JEE Main 2023 8th April Morning Shift Question Paper with Solutions

The phenomenon of frequency change due to relative motion between a source and an observer is called Doppler effect. However, if the observer is at rest relative to the source (as in this case where the passenger is inside the train), then the frequency heard by the observer is the same as the frequency produced by the source.

This is because the relative velocity between the source (whistle) and the observer (passenger in the train) is zero. The Doppler effect only applies when there is a relative velocity between the source and the observer.

Therefore, the frequency heard by the passenger inside the train is the same as the frequency of the whistle, i.e., 400 Hz.

So, the correct answer is:

400 Hz.

The volume of the air bubble changes due to the change in pressure as it rises from the bottom of the lake to the surface. We can use Boyle's Law to calculate the change in volume, which states that the product of pressure and volume is constant for a given mass of confined gas held at a constant temperature:

$P_1V_1 = P_2V_2$

where $P_1$ and $V_1$ are the pressure and volume at the bottom of the lake and $P_2$ and $V_2$ are the pressure and volume at the surface of the lake.

At the bottom of the lake, the pressure is the atmospheric pressure plus the pressure due to the water column above the bubble:

$P_1 = P_{\text{atm}} + \rho gh$

where $\rho$ is the density of water, $g$ is the acceleration due to gravity, and $h$ is the height of the water column. Substituting the given values, we get:

$P_1 = 1 \times 10^{5} \text{ Pa} + 1000 \text{ kg/m}^3 \times 10 \text{ m/s}^2 \times 40 \text{ m} = 5 \times 10^{5} \text{ Pa}$

At the surface of the lake, the pressure is the atmospheric pressure:

$P_2 = P_{\text{atm}} = 1 \times 10^{5} \text{ Pa}$

The initial volume of the bubble is:

$V_1 = 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3$

Substituting these values into Boyle's Law and solving for $V_2$, we get:

$V_2 = \frac{P_1V_1}{P_2} = \frac{5 \times 10^{5} \text{ Pa} \times 1 \times 10^{-6} \text{ m}^3}{1 \times 10^{5} \text{ Pa}} = 5 \times 10^{-6} \text{ m}^3 = 5 \text{ cm}^3$

So, the volume of the air bubble when it reaches the surface is 5 cm³.

Reflecting telescopes, also known as reflectors, use a set of mirrors instead of lenses to gather and focus light. The primary mirror (usually a concave mirror) gathers the incoming light and reflects it to a focus point. The secondary mirror is used to redirect this focused light out to where it can be conveniently observed.

In other words, the secondary mirror in a reflecting telescope is used to move the eyepiece outside the telescopic tube, where the image can be comfortably viewed. This is especially important for large telescopes, where the focus point may be inside the telescope tube and inaccessible without a secondary mirror.

An organ pipe that is open at both ends resonates at all harmonics, including the fundamental (first harmonic), second harmonic, third harmonic, etc.

The frequency $f$ of the $n$-th harmonic for a pipe open at both ends is given by:

$f_n = \frac{n v}{2L}$,

where:

• $n$ is the number of the harmonic,
• $v$ is the speed of sound, and
• $L$ is the length of the pipe.

To find the frequency of the second harmonic ($n = 2$), we can substitute the given values into the formula:

$f_2 = \frac{2 \times 360}{2 \times 0.4} = 900 \, \text{Hz}$.

Therefore, the frequency of the second harmonic is $900 \, \text{Hz}$.

$$\begin{aligned} &\text {For the given circuit the truth table would be : }\\ &\begin{array}{|c|c|c|} \hline \mathbf{A} & \mathbf{B} & \mathbf{Y} \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{array} \end{aligned}$$

The truth table shows that the output graph is given by :

The work done $W$ in rotating an electric dipole in a uniform electric field is given by:

$W = pE(1 - \cos\theta)$,

where $p$ is the dipole moment, $E$ is the strength of the electric field, and $\theta$ is the angle the dipole is rotated through.

In this case, the dipole moment $p$ is $6.0 \times 10^{-6} ~\mathrm{C m}$, the electric field $E$ is $1.5 \times 10^{3} ~\mathrm{NC}^{-1}$, and the angle $\theta$ is $180^{\circ}$.

Substituting these values into the formula gives:

$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times (1 - \cos180^{\circ})$.

Since $\cos180^{\circ} = -1$, the equation becomes:

$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times (1 - (-1))$,

$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times 2$,

$W = 18.0 \times 10^{-3} ~\mathrm{J} = 18.0 ~\mathrm{mJ}$.

So the work done in rotating the dipole by $180^{\circ}$ in this field is 18.0 millijoules.

When a charged particle moves in a magnetic field, its motion is affected by the components of its velocity that are parallel and perpendicular to the magnetic field.

1. The component of velocity that is parallel to the magnetic field doesn't get affected by the magnetic field. It causes the particle to move along the magnetic field lines in a straight line.

2. The component of velocity that is perpendicular to the magnetic field causes the charged particle to move in a circular path around the magnetic field lines due to the magnetic Lorentz force.

Combining these two effects, the charged particle follows a helical path, where the axis of the helix is aligned with the magnetic field.

The momentum (p) and kinetic energy (K) of a body are related by the equations:

$p = mv$,

$K = \frac{1}{2}mv^2$,

where m is the mass and v is the velocity of the body.

We can express v in terms of p and m:

$v = \frac{p}{m}$,

and substitute this into the equation for K to get:

$K = \frac{p^2}{2m}$.

So, the kinetic energy is proportional to the square of the momentum.

If the momentum is increased by 50%, the new momentum is 1.5p, and the new kinetic energy is:

$K' = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m} = 2.25K$.

The percentage increase in the kinetic energy is then:

$\frac{K'-K}{K} \times 100 = \frac{2.25K - K}{K} \times 100 = 1.25 \times 100 = 125\%$.

So, the percentage increase in the kinetic energy of the body is 125%.

The maximum current in an LC circuit can be found using the following formula related to simple harmonic motion:

$I_{\text{max}} = \omega Q_{\text{max}}$,

where:

• $I_{\text{max}}$ is the maximum current,
• $\omega$ is the angular frequency, and
• $Q_{\text{max}}$ is the maximum charge on the capacitor.

The angular frequency $\omega$ for an LC circuit is given by:

$\omega = \frac{1}{\sqrt{LC}}$,

where:

• $L$ is the inductance, and
• $C$ is the capacitance.

Given that $L = 75 \, \text{mH} = 75 \times 10^{-3} \, \text{H}$, $C = 1.2 \, \mu \text{F} = 1.2 \times 10^{-6} \, \text{F}$,

and $Q_{\text{max}} = 2.7 \, \mu \text{C} = 2.7 \times 10^{-6} \, \text{C}$,

we can substitute these values into the formulas to find $I_{\text{max}}$:

$\omega = \frac{1}{\sqrt{(75 \times 10^{-3})(1.2 \times 10^{-6})}} = 3333.33 \, \text{rad/s}$,

$I_{\text{max}} = \omega Q_{\text{max}} = 3333.33 \times 2.7 \times 10^{-6} = 0.009 \, \text{A}$.

Therefore, the maximum current in the circuit is $0.009 \, \text{A}$, or equivalently, $9 \, \text{mA}$.

A satellite in orbit around a planet is subject to two main forces: gravitational force, which is trying to pull it towards the planet, and its own kinetic energy or inertia, which is trying to keep it moving in a straight line. The balance of these two forces results in the satellite moving in a circular or elliptical orbit.

The gravitational potential energy ($U$) of the satellite is given by the formula:

$U = -\frac{GMm}{R}$

where $G$ is the gravitational constant, $M$ is the mass of the Earth, $m$ is the mass of the satellite, and $R$ is the radius of the orbit. The negative sign indicates that work would have to be done to remove the satellite from the Earth's gravitational influence.

The kinetic energy ($K$) of the satellite is given by the formula:

$K = \frac{GMm}{2R}$

This is obtained from the fact that for a satellite in stable orbit, the gravitational force must be equal to the centripetal force required to keep the satellite moving in a circle. From this, we can derive an expression for the velocity of the satellite, and hence its kinetic energy.

The total mechanical energy ($E$) of the satellite, which is the sum of its kinetic and potential energy, is therefore:

$E = K + U = \frac{GMm}{2R} - \frac{GMm}{R} = -\frac{GMm}{2R}$

So the potential energy $U$ is $-2E$, and the kinetic energy $K$ is $-E$. Thus, the statement "If $E$ be the total energy of a satellite moving around the earth, then its potential energy will be $2E$" is correct, and the statement "The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy $E$" is incorrect.

The function of the non-magnetic metallic core in a galvanometer is to provide a path for the induced current generated by the moving coil in the magnetic field. This induced current opposes the motion of the coil and brings it to rest quickly due to the effect known as eddy current damping.

When the coil swings past its equilibrium position, a change in magnetic flux occurs. According to Faraday's law of electromagnetic induction, this change in magnetic flux generates an induced current, known as an eddy current. The eddy current creates its own magnetic field which opposes the original change in flux, causing the coil to quickly come to rest.

The total binding energy of a nucleus is the binding energy per nucleon multiplied by the number of nucleons (protons and neutrons), which is the mass number.

The initial total binding energy of the nucleus is $242 \times 7.6 \, \text{MeV}$.

After the break, each fragment has a total binding energy of $121 \times 8.1 \, \text{MeV}$.

Since there are two such fragments, the final total binding energy is $2 \times 121 \times 8.1 \, \text{MeV}$.

The gain in binding energy is the final total binding energy minus the initial total binding energy. Therefore, the gain in binding energy is:

$2 \times 121 \times 8.1 \, \text{MeV} - 242 \times 7.6 \, \text{MeV} = 1960.2 \, \text{MeV} - 1839.2 \, \text{MeV} = 121 \, \text{MeV}$.

Therefore, the total gain in binding energy is $121 \, \text{MeV}$.

The resultant of two perpendicular vectors is given by the Pythagorean theorem.

If we have two vectors of magnitudes $A$ and $\frac{A}{2}$, the resultant $R$ is:

$R = \sqrt{A^{2} + \left(\frac{A}{2}\right)^{2}} = \sqrt{A^{2} + \frac{A^{2}}{4}} = \sqrt{\frac{5A^{2}}{4}} = \frac{\sqrt{5} A}{2}$.

At a steady state, no current would be flowing in capacitor circuit.

Required $=6+2+8=16 \Omega$

From Ohm's law

$i=\frac{\mathrm{V}}{\mathrm{R}}=\frac{4}{16}=\frac{1}{4} \mathrm{~A}$

Voltage across AC

$\mathrm{V}_{\mathrm{AC}}=i(6+2)=\frac{1}{4} \times 8=2 \mathrm{~V}$

Voltage across BD

$\mathrm{V}_{\mathrm{BD}}=i(2+8)=\frac{1}{4} \times 10=2.5 \mathrm{~V}$

Now $\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{V}_{\mathrm{BD}}}=\frac{2}{2.5}=\frac{4}{5}$

Electric field due to the uniformly charged solid sphere is given by

$$\begin{array}{ll} \mathrm{E}=\frac{\mathrm{Q}}{4 \pi \varepsilon_0 r^2} \quad r \geq \mathrm{R} \\ \& & \\ \mathrm{E}=\frac{\mathrm{Q} r}{4 \pi \varepsilon_0 \mathrm{R}^3} \quad r \leq \mathrm{R} \end{array}$$

Therefore $\mathrm{E} \propto \mathrm{r}$ when

$\mathrm{r} \leq \mathrm{R}$ and $\mathrm{E} \propto \frac{1}{r^2}$ when $r \geq \mathrm{R}$

So

The range (or distance to the horizon) for line-of-sight communication is given by the formula:

$d = \sqrt{2hR}$,

where:

• $d$ is the range,
• $h$ is the height of the antenna, and
• $R$ is the radius of the Earth.

Given that $h = 98 \, \text{m} = 98 \times 10^{-3} \, \text{km}$ and $R = 6400 \, \text{km}$, we can substitute these values into the formula to find $d$:

$d = \sqrt{2 \times 98 \times 10^{-3} \times 6400} \approx 35.42 \, \text{km}$.

The surface area $A$ covered by the transmitting antenna is approximately a circle with radius equal to the range $d$. The area of a circle is given by the formula $A = \pi r^2$, where $r$ is the radius. Therefore, we can substitute $d$ into this formula to find $A$:

$A = \pi d^2 = \pi (35.42)^2 \approx 3940 \, \text{km}^{2}$.

Therefore, the surface area covered by the transmitting antenna is approximately $3940 \, \text{km}^{2}$.

The terminal velocity of a small spherical object moving under the action of gravity through a fluid medium is given by Stokes' Law, which is stated as:

$v = \frac{2}{9} \frac{r^2 g (\rho_p - \rho_f)}{\eta}$,

where:

• $v$ is the velocity of the object (in this case, the air bubble),
• $r$ is the radius of the object,
• $g$ is the acceleration due to gravity,
• $\rho_p$ is the density of the object (negligible in this case, as it's an air bubble),
• $\rho_f$ is the density of the fluid (the solution), and
• $\eta$ is the coefficient of viscosity of the fluid.

Since we are neglecting the density of the air bubble, the formula simplifies to:

$v = \frac{2}{9} \frac{r^2 g \rho_f}{\eta}$.

Rearranging for $\eta$, we get:

$\eta = \frac{2}{9} \frac{r^2 g \rho_f}{v}$.

Given that $r = \frac{6 \, \text{mm}}{2} = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}$, $g = 10 \, \text{ms}^{-2}$, $\rho_f = 1750 \, \text{kg/m}^{3}$, and $v = 0.35 \, \text{cm/s} = 0.35 \times 10^{-2} \, \text{m/s}$, we can substitute these values into the formula to find $\eta$:

$\eta = \frac{2}{9} \frac{(3 \times 10^{-3})^2 \times 10 \times 1750}{0.35 \times 10^{-2}} = 10 \, \text{Pas}$.

Therefore, the coefficient of viscosity of the solution is $10 \, \text{Pas}$.

In the semi-empirical mass formula, the terms have the following forms:

• The volume term (binding energy) is proportional to the volume of the nucleus, and therefore to the number of nucleons $A$.


• The surface term takes into account that the nucleons at the surface of the nucleus have fewer nearest neighbors than those in the interior, and so the binding energy is less for surface nucleons. This term is usually proportional to $A^{2/3}$.


• The Coulomb term arises because the protons in the nucleus repel each other. This term is usually proportional to $\frac{Z(Z-1)}{A^{1/3}}$.


• The symmetry term arises because of the Pauli exclusion principle, but it's not mentioned here.
• The pairing term arises because of the tendency of protons and neutrons to pair up, but it's not mentioned here.

Now, let's check the statements:

• A: The surface energy per nucleon $b_{s}=-a_{1} A^{2 / 3}$ is incorrect. The correct form should be proportional to $A^{2/3}$, but it should not be per nucleon.


• B: The Coulomb contribution to the binding energy $b_{c}=-a_{2} \frac{Z(Z-1)}{A^{4 / 3}}$ is incorrect. The correct form should be proportional to $\frac{Z(Z-1)}{A^{1/3}}$.


• C: The volume energy $b_{v}=a_{3} A$ is correct.


• D: Decrease in the binding energy is proportional to surface area is correct. This represents the surface term in the semi-empirical mass formula.


• E: While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons is incorrect. While the exact number can vary depending on the model used, it's generally agreed that each nucleon interacts with a few nearest neighbors, not necessarily 12.

So, the correct answer is C, D only.

To solve this problem, we need to understand the concept of moment of inertia. Moment of inertia is a measure of an object's resistance to rotational motion. It depends on the mass distribution of the object and the axis of rotation.

For a continuous object like a semicircular ring, we can calculate the moment of inertia by integrating over the entire object. Here's how we can approach this problem:

1. Divide the semicircular ring into small mass elements: Imagine the semicircular ring divided into infinitesimally small mass elements, each with mass $dm$.

2. Calculate the moment of inertia of each element: The moment of inertia of each element about the axis passing through the center and perpendicular to the plane of the ring is given by $dI = dmR^2$, where R is the radius of the ring.

3. Integrate to find the total moment of inertia: To find the total moment of inertia, we need to integrate $dI$ over the entire ring. This means integrating from $0$ to $\pi$ (the angle spanned by the semicircle) with respect to the angle $\theta$.

4. Relate $dm$ to the total mass: Since the ring has a uniform mass distribution, we can express the mass of each element $dm$ as a fraction of the total mass $M$: $dm = \frac{M}{πR} Rd\theta = \frac{M}{\pi} d\theta$.

Now, let's perform the integration:

$I = \int_{0}^{\pi} dI = \int_{0}^{\pi} dmR^2 = \int_{0}^{\pi} \frac{M}{\pi} d\theta R^2$

$I = \frac{MR^2}{\pi} \int_{0}^{\pi} d\theta = \frac{MR^2}{\pi} [\theta]_{0}^{\pi}$

$I = \frac{MR^2}{\pi} [\pi - 0] = MR^2$

Therefore, the moment of inertia of the semicircular ring about the given axis is $MR^2$. Comparing this to the given formula, we find that $x = \boxed{1}$.

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

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

where $g$ is the acceleration due to gravity.

Let's calculate the ranges of projectiles A and B:

For projectile A, $v = 40 \, \text{m/s}$ and $\theta = 30^\circ$, so:

$R_A = \frac{(40)^2 \sin(2 \times 30)}{10} = 4 \times 40 = 160 \, \text{m}$.

For projectile B, $v = 60 \, \text{m/s}$ and $\theta = 60^\circ$, so:

$R_B = \frac{(60)^2 \sin(2 \times 60)}{10} = 6 \times 60 = 360 \, \text{m}$.

Therefore, the ratio of their ranges is $R_A : R_B = 160 : 360 = 4 : 9$.

Given the velocity vector of a particle $v = (2t \hat{i}+3 t^{2} \hat{j}) \, \text{ms}^{-1}$, the acceleration $a$ is the derivative of the velocity vector with respect to time. So, we have:

$a = \frac{dv}{dt} = (2 \hat{i} + 6t \hat{j}) \, \text{ms}^{-2}$.

At $t=1 \, \text{s}$, the acceleration $a$ is $(2 \hat{i} + 6 \hat{j}) \, \text{ms}^{-2}$.

According to Newton's second law, the force $F$ is equal to the mass $m$ times acceleration $a$. The mass $m$ is given as $500 \, \text{g}$, or equivalently, $0.5 \, \text{kg}$.

Therefore, the force $F$ on the particle at $t=1 \, \text{s}$ is:

$F = m \cdot a = 0.5 \cdot (2 \hat{i} + 6 \hat{j}) = (1 \hat{i} + 3 \hat{j}) \, \text{N}$.

So, the force acting on the particle at $t=1 \, \text{s}$ is $(\hat{i} + x \hat{j}) \, \text{N}$, where $x=3$.

Therefore, the answer is $x=3$.

Statement I: If heat is added to a system, its temperature must increase. This statement is not necessarily true. For example, in a phase transition (like melting or boiling), heat can be added to a system without increasing its temperature. The added heat energy is used to break intermolecular bonds and change the phase of the substance, not to increase the kinetic energy of the particles (which would raise the temperature).

Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase. This statement is generally true, as positive work being done by a system often involves expansion against an external pressure, thus increasing its volume.

The drift velocity $v_d$ can be found using the formula for current $I$ in a conductor:

$I = nqAv_d$,

where:

• $n$ is the number density of free electrons (number of free electrons per unit volume),
• $q$ is the charge of an electron,
• $A$ is the cross-sectional area of the conductor, and
• $v_d$ is the drift velocity of the electrons.

We can rearrange the above formula to solve for $v_d$:

$v_d = \frac{I}{nqA}$.

Given that $I = 2 \, \text{A}$, $n = 2.0 \times 10^{28} \, \text{m}^{-3}$, $q = 1.6 \times 10^{-19} \, \text{C}$, and $A = 25.0 \, \text{mm}^{2} = 25.0 \times 10^{-6} \, \text{m}^{2}$, we can substitute these values into the formula to find $v_d$:

$v_d = \frac{2}{(2.0 \times 10^{28})(1.6 \times 10^{-19})(25.0 \times 10^{-6})}$

$v_d = 25 \times 10^{-6} \, \text{ms}^{-1}$.

Therefore, the drift velocity of the electrons is $25 \times 10^{-6} \, \text{ms}^{-1}$.

The gravitational field inside a uniform spherical body varies linearly with distance from the center. If we consider the Earth to be such a body, then the gravitational field strength (and hence weight) of an object would decrease linearly as we go deeper inside the Earth.

The weight of an object at a depth $d$ from the Earth's surface is given by:

$W_d = W_e (1 - \frac{d}{R})$,

where:

• $W_d$ is the weight at depth $d$,
• $W_e$ is the weight at the Earth's surface (i.e., the weight of the object), and
• $R$ is the radius of the Earth.

In this case, we're given that $W_e = 400 \, \text{N}$, and we're asked to find the weight at a depth of $d = \frac{R}{2}$. Substituting these values into the formula, we get:

$W_d = 400 \, \text{N} (1 - \frac{1}{2}) = 200 \, \text{N}$.

Therefore, the weight of the body when taken to a depth half of the radius of the Earth is 200 N.

The density of a cylindrical wire is given by the formula:

$\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}$

where $m$ is the mass, $r$ is the radius, and $l$ is the length.

The relative error in a calculated quantity is the sum of the relative errors in the quantities it depends on. For the density, this is given by:

$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$

Given that $\Delta m = 0.01$ g, $m = 0.4$ g, $\Delta r = 0.03$ mm, $r = 6$ mm, $\Delta l = 0.04$ cm, and $l = 8$ cm, we can substitute these values into the formula to find the relative error in the density:

$\frac{\Delta \rho}{\rho} = \frac{0.01}{0.4} + 2\frac{0.03}{6} + \frac{0.04}{8} = 0.025 + 0.01 + 0.005 = 0.04$

So the relative error in the density is 0.04, or 4%.

The term $\frac{1}{\mu_{0} \epsilon_{0}}$ appears in the formula for the speed of light $c$, which is:

$c = \sqrt{\frac{1}{\mu_{0} \epsilon_{0}}}$

where $\mu_{0}$ is the permeability of free space and $\epsilon_{0}$ is the permittivity of free space.

The speed of light $c$ has dimensions of length over time ($\mathrm{L} / \mathrm{T}$). Therefore, the term $\frac{1}{\mu_{0} \epsilon_{0}}$ has dimensions equal to the square of the speed of light, which is $\mathrm{L}^{2} / \mathrm{T}^{2}$.

In a solenoid, the magnetic field intensity ($H$) is given by the product of the number of turns per unit length ($n$) and the current ($I$) flowing through the solenoid. This can be represented mathematically as:

$H = nI$

This is actually derived from Ampere's law applied to the special case of a solenoid, where the magnetic field is uniform and directed along the axis of the solenoid.

If we rearrange this equation to solve for the current ($I$), we get:

$I = \frac{H}{n}$

In this problem, we're given that the magnetic field intensity ($H$) at the center of the solenoid is $1.6 \times 10^{3} \, \text{Am}^{-1}$ and the number of turns per unit length ($n$) is 8 per cm (which is equal to $8 \times 10^{-2}$ per meter, as there are 100 cm in a meter).

Substituting these values into the equation gives:

$I = \frac{1.6 \times 10^{3} \, \text{Am}^{-1}}{8 \times 10^{-2} \, \text{turns/m}} = 2 \, \text{A}$

So, the current flowing through the solenoid is $2 \, \text{A}$.

Object distance $=8 \mathrm{~cm}$ [For image by $B]$

$$\begin{aligned} \Rightarrow \text { Required distance } & =8+8+2 \mathrm{~cm} \\ & =18 \mathrm{~cm} \end{aligned}$$

The strain energy stored per unit volume in a material under stress can be calculated using the following formula:

$U = \frac{1}{2} \sigma \epsilon$

where $\sigma$ is the stress and $\epsilon$ is the strain.

For an elastic material, stress is proportional to strain (Hooke's law), and the constant of proportionality is the Young's modulus (Y). So we can write:

$\sigma = Y \epsilon$

Substituting this into the energy density equation we get:

$U = \frac{1}{2} Y \epsilon^2$

The strain given in the problem is 0.04%, which needs to be converted to a decimal for use in this formula. Therefore, $\epsilon = 0.04/100 = 0.0004$.

Substituting the values into the equation gives:

$U = \frac{1}{2} \times 7.0 \times 10^{10} N/m^2 \times (0.0004)^2 = 5600 ~J/m^3$

The de Broglie wavelength of a particle is given by the formula:

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

where $h$ is Planck's constant and $p$ is the momentum of the particle.

If the de Broglie wavelengths of the proton and electron are the same, then:

$\frac{h}{p_p} = \frac{h}{p_e}$

where $p_p$ and $p_e$ are the momenta of the proton and electron, respectively.

Solving this equation for the ratio of their momenta gives:

$\frac{p_p}{p_e} = 1$

So, the ratio of their momenta is 1:1

Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as we move from left to right across a period due to increase in effective nuclear charge.

Here are the Pauling electronegativities for the elements:

• Carbon (C): 2.55


• Phosphorus (P): 2.19


• Astatine (At): 2.2


• Bromine (Br): 2.96

Based on these values, the correct order of electronegativity from highest to lowest is Br > C > At > P.

Therefore, Option B is the correct answer:

Br > C > At > P

(A) Coupling Reaction:


(B) Balz-Schiemann reaction:


(C) Gattermann Reaction $\rightarrow$ In this reaction, chlorine or bromine can be introduced in the benzene ring by treating the benzene diazonium salt solution with corresponding halogen acid in the presence of copper powder.


(D) In the given reaction when benzene diazonium salt is treated with KCN and copper powder, it forms aryl nitrile or benzyl nitrile as a product.

The reaction given is a disproportionation reaction where iodine ($\mathrm{I}_2$) undergoes both oxidation and reduction. Disproportionation reactions are a type of redox reaction where an element is simultaneously oxidized and reduced.

In the process, iodine gets oxidized from 0 oxidation state (in $\mathrm{I}_2$) to +5 oxidation state (in $\mathrm{IO}_3^{-}$), and reduced from 0 oxidation state (in $\mathrm{I}_2$) to -1 oxidation state (in $\mathrm{I}^{-}$).

The n-factor is the total change in oxidation state per molecule that undergoes the redox reaction. In this case, the n-factor for $\mathrm{IO}_3^{-}$ is 5 (as iodine goes from 0 to +5) and for $\mathrm{I}^{-}$, it's 1 (as iodine goes from 0 to -1).

Now, to balance the redox reaction, the total increase in oxidation state (total oxidation) must equal the total decrease in oxidation state (total reduction). Hence, the molar ratio of $\mathrm{IO}_3^{-}$ to $\mathrm{I}^{-}$ must be 1:5.

So, the balanced reaction would be:

$\mathrm{IO}_3^{-}+6 \mathrm{H}^{+}+5 \mathrm{I}^{-} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}$

This equation yields 3 moles of $\mathrm{I}_2$, but the original equation needs to produce 6 moles of $\mathrm{I}_2$, so the entire equation is multiplied by 2:

$2 \mathrm{IO}_3^{-}+12 \mathrm{H}^{+}+10 \mathrm{I}^{-} \rightarrow 6 \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O}$

This tells us that to get 6 moles of $\mathrm{I}_2$, you need 10 moles of $\mathrm{I}^{-}$. So, $x = 10$.

The spin-only magnetic moment of a complex ion is calculated using the formula $\mu = \sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons in the complex ion. In the case of transition metal complexes, the number of unpaired electrons and hence the magnetic moment can be influenced by the field strength of the ligands (the Ligand Field Theory).

Here, CN$^-$ is a strong-field ligand (or high-spin ligand) that can cause pairing of electrons, while F$^-$ and Br$^-$ are considered weak-field ligands (or low-spin ligands) that do not promote electron pairing.

1. $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$: In this complex, Fe is in +3 oxidation state and has 5 d-electrons. Due to the strong field ligand, CN$^-$, all the 5 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{0(0+2)} = 0$.




2. $\left[\mathrm{CoF}_{6}\right]^{3-}$: Here, Co is in +3 oxidation state and has 6 d-electrons. As F$^-$ is a weak field ligand, no electron pairing occurs, and hence there are 4 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{4(4+2)} = \sqrt{24}$.




3. $\left[\mathrm{MnBr}_{4}\right]^{2-}$: Here, Mn is in +2 oxidation state and has 5 d-electrons. Since Br$^-$ is a weak field ligand, no electron pairing occurs, and hence there are 5 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35}$.




4. $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$: In this complex, Mn is in +3 oxidation state and has 4 d-electrons. Due to the strong field ligand, CN$^-$, all the 4 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{0(0+2)} = 0$.

So the correct order of spin-only magnetic moments is:

$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}<\left[\mathrm{CoF}_{6}\right]^{3-}<\left[\mathrm{MnBr}_{4}\right]^{2-}$