, the only possible case which gives integral solution, is and $\quad \begin{aligned} x^2+y^2 & =25 \\ x^2-y^2 & =7\end{aligned}$
From Eqs. (i) and (ii), we get $x^2=16 ; y^2=9 \Rightarrow x=\pm 4 ; y=\pm 3$
$\therefore$ Area of rectangle $=8 \times 6=48$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2009 (Paper 1)
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ equals

Options:
A
$\frac{8}{\pi} f(2)$

B
$\frac{2}{\pi} f(2)$

C
$\frac{2}{\pi} f\left(\frac{1}{2}\right)$

D
$4 f(2)$

The correct answer is:
$\frac{8}{\pi} f(2)$

Solution:
$\lim _{x \rightarrow \frac{x}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ $\left[\therefore \frac{0}{0}\right.$ form $]$
Let
$$
\begin{array}{ll}
\text { Let } & L=\lim _{x \rightarrow \frac{x}{4}} \frac{f\left(\sec ^2 x\right) 2 \sec x \sec x \tan x}{2 x} \\
\therefore & L=\frac{2 f(2)}{\pi / 4}=\frac{8 f(2)}{\pi}
\end{array}
$$

Topic: Mathematics, Limits, JEE Advanced, JEE Advanced 2007 (Paper 1)
Lines $L_1: y-x=0$ and $L_2: 2 x+y=0$ intersect the line $L_3: y+2=0$ at $P$ and $Q$, respectively. The bisector of the acute angle between $L_1$ and $L_2$ intersects $L_3$ at $R$.
Statement I The ratio $P R: R Q$ equals $2 \sqrt{2}: \sqrt{5}$.
Statement II In any triangle, bisector of an angle divides the triangle into two similar triangles.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

Solution:
In $\triangle O P Q$
Clearly,
$$
\frac{P R}{R Q}=\frac{O P}{O Q}=\frac{2 \sqrt{2}}{\sqrt{5}}
$$

Topic: Mathematics, Straight Lines, JEE Advanced, JEE Advanced 2007 (Paper 2)
List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in m s-1) at time t=π3 s.

List-I	List-II

(I)

A and B are moving on a horizontal circle of radius 1 m with uniform angular speed ω=1 rad s-1. The initial angular positions of A and B at time t=0 are θ=0 and θ=π2 respectively.

(P)	3+12

(II)

Projectiles A and B are fired (in the same vertical plane) at t=0 and t=0.1 s respectively, with the same speed v=5π2 m s-1 and at 45° from the horizontal plane. The initial separation between A and B is large enough so that they do not collide, g=10 m s-2.

(Q)	3-12

(III)

Two harmonic oscillators A and B moving in the x direction according to xA=x0sintt0 and xB=x0sintt0+π2 respectively, starting from t=0. Take x0=1 m,t0=1 s.

(R)	10

(IV)

Particle A is rotating in a horizontal circular path of radius 1 m on the xy plane, with constant angular speed ω=1 rad s-1. Particle B is moving up at a constant speed 3 m s-1 in the vertical direction as shown in the figure. (Ignore gravity.)

(S)	2

(T)	25π2+1

Which one of the following options is correct?

Options:
A I→R,II→T,III→P,IV→S
B I→S,II→P,III→Q,IV→R
C I→S,II→T,III→P,IV→R
D I→T,II→P,III→R,IV→S

The correct answer is:
I→S,II→T,III→P,IV→R

Solution:
(I)

Angular velocity of both the particles is same. Therefore, angle between them will remain 90°.

vrel=2ωR=2 m s-1

Hence, I→S

(II)

The velocity of A at any time t can be written as

vAt=5π2cos45°i^+5π2×sin45°-g×tj^

=5π2i^+5π2-gtj^

Velocity of B at any time t

vBt=-5π2i^+5π2-gt-0.1j^

Relative velocity at time t will be,

v→rel=5πi^-g×0.1j^

v→rel=25π2+1 m s-1

Hence, II→T

(III)

Relative position of A wrt B can be written as

x=xA-xB=x0sint-x0sint+π2

Using trigonometric relation

x=2x0sint-π4

Therefore,

vrel=dxdt=2x0cost-π4

at t=π3

vrel=2cosπ3-π4=2×3+122=3+12 m s-1

Hence, III→P

(IV)

The velocity of A is in xy plane and it is vA=ωr=1 m s-1 and the velocity of B is along z axis and vB=3 m s-1. Therefore,

vrel=32+12=10 m s-1

Hence, IV→R

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2022 (Paper 1)
Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness.
The mass of ink used to draw each of the two inner circles, and each of the two line segments is $m$. The mass of the ink used to draw the outer circle is $6 \mathrm{~m}$. The coordinates of the centres of the different parts are, outer circle 10 , $0)$, left inner circle $(0,0)$, right inner circle $(a, a)$, vertical line $(0,0)$ and horizontal line $(0,-a)$. The $y$-coordinate of the centre of mass of the ink in this drawing is

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

B
$\frac{a}{8}$

C
$\frac{a}{12}$

D
$\frac{a}{3}$

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

Solution:
$y_{C M}$
$$
\begin{gathered}
=\frac{m_1 y_1+m_2 y_2+m_3 y_3+m_4 y_4+m_5 y_5}{m_1+m_2+m_3+m_4+m_5} \\
=\frac{(6 m)(0)+(m)(a)+m(a)+m(0)}{6 m+m+m+m+m}=\frac{a}{10}
\end{gathered}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2009 (Paper 1)
Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness.
The mass of ink used to draw each of the two inner circles, and each of the two line segments is $m$. The mass of the ink used to draw the outer circle is $6 \mathrm{~m}$. The coordinates of the centres of the different parts are, outer circle 10 , $0)$, left inner circle $(0,0)$, right inner circle $(a, a)$, vertical line $(0,0)$ and horizontal line $(0,-a)$. The $y$-coordinate of the centre of mass of the ink in this drawing is

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

B
$\frac{a}{8}$

C
$\frac{a}{12}$

D
$\frac{a}{3}$

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

Solution:
$y_{C M}$
$$
\begin{gathered}
=\frac{m_1 y_1+m_2 y_2+m_3 y_3+m_4 y_4+m_5 y_5}{m_1+m_2+m_3+m_4+m_5} \\
=\frac{(6 m)(0)+(m)(a)+m(a)+m(0)}{6 m+m+m+m+m}=\frac{a}{10}
\end{gathered}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Mark the incorrect option.

Options:
A The larva of Scoliodon exists.
B Lateral line is absent in Bangarus.
C Air bladder is present in Betta.
D All chordates are deuterostomes.

The correct answer is:
The larva of Scoliodon exists.

Solution:
Scoliodon is a predator of small bony fishes and invertebrates. This species exhibits the most advanced mode of viviparity of any fish, in which the developed embryos form a highly complex placental connection to the mother at a very small size.

Topic: Biology, Animal Kingdom, NEET
Match each of the compounds given in Column I with the reaction (s) that they can undergo, given in Column II.

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

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

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

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

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

Solution:
No Solution Available

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2009 (Paper 2)
$$
\text { Match each of the compounds is Column I with its characteristic reaction(s) is Column II. }
$$

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

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

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

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

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

Solution:
No Solution Available

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Advanced, JEE Advanced 2009 (Paper 1)
Match each of the diatomic molecules in Column I with its property/properties in Column II.

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

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

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

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

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

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Match each of the diatomic molecules in Column I with its property/properties in Column II.

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

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

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

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

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

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Advanced, JEE Advanced 2009 (Paper 1)
Match each of the reactions given in Column I with the corresponding products(s) given in Column II.

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

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

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

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

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

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Match each of the reactions given in Column I with the corresponding products(s) given in Column II.

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

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

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

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

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

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2009 (Paper 2)
Match the chemical substance in Column I with type of polymers/type of bond in Column II. Indicate your answer by darkening the appropriate bubbles of $4 \times 4$ matrix given in the ORS.

Options:
A
A-p; B-q; C-p; D-q, r

B
A-p, s; B-q, r; C-p, r; D-s

C
A-s; B-p; C-p; D-q

D
A-p, s; B-r; C-p, q; D-s

The correct answer is:
A-p, s; B-q, r; C-p, r; D-s

Solution:
(A) $p, s$
(B) $q, r$
(C) $p, r$
(D) $s$
Sucrose (diasaccharide) < smiles>OCC1OC(CO)C(O)C(O)C1O < /smiles>
Sucrose/glucosidic as well as fructosidic linkage

Topic: Chemistry, Polymers, JEE Advanced, JEE Advanced 2007 (Paper 1)
Match the chemical substance in Column I with type of polymers/type of bonds in Column II. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

Options:
A
A-p, r; B-p, q; C-p; D-q, r

B
A-p, s; B-p; C-p, s; D-q, r

C
A-p, q, s; B-q; C-r, s, s; D-r

D
A-r, s; B-q; C-p; D-p, q

The correct answer is:
A-p, q, s; B-q; C-r, s, s; D-r

Solution:
A-p, $q, s, \quad \mathrm{~B}-q, \quad \mathrm{C}-r, s, \quad \mathrm{D}-r$

Topic: Chemistry, General Organic Chemistry, JEE Main
Match the chemical substance in Column I with type of polymers/type of bonds in Column II. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

Options:
A
A-p, r; B-p, q; C-p; D-q, r

B
A-p, s; B-p; C-p, s; D-q, r

C
A-p, q, s; B-q; C-r, s, s; D-r

D
A-r, s; B-q; C-p; D-p, q

The correct answer is:
A-p, q, s; B-q; C-r, s, s; D-r

Solution:
A-p, $q, s, \quad \mathrm{~B}-q, \quad \mathrm{C}-r, s, \quad \mathrm{D}-r$

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 2)
Match the column.
Normals at $P, Q, R$ are drawn to $y^2=4 x$ which intersect at $(3,0)$. Then,

Options:
A
(i) (a), (ii) (b), (iii) (d), (iv) (c)

B
(i) (b), (ii) (a), (iii) (d), (iv) (c)

C
(i) (a), (ii) (b), (iii) (c), (iv) (d)

D
(i) (b), (ii) (a), (iii) (c), (iv) (d)

The correct answer is:
(i) (a), (ii) (b), (iii) (d), (iv) (c)

Solution:
Since, equation of normal is $y+x t=2 a t+a t^3$ passes through $(3,0)$.
$$
\Rightarrow \quad 3 t=2 t+t^3 \Rightarrow t=0,1,-1 \text {. }
$$
$\therefore P(1,2), Q(0,0), R(1,-2)$. Thus,
(i) Area of $\triangle P Q R=\frac{1}{2} \times 1 \times 4=2$
(ii) Centroid of $\triangle P Q R=\left(\frac{2}{3}, 0\right)$
Equation of circle passing through $P, Q, R$ is
$$
\begin{aligned}
& \quad(x-1)(x-1)+(y-2)(y+2)+\lambda(x-1)=0 \\
& \Rightarrow \quad 1-4-\lambda=0 \Rightarrow \lambda=-3 \\
& \therefore \text { Required equation of circle is } x^2+y^2-5 x-1=0 \\
& \therefore \text { Centre }\left(\frac{5}{2}, 0\right) \text { and radius } \frac{5}{2} .
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2006
Match the complexes in Column I with their properties listed in Column II. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

Options:
A
A-p; B-q; C-p; D-q, r

B
A-p, s; B-q, r; C-p, r; D-r, s

C
A-s; B-p; C-p; D-q

D
A-p, s; B-r; C-p, q; D-p, s

The correct answer is:
A-p, s; B-r; C-p, q; D-p, s

Solution:
A-p,s B-r, C-p,q, D-p,s,
$$
\begin{aligned}
2 \mathrm{CrO}_4^{2-} \longrightarrow \mathrm{Cr}_2 \mathrm{O}_7^{2-} \\
\mathrm{MnO}_4^{-}+\mathrm{NO}_2^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{O} \\
\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{NO}+\mathrm{SO}_4^{2-}+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, JEE Main
Match the complexes in Column I with their properties listed in Column II. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

Options:
A
A-p; B-q; C-p; D-q, r

B
A-p, s; B-q, r; C-p, r; D-r, s

C
A-s; B-p; C-p; D-q

D
A-p, s; B-r; C-p, q; D-p, s

The correct answer is:
A-p, s; B-r; C-p, q; D-p, s

Solution:
A-p,s B-r, C-p,q, D-p,s,
$$
\begin{aligned}
2 \mathrm{CrO}_4^{2-} \longrightarrow \mathrm{Cr}_2 \mathrm{O}_7^{2-} \\
\mathrm{MnO}_4^{-}+\mathrm{NO}_2^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{O} \\
\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{NO}+\mathrm{SO}_4^{2-}+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2007 (Paper 2)
Match the complexes in Column I with their properties listed in Column II. Indicates your answer by darkening the appropriate bubbles of $4 \times 4$ matrix given in the ORS.

Options:
A
A-p, r; B-p, q, s; C-p; D-q, r

B
A-p, s; B-p, r, s; C-p, s; D-q, r

C
A-p, q, s; B-p, r, s; C-q, s; D-q, s

D
A-r, s; B-q; C-p; D-p, q, s

The correct answer is:
A-p, q, s; B-p, r, s; C-q, s; D-q, s

Solution:
(A) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left[\left(\mathrm{H}_2 \mathrm{O}\right)_2\right] \mathrm{Cl}_2 p, q, s\right.$
Oxidation state $\mathrm{Co}$ is $+2$
$\Rightarrow$ Paramagnetic, exhibits geometrical isomerism.
(B) $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] p, r, s$
$\mathrm{dsp}{ }^2$-hybridisation
$\mathrm{Pt}^{2+}$, diamagnetic square planar, geometrical isomerism
(C) $q, s$ (D) $q, s$

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2007 (Paper 1)
Match the compounds in Column I with their characteristic test(s)/reactions(s) given in Column II. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
A-r; B-p, q; C- p, q, r; D-p
Sodium fusion extract gives Prussian blue when nitrogen is present alongwith carbon. Phenolic group and carboxylate gives positive $\mathrm{FeCl}_3$ test - white precipitate comes with $\mathrm{AgCl}$.
$$
\text { Hydrazone formation occurs effectively at } \mathrm{pH} \approx 4.5 \text {. The reaction proceeds in that }
$$
condition only when $\mathrm{H}^{+}$concentration is merely sufficient to activate. As $\mathrm{H}^{+}$concentration raises, sufficient molecule of hydrazone gets converted into hydrazonium which is not nucleophilic and reaction becomes impossible. Further, low concentration of $\mathrm{H}^{+}$(in the case of hydrolysis of 2, 4-dinitrophenyl hydrazonium bromide) is not effective to proceed elimination (rate determining step).
If it is assumed that the given hydrazonian bromiate is the result as salt main $\mathrm{HBr}$ is used as acid catalyst. Remember! hydrazone formation involves first addition and then elimination (res). In the second step substitution dominates over elimination ( $\mathrm{HBr}$ is not a dehydrating reagent mutually).

Topic: Chemistry, Hydrocarbons, JEE Main
Match the compounds in Column I with their characteristic test(s)/reactions(s) given in Column II. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
A-r; B-p, q; C- p, q, r; D-p
Sodium fusion extract gives Prussian blue when nitrogen is present alongwith carbon. Phenolic group and carboxylate gives positive $\mathrm{FeCl}_3$ test - white precipitate comes with $\mathrm{AgCl}$.
$$
\text { Hydrazone formation occurs effectively at } \mathrm{pH} \approx 4.5 \text {. The reaction proceeds in that }
$$
condition only when $\mathrm{H}^{+}$concentration is merely sufficient to activate. As $\mathrm{H}^{+}$concentration raises, sufficient molecule of hydrazone gets converted into hydrazonium which is not nucleophilic and reaction becomes impossible. Further, low concentration of $\mathrm{H}^{+}$(in the case of hydrolysis of 2, 4-dinitrophenyl hydrazonium bromide) is not effective to proceed elimination (rate determining step).
If it is assumed that the given hydrazonian bromiate is the result as salt main $\mathrm{HBr}$ is used as acid catalyst. Remember! hydrazone formation involves first addition and then elimination (res). In the second step substitution dominates over elimination ( $\mathrm{HBr}$ is not a dehydrating reagent mutually).

Topic: Chemistry, Hydrocarbons, JEE Advanced, JEE Advanced 2008 (Paper 2)
$$
\text { Match the conics in Column I with the statements/expressions in Column II. }
$$

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

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

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

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

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

Solution:
(p) $\frac{1}{\sqrt{h^2+k^2}}=2 \Rightarrow h^2+k^2=\frac{1}{4}$
Hence, locus is a circle.
(q) ||$z+2|-| z-2||=3$
and $\quad 2-(-2)=4>3$
Hence, locus is hyperbola.
(r) Let $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$
Let $\quad \tan \theta=t$ $\Rightarrow x=\sqrt{3} \cos 2 \theta, y=\sin 2 \theta$
$\therefore \quad \frac{x^2}{3}+y^2=1$
Hence, locus is an ellipse.
(s) Eccentricity $x=1 \Rightarrow$ Parabola, $1 < x < \infty \Rightarrow$ Hyperbola
(t) Let $z=x+i y$
Since, $\operatorname{Re}(z+1)^2=|z|^2+1$
$\Rightarrow \quad(x+1)^2-y^2=x^2+y^2+1$
$\Rightarrow \quad 2 x=2 y^2$
$\Rightarrow \quad x=y^2$
Hence, locus is parabola.

Topic: Mathematics, Hyperbola, JEE Advanced, JEE Advanced 2009 (Paper 1)
Match the conversions in Column I with the type(s) given in Column II. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2008 (Paper 2)
Match the entries in Column I with the correctly related quantum number(s) in Column II. Indicate your answer by darkening the appropriate bubbles of the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
A-q; B-p, q, r, s; C-p, q, r; D-p, q, r
(B) A hydrogen-like one electron wave function obeying Pauli's principle First, as it is the wave function so it will be characterised by $n, l$ and $m$. Further, it obeys the Pauli's principle, that's why it will be characterised completely by adding spin quantum number also.
(D) Radial probability density of electron at the nucleus basically depends on azimuthal quantum number, but as it is the function of radius depends on principal quantum number, When angular probability density is included, that depends on magnetic quantum number.

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2008 (Paper 2)