datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
A small block of mass $M$ moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60^{\circ}$ to $30^{\circ}$ at point $B$. The block is initially at rest at $A$. Assume that collisions between the block and the incline are totally inelastic $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
Question:
If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point $B$, immediately after it strikes the second incline is

Options:
A
$\sqrt{30} \mathrm{~m} / \mathrm{s}$

B
$\sqrt{15} \mathrm{~m} / \mathrm{s}$

C
zero

D
$-\sqrt{15} \mathrm{~m} / \mathrm{s}$

The correct answer is:
zero

Solution:
In elastic collision, component of $v_1$ parallel to $B C$ will remain unchanged, while component perpendicular to $B C$ will remain unchanged in magnitude but its direction will be reversed.
$$
\begin{aligned}
& v_{\|}=v_1 \cos 30^{\circ}=(\sqrt{60})\left(\frac{\sqrt{3}}{2}\right)=\sqrt{45} \mathrm{~ms}^{-1} \\
& v_{\perp}=v_1 \sin 30^{\circ}=(\sqrt{60})\left(\frac{1}{2}\right)=\sqrt{15} \mathrm{~ms}^{-1}
\end{aligned}
$$
Now vertical component of velocity of block :
$$
\begin{aligned}
v & =v_{\perp} \cos 30^{\circ}-v_{\|} \cos 60^{\circ} \\
& =(\sqrt{15})\left(\frac{\sqrt{3}}{2}\right)-(\sqrt{45})\left(\frac{1}{2}\right)=0
\end{aligned}
$$
$\therefore$ correct option is (c).

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
A small block of mass $M$ moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60^{\circ}$ to $30^{\circ}$ at point $B$. The block is initially at rest at $A$. Assume that collisions between the block and the incline are totally inelastic $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
Question:
The speed of the block at point $B$ immediately after it strikes the second incline is

Options:
A
$\sqrt{60} \mathrm{~m} / \mathrm{s}$

B
$\sqrt{45} \mathrm{~m} / \mathrm{s}$

C
$\sqrt{30} \mathrm{~m} / \mathrm{s}$

D
$\sqrt{15} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$\sqrt{45} \mathrm{~m} / \mathrm{s}$

Solution:
Between $A$ and $B$, height fallen by block $h_1=\sqrt{3} \tan 60^{\circ}=3 \mathrm{~m}$.
$\therefore$ speed of block just before striking the second incline,
$$
v_1=\sqrt{2 g h_1}=\sqrt{2 \times 10 \times 3}=\sqrt{60} \mathrm{~ms}^{-1}
$$
In perfectly inelastic collision, component of $v_1$ perpendicular to $B C$ will become zero, while component of $v_1$ parallel to $B C$ will remain unchanged.
$\therefore$ speed of block $B$ immediately after it strikes the incline is,
$$
\begin{aligned}
v_2 & =\text { component of } v_1 \text { along } B C \\
& =v_1 \cos 30^{\circ} \\
& =(\sqrt{60})\left(\frac{\sqrt{3}}{2}\right)=\sqrt{45} \mathrm{~ms}^{-1}
\end{aligned}
$$
$\therefore$ correct option is (b)

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
A small block of mass $M$ moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60^{\circ}$ to $30^{\circ}$ at point $B$. The block is initially at rest at $A$. Assume that collisions between the block and the incline are totally inelastic $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
Question:
If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point $B$, immediately after it strikes the second incline is

Options:
A
$\sqrt{30} \mathrm{~m} / \mathrm{s}$

B
$\sqrt{15} \mathrm{~m} / \mathrm{s}$

C
zero

D
$-\sqrt{15} \mathrm{~m} / \mathrm{s}$

The correct answer is:
zero

Solution:
In elastic collision, component of $v_1$ parallel to $B C$ will remain unchanged, while component perpendicular to $B C$ will remain unchanged in magnitude but its direction will be reversed.
$$
\begin{aligned}
& v_{\|}=v_1 \cos 30^{\circ}=(\sqrt{60})\left(\frac{\sqrt{3}}{2}\right)=\sqrt{45} \mathrm{~ms}^{-1} \\
& v_{\perp}=v_1 \sin 30^{\circ}=(\sqrt{60})\left(\frac{1}{2}\right)=\sqrt{15} \mathrm{~ms}^{-1}
\end{aligned}
$$
Now vertical component of velocity of block :
$$
\begin{aligned}
v & =v_{\perp} \cos 30^{\circ}-v_{\|} \cos 60^{\circ} \\
& =(\sqrt{15})\left(\frac{\sqrt{3}}{2}\right)-(\sqrt{45})\left(\frac{1}{2}\right)=0
\end{aligned}
$$
$\therefore$ correct option is (c).

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Paragraph:
A small block of mass $M$ moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60^{\circ}$ to $30^{\circ}$ at point $B$. The block is initially at rest at $A$. Assume that collisions between the block and the incline are totally inelastic $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
Question:
The speed of the block at point $B$ immediately after it strikes the second incline is

Options:
A
$\sqrt{60} \mathrm{~m} / \mathrm{s}$

B
$\sqrt{45} \mathrm{~m} / \mathrm{s}$

C
$\sqrt{30} \mathrm{~m} / \mathrm{s}$

D
$\sqrt{15} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$\sqrt{45} \mathrm{~m} / \mathrm{s}$

Solution:
Between $A$ and $B$, height fallen by block $h_1=\sqrt{3} \tan 60^{\circ}=3 \mathrm{~m}$.
$\therefore$ speed of block just before striking the second incline,
$$
v_1=\sqrt{2 g h_1}=\sqrt{2 \times 10 \times 3}=\sqrt{60} \mathrm{~ms}^{-1}
$$
In perfectly inelastic collision, component of $v_1$ perpendicular to $B C$ will become zero, while component of $v_1$ parallel to $B C$ will remain unchanged.
$\therefore$ speed of block $B$ immediately after it strikes the incline is,
$$
\begin{aligned}
v_2 & =\text { component of } v_1 \text { along } B C \\
& =v_1 \cos 30^{\circ} \\
& =(\sqrt{60})\left(\frac{\sqrt{3}}{2}\right)=\sqrt{45} \mathrm{~ms}^{-1}
\end{aligned}
$$
$\therefore$ correct option is (b)

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Paragraph:
A small block of mass $M$ moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60^{\circ}$ to $30^{\circ}$ at point $B$. The block is initially at rest at $A$. Assume that collisions between the block and the incline are totally inelastic $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$
Question:
The speed of the block at point $C$, immediately before it leaves the second incline is

Options:
A
$\sqrt{120} \mathrm{~m} / \mathrm{s}$

B
$\sqrt{105} \mathrm{~m} / \mathrm{s}$

C
$\sqrt{90} \mathrm{~m} / \mathrm{s}$

D
$\sqrt{75} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$\sqrt{105} \mathrm{~m} / \mathrm{s}$

Solution:
Height fallen by the block from $B$ to $C$. $h_2=3 \sqrt{3} \tan 30^{\circ}=3 \mathrm{~m}$
Let $v_3$ be the speed of block, at point $C$, just before it leaves the second incline, then
$$
\begin{aligned}
v_3 & =\sqrt{v_2^2+2 g h_2} \\
& =\sqrt{45+2 \times 10 \times 3} \\
& =\sqrt{105} \mathrm{~ms}^{-1}
\end{aligned}
$$
$\therefore$ correct option is (b).

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
A small spherical monoatomic ideal gas bubble $\left(\gamma=\frac{5}{3}\right)$ is trapped inside a liquid of density $\rho_1$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_0$, the height of the liquid is $H$ and the atmospheric pressure is $p_0$ (Neglect surface tension).
Question:
As the bubble moves upwards, besides the buoyancy force the following forces are acting on it.

Options:
A
Only the force of gravity

B
The force due to gravity and the force due to pressure of the liquid

C
The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid

D
The force due to gravity and the force due to viscosity of the liquid

The correct answer is:
The force due to gravity and the force due to viscosity of the liquid

Solution:
As the bubble moves upwards, besides the buoyancy force (cause of which is pressure difference) only force of gravity and force of viscosity will act.
$\therefore$ correct option is (d).

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
A small spherical monoatomic ideal gas bubble $\left(\gamma=\frac{5}{3}\right)$ is trapped inside a liquid of density $\rho_1$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_0$, the height of the liquid is $H$ and the atmospheric pressure is $p_0$ (Neglect surface tension).
Question:
The buoyancy force acting on the gas bubble is (Assume $R$ is the universal gas constant)

Options:
A
$\rho_l n R g T_0 \frac{\left(\rho_0+\rho_l g H\right)^{\frac{2}{5}}}{\left(\rho_0+\rho_l g y\right)^{\frac{7}{5}}}$

B
$\frac{\rho_l n R g T_0}{\left(\rho_0+\rho_l g H\right)^{\frac{2}{5}}\left[\rho_0+\rho_l g(H-y)\right]^{\frac{3}{5}}}$

C
$\rho n R g T_0 \frac{\left(\rho_0+\rho_l g H\right)^{\frac{3}{5}}}{\left(\rho_0+\rho_l g y\right)^{\frac{8}{5}}}$

D
$\frac{\rho_l n R g T_0}{\left(\rho_0+\rho_l g H\right)^{\frac{3}{5}}\left[\rho_0+\rho_l g(H-y)\right]^{\frac{2}{5}}}$

The correct answer is:
$\frac{\rho_l n R g T_0}{\left(\rho_0+\rho_l g H\right)^{\frac{2}{5}}\left[\rho_0+\rho_l g(H-y)\right]^{\frac{3}{5}}}$

Solution:
Buoyancy force
Here, and
$$
\begin{aligned}
F & =(\text { volume of bubble })\left(\rho_l\right) g \\
& =\left(\frac{n R T_2}{p_2}\right) p_l g \\
T_2 & =T_0\left[\frac{p_0+p_l g(H-y)}{p_0+\rho_l g H}\right] \\
p_2 & =p_0+p_l(H-y)
\end{aligned}
$$
Substituting the values we get,
$$
F=\frac{\rho_n n g T_0}{\left(p_0+\rho_l g H\right)^{2 / 5}\left[p_0+\rho_l g(H-y)\right]^{3 / 5}}
$$
$\therefore$ correct option is (b).

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
A small spherical monoatomic ideal gas bubble $\left(\gamma=\frac{5}{3}\right)$ is trapped inside a liquid of density $\rho_1$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_0$, the height of the liquid is $H$ and the atmospheric pressure is $p_0$ (Neglect surface tension).
Question:
When the gas bubble is at height $y$ from the bottom, its temperature is

Options:
A
$T_0\left(\frac{p_0+\rho \lg H}{p_0+\rho \lg y}\right)^{\frac{2}{5}}$

B
$T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}$

C
$T_0\left(\frac{p_0+\rho l g H}{p_0+\rho \lg y}\right)^{\frac{3}{5}}$

D
$T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{3}{5}}$

The correct answer is:
$T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}$

Solution:
As there is no exchange of heat. Therefore, process is adiabatic. Applying,
$$
\begin{array}{rlrl}
& T p^{\frac{1-\gamma}{\gamma}}=\text { constant } \\
\therefore \quad & T_2 p_2^{\frac{1-\gamma}{\gamma}}=T_1 p_1^{\frac{1-\gamma}{\gamma}} \\
\text { or } & T_2 & =T_1\left(\frac{p_1}{p_2}\right)^{\frac{1-\gamma}{\gamma}}=T_1\left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}
\end{array}
$$
Substituting the values we have,
$$
\begin{aligned}
T_2 & =T_0\left[\frac{p_0+\rho \lg (H-y)^T}{p_0+\rho \lg H}\right]^{\frac{5 / 3-1}{5 / 3}} \\
& =T_0\left[\frac{p_0+\rho \lg (H-y)}{p_0+\rho \lg H}\right]^{2 / 5}
\end{aligned}
$$
$\therefore$ correct option is (b).

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Paragraph:
A small spherical monoatomic ideal gas bubble $\left(\gamma=\frac{5}{3}\right)$ is trapped inside a liquid of density $\rho_1$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_0$, the height of the liquid is $H$ and the atmospheric pressure is $p_0$ (Neglect surface tension).
Question:
When the gas bubble is at height $y$ from the bottom, its temperature is

Options:
A
$T_0\left(\frac{p_0+\rho \lg H}{p_0+\rho \lg y}\right)^{\frac{2}{5}}$

B
$T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}$

C
$T_0\left(\frac{p_0+\rho l g H}{p_0+\rho \lg y}\right)^{\frac{3}{5}}$

D
$T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{3}{5}}$

The correct answer is:
$T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}$

Solution:
As there is no exchange of heat. Therefore, process is adiabatic. Applying,
$$
\begin{array}{rlrl}
& T p^{\frac{1-\gamma}{\gamma}}=\text { constant } \\
\therefore \quad & T_2 p_2^{\frac{1-\gamma}{\gamma}}=T_1 p_1^{\frac{1-\gamma}{\gamma}} \\
\text { or } & T_2 & =T_1\left(\frac{p_1}{p_2}\right)^{\frac{1-\gamma}{\gamma}}=T_1\left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}
\end{array}
$$
Substituting the values we have,
$$
\begin{aligned}
T_2 & =T_0\left[\frac{p_0+\rho \lg (H-y)^T}{p_0+\rho \lg H}\right]^{\frac{5 / 3-1}{5 / 3}} \\
& =T_0\left[\frac{p_0+\rho \lg (H-y)}{p_0+\rho \lg H}\right]^{2 / 5}
\end{aligned}
$$
$\therefore$ correct option is (b).

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph
A special metal $S$ conducts electricity without any resistance. A closed wire loop, made of $S$, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius $a$, with its center at the origin. A magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $r(\gg a)$ from the center of the loop with its north pole always facing the loop, as shown in the figure below.
The magnitude of magnetic field of a dipole $m$, at a point on its axis at distance $r$, is $\frac{\mu_{0}}{2 \pi} \frac{m}{r^{3}}$, where $\mu_{0}$ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, $m_{1}$ and $m_{2}$, separated by a distance $r$ on the common axis, with their north poles facing each other, is $\frac{k m_{1} m_{2}}{r^{4}}$, where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.
QuestionWhen the dipole m is placed at a distance r from the centre of the loop (as shown in the figure), the current induced in the loop will be proportional to:

Options:
A mr3
B m2r2
C mr2
D m2r

The correct answer is:
mr3

Solution:
For a super conducting loop, net flux passing through it will remain constant.

ϕtotal i=ϕtotal f

As area is same, therefore,

μ0m2πr3=μ0I2a

⇒I=amπr3

⇒I∝mr3

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2021 (Paper 2)
Paragraph
A special metal $S$ conducts electricity without any resistance. A closed wire loop, made of $S$, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius $a$, with its center at the origin. A magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $r(\gg a)$ from the center of the loop with its north pole always facing the loop, as shown in the figure below.
The magnitude of magnetic field of a dipole $m$, at a point on its axis at distance $r$, is $\frac{\mu_{0}}{2 \pi} \frac{m}{r^{3}}$, where $\mu_{0}$ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, $m_{1}$ and $m_{2}$, separated by a distance $r$ on the common axis, with their north poles facing each other, is $\frac{k m_{1} m_{2}}{r^{4}}$, where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.
QuestionThe work done in bringing the dipole from infinity to a distance r from the centre of the loop by the given process is proportional to:

Options:
A mr5
B m2r5
C m2r6
D m2r7

The correct answer is:
m2r6

Solution:
For a super conducting loop, net flux passing through it will remain constant.

ϕtotal i=ϕtotal f

As area is same, therefore,

μ0m2πr3=μ0I2a

⇒I=amπr3

⇒I∝mr3

The current carrying superconducting loop will also behave like a magnet, whose magnetic dipole moment is given by,m1=NIA=(1)amπr3×πa2=ma3r3

⇒ m1∝mr3

The repulsive force felt on the magnet will be

F=Km1m2r4=K'mr3mr4∝m2r7

Wext=-∫Fdr=-K'∫m2r7dr

Using ∫xndx=xn+1n+1

We get,

Wext∝m2r6

Topic: Physics, Magnetic Effects of Current, JEE Main
Paragraph
A special metal $S$ conducts electricity without any resistance. A closed wire loop, made of $S$, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius $a$, with its center at the origin. A magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $r(\gg a)$ from the center of the loop with its north pole always facing the loop, as shown in the figure below.
The magnitude of magnetic field of a dipole $m$, at a point on its axis at distance $r$, is $\frac{\mu_{0}}{2 \pi} \frac{m}{r^{3}}$, where $\mu_{0}$ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, $m_{1}$ and $m_{2}$, separated by a distance $r$ on the common axis, with their north poles facing each other, is $\frac{k m_{1} m_{2}}{r^{4}}$, where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.
QuestionThe work done in bringing the dipole from infinity to a distance r from the centre of the loop by the given process is proportional to:

Options:
A mr5
B m2r5
C m2r6
D m2r7

The correct answer is:
m2r6

Solution:
For a super conducting loop, net flux passing through it will remain constant.

ϕtotal i=ϕtotal f

As area is same, therefore,

μ0m2πr3=μ0I2a

⇒I=amπr3

⇒I∝mr3

The current carrying superconducting loop will also behave like a magnet, whose magnetic dipole moment is given by,m1=NIA=(1)amπr3×πa2=ma3r3

⇒ m1∝mr3

The repulsive force felt on the magnet will be

F=Km1m2r4=K'mr3mr4∝m2r7

Wext=-∫Fdr=-K'∫m2r7dr

Using ∫xndx=xn+1n+1

We get,

Wext∝m2r6

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2021 (Paper 2)
Paragraph:
A tertiary alcohol $H$ upon acid catalysed dehydration gives a product $I$. Ozonolysis of I leads to compounds $\mathrm{J}$ and $\mathrm{K}$. Compound $\mathrm{J}$ upon reaction with
$\mathrm{KOH}$ gives benzyl alcohol and a compound $\mathrm{L}$, where $\mathrm{K}$ on reaction with $\mathrm{KOH}$ gives only $M$.
Question:
$$
\text { Compound } H \text { is formed by the reactions of }
$$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
A tertiary alcohol $H$ upon acid catalysed dehydration gives a product $I$. Ozonolysis of I leads to compounds $\mathrm{J}$ and $\mathrm{K}$. Compound $\mathrm{J}$ upon reaction with
$\mathrm{KOH}$ gives benzyl alcohol and a compound $\mathrm{L}$, where $\mathrm{K}$ on reaction with $\mathrm{KOH}$ gives only $M$.
Question:
The structures of compounds $J, K$ and $L$, respectively, are

Options:
A
$\mathrm{PhCOCH}_3, \mathrm{PhCH}_2 \mathrm{COCH}_3$ and $\mathrm{PhCH}_2 \mathrm{COO}^{-} \mathrm{K}^{+}$

B
$\mathrm{PhCHO}, \mathrm{PhCH}_2 \mathrm{CHO}$ and $\mathrm{PhCH}_2 \mathrm{COO}^{-} \mathrm{K}^{+}$

C
$\mathrm{PhCOCH}_3, \mathrm{PhCH}_2 \mathrm{CHO}$ and $\mathrm{CH}_3 \mathrm{COO}^{-} \mathrm{K}^{+}$

D
$\mathrm{PhCHO}, \mathrm{PhCOCH}_3$ and $\mathrm{PhCOO}^{-} \mathrm{K}^{+}$

The correct answer is:
$\mathrm{PhCHO}, \mathrm{PhCOCH}_3$ and $\mathrm{PhCOO}^{-} \mathrm{K}^{+}$

Solution:
$$
J=\mathrm{PhCH}=\mathrm{O} K=\mathrm{PhCOCH}_3 \quad L=\mathrm{PhCOO}^{-} \mathrm{K}^{+}
$$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
A tertiary alcohol $H$ upon acid catalysed dehydration gives a product $I$. Ozonolysis of I leads to compounds $\mathrm{J}$ and $\mathrm{K}$. Compound $\mathrm{J}$ upon reaction with
$\mathrm{KOH}$ gives benzyl alcohol and a compound $\mathrm{L}$, where $\mathrm{K}$ on reaction with $\mathrm{KOH}$ gives only $M$.
Question:
The structures of compounds $J, K$ and $L$, respectively, are

Options:
A
$\mathrm{PhCOCH}_3, \mathrm{PhCH}_2 \mathrm{COCH}_3$ and $\mathrm{PhCH}_2 \mathrm{COO}^{-} \mathrm{K}^{+}$

B
$\mathrm{PhCHO}, \mathrm{PhCH}_2 \mathrm{CHO}$ and $\mathrm{PhCH}_2 \mathrm{COO}^{-} \mathrm{K}^{+}$

C
$\mathrm{PhCOCH}_3, \mathrm{PhCH}_2 \mathrm{CHO}$ and $\mathrm{CH}_3 \mathrm{COO}^{-} \mathrm{K}^{+}$

D
$\mathrm{PhCHO}, \mathrm{PhCOCH}_3$ and $\mathrm{PhCOO}^{-} \mathrm{K}^{+}$

The correct answer is:
$\mathrm{PhCHO}, \mathrm{PhCOCH}_3$ and $\mathrm{PhCOO}^{-} \mathrm{K}^{+}$

Solution:
$$
J=\mathrm{PhCH}=\mathrm{O} K=\mathrm{PhCOCH}_3 \quad L=\mathrm{PhCOO}^{-} \mathrm{K}^{+}
$$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Paragraph:
A tertiary alcohol $H$ upon acid catalysed dehydration gives a product $I$. Ozonolysis of I leads to compounds $\mathrm{J}$ and $\mathrm{K}$. Compound $\mathrm{J}$ upon reaction with
$\mathrm{KOH}$ gives benzyl alcohol and a compound $\mathrm{L}$, where $\mathrm{K}$ on reaction with $\mathrm{KOH}$ gives only $M$.
Question:
$$
\text { The structure of compound } I \text { is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
A uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is $L$. The disk is initially at its equilibrium position with its centre of mass $(C M)$ at a distance Lfrom the wall. The disk rolls without slipping with velocity $\mathbf{v}_0=v_0 \hat{\mathbf{i}}$ The coefficient of friction is $\mu$.
Question:
The maximum value of $v_0$ for which the disk will roll without slipping is

Options:
A
$\mu g \sqrt{\frac{M}{k}}$

B
$\mu g \sqrt{\frac{M}{2 k}}$

C
$\mu g \sqrt{\frac{3 M}{k}}$

D
$\mu g \sqrt{\frac{5 M}{2 k}}$

The correct answer is:
$\mu g \sqrt{\frac{3 M}{k}}$

Solution:
In case of pure rolling, mechanical energy will remain conserved.
$$
\begin{array}{rlrl}
& \therefore & \frac{1}{2} M v_0^2+\frac{1}{2}\left(\frac{1}{2} M R^2\right)\left(\frac{v_0}{R}\right)^2 & =2\left[\frac{1}{2} k x_{\max }^2\right] \\
& \therefore & x_{\max } & =\sqrt{\frac{3 M}{4 k}} v_0 \\
& \text { As } & & =\frac{2 k x}{3} \\
& f_{\max } & =\mu M g=\frac{2 k x_{\max }}{3}=\frac{2 k}{3} \sqrt{\frac{3 M}{4 k}} v_0 \\
& v_0 & =\mu g \sqrt{\frac{3 M}{k}}
\end{array}
$$
$\therefore$ correct option is (c).

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2008 (Paper 2)