datavorous/entrance-exam-dataset huggingface dataset:
Plots showing the variation of the rate constant $(k)$ with temperature $(T)$ are given below. The plot that follows Arrhenius equation is

Options:
A

B

C

D

The correct answer is:

Solution:
$k=A e^{-E_a / R T}, \quad$ As temperature increases $k$ increases exponentially. Chemical kinetics Conceptual mathematical analysis of data I

Topic: Chemistry, Chemical Kinetics, JEE Main
Plots showing the variation of the rate constant $(k)$ with temperature $(T)$ are given below. The plot that follows Arrhenius equation is

Options:
A

B

C

D

The correct answer is:

Solution:
$k=A e^{-E_a / R T}, \quad$ As temperature increases $k$ increases exponentially. Chemical kinetics Conceptual mathematical analysis of data I

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2010 (Paper 1)
Positive and negative point charges of equal magnitude are kept at $\left(0,0, \frac{a}{2}\right)$ and $\left(0,0, \frac{-a}{2}\right)$, respectively. The work done by the electric field when another positive point charge is moved from $(-a, 0,0)$ to $(0, a, 0)$ is

Options:
A
positive

B
negative

C
zero

D
depends on the path connecting the initial and final positions

The correct answer is:
zero

Solution:
$A \equiv(-a, 0,0) B \equiv(0, a, 0)$
Point charge is moved from $A$ to $B$
$$
\begin{aligned}
V_A & =V_B=0 \\
W & =0
\end{aligned}
$$
$$
\therefore \quad W=0
$$
or the correct option is (c).

Topic: Physics, Electrostatics, JEE Main
Positive and negative point charges of equal magnitude are kept at $\left(0,0, \frac{a}{2}\right)$ and $\left(0,0, \frac{-a}{2}\right)$, respectively. The work done by the electric field when another positive point charge is moved from $(-a, 0,0)$ to $(0, a, 0)$ is

Options:
A
positive

B
negative

C
zero

D
depends on the path connecting the initial and final positions

The correct answer is:
zero

Solution:
$A \equiv(-a, 0,0) B \equiv(0, a, 0)$
Point charge is moved from $A$ to $B$
$$
\begin{aligned}
V_A & =V_B=0 \\
W & =0
\end{aligned}
$$
$$
\therefore \quad W=0
$$
or the correct option is (c).

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2007 (Paper 2)
Reaction of $\mathrm{Br}_2$ with $\mathrm{Na}_2 \mathrm{CO}_3$ in aqueous solution gives sodium bromide and sodium bromate with evolution of $\mathrm{CO}_2$ gas. The number of sodium bromide molecules involved in the balanced chemical equation is

Options:

The correct answer is:
5

Solution:
$\mathrm{Br}_2$ is disproportionated in basic medium as
$$
3 \mathrm{Br}_2+3 \mathrm{Na}_2 \mathrm{CO}_3 \longrightarrow 5 \mathrm{NaBr}+\mathrm{NaBrO}_3+3 \mathrm{CO}_2
$$

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2011 (Paper 1)
Reduction of the metal centre in aqueous permanganate ion involves

Options:
A
three electrons in neutral medium

B
five electrons in neutral medium

C
three electrons in alkaline medium

D
five electrons in acidic medium

The correct answers are:
three electrons in neutral medium
,
three electrons in alkaline medium
,
five electrons in acidic medium

Solution:
In neutral medium: $\quad \mathrm{MnO}_4^{-} \rightarrow \mathrm{MnO}_2\left(\mathrm{Mn}^{7+}+3 e^{-} \rightarrow \mathrm{Mn}^{4+}\right)$
In alkaline medium : $\quad \mathrm{MnO}_4^{-} \rightarrow \mathrm{MnO}_2\left(\mathrm{Mn}^{7+}+3 \varepsilon^{-} \rightarrow \mathrm{Mn}^{4+}\right)$
In acidic medium : $\quad \mathrm{Mn}_4^{-} \rightarrow \mathrm{Mn}^{2+}\left(\mathrm{Mn}^{7+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}\right)$

Topic: Chemistry, d and f Block Elements, JEE Advanced, JEE Advanced 2011 (Paper 2)
Retrogressive metamorphosis occurs in

Options:
A Hemichordata
B Cephalochordata
C Urochordata
D Vertebrata.

The correct answer is:
Urochordata

Solution:
In retrogressive metamorphosis, degeneration is shown by larva to form less developed adult. It is a type of metamorphosis seen in Herdmania (Urochordate). It involves transformation of an active, free swimming larva with advanced characters into sedentary and simple adult.

Topic: Biology, Animal Kingdom, AIIMS, AIIMS 2016
Selaginella can be regarded as an evolutionary advanced Pteridophyte because of

Options:
A Rhizophore
B Ligule
C Heterospory
D Apospory

The correct answer is:
Heterospory

Solution:
Heterospory is the occurrence of two types of spore, smaller microspore, and larger macrospore. Heterospory has led to the evolution of seed habit in plants. Heterospory first evolved in Pteridophyta in which megaspore was retained on parent sporophyte which is considered as a precursor to seed habit.

Topic: Biology, Plant Kingdom, NEET
Select the animal that exhibits retrogressive metamorphosis.

Options:
A Bufo
B Amphioxus
C Limulus
D Herdmania

The correct answer is:
Herdmania

Solution:
Retrogressive metamorphosis occurs in Urochordates (e.g., Herdmania, Ascidia, Doliolum etc.), in which better developed (advanced) larva changes to less developed (primitive) adult.

Topic: Biology, Animal Kingdom, AIIMS, AIIMS 2015
Sexual reproduction in Spirogyra is an advanced feature because it shows:

Options:
A different sizes of motile sex organs
B same size of motile sex organs
C morphologically different sex organs
D physiologically differentiated sex organs

The correct answer is:
physiologically differentiated sex organs

Solution:
In Spirogyra, the sexual reproduction involves the fusion of two morphologically identical isogametes and physiologically dissimilar anisogametes. This is a case of primitive anisogamy.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2003
Silver (atomic weight $=108 \mathrm{~g} \mathrm{~mol}^{-1}$ ) has a density of $10.5 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of silver atoms on a surface of area $10^{-12} \mathrm{~m}^2$ can be expressed in scientific notation as $y \times 10^x$. The value of $x$ is

Options:

The correct answer is:
7

Solution:
Volume of one mole of silver atoms $=\frac{108}{10.5} \mathrm{~cm}^3 /$ mole Volume of one silver atom $=\frac{108}{10.5} \times \frac{1}{6.022 \times 10^{23}} \mathrm{~cm}^3$
So,
$$
\begin{aligned}
\frac{4}{3} \pi R^3 & =\frac{108}{10.5} \times \frac{1}{6.022 \times 10^{23}}=1.708 \times 10^{-23} \\
R^3 & =0.407 \times 10^{-23} \mathrm{~cm}^3 \\
& =0.407 \times 10^{-29} \mathrm{~m}^3
\end{aligned}
$$
Area of each silver atom
$$
\pi R^2=\pi \times\left(0.407 \times 10^{-29} \mathrm{~m}^3\right)^{2 / 3}
$$
So number of silver atoms in given area
$$
=\frac{10^{-12}}{\left(0.407 \times 10^{-29} \mathrm{~m}^3\right)^{2 / 3}}=\frac{10^8}{(\pi \times 2)}=1.6 \times 10^7=y \times 10^x \quad \text { so, } \quad x=7
$$
Solid state
Mathematical understanding
III

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2010 (Paper 2)
Six charges are placed around a regular hexagon of side length a as shown in the figure. Five of them have charge q, and the remaining one has charge x. The perpendicular from each charge to the nearest hexagon side passes through the center O of the hexagon and is bisected by the side.

Which of the following statement(s) is(are) correct in SI units?

Options:
A

When x=q, the magnitude of the electric field at O is zero.

B When x=-q, the magnitude of the electric field at O is q6π∈0a2.
C When x=2q, the potential at O is 7q43π∈0a.
D When x=-3q, the potential at O is 3q43πϵ0a.

The correct answers are:

When x=q, the magnitude of the electric field at O is zero.

, When x=-q, the magnitude of the electric field at O is q6π∈0a2., When x=2q, the potential at O is 7q43π∈0a.

Solution:
âââââââ

The given arrangements of the charges also form a regular hexagon with side length equal to 3a

When x=q, the situation is symmetric

⇒ Electric field at O would be zero.

⇒A is correct.

When x=-q, we can think of x as q+-2q. Therefore, the magnitude of the electric field will be equivalent to that if only charge of -2q is kept at the the bottom point. Hence,

EO=14πÏµ02q3a2=14πÏµ02q3a2=q6πÏµ0a2

⇒B is correct

For x=2q, potential at O is

VO=5×14πÏµ0q3a+14πÏµ02q3a=7q43πÏµ0a

⇒C is correct

For x=-3q,

VO=5×14πÏµ0q3a+14πÏµ0-3q3a=q23πÏµ0a

⇒Dis not correct.

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2022 (Paper 1)
Six point charges are kept at the vertices of a regular hexagon of side $\mathrm{L}$ and centre $O$, as shown in the figure.

Given that $K=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{L^{2}}$, which of the following statement(s) is (are) correct?

Options:
A The electric field at $O$ is $6 \mathrm{~K}$ along $O D$
B The potential at $O$ is zero
C The potential at all points on the line $P R$ is same
D The potential at all points on the line $S T$ is same

The correct answers are:
The electric field at $O$ is $6 \mathrm{~K}$ along $O D$, The potential at $O$ is zero, The potential at all points on the line $P R$ is same

Solution:
Here

$\begin{aligned}

& \frac{\left|\overrightarrow{E_{A}}\right|}{2}=\left|\overrightarrow{E_{B}}\right|=\left|\overrightarrow{E_{C}}\right|=\frac{\left|\overrightarrow{E_{D}}\right|}{2}=\left|\overrightarrow{E_{E}}\right|=\left|\overrightarrow{E_{F}}\right|=K \\

\therefore & E_{O}=E_{A}+E_{D}+\left(E_{F}+E_{C}\right) \cos 60^{\circ}+\left(E_{B}+E_{C}\right) \cos 60^{\circ} \\

=& 2 \mathrm{~K}+2 \mathrm{~K}+(\mathrm{K}+\mathrm{K}) \times \frac{1}{2}+(\mathrm{K}+\mathrm{K}) \times \frac{1}{2}=6 \mathrm{~K}

\end{aligned}$

Electric potential at $O$

$V_{O}=\frac{1}{4 \pi \varepsilon_{0} L}[2 q+q+q-q-q-2 q]=0$

Potential at all points on the line PR is same not on line ST. $P R$ is perpendicular bisector (the equatorial line) for the electric dipoles $A B, F E$ and $B C$. Therefore the electric potential will be zero at any point on $P R$.

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2012 (Paper 2)
Six point charges, each of the same magnitude $q$, are arranged in different manners as shown in Column-II. In each case, a point $M$ and a line $P Q$ passing through $M$ are shown. Let $E$ be the electric field and $V$ be the electric potential at $M$ (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line $P Q$. Let $B$ the magnetic field at $M$ and $\mu$ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Electrostatics, JEE Main
Six point charges, each of the same magnitude $q$, are arranged in different manners as shown in Column-II. In each case, a point $M$ and a line $P Q$ passing through $M$ are shown. Let $E$ be the electric field and $V$ be the electric potential at $M$ (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line $P Q$. Let $B$ the magnetic field at $M$ and $\mu$ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2009 (Paper 1)
Solubility product constant $\left(K_{\mathrm{sp}}\right)$ of salts of types $M X, M X_2$ and $M_3 X$ at temperature ' $T$ ' are $4.0 \times 10^{-8}, 3.2 \times 10^{-14}$ and $2.7 \times 10^{-15}$, respectively. Solubilities $\left(\mathrm{mol} \mathrm{dm}^{-3}\right)$ of the salts at temperature ' $T$ ' are in the order

Options:
A
$M X>M X_2>M_3 X$

B
$M_3 X>M X_2>M X$

C
$M X_2>M_3 X < M X$
D
$M X>M_3 X>M X_2$

The correct answer is:
$M X>M_3 X>M X_2$

Solution:
$\begin{array}{lcl}\text { Salt } & \text { Solubility product } & \text { Solubility } \\ M X & S_1^2=4.0 \times 10^{-8} & S_1=2 \times 10^{-4} \\ M X_2 & 4 S_2^3=3.2 \times 10^{-14} & S_2=2 \times 10^{-5} \\ M_3 X & 27 S_3^4=2.7 \times 10^{-15} & S_3=1 \times 10^{-4}\end{array}$
Thus, solubility order $=M X>M_3 X>M X_2$.

Topic: Chemistry, Ionic Equilibrium, JEE Advanced, JEE Advanced 2008 (Paper 2)
Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Nuclear Physics, JEE Main
Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2007 (Paper 1)
Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Units and Dimensions, JEE Main
Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Units and Dimensions, JEE Advanced, JEE Advanced 2007 (Paper 1)
Statement 1 An astronaut in an orbiting space station above the earth experiences weightlessness.
and Statement 2 An object moving around the earth under the influence of earth's gravitational force is in state of 'free fall'.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
Force acting on astronaut is utilised in providing necessary centripetal force; thus he feels weightlessness or he is in a state of free fall.
$$
\therefore \text { correct option is (a) }
$$

Topic: Physics, Gravitation, JEE Advanced, JEE Advanced 2008 (Paper 1)
Statement 1 Aniline on reaction with $\mathrm{NaNO}_2 / \mathrm{HCl}$ at $0^{\circ} \mathrm{C}$ followed by coupling with $\beta$-naphthol gives a dark coloured precipitate.
Statement 2 The colour of the compound formed in the reaction of aniline with $\mathrm{NaNO}_2 / \mathrm{HCl}$ at $0^{\circ} \mathrm{C}$ followed by coupling with $\beta$-naphthol is due to the extended conjugation.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
Statement II is correct. Statement I appears correct in the light of Statement II. What is the colour of a dye that dye is not in included in JEE syllabus but what is the cause and understanding behind a colour could be essentially expected.
Concept Usually a certain proportion of the flight is absorbed and the rest reflected, the substance has the colour of the reflected light.
If only a single band is absorbed, the substance has the complementary colour.
- The wavelength of photon absorbed depends on the energy difference between the excited and ground states. The smaller the difference, the longer being the wavelength.
- The larger the number of electrons involved in resonance, the smaller is the energy difference between the ground and excited state. Hence, the more extended the conjugation in a molecule and the greater the contribution of charged structures, the longer is the wavelength of the photon required to excite the molecule.
- Especially, the presence of $+I /+M$ and $-I /-M$ groups in the conjugated system will both extend the conjugation and increase the contributions of charged structures to the resonance hybrid. Therefore, in these circumstances, the wavelength should become progressively longer.
- Depending on this discussion it is apparent that the concerned resultant azodye will absorb higher wavelength.
$\begin{array}{ccc}\text { Wavelength } & \text { Colour absorbed } & \text { Complimentary colour } \\ 490-500 & \text { Blue-green } & \text { Red } \\ 580-595 & \text { Yellow } & \text { Blue } \\ 605-750 & \text { Red } & \text { Blue-green }\end{array}$

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2008 (Paper 2)
Statement 1 Bromobenzene, upon reaction with $\mathrm{Br}_2 / \mathrm{Fe}$ gives 1, 4-dibromobenzene as the major product.
Statement 2 In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in directing the incoming electrophile.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
Bromine orientation is controlled by its mesomeric effect, its stabilisation of arenium ion by $+M$ activity
In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in the activity of ring.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2008 (Paper 1)
Statement $\mathbf{1}\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4$ is paramagnetic.
Statement 2 The $\mathrm{Fe}$ in $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4$ has three unpaired electrons.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
The oxidation number of $\mathrm{Fe}$ in the complex, $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right] \mathrm{SO}_4$ is $+1$ [ $\mathrm{NO}$ has $+1$ charge ]
$$
\mathrm{Fe}^{+}=[\mathrm{Ar}] 3 d^6 4 s^1
$$
$\mathrm{NO}^{+}$causes pairing of 4 s electron inside .
Thus, the configuration is $3 d^7$ and number of unpaired electrons $=3$

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2008 (Paper 2)
Statement 1 For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.
and Statement 2 If the observer and the object are moving at velocities $\overrightarrow{\mathbf{v}}_{\mathbf{1}}$ and $\overrightarrow{\mathbf{v}_2}$ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is $\overrightarrow{\mathbf{v}_2}-\overrightarrow{\mathbf{v}_1}$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Solution:
Statement-2, is formula of relative velocity. But it does not explain statement-1 correctly. The correct explanation of statement-1 is due to visual perception of motion. The object appears to be moving faster, when its angular velocity is greater w.r.t observer.
$\therefore$ correct answer is (b).

Topic: Physics, Motion In Two Dimensions, JEE Main
Statement 1 For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.
and Statement 2 If the observer and the object are moving at velocities $\overrightarrow{\mathbf{v}}_{\mathbf{1}}$ and $\overrightarrow{\mathbf{v}_2}$ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is $\overrightarrow{\mathbf{v}_2}-\overrightarrow{\mathbf{v}_1}$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Solution:
Statement-2, is formula of relative velocity. But it does not explain statement-1 correctly. The correct explanation of statement-1 is due to visual perception of motion. The object appears to be moving faster, when its angular velocity is greater w.r.t observer.
$\therefore$ correct answer is (b).

Topic: Physics, Motion In Two Dimensions, JEE Advanced, JEE Advanced 2008 (Paper 2)
Statement 1 For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero.
Statement 2 At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is false, Statement 2 is true

Solution:
At equilibrium, its free energy change, is zero and for a spontaneous process, $\Delta G < 0$ but at equilibrium standard free energy is not zero.
Analysis of the Paragraph

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2008 (Paper 1)