datavorous/entrance-exam-dataset huggingface dataset:
Match the extraction processes listed in column I with metals listed in column II.

Options:
A
(A) P,S, (B) R, (C) P,Q, (D) P,S

B
(A) Q,S, (B) R, (C) S, (D) S

C
(A) P,R, (B) P, R, (C) Q, (D) R

D
(A) P,R, (B) P, (C) Q, (D) S

The correct answer is:
(A) P,R, (B) P, (C) Q, (D) S

Solution:
Column I
Column II
(A) Self reduction $\left[\begin{array}{c}P \text { - Lead } \\ R \text { - Copper }\end{array}\right]$
$$
\begin{aligned}
& \text { i.e., } A \rightleftarrows P \text { and } R \\
& \text { In lead metal } \\
& \qquad\left\{\begin{array}{c}
2 \mathrm{PbS}+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{PbO}+2 \mathrm{SO}_2 \\
2 \mathrm{PbO}+\mathrm{PbS} \longrightarrow 3 \mathrm{~Pb}+\mathrm{SO}_2
\end{array}\right.
\end{aligned}
$$
(self reduction)
In copper metal
$$
\begin{aligned}
& 2 \mathrm{Cu}_2 \mathrm{~S}+3 \mathrm{O}_2 \longrightarrow 2 \mathrm{Cu}_2 \mathrm{O}+2 \mathrm{SO}_2 \\
& \mathrm{Cu}_2 \mathrm{~S}+2 \mathrm{Cu}_2 \mathrm{O} \longrightarrow 6 \mathrm{Cu}+\mathrm{SO}_2
\end{aligned}
$$
(self reduction)
(B) Carbon reduction $P$-lead
$$
\begin{aligned}
& \text { i.e., } \mathrm{B}-\mathrm{P} \\
& \text { In lead metal } \\
& \qquad \mathrm{PbO}+\mathrm{C} \longrightarrow \mathrm{Pb}+\mathrm{CO} \\
& \mathrm{PbO}+\mathrm{CO} \longrightarrow \mathrm{Pb}+\mathrm{CO}_2
\end{aligned}
$$
Carbon reduction process is used for the reduction of oxides of electropositive metals like $\mathrm{Fe}, \mathrm{Pb}$ etc. This reduction is carried out with coal or coke on strongly heating.
(C) Complex formation and displacement by metal.
i.e., $C-Q$ (silver)
In silver metal
Argentite mineral of $\mathrm{Ag}\left(\mathrm{Ag}_2 \mathrm{~S}\right)$ is treated with $0.7 \% \mathrm{NaCN}$ solution in presence of air, therefore firstly sodium argento cyanide complex is formed which on heating with zinc metal, therefore by displacement silver is isolated.
$$
\begin{aligned}
& \mathrm{Ag}_2 \mathrm{~S}+4 \mathrm{NaCN} \longrightarrow 2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+\mathrm{Na}_2 \mathrm{~S} \\
& 2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+\mathrm{Zn} \longrightarrow \mathrm{Na}\left[\mathrm{Zn}(\mathrm{CN})_4\right]+\underset{\text { Soluble }}{2 \mathrm{Ag}} \downarrow
\end{aligned}
$$
(D) Decomposition of iodide-(S) Boron
$$
\text { i.e., }(D)-(S)
$$
By decomposition of iodide boron metal is isolated.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2006
Match the following

Options:
A
(A) Q,S, (B) S, (C) R, (D) Q,R

B
(A) Q,S, (B) R, (C) S, (D) Q

C
(A) P,R, (B) P, R, (C) R, (D) Q,R

D
(A) P,R, (B) P, (C) R, (D) Q,R

The correct answer is:
(A) Q,S, (B) S, (C) R, (D) Q,R

Solution:
A match with $\boldsymbol{Q}, \boldsymbol{S} \quad \mathrm{Bi}^{3+}+\mathrm{H}_2 \mathrm{O} \longrightarrow[\mathrm{BiO}]^{+}+2 \mathrm{H}^{+}$ B match with $\boldsymbol{S} \mathrm{NaAlO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Al}(\mathrm{OH})_3+\mathrm{NaOH}$ C match with $\boldsymbol{R} \quad 2 \mathrm{SiO}_4^{4-}+2 \mathrm{H}^{+} \longrightarrow \mathrm{Si}_2 \mathrm{O}_7^{6-}+\mathrm{H}_2 \mathrm{O}$ Pyrosilicate is formed by treating orthosilicate with acid. D match with $\boldsymbol{Q}, \boldsymbol{R} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \underset{\text { or } \mathrm{H}_2 \mathrm{SO}_4}{\stackrel{\mathrm{HCl}}{\mathrm{O}_3}} \mathrm{H}_3 \mathrm{BO}_3$
$$
\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_4 \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{H}_3 \mathrm{BO}_3
$$

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Advanced, JEE Advanced 2006
Match the following

Options:
A
(A) Q, (B) P, (C) R,S, (D) Q,S

B
(A) S, (B) R, (C) R, (D) P,S

C
(A) R, (B) Q, (C) S, (D) Q,S

D
(A) Q, (B) Q, (C) R,S, (D) P,S

The correct answer is:
(A) Q, (B) Q, (C) R,S, (D) P,S

Solution:
$$
\text { (A) Match with (Q) }
$$
The formation of $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CD}_3$ can be explained on the basis of the fact that $\mathrm{C}-\mathrm{D}$ bond is much stronger than $\mathrm{C}-\mathrm{H}$ bond.
(B) match with (Q)
Reactivity of $\mathrm{PhCHBrCH}_3$ is greater than
$\mathrm{Ph} \mathrm{CHBrCD}_3$ because $\mathrm{C}-\mathrm{D}$ bond is more stronger than $\mathrm{C}-\mathrm{H}$ bond.
$$
\text { (C) match with (R) and (S) }
$$
In the step (II), a slow unimolecular elimination occurs in the conjugate base of the reactant and hence this mechanism is called $E_1 C B$ or carbanion mechanism. Since step (I) must be reversible, if ethanol containing $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD}$ is used as solvent, it would be expected that the original bromide would incorporate deuterium (D).
$$
\text { (D) Match with (P) and (S) }
$$
Step I. $\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br} \stackrel{\text { Slow }}{\longrightarrow} \mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}} \mathrm{H}_2+\mathrm{Br}^{-}$
Step II. $\mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}_2} \mathrm{H}_2 \stackrel{\text { Fast }}{\longrightarrow} \mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_2+\mathrm{H}^{-}$
Rate $\propto\left[\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br}\right]$
Similarly
Step II. $\mathrm{PhCD}_2 \stackrel{+}{\mathrm{C}_2} \mathrm{H}_2 \stackrel{\text { Fast }}{\longrightarrow} \mathrm{PhCD}=\mathrm{CHD}+\mathrm{H}^{+}$
Rate $\propto\left[\mathrm{PhCD}_2-\mathrm{CH}_2 \mathrm{Br}\right]$
Hence, $E_1$ reaction and first order kinetics.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Advanced, JEE Advanced 2006
Match the following

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

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

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

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

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

Solution:
(i) Solved the two equations, say, i.e. $x+y=|a|$ and $a x-y=1$, we get $x=\frac{|a|+1}{a+1}>0$ and $y=\frac{|a|-1}{a+1}>0$
when $a+1>0$; we get $a>1$
$\therefore \quad a_0=1$
(ii) We have,
$\mathbf{a}=\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+\gamma \hat{\mathbf{k}}$
$\Rightarrow \quad \mathbf{a} \cdot \hat{\mathbf{k}}=\gamma$
Now, $\hat{\mathbf{k}} \times(\hat{\mathbf{k}} \times \mathbf{a})=(\hat{\mathbf{k}} \cdot \mathbf{a})-(\hat{\mathbf{k}} \cdot \hat{\mathbf{k}}) \mathbf{a}=\gamma \hat{\mathbf{k}}-(\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}+\gamma \hat{\mathbf{k}})$ $\Rightarrow \quad \alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}}=0 \Rightarrow \alpha=\beta=0$
Also, $\quad \alpha+\beta+\gamma=2 \Rightarrow \gamma=2$
(iv) $\sin A \sin B \sin C+\cos A \cos B \leq \sin A \sin B+\cos A \cos B=\cos (A-B)$
$$
\begin{array}{lc}
\Rightarrow & \cos (A-B) \geq 1 \Rightarrow \cos (A-B)=1 \\
\Rightarrow & \sin C=1
\end{array}
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2006
Match the following.

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

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

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

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

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

Solution:
$$
\begin{aligned}
& \text { (i) } \sum_{i=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 i^2}\right)=t \Rightarrow \sum_{i=1}^{\infty} \tan ^{-1}\left(\frac{2}{4 i^2-1+1}\right) \\
& \Rightarrow \sum_{i=1}^{\infty} \tan ^{-1}\left\{\frac{(2 i+1)-(2 i-1)}{1+(2 i-1)(2 i+1)}\right\} \\
& \Rightarrow\left(\tan ^{-1} 3-\tan ^{-1} 1\right)+\left(\tan ^{-1} 5-\tan ^{-1} 3\right)+\ldots+\left\{\tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)\right\} \\
& \therefore \quad \quad t=\lim _{n \rightarrow \infty}\left(\tan ^{-1}(2 n+1)-\tan ^{-1} 1\right)=\lim _{n \rightarrow \infty} \tan ^{-1}\left(\frac{2 n}{1+2 n+1}\right)=\frac{\pi}{4} \\
& \therefore \quad \tan t=1 .
\end{aligned}
$$
(ii) We have, $\cos \theta_1=\frac{1-\tan ^2 \frac{\theta_1}{2}}{1+\tan ^2 \frac{\theta_2}{2}}=\frac{a}{b+c}$
$$
\Rightarrow \quad \tan ^2 \frac{\theta}{2}=\frac{b+c-a}{b+c+a}
$$
Also, $\quad \cos \theta_3=\frac{1-\tan ^2 \frac{\theta_3}{2}}{1+\tan ^2 \frac{\theta_3}{2}}=\frac{c}{a+b}$
$$
\Rightarrow \quad \tan ^2 \frac{\theta_3}{2}=\frac{a+b-c}{a+b+c}
$$
$\Rightarrow \tan ^2 \frac{\theta_1}{2}+\tan ^2 \frac{\theta_3}{2}=\frac{2 b}{a+b+c}=\frac{2 b}{3 b}=\frac{2}{3}, \quad$ [as, $a, b, c$ are in AP $\Rightarrow 2 b=a+c$ ]
(iii) Line through $(0,1,0)$ and perpendicular to plane $x+2 y+2 z=0$ is given by $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-0}{2}=r$
$\therefore P(r, 2 r+1,2 r)$ be the foot of perpendicular on the straight line, then
$$
\begin{array}{ll}
& r \cdot 1+(2 r+1) \cdot 2+(2 r) \cdot 2=0 \\
\Rightarrow \quad r= & \quad r\left(-\frac{2}{9}, \frac{5}{9},-\frac{4}{9}\right) \\
\therefore & \quad P(r) \\
\therefore & \text { Required perpendicular distance }=\sqrt{\frac{4+25+16}{81}}=\frac{\sqrt{5}}{3} \text { unit. }
\end{array}
$$

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2006
Match the following.

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

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

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

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

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

Solution:
(i) $I=\int_0^{\pi / 2}(\sin x)^{\cos x}\left(\cos x \cot x-\log (\sin x)^{\sin x}\right) d x=\int_0^{\pi / 2} \frac{d}{d x}(\sin x)^{\cos x} d x=1$
(ii) The point of intersection of $-4 y^2=x$ and $x-1=-5 y^2$ is $(-4,-1)$ and $(-4,1)$.
Hence, required area $=2\left\{\int_0^1\left(1-5 y^2\right) d y-\int_0^1-4 y^2 d y\right\}=\frac{4}{3}$.
(iii) The point intersection of $y=3^{x-1} \log x$ and $y=x^x-1$ is $(1,0)$.
Hence, $\quad \frac{d y}{d x}=\frac{3^{x-1}}{x}+3^{x-1} \cdot \log 3 \cdot \log x$
$\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1$
For
$$
\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1
$$
If $\theta$ is angle between the curves, then $\tan \theta=0$.
$$
\because \quad \theta=0^{\circ}
$$
(iv)
$$
\begin{aligned}
& \frac{d y}{d x}=\frac{2}{x+y} \Rightarrow \frac{d x}{d y}-\frac{x}{2}=\frac{y}{2} \\
& \Rightarrow \quad x e^{-y / 2}=\frac{1}{2} \cdot \int y \cdot e^{-y / 2} d y \\
& \Rightarrow \quad x+y+2=k e^{y / 2} \\
& \Rightarrow \quad k=3 \\
& \therefore \quad x+y+2=3 e^{y / 2} \\
&
\end{aligned}
$$
[passing through $(1,0)$ ]

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2006
Match the following

Options:
A
(A) Q, (B) P, (C) R,S, (D) Q,S

B
(A) S, (B) R, (C) R, (D) P,S

C
(A) R, (B) Q, (C) S, (D) Q,S

D
(A) Q, (B) Q, (C) R,S, (D) P,S

The correct answer is:
(A) Q, (B) Q, (C) R,S, (D) P,S

Solution:
$$
\text { (A) Match with (Q) }
$$
The formation of $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CD}_3$ can be explained on the basis of the fact that $\mathrm{C}-\mathrm{D}$ bond is much stronger than $\mathrm{C}-\mathrm{H}$ bond.
(B) match with (Q)
Reactivity of $\mathrm{PhCHBrCH}_3$ is greater than
$\mathrm{Ph} \mathrm{CHBrCD}_3$ because $\mathrm{C}-\mathrm{D}$ bond is more stronger than $\mathrm{C}-\mathrm{H}$ bond.
$$
\text { (C) match with (R) and (S) }
$$
In the step (II), a slow unimolecular elimination occurs in the conjugate base of the reactant and hence this mechanism is called $E_1 C B$ or carbanion mechanism. Since step (I) must be reversible, if ethanol containing $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD}$ is used as solvent, it would be expected that the original bromide would incorporate deuterium (D).
$$
\text { (D) Match with (P) and (S) }
$$
Step I. $\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br} \stackrel{\text { Slow }}{\longrightarrow} \mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}} \mathrm{H}_2+\mathrm{Br}^{-}$
Step II. $\mathrm{PhCH}_2-\stackrel{+}{\mathrm{C}_2} \mathrm{H}_2 \stackrel{\text { Fast }}{\longrightarrow} \mathrm{Ph}-\mathrm{CH}=\mathrm{CH}_2+\mathrm{H}^{-}$
Rate $\propto\left[\mathrm{PhCH}_2-\mathrm{CH}_2-\mathrm{Br}\right]$
Similarly
Step II. $\mathrm{PhCD}_2 \stackrel{+}{\mathrm{C}_2} \mathrm{H}_2 \stackrel{\text { Fast }}{\longrightarrow} \mathrm{PhCD}=\mathrm{CHD}+\mathrm{H}^{+}$
Rate $\propto\left[\mathrm{PhCD}_2-\mathrm{CH}_2 \mathrm{Br}\right]$
Hence, $E_1$ reaction and first order kinetics.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Match the following.

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

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

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

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

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

Solution:
(i) $I=\int_0^{\pi / 2}(\sin x)^{\cos x}\left(\cos x \cot x-\log (\sin x)^{\sin x}\right) d x=\int_0^{\pi / 2} \frac{d}{d x}(\sin x)^{\cos x} d x=1$
(ii) The point of intersection of $-4 y^2=x$ and $x-1=-5 y^2$ is $(-4,-1)$ and $(-4,1)$.
Hence, required area $=2\left\{\int_0^1\left(1-5 y^2\right) d y-\int_0^1-4 y^2 d y\right\}=\frac{4}{3}$.
(iii) The point intersection of $y=3^{x-1} \log x$ and $y=x^x-1$ is $(1,0)$.
Hence, $\quad \frac{d y}{d x}=\frac{3^{x-1}}{x}+3^{x-1} \cdot \log 3 \cdot \log x$
$\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1$
For
$$
\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1
$$
If $\theta$ is angle between the curves, then $\tan \theta=0$.
$$
\because \quad \theta=0^{\circ}
$$
(iv)
$$
\begin{aligned}
& \frac{d y}{d x}=\frac{2}{x+y} \Rightarrow \frac{d x}{d y}-\frac{x}{2}=\frac{y}{2} \\
& \Rightarrow \quad x e^{-y / 2}=\frac{1}{2} \cdot \int y \cdot e^{-y / 2} d y \\
& \Rightarrow \quad x+y+2=k e^{y / 2} \\
& \Rightarrow \quad k=3 \\
& \therefore \quad x+y+2=3 e^{y / 2} \\
&
\end{aligned}
$$
[passing through $(1,0)$ ]

Topic: Mathematics, Definite Integration, JEE Main
Match the following columns

Options:
A
(a) p,q, (b) p, (c) p, (d) p,q

B
(a) p,q,r, (b) p, s, (c) p,r, (d) p,q,r

C
(a) p,q, (b) p, r, (c) p,s, (d) p,q,r

D
(a) p,s, (b) p, s, (c) p, (d) p,q

The correct answer is:
(a) p,q, (b) p, r, (c) p,s, (d) p,q,r

Solution:
(a) $\rightarrow$ p, q
(b) $\rightarrow \mathrm{p}, \mathrm{r}$
(c) $\rightarrow$ p, $s$
(d) $\rightarrow$ p, q, r

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2006
Match the following columns

Options:
A
(a) p,q, (b) p,q, (c) p, (d) q,r

B
(a) p, (b) p,q,s, (c) p,s, (d) q,r,s

C
(a) p,q, (b) p, q, (c) p,s, (d) p,q,r

D
(a) p, (b) p, s, (c) p, (d) p,q,s

The correct answer is:
(a) p, (b) p,q,s, (c) p,s, (d) q,r,s

Solution:
(a) $\rightarrow$ p
(b) $\rightarrow$ p, q, s
(c) $\rightarrow$ q, s
(d) $\rightarrow$ q, r, s

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2006
Match the following columns

Options:
A
(a) p,q, (b) p,q, (c) p, (d) q,r

B
(a) p, (b) p,q,s, (c) p,s, (d) q,r,s

C
(a) p,q, (b) p, q, (c) p,s, (d) p,q,r

D
(a) p, (b) p, s, (c) p, (d) p,q,s

The correct answer is:
(a) p, (b) p,q,s, (c) p,s, (d) q,r,s

Solution:
(a) $\rightarrow$ p
(b) $\rightarrow$ p, q, s
(c) $\rightarrow$ q, s
(d) $\rightarrow$ q, r, s

Topic: Physics, Electromagnetic Induction, JEE Main
Match the following columns

Options:
A
(a) p,q, (b) p, (c) p, (d) p,q

B
(a) p,q,r, (b) p, s, (c) p,r, (d) p,q,r

C
(a) p,q, (b) p, r, (c) p,s, (d) p,q,r

D
(a) p,s, (b) p, s, (c) p, (d) p,q

The correct answer is:
(a) p,q, (b) p, r, (c) p,s, (d) p,q,r

Solution:
(a) $\rightarrow$ p, q
(b) $\rightarrow \mathrm{p}, \mathrm{r}$
(c) $\rightarrow$ p, $s$
(d) $\rightarrow$ p, q, r

Topic: Physics, Nuclear Physics, JEE Main
Match the following columns and select the correct option.

Column-I	Column-II

(a)	Gregarious, polyphagous pest	(i)	Asterias

(b)	Adult with radial symmetry and larva with bilateral symmetry	(ii)	Scorpion

(c)	Book lungs	(iii)	Ctenoplana

(d)	Bioluminescence	(iv)	Locusta

Options:
A (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
B (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
C (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
D (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)

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

Solution:
Gregarious pest: A pest is an animal that is detrimental or harmful to humans or human concerns. Locusta (locust) is a pest that is dangerous to crops. It causes heavy damage to the vegetation. As it lives in groups of the same individuals but does not help one another, hence called gregarious.

Body symmetry of Echinoderms: The adult echinoderms are radially symmetrical, but their larvae are bilaterally symmetrical. The echinoderms are unique in this feature that the larvae are bilaterally symmetrical (an advanced feature) while the adults developing from them are radially symmetrical (a primitive feature). Examples: Asterias (Starfish), Echinus (Sea urchin), Antedon (sea lily), Cucumaria (sea cucumber), and Ophiura (Brittle star).

Book-lungs: Present in scorpions, spiders, etc. These are the modified book-gills.

Bioluminescence: Bios means living and lumen means light. Bioluminescence is the property of production and emission of light by a living organism and hence shines in the dark background. It is a naturally occurring phenomenon. This property is well-marked in Ctenophores as they emit light and hence shine against the dark backgrounds. Examples: Pleurobrachia, Ctenoplana.

Topic: Biology, Animal Kingdom, NEET, NEET 2020 (Phase 1)
Match the following for the given process

Options:
A
(a) s, (b) p, r, (c) r, (d) q,s

B
(a) r, (b) p, s, (c) q, (d) q

C
(a) s, (b) p, r, (c) q, (d) q

D
(a) r, (b) p, s, (c) r, (d) q,s

The correct answer is:
(a) s, (b) p, r, (c) r, (d) q,s

Solution:
(a) $\rightarrow \mathrm{s}$ (b) $\rightarrow \mathrm{p}, \mathrm{r}$, (c) $\rightarrow \mathrm{r}$ (d) $\rightarrow \mathrm{q}, \mathrm{s}$
In process $\mathbf{J} \rightarrow \mathbf{K}: V$ is constant whereas $P$ is decreasing, Therefore, $T$ should also decrease.
$$
\therefore \quad W=0, \Delta U=-\text { ve and } Q < 0
$$
In process $\mathbf{K} \rightarrow \mathbf{L}: P$ is constant while $V$ is increasing. Therefore, temperature should also increase.
$$
\therefore \quad W>0, \Delta U>0 \text { and } Q>0
$$
In process $\mathbf{L} \rightarrow \mathbf{M}$ : This is inverse of process $J \rightarrow K$.
$$
\therefore \quad W=0, \Delta U>0 \text { and } Q>0
$$
In process $\mathbf{M} \rightarrow \mathbf{J}: V$ is decreasing. Therefore, $W < 0$
$$
\begin{array}{lcl}
& (P \mathrm{~V})_J & < (P V)_M \\
\therefore & T_J & < T_M \text { or } \quad \Delta U < 0 \\
\text { Therefore, } & Q & < 0 .
\end{array}
$$
No, solution is required.

Topic: Physics, Thermodynamics, JEE Main
Match the following for the given process

Options:
A
(a) s, (b) p, r, (c) r, (d) q,s

B
(a) r, (b) p, s, (c) q, (d) q

C
(a) s, (b) p, r, (c) q, (d) q

D
(a) r, (b) p, s, (c) r, (d) q,s

The correct answer is:
(a) s, (b) p, r, (c) r, (d) q,s

Solution:
(a) $\rightarrow \mathrm{s}$ (b) $\rightarrow \mathrm{p}, \mathrm{r}$, (c) $\rightarrow \mathrm{r}$ (d) $\rightarrow \mathrm{q}, \mathrm{s}$
In process $\mathbf{J} \rightarrow \mathbf{K}: V$ is constant whereas $P$ is decreasing, Therefore, $T$ should also decrease.
$$
\therefore \quad W=0, \Delta U=-\text { ve and } Q < 0
$$
In process $\mathbf{K} \rightarrow \mathbf{L}: P$ is constant while $V$ is increasing. Therefore, temperature should also increase.
$$
\therefore \quad W>0, \Delta U>0 \text { and } Q>0
$$
In process $\mathbf{L} \rightarrow \mathbf{M}$ : This is inverse of process $J \rightarrow K$.
$$
\therefore \quad W=0, \Delta U>0 \text { and } Q>0
$$
In process $\mathbf{M} \rightarrow \mathbf{J}: V$ is decreasing. Therefore, $W < 0$
$$
\begin{array}{lcl}
& (P \mathrm{~V})_J & < (P V)_M \\
\therefore & T_J & < T_M \text { or } \quad \Delta U < 0 \\
\text { Therefore, } & Q & < 0 .
\end{array}
$$
No, solution is required.

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2006
Match the gases under specified conditions listed in Column I with their properties/laws 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, q; B-r; C-r, s; D-p, q

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

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

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

Solution:
(A) $-p, s(\mathrm{~B})-r$ (C) $-p, q$ (D) $-r$
Explanation van der Waals' equation
$$
\left(p+\frac{n^2 a}{V_m^2}\right)\left(V_m-b\right)=n R T
$$
For hydrogen gas $(p=200 \mathrm{~atm}, T=273 \mathrm{~K})$
As pressure is large $V_m$ can be assumed small, thus ' $b$ ' can not be ignored, while due to high pressure $a / V_m^2$ can be considered negligible in comparison to $p$.
$$
p\left(V_m-b\right)=R T \quad \text { and } \quad Z=1+\frac{p b}{R T}
$$
For hydrogen gas $(p \sim 0, T=273 \mathrm{~K})$
when pressure occurs of low about $1 \mathrm{~atm}$ or less and temperature is not very close to the point of liquification $\left[T_c\left(\mathrm{H}_2\right)=33.3 \mathrm{~K}\right]$ gas behaves ideally.
For
$$
P V=n R T
$$
Temperature is close to the point of liquification $\left[T_C\left(\mathrm{CO}_2\right)=304.2\right]$ thus, deviation from ideality appears very high (due to high attractive force of attraction).
For real gas with very large molar volume.
As molar volume is very large $a / V_m^2$ will be negligible and at the same time ' $b$ ' in comparison to $V_m$ is also considered negligible, thus, $p V_m=n R T$

Topic: Chemistry, States of Matter, JEE Advanced, JEE Advanced 2007 (Paper 1)
Match the integrals in Column I with the values in Column II.

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

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

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

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

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

Solution:
(A)
(B)
(C)
(D)

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2007 (Paper 1)
$$
\text { Match the reaction in Column I with appropriate options in Column II. }
$$

Options:
A
(A) R,S, (B) R,T, (C) P, (D) T

B
(A) R,S,T, (B) R,T, (C) P,Q, (D) R

C
(A) R,S, (B) T, (C) P, (D) T

D
(A) R,S,T, (B) T, (C) P,Q, (D) R

The correct answer is:
(A) R,S,T, (B) T, (C) P,Q, (D) R

Solution:
Organic chemistry
Conceptual understanding of reaction mechanism IV

Topic: Chemistry, Hydrocarbons, JEE Main
$$
\text { Match the reaction in Column I with appropriate options in Column II. }
$$

Options:
A
(A) R,S, (B) R,T, (C) P, (D) T

B
(A) R,S,T, (B) R,T, (C) P,Q, (D) R

C
(A) R,S, (B) T, (C) P, (D) T

D
(A) R,S,T, (B) T, (C) P,Q, (D) R

The correct answer is:
(A) R,S,T, (B) T, (C) P,Q, (D) R

Solution:
Organic chemistry
Conceptual understanding of reaction mechanism IV

Topic: Chemistry, Hydrocarbons, JEE Advanced, JEE Advanced 2010 (Paper 2)
Match the reactions in Column I with appropriate types of steps/reactive intermediate involved in these reactions as given in Column II.

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

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

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

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

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

Solution:
(a)
(b)
(c)
(d)

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2011 (Paper 2)
Match the specified conditions in solids listed in Column I with their properties 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-p, r

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

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

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

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

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

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2007 (Paper 2)