IIT-JEE 2012 Paper 2 Question Paper with Solutions

For meta-material, the refractive index is negative. Let n₁ is refractive index of air and n₂ is refractive index of meta-material.

$\therefore$ From Snell's law, ${{\sin {\theta _1}} \over {\sin {\theta _2}}} = {{{n_2}} \over {{n_1}}}$

Since, n₂ is negative, therefore $\theta$₂ is also negative. Hence, appropriate diagram (c) is correct.

The magnetic field due to an infinitely long conductor is circumferential.

Since this conductor is placed along the diameter of circular loop (see figure), the total magnetic flux through the circular wire loop is always zero.

Here's a breakdown of how to solve this problem:

Understanding the Concepts

• Ideal Gas: An ideal gas is a theoretical gas that follows the ideal gas law perfectly. The ideal gas law is a relationship between pressure ($𝑃$), volume ($𝑉$), temperature ($𝑇$), and the number of moles ($𝑛$) of a gas:

$𝑃𝑉=𝑛𝑅𝑇$, where $𝑅$ is the ideal gas constant.

• Heat Capacity: Heat capacity is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius (or 1 Kelvin). For an ideal gas, we can use the specific heat capacity at constant volume (${𝐶}_{𝑣}$) or at constant pressure (${𝐶}_{𝑝}$).


• Constant Volume Process: In a constant volume process, the volume of the gas remains constant. The heat required to raise the temperature is given by:

$𝑄=𝑛{𝐶}_{𝑣}\Delta 𝑇$

• Constant Pressure Process: In a constant pressure process, the pressure of the gas remains constant. The heat required to raise the temperature is given by:

$𝑄=𝑛{𝐶}_{𝑝}\Delta 𝑇$

• Molar Heat Capacities of Helium: Helium is a monatomic gas. For monatomic gases, we have:


• ${𝐶}_{𝑣}=\frac{3}{2}𝑅$


• ${𝐶}_{𝑝}=\frac{5}{2}𝑅$

Solving the Problem

1. Identify the Process: Since the balloon is fully expandable, the pressure remains constant. This is a constant pressure process.


2. Calculate the Heat: Use the formula for heat at constant pressure:

$𝑄=𝑛{𝐶}_{𝑝}\Delta 𝑇$

1. Substitute the Values:

• $𝑛=2\text{ moles}$


• ${𝐶}_{𝑝}=\frac{5}{2}𝑅=\frac{5}{2}\left(8.31\text{ J/mol.K}\right)$


• $\Delta 𝑇={35}^{\circ }𝐶-{30}^{\circ }𝐶=5𝐾$

1. Solve for Q:

$𝑄=\left(2\text{ moles}\right)\times (\frac{5}{2}\times 8.31\text{ J/mol.K})\times \left(5𝐾\right)$

$𝑄=207.75\text{ J}$

Answer:

The amount of heat required to raise the temperature of the helium gas is approximately 208 J. Therefore, the correct answer is Option D.

As depicted in the figure shown here, when the system, as a whole, turns by 180$^\circ$, the disc also turns by 180$^\circ$ about its vertical axis. Hence, the instantaneous axis that passes through the centre of mass is vertical in both cases (a) and (b).

Escape velocity, expressed as

${𝑣}_{𝑒}=\sqrt{\surd \frac{2\mathrm{GM}}{R}}=\sqrt{\surd \frac{2G\cdot \frac{4}{3}𝜋R^3𝜌}{R}}$

can be simplified to

$\sqrt{\surd \frac{8𝜋G𝜌}{3}}R.$

The surface area, ${𝐴}_{𝑠}$, is given by

${𝐴}_{𝑠}=4𝜋{𝑅}^2.$

Considering planet Q's surface area is 4 times that of planet P:

$A_Q=4A=4𝜋R_2^Q$

$A_P=A=4𝜋R_2^P$

therefore,

$4R_2^P=R_2^Q.$

Solving for ${𝑅}_{𝑄}$ :

$\Rightarrow {𝑅}_Q=2R_P\text{........(i)}.$

For planet R, with mass $M_R=M_P+M_Q$:

$𝜌\cdot \frac{4}{3}𝜋R_3^R=𝜌\cdot \frac{4}{3}𝜋(R_3^P+R_3^Q)$

Since

$R_Q=2R_P,$

we have :

$R_3^R=R_3^P+8R_3^P=9R_3^P.$

Thus,

$R_R=(9{)}^{1/3}{R}_P\text{........(ii)}.$

From (i) and (ii), it follows that :

${𝑅}_{𝑅}>{𝑅}_{𝑄}>{𝑅}_{𝑃}.$

Since

${𝑉}_{𝑒}\propto R,$

it implies that :

$V_R>V_Q>V_P.$

Also,

$\frac{V_R}{V_P}=\frac{R_R}{R_P}=9^{1/3},$

and

$\frac{V_P}{V_Q}=\frac{R_P}{R_Q}=\frac{1}{2}.$

Consider a thin, uniform cylindrical shell that is closed at both ends and partially filled with water. This cylindrical shell is floating vertically in water, with exactly half of its height submerged. The relative density of the shell's material compared to water is denoted by ${𝜌}_{𝑐}$.

The inner radius of the cylindrical shell is ${𝑟}_1$, the outer radius is ${𝑟}_2$, the height is $ℎ$, and the material density is $𝜌$. The shell is filled with water, which has a density of ${𝜌}_{𝑤}$, up to a height of $𝑥$. Given ${𝜌}_{𝑐}=\frac{𝜌}{{𝜌}_{𝑤}}$, we know that the cylinder is in a half-submerged state, meaning $ℎ/2$ of the cylinder's height is submerged in water.

In equilibrium, the weight of the cylindrical shell, including the water inside it, must equal the buoyant force (upthrust) acting on it. This relationship can be expressed as follows:

$(𝜋{𝑟}_2^2-𝜋{𝑟}_2^1)ℎ𝜌𝑔+𝜋{𝑟}_2^1𝑥{𝜌}_{𝑤}𝑔=𝜋{𝑟}_2^2\left(\frac{ℎ}{2}\right){𝜌}_{𝑤}𝑔$

By simplifying this equation, we get:

$𝑥=ℎ[\frac{{𝑟}_2^2}{{𝑟}_2^1}(0.5-{𝜌}_{𝑐})+{𝜌}_{𝑐}]=ℎ[0.5+\frac{{𝑟}_2^2-{𝑟}_2^1}{{𝑟}_2^1}(0.5-{𝜌}_{𝑐})]...........\left(𝑖\right)$

From equation (i), we can determine the condition under which $𝑥$ is greater than $0.5ℎ$. This occurs if ${𝜌}_{𝑐}<0.5(\because {𝑟}_2>{𝑟}_1)$.

Area of the loop

$A = \left[ {{a^2} + 4 \times {{\pi {{\left( {{a \over 2}} \right)}^2}} \over 2}} \right]\widehat k = \left[ {{a^2} + {{\pi {a^2}} \over 2}} \right]\widehat k$

Therefore, the magnetic moment of the current loop is

$M = I \times A = I\left[ {{a^2} + {{\pi {a^2}} \over 2}} \right]\widehat k = \left[ {1 + {\pi \over 2}} \right]I{a^2}\widehat k$

${\left(\frac{1}{8}\right)}^{\mathrm{th}}$ rotation of the disc with a constant $\Omega$ implies that the disc has turned through

$\frac{1}{8}\times {360}^{\circ }={45}^{\circ }$

The orientation of the disc now is

For the particle thrown from $𝑄$ towards $𝑅$, will have only the velocity given to it with respect to disc i.e., it possess a velocity only in the Y-Z plane and covers a range along OR. Hence, it can be seen to land on the disc in the unshaded region. Actually point $𝑂$ of the disc has zero velocity and $𝑄$ being very close to $𝑂$, will have only the velocity given to it w.r.t disc.

For the particle thrown from P , will have an additional velocity $=𝑅\Omega$ along +X direction i.e., it will cover a range $=\frac{𝑅}{2}$ along PO as well as a distance $=\left(𝑅𝜔\right)\frac{𝑇}{8}=\frac{𝑅}{8}\times 2𝜋=\frac{𝜋𝑅}{4}$ along $+𝑋$ direction

The resultant displacement of particle P is thus obtained as

$\begin{array}{cc}\Rightarrow 𝑑& =\frac{𝑅}{2}\sqrt{\surd 1+\frac{{𝜋}^2}{4}}\\ \Rightarrow {𝜋}^2& =10\\ 𝑑& =\frac{R}{4}\sqrt{\surd 14}\\ \tan 𝜃& =\frac{𝜋R/4}{R/2}=\frac{𝜋}{2}\end{array}$

In the diagram shown,

$\begin{array}{cc}𝑃𝑆& =𝑅\cos 𝜃\\ & =𝑅\times \frac{2}{\sqrt{\surd 14}}\\ & =\frac{2𝑅}{\sqrt{\surd 14}}\end{array}$

Since, $𝑑>$ PS $$ PS $=\frac{𝑅}{7}\sqrt{\surd 14}$

Hence, the particle from P also lands in the unshaded region.

Since Cylinder P has most of its mass concentrated near its surface, its moment of inertia (about the cylinder axis) is greater than that of Cylinder Q, i.e., ${𝐼}_{𝑃}>{𝐼}_{𝑄}$. The forces acting on the cylinder include its weight $𝑚𝑔$, the normal reaction $𝑁$, and the frictional force $𝑓$.

In the case of rolling without slipping, we have :

$𝑣=𝜔𝑟$

$𝑎=𝛼𝑟$

The torque about the center of mass is related to $𝛼$ by :

$𝜏=𝑟𝑓=𝐼𝛼\left(1\right)$

The force along the plane is related to $𝑎$ by :

$𝑚𝑔\sin 𝜃-𝑓=𝑚𝑎\left(2\right)$

By solving equations (1) and (2), we get :

$𝑎=\frac{𝑔\sin 𝜃}{1+\frac{𝐼}{𝑚{𝑟}^2}}\left(3\right)$

From equation (3), it follows that ${𝑎}_{𝑃}<{𝑎}_{𝑄}$ (since ${𝐼}_{𝑃}>{𝐼}_{𝑄}$). Therefore, Cylinder P reaches the ground later, has a lower velocity, lower angular velocity ($𝜔=\frac{𝑣}{𝑟}$), and lower translational kinetic energy.

For verification, the conservation of energy can be used :

$𝑚𝑔ℎ=\frac{1}{2}𝑚{𝑣}^2+\frac{1}{2}𝐼{𝜔}^2$

This yields :

$\frac{1}{2}𝑚{𝑣}_2^{𝑃}<\frac{1}{2}𝑚{𝑣}_2^{𝑄}$

Thus :

${𝑣}_{𝑃}<{𝑣}_{𝑄},{𝜔}_{𝑃}<{𝜔}_{𝑄},{𝑎}_{𝑃}<{𝑎}_{𝑄}$

And the time relationships :

${𝑡}_{𝑃}>{𝑡}_{𝑄}$

Total energy remains conserved. Energy is shared by antineutrino, proton and electron. Kinetic energy of electron has continuous spectrum and it is maximum when antineutrino does not share any kinetic energy.

So total energy is shared with proton and electron only.

$\therefore$ K $\le$ 0.8 $\times$ 10⁶ eV

and kinetic energy of electron will be minimum or zero when total energy is shared by proton and antineutrino.

$\therefore$ 0 $\le$ K $\le$ 0.8 $\times$ 10⁶ eV

Refractive index for a medium

$n = \left( {{c \over v}} \right)$

For meta-material, $n = \,|n|$

$\therefore$ $v = {c \over {|n|}}$

Here, $\Omega$ = 100 rad/s, L = 0.5 H, C = 100 $\mu$F, V = 20 V

$\therefore$ ${X_L} = \omega L = 100 \times 0.5 = 50\,\Omega$

${X_C} = {1 \over {\omega C}} = {1 \over {100 \times 100 \times {{10}^{ - 6}}}} = 100\,\Omega$

Impedance across capacitor,

${Z_1} = \sqrt {{R^2} + X_C^2}$

$= \sqrt {{{(100)}^2} + {{(100)}^2}}$

${Z_1} = 100\sqrt 2 \,\Omega$

$\therefore$ ${I_1} = {{20} \over {100\sqrt 2 }} = {1 \over {5\sqrt 2 }}\,A$

Voltage across 100 $\Omega$

$V = {I_1} \times 100 = {1 \over {5\sqrt 2 }} \times 100 = 10\sqrt 2 \,V$

Impedance across inductance,

${Z_2} = \sqrt {{R^2} + {{({X_L})}^2}} = \sqrt {{{(50)}^2} + {{(50)}^2}}$

${Z_2} = 50\sqrt 2 \,\Omega$

$\therefore$ ${I_2} = {{20} \over {50\sqrt 2 }} = {2 \over {5\sqrt 2 }} = {{\sqrt 2 } \over 5}$

Now, voltage across $50\,\Omega = {{\sqrt 2 } \over 5} \times 50 = 10\sqrt 2$

${I_1} = {1 \over {5\sqrt 2 }}A$ at 45$^\circ$ leading

${I_2} = {{\sqrt 2 } \over 5}A$ at 45$^\circ$ lagging

$\therefore$ Current through circuit

${I_{net}} = \sqrt {I_1^2 + I_2^2} = \sqrt {{{\left( {{1 \over {5\sqrt 2 }}} \right)}^2} + \left( {{{\sqrt 2 } \over 5}} \right)} = 0.3\,A$

As depicted in the figure shown here, when the system, as a whole, turns by 180$^\circ$, the disc also turns by 180$^\circ$ about vertical axis. Hence, for both cases (a) and (b), the angular speed about the instantaneous axis that passes through the centre of mass is $\omega$.

A student is conducting a resonance column experiment. The tube has a diameter of 4 cm, and the tuning fork vibrates at 512 Hz. The air temperature is 38^oC, where the speed of sound is 336 m/s. The zero mark on the meter stick aligns with the top of the tube. For the first resonance, the water level reads :

During the first resonance :

$\frac{𝜆}{4}={𝑙}_1+𝑒$

where ${𝑙}_1$ is the length of the air column at resonance

Given that :

$\text{end correction}=𝑒=0.6𝑟=0.6\times 2=1.2\text{cm}$

where $𝑟$ is the radius of the tube

The wavelength $𝜆$ can be expressed as :

$𝜆=4({𝑙}_1+𝑒)$

The frequency $𝑓$ is related to the wavelength $𝜆$ by :

$𝑓=\frac{𝑣}{𝜆}=\frac{𝑣}{4({𝑙}_1+𝑒)}$

Rearranging for ${𝑙}_1$ :

$4({𝑙}_1+𝑒)=\frac{𝑣}{𝑓}$

Solving for ${𝑙}_1$ :

${𝑙}_1=\frac{𝑣}{4𝑓}-𝑒$

Substitute the given values :

${𝑙}_1=\frac{336\text{m/s}}{4\times 512\text{Hz}}-0.012\text{m}$

${𝑙}_1=\frac{336}{2048}-0.012$

${𝑙}_1=0.164\text{m}-0.012\text{m}$

${𝑙}_1=0.152\text{m}$

Converting to cm :

${𝑙}_1=15.2\text{cm}$

${K_p} + {K_{\overline e }} + {K_{\overline v }}$ = 0.8 $\times$ 10⁶ eV

When electron has zero kinetic energy is shared by antineutrino and proton.

Then, ${K_p} + {K_{\overline v }}$ = 0.8 $\times$ 10⁶ eV

As antineutrino is very light mass in comparison to proton so it will have almost contribution in total energy.

$\therefore$ Its energy is almost 0.8 $\times$ 10⁶ eV

The outer ring rolls without slipping, meaning the point on the ring in contact with the ground (denoted as point C) is stationary, i.e., ${\overrightarrow{𝑣}}_C=\overrightarrow{0}$. Therefore, the velocity of point O is :

${\overrightarrow{𝑣}}_O={\overrightarrow{𝑣}}_C+{\overrightarrow{𝜔}}_o\times {\overrightarrow{𝑟}}_{\mathrm{CO}}=\overrightarrow{0}+𝜔\hat{ȷ}\times 3𝑅\widehat{𝑘}=3𝑅𝜔\hat{ı}$

Point P is located on the inner disc, which has an angular velocity ${\overrightarrow{𝜔}}_i=-𝜔/2\hat{ȷ}$. The position vector from O to P is given by :

${\overrightarrow{𝑟}}_{\mathrm{OP}}=𝑅\cos {30}^{\circ }\hat{ı}+𝑅\sin {30}^{\circ }\widehat{𝑘}=\frac{\sqrt{\surd 3}𝑅}{2}\hat{ı}+\frac{𝑅}{2}\widehat{𝑘}$

Thus, the velocity of point P is calculated as follows :

$\begin{array}{cc}{\overrightarrow{𝑣}}_P& ={\overrightarrow{𝑣}}_O+{\overrightarrow{𝜔}}_i\times {\overrightarrow{𝑟}}_{\mathrm{OP}}\\ & =3𝑅𝜔\hat{ı}+(-\frac{𝜔}{2}\hat{ȷ})\times (\frac{\sqrt{\surd 3}𝑅}{2}\hat{ı}+\frac{𝑅}{2}\widehat{𝑘})\\ & =\frac{11𝜔𝑅}{4}\hat{ı}+\frac{\sqrt{\surd 3}𝜔𝑅}{4}\widehat{𝑘}.\end{array}$

The relative velocity of the point $𝑃$ w.r.t. the point $𝑄$ is given by

${\overrightarrow{𝑣}}_{𝑟}={\overrightarrow{𝑣}}_{𝑃}-{\overrightarrow{𝑣}}_{𝑄}$ .........(1)

It is easy to see that $\mid {\overrightarrow{𝑣}}_{𝑃}\mid =\mid {\overrightarrow{𝑣}}_{𝑄}\mid =𝜔𝑅$ and angle traversed in time $𝑡$ is $𝜔𝑡$. Thus, velocities of $𝑃$ and $𝑄$ are

$\begin{array}{cc}& {\overrightarrow{𝑣}}_{𝑃}=𝜔𝑅(-\sin 𝜔𝑡\hat{ı}-\cos 𝜔𝑡\hat{ȷ})\\ & {\overrightarrow{𝑣}}_{𝑄}=𝜔𝑅(\sin 𝜔𝑡\hat{ı}-\cos 𝜔𝑡\hat{ȷ})\end{array}$

Substitute ${\overrightarrow{𝑣}}_{𝑃}$ and ${\overrightarrow{𝑣}}_{𝑄}$ in equation (1) to get ${\overrightarrow{𝑣}}_{𝑟}=$ $-2𝜔𝑅\sin 𝜔𝑡\hat{ı}$ and thus ${𝑣}_{𝑟}=2𝜔𝑅|\sin 𝜔𝑡|$.

Let the charges on the $3𝜇F$ and $2𝜇F$ capacitors be $𝑞$ and ${𝑞}^{\prime }$ respectively. The charge on the lower plate of the $4𝜇F$ capacitor is $-80𝜇C$. The lower plate of the $4𝜇F$ capacitor and the upper plates of the $2𝜇F$ and $3𝜇F$ capacitors form an isolated system. Therefore, the net charge on this system must be zero, which can be represented as :

$𝑞+{𝑞}^{\prime }-80𝜇C=0\text{..........(i)}$

The potentials ($𝑉=𝑞/𝐶$) across these two capacitors are equal, which gives :

$\frac{𝑞}{3}=\frac{{𝑞}^{\prime }}{2}\text{..........(ii)}$

By substituting ${𝑞}^{\prime }$ from equation (ii) into equation (i), we can solve for $𝑞$ and find :

$𝑞=48𝜇C$

The electric field at point $𝑂$ due to the charges at vertices $𝐴$ and $𝐷$ is $4𝐾$ along the direction OD. Similarly, due to the charges at vertices $𝐵$ and $𝐸$, the electric field is $2𝐾$ along the direction OE, and due to the charges at vertices $𝐶$ and $𝐹$, it is $2𝐾$ along the direction OC. Given the uniform geometry of this setup, the resulting electric field is $6𝐾$ along OD.

The potential at point $𝑂$ is calculated as :

${𝑉}_O=\sum \frac{1}{4𝜋{𝜖}_0}\frac{{𝑞}_{𝑖}}{𝐿}=\frac{1}{4𝜋{𝜖}_0𝐿}\sum {𝑞}_{𝑖}=0$

For any point on the line PR, we observe that there are pairs of equal and opposite charges equidistant from these points, making the potential at any point on PR zero. If we consider points on OS, the potential is positive, while for points on OT, the potential is negative. The potential at points on the line ST, at a distance $𝑥$ from $𝑂$ (with $𝑥$ considered positive towards the right), can be shown to be :

$\begin{array}{c}𝑉\left(𝑥\right)=\frac{𝑞}{4𝜋{𝜖}_0}[\frac{2}{\sqrt{\surd {𝐿}^2+{𝑥}^2+𝑥𝐿}}-\frac{2}{\sqrt{\surd {𝐿}^2+{𝑥}^2-𝑥𝐿}}-\frac{4𝑥}{{𝐿}^2-{𝑥}^2}].\end{array}$

r = distance of a point from centre.

For r $\le$ R/2 Using Ampere's circuital law,

$\oint {B.\,dl}$

or, $Bl = {\mu _0}({I_{in}})$

$B(2\pi r) = {\mu _0}({I_{in}})$

$B = {{{\mu _0}} \over {2\pi }}{{{I_{in}}} \over r}$ ...... (i)

Since, ${I_{in}} = 0$ $\therefore$ $B = 0$

For ${R \over 2} \le r \le R$

${I_{in}} = \left[ {\pi {r^2} - \pi {{\left( {{R \over 2}} \right)}^2}} \right]\sigma$

Here $\sigma$ = current per unit area.

Substituting in Eq. (i), we have

$B = {{{\mu _0}} \over {2\pi }}{{\left[ {\pi {r^2} - \pi {{{R^2}} \over 2}} \right]\sigma } \over r}$

$= {{{\mu _0}\sigma } \over {2r}}\left( {{r^2} - {{{R^2}} \over 4}} \right)$

At $r = {R \over 2},\,B = 0$

At $r = R,\,B = {{3{\mu _0}\sigma R} \over 8}$

For r $\ge$ R

${I_{in}} = {I_{Total}} = I$ (say)

Therefore, substituting in Eq. (i), we have

$B = {{{\mu _0}} \over {2\pi }}.\,{I \over r}$

or, $B \propto {1 \over r}$

Diamond is hard and graphite is soft. Graphite is a good conductor of electricity as it has one free electron which is responsible for the conduction.

Diamond has higher thermal conductivity than graphite because the structure of diamond is precise and thus the transfer of heat is faster in it.

In case of graphite, the C-C bond has a double bond character so its bond order becomes higher than that of diamond which has only single C-C bonds.

The reactions involved are as follows:

The reaction is

The given electrochemical cell is a concentration cell where both the electrodes are of the same metal but immersed in solutions of different concentrations of the same metal ion. The emf (E) generated by this cell can be calculated using the Nernst equation, which for this cell at 298 K (25°C) is given by:

$E = E^\circ - \frac{0.059}{n} \log \frac{[C_1]}{[C_2]}$

Where:

• $E^\circ$ is the standard electrode potential which is zero in a concentration cell because both electrodes are same.


• $n$ is the number of moles of electrons transferred in the redox reaction (2 in this case, as it involves $M^{2+}$ ions).


• $[C_1]$ and $[C_2]$ are the concentrations of $M^{2+}$ at the two electrodes.


• $E$ is given to be 0.059 V.

Assuming the more concentrated solution is at the left hand electrode and the less concentrated solution (0.001 M) is at the right hand electrode, the Nernst equation simplifies to:

$E = - \frac{0.059}{2} \log \frac{0.001}{[C_1]}$

To find $[C_1]$, we solve for the argument of log such that the calculated emf matches the given emf (0.059 V):

$0.059 = - \frac{0.059}{2} \log \frac{0.001}{[C_1]}$

$-2 = \log \frac{0.001}{[C_1]}$

$10^{-2} = \frac{0.001}{[C_1]}$

$[C_1] = 0.1 \text{ M}$

This $[C_1]$ value supports the direction of the redox reactions assumed. Now, to find the change in Gibbs free energy $\Delta G$ for this cell, we use the relationship:

$\Delta G = -nFE$

But $E$ should be positive for the spontaneous reaction, hence:

$\Delta G = -2 \times 96500 \times 0.059 \, \text{V} = -11381 \, \text{J/mol} = -11.381 \, \text{kJ/mol}$

Therefore, $\Delta G$ for the cell is approximately -11.4 kJ/mol. The correct answer is Option D: -11.4.

In the given reaction, an aromatic aldehyde is condensed with an acid anhydride (Perkin condensation) in the presence of a base to form $\alpha$,$\beta$-unsaturated acid (cinnamic acid). This gives effervescence on reaction with NaHCO₃ and positive test with Bayer's reagent (1% alkaline KMnO₄).

Consider the titration reaction

CaOCl₂ + 2KI $\to$ I₂ + Ca(OH)₂ + KCl

According to this reaction, 25 mL of CaOCl₂ reacts with 30 mL of 0.50 M KI

I₂ + 2Na₂S₂O₃ $\to$ Na₂S₄O₆ + 2NaI

Given that 48 mL of 0.25 N Na₂S₂O₃ was used to reach the end point. So, the number of moles of I₂ produced = 48 $\times$ 0.25/2 = 6.

According to the reaction,

Number of millimoles of bleaching powder = Number of moles of I₂ = (1/2) $\times$ Number of moles of Na₂S₂O₃ = 6

So, the molarity of CaOCl₂ is

${{Number\,of\,moles\,of\,bleaching\,powder} \over {Volume\,of\,solution}} = {{6\,mmol} \over {25\,mL}} = 0.24M$

The electronic configuration of Xe is 5s² 5p⁶; and that of Xe in excited state is 5s² 5p⁵ 5d¹. The hybridisation is sp³d and geometry is see-saw. The actual shape is trigonal bipyramidal but due to the preaence of lone pair it gets distorted to see-saw structure.

The $\beta$-keto acid given in option (B) undergoes decarboxylation most easily because it results in formation of stable product on rearrangement. The reaction taking place is

Silver ore is oxidized using oxygen from the air, as shown in the reaction below :

$4\text{Ag}+8\text{NaCN}+2{\text{H}}_2\text{O}+{\text{O}}_2\left(\text{air}\right)⟶4{\text{NaAg(CN)}}_2+4\text{NaOH}$

Here, silver ($\text{Ag}\left(0\right)$) is undergoing oxidation to form $\text{Ag}(+1)$ :

$\text{Ag}\left(0\right)\stackrel{\text{oxidation}}{\to \leftarrow }\text{Ag}(+1)$

Silver is then precipitated from the solution by adding zinc dust in a finely divided form :

$2{\text{NaAg(CN)}}_2+\text{Zn}⟶{\text{Na}}_2{\text{Zn(CN)}}_4+2\text{Ag}$

In this reaction, $\text{Ag}(+1)$ is reduced back to $\text{Ag}\left(0\right)$ :

$\text{Ag}(+1)\stackrel{\text{reduction}}{\to \leftarrow }\text{Ag}\left(0\right)$

In summary, oxygen from the air acts as the oxidizing agent, and zinc dust serves as the reducing agent.

The reaction sequence is

Bleaching powder is CaOCl₂ which contains HOCl as an oxoacid and the anhydride of which is Cl₂O.

The relation between the given compounds can be determined assigning them R and S configuration. The given structures can be represented as

M and N are diastereomers.

M and O are identical.

M and P are enantiomers (non-superimposable mirror images).

M and Q are diastereomers.

The configuration of Ni = 3d⁸ 4s² and that of Ni²⁺ = 3d⁸.

In paramagnetic state, the hybridisation is sp³ and geometry is tetrahedral.

In diamagnetic state, the hybridisation is dsp² and geometry is square planar.

We know that

Binding energy = Total energy of reactants $-$ Total energy of products

So, we have

225 = 2(C $-$ C) + 1(H $-$ H) $-$ 2(C $-$ H) + 1(C $\equiv$ C)

225 = 1410 + 330 $-$ 2 $\times$ 350 + (C $\equiv$ C)

Solving, we get the value of C $\equiv$ C as 815 kJ mol^$-$1.

For isothermal process T₁ = T₂. Work done in isothermal process is less than adiabatic process. In case of isothermal process, the temperature remains constant so there is no change in the internal energy, whereas in case of adiabatic process expansion occurs through internal energy.

In the case of physiorption, with the increase of temperature and pressure the rate of adsorption decreases because according to Le Chatelier's principle, increase of temperature and pressure will shift the equilibrium to the left

Adsorbate + Adsorbent $\rightleftharpoons$ Adsorption + Heat

This is shown in Graphs I and III; whereas in the case of chemisorption, there is a formation of strong bond between the adsorbate and the adsorbent and so the rate of adsorption increases with increase in temperature (Graphs II and IV).

To find the solubility product (K_sp) for the sparingly soluble salt, MX₂, in a concentration cell setup, we first analyze how the emf is related to the concentration differences across the cell. The reaction at each electrode involves the metal ion M²⁺ and the metal M, whereas the net cell reaction has no change in number of moles on both sides of the equation due to the symmetry of the cell. Thus, the notation for the cell reaction is:

$M(s) \longleftrightarrow M^{2+}(aq) + 2e^{-}$

Given that it's a concentration cell, the emf generated is due to the concentration difference of the M²⁺ ions at the two electrodes. The emf of the cell can be calculated by the Nernst equation:

$E = E^{\circ} - \frac{RT}{nF} \ln\left(\frac{[M^{2+}]_{\text{cathode}}}{[M^{2+}]_{\text{anode}}}\right)$

Since it's a concentration cell, $E^\circ = 0$. At 298 K, substituting from the provided conversion (2.303 $\times$ R $\times$ 298/F = 0.059 V) and given n = 2 (because two electrons are transferred per metal ion):

$E = - \frac{(0.059)}{2} \log\left(\frac{[M^{2+}]_{saturated}}{[M^{2+}]_{0.001 \text{ M}}}\right)$

Given $E = 0.059$ V, we can solve for the concentration of $M^{2+}$ in the saturated solution:

$0.059 = -0.0295 \log\left(\frac{[M^{2+}]_{saturated}}{0.001}\right)$

$\log\left(\frac{[M^{2+}]_{saturated}}{0.001}\right) = -2$

$[M^{2+}]_{saturated} = 0.001 \times 10^{-2}$

$[M^{2+}]_{saturated} = 0.00001 \text{ M}$

The solubility product, K_sp, of MX₂ can now be computed. Let s be the solubility of MX₂ in mol/L. The dissolution of MX₂ is given by:

$MX_2 \rightleftharpoons M^{2+} + 2 X^-$

$\quad s \quad\quad\quad s \quad \quad \, 2s$

The solubility product expression is:

$K_{sp} = [M^{2+}][X^-]^2 = (s)(2s)^2$

$K_{sp} = 4s^3$

Since $s = [M^{2+}]_{saturated} = 0.00001$ M,

$K_{sp} = 4(0.00001)^3$

$K_{sp} = 4 \times 10^{-15}$

Thus, the solubility product of MX₂ at 298 K is $4 \times 10^{-15}$ mol³ dm^-9, which corresponds to Option B.

Using the boiling point elevation formula:

$\Delta {𝑇}_{𝑏}={𝐾}_{𝑏}\cdot 𝑚$,

where:

$\Delta {𝑇}_{𝑏}$ is the elevation in boiling point,

${𝐾}_{𝑏}$ is the ebullioscopic constant,

$𝑚$ is the molality.

Given:

${𝐾}_{𝑏}=0.76\text{K}\text{kg}{\text{mol}}^{-1}$,

$\Delta {𝑇}_{𝑏}=2^{\circ }\text{C}$,

mass of solute $=2.5\text{g}$,

mass of solvent (water) $=100\text{g}=0.1\text{kg}$.

First, find the molality $𝑚$:

$𝑚=\frac{\Delta {𝑇}_{𝑏}}{{𝐾}_{𝑏}}=\frac{2}{0.76}=\frac{200}{76}\approx 2.63\text{mol}{\text{kg}}^{-1}$.

Now calculate the moles of solute:

Given the molality $𝑚$,

$𝑚=\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.

Thus,

$\text{moles of solute}=𝑚\times \text{mass of solvent in kg}=2.63\times 0.1=0.263\text{mol}$.

Next, determine the molecular weight $𝑀$ of the solute:

$𝑀=\frac{\text{mass of solute}}{\text{moles of solute}}=\frac{2.5\text{g}}{0.263\text{mol}}\approx 9.51\text{g/mol}$.

Using Raoult's Law for the vapour pressure of the solution:

${𝑃}_{\text{solution}}={𝑃}_0\cdot (1-{𝜒}_{\text{solute}})$,

where:

${𝑃}_0$ is the vapour pressure of pure solvent,

${𝜒}_{\text{solute}}$ is the mole fraction of the solute.

Approximate the mole fraction of the solute:

${𝜒}_{\text{solute}}=\frac{\text{moles of solute}}{{𝜈}_{\text{solvent}}+{𝜈}_{\text{solute}}}\approx \frac{0.263}{5.55+0.263}\approx \frac{0.263}{5.813}\approx 0.0452$,

where ${𝜈}_{\text{solvent}}$ are the moles of water (approximately 5.55 mol for 100 g of water).

Therefore, the vapour pressure of the solution:

${𝑃}_{\text{solution}}=760\times (1-0.0452)\approx 760\times 0.9548\approx 724\text{mm Hg}$.

Thus, the vapour pressure of the solution is 724 mm of Hg, corresponding to Option A.

The first reaction is

${K_3}[Fe{(CN)_6}] + \mathop {KI}\limits_{excess} \to {K_4}[Fe{(CN)_6}] + \mathop {K{I_3}}\limits_{brownish\,yellow\,solution}$

This is a redox reaction as both oxidation and reduction are taking place.

$\mathop {{K_4}[Fe{{(CN)}_6}]}\limits_{soluble} \, + ZnS{O_4} \to \mathop {{K_2}Z{n_3}{{[Fe{{(CN)}_6}]}_3}}\limits_{white\,ppt} + NaOH \to \mathop {N{a_2}{{[Zn{{(OH)}_4}]}^{2 - }}}\limits_{soluble}$

$2I_3^ - + 2N{a_2}{S_2}{O_3} \to \mathop {N{a_2}{S_4}{O_6}}\limits_{clear\,solution} + 2NaI + \mathop {2{I_2}}\limits_{(blue\,colour)}$