JEE Main 2024 27th January Evening Shift Question Paper with Solutions

$$\begin{aligned} & \frac{V_1}{V_2}=\frac{N_1}{N_2} \\ & \frac{230}{V_2}=\frac{10}{1} \\ & V_2=23 \mathrm{~V} \end{aligned}$$

Power consumed $=\frac{\mathrm{V}_2^2}{\mathrm{R}}$

$=\frac{23 \times 23}{46}=11.5 \mathrm{~W}$

Let $\mathrm{I}_0$ be intensity of unpolarised light incident on first polaroid.

$\mathrm{I}_1=$ Intensity of light transmitted from $1^{\text {st }}$ polaroid $=\frac{\mathrm{I}_0}{2}$

$\theta$ be the angle between $1^{\mathrm{st}}$ and $2^{\text {nd }}$ polaroid

$\phi$ be the angle between $2^{\text {nd }}$ and $3^{\text {rd }}$ polaroid

$\theta+\phi=90^{\circ}$ (as $1^{\text {st }}$ and $3^{\text {rd }}$ polaroid are crossed)

$\phi=90^{\circ}-\theta$

$\mathrm{I}_2=$ Intensity from $2^{\text {nd }}$ polaroid

$\mathrm{I}_2=\mathrm{I}_1 \cos ^2 \theta=\frac{\mathrm{I}_0}{2} \cos ^2 \theta$

$\mathrm{I}_3=$ Intensity from $3^{\text {rd }}$ polaroid

$\mathrm{I}_3=\mathrm{I}_2 \cos ^2 \phi$

$\mathrm{I}_3=\mathrm{I}_1 \cos ^2 \theta \cos ^2 \phi$

$\mathrm{I}_3=\frac{\mathrm{I}_0}{2} \cos ^2 \theta \cos ^2 \phi$

$\phi=90-\theta$

$$\begin{aligned} & I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta \\ & I_3=\frac{I_0}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^2 \\ & I_3=\frac{I_0}{8} \sin ^2 2 \theta \end{aligned}$$

$\mathrm{I}_3$ will be maximum when $\sin 2 \theta=1$

$$\begin{aligned} & 2 \theta=90^{\circ} \\ & \theta=45^{\circ} \end{aligned}$$

To find the distance of the upper point A from the starting point, we need to first understand the motion of a freely falling body under the influence of gravity. The body falling under gravity is an example of uniformly accelerated motion with the acceleration equal to the acceleration due to gravity, which is given as $g = 10 \textrm{m/s}^2$.

We will consider point A where the body was at time $t_1$ and point B where the body was at time $t_2 = t_1 + 2 \textrm{s}$. The displacement in these 2 seconds is given to be 80 meters.

For an object under constant acceleration, the displacement $s$ can be found using the equation:

$s = ut + \frac{1}{2}at^2$

where

$u$ is the initial velocity,

$t$ is the time,

$a$ is the acceleration,

$s$ is the displacement.

Since the body falls under gravity, its initial velocity at the starting point is 0 ($u = 0$), thus the equation simplifies to:

$s = \frac{1}{2}gt^2$

Let's denote the distance of point A from the starting point as $s_A$ and the distance of point B as $s_B$. We know the body covers $80$ meters in $2$ seconds from A to B, so we can write:

$s_B - s_A = 80 \textrm{m}$

We can calculate the distance covered till point B (in time $t_2 = t_1 + 2$) as:

$s_B = \frac{1}{2}g(t_1 + 2)^2$

And the distance covered till point A (in time $t_1$) as:

$s_A = \frac{1}{2}gt_1^2$

Now, substituting $s_B$ and $s_A$ in our difference equation:

$\frac{1}{2}g(t_1 + 2)^2 - \frac{1}{2}gt_1^2 = 80$

$g \left( \frac{1}{2}(t_1 + 2)^2 - \frac{1}{2}t_1^2 \right) = 80$

$5((t_1 + 2)^2 - t_1^2) = 80$

$(t_1 + 2)^2 - t_1^2 = 16$

$t_1^2 + 4t_1 + 4 - t_1^2 = 16$

$4t_1 + 4 = 16$

$4t_1 = 16 - 4$

$4t_1 = 12$

$t_1 = 3 \textrm{s}$

Therefore, time $t_1$ at point A is $3$ seconds. To find $s_A$, we can use the above simplified motion equation:

$s_A = \frac{1}{2}gt_1^2$

$s_A = \frac{1}{2} \times 10 \times 3^2$

$s_A = 5 \times 9$

$s_A = 45 \textrm{m}$

So, the distance of the upper point A from the starting point is $\mathbf{45 \textrm{m}}$.

Alternate Method :

Longest wavelength corresponds to transition between $\mathrm{n}=3$ and $\mathrm{n}=4$

$$\begin{aligned} & \frac{1}{\lambda}=\mathrm{RZ}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\mathrm{RZ}^2\left(\frac{1}{9}-\frac{1}{16}\right) \\ & =\frac{7 \mathrm{RZ}^2}{9 \times 16} \\ & \Rightarrow \lambda=\frac{144}{7 \mathrm{R}} \text { for } \mathrm{Z}=1 \quad \therefore \alpha=144 \end{aligned}$$

$\mathrm{Y}=\mathrm{A} \cdot \overline{\mathrm{B}}+\overline{\mathrm{A}} \cdot \mathrm{B}$

This is XOR GATE

$$\begin{aligned} & \frac{\mu_0 \mathrm{i}}{2 \mathrm{R}_2}\left(\frac{\pi}{2 \pi}\right) \otimes+\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}_1}\left(\frac{\pi}{2 \pi}\right) \otimes \\ & \left(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}_2}+\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}_1}\right) \otimes \\ & \frac{4 \pi \times 10^{-7} \times 4}{4 \times 4 \pi}+\frac{4 \pi \times 10^{-7} \times 4}{4 \times 2 \pi} \\ & =3 \times 10^{-7}=\alpha \times 10^{-7} \\ & \alpha=3 \end{aligned}$$

The specific resistance (or resistivity) of a material is a fundamental property that describes how much the material resists the flow of electric current. The resistivity is typically denoted by the symbol $\rho$ (rho), and it can be calculated by using the resistance $X$ of a uniform specimen of the material, along with its physical dimensions. In the case of a wire, the resistivity formula in terms of its resistance $X$, length $L$, and cross-sectional area $A=\pi r^2$ is given by:

This formula is a reinterpretation of Ohm's law, and it states that the specific resistance is proportional to the area of the cross-section of the wire and inversely proportional to its length.

Given that the specific resistance of the wire $S_1$ is determined using the formula:

Now, let's see what happens to the specific resistance if the length of the wire is doubled. If we denote the new length as $2L$, the resistance of the wire with the new length will change because resistance is directly proportional to the length of the wire. However, resistivity (specific resistance) is an intrinsic property of the material and does not depend on its length or shape, only on its temperature. Therefore, even if we change the length of the wire, the specific resistance should remain the same.

If we calculate the new resistance $X'$ with the doubled length, it would be:

Thus, we would use the new resistance $X'$ and the new length $2L$ to calculate the specific resistance again:

Since this formula is essentially the same as our original formula for $S_1$, we can conclude that:

Therefore, the value of the specific resistance $S_1$ will remain the same even if the length of the wire gets doubled. The correct answer to the question is:

Option D: $S_1$

However, weight is conventionally measured in kilograms.

So, to convert the force of 30 N into weight, we divide by 10 (since 1 kg is roughly equal to 10 N under gravity).

This gives us the final answer: the weight experienced by the man is 3 kg.

$$\begin{aligned} & \text { Potential }=\frac{\mathrm{kQ}}{\mathrm{R}}=\frac{\mathrm{k} . \mathrm{Ze}}{\mathrm{R}} \\ & =\frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}} \\ & =8 \times 10^6 \mathrm{~V} \end{aligned}$$

The provided statements (A) Assertion and (R) Reason have to be analyzed to determine their correctness and also whether Reason (R) correctly explains Assertion (A).

Assertion (A) is addressing a scenario where a positive zero error occurs in a Vernier caliper. Positive zero error refers to the condition when the zero mark on the Vernier scale is to the right of the zero mark on the main scale when the jaws of the caliper are completely closed. In this case, even when measuring a zero length, the Vernier caliper will show a positive reading, indicating an error. This error is constant and will get added to actual measurements, causing the instrument to give readings that are more than the actual measurement. So, Assertion (A) is true.

Reason (R) explains the possible causes of zero error in Vernier calipers. Zero errors can indeed occur due to imperfect manufacturing processes, where the scales are not perfectly aligned. They can also happen due to rough handling, for instance, if the instrument is dropped, which might cause a permanent deformation leading to a continuous zero error. Hence, Reason (R) is true as well.

Finally, we need to analyze whether Reason (R) is the correct explanation of Assertion (A). Although both statements are correct, the Reason (R) does not explain why a positive zero error would lead to the reading being more than the actual reading, it merely states the possible causes of zero errors. Therefore, the correct relationship between the statements is that they are both true, but (R) does not provide the correct explanation for (A).

The correct option is :

$$\begin{aligned} & {[\mathrm{P}]=\left[\frac{\mathrm{a}}{\mathrm{V}^2}\right] \Rightarrow[\mathrm{a}]=\left[\mathrm{PV}^2\right]} \\ & \text { And }[\mathrm{V}]=[\mathrm{b}] \\ & \frac{[\mathrm{a}]}{\left[\mathrm{b}^2\right]}=\frac{\left[\mathrm{PV}^2\right]}{\left[\mathrm{V}^2\right]}=[\mathrm{P}] \end{aligned}$$

$$\begin{aligned} & v^2-u^2=2 a S \\ & \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \\ & \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \\ & -\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \ldots(1) \\ & 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \\ & -\frac{4 u^2}{9}=-2 a x \ldots(2) \end{aligned}$$

$(1)/(2)$

$$\begin{aligned} & \frac{5}{4}=\frac{4 \times 10^{-2}}{\mathrm{x}} \\ & \mathrm{x}=\frac{16}{5} \times 10^{-2} \\ & \mathrm{x}=3 \cdot 2 \times 10^{-2} \mathrm{~m} \\ & \mathrm{x}=32 \times 10^{-3} \mathrm{~m} \end{aligned}$$

The statement given as Assertion (A) is indeed true. Elasticity is a physical property of a material whereby it can restore it to its original shape or size after deformation (stretching, compressing, or twisting) provided the deformation is within the material's elastic limit. Once the external force is removed, an elastic material will return to its initial state due to its intrinsic property. So, we can affirm the validity of Assertion (A).

The statement labelled as Reason (R) is also true. When a material is deformed due to an applied external force, the molecules within the material are displaced from their equilibrium positions. The intermolecular and interatomic forces create a restoring force that attempts to bring the molecules back to their original, or equilibrium, positions. This restoring force is what allows the material to regain its shape when the external force is removed, assuming the material has not been deformed beyond its elastic limit. Hence, the restoring force indeed depends on the bonded interatomic and intermolecular forces within a solid.

Furthermore, Reason (R) is not just a true statement in isolation, but it is the correct explanation for Assertion (A). The reason that a material exhibits elasticity (Assertion A) is directly tied to the nature of the restoring force (Reason R), which is a result of the bonded interatomic and intermolecular forces within the material. Without these restoring forces, the material would not be able to return to its original shape, and thus would not exhibit elasticity.

Therefore, the correct answer to the given question is:

For an adiabatic process, the following relation holds:

$P V^{\gamma} = \text{constant}$

where P is the pressure, V is the volume, and $\gamma = \frac{C_p}{C_v}$.

We are given that the pressure is proportional to the cube of the absolute temperature:

$P \propto T^3$.

Using the ideal gas law, $PV = nRT$, we can rewrite this as:

$V \propto \frac{T}{P} \propto \frac{T}{T^3} \propto \frac{1}{T^2}$.

Substituting this into the adiabatic relation, we get:

$P \left( \frac{1}{T^2} \right)^{\gamma} = \text{constant}$.

Simplifying, we have:

$P^{1-\gamma} T^{2\gamma} = \text{constant}$.

Since P is proportional to $T^3$, we can write:

$(T^3)^{1-\gamma} T^{2\gamma} = \text{constant}$.

This simplifies to:

$T^{3-3\gamma + 2\gamma} = \text{constant}$.

For this equation to hold, the exponent of T must be zero. Therefore:

$3 - 3\gamma + 2\gamma = 0$.

Solving for $\gamma$, we get:

$\gamma = \frac{C_p}{C_v} = \boxed{\frac{3}{2}}$.

Therefore, the correct answer is Option B.

$$\begin{aligned} & \text { Radiation pressure }=\frac{I}{\mathrm{~V}} \\ & =\frac{\mathrm{I} \cdot \mu}{\mathrm{c}} \\ & =\frac{6 \times 10^8 \times 3}{3 \times 10^8} \\ & =6 \mathrm{~N} / \mathrm{m}^2 \end{aligned}$$

The angular speed $\omega$ of an object in circular motion is given by the equation

$\omega = \frac{2\pi}{T}$

where $T$ is the period of the motion - the time it takes to make one complete revolution.

Assertion (A) states that the angular speed of the Moon in its orbit around the Earth is more than the angular speed of the Earth in its orbit around the Sun. We can compare these angular speeds by comparing their periods.

The Moon takes approximately 27.3 days to orbit the Earth (this is its sidereal period, not its synodic period, which accounts for the Earth's motion around the Sun as well). The Earth takes approximately 365.25 days to orbit the Sun. Since the period of the Moon’s orbit is much less than the period of the Earth’s orbit, the angular speed of the Moon is much larger than that of the Earth. This makes Assertion (A) correct.

Reason (R) states that the Moon takes less time to move around the Earth than the time taken by the Earth to move around the Sun. This is also correct, as the previously mentioned periods demonstrate: 27.3 days for the Moon to orbit Earth vs. 365.25 days for Earth to orbit the Sun.

Furthermore, the reason (R) directly explains why assertion (A) is true. Since angular speed is inversely proportional to the period of orbit, the fact that the Moon takes less time to complete an orbit implies it has a higher angular speed relative to the Earth’s angular speed in its orbit around the Sun.

Therefore, the correct answer is:

Option C Both (A) and (R) are correct and (R) is the correct explanation of (A)

The threshold frequency, $\nu_0$, corresponds to the minimum frequency of light required to eject electrons from the surface of a metal, a phenomenon known as the photoelectric effect. The work function, represented by $\phi$, is the minimum energy needed to remove an electron from the surface of the metal.

The energy of a photon is given by the equation $E = h\nu$, where:

• $E$ is the energy of the photon,
• $h$ is Planck's constant ($h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$), and
• $\nu$ is the frequency of the photon.

To find the threshold frequency for a metal with a work function of $6.63 \, \text{eV}$, we must first express the work function in joules (since Planck's constant is in joules per second). To convert electron volts to joules, use the conversion factor $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$:

The energy of the photon at the threshold frequency is equal to the work function:

Solve for $\nu_0$:

Thus, the correct answer is:

The total kinetic energy of a mole of an ideal gas can be determined by using the equipartition theorem, which states that the energy is equally distributed among degrees of freedom. For a diatomic molecule such as oxygen ($O_2$), there are 5 degrees of freedom (3 translational and 2 rotational - assuming the vibrational modes are not excited at room temperature), so each degree of freedom has an average energy of $\frac{1}{2} kT$ per molecule, where $k$ is the Boltzmann constant and $T$ is the temperature in kelvins.

However, since we are dealing with moles, we'll use the universal gas constant $R$ instead of the Boltzmann constant $k$, because $R = k \cdot N_A$ where $N_A$ is the Avogadro constant (the number of molecules in a mole). Therefore, the average energy per mole for each degree of freedom is $\frac{1}{2}RT$.

To find the total energy, we multiply the energy per degree of freedom by the number of degrees of freedom for the diatomic gas:

For diatomic oxygen:

Given that temperature $T$ is $27^{\circ} \mathrm{C}$, we first convert it to kelvins:

Now we plug in the values for $R$ and $T_{\text{K}}$:

When we calculate this, we find:

So the total kinetic energy of 1 mole of oxygen at $27^{\circ} \mathrm{C}$ is approximately 6232.5 J.

$\mathrm{i} \propto \theta$ (angle of deflection)

$$\begin{aligned} & \therefore \frac{\mathrm{i}_2}{\mathrm{i}_1}=\frac{\theta_2}{\theta_1} \Rightarrow \frac{\mathrm{i}_2}{200 \mu \mathrm{A}}=\frac{\pi / 10}{\pi / 3}=\frac{3}{10} \\ & \Rightarrow \mathrm{i}_2=60 \mu \mathrm{A} \end{aligned}$$

$$\begin{aligned} & \mathrm{f}_{\mathrm{c}}=\frac{\mathrm{v}}{4 \ell_1} \quad \mathrm{f}_{\mathrm{o}}=\frac{\mathrm{v}}{2 \ell_2} \\ & \left|\mathrm{f}_{\mathrm{c}}-\mathrm{f}_0\right|=7 \\ & \frac{\mathrm{v}}{4 \times 150}-\frac{\mathrm{v}}{2 \times 350}=7 \\ & \frac{\mathrm{v}}{600 \mathrm{~cm}}-\frac{\mathrm{v}}{700 \mathrm{~cm}}=7 \\ & \frac{\mathrm{v}}{6 \mathrm{~m}}-\frac{\mathrm{v}}{7 \mathrm{~m}}=7 \\ & \mathrm{v}\left(\frac{1}{42}\right)=7 \\ & \mathrm{v}=42 \times 7 \\ & =294 \mathrm{~m} / \mathrm{s} \end{aligned}$$

The most appropriate answer from the options given would be Option A: Both (A) and (R) are correct, and (R) is the correct explanation of (A).

Here is the reasoning for this answer:

Assertion (A) states that the work done by an electric field on moving a positive charge on an equipotential surface is always zero. This statement is true because by definition, an equipotential surface is a surface over which the electric potential is constant. When a charge moves along an equipotential surface, there is no change in its electric potential energy since potential difference $\Delta V$ is zero. Work done ($W$) is defined as the product of charge ($q$), potential difference ($\Delta V$), and the cosine of the angle between the field and direction of motion ($\cos \theta$), which can be written as:

$W = q \Delta V \cos \theta$

Because $\Delta V = 0$ on an equipotential surface, irrespective of the value of $\cos \theta$, the work $W$ will be zero. Hence, the Assertion (A) is correct.

Reason (R) says that electric lines of forces are always perpendicular to equipotential surfaces. This statement is also correct as the electric field lines, by definition, are directed such that they are tangent to the electric field vector at any point in space. Since the electric potential is constant on an equipotential surface, there can be no component of the electric field parallel to the surface, as that would imply a force and potential change along the surface. The electric field thus must be perpendicular to the equipotential surface, which means that the electric field lines must also be perpendicular to the equipotential surface. In other words, the electric field does no work when a charge moves along an equipotential surface because the motion is perpendicular to the force.

Therefore, Reason (R) is not only correct, but it is also the correct explanation for Assertion (A), making Option A the right choice.

For first minima

$$\begin{aligned} & \operatorname{asin} \theta=\lambda \\ & \Rightarrow \sin \theta=\frac{\lambda}{a}=\frac{5000 \times 10^{-10}}{1 \times 10^{-6}}=\frac{1}{2} \\ & \Rightarrow \theta=30^{\circ} \end{aligned}$$

$$\begin{aligned} & { }_6 \mathrm{C}^{13}+\text { Energy } \rightarrow{ }_6 \mathrm{C}^{12}+{ }_0 \mathrm{n}^1 \\ & \Delta \mathrm{m}=(12.000000+1.008665)-13.003354 \\ & =-0.00531 \mathrm{u} \\ & \therefore \text { Energy required }=0.00531 \times 931.5 \mathrm{~MeV} \\ & =4.95 \mathrm{~MeV} \end{aligned}$$

$$\begin{aligned} & \vec{\tau}=\vec{p} \times \vec{E} \\ & \vec{p}=q \vec{\ell} \\ & \overrightarrow{\mathrm{E}}=0.2 \frac{\mathrm{V}}{\mathrm{cm}}=20 \frac{\mathrm{V}}{\mathrm{m}} \\ & \overrightarrow{\mathrm{p}}=4 \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & =(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \mu \mathrm{C}-\mathrm{m} \\ & \vec{\tau}=(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times(20 \hat{\mathrm{i}}) \times 10^{-6} \mathrm{Nm} \\ & =(8 \hat{\mathrm{k}}+8 \hat{\mathrm{j}}) \times 10^{-5}=8 \sqrt{2} \times 10^{-5} \\ & \alpha=2 \end{aligned}$$

Loss in kinetic energy $=$ Gain in potential energy

$$\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos \theta) \\ & \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos \theta) \end{aligned}$$

Acceleration at lowest point $=\frac{\mathrm{v}^2}{\ell}$

Acceleration at extreme point $=g \sin \theta$

$$\begin{aligned} & \text { Hence, } \frac{\mathrm{v}^2}{\ell}=\mathrm{g} \sin \theta \\ & \therefore \sin \theta=2(1-\cos \theta) \\ & \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}$$

$$\begin{aligned} & \mathrm{X}_{\mathrm{c}}=\frac{1}{\omega \mathrm{C}}=\frac{\pi}{2 \pi \times 50 \times 10^{-3}}=10 \Omega \\ & \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3} \\ & =10 \Omega \\ & \because \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}, \text { Hence, circuit is in resonance } \\ & \therefore \text { power factor }=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{R}}=1 \end{aligned}$$

$$\begin{aligned} & \text { Change in pressure }=\frac{1}{2} \rho \mathrm{v}^2 \\ & 4.5 \times 10^4-2.0 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \\ & 2.5 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \\ & \mathrm{v}^2=50 \\ & \mathrm{v}=\sqrt{50} \\ & \text { Velocity of water }=\sqrt{\mathrm{V}}=\sqrt{50} \\ & =\mathrm{V}=50 \end{aligned}$$

From KVL,

$\mathbf{V}_{\mathbf{1}}+\mathbf{V}_{\mathbf{2}}-\mathbf{V}_{\mathbf{3}}=\mathbf{0} \Rightarrow \mathrm{V}_1+\mathrm{V}_2=\mathrm{V}_3$

Let's analyze both statements:

Statement (I): The limiting force of static friction depends on the area of contact and independent of materials.

This statement is incorrect. The limiting force of static friction does not depend on the area of contact but is dependent on the materials in contact. According to the law of static friction, the maximum static frictional force $f_s$ that can occur before motion commences is given by the product of the coefficient of static friction $\mu_s$ and the normal reaction force $N$:

$f_s = \mu_s \times N$

The coefficient $\mu_s$ is a property that depends on the materials in contact, not on the area of contact. The normal force $N$ is the force perpendicular to the surfaces in contact, influenced by the weight of the object and any other perpendicular forces acting upon it.

Statement (II): The limiting force of kinetic friction is independent of the area of contact and depends on materials.

This statement is correct. Once an object is in motion, the kinetic frictional force $f_k$ opposing its motion is given by the product of the coefficient of kinetic friction $\mu_k$ and the normal force $N$:

$f_k = \mu_k \times N$

The coefficient $\mu_k$, like $\mu_s$, is also a property dependent on the materials of the surfaces in contact. It generally has a lower value than $\mu_s$, which is why objects tend to be easier to keep moving once they've started. The kinetic frictional force is independent of the area of contact between the two surfaces.

Given these explanations, the correct answer would be:

Option A: Statement I is incorrect but Statement II is correct.

(A) $\rightarrow$ Kolbe Schmidt Reaction

(B) $\rightarrow$ Reimer Tiemann Reaction

(C) $\rightarrow$ Oxidation of phenol to p-benzoquinone

(D) $\rightarrow \mathrm{PhOH}+\mathrm{NaOH} \rightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{PhO}^{-}$