JEE Main 2023 10th April Morning Shift Question Paper with Solutions

The initial condition states that the particle is at position $x=\frac{A}{2}$ at $t=0$.

If we substitute these initial conditions into the equation for the displacement of the particle:

$x = A \sin(wt + \delta)$

We have:

$\frac{A}{2} = A \sin(\delta)$

Dividing both sides by $A$ gives us:

$\frac{1}{2} = \sin(\delta)$

The angle whose sine is $\frac{1}{2}$ is $\delta = \frac{\pi}{6}$ radians (or 30 degrees).

Therefore, the correct answer is $\delta = \frac{\pi}{6}$.

The maximum and minimum amplitude of a modulated wave are given by:

Maximum amplitude = carrier wave amplitude + base band signal magnitude = 15V + 3V = 18V

Minimum amplitude = carrier wave amplitude - base band signal magnitude = 15V - 3V = 12V

Therefore, the ratio of maximum amplitude to minimum amplitude in an amplitude modulated wave is 18/12 = 1.5 or $\frac{3}{2}$.

The range $R$ of a projectile launched with an initial speed $v$ and at an angle $\theta$ to the horizontal is given by:

$R = \frac{v^2}{g} \sin(2\theta)$

where $g$ is the acceleration due to gravity.

From this equation, we can see that the range is dependent on the sine of twice the launch angle.

Given that the range at $15^\circ$ is $50$ m, if we launch the projectile at $45^\circ$ with the same velocity, we can compare the ranges by comparing $\sin(2 \times 15^\circ)$ and $\sin(2 \times 45^\circ)$:

$\sin(30^\circ) = \frac{1}{2}$

$\sin(90^\circ) = 1$

Therefore, the range at $45^\circ$ will be twice the range at $15^\circ$, because $\sin(90^\circ)$ is twice as large as $\sin(30^\circ)$.

So, the range when the projectile is launched at $45^\circ$ will be $2 \times 50$ m = $100$ m.

So, $100$ m is the correct answer.

Rearranging the above circuit we get



The above circuit is a balanced wheatstone bridge there is no current in the branch $c d$.

So we have two $8 \Omega$ and one $16 \Omega$ resistor in parallel. If $R_{\mathrm{eq}}$ is equivalent resistance, then

$$\begin{aligned} &\frac{1}{R_{\mathrm{eq}}}=\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\frac{5}{16} \\ &\Rightarrow R_{\mathrm{eq}}=\frac{16}{5}=3.2 \Omega \end{aligned}$$

Monoatomic gases possess only translational degrees of freedom. So, option (I) matches with (A).

Polyatomic gases have translational and rotational degrees of freedom. So, option (II) matches with (D).

Rigid diatomic gases have translational, rotational and vibrational degrees of freedom. So, option (III) matches with (B).

Non-rigid diatomic gases possess translational, rotational and vibrational degrees of freedom. So, option (IV) matches with (C).

From the given graphs, it can be seen

$\begin{gathered} \text { slope }_{\mathrm{B}}>\text { slope }_{\mathrm{A}} \\ v_{\mathrm{B}}>v_{\mathrm{A}} \end{gathered}$

$\therefore$ Statement (E) is correct.

Also, $x_{\mathrm{A}} < x_{\mathrm{B}}$, A lives closer to school.

$\therefore$ Statement $(\mathrm{A})$ is correct.

Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.

This statement is true. The formula for current sensitivity ($I_s$) of a moving coil galvanometer is given by:

$I_s = \frac{NAB}{k}$

where:

• $N$ is the number of turns in the coil,
• $A$ is the area of the coil,
• $B$ is the magnetic field strength, and
• $k$ is the spring constant of the coil.

From this formula, you can see that the current sensitivity is directly proportional to the number of turns (N). If $N$ is doubled, then the current sensitivity will also double.

Statement II: Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

This statement is false. The formula for voltage sensitivity ($V_s$) of a moving coil galvanometer is given by:

$V_s = I_s R = \frac{NAB}{k} R$

where:

• $R$ is the resistance of the coil.

From this formula, you can see that the voltage sensitivity is proportional to the number of turns ($N$) but also inversely proportional to the coil resistance ($R$). If you double the number of turns ($N$), you also double the length of the wire making up the coil, and thus, you double the resistance ($R$) of the coil. The doubling of $N$ is offset by the doubling of $R$, so the overall voltage sensitivity remains the same.

Therefore, increasing the current sensitivity by only increasing the number of turns in the coil will not increase the voltage sensitivity in the same ratio.

The final pressure of gas in container A after isothermal compression can be found using the equation of state for an ideal gas, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature. For an isothermal process, the temperature $T$ is constant, so the equation becomes $P_1V_1 = P_2V_2$.

The final volume is $\frac{1}{8}$ of the initial volume, so $P_2 = P_1 \times \frac{V_1}{V_2} = P_1 \times 8 = 8P_1$.

The final pressure of gas in container B after adiabatic compression can be found using the adiabatic equation for an ideal gas, $PV^\gamma = \text{constant}$, where $\gamma$ is the ratio of the heat capacities, which is $\frac{5}{3}$ for a monoatomic gas.

Since $V_2 = \frac{V_1}{8}$, we have $P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) ^ \gamma = P_1 \times 8^\frac{5}{3} = P_1 \times 2^5 = 32P_1$.

The ratio of the final pressure of gas in B to that of gas in A is therefore $\frac{32P_1}{8P_1} = 4$.

The energy of an electromagnetic wave contained in a small volume oscillates with double the frequency of the wave.

Here's why: The electromagnetic wave consists of oscillating electric and magnetic fields. The energy density of the wave is proportional to the square of the amplitude of these fields. Since the square of a sinusoidal function oscillates with twice the frequency of the original function, the energy density of the electromagnetic wave also oscillates with twice the frequency of the wave.

The orbital speed of an object moving in a circular orbit around Earth (or any other celestial body) is given by the formula:

$v = \sqrt{\frac{GM}{r}}$

where (v) is the orbital speed, (G) is the gravitational constant, (M) is the mass of the central body (Earth, in this case), and (r) is the radius of the orbit.

For the two satellites of masses (m) and (3m) in orbits of radii (r) and (3r) respectively, the ratio of their orbital speeds ($v_1/v_2$) is:

$\frac{v_1}{v_2} = \sqrt{\frac{\frac{GM}{r}}{\frac{GM}{3r}}} = \sqrt{\frac{3r}{r}} = \sqrt{3}$

So, the ratio of the orbital speeds of the satellites is ($\sqrt{3} : 1$), which corresponds to Option B.

The inclusion of two different masses for the satellites, $m$ and $3m$, in the problem might initially seem to suggest that the masses would influence their orbital speeds. However, when it comes to circular orbital motion, especially around a much larger body like the Earth, the mass of the orbiting satellite does not directly affect its orbital speed. This is because the orbital speed equation:

$v = \sqrt{\frac{GM}{r}}$

only takes into account the mass of the central body (in this case, Earth's mass $M$), and the radius of the orbit $(r)$, where $G$ is the gravitational constant.

Statement I: Pressure in a reservoir of water is same at all points at the same level of water.

This statement is true. According to the principle of fluid statics, in a body of static fluid, the pressure is the same at all points at the same horizontal level. This is because the pressure at any point in a static fluid is determined by the weight of the fluid above it. Therefore, at any given level in the reservoir, the pressure is the same because the weight of the water above each point is the same.

Statement II: The pressure applied to enclosed water is transmitted in all directions equally.

This statement is also true. It is a direct statement of Pascal's law, which states that any change in pressure applied at any point in a fluid in a closed system is transmitted undiminished to all points in the fluid and acts in all directions.

Therefore, both Statement I and Statement II are true.

The de Broglie wavelength of a particle is given by:

$\lambda = \frac{h}{p}$

where h is Planck's constant and p is the momentum of the particle. The momentum of a gas molecule can be related to its kinetic energy (which is related to the temperature of the gas) by:

$p = \sqrt{2mK}$

where m is the mass of the molecule and K is the kinetic energy of the molecule.

At a given temperature T, the average kinetic energy of a molecule in a gas is given by:

$K = \frac{3}{2} kT$

where k is Boltzmann's constant.

Therefore, the de Broglie wavelength of a molecule in a gas is given by:

$\lambda = \frac{h}{\sqrt{2m(3/2)kT}} = \frac{h}{\sqrt{3mkT}}$

If the temperature of the gas is increased from T = 300 K to T = 600 K, the new de Broglie wavelength becomes:

$\lambda' = \frac{h}{\sqrt{3mk(2T)}} = \frac{h}{\sqrt{2} \sqrt{3mkT}} = \frac{1}{\sqrt{2}} \lambda$

So, the de Broglie wavelength of the gas molecule decreases by a factor of $\sqrt{2}$ when the temperature of the gas is doubled.

Therefore, the correct answer is $\lambda' = \frac{1}{\sqrt{2}} \lambda_1$

$\therefore$ Shifting of image will be 8 cm towards mirror.

The percentage error in a quantity that is a product or quotient of other quantities is given by the sum of the percentage errors in those quantities, each multiplied by the power to which it is raised in the expression for the quantity.

Given the physical quantity P as

$P = \frac{a^2b^3}{c\sqrt{d}}$

The percentage error in P, denoted as $\Delta P/P$, is given by:

$\frac{\Delta P}{P} = 2 \left(\frac{\Delta a}{a}\right) + 3 \left(\frac{\Delta b}{b}\right) + \left(\frac{\Delta c}{c}\right) + \frac{1}{2} \left(\frac{\Delta d}{d}\right)$

Substituting the given percentage errors for a, b, c, and d:

$\frac{\Delta P}{P} = 2(0.01) + 3(0.02) + 0.03 + \frac{1}{2}(0.04) = 0.02 + 0.06 + 0.03 + 0.02 = 0.13$

Therefore, the percentage error in P is 13%.

Capacitors numbered (3) and (4) are short circuit as their both plates are connected to terminal B which makes the potential difference across both to be zero. The equivalent circuit is as shown :

Hence $C_{A B}=2 C$ as capacitors numbered (1) and (2) are connected in parallel.

According to Bohr's model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck's constant divided by $2\pi$ (or $h/2\pi$, where $h$ is the Planck's constant). This can be expressed as:

$L = n \frac{h}{2\pi}$

where $n$ is the principal quantum number or the orbit number.

So, for the first orbit ($n=1$), the angular momentum $L_1$ is:

$L_1 = 1 \times \frac{h}{2\pi} = \frac{h}{2\pi}$

And for the second orbit ($n=2$), the angular momentum $L_2$ is:

$L_2 = 2 \times \frac{h}{2\pi} = \frac{h}{\pi}$

The change in angular momentum when moving from the first to the second orbit is the difference between $L_2$ and $L_1$:

$\Delta L = L_2 - L_1 = \frac{h}{\pi} - \frac{h}{2\pi} = \frac{h}{2\pi}$

Since $\frac{h}{2\pi}$ is equal to the initial angular momentum $L_1$, the change in angular momentum when the electron moves to the second orbit is $L$.

Therefore, the correct answer is $L$.

From the conservation of angular momentum, we have:

$ \frac{2}{5}MR^2\omega_1 = \frac{2}{5}M\left(\frac{R}{4}\right)^2\omega_2 $

This simplifies to:

$ MR^2\omega_1 = \frac{MR^2}{16}\omega_2 $

From this, we can derive the ratio of the initial and final angular velocities:

$ \frac{\omega_1}{\omega_2} = \frac{1}{16} $

Since the angular velocity (\omega) is inversely proportional to the period of rotation (T) ((\omega = \frac{2\pi}{T})), we can write:

$ \frac{T_2}{T_1} = \frac{1}{16} $

We can express this ratio in terms of the variable (x):

$ \frac{T_1}{T_2} = \frac{16}{1} = \frac{24}{x} $

Solving this equation for (x) gives:

$ x = 16 $

So, if the Earth suddenly shrinks to ( $\frac{1}{64}$ )th of its original volume with its mass remaining the same, the period of rotation of Earth becomes ( $\frac{24}{16}$ )h, or 1.5 hours. Therefore, the value of (x) is 16.

We know that the magnetizing field (H) inside a solenoid is given by the formula :

$H = \frac{N \cdot I}{L}$

where (N) is the number of turns, (I) is the current in Amperes, and (L) is the length of the solenoid in meters.

To demagnetize a bar magnet that has a magnetic intensity (H) of ($2.4 \times 10^3 \, \text{Am}^{-1}$), you need to create a magnetizing field in the solenoid that is equal in magnitude but opposite in direction.

Given:

• ($H = 2.4 \times 10^3 \, \text{Am}^{-1}$)
• (N = 60) turns
• ($L = 15 \, \text{cm} = 0.15 \, \text{m}$) (since $(1\, \text{m} = 100 \, \text{cm})$)

You can rearrange the formula for (H) to solve for (I) :

$I = \frac{H \cdot L}{N}$

Substituting the given values, you get :

$I = \frac{(2.4 \times 10^3 \, \text{Am}^{-1}) \cdot 0.15 \, \text{m}}{60}$

$I = 6 \, \text{Amperes}$

So, the current required to be passed through the solenoid to demagnetize the bar magnet is (6 $\, \text{A}$).

When two particles collide, the total momentum before the collision is equal to the total momentum after the collision, according to the law of conservation of momentum. Therefore, we can write:

$mv = (m+2m) v_f$

where $v_f$ is the final velocity of the combined particles.

Simplifying the above equation, we get:

$v_f = \frac{mv}{3m} = \frac{v}{3}$

So, the correct option is $\frac{v}{3}$.

$\begin{aligned} & r_1 =r_2=0.2 \mathrm{~cm}=2 \times 10^{-3} \mathrm{~m} \\ & y_1 \text { (steel) } =2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} \\ & l_1 =1.6 \mathrm{~m}\end{aligned}$

At equilibrium,

$$\begin{aligned} &T_2 =1.14 \mathrm{~g}=11.4 \mathrm{~N} \\ &T_2+2 g =T_1 \\ &\therefore T_1 =11.4+20=31.4 \mathrm{~N} \end{aligned}$$

$$\begin{aligned} & \Delta l_1=\frac{T_1 l_1}{Y_1 A_1} \\ = & \frac{31.4 \times 1.6}{2 \times 10^{11} \times \pi\left(2 \times 10^{-3}\right)^2} 1.1 \\ = & \frac{16}{8} \times 10^{-5} \\ = & 2 \times 10^{-5} \mathrm{~m}=20 \times 10^{-6} \mathrm{~m} \end{aligned}$$

When dealing with three polaroids, the intensity after the second polaroid will be given by Malus' law as $I_0 \cos^2 \theta$, where $\theta$ is the angle between the pass axes of the first two polaroids. However, as the third polaroid is orthogonal to the first, no light from the first polaroid passes through, only light from the second polaroid. So the final intensity is also modulated by a $\sin^2 \theta$ term (as the second and third polaroids are orthogonal).

So if we set up the equation for the final intensity $I_{\text{net}}$:

$I_{\text{net}} = I_0 \cos^2 \theta \sin^2 \theta$

And we substitute the given values $I_{\text{net}} = 3 \, \text{W/m}^2$ and $I_0 = \frac{32 \, \text{W/m}^2}{2} = 16 \, \text{W/m}^2$:

$3 = 16 \cos^2 \theta \sin^2 \theta$

This simplifies to:

$\frac{3}{16} = \sin^2 \theta \cos^2 \theta = \left(\frac{1}{2} \sin 2\theta\right)^2$

Taking the square root of both sides gives:

$\frac{\sqrt{3}}{2} = \left|\sin 2\theta\right|$

The solutions for this are $\theta = 30^\circ$ and $\theta = 60^\circ$.

So, the angle between the pass axes of the first two polaroids is either $30^\circ$ or $60^\circ$.

The general equation for a transverse harmonic wave on a string is given by:

$y(x,t) = A \sin(kx - \omega t + \phi)$

where $A$ is the amplitude of the wave, $k$ is the wave number, $\omega$ is the angular frequency, and $\phi$ is the phase constant. The wave velocity $v$ is related to the wave number and angular frequency by the formula:

$v = \frac{\omega}{k}$

Comparing the given equation with the general equation, we can see that:

$A = 5 \, \text{cm}$

$k = 0.003 \, \text{cm}^{-1}$

$\omega = 6 \, \text{rad/s}$

Therefore, the wave velocity is:

$v = \frac{\omega}{k} = \frac{6}{0.003} = 2000 \, \text{cm/s} = \boxed{20 \, \text{m/s}}$

The activity of a radioactive substance is defined as the rate of decay or disintegration of the substance. It is given by the following formula:

$A = \lambda N$

where $A$ is the activity, $\lambda$ is the decay constant, and $N$ is the number of radioactive atoms present.

We can use this formula to find the activity of 1.0 $\mu$g (or $10^{-6}$ g) of the substance. First, we need to find the number of radioactive atoms present in 1.0 $\mu$g of the substance. We can use the following formula to do this:

$N = \frac{m}{M} N_A$

where $m$ is the mass of the substance, $M$ is its molar mass, and $N_A$ is Avogadro's number.

Substituting the given values, we get:

$N = \frac{1.0 \times 10^{-6} \, \text{g}}{60 \, \text{g/mol}} \times 6 \times 10^{23} \approx 10^{16} \text{ atoms}$

Now, we can use the formula for activity:

$A = \lambda N = (1.5 \times 10^{-5} \, \text{s}^{-1}) (10^{16}) = 1.5 \times 10^{11} \, \text{Bq}$

Therefore, the activity of 1.0 $\mu$g of the substance is 15 $\times$ 10$^{10}$ Bq.

Power rating of zener $=1.6 \mathrm{~W}$

Maximum voltage across zener $=8 \mathrm{~V}$

Using $P=V I$, for zenner diode,

$I=\frac{P}{V}=\frac{1.6}{8}=0.2 \mathrm{~A}$

Hence, for safe operation, $2 \mathrm{~V}$ (excess voltage of $10 \mathrm{~V}$ from $8 \mathrm{~V}$ ), potential drop must occur in series resistor $R_S$.

For $R_S, V=2 \mathrm{~V}$ and $I=0.2 \mathrm{~A}$

Using $V=I R$,

$R=\frac{2}{0.2}=10 \Omega$

Given three concentric spherical shells X, Y, and Z with radii a, b, and c respectively, and with surface charge densities ( $\sigma$ ), ( $-\sigma$ ), and ( $\sigma$ ) respectively, we know that the potential at the surface of a sphere due to a uniform surface charge is given by:

$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$

where ( $\epsilon_0$ ) is the permittivity of free space, ( Q ) is the total charge on the sphere, and ( r ) is the radius of the sphere.

However, in this case, the total charge on each sphere is given by its surface charge density ( $\sigma$ ) times its surface area ( $4\pi r^2$ ). Substituting this into the formula for ( Q ) gives:

$Q = \sigma 4\pi r^2$

So the potential at the surface of each sphere is given by:

$V = \frac{1}{4\pi\epsilon_0} \frac{\sigma 4\pi r^2}{r} = \frac{\sigma r}{\epsilon_0}$

We are given that the potential at X and Z are the same. Thus:

$V_X = V_Z$

Substituting the formula for the potential into this equation gives:

$\frac{\sigma a}{\epsilon_0} = \frac{\sigma c}{\epsilon_0}$

This simplifies to:

$a = c$

However, we also need to take into account the effect of the charge on shell Y on the potentials at X and Z. The potential at any point due to a charged shell is the same everywhere outside the shell, so we can add the potential due to shell Y at X to both sides of the equation. This gives:

$\frac{\sigma a}{\epsilon_0} - \frac{\sigma b}{\epsilon_0} + \frac{\sigma c}{\epsilon_0} = \frac{\sigma a}{\epsilon_0} + \frac{\sigma c}{\epsilon_0}$

This simplifies to:

$c(a - b + c) = a^2 - b^2 + c^2$

Further simplification gives:

$c(a - b) = a^2 - b^2$

So:

$c = a + b$

Given that the radii of X & Y are 2 cm and 3 cm, respectively, we have:

$c = 2\, \text{cm} + 3\, \text{cm} = 5\, \text{cm}$

Therefore, the radius of shell Z is 5 cm.

When resistors are connected in series, the equivalent resistance is the sum of individual resistances. Therefore, the maximum equivalent resistance will be when all 10 resistors are connected in series, giving a total resistance of 100 $\Omega$.

When resistors are connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. Therefore, the minimum equivalent resistance will be when all 10 resistors are connected in parallel, giving a total resistance of 1 $\Omega$.

The ratio of maximum to minimum equivalent resistance is thus $\frac{100}{1}=100$.

To move the rod with a constant velocity $v=4 \mathrm{~m} \mathrm{~s}^{-1}$

$\mathrm{F}_{\text {net }}$ on the rod should be zero.

$$\begin{aligned} F & =B i l=0.15\left(\frac{B l v}{R}\right) l \\ & =0.15\left(\frac{0.15 \times 1 \times 4}{5}\right) \times 1 \\ & =0.03 \times 0.15 \times 4 \\ & =180 \times 10^{-4} \mathrm{~N} \\ & =18 \times 10^{-3} \mathrm{~N} \end{aligned}$$

In a series LCR circuit connected to an AC source, resonance occurs at a particular frequency at which the inductive reactance is equal to the capacitive reactance, resulting in the minimum impedance of the circuit. At this frequency, the circuit draws maximum current from the source, and thus, the maximum power is dissipated in the circuit. Therefore, Statement I is true.

In a circuit containing only a resistor, the power dissipated is given by P = VI = I$^2$R, where V is the voltage across the resistor, I is the current flowing through the resistor, and R is the resistance of the resistor. The voltage and current are in phase in a purely resistive circuit, which means that the power is maximized. Therefore, Statement II is also true.

From work energy theorem

$$\begin{aligned} &W_{\text {gravity }}+W_{\text {friction }} =\Delta(K E)=K E_f-K E_i \\ &W_{\text {gravity }} =m g h=1 \times 10 \times 0.3=3 \mathrm{~J} \\ &W_{\text {friction }} =0-\frac{1}{2} \times(22)^2-3 \\ & =-(242+3)=-245 \mathrm{~J} \end{aligned}$$

The negative sign indicates that the work done by frictional force (the tube) is in the direction opposite to the displacement of the block. In conclusion, the work done by the tube on the block is 245 J.

The gravitational field strength, or equivalently, the weight of an object, decreases linearly from its surface value to zero at the center of a sphere of uniform mass density. This is because only the mass inside the radius at which the object is located contributes to the gravitational force at that location.

If ( $d = \frac{R}{2}$ ) is the depth below the surface of the Earth, then the radius of the sphere contributing to the gravitational force at that depth is ( $R - d = R - \frac{R}{2} = \frac{R}{2}$ ).

Since the gravitational force (or weight) decreases linearly with the radius in a sphere of uniform density, the weight of the object at depth ( $d = \frac{R}{2}$) is half its weight at the surface of the Earth.

So, if the weight of the body on the surface of the Earth is 200 N, its weight at depth ( $d = \frac{R}{2}$) is half of that, or 100 N.

Therefore, the correct answer is 100 N.

Statement I: Potassium dichromate (K$_2$Cr$_2$O$_7$) is indeed commonly used as a primary standard in volumetric (titrimetric) analysis. It is a powerful oxidizing agent and has a high equivalent weight. Furthermore, it is readily available, it can be obtained in a high state of purity, it is not hygroscopic, and its solutions are stable. However, the reason it is preferred over sodium dichromate (Na$_2$Cr$_2$O$_7$) is not due to its solubility, but rather its stability and molar mass.

Statement II: Typically, the solubility of salts increases with the size of the cation. Sodium ions (Na$^+$) are smaller than potassium ions (K$^+$), so sodium salts tend to be more soluble in water than corresponding potassium salts. Thus, we would expect sodium dichromate (Na$_2$Cr$_2$O$_7$) to be more soluble in water than potassium dichromate (K$_2$Cr$_2$O$_7$).

Therefore, the correct answer is Option A: Statement I is true but Statement II is false.

Prolonged heating is avoided during the preparation of ferrous ammonium sulfate to prevent oxidation. Ferrous ammonium sulfate is a green crystalline solid that is used as a reducing agent in various chemical reactions. When heated, it can undergo oxidation to form ferric ammonium sulfate, which is a brown crystalline solid.

Here is a chemical equation for the oxidation of ferrous ammonium sulfate:

\begin{equation} \ce{FeSO4(NH4)2SO4.6H2O + O2 -> Fe2(SO4)3(NH4)2SO4.12H2O} \end{equation}

As you can see, the oxidation of ferrous ammonium sulfate results in the formation of ferric ammonium sulfate, which is a brown crystalline solid. This is why prolonged heating is avoided during the preparation of ferrous ammonium sulfate.

Here are the other options and why they are incorrect:

• Option A: Hydrolysis is the process of a chemical compound reacting with water. Ferrous ammonium sulfate is not hydrolyzed by water, so prolonged heating is not necessary to prevent hydrolysis.


• Option B: Breaking is the process of a chemical compound being physically broken apart. Ferrous ammonium sulfate is not broken apart by heat, so prolonged heating is not necessary to prevent breaking.


• Option D: Reduction is the process of a chemical compound gaining electrons. Ferrous ammonium sulfate is a reducing agent, so it is not reduced by heat.

The standard enthalpy change for the reaction

$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}$

can be calculated using Hess's Law, which states that the total enthalpy change for a reaction is the same whether it occurs in one step or in many steps.

If we modify the given reactions to reflect the formation of CO, we can add the modified reactions together to get the reaction of interest.

Consider the reactions:

(A) $\mathrm{2CO(g)+O_2(g)\to 2CO_2(g)} \quad \Delta H_1^0=-x~kJ~mol^{-1}$

(B) $\mathrm{C(graphite)+O_2(g)\to CO_2(g)} \quad \Delta H_2^0=-y~kJ~mol^{-1}$

First, divide reaction (A) by 2 to get the reaction for 1 mol of CO:

(A/2) $\mathrm{CO(g)+\frac{1}{2}O_2(g)\to CO_2(g)} \quad \Delta H_1^0=-\frac{x}{2}~kJ~mol^{-1}$

Then, subtract the divided reaction (A) from reaction (B) to get the reaction for the formation of CO:

$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}$

$\Delta H^0=\Delta H_2^0-\Delta H_1^0=-y-(-\frac{x}{2})= \frac{x}{2} - y~kJ~mol^{-1}$

Therefore, the correct answer is Option A: $\frac{x-2y}{2}$

In the froth floatation process pine oils, fatty acids, and xanthates are collectors, and froth stabilizers cresols, and aniline are used.

In column chromatography, the compound that elutes first is the one that is less strongly adsorbed to the stationary phase. Hence, 'A' has weaker adsorption than 'B'. Therefore, 'B' has stronger adsorption.

The retention factor (R_f) is a measure of how much a compound moves up the column; a higher Rf means the compound travels further and is less strongly adsorbed to the stationary phase. Since 'A' elutes first, 'A' has a higher R_f than 'B'. Hence, 'B' has a lower R_f.

So, the correct answer is 'Option D: low R_f, stronger adsorption'.

In the complex [Co(NH$_3$)$_6$]$^{3+}$, Co is in the +3 oxidation state. This means it has a configuration of $d^6$. NH$_3$ is a strong field ligand, which causes the d-electrons to pair up. Therefore, this complex is a low-spin (also referred to as spin-paired) complex.

The term "low-spin" means that the electrons prefer to pair up in the lower energy d-orbitals rather than occupy the higher energy orbitals. When all the electrons are paired, the complex is diamagnetic, which means it's not attracted to an external magnetic field. The hybridization of this complex is $d^2sp^3$, also known as inner orbital complex, because the inner d-orbitals are used in the hybridization.

The complexes [CoF$_6$]$^{3-}$ and [CoCl$_6$]$^{3-}$ are both octahedral, but they do have unpaired electrons because $F^-$ and $Cl^-$ are not strong enough field ligands to cause pairing of all the electrons in the d-orbitals. Therefore, they are not low-spin or diamagnetic.

Lastly, the complex [NiCl$_4$]$^{2-}$ is not octahedral. Because $Cl^-$ is a weak field ligand, the electrons in the d-orbitals do not pair up, and the $Ni^{2+}$ ion, with a $d^8$ configuration, undergoes sp^3 hybridization, resulting in a tetrahedral complex.

Therefore, the only octahedral, diamagnetic, and low-spin complex in the options provided is [Co(NH$_3$)$_6$]$^{3+}$ (Option A).

The reduction electrode potential of $\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}$ is $+1.57 \mathrm{~V}$ while that of $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ is $+0.77 \mathrm{~V}$, hence $\mathrm{A}$ is wrong. Higher oxidation state of smaller $d$-block elements is stabilized (or say form compounds) with smaller anion oxide that can be explained by stearic reason hence $B$ is correct.

The oxidation electrode potential of $\mathrm{Cr}^{2+} / \mathrm{Cr}^{3+}$ is $+0.41 \mathrm{~V}$ hence it can reduce $\mathrm{H}^{+}$and so liberate $\mathrm{H}_2$. The unpaired electrons in $\mathrm{V}^{2+}$ are 3 hence the magnetic moment of $\mathrm{V}^{2+}$ will be lesser than 4.4 BM.

Hence, only $B$ and $C$ are correct.

Lime (Calcium oxide, CaO) reacts exothermally with water to form slaked lime (Calcium hydroxide, Ca(OH)$_2$):

CaO (s) + H$_2$O (l) $\rightarrow$ Ca(OH)$_2$ (s)

This slaked lime has low solubility in water and is used for the test of carbon dioxide (CO$_2$). In this test, the carbon dioxide reacts with the calcium hydroxide to form calcium carbonate (CaCO$_3$), an insoluble white precipitate:

Ca(OH)$_2$ (aq) + CO$_2$ (g) $\rightarrow$ CaCO$_3$ (s) + H$_2$O (l)

Further reaction of calcium carbonate with carbon dioxide forms calcium bicarbonate [Ca(HCO$_3$)$_2$], a soluble compound:

CaCO$_3$ (s) + CO$_2$ (g) + H$_2$O (l) $\rightarrow$ Ca(HCO$_3$)$_2$ (aq)

Thus, 'A' in this case is slaked lime. So, the correct option is:

Option D : Slaked lime

• Nylon-2-Nylon-6 is a biodegradable polymer. It is made from two monomers, hexamethylenediamine and adipic acid. The polymer is strong and durable, but it can also be broken down by bacteria and other microorganisms. This makes it a good choice for use in products that are designed to be biodegradable, such as food packaging and medical implants.


• Buna N is a synthetic rubber. It is made from a mixture of 1,3-butadiene and acrylonitrile. The polymer is strong and flexible, and it is resistant to wear and tear. Buna N is used in a variety of applications, including tires, hoses, and belts.


• Urea-formaldehyde is a thermosetting resin. It is made from a mixture of urea and formaldehyde. The polymer is hard and strong, and it is resistant to heat, water, and chemicals. Urea-formaldehyde is used in a variety of applications, including adhesives, coatings, and foams.


• Dacron is a polyester. It is made from a mixture of terephthalic acid and ethylene glycol. The polymer is strong, durable, and lightweight. Dacron is used in a variety of applications, including clothing, fabrics, and bottles.

Sodium superoxide (NaO$_2$) is not a stable compound. It is not found in normal conditions due to its high reactivity.

A - Steel plants : They generate waste in the form of slag (III), which is a by-product of the smelting process used in steel production.

B - Thermal power plants : These plants generate fly ash (II) as a by-product of coal combustion.

C - Fertilizer industries : The manufacture of phosphate fertilizers can produce gypsum (I) as a waste product.

D - Paper mills : They generate bio-degradable wastes (IV) as a result of the pulp and papermaking processes.

Therefore, the correct option is A : (A)-(III), (B)-(II), (C)-(I), (D)-(IV).

$\mathrm{KMnO}_4$ oxidises benzylic carbon containing atleast one $\alpha$-hydrogen atom to $-\mathrm{COOH}$.