datavorous/entrance-exam-dataset huggingface dataset:
Column I gives some devices and Column II gives some processes on which the functioning of these devices depend. Match the devices in Column I with the processes in Column II and indicate your answer by darkening appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Thermal Properties of Matter, JEE Advanced, JEE Advanced 2007 (Paper 2)
Column I showns four systems, each of the same length $L$, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as $\lambda_f$. Match each system with statements given in Column II describing the nature and wave length of the standing waves.

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

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

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

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

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

Solution:
In organ pipes, longitudinal waves are formed. In string, transverse waves are formed. Open end of pipe is displacement antinode and closed end is displacement node. In case of string fixed end of a string is node.
Further, least distance between a node and an antinode is $\frac{\lambda}{4}$ and between two nodes is $\frac{\lambda}{2}$. Keeping these points in mind answer to this question is as under;
$(\mathrm{A}) \rightarrow(\mathrm{p}, \mathrm{t})$
(B) $\rightarrow(\mathrm{p}, \mathrm{s})$
$(\mathrm{C}) \rightarrow(\mathrm{q}, \mathrm{s})$
(D) $\rightarrow$ (q, r)
Analysis of Question
(i) Unlike the previous years's few match the column problems, this question seems to be more logical. If you know the subject, you can select correct matching without confusion.
(ii) In option (d) of column I, string is fixed from centre, therefore at lowest frequency two loops will be formed as shown below.

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2011 (Paper 2)
Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits $S_1$ and $S_2$. In each of these cases $S_1 P_0=S_2 P_0$, $S_1 P_1-S_2 P_1=\frac{\lambda}{4}$ and $S_1 P_2-S_2 P_2=\frac{\lambda}{3}$, where $\lambda$ is the wavelength of the light used. In the cases $\mathrm{B}, \mathrm{C}$ and $\mathrm{D}$, a transparent sheet of refractive index $\mu$ and thickness $t$ is pasted on slit $S_2$. The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point $P$ on the screen from the two slits is denoted by $\delta(P)$ and the intensity by $I(P)$. Match each situation given in Column-I with the statement(s) in Column-II valid for that situation.

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

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

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

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

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

Solution:
(A) $\rightarrow(\mathrm{p}, \mathrm{s}) \rightarrow$ Intensity at $P_0$ is maximum. It will continuously decrease from $P_0$ towards $P_2$.
$(B) \rightarrow(q) \rightarrow$ Path difference due to slap will be compensated by geometrical path difference. Hence, $\delta\left(P_1\right)=0$.
$(\mathrm{C}) \rightarrow(\mathrm{t}) \rightarrow \delta\left(P_0\right)=\frac{\lambda}{2}, \delta\left(P_1\right)=\frac{\lambda}{2}-\frac{\lambda}{4}=\frac{\lambda}{4}$ and $\delta\left(P_2\right)=\frac{\lambda}{2}-\frac{\lambda}{3}=\frac{\lambda}{6}$. When path difference increases from 0 to $\frac{\lambda}{2}$, intensity will decrease from maximum to zero. Hence, in this case,
$$
I\left(P_2\right)>I\left(P_1\right)>I\left(P_0\right)
$$
$(\mathrm{D}) \rightarrow(\mathrm{r}, \mathrm{s}, \mathrm{t})$
$$
\delta\left(P_0\right)=\frac{3 \lambda}{4}, \delta\left(P_1\right)=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2}
$$
and $\delta\left(P_2\right)=\frac{3 \lambda}{4}-\frac{\lambda}{3}=\frac{5 \lambda}{12}$
In this case $I\left(P_1\right)=0$

Topic: Physics, Wave Optics, JEE Advanced, JEE Advanced 2009 (Paper 2)
Column II shows five systems in which two objects are labelled as $X$ and $Y$. Also in each case a point $P$ is shown. Column-I gives some statements about $X$ and/or $Y$. Match these statements to the appropriate system(s) from Column-II.

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

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

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

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

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

Solution:
$(\mathrm{A}) \rightarrow(\mathrm{p}, \mathrm{t}) \rightarrow$ Net force on $Y$ is ero.
(B) $\rightarrow(\mathrm{q}, \mathrm{s}, \mathrm{t}) \rightarrow \ln (\mathrm{s})$ and $(\mathrm{t})$ gravitational potential energy of $Y$ is decreasing and that of $X$ is increasing.
(C) $\rightarrow(\mathrm{p}, \mathrm{r}, \mathrm{t}) \rightarrow \operatorname{In}(\mathrm{t}), \mathrm{Y}$ is moving with constant speed.
(D) $\rightarrow$ (q) $\rightarrow$ No explanation is required

Topic: Physics, Work Power Energy, JEE Main
Column II shows five systems in which two objects are labelled as $X$ and $Y$. Also in each case a point $P$ is shown. Column-I gives some statements about $X$ and/or $Y$. Match these statements to the appropriate system(s) from Column-II.

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

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

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

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

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

Solution:
$(\mathrm{A}) \rightarrow(\mathrm{p}, \mathrm{t}) \rightarrow$ Net force on $Y$ is ero.
(B) $\rightarrow(\mathrm{q}, \mathrm{s}, \mathrm{t}) \rightarrow \ln (\mathrm{s})$ and $(\mathrm{t})$ gravitational potential energy of $Y$ is decreasing and that of $X$ is increasing.
(C) $\rightarrow(\mathrm{p}, \mathrm{r}, \mathrm{t}) \rightarrow \operatorname{In}(\mathrm{t}), \mathrm{Y}$ is moving with constant speed.
(D) $\rightarrow$ (q) $\rightarrow$ No explanation is required

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2009 (Paper 1)
Column-II gives certain systems undergoing a process. Column-I suggests changes in some of the parameters related to the system. Match the statements in Column-I to the appropriate process (es) from Column-II.

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

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

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

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

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

Solution:
No Solution Available

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2009 (Paper 2)
Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$, then the area of the $\triangle A B C$ is

Options:
A
$1-\sqrt{\frac{2}{3}}$ sq unit

B
$\sqrt{\frac{3}{2}}-1$ sq unit

C
$1+\sqrt{\frac{2}{3}}$ sq unit

D
$\sqrt{\frac{3}{2}}+1$ sq unit

The correct answer is:
$\sqrt{\frac{3}{2}}-1$ sq unit

Solution:
The given equation can be rewritten as $\frac{(x-\sqrt{2})^2}{1}-\frac{(y+\sqrt{2})^2}{2}=1$ for $A(x, y)$,
$$
\begin{aligned}
& e=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}} \\
& \therefore x-\sqrt{2}=2 \Rightarrow x=2+\sqrt{2} \\
& \text { For } C(x, y), x-\sqrt{2}=a e=\sqrt{6} \\
& \Rightarrow \quad x=\sqrt{6}+\sqrt{2} \\
& \text { Now, } A C=\sqrt{6}+\sqrt{2}-2-\sqrt{2}=\sqrt{6}-2 \\
& \qquad B C=\frac{b^2}{a}=\frac{2}{2}=1 \\
& \text { Area of } \Delta A B C=\frac{1}{2} \times(\sqrt{6}-2) \times B C \\
& =\frac{1}{2} \times(\sqrt{6}-2) \times 1=\sqrt{\frac{3}{2}}-1
\end{aligned}
$$

Topic: Mathematics, Hyperbola, JEE Advanced, JEE Advanced 2008 (Paper 2)
Consider a disc rotating in the horizontal plane with a constant angular speed $\omega$ about its centre $\mathrm{O}$. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles $P$ and $Q$ are simultaneously projected at an angle towards $R$. The velocity of projection is in the $y$ - $z$ plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed $1 / 8$ rotation, (ii) their range is less than half the disc radius, and (iii) $\omega$ remains constant throughout. Then

Options:
A $P$ lands in the shaded region and $Q$ in the unshaded region.
B $P$ lands in the unshaded region and $Q$ in the shaded region.
C Both $P$ and $Q$ land in the unshaded region.
D Both $P$ and $Q$ land in the shaded region.

The correct answer is:
Both $P$ and $Q$ land in the unshaded region.

Solution:
For $\frac{1}{8}$ of rotation of disc,

$\begin{aligned}

t=\frac{1}{8} \times \frac{2 \pi}{\omega}=\frac{\pi}{4 \omega} & \\

x \text {-coordinate of } P &=\omega R T \\

&=\frac{\pi R}{4}>R \sin 45^{\circ}

\end{aligned}$

To reach the unshaded part, particle$P$ needs to travel horizontal range $>$ $R \sin 45^{\circ} \simeq 0.7 \mathrm{R}$

But its range is less than $\mathrm{R} / 2$ so it will land on shaded part. $Q$ is near the origin, its velocity will be nearly along QR so it will land in unshaded part.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2012 (Paper 2)
Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then

Options:
A
negative and distributed uniformly over the surface of the sphere

B
negative and appears only at the point on the sphere closest to the point charge

C
negative and distributed non-uniformly over the entire surface of the sphere

D
zero

The correct answer is:
zero

Solution:
$$
\text { Charge will be induced in the conducting sphere, but net charge on it will be zero. }
$$
$$
\therefore \text { Option (d) is correct. }
$$

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2007 (Paper 1)
Consider a reaction $a G+b H \rightarrow$ products. When concentration of both the reactants $G$ and $H$ is doubled, the rate increases by eight times. However, when concentration of $G$ is doubled keeping the concentration of $H$ fixed, the rate is doubled. The overall order of the reaction is

Options:
A
0

B
1

C
2

D
3

The correct answer is:
3

Solution:
$a G+b H \longrightarrow$ Product
Suppose order of reaction $=n$
When concentration of both $G$ and $H$ is double than rate increase by eight times.
$$
\begin{aligned}
\text { rate } & =k(\text { reactants })^n \\
(2)^8 & =k(2)^n \\
(2)^3 & =k(2)^n \\
n & =3
\end{aligned}
$$
when concentration of $G$ is doubled than concentration $H$ is fixed, the rate is double.
than $\quad$ rate $\propto[G]^1[H]^2$

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2007 (Paper 2)
Consider a system of three charges $\frac{q}{3}, \frac{q}{3}$ and $-\frac{2 q}{3}$ placed at points $A, B$ and $C$, respectively, as shown in the figure. Take $O$ to be the centre of the circle of radius $R$ and angle $C A B=60^{\circ}$.

Options:
A
The electric field at point $O$ is $\frac{q}{8 \pi \varepsilon_0 R^2}$ directed along the negative $x$-axis

B
The potential energy of the system is zero

C
The magnitude of the force between the charges at $C$ and $B$ is $\frac{q^2}{54 \pi \varepsilon_0 R^2}$

D
The potential at point $O$ is $\frac{q}{12 \pi \varepsilon_0 R}$

The correct answer is:
The magnitude of the force between the charges at $C$ and $B$ is $\frac{q^2}{54 \pi \varepsilon_0 R^2}$

Solution:
Distance $B C=A B \sin 60^{\circ}=(2 R) \frac{\sqrt{3}}{2}=\sqrt{3} R$
$\therefore \quad\left|F_{B C}\right|=\frac{1}{4 \pi \varepsilon_0} \frac{(q / 3)(2 q / 3)}{(\sqrt{3} R)^2}=\frac{q^2}{54 \pi \varepsilon_0 R^2}$
$\therefore$ correct option is (c).

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2008 (Paper 2)
Consider a thin spherical shell of radius $R$ with centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field $|\vec{E}(r)|$ and the electric potential $V(r)$ with the distance $r$ from the centre, is best represented by which graph?

Options:
A
B
C
D

The correct answer is:

Solution:
For a thin uniformly positive charged spherical shell

(i) Inside the shell at any point

$E=O$ and $V=\frac{1}{4 \pi \in_{0}} \quad \frac{q}{R}=$ constt.

where $q=$ charge on sphere

$R=$ Radius of sphere

(ii) Outside the shell at any point at any distance $r$ from

the centre $E \propto \frac{1}{r^{2}}$ and $V \propto \frac{1}{r}$

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2012 (Paper 1)
Consider a thin square sheet of side $L$ and thickness $t$, made of a material of resistivity $\rho$. The resistance between two opposite faces, shown by the shaded areas in the figure is

Options:
A
directly proportional to $L$

B
directly proportional to $t$

C
independent of $L$

D
independent of $t$

The correct answer is:
independent of $L$

Solution:
$R=\frac{\rho(L)}{A}=\frac{\rho L}{t L}=\frac{\rho}{t}$ ie, $R$ is independent of $L$. Hence the correct option is (c).

Topic: Physics, Current Electricity, JEE Main
Consider a thin square sheet of side $L$ and thickness $t$, made of a material of resistivity $\rho$. The resistance between two opposite faces, shown by the shaded areas in the figure is

Options:
A
directly proportional to $L$

B
directly proportional to $t$

C
independent of $L$

D
independent of $t$

The correct answer is:
independent of $L$

Solution:
$R=\frac{\rho(L)}{A}=\frac{\rho L}{t L}=\frac{\rho}{t}$ ie, $R$ is independent of $L$. Hence the correct option is (c).

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2010 (Paper 1)
Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is

Options:
A
3

B
4

C
5

D
6

The correct answer is:
6

Solution:
Mole ratio is reverse of $x$-factor ratio.

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2007 (Paper 2)
Consider a triangle $A B C$ and let $a, b$ and $c$ denote the lengths of the sides opposite to vertices $A, B$ and $C$ respectively. Suppose $a=6, b=10$ and the area of the triangle is $15 \sqrt{3}$. If $\angle A C B$ is obtuse and if $r$ denotes the radius of the incircle of the triangle, then $r^2$ is equal to

Options:

The correct answer is:
3

Solution:
$\sin C=\frac{\sqrt{3}}{2}$ and $C$ is given to be obtuse
$$
\begin{aligned}
& \Rightarrow \quad C=\frac{2 \pi}{3}=\sqrt{a^2+b^2-2 a b \cos C} \\
& =\sqrt{6^2+10^2-2 \times 6 \times 10 \times \cos \frac{2 \pi}{3}}=14 \\
& \therefore \quad r=\frac{\Delta}{s} \Rightarrow r^2=\frac{225 \times 3}{\left(\frac{6+10+14}{2}\right)^2}=3
\end{aligned}
$$

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2010 (Paper 2)
Consider all possible permutations of the letters of the word ENDEANOEL. Match the Statements/Expressions in Column I with the Statements/Expressions in Column II.

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

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

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

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

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

Solution:
(A) If ENDEA is fixed word, then assume this as a single letter.
Total number of letters $=5$ and total number of arrangements $=5$ ! .
(B) If $E$ is at first and last places, then total number of permutations
$$
\frac{7 !}{2 !}=21 \times 5 !
$$
(C) If $\mathrm{D}, \mathrm{L}, \mathrm{N}$ are not in last five positions
$$
\leftarrow \mathrm{D}, \mathrm{L}, \mathrm{N}, \mathrm{N} \rightarrow \leftarrow \text { E, E, E, A, O } \rightarrow
$$
Total number of permutations $=\frac{4 !}{2 !} \times \frac{5 !}{3 !}=2 \times 5$ ! .
(D) Total number of odd positions $=5$
Permutations of AEEEO are $\frac{5 !}{3 !}$
Total number of even positions $=4$
Number of permutations of $\mathrm{N}, \mathrm{N}, \mathrm{D}, \mathrm{L}=\frac{4 !}{2 !}$
Hence, total number of permutations $=\frac{5 !}{3 !} \times \frac{4 !}{2 !}=2 \times 5 !$.

Topic: Mathematics, Permutation Combination, JEE Main
Consider all possible permutations of the letters of the word ENDEANOEL. Match the Statements/Expressions in Column I with the Statements/Expressions in Column II.

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

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

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

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

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

Solution:
(A) If ENDEA is fixed word, then assume this as a single letter.
Total number of letters $=5$ and total number of arrangements $=5$ ! .
(B) If $E$ is at first and last places, then total number of permutations
$$
\frac{7 !}{2 !}=21 \times 5 !
$$
(C) If $\mathrm{D}, \mathrm{L}, \mathrm{N}$ are not in last five positions
$$
\leftarrow \mathrm{D}, \mathrm{L}, \mathrm{N}, \mathrm{N} \rightarrow \leftarrow \text { E, E, E, A, O } \rightarrow
$$
Total number of permutations $=\frac{4 !}{2 !} \times \frac{5 !}{3 !}=2 \times 5$ ! .
(D) Total number of odd positions $=5$
Permutations of AEEEO are $\frac{5 !}{3 !}$
Total number of even positions $=4$
Number of permutations of $\mathrm{N}, \mathrm{N}, \mathrm{D}, \mathrm{L}=\frac{4 !}{2 !}$
Hence, total number of permutations $=\frac{5 !}{3 !} \times \frac{4 !}{2 !}=2 \times 5 !$.

Topic: Mathematics, Permutation Combination, JEE Advanced, JEE Advanced 2008 (Paper 2)
Consider an electric field $\mathbf{E}=E_0 \hat{\mathbf{x}}$ where $\bar{E}_0$ is a constant. The flux through the shaded area ( as shown in the figure) due to this field is

Options:
A
$2 E_0 a^2$

B
$\sqrt{2} E_0 a^2$

C
$E_0 a^2$

D
$\frac{E_0 a^2}{\sqrt{2}}$

The correct answer is:
$E_0 a^2$

Solution:
Electric flux, $\mathbf{E} \cdot \mathbf{S}$, or $\phi=E S \cos \theta$ Here, $\theta$ is the angle between $\mathbf{E}$ and $\mathbf{S}$. In this question $\theta=45^{\circ}$, because $\mathbf{S}$ is perpendicular to the surface.
$$
\begin{gathered}
E=E_0 \\
\Rightarrow \quad S=(\sqrt{2 a})(a)=\sqrt{2} a^2 \\
\therefore \phi=\left(E_0\right)\left(\sqrt{2} a^2\right) \cos 45^{\circ}=E_0 a^2
\end{gathered}
$$
$\therefore$ Correct option is (c).
Analysis of Question
(i) Question is moderately tough.
(ii) The given shaded area is a rectangle not a square. One side of this rectangle is $a$ and other side is $\sqrt{2} a$.
(iii) Electric field is uniform, whose magnitude is $E_0$ and direction is positive $x$. In uniform electric field we can use, $\phi=E S \cos \theta$

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2011 (Paper 1)
Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional change in density 1ρdρdt is constant. The velocity v of any point on the surface of the expanding sphere is proportional to

Options:
A R3
B 1R
C R
D R23

The correct answer is:
R

Solution:
Density of sphere is ρ=mv=3m4πR3

⇒ 1ρdρdt= -3R dRdt

Since ⇒ 1ρdρdt is constant

∴ dRdt ∝R

Topic: Physics, Units and Dimensions, JEE Advanced, JEE Advanced 2017 (Paper 2)
Consider $L_1: 2 x+3 y+p-3=0 ; L_2: 2 x+3 y+p+3=0$ where $p$ is a real number and $C: x^2+y^2+6 x-10 y+30=0$.
Statement 1 If line $L_1$ is a chord of circle $C$, then line $L_2$ is not always a diameter of circle $C$.
Statement 2 If line $L_1$ is a diameter of circle $C$, then line $L_2$ is not a chord of circle $C$.

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:
Equation of circle $C$ is
$$
\begin{aligned}
(x+3)^2+(y-5)^2 & =9+25-30=4 \\
\Rightarrow \quad \quad(x+3)^2+(y-5)^2 & =2^2 \\
\text { Centre } & =(3,-5)
\end{aligned}
$$
If $L_1$ is diameter, then $2(3)+3(-5)+p-3=0 \Rightarrow p=12$
$\therefore \quad L_1$ is $2 x+3 y+9=0$
and
$L_2$ is $2 x+3 y+15=0$
Distance of centre of circle from $L_2=\left|\frac{2(3)+3(-5)+15}{\sqrt{2^2+3^2}}\right|=\frac{6}{\sqrt{13}} < 2 \quad$ [radius of circle]
$\therefore L_2$ is a chord of circle $C$.
Statement 2 is false.

Topic: Mathematics, Straight Lines, JEE Advanced, JEE Advanced 2008 (Paper 2)
Consider the following cell reaction,
$$
\begin{array}{cc}
& 2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q) \longrightarrow \\
& 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l), E^{\circ}=1.67 \mathrm{~V} \\
\text { At } & {\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_2\right)=0.1 \mathrm{~atm}}
\end{array}
$$
and $\mathrm{pH}=3$, the cell potential at $25^{\circ} \mathrm{C}$ is

Options:
A
$1.47 \mathrm{~V}$

B
$1.77 \mathrm{~V}$

C
$1.87 \mathrm{~V}$

D
$1.57 \mathrm{~V}$

The correct answer is:
$1.57 \mathrm{~V}$

Solution:
The half reactions are $\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 e^{-} \times 2$
$$
\begin{array}{r}
\mathrm{O}_2(g)+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Fe}(s) \\
+\mathrm{O}_2(g)+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}{ }_2 \mathrm{O}(l) \\
E=E^{\circ}-\frac{0.059}{4} \log \frac{\left(10^{-3}\right)^2}{\left(10^{-3}\right)^4(0.1)}=1.57 \mathrm{~V}
\end{array}
$$

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2011 (Paper 2)
Consider the following four statements whether they are correct of wrong.
I. The sporophyte in liverworts is more elaborate than that in mosses
II. Salvinia is heterosporous
III. The life-cycle in all seed-bearing plants is diplontic
IV. In Pinus male and female cones are borne on different tress

Options:
A Statements (I) and (III)
B Statements (I) and (IV)
C Statements (II) and (III)
D Statements (I) and (II)

The correct answer is:
Statements (I) and (IV)

Solution:
The sporophyte is more developed in mosses rather than liverwort. The simplest sporophyte is found in Riccia, a liverwort, where it is represented by capsule only. The most complex sporophyte is found in Funaria, a moss. However, the most advanced sporophyte is found in Anthoceros (a hornwort) due to the presence of intercalary meristem.
Pinus (a gymnosperm) is monoecious plant in which male and female cones are borne on different branches of the same plant.

Topic: Biology, Plant Kingdom, NEET, NEET 2011 (Mains)
Consider the following linear equations
$$
a x+b y+c z=0, b x+c y+a z=0, c x+a y+b z=0
$$
Match the conditions/expressions in Column I with statements in Column II.

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

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

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

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

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

Solution:
Let $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=-\frac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$
(A) If $a+b+c \neq 0$ and $a^2+b^2+c^2=a b+b c+c a$
$$
\Rightarrow \quad \Delta=0 \text { and } a=b=c \neq 0
$$
The equations represents identical planes.
(B) $a+b+c=0$ and $a^2+b^2+c^2 \neq a b+b c+c a$
$$
\Rightarrow \quad \Delta=0
$$
The equations have infinitely many solutions.
$$
\begin{array}{rlrl}
& & a x+b y & =(a+b) z \\
\Rightarrow & b x+c y & =(b+c) z \\
\Rightarrow & & \left(b^2-a c\right) y & =\left(b^2-a c\right) z \Rightarrow y=z \\
\Rightarrow & a x+b y+c y & =0 \\
& a x & =a y \Rightarrow x=y=z
\end{array}
$$
(C) $a+b+c \neq 0$ and $a^2+b^2+c^2 \neq a b+b c+c a$
$$
\Rightarrow \quad \Delta \neq 0
$$
The equation represent planes meeting at only one point.
(D) $a+b+c=0$ and $a^2+b^2+c^2=a b+b c+c a$
$$
\Rightarrow \quad a=b=c=0
$$
The equations represent whole of the three dimensional space.

Topic: Mathematics, Determinants, JEE Main