JEE Main 2024 5th April Evening Shift Question Paper with Solutions

To begin with, we're told that during an adiabatic process, the pressure $P$ of a gas is directly proportional to the cube of its absolute temperature $T$, that is, $P \propto T^3$. From this, we can express the relation as $P = kT^3$, where $k$ is a constant.

In an adiabatic process, $PV^\gamma = \text{constant}$, where $\gamma = \frac{C_P}{C_V}$, $P$ is the pressure, $V$ is the volume, $C_P$ is the heat capacity at constant pressure, and $C_V$ is the heat capacity at constant volume.

Also, the ideal gas equation is $PV = nRT$, where $n$ is the amount of substance, $R$ is the ideal gas constant, and $T$ is the absolute temperature. This can be rearranged to express $P$ in terms of $T$ and $V$, resulting in $P = \frac{nRT}{V}$.

Given that $P = kT^3$, we can equate this to the expression we got from the ideal gas law: $\frac{nRT}{V} = kT^3$. Simplifying, we get $V = \frac{nR}{kT^2}$, which shows that $V$ is inversely proportional to $T^2$.

Now, let's recall the adiabatic condition $PV^\gamma = \text{constant}$. Substituting the proportionalities $P \propto T^3$ and $V \propto T^{-2}$ into this equation, it becomes $T^3(T^{-2})^\gamma = \text{constant}$, simplifying to $T^{3-2\gamma} = \text{constant}$.

For the expression $T^{3-2\gamma} = \text{constant}$ to hold true in an adiabatic process, where the only variable is temperature $T$, the exponent must equal to zero (as the quantity of $T$ to some power equals to a constant suggests that the change in $T$ does not alter the value of the expression). This implies $3 - 2\gamma = 0$, solving for $\gamma$:

$3 - 2\gamma = 0 \implies 2\gamma = 3 \implies \gamma = \frac{3}{2}$

Therefore, the ratio of $\frac{C_P}{C_V}$ for the gas is $\frac{3}{2}$, which matches with Option B $\frac{3}{2}$.

When a body moves under the influence of a constant power, the relationship between displacement and time can be established through the concept of power. Power (P) is defined as the rate at which work is done, and it can also be expressed in terms of force (F) and velocity (v) as $P = F \cdot v$.

For a constant power P and assuming the force acts in the direction of the velocity, we can analyze how displacement (s) changes with time (t). Since force can also be written as $F = \frac{d(mv)}{dt}$ for a constant mass m, this simplifies to $F = m \frac{dv}{dt}$, because mass doesn't change with time for most cases. Integrating force over a distance gives work (W), and power is the rate of doing work, thus we can connect these concepts.

The kinetic energy (K.E) of the body is given by $K.E = \frac{1}{2}mv^2$, and the work done by the force is equal to the change in kinetic energy. Considering power is constant, $P = \frac{dW}{dt} = \frac{d(\frac{1}{2}mv^2)}{dt}$. Rearranging terms to focus on velocity and integrating with respect to time will give us a relation involving velocity and time.

For a constant mass system, and using $P = F \cdot v = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v$, and knowing that $P = \text{constant}$, we rearrange to find the relationship between velocity and time.

Given $P = m \cdot v \cdot \frac{dv}{dt}$, we rearrange to $\frac{P}{m} dt = v dv$. Integrating both sides where the initial condition is when $t = 0, v = 0$, we get $\frac{P}{m} t = \frac{1}{2} v^2$, solving for $v$ gives $v \propto t^{1/2}$, so $v = k \cdot t^{1/2}$ for some constant $k$.

The displacement $s$ is obtained by integrating the velocity with respect to time, $s = \int v dt = \int k \cdot t^{1/2} dt = \frac{2}{3}k \cdot t^{3/2}$. Therefore, the displacement $s$ is proportional to $t^{3/2}$.

The correct answer is Option B, $t^{3/2}$.

First, let's calculate the centripetal acceleration experienced by the monkey while the man does cycling smoothly on a circular track. The formula for centripetal acceleration ($a_c$) is given by:

$a_c = \frac{v^2}{r}$

where

• $v$ is the velocity of the monkey and the man cycling,


• $r$ is the radius of the circular track.

To find $v$, we first need to find the circumference of the circle, which is given by:

$C = 2\pi r$

Given the radius $r = 9 \, \mathrm{m}$, the circumference $C$ is:

$C = 2\pi \times 9 = 18\pi \, \mathrm{m}$

Then, we determine the total distance travelled by calculating how many times they complete the circle in the given time. With 120 revolutions in 3 minutes (180 seconds), the total distance $D$ travelled is:

$D = 120 \times C = 120 \times 18\pi = 2160\pi \, \mathrm{m}$

To find the speed $v$, which is distance over time, we divide the total distance by the total time in seconds:

$v = \frac{D}{t} = \frac{2160\pi}{180} = 12\pi \, \mathrm{m/s}$

Now, using the formula for centripetal acceleration $a_c = \frac{v^2}{r}$, we can plug in the values:

$a_c = \frac{(12\pi)^2}{9} = \frac{144\pi^2}{9} = 16\pi^2 \mathrm{~m/s}^2$

Therefore, the magnitude of the centripetal acceleration of the monkey is $16\pi^2 \, \mathrm{m/s}^2$, which corresponds to Option B.

To find the force of kinetic friction acting on the box, we can use the formula for kinetic friction, which is given by:

$F_{\text{friction}} = \mu_{k} \cdot N$

where:

• $F_{\text{friction}}$ is the force of friction,
• $\mu_{k}$ is the coefficient of kinetic friction,
• $N$ is the normal force exerted by the surface onto the box.

Since the box is moving on a horizontal surface, the normal force $N$ would be equal to the weight of the box, which is calculated by $mg$, where $m$ is the mass of the box and $g$ is the acceleration due to gravity. For most calculations, $g$ is approximated as $9.8 \, \text{m/s}^2$.

Substituting the given values:

• $m = 50 \, \text{kg}$,
• $\mu_k = 0.3$,
• $g = 9.8 \, \text{m/s}^2$.

First, calculate the weight of the box, which is the normal force:

$N = mg = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N}$

Then, use this value to find the force of kinetic friction:

$F_{\text{friction}} = \mu_{k} \cdot N = 0.3 \times 490 \, \text{N} = 147 \, \text{N}$

Thus, the force of kinetic friction acting on the box is 147 N. The correct answer is Option B.

To find the dimensional formula of $a b^{-1}$ in the equation given by $\left( P + \frac{a}{V^2}\right)(V-b) = RT$, where $P$ is the pressure, $V$ is the volume, and $T$ is the temperature, we will first understand the dimensions of each term in the equation. The variables mentioned have their usual meanings in the context of physics and chemistry, associated with the Ideal gas laws and the Van der Waals equation.

The dimensional formula for pressure ($P$) is $[M^1 L^{-1} T^{-2}]$ assuming $P = \frac{Force}{Area}$ and Force = $Mass \times Acceleration$.

The volume ($V$) has a dimensional formula of $[L^3]$.

Temperature ($T$) typically does not factor into the dimensional analysis directly in this context as we are considering the units it would be measured in (e.g., Kelvin), which doesn't directly convert into mass, length, and time. However, $RT$ suggests energy, and since the gas constant $R$ has dimensions including time, we consider energy's dimensions, $[M^1 L^2 T^{-2}]$, but this will not directly affect the dimension we are solving for.

Given these, let's analyze the term $\frac{a}{V^2}$ to deduce $a$'s dimensions:

• $\frac{a}{V^2}$ must have the same dimensions as pressure ($P$) for the equation to be dimensionally consistent, so

$[a] [L^{-6}] = [M^1 L^{-1} T^{-2}]$

Thus, $a$ has the dimensional formula $[M^1 L^5 T^{-2}]$.

The term we're interested in is $a b^{-1}$, which requires finding the dimension of $b$ as it's being subtracted from $V$, implying it shares dimensions with $V$:

• Hence, $b$ has the dimensional formula $[L^3]$.

Putting it all together for $a b^{-1}$:

$[a b^{-1}] = [M^1 L^5 T^{-2}] [L^{-3}] = [M^1 L^2 T^{-2}]$

Thus, the correct answer is:

Option D $[\mathrm{ML}^2 \mathrm{~T}^{-2}]$.

To find the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$, we first need to understand the relationship between force and momentum. The force $\vec{F}$ acting on a particle is related to the rate of change of its linear momentum $\overrightarrow{\mathrm{p}}$ with respect to time, as described by Newton's second law of motion:

$\vec{F} = \frac{d\overrightarrow{\mathrm{p}}}{dt}$

Given the expression for the momentum $\overrightarrow{\mathrm{p}}(t) = \hat{i} \cos (kt) - \hat{j} \sin (kt)$, we can find $\vec{F}$ by differentiating $\overrightarrow{\mathrm{p}}$ with respect to $t$:

$\frac{d\overrightarrow{\mathrm{p}}}{dt} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)$

So, $\vec{F} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)$.

Now, to find the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$, we use the dot product formula:

$\vec{F} \cdot \overrightarrow{\mathrm{p}} = |\vec{F}| |\overrightarrow{\mathrm{p}}| \cos(\theta)$,

where $\theta$ is the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$. However, in this case, it's more insightful to see if $\vec{F}$ and $\overrightarrow{\mathrm{p}}$ are orthogonal (at a $\frac{\pi}{2}$ angle to each other), because the dot product of two perpendicular vectors is zero.

The dot product of $\vec{F}$ and $\overrightarrow{\mathrm{p}}$ is:

$(-k\hat{i} \sin (kt) - k\hat{j} \cos (kt)) \cdot (\hat{i} \cos (kt) - \hat{j} \sin (kt)) =$

$- k \sin (kt) \cos (kt) + k \cos (kt) \sin (kt) = 0$

The result is zero, indicating that the angle between $\vec{F}$ and $\overrightarrow{\mathrm{p}}$ is indeed $\frac{\pi}{2}$.

Therefore, the correct option is:

Option A $\frac{\pi}{2}$.

The correct option is Option D: To conduct excess charge due to air friction to ground and prevent sparking.

This method is grounded in the principles of physics, particularly relating to static electricity and grounding. As vehicles move through the air, especially at high speeds, friction between the air and the vehicle can lead to the accumulation of static electricity on the vehicle. This is a common phenomenon and is more pronounced in dry conditions where humidity is low, as moisture in the air can help dissipate static charge more effectively.

In the case of vehicles carrying inflammable fluids, the presence of static electricity poses a significant risk. This is because a static discharge (or sparking) in the presence of flammable vapors can ignite those vapors, leading to potentially catastrophic fires or explosions. The metallic chains that touch the ground serve a crucial safety function by providing a path for the static electrical charge to safely dissipate into the earth, a process known as grounding. The earth acts as a vast reservoir that can absorb large amounts of electrical charge. By grounding the vehicle in this way, the risk of static discharge into the surrounding atmosphere is significantly reduced, enhancing safety by preventing ignition of inflammable materials.

The effectiveness of this safety measure hinges on the electrical conductivity of the chain and its contact with the ground. The chain must be made of a material with good electrical conductivity (such as metal) and maintain adequate contact with the ground to ensure that the static charge can be continuously dissipated. This safety precaution is a practical application of electrostatic principles, showcasing how understanding and harnessing the laws of physics can provide effective solutions to real-world problems.

For a hollow sphere rolling on a plane surface without slipping, its total kinetic energy (K.E.) is the sum of its translational kinetic energy and rotational kinetic energy. The translational kinetic energy results from the motion of the center of mass of the sphere, and the rotational kinetic energy is due to its rotation about an axis through its center of mass (in this case, the axis of symmetry).

The translational kinetic energy (TKE) can be expressed as:

$\text{TKE} = \frac{1}{2}mv^2$

Where:

• m is the mass of the sphere,


• v is the linear velocity of the center of mass.

The rotational kinetic energy (RKE) for a rolling object can be given by:

$\text{RKE} = \frac{1}{2}I\omega^2$

For a hollow sphere, the moment of inertia (I) about its axis of symmetry is:

$I = \frac{2}{3}mr^2$

where r is the radius of the sphere. The angular velocity, $\omega$, can be related to the linear velocity, v, by the relation $v = r\omega$, for an object rolling without slipping. We substitute $\omega = \frac{v}{r}$ into the expression for RKE:

$\text{RKE} = \frac{1}{2} \cdot \frac{2}{3}mr^2 \cdot \left(\frac{v}{r}\right)^2$

This simplifies to:

$\text{RKE} = \frac{1}{3}mv^2$

The total kinetic energy (Total K.E.) of the rolling hollow sphere is the sum of its translational and rotational kinetic energies:

$\text{Total K.E.} = \text{TKE} + \text{RKE} = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2$

Now, we want to find the ratio of the rotational kinetic energy to the total kinetic energy:

$\frac{\text{RKE}}{\text{Total K.E.}} = \frac{\frac{1}{3}mv^2}{\frac{5}{6}mv^2}$

Since the mass and velocity are common in both the numerator and the denominator, they will cancel out, leaving:

$\frac{\frac{1}{3}}{\frac{5}{6}} = \frac{1}{3} \cdot \frac{6}{5} = \frac{2}{5}$

Therefore, the value of $x$, representing the ratio of the rotational kinetic energy to the total kinetic energy for a hollow sphere rolling on a plane surface about its axis of symmetry, is $2$. Thus, $x = 2$.

$$\begin{aligned} & L=m v r \propto n \\ & \therefore \quad m v r \propto \sqrt{r} \end{aligned}$$

The mean free path $\lambda$ is the average distance covered by a molecule between two successive collisions. It is given by the formula:

$\lambda = \frac{1}{\sqrt{2} \mathrm{n} \pi \mathrm{d}^2}$

Where:

• $n$ is the number density of the molecules, i.e., the number of molecules per unit volume,
• $d$ is the diameter of the molecules, and
• $\sqrt{2}$ arises from considering the relative motion of a pair of particles in a gas, taking into account that either particle can be moving towards the other, which effectively doubles the cross-sectional area through which they can collide.

This formula shows that the mean free path is inversely proportional to the number density $n$ of the molecules and the square of the diameter $d$ of the molecules. It also takes into account that collisions are more likely as the cross-sectional area of the molecules increases, or as the density of the molecules increases.

Therefore, the correct answer is:

Option A: $\frac{1}{\sqrt{2} \mathrm{n} \pi \mathrm{d}^2}$

$$\begin{aligned} y & =\frac{n \lambda D}{a} \\ & =\frac{1 \times\left(550 \times 10^{-9}\right)(1)}{\left(0.2 \times 10^{-3}\right)} \\ & =275 \times 10^{-5} \mathrm{~m} \end{aligned}$$

The least count (LC) of a vernier caliper is calculated using the formula:

Given that the least count of the instrument is $0.1 \mathrm{~mm}$ and the vernier scale has 20 divisions, we can plug these values into the formula to find the value of one main scale division (MSD).

To find the value of MSD, we solve for MSD:

So, one main scale division is equal to $2 \, \mathrm{mm}$. Therefore, the correct answer is Option B 2.

$$\begin{aligned} &V_L=I(\omega L)=31.4\\ &\begin{aligned} \Rightarrow \quad I & =\frac{31.4}{2 \times 3.14 \times 50 \times 10 \times 10^{-3}} \\ & =10 \mathrm{~A} \end{aligned} \end{aligned}$$

To match the given List I (EM-Wave) with List II (Wavelength Range), we need to understand the typical wavelength ranges associated with each type of electromagnetic (EM) wave mentioned in List I.

Here's the matching based on known wavelength ranges:

• Infra-red: Infrared radiation lies between the visible and microwave portions of the electromagnetic spectrum. The wavelength range for infrared is typically from about 700 nm (nanometers) to 1 mm (millimeter), which means it matches with option (III) - 1 mm to 700 nm.


• Ultraviolet: Ultraviolet (UV) light sits between visible light and X-rays along the electromagnetic spectrum, with wavelengths shorter than visible light but longer than X-rays. The typical wavelength range for UV light is from 400 nm down to 10 nm, so it matches with option (II) - 400 nm to 1 nm (even though the strict lower bound should be 10 nm, this is the closest option provided).


• X-rays: X-rays have a wavelength range shorter than UV rays and longer than gamma rays, typically ranging from 10 nm down to 0.01 nm (or $10^{-2}$ nm). Thus, X-rays match with option (IV) - 1 nm to $10^{-3}$ nm, as this is the closest range provided that fits the typical scope of X-ray wavelengths.


• Gamma rays: Gamma rays have the shortest wavelength on the electromagnetic spectrum, with wavelength less than 10 picometers (pm), which equates to less than $10^{-3}$ nm. This corresponds to option (I) - $<10^{-3}$ nm.

So, based on the above analysis, the correct option for the matching would be:

• (A)-(III)


• (B)-(II)


• (C)-(IV)


• (D)-(I)

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

To determine the ratio of heat dissipated per second through resistances $5 \Omega$ and $10 \Omega$ in the given parallel circuit, let's break it down step by step.

Understanding the Problem

When resistors are connected in parallel:

1. The voltage across each resistor is the same.


2. The power dissipated in each resistor can be found using the formula:

$P = \frac{V^2}{R}$

Applying the Power Formula in Parallel Resistors

Given:

• $R_1 = 5 \Omega$


• $R_2 = 10 \Omega$

Let $V$ be the voltage across the resistors.

Power Dissipated in Each Resistor

For the $5 \Omega$ resistor:

$P_{5} = \frac{V^2}{5}$

For the $10 \Omega$ resistor:

$P_{10} = \frac{V^2}{10}$

Finding the Ratio of Powers

Now, let's find the ratio of the power dissipated through the $5 \Omega$ resistor to the power dissipated through the $10 \Omega$ resistor:

$\frac{P_{5}}{P_{10}} = \frac{\frac{V^2}{5}}{\frac{V^2}{10}} = \frac{10}{5} = 2$

So, the ratio of power dissipated through the $5 \Omega$ resistor to the $10 \Omega$ resistor is $2:1$.

Let's start by understanding the process of dividing the wire and recombining its parts to form the final configuration. Initially, we have a wire with a resistance of $20 \Omega$. This wire is divided into 10 equal parts, each part then has a resistance of:

$\frac{20 \Omega}{10} = 2 \Omega$

Since each part has the same length and presumably the same material and cross-sectional area, then each part will have the same resistance of $2 \Omega$.

When two parts are connected in parallel, the equivalent resistance, $R_{\text{parallel}}$, of this configuration can be calculated using the formula for two resistors in parallel:

$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2}$

Given that $R_1 = R_2 = 2 \Omega$ (since the parts are identical), we have:

$\frac{1}{R_{\text{parallel}}} = \frac{1}{2 \Omega} + \frac{1}{2 \Omega} = \frac{2}{2 \Omega}$

This simplifies to:

$\frac{1}{R_{\text{parallel}}} = \frac{2}{2 \Omega} = \frac{1}{\Omega}$

From which it follows that:

$R_{\text{parallel}} = 1 \Omega$

Now, since the original wire was divided into 10 equal parts, and pairs of these parts are connected in parallel, this results in $\frac{10}{2} = 5$ pairs. Each of these pairs has an equivalent resistance of $1 \Omega$.

Finally, these pairs are all connected in series. The total resistance of resistors in series is simply the sum of their individual resistances. Therefore, the equivalent resistance of the final configuration, $R_{\text{series}}$, is:

$R_{\text{series}} = 5 \times R_{\text{parallel}} = 5 \times 1 \Omega = 5 \Omega$

So, the equivalent resistance of the final combination is $5 \Omega$.

$$\begin{aligned} & E=\frac{2 k p}{r^3} \\ & E_R=\frac{k p}{(2 r)^3}=\frac{1}{8}\left(\frac{E}{2}\right) \\ & =\frac{E}{16} \\ & \therefore \quad x=16 \\ & \end{aligned}$$

The magnetic field inside a solenoid can be calculated using the formula:

$B = \mu_0 n I$

where:

• $B$ is the magnetic field in teslas (T),
• $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \mathrm{~Tm/A}$),
• $n$ is the number of turns per unit length of the solenoid (turns/m),
• $I$ is the current in amperes (A).

Given:

• The magnetic field $B = 6.28 \times 10^{-3} \mathrm{~T}$,
• The current $I = 5 \mathrm{~A}$,
• The length of the solenoid $L = 0.5 \mathrm{~m}$,

First, let's calculate the number of turns per unit length $n$, which is $n = \frac{m}{L}$ where $m$ is the total number of turns and $L$ is the length of the solenoid.

Rearrange the formula for $B$ to solve for $m$:

$B = \mu_0 \frac{m}{L} I$

Therefore,

$m = \frac{B L}{\mu_0 I}$

Substituting the values we have:

$m = \frac{(6.28 \times 10^{-3} \mathrm{T}) (0.5 \mathrm{m})}{(4\pi \times 10^{-7} \mathrm{Tm/A}) (5 \mathrm{A})}$

$m = \frac{6.28 \times 10^{-3} \times 0.5}{4\pi \times 10^{-7} \times 5}$

$m = \frac{6.28 \times 0.5 \times 10^{-3}}{20\pi \times 10^{-7}}$

$m = \frac{3.14 \times 10^{-3}}{20 \pi \times 10^{-7}}$

$m = \frac{3.14 \times 10^{-3}}{20 \times 3.14 \times 10^{-7}}$

$m = \frac{1}{20 \times 10^{-4}}$

$m = \frac{1 \times 10^4}{20}$

$m = 500$

Therefore, the value of $m$ is 500 turns.

To solve this problem, we first need to understand the formula that relates the maximum height $H$ reached by a projectile to its initial velocity $v_0$ and the acceleration due to gravity $g$:

$H = \frac{v_0^2 \sin^2(\theta)}{2g}$

where:

• $H$ is the maximum height,


• $v_0$ is the initial velocity of the projectile,


• $\theta$ is the angle of projection with the horizontal, and


• $g$ is the acceleration due to gravity, which is approximately $9.8 \, \mathrm{m/s^2}$.

Given that the maximum height attained by the projectile is $64 \, \mathrm{m}$, we can write:

$64 = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 9.8}$

Now, we are asked to find the new maximum height if the initial velocity is halved. Let's denote the new initial velocity as $v'_0 = \frac{v_0}{2}$. Using the formula for maximum height again, we get:

$H' = \frac{{v'_0}^2 \sin^2(\theta)}{2g}$

Substituting $v'_0 = \frac{v_0}{2}$ into this equation:

$H' = \frac{(\frac{v_0}{2})^2 \sin^2(\theta)}{2g} = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 4g} = \frac{1}{4} \cdot \frac{v_0^2 \sin^2(\theta)}{2g}$

Since we know the original height:

$64 = \frac{v_0^2 \sin^2(\theta)}{2 \cdot 9.8}$

Substituting this value into our equation for $H'$, we find:

$H' = \frac{1}{4} \cdot 64 = 16 \, \mathrm{m}$

Therefore, if the initial velocity of the projectile is halved, the new maximum height reached by the projectile would be $16 \, \mathrm{m}$.

The induced emf ($\varepsilon$) in an inductor is given by Faraday's law of electromagnetic induction, which in its differential form for an inductor can be expressed as:

$\varepsilon = L \frac{dI}{dt}$

where:

• $\varepsilon$ is the induced emf in the inductor,
• $L$ is the inductance of the inductor,
• $\frac{dI}{dt}$ is the rate of change of current through the inductor.

Given that the current $I = (3t + 8)$, where $t$ is in seconds, we can find the rate of change of current by differentiating $I$ with respect to $t$.

$\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3$

The given magnitude of induced emf is $12 \, \text{mV} = 12 \times 10^{-3} \, \text{V}$ (since $1\,\text{mV} = 10^{-3} \, \text{V}$).

Now, plug these values into the formula to find $L$:

$12 \times 10^{-3} = L \cdot 3$

Solving for $L$ gives:

$L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H} = 4 \, \text{mH}$

Therefore, the self-inductance of the inductor is 4 mH.

To match List I with List II, we must understand the definitions or concepts described in List I and associate them with the correct terms in List II.

• A. A force that restores an elastic body of unit area to its original state: This definition describes Stress, as stress is the force applied over a unit area of a material that results in a deformation. Stress is involved in restoring the body to its original state.
• B. Two equal and opposite forces parallel to opposite faces: This is a description of Shear Modulus (also known as Modulus of Rigidity), which measures the material's response to shear stress (like forces applied parallel to one of its faces).
• C. Forces perpendicular everywhere to the surface per unit area same everywhere: This definition resembles the concept of Bulk Modulus, which is concerned with uniform pressure applied in all directions at points on the surface of a body and relates to volume change.
• D. Two equal and opposite forces perpendicular to opposite faces: This describes Young's Modulus (also known as Modulus of Elasticity), which involves tensile stress (force applied perpendicularly) leading to changes in length of a material.

Therefore, matching the lists results in the following:

• (A)-(III), because stress is the general concept applicable to the force per unit area described in restoring the body to its original state.
• (B)-(IV), because shear modulus is specifically related to parallel forces applied on opposite faces.
• (C)-(I), because bulk modulus deals with situations where force is applied uniformly over the surface of an object, affecting its volume.
• (D)-(II), because Young's modulus applies to forces perpendicular to the faces, affecting the length of a material.

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

$$\begin{aligned} & \frac{T_1}{T_2}=\left(\frac{r_1}{r_2}\right)^{3 / 2} \sqrt{\frac{m_2}{m_1}} \\ & \Rightarrow \quad \frac{24}{6}=\left(\frac{4.2 \times 10^4}{r_2}\right)^{3 / 2} \sqrt{\frac{m / 4}{m}} \\ & \Rightarrow r_2=1.05 \times 10^4 \mathrm{~km} \end{aligned}$$

The fundamental frequency of a string (in this case, a sonometer wire) when it is vibrating, is inversely proportional to its length, provided the tension in the string and the linear mass density (mass per unit length) remain constant. This relationship is given by the formula:

$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

where:

• $f$ is the frequency of the string,
• $L$ is the length of the string,
• $T$ is the tension in the string, and
• $\mu$ is the linear mass density of the string.

Given that the sonometer wire has a resonating length of 90 cm (0.9 m) with a fundamental frequency of $400 \mathrm{Hz}$, we can set up our first equation but note that we are comparing two states of the same string under the same tension and hence can eliminate the tension and density terms for comparison purposes:

$400 = \frac{1}{2 \times 0.9} \sqrt{\frac{T}{\mu}}$

For the second scenario where the fundamental frequency is $600 \mathrm{Hz}$, we are asked to find the new length $L_2$. We can set up the equation in a similar manner:

$600 = \frac{1}{2L_2} \sqrt{\frac{T}{\mu}}$

Since the tension $T$ and the linear density $\mu$ are constants, and they do not change between the two states, we can set up a proportion between the two states by dividing the second equation by the first, which yields:

$\frac{600}{400} = \frac{\frac{1}{2L_2}}{\frac{1}{2 \times 0.9}}$

Simplifying this equation gives:

$\frac{600}{400} = \frac{0.9}{L_2}$

or

$\frac{3}{2} = \frac{0.9}{L_2}$

Solving for $L_2$ gives:

$L_2 = \frac{0.9 \times 2}{3}$

Calculating the value:

$L_2 = \frac{1.8}{3} = 0.6 \hspace{1mm} \text{meters}$

Converting meters to centimeters (since 1 meter = 100 centimeters), we find:

$L_2 = 0.6 \times 100 = 60 \hspace{1mm} \text{centimeters}$

Therefore, the resonating length of the wire with a fundamental frequency of $600 \mathrm{Hz}$ under the same tension is 60 cm.

The correct expressions for the electrostatic force, $\vec{F}_1$, and the magnetic force, $\vec{F}_2$, acting on a charge $q$ moving with velocity $\vec{v}$, are given by the Lorentz force law. This law states that the total force acting on a charged particle in both an electric field and a magnetic field is the sum of an electrostatic force due to the electric field and a magnetic force due to the magnetic field.

The electrostatic force is given by:

$\vec{F}_1 = q \vec{E}$

where $\vec{E}$ is the electric field.

The magnetic force is given by:

$\vec{F}_2 = q(\vec{v} \times \vec{B})$

where $\vec{B}$ is the magnetic field, and $\times$ denotes the cross product, indicating that the magnetic force is perpendicular both to the direction of the velocity of the charge and the direction of the magnetic field.

Therefore, the correct option is:

Option C: $\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{V} \times \vec{B})$

Options A, B, and D are incorrect because they do not accurately reflect the definitions of electrostatic and magnetic forces as described by the Lorentz force law.

$\frac{1}{915}=R_H\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)\quad$ (For Lyman)

$\Rightarrow \frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\quad$ (For Balmer)

$\Rightarrow \lambda=6588$ $\mathop A\limits^o$

$$\begin{aligned} Y & =(A+B)+(B \cdot 1) \\ & =A+B+B \\ & =A+B \end{aligned}$$

The correct answer is Option A: Both Statement I and Statement II are true.

Explanation:

Statement I discusses the dispersion of white light through a prism, which is a phenomenon where white light splits into its component colors when passed through a prism. This happens because different colors (or wavelengths) of light bend (refract) by different amounts upon passing through a prism. Red light bends the least while violet bends the most, with yellow falling somewhere in between. This is why the statement, "When the white light passed through a prism, the red light bends lesser than yellow and violet," is true.

Statement II addresses the underlying cause of the dispersion of light, which is that the refractive index of a medium varies with wavelength (or color) of light. This property of the medium is known as dispersion. A refractive index determines how much light bends when entering a medium. Since the refractive index varies with the wavelength, different colors of light bend by different amounts when they pass through a dispersive medium like a glass prism. This is why violet light bends more than red light - because the refractive index for violet light is higher than that for red light in glass. Therefore, the statement, "The refractive indices are different for different wavelengths in dispersive medium," is also true.

Thus, both statements I and II accurately describe the principles behind the phenomenon of light dispersion through a prism, making Option A the correct choice.

Stopping potential is independent of intensity of light. It depends on frequency of light.

$\frac{F_1}{A_1}=\frac{F_2}{A_2}\quad$ (By Pascal's law)

$$\begin{aligned} \Rightarrow F_1 & =10\left(\frac{14^2}{1.4^2}\right) \\ & =10 \times 100 \\ & =1000 \mathrm{~N} \end{aligned}$$

To convert a galvanometer into an ammeter to measure larger currents, a low resistance known as the shunt resistance ($R_{\text{sh}}$) is connected in parallel with the galvanometer. The value of this shunt resistance can be calculated using the principles of parallel circuits and the desired maximum current the ammeter should read.

The original configuration of the galvanometer allows it to measure up to $10\,\text{V}$, and it has a resistance of $100\,\Omega$. When connected in series with a $400\,\Omega$ resistor, the total resistance in the circuit is $100\,\Omega + 400\,\Omega = 500\,\Omega$. Given this configuration measures up to $10\,\text{V}$, we can calculate the maximum current it is designed to measure using Ohm's law:

$I = \frac{V}{R} = \frac{10\,\text{V}}{500\,\Omega} = 0.02\,\text{A}$

Now, to recalibrate the device to measure up to $10\,\text{A}$, we require the calculation of the shunt resistor $R_{\text{sh}}$ that needs to be connected in parallel with the galvanometer. The total current $I$ will now be $10\,\text{A}$, and the part of this current flowing through the galvanometer ($I_g$) remains $0.02\,\text{A}$ (as before, to ensure we do not exceed the device's original maximum measuring capability), leaving the rest to flow through the shunt. Thus, $I - I_g$ flows through the shunt.

Since the voltage drop across both the shunt and the galvanometer must be the same for parallel components, we use Ohm’s Law $V = IR$ for both and set up an equation to calculate $R_{\text{sh}}$:

$I_g R_g = (I - I_g) R_{\text{sh}}$

Substituting known values ($R_g = 100\,\Omega$, $I = 10\,\text{A}$, and $I_g = 0.02\,\text{A}$):