datavorous/entrance-exam-dataset huggingface dataset:
If $y(x)$ satisfies the differential equation $y^{\prime}-y \tan x$ $=2 x \sec x$ and $y(0)=0$, then

Options:
A $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}$
B $y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{18}$
C $y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9}$
D $y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}$

The correct answers are:
$y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}$, $y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}$

Solution:
$\frac{d y}{d x}-y \tan x=2 x \sec x$

I.F. $=e^{-\int \tan x d x}=\cos x$

$\therefore \quad y \cdot \cos x=\int 2 x d x=x^{2}+c$

Now, $y(0)=0 \Rightarrow c=0, \therefore y=x^{2} \sec x$

$\Rightarrow y^{\prime}=2 x \sec x+x^{2} \sec x \tan x$

Now, $y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{16} \times \sqrt{2}=\frac{\pi^{2}}{8 \sqrt{2}}$

$y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9} \times 2=\frac{2 \pi^{2}}{9}$

$y^{\prime}\left(\frac{\pi}{4}\right)=\frac{2 \pi}{4} \times \sqrt{2}+\frac{\pi^{2}}{8 \sqrt{2}} \times 1=\frac{\pi^{2}}{8 \sqrt{2}}+\frac{\pi}{\sqrt{2}}$

$y^{\prime}\left(\frac{\pi}{3}\right)=\frac{2 \pi}{3} \times 2+\frac{2 \pi^{2}}{9} \times \sqrt{3}=\frac{2 \pi^{2}}{3 \sqrt{3}}+\frac{4 \pi}{3}$

Topic: Mathematics, Differential Equations, JEE Advanced, JEE Advanced 2012 (Paper 1)
If $|z|=1$ and $z \neq \pm 1$, then all the values of $\frac{z}{1-z^2}$ lie on

Options:
A
a line not passing through the origin

B
$|z|=\sqrt{2}$

C
the X-axis

D
the Y-axis

The correct answer is:
the Y-axis

Solution:
Let $z=\cos \theta+i \sin \theta$
$$
\begin{aligned}
\frac{z}{1-z^2} & =\frac{\cos \theta+i \sin \theta}{1-(\cos 2 \theta+i \sin 2 \theta)} \\
& =\frac{\cos \theta+i \sin \theta}{2 \sin ^2 \theta-2 i \sin \theta \cos \theta} \\
& =\frac{\cos \theta+i \sin \theta}{-2 i \sin \theta(\cos \theta+i \sin \theta)} \\
& =\frac{i}{2 \sin \theta}
\end{aligned}
$$
Hence, $\frac{z}{1-z^2}$ lies on the imaginary axis, i.e. $x=0$.
ALITER
Let
$E=\frac{z}{1-z^2}=\frac{z}{z \bar{z}-z^2}=\frac{1}{\bar{z}-z}$ which is imaginary.

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2007 (Paper 2)
If $z$ is any complex number satisfying $|z-3-2 i| \leq 2$, then the maximum value of $|2 z-6+5 i|$ is

Options:

The correct answer is:
5

Solution:
Given, $|z-3-2 i| \leq 2$
To find minimum of $|2 z-6+5 i|$ or $\quad 2\left|z-3+\frac{5}{2} i\right|$
Using triangle inequality, i.e. ||$z_1|-| z_2|| \leq\left|z_1+z_2\right|$ $\therefore\left|z-3+\frac{5}{2} i\right|$ $=\left|z-3-2 i+2 i+\frac{5}{2} i\right|$ $=\left|(z-3-2 i)+\frac{9}{2} i\right| \geq|| z-3-2 i\left|-\frac{9}{2}\right|$ $\geq\left|2-\frac{9}{2}\right| \geq \frac{5}{2} \Rightarrow\left|z-3+\frac{5}{2} i\right| \geq \frac{5}{2}$ or $\quad|2 z-6+5 i| \geq 5$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2011 (Paper 1)
Image of an object approaching a convex mirror of radius of curvature $20 \mathrm{~m}$ along its optical axis is observed to move from $\frac{25}{3} \mathrm{~m}$ to $\frac{50}{7} \mathrm{~m}$ in $30 \mathrm{~s}$. What is the speed of the object in $\mathrm{km} \mathrm{h}^{-1}$ ?

Options:

The correct answer is:
3

Solution:
Using mirror formula twice,
$\frac{1}{+25 / 3}+\frac{1}{-u_1}=\frac{1}{+10}$
or $\quad \frac{1}{u_1}=\frac{3}{25}-\frac{1}{10}$ or $u_1=50 \mathrm{~m}$
and, $\quad \frac{1}{(+50 / 7)}+\frac{1}{-u_2}=\frac{1}{+10}$
$\therefore \quad \frac{1}{u_2}=\frac{7}{50}-\frac{1}{10}$ or $u_2=25 \mathrm{~m}$
Speed of object $=\frac{u_1-u_2}{\text { time }}$
$=\frac{25}{30} \mathrm{~ms}^{-1}$
$=3 \mathrm{kmh}^{-1}$
$\therefore$ The answer is 3 .

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2010 (Paper 2)
In $1 \mathrm{~L}$ saturated solution of $\mathrm{AgCl}\left[K_{\mathrm{sp}} \mathrm{AgCl}=1.6 \times 10^{-10}\right], 0.1$ mole of $\mathrm{CuCl}\left[K_{\mathrm{sp}}(\mathrm{CuCl})=1.0 \times 10^{-6}\right]$ is added. The resultant concentration of $\mathrm{Ag}^{+}$in the solution is $1.6 \times 10^{-x}$. The value of $x$ is

Options:

The correct answer is:
7

Solution:
It is a case of simultaneous solubility of salts with a common ion. Here solubility product of $\mathrm{CuCl}$ is much greater than that of $\mathrm{AgCl}$, it can be assumed that $\mathrm{Cl}^{-}$in solution comes mainly from $\mathrm{CuCl}$.
$$
\Rightarrow \quad\left[\mathrm{Cl}^{-}\right]=\sqrt{K_{s p}(\mathrm{CuCl})}=10^{-3} \mathrm{M}
$$
Now for $\mathrm{AgCl}: K_{s p}=1.6 \times 10^{-10}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\left[\mathrm{Ag}^{+}\right] \times 10^{-3} \Rightarrow\left[\mathrm{Ag}^{+}\right]=1.6 \times 10^{-7}$

Topic: Chemistry, Ionic Equilibrium, JEE Advanced, JEE Advanced 2011 (Paper 2)
In a $\triangle A B C$ with fixed base $B C$, the vertex $A$ moves such that $\cos B+\cos C=4 \sin ^2 \frac{A}{2}$. If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$ respectively, then

Options:
A
$b+c=4 a$

B
$b+c=2 a$

C
locus of point $A$ is an ellipse

D
locus of point $A$ is a pair of straight line

The correct answers are:
$b+c=2 a$
,
locus of point $A$ is an ellipse

Solution:
Given, $\cos B+\cos C=4 \sin ^2 \frac{A}{2}$
$$
\begin{aligned}
& \Rightarrow 2 \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)=4 \sin ^2 \frac{A}{2} \\
& \Rightarrow 2 \sin \frac{A}{2}\left[\cos \left(\frac{B-C}{2}\right)-2 \sin \frac{A}{2}\right]=0 \\
& \Rightarrow \cos \left(\frac{B-C}{2}\right)-2 \cos \left(\frac{B+C}{2}\right)=0 \\
& \Rightarrow-\cos \frac{B}{2} \cos \frac{C}{2}+3 \sin \frac{B}{2} \sin \frac{C}{2}=0 \\
& \Rightarrow \quad \tan \frac{B}{2} \tan \frac{C}{2}=\frac{1}{3} \\
& \Rightarrow \quad \sqrt{\frac{(s-a)(s-c)}{s(s-b)} \cdot \frac{(s-b)(s-a)}{s(s-c)}}=\frac{1}{3} \\
& \Rightarrow \quad \frac{s-a}{s}=\frac{1}{3} \Rightarrow 2 s=3 a \\
& \Rightarrow \quad b+c=2 a
\end{aligned}
$$
$\therefore$ Locus of $A$ is an ellipse.

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2009 (Paper 1)
In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance $\mathrm{C} \mu F$ across a $200 \mathrm{~V}, 50 \mathrm{~Hz}$ supply. The power consumed by the lamp is $500 \mathrm{~W}$ while the voltage drop across it is $100 \mathrm{~V}$. Assume that there is no inductive load in the circuit. Take $r m s$ values of the voltages. The magnitude of the phaseangle (in degrees) between the current and the supply voltage is $\varphi$. Assume, $\pi \sqrt{3} \approx 5$.
The value of ϕ is _____.

Options:

The correct answer is:
60

Solution:
Given,

P=500 W

VR=100 V

f=50 Hz

Vrms=200 V

P=VR2R⇒R=VR2P⇒R=1002500

∴ R=20 Ω

Now current in the circuit,

i=irms=voltage across resistorresistancei=10020i=5 A

Phasor diagram of the given circuit for voltage,

Now,

V2=VC2+VR2⇒2002=VC2+1002⇒VC=1003 V

Now,

VC=1003 V=IXC⇒XC=10035=203 Ω

and

XC=12πfC⇒203=12π×100C⇒C=100 μF

From the phasor diagram,

tanϕ=XCR=20320=3⇒ϕ=60°

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2021 (Paper 2)
In a constant volume calorimeter, $3.5 \mathrm{~g}$ of a gas with molecular weight $=28$ was burnt in excess oxygen at $298.0 \mathrm{~K}$. The temperature of the calorimeter was found to increases from $298.0 \mathrm{~K}$ to $298.45 \mathrm{~K}$ due to the combustion process. Given, that the heat capacity of the calorimeter is $2.5$ $\mathrm{kJ} \mathrm{K}^{-1}$, the numerical value for the enthalpy of combustion of the gas in $\mathrm{kJ} \mathrm{mol}^{-1}$ is

Options:

The correct answer is:
9

Solution:
The temperature rise is : $\Delta T=T_2-T_1=298.45-298=0.45 \mathrm{~K}$
This indicates that heat produced from combustion of $3.5 \mathrm{~g}$ of compound rises temperature of calorimeter by $0.45 \mathrm{~K}$.
Heat produced $=0.45 \mathrm{~K} \times 2.5 \mathrm{k} \mathrm{JK}^{-1}=1.125 \mathrm{~kJ}$
$\Rightarrow$ Heat produced from $28 \mathrm{~g}$ of compound $(1.0 \mathrm{~mol})=\frac{1.125}{3.5} \times 28=9 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2009 (Paper 2)
In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $\lambda$. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2 . Choose the correct choice(s),

Options:
A
If $d=\lambda$, the screen will contain only one maximum

B
If $\lambda < d < 2 \lambda$, at least one more maximum (besides the central maximum) will be observed on the screen
C
If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2 , the intensities of the observed dark and bright fringes will increase

D
If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1 , the intensities of the observed dark and bright fringes will increase

The correct answers are:
If $d=\lambda$, the screen will contain only one maximum
,
If $\lambda < d < 2 \lambda$, at least one more maximum (besides the central maximum) will be observed on the screen

Solution:
For $d=\lambda$, there will be only one, acentral maxima.
For $\lambda < d < 2 \lambda$, there will be three maximas on the screen corresponding to path difference, $\Delta x=0$ and $\Delta x=\pm \lambda$ correct options are (a) and (b).

Topic: Physics, Wave Optics, JEE Advanced, JEE Advanced 2008 (Paper 1)
In allene $\left(\mathrm{C}_{3} \mathrm{H}_{4}\right)$, the type(s) of hybridisation of the carbon atoms is (are):

Options:
A $s p$ and $s p^{3}$
B $s p$ and $s p^{2}$
C only $s p^{3}$
D $s p^{2}$ and $s p^{3}$

The correct answer is:
$s p$ and $s p^{2}$

Solution:
Allene $\left(\mathbf{C}_{3} \mathbf{H}_{4}\right)$ is $\stackrel{s p^{2}}{\mathrm{H}_{2}} \mathrm{C}=\stackrel{s p}{\mathrm{C}}=\stackrel{s p^{2}}{\mathrm{CH}_{2}}$

Topic: Chemistry, Hydrocarbons, JEE Advanced, JEE Advanced 2012 (Paper 1)
In an experiment to determine the focal length $(f)$ of a concave mirror by the $u-v$ method, a student places the object pin $A$ on the principal axis at a distance $x$ from the pole $P$. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then

Options:
A
$x < f$
B
$f < x < 2 f$
C
$x=2 f$

D
$x>2 f$

The correct answer is:
$f < x < 2 f$

Solution:
Since object and image move in opposite directions, the positioning should be as shown in the figure. Object lies between focus and centre of curvature
$$
f < x < 2 f \text {. }
$$
$\therefore$ Correct option is (b).

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2007 (Paper 1)
In an insulated vessel, $0.05 \mathrm{~kg}$ steam at $373 \mathrm{~K}$ and $0.45 \mathrm{~kg}$ of ice at $253 \mathrm{~K}$ are mixed. Find the final temperature of the mixture (in kelvin).
$$
\text { Given, } \begin{aligned}
L_{\text {fusion }} & =80 \mathrm{cal} / \mathrm{g}=336 \mathrm{~J} / \mathrm{g}, L_{\text {vaporization }}=540 \mathrm{cal} / \mathrm{g}=2268 \mathrm{~J} / \mathrm{g}, \\
S_{\text {ice }} & =2100 \mathrm{~J} / \mathrm{kg}, K=0.5 \mathrm{cal} / \mathrm{gK} \text { and } S_{\text {water }}=4200 \mathrm{~J} / \mathrm{kg}, K=1 \mathrm{cal} / \mathrm{gK} .
\end{aligned}
$$

Options:

The correct answer is:
273

Solution:
$0.05 \mathrm{~kg}$ steam at $373 \mathrm{~K} \stackrel{Q_1}{\longrightarrow} 0.05 \mathrm{~kg}$ water at $373 \mathrm{~K}$
$0.05 \mathrm{~kg}$ water at $373 \mathrm{~K} \stackrel{Q_2}{\longrightarrow} 0.05$ kg water at $273 \mathrm{~K}$
$0.45 \mathrm{~kg}$ ice at $253 \mathrm{~K} \stackrel{Q_3}{\longrightarrow} 0.45 \mathrm{~kg}$ ice at $273 \mathrm{~K}$
$0.45 \mathrm{~kg}$ ice at $273 \mathrm{~K} \stackrel{Q_4}{\longrightarrow} 0.45 \mathrm{~kg}$ water at $273 \mathrm{~K}$
$$
\begin{aligned}
& Q_1=(50)(540)=27,000 \mathrm{cal}=27 \mathrm{kcal} \\
& Q_2=(50)(1)(100)=5000 \mathrm{cal}=5 \mathrm{kcal} \\
& Q_3=(450)(0.5)(20)=4500 \mathrm{cal}=4.5 \mathrm{kcal} \\
& Q_4=(450)(80)=36000 \mathrm{cal}=36 \mathrm{kcal}
\end{aligned}
$$
Now since $Q_1+Q_2>Q_3$ but $Q_1+Q_2 < Q_3+Q_4$ ice will come to $273 \mathrm{~K}$ from $253 \mathrm{~K}$, but whole ice will not melt. Therefore temperature of the mixture is $273 \mathrm{~K}$.

Topic: Physics, Thermal Properties of Matter, JEE Advanced, JEE Advanced 2006
In animals, there are special glands secreting hormones, whereas there are no glands in plants. Where are plant hormones formed? How are the hormones translocated to the site of activity?

Options:

None

Solution:
The plant hormones are synthesised by the plant cells needed. Few hormones are specifically synthesised at a particular part of the plant like auxin synthesised in growing shoot apices and ethylene is secretes by ripened fruits. Cytokinin is found in dividing cells. Unlike plants animal being more advanced, and organised they have proper hormone secreting glands and organs. These are transported through the transport system of their body in both plant and animals. In plants, hormone are translocated via xylem and phloem to the site of activity.

Topic: Biology, Plant - Growth and Development
In dicot stem, the xylem is

Options:
A Exarch
B Mesarch
C Centarch
D Endarch

The correct answer is:
Endarch

Solution:
There are four types of primary xylem development: exarch, endarch, centrarch, and mesarch.
In endarch development, the protoxylem begins its development from the innermost procambial cells located adjacent to the pith and development progresses outward. Therefore, the protoxylem is found toward the inside and metaxylem toward the outside of the stem. Endarch development is considered the most highly advanced type of primary xylem development.

Topic: Biology, Anatomy of Flowering Plants, NEET
In the cyanide extraction process of silver from argentite ore, the oxidising and reducing agents used are

Options:
A $\mathrm{O}_{2}$ and $\mathrm{CO}$ respectively
B $\mathrm{O}_{2}$ and $\mathrm{Zn}$ dust respectively
C $\mathrm{HNO}_{3}$ and $\mathrm{Zn}$ dust respectively
D $\mathrm{HNO}_{3}$ and $\mathrm{CO}$ respectively

The correct answer is:
$\mathrm{O}_{2}$ and $\mathrm{Zn}$ dust respectively

Solution:
The reactions involved in cyanide extraction process are:

$\underset{\text { (argentite) }}{\mathrm{Ag}_{2} \mathrm{~S}}+4 \mathrm{NaCN} \rightarrow 2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]+\mathrm{Na}_{2} \mathrm{~S}$

$4 \mathrm{Na}_{2} \mathrm{~S}+\underset{\text { Oxiding agent }}{5 \mathrm{O}_{2}}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Na}_{2} \mathrm{SO}_{4}+4 \mathrm{NaOH}+2 \mathrm{~S}$

$2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]+\underset{\begin{array}{c}\text { (reducing } \\ \text { agent) }\end{array}}{\mathrm{Zn}} \rightarrow \mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]+2 \mathrm{Ag} \downarrow$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2012 (Paper 2)
In the determination of Young's modulus $\left(Y=\frac{4 M L g}{\pi l d^{2}}\right)$ by using Searle's method, a wire of length $L=2 \mathrm{~m}$ and diameter $d=0.5 \mathrm{~mm}$ is used. For a load $M=2.5 \mathrm{~kg}$, an extension $l=0.25 \mathrm{~mm}$ in the length of the wire is observed. Quantities $d$ and $l$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of $0.5 \mathrm{~mm}$. The number of divisions on their circular scale is 100 . The contributions to the maximum probable error of the $Y$ measurement

Options:
A due to the errors in the measurements of $d$ and $l$ are the same.
B due to the error in the measurement of $d$ is twice that due to the error in the measurement of $l$.
C due to the error in the measurement of $l$ is twice that due to the error in the measurement of $d$.
D due to the error in the measurement of $d$ is four times that due to the error in the measurement of $l$.

The correct answer is:
due to the errors in the measurements of $d$ and $l$ are the same.

Solution:
The maximum possible error in $Y$ due to $l$ and $d$ $\frac{\Delta Y}{Y}=\frac{\Delta l}{l}+\frac{2 \Delta d}{d}$ Least count $=\frac{\text { Pitch }}{\text { No. of division on circular scale }}$ $=\frac{0.5}{100} \mathrm{~mm}=0.005 \mathrm{~mm}$

Here, $\Delta d=\Delta l=0.005 \mathrm{~mm}$

Error contribution of $l=\frac{\Delta l}{l}=\frac{0.005 \mathrm{~mm}}{0.25 \mathrm{~mm}}=\frac{1}{50}$

Error contribution of

$d=\frac{2 \Delta d}{d}=\frac{2 \times 0.005 \mathrm{~mm}}{0.5 \mathrm{~mm}}=\frac{1}{50}$

Hence contribution to the maximum possible error in the measurement of $y$ due to $l$ and $d$ is the same.

Topic: Physics, Mathematics in Physics, JEE Advanced, JEE Advanced 2012 (Paper 1)
In the experiment to determine the speed of sound using a resonance column:

Options:
A
Prongs of the tuning fork are kept in a vertical plane

B
Prongs of the tuning fork are kept in a horizontal plane

C
In one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air

D
In one of the two resonances observed, the length of the resonating air column is close to half of the wavelength of sound in air

The correct answer is:
Prongs of the tuning fork are kept in a vertical plane

Solution:
Length of air column in resonance is odd integer multiple of $\frac{\lambda}{4}$. Correct option is (a).

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2007 (Paper 2)
In the following carbocation, $\mathrm{H} / \mathrm{CH}_3$ that is most likely to migrate to the positively charged carbon is

Options:
A
$\mathrm{CH}_3$ at $\mathrm{C}-4$

B
$\mathrm{H}$ at $\mathrm{C}-4$

C
$\mathrm{CH}_3$ at $\mathrm{C}-2$

D
$\mathrm{H}$ at $\mathrm{C}-2$

The correct answer is:
$\mathrm{H}$ at $\mathrm{C}-2$

Solution:
Hydride shift from $\mathrm{C}-2$ will give most stable resonance stabilised carbocation as Fig.

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2009 (Paper 2)
In the following reaction sequence, the compound $\mathrm{J}$ is an intermediate.

$\mathbf{J}\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{2}\right)$ gives effervescence on treatment with $\mathrm{NaHCO}_{3}$ and a positive Baeyer's test. Question: The compound $\mathbf{K}$ is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2012 (Paper 2)
In the following reaction sequence, the compound $\mathrm{J}$ is an intermediate.

$\mathbf{J}\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{2}\right)$ gives effervescence on treatment with $\mathrm{NaHCO}_{3}$ and a positive Baeyer's test. Question: The compound $\mathbf{I}$ is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the following reaction sequence, the compound $\mathrm{J}$ is an intermediate.

$\mathbf{J}\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{2}\right)$ gives effervescence on treatment with $\mathrm{NaHCO}_{3}$ and a positive Baeyer's test. Question: The compound $\mathbf{I}$ is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2012 (Paper 2)
In the following reaction sequence, the correct structures of $E, F$ and $G$ are

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the following reaction sequence, the correct structures of $E, F$ and $G$ are

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2008 (Paper 2)
In the following reaction,
$$
\text { The structure of the major product } X \text { is }
$$

Options:
A

B

C

D
< smiles>O=C(Nc1cccc([N+](=O)[O-])c1)c1ccccc1 < /smiles>

The correct answer is:

Solution:
Ring 1 is more active, electrophilic substitution takes place over ring 1. - $\mathrm{NH}-\mathrm{C}-\mathrm{Ph}$ is ortho-para directing, para product is predominating.

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2007 (Paper 1)
In the following reaction,
$$
\text { The structure of the major product } X \text { is }
$$

Options:
A

B

C

D
< smiles>O=C(Nc1cccc([N+](=O)[O-])c1)c1ccccc1 < /smiles>

The correct answer is:

Solution:
Ring 1 is more active, electrophilic substitution takes place over ring 1. - $\mathrm{NH}-\mathrm{C}-\mathrm{Ph}$ is ortho-para directing, para product is predominating.

Topic: Chemistry, Amines, JEE Main
In the following $[x]$ denotes the greatest integer less than or equal to $x$. Match the functions in Column I with the properties Column II.

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

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

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

Solution:
(A) $x|x|$ is continuous, differentiable and strictly increasing in $(-1,1)$.
(B) $\sqrt{|x|}$ is continuous in $(-1,1)$ and not differentiable at $x=0$.
(C) $x+[x]$ is strictly increasing in $(-1,1)$ and discontinuous at $x=0$ $\Rightarrow$ not differentiable at $x=0$.
(D) $|x-1|+|x+1|=2$ in $(-1,1)$
The function is continuous and differentiable in $(-1,1)$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
In the following $[x]$ denotes the greatest integer less than or equal to $x$. Match the functions in Column I with the properties Column II.

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

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

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

Solution:
(A) $x|x|$ is continuous, differentiable and strictly increasing in $(-1,1)$.
(B) $\sqrt{|x|}$ is continuous in $(-1,1)$ and not differentiable at $x=0$.
(C) $x+[x]$ is strictly increasing in $(-1,1)$ and discontinuous at $x=0$ $\Rightarrow$ not differentiable at $x=0$.
(D) $|x-1|+|x+1|=2$ in $(-1,1)$
The function is continuous and differentiable in $(-1,1)$.

Topic: Mathematics, Continuity and Differentiability, JEE Advanced, JEE Advanced 2007 (Paper 1)
In the given circuit, a charge of $+80 \mu C$ is given to the upper plate of the $4 \mu F$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \mu F$ capacitor is

Options:
A $+32 \mu \mathrm{C}$
B $+40 \mu \mathrm{C}$
C $+48 \mu \mathrm{C}$
D $+80 \mu \mathrm{C}$

The correct answer is:
$+48 \mu \mathrm{C}$

Solution:
The total charge on plate $A$ will be $80 \mu \mathrm{C}$.

$2 \mu F$ and $3 \mu F$ capacitors are in parallel. Therefore, $C_{e q}=2+3=5 \mathrm{HF}$

Charge on capacitor of $3 \mu \mathrm{F}$ capacitance $q=\frac{3}{5} \times 80=48 \mu C$

Topic: Physics, Capacitance, JEE Main
In the given circuit, a charge of $+80 \mu C$ is given to the upper plate of the $4 \mu F$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \mu F$ capacitor is

Options:
A $+32 \mu \mathrm{C}$
B $+40 \mu \mathrm{C}$
C $+48 \mu \mathrm{C}$
D $+80 \mu \mathrm{C}$

The correct answer is:
$+48 \mu \mathrm{C}$

Solution:
The total charge on plate $A$ will be $80 \mu \mathrm{C}$.

$2 \mu F$ and $3 \mu F$ capacitors are in parallel. Therefore, $C_{e q}=2+3=5 \mathrm{HF}$

Charge on capacitor of $3 \mu \mathrm{F}$ capacitance $q=\frac{3}{5} \times 80=48 \mu C$

Topic: Physics, Capacitance, JEE Advanced, JEE Advanced 2012 (Paper 2)