CBSE Class 12 Physics Question Paper 2024 (Set 2- 55/3/2) Available- Download Solution PDF with Answer Key

Step 1: Understanding the Charge Distribution Inside the Conductor.

In electrostatic equilibrium, the electric field inside the conductor and thus within the conductor's material is zero. This condition implies that any excess charge resides on the surface of the conductor. When a point charge +q is placed inside a cavity of the conductor, it induces an equal and opposite charge of \(-q\) on the inner surface of the cavity to ensure the electric field inside the conductor's material remains zero.


Step 2: Determining the Charge on the Outer Surface.

Since the conductor itself has a net charge of +Q, and an additional \(-q\) charge has been induced on the inner surface, the outer surface must compensate for both the original charge and the induced charge. Therefore, the outer surface carries a charge of:( Q + q )to maintain the net charge of the conductor as \(+Q\) after accounting for the induced charge on the cavity wall. Quick Tip: Remember that in a conductor, charges will redistribute themselves to maintain electrostatic equilibrium, where the electric field inside the conducting material is zero. This redistribution leads to induced charges on surfaces enclosing cavities.

Step 1: Calculating the potential difference.

The potential difference \( V \) between points \(P_1\) and \(P_2\) is calculated as: \[ V = V_{\(P_2\)} - V_{\(P_1\)} = 5 \, V - (-5 \, V) = 10 \, V \]

Step 2: Calculating the work done.

The work done \( W \) on a charge \( q \) moving through a potential difference \( V \) is given by: \[ W = qV \]
For a proton, the charge \( q = 1.602 \times 10^{-19} \, Coulombs \). Thus, the work done on the proton when moving from \(P_1\) to \(P_2\) is: \[ W = 1.602 \times 10^{-19} \, C \times 10 \, V = 1.602 \times 10^{-18} \, J \]
Rounding off slightly, the work done matches the value given in option (B). Quick Tip: Remember, the work done in moving a charge in an electric field is dependent on the potential difference and the charge itself. The direction of the electric field and the initial and final potentials will determine whether the work is done by the field or against it.

Step 1: Application of the Biot-Savart Law.
The magnetic field due to a segment of current-carrying wire is determined by the Biot-Savart Law, which states: \[ dB = \frac{\mu_0}{4\pi} \frac{I \, dl \times \hat{r}}{r^2} \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, N/A^2 \) is the magnetic constant.
- \( I = 5.0 \, A \) is the current.
- \( dl = 0.02 \, m \, \hat{j} \) is the length vector of the wire segment.
- \( \hat{r} \) is the unit vector from the segment to the observation point.

Step 2: Calculation of \( r \) and \( \hat{r} \).
From the wire segment to the point (3 m, 4 m, 0): \[ r = \sqrt{3^2 + 4^2} = 5 \, m \] \[ \hat{r} = \left(\frac{3}{5}, \frac{4}{5}, 0\right) \]

Step 3: Calculation of \( dl \times \hat{r} \) and \( dB \).
Using the right-hand rule for the cross product: \[ dl \times \hat{r} = 0.02 \, \hat{j} \times \left(\frac{3}{5}, \frac{4}{5}, 0\right) = 0.02 \left( 0, 0, -\frac{3}{5} \right) = \left( 0, 0, -0.012 \right) \, \hat{k} \]
Thus, \( dB \) points in the negative \( \hat{k} \) direction. The magnitude of \( dB \) is: \[ dB = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{5 \cdot 0.012}{25} = \frac{10^{-7} \cdot 0.06}{25} = 2.4 \times 10^{-9} \, T = 0.24 \, nT \]
Considering the direction, the field is \( -0.24 \, nT \, \hat{k} \), matching option (C). Quick Tip: When dealing with the magnetic field calculation, always consider the right-hand rule for determining the direction of the magnetic field vector resulting from a current-carrying wire.

The magnetic force between two parallel current-carrying conductors is given by Ampere’s law. The force per unit length between two long parallel wires carrying currents in opposite directions is:
\[ F = \frac{\mu_0 I_1 I_2}{2 \pi a} \]

where:
- \( \mu_0 \) is the permeability of free space,
- \( I_1 \) and \( I_2 \) are the currents in the wires (here, both are equal to \( I \)),
- \( a \) is the distance between the wires.

Since the currents are in opposite directions, the wires will repel each other. The force is calculated as:
\[ F = \frac{\mu_0 I^2}{2 \pi a} \]

Thus, the wires will repel each other with a force \( \frac{\mu_0 I^2}{2 \pi a} \), per unit length. Quick Tip: For two parallel wires carrying currents in opposite directions, the magnetic forces between them will be repulsive. The magnitude of the force depends on the current and the distance between the wires.

Step 1: Understanding the magnetic potential energy.
The potential energy \( U \) of a magnetic dipole in a magnetic field is given by \( U = -\vec{M} \cdot \vec{B} \). Initially, the plane of the loop is normal to \( \vec{B} \), meaning \( \vec{M} \) is aligned with \( \vec{B} \), so the initial potential energy is \( U_i = -M B \).

Step 2: Calculating the final potential energy.
When the loop is rotated by 45 degrees, the angle \( \theta \) between \( \vec{M} \) and \( \vec{B} \) changes to \( 45^\circ \). The cosine of 45 degrees is \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \). Thus, the final potential energy \( U_f \) becomes: \[ U_f = -MB \cos(45^\circ) = -MB \frac{\sqrt{2}}{2} \]

Step 3: Calculating the work done.
The work done \( W \) in rotating the dipole is the change in potential energy: \[ W = U_f - U_i = \left(-MB \frac{\sqrt{2}}{2}\right) - (-MB) \] \[ W = MB \left(1 - \frac{\sqrt{2}}{2}\right) \]

Given \( \sqrt{2} \approx 1.414 \), this simplifies to: \[ W = MB \left(1 - 0.707\right) \] \[ W = MB (0.293) \]
Approximating for simplicity and clarity in the answer choices: \[ W \approx 0.3 MB \] Quick Tip: Remember, the work done on the system in magnetic fields involves changes in potential energy that reflect the orientation of the magnetic moment relative to the magnetic field direction.

The mutual inductance \( M \) between two coils is related to the induced emf in one coil due to the change in current in the other coil. The induced emf \( \varepsilon \) in coil \( C_2 \) is given by: \[ \varepsilon = - M \frac{\Delta I}{\Delta t} \]
where:
- \( M = 20 \, mH = 20 \times 10^{-3} \, H \) is the mutual inductance,
- \( \Delta I = 4 \, A \) is the change in current in coil \( C_1 \),
- \( \Delta t = 0.2 \, s \) is the time in which the current changes.

Substitute the values into the formula: \[ \varepsilon = - 20 \times 10^{-3} \times \frac{4}{0.2} = - 20 \times 10^{-3} \times 20 = - 0.4 \, V \]
The negative sign indicates the direction of the induced emf (which we do not need for the magnitude).

Now, the current \( I_2 \) induced in coil \( C_2 \) due to the emf is: \[ I_2 = \frac{\varepsilon}{R} \]
where \( R = 4 \, \Omega \) is the resistance of coil \( C_2 \). Substituting the values: \[ I_2 = \frac{0.4}{4} = 0.1 \, A \]
The charge \( Q \) flowing through coil \( C_2 \) per second (which is the current) is: \[ Q = I_2 \times \Delta t = 0.1 \, A \times 1 \, s = 0.1 \, C \]

Thus, the charge that flows through coil \( C_2 \) per second is \( 0.1 \, C \). Quick Tip: Remember, the mutual inductance relates the change in current in one coil to the induced emf in the other coil. Use Ohm's law to calculate the current and then use \( Q = I \times t \) to find the charge flow.

Step 1: Understanding self-inductance.

The self-inductance \( L \) of a solenoid is given by:
\[ L = \frac{\mu_0 N^2 A}{l} \]
where:

\( \mu_0 \) is the permeability of free space,

\( N \) is the number of turns,

\( A \) is the cross-sectional area,

\( l \) is the length of the solenoid.


Step 2: Analyzing how \( L \) changes with \( l \) and \( A \).

From the formula, it is clear that:

Increasing \( A \) increases \( L \) because \( A \) is in the numerator.

Decreasing \( l \) increases \( L \) because \( l \) is in the denominator.


Step 3: Evaluating the options.

Increasing both \( l \) and \( A \) would have conflicting effects on \( L \), making the net effect less predictable.

Decreasing \( l \) and increasing \( A \) simultaneously provides the most direct increase in \( L \), as both changes contribute positively to increasing \( L \).

Increasing \( l \) and decreasing \( A \) would lead to a decrease in \( L \).

Decreasing both \( l \) and \( A \) would also have conflicting effects, but the decrease in \( A \) would be more detrimental.
Quick Tip: When evaluating the effect of physical changes on the inductance of a solenoid, remember that increasing the area of cross-section and decreasing the length will enhance the inductance, benefiting from the direct proportionality to \( A \) and inverse proportionality to \( l \).

The electromagnetic spectrum is divided into different categories based on frequency and wavelength. The frequency \( 1.0 \times 10^{18} \, Hz \) falls in the range of X-rays, which typically have frequencies between \( 10^{17} \, Hz \) and \( 10^{19} \, Hz \).

The order of increasing frequency for electromagnetic waves is as follows:

Radio waves: \( 10^4 - 10^9 \, Hz \)

Microwaves: \( 10^9 - 10^{12} \, Hz \)

Infrared rays: \( 10^{12} - 10^{14} \, Hz \)

Visible light: \( 10^{14} - 10^{15} \, Hz \)

Ultraviolet rays: \( 10^{15} - 10^{17} \, Hz \)

X-rays: \( 10^{17} - 10^{19} \, Hz \)

Gamma rays: \( 10^{19} \, Hz \) and higher.


Thus, electromagnetic waves with a frequency of \( 1.0 \times 10^{18} \, Hz \) are categorized as X-rays. Quick Tip: X-rays have frequencies ranging from \( 10^{17} \, Hz \) to \( 10^{19} \, Hz \), making them distinct from both ultraviolet rays and gamma rays.

Step 1: Analyzing the forces and energies involved.
The distance of closest approach can be determined by equating the kinetic energy to the electrostatic potential energy at the point of closest approach: \[ \frac{1}{2} mv^2 = \frac{Zke^2}{r} \]
where \( m \) is the mass, \( v \) is the velocity, \( Z \) is the atomic number of the target nucleus, \( k \) is Coulomb's constant, \( e \) is the charge, and \( r \) is the distance of closest approach.

Step 2: Comparing proton and alpha particle.
The alpha particle has twice the charge of a proton (since it contains two protons) and four times the mass. Since both particles have equal velocities, their kinetic energies differ but the potential energy of the alpha particle is higher due to its greater charge.

Step 3: Calculating the ratio of distances.
The formula rearranges to \( r = \frac{Zke^2}{mv^2} \). For the alpha particle, both the charge and mass affect the distance: \[ r_{proton} = \frac{Zke^2}{m_pv^2}, \quad r_{alpha} = \frac{Zke^2}{4m_pv^2} \]
Thus, \( r_{proton} = 2r_{alpha} \), giving the ratio of 2. Quick Tip: For electrostatic potential energy problems involving charged particles, remember that the charge affects the potential energy directly and the mass indirectly through kinetic energy.

In the photoelectric effect, the stopping potential \( V_0 \) is related to the work function \( \phi \) and the frequency \( \nu \) of the incident radiation by the equation: \[ V_0 = \frac{h \nu}{e} - \phi \]
where:
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident radiation,
- \( e \) is the charge of the electron,
- \( \phi \) is the work function of the material.

From the graph, we observe that the stopping potential increases with the frequency, and the line with the highest slope corresponds to the material with the highest work function. Since \( M_3 \) has the largest slope, it must have the largest work function \( \phi_3 \), followed by \( M_2 \) with \( \phi_2 \), and \( M_1 \) with \( \phi_1 \).

Thus, \( \phi_3 > \phi_2 > \phi_1 \). Quick Tip: In the photoelectric effect, a higher work function corresponds to a higher intercept on the \( V_0 \)-axis, as more energy is required to release electrons from the material.

To determine how the period of revolution of an electron changes when it transitions from the \( n = 2 \) level to the \( n = 1 \) level in the Bohr model of a hydrogen atom, we can follow these steps:

1. Bohr Model Basics

In the Bohr model, the period of revolution \( T \) of an electron in the \( n \)-th energy level is given by:
\[ T_n \propto n^3 \]
This means that the period of revolution is proportional to the cube of the principal quantum number \( n \).


2. Initial and Final Periods


For \( n = 2 \):
\[ T_2 \propto 2^3 = 8 \]
For \( n = 1 \):
\[ T_1 \propto 1^3 = 1 \]


3. Change in Period

The change in the period of revolution when the electron transitions from \( n = 2 \) to \( n = 1 \) is:
\[ \Delta T = T_1 - T_2 = 1 - 8 = -7 \]
The negative sign indicates a decrease in the period.


4. Percentage Change

The percentage change in the period is calculated as:
\[ Percentage Change = \left( \frac{\Delta T}{T_2} \right) \times 100% = \left( \frac{-7}{8} \right) \times 100% = -87.5% \]
This means the period of revolution decreases by 87.5%.

Therefore, the correct answer is:
\[ \boxed{(B) decreases by 87·5%} \] Quick Tip: Remember that in the Bohr model, the period of revolution of the electron around the nucleus is directly proportional to the cube of the principal quantum number \( n \).

Step 1: Understanding doping in semiconductors.
Doping Silicon with a pentavalent element introduces extra electrons in the conduction band, effectively creating n-type semiconductors.

Step 2: Energy required for electron mobility.
The energy required to free these additional electrons is generally very low since they are loosely bound compared to electrons in the valence band.

Step 3: Estimating the ionization energy.
For typical pentavalent dopants like phosphorus in silicon, the energy required to ionize the dopant and release an electron into the conduction band is about 0.05 eV. Quick Tip: When considering the energy levels in doped semiconductors, it's essential to remember that the extra electrons or holes introduced by the dopant have much lower ionization energies compared to the intrinsic semiconductor.

Analysis of Assertion (A):

The assertion is false because in a semiconductor, the electrons in the conduction band have higher energy than those in the valence band, which is why they are free to move and conduct electricity.

Analysis of Reason (R):

The reason is also false because the donor energy level is just below the conduction band, not above the valence band. Donor levels donate electrons to the conduction band, enhancing conductivity. Quick Tip: Understanding the energy band structure of semiconductors is crucial. Remember, the conduction band houses higher energy electrons compared to those in the valence band.

Step 1: Analyzing Assertion (A).
The photoelectric effect demonstrates the particle nature of light because it involves the interaction of light with matter in such a way that light is treated as a collection of particles (photons). According to Einstein's explanation, each photon carries energy proportional to its frequency. This energy is transferred to electrons, allowing them to escape from the surface of the material, thus demonstrating the particle nature of light.

Step 2: Analyzing Reason (R).
The photoelectric current is not proportional to the frequency of the incident radiation. In fact, the current is proportional to the intensity of the incident radiation (i.e., the number of photons striking the surface per unit time). The frequency of the radiation determines the energy of the emitted photoelectrons, but it is the intensity (not the frequency) that controls the photoelectric current. The reason provided in Statement (R) is incorrect.

Conclusion:
Assertion (A) is true because the photoelectric effect demonstrates the particle nature of light. However, Reason (R) is false because the photoelectric current is proportional to the intensity (not the frequency) of the incident radiation. Quick Tip: For clarity, photoelectric current's dependence on light intensity is a cornerstone in understanding how light's particle nature influences photoelectric effect outcomes. Frequency affects energy, not the number of emitted electrons.

Assertion (A) is true: Both a proton and an electron entering a uniform magnetic field with the same momentum and perpendicular to the magnetic field will follow circular paths of the same radius, as the magnetic force provides the centripetal force required for circular motion.


Reason (R) is true: The radius of the circular path for a charged particle in a magnetic field is given by \( r = \frac{p}{qB} \), where \( p \) is the momentum, \( q \) is the charge, and \( B \) is the magnetic field strength. Since the momentum and magnetic field are the same for both the proton and the electron, they will describe circular paths of the same radius.

Thus, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A). Quick Tip: For charged particles moving in a magnetic field, the radius of the circular path is given by \( r = \frac{p}{qB} \), which depends on the momentum, charge, and magnetic field strength.

Analysis of Assertion (A):

The assertion is true. A convex lens can indeed seem to disappear when immersed in a liquid if the refractive indices of the lens and the liquid are the same, because there is no change in the speed of light at the boundary.

Analysis of Reason (R):

The reason is true and provides the correct explanation for the assertion. If the refractive indices of the lens and the liquid are equal, light does not bend at the interface, making the lens invisible in the liquid. Quick Tip: When studying optics, it's essential to understand how the refractive index affects light transmission and bending at interfaces. Similar refractive indices can make objects visually disappear in mediums.

% Explanation of Relaxation Time
Relaxation Time:
Relaxation time (\( \tau \)) is the average time interval between consecutive collisions of an electron as it moves through a conductor. It is a measure of how long an electron travels freely without interaction, influencing the electrical conductivity of the material.



% Derivation of the Resistance Formula
Derivation:

The resistance \( R \) of a conductor can be derived using the Drude model, which relates electrical properties to the behavior of electrons in a material. The resistivity \( \rho \) according to the Drude model is given by:
\[ \rho = \frac{m}{n e^2 \tau} \]
where \( m \) is the electron mass, \( n \) is the density of charge carriers, \( e \) is the electron charge, and \( \tau \) is the relaxation time. The formula for resistance \( R \), involving the geometry of the conductor, is:
\[ R = \rho \frac{L}{A} \]
Substituting the expression for \( \rho \) gives:
\[ R = \frac{m}{n e^2 \tau} \frac{L}{A} = \frac{mL}{n e^2 \tau A} \]
This equation illustrates that resistance is inversely proportional to the density of charge carriers and their relaxation time, and directly proportional to the conductor's length.
Quick Tip: Understanding the concept of relaxation time helps in comprehending how material impurities and temperature affect the resistance and overall electrical conductivity.

This balance condition implies that the potential drop across the bridge is symmetrically distributed, resulting in zero potential difference across the galvanometer, thus no current flow. Quick Tip: The Wheatstone bridge is a fundamental tool in measuring unknown electrical resistances and is extensively used in sensors and other measuring devices.

Given that the distance between the lenses, \( L = 150 \, cm \), and the magnifying power, \( M = 24 \), we have: \[ L = f_o + f_e \]
and \[ M = \frac{f_o}{f_e} \]
By rearranging and substituting from the magnification formula: \[ f_e = \frac{f_o}{M} = \frac{f_o}{24} \]
Substituting into the lens distance equation gives: \[ 150 = f_o + \frac{f_o}{24} \] \[ 150 = \frac{25f_o}{24} \] \[ f_o = \frac{150 \times 24}{25} = 144 \, cm \]
Therefore, the focal length of the objective lens is 144 cm. Quick Tip: When solving problems involving optical instruments like telescopes, always check if the system is in normal adjustment as it simplifies the use of formulae.

Nature of the Phenomenon:

Interference is the phenomenon where two or more light waves superpose to form a resultant wave, leading to constructive or destructive interference.


Diffraction, on the other hand, refers to the bending of light waves around obstacles and the spreading of light as it passes through small apertures.

Conditions for Occurrence:

Interference requires at least two coherent sources of light (sources that maintain a constant phase relationship), while diffraction occurs when light encounters obstacles or slits comparable in size to its wavelength, and it does not require coherence between the sources. Quick Tip: Interference depends on the superposition of coherent light waves, while diffraction is the spreading of light due to obstacles or slits, showcasing wave nature.

Step 1: Energy of the incident photons.

The energy of the incident photons \( E_{photon} \) is given by the equation: \[ E_{photon} = \frac{hc}{\lambda} \]
where:
- \( h = 6.626 \times 10^{-34} \, J \cdot s \) (Planck's constant),
- \( c = 3.0 \times 10^8 \, m/s \) (speed of light),
- \( \lambda = 500 \, nm = 500 \times 10^{-9} \, m \) (wavelength of the light).

Substituting the known values: \[ E_{photon} = \frac{1240 \, eV \cdot nm}{500 \, nm} = 2.48 \, eV \]
So, the energy of the incident photons is 2.48 eV.


Step 2: Kinetic Energy of the Fastest Electrons.

The kinetic energy \( E_{kinetic} \) of the fastest electrons is given by the photoelectric equation: \[ E_{kinetic} = E_{photon} - \phi \]
where \( \phi = 2.14 \, eV \) is the work function of the caesium metal.

Substituting the values: \[ E_{kinetic} = 2.48 \, eV - 2.14 \, eV = 0.34 \, eV \]
Thus, the kinetic energy of the fastest electrons is \( 0.34 \, eV \).


Step 3: Stopping Potential.

The stopping potential \( V_{stop} \) is related to the kinetic energy of the fastest electrons by the equation: \[ E_{kinetic} = eV_{stop} \]
where \( e = 1 \, e \) is the charge of the electron.

Rearranging to find the stopping potential: \[ V_{stop} = \frac{E_{kinetic}}{e} = 0.34 \, V \]
Thus, the stopping potential is \( 0.34 \, V \). Quick Tip: The photoelectric effect equation is \( E_{kinetic} = E_{photon} - \phi \), where \( \phi \) is the work function of the material. The stopping potential is directly related to the kinetic energy of the fastest electrons.

First, calculate the concentration of boron dopants: \[ Boron concentration = \frac{1 \, ppm}{10^6} \times 5 \times 10^{28} = 5 \times 10^{22} \, m^{-3} \]
Since boron adds holes, the concentration of holes \( p \) will be approximately equal to the concentration of boron dopants. Assuming intrinsic carrier concentration \( n_i \) remains much smaller compared to the hole concentration due to doping: \[ p \approx 5 \times 10^{22} \, m^{-3} \]
Using the mass action law \( n \times p = n_i^2 \), calculate the concentration of electrons \( n \): \[ n = \frac{n_i^2}{p} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = 4.5 \times 10^{9} \, m^{-3} \]
Thus, the concentration of holes \( p \) is \(5 \times 10^{22} \, m^{-3}\) and electrons \( n \) is \(4.5 \times 10^{9} \, m^{-3}\), indicating a p-type semiconductor. Quick Tip: Remember, doping a semiconductor with trivalent impurities (like Boron) increases the hole concentration, making it p-type. Doping with pentavalent impurities (like Phosphorus) would increase electron concentration, making it n-type.

Step 1: Calculate emf of the battery.

When the key (K) is open, no current flows, and the voltage measured by the voltmeter is the emf of the battery. Thus: \[ E = 10.0 \, V \]
So, the emf of the battery is \( E = 10.0 \, V \).


Step 2: Use the closed circuit data to find internal resistance \( r \).

When the key (K) is closed, the current flows, and the voltmeter reads 6.0 V while the ammeter reads 2.0 A. In this case, the total voltage across the battery is used to overcome both the internal resistance and the external resistance. According to Ohm’s Law: \[ V = I R_{total} \]
where \( V \) is the voltage across the external resistance \( R \), \( I \) is the current, and \( R_{total} = R + r \) is the total resistance of the circuit.

The voltage drop across the internal resistance \( r \) is \( I r \). Therefore, the emf \( E \) is given by: \[ E = V + I r \]
Substituting the values: \[ 10.0 \, V = 6.0 \, V + 2.0 \, A \times r \] \[ r = \frac{10.0 \, V - 6.0 \, V}{2.0 \, A} = \frac{4.0 \, V}{2.0 \, A} = 2.0 \, \Omega \]
Thus, the internal resistance of the battery is \( r = 2.0 \, \Omega \).

(c) external resistance \( R \):
Now, we calculate the external resistance using the fact that the total resistance is the sum of the internal and external resistances: \[ R_{total} = r + R = 3.0 \, \Omega \]
Since \( r = 0 \, \Omega \), we get: \[ R = 3.0 - 0 = 3.0 \, \Omega \]

Thus, the external resistance \( R \) is \( 3.0 \, \Omega \). Quick Tip: When the key is open, the voltmeter reading gives the emf of the battery. In a closed circuit, the voltage drop across the internal resistance can be calculated using the difference between the emf and the voltmeter reading.

Consider a rectangular current loop with area \( \vec{A} \), carrying a current \( I \), placed in a uniform magnetic field \( \vec{B} \). The loop experiences a torque due to the magnetic field, which tends to rotate it.

Step 1: Magnetic Force on a Current-Carrying Wire.

The magnetic force on a segment of current-carrying wire is given by: \[ d\vec{F} = I \, d\vec{l} \times \vec{B} \]
where \( d\vec{l} \) is the length element of the wire, \( I \) is the current, and \( \vec{B} \) is the magnetic field.

Step 2: Net Torque on the Loop.

The total torque \( \vec{\tau} \) on the loop is the vector sum of the torques on each segment of the loop. The torque due to a force \( d\vec{F} \) acting at a position vector \( \vec{r} \) from the center of the loop is given by: \[ d\vec{\tau} = \vec{r} \times d\vec{F} \]
Thus, the total torque on the loop is the integral of these infinitesimal torques over the entire loop: \[ \vec{\tau} = \int \vec{r} \times d\vec{F} \]

Step 3: Magnetic Moment of the Loop.

The magnetic moment \( \vec{m} \) of a current loop is given by: \[ \vec{m} = I \vec{A} \]
where \( \vec{A} \) is the vector area of the loop, which is perpendicular to the plane of the loop. The magnitude of \( \vec{A} \) is the area of the loop, and the direction of \( \vec{A} \) is given by the right-hand rule (pointing perpendicular to the plane of the loop in the direction of the current flow).

Step 4: Expression for Torque.

The torque \( \vec{\tau} \) acting on a current loop in a uniform magnetic field \( \vec{B} \) is related to the magnetic moment \( \vec{m} \) by the following equation: \[ \vec{\tau} = \vec{m} \times \vec{B} \]
where \( \vec{m} = I \vec{A} \) is the magnetic moment of the loop.

Thus, the torque acting on the rectangular loop in the uniform magnetic field is given by: \[ \vec{\tau} = I \vec{A} \times \vec{B} \] Quick Tip: The torque on a current loop in a magnetic field is maximized when the magnetic moment is perpendicular to the magnetic field, and the magnitude of the torque depends on the angle between \( \vec{A} \) and \( \vec{B} \).

(a) Difference between Reactance and Impedance:


Reactance: Reactance is the opposition to the change in current due to either inductance or capacitance in an AC circuit. It is a scalar quantity and is represented by \( X \). There are two types of reactance:

Inductive Reactance (\( X_L \)) due to an inductor:
\[ X_L = \omega L \]
where \( \omega = 2\pi f \) is the angular frequency and \( L \) is the inductance.
Capacitive Reactance (\( X_C \)) due to a capacitor:
\[ X_C = \frac{1}{\omega C} \]
where \( C \) is the capacitance.


Impedance: Impedance is the total opposition to the flow of alternating current in a circuit, considering both resistance and reactance. It is a complex quantity, represented by \( Z \), and is given by:
\[ Z = R + jX \]
where \( R \) is the resistance, \( X \) is the reactance (which can be either inductive or capacitive), and \( j \) is the imaginary unit. The magnitude of impedance is given by:
\[ |Z| = \sqrt{R^2 + X^2} \]
Impedance combines both the real and imaginary components of opposition in an AC circuit.


Thus, reactance is the opposition due to either capacitance or inductance alone, while impedance includes both the resistance and reactance in an AC circuit.


(b) Power Dissipation in an Ideal Inductor:

An ideal inductor in an AC circuit has the reactance \( X_L = \omega L \), but no resistance. The current \( I(t) \) in an inductive AC circuit is related to the voltage \( V(t) \) across the inductor by: \[ V(t) = L \frac{dI(t)}{dt} \]
The instantaneous power \( P(t) \) dissipated in the inductor is given by the product of the voltage and current: \[ P(t) = V(t) \cdot I(t) \]
Since the voltage across the inductor and the current are 90 degrees out of phase, the instantaneous power alternates between positive and negative values. Therefore, over one complete cycle, the net power dissipated is zero: \[ P_{avg} = \frac{1}{T} \int_0^T P(t) \, dt = 0 \]
Hence, an ideal inductor does not dissipate any power in an AC circuit, even though it stores and releases energy in each cycle. Quick Tip: In an ideal inductor, the power alternates between being stored in the magnetic field and released back to the circuit, leading to no net energy dissipation over time.

We know that the wave number \( k \) is related to the wavelength \( \lambda \) by: \[ k = \frac{2 \pi}{\lambda} \]

So, rearranging: \[ \lambda = \frac{2 \pi}{k} = \frac{2 \pi}{1.5} = 4.18 \, m \]

Next, the angular frequency \( \omega \) is related to the frequency \( \nu \) by: \[ \omega = 2 \pi \nu \]
Rearranging: \[ \nu = \frac{\omega}{2 \pi} = \frac{4.5 \times 10^8}{2 \pi} \approx 7.16 \times 10^7 \, Hz \]

Thus, the wavelength \( \lambda \) is approximately \( 4.18 \, m \) and the frequency \( \nu \) is approximately \( 7.16 \times 10^7 \, Hz \).


(b) What is the amplitude of the magnetic field of the wave?

Solution:

For an electromagnetic wave, the relation between the electric field amplitude \( E_0 \) and the magnetic field amplitude \( B_0 \) is: \[ E_0 = c B_0 \]
where \( c = 3 \times 10^8 \, m/s \) is the speed of light.

Given \( E_0 = 6.3 \, N/C \), we can solve for \( B_0 \): \[ B_0 = \frac{E_0}{c} = \frac{6.3}{3 \times 10^8} = 2.1 \times 10^{-8} \, T \]

Thus, the amplitude of the magnetic field is \( B_0 = 2.1 \times 10^{-8} \, T \).


(c) Write an expression for the magnetic field of this wave.

Solution:

The magnetic field \( \vec{B} \) in an electromagnetic wave is given by: \[ \vec{B} = B_0 \cos(k y - \omega t) \hat{j} \]

Substituting \( B_0 = 2.1 \times 10^{-8} \, T \) and the given values for \( k \) and \( \omega \): \[ \vec{B} = 2.1 \times 10^{-8} \left[\cos \left(1.5 \, rad/m \, y - (4.5 \times 10^8 \, rad/s) \, t \right)\right] \hat{j} \]

This is the expression for the magnetic field of the wave. Quick Tip: Remember, in electromagnetic waves, the electric and magnetic fields oscillate perpendicular to each other and the direction of wave propagation, following a right-handed coordinate system. The magnitudes of \( \vec{E} \) and \( \vec{B} \) are related through the speed of light in vacuum.

Bohr’s theory of the hydrogen atom explains the spectral lines observed in the emission and absorption spectra of hydrogen. According to Bohr's theory:

- The electron in a hydrogen atom revolves in fixed orbits around the nucleus without radiating energy, which are called stationary orbits.
- The electron in the \(n\)-th orbit has a specific energy: \[ E_n = -\frac{13.6 \, eV}{n^2} \]
where \(n\) is the principal quantum number.
- When an electron moves from a higher energy orbit \(n_2\) to a lower energy orbit \(n_1\), it emits radiation with energy equal to the difference between the two orbits: \[ \Delta E = E_{n_1} - E_{n_2} = h \nu \]
where \(h\) is Planck's constant and \( \nu \) is the frequency of the emitted radiation.
- The frequency of the emitted radiation can be expressed as: \[ \nu = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant, and \(n_1\) and \(n_2\) are the initial and final quantum numbers of the electron.


Spectral Series:


1. Lyman Series: Transitions from \(n_2 = 2, 3, 4, \dots\) to \(n_1 = 1\) (ultraviolet region).

2. Balmer Series: Transitions from \(n_2 = 3, 4, 5, \dots\) to \(n_1 = 2\) (visible region).

3. Paschen Series: Transitions from \(n_2 = 4, 5, 6, \dots\) to \(n_1 = 3\) (infrared region).

4. Brackett Series: Transitions from \(n_2 = 5, 6, 7, \dots\) to \(n_1 = 4\) (infrared region).

5. Pfund Series: Transitions from \(n_2 = 6, 7, 8, \dots\) to \(n_1 = 5\) (infrared region).



% Energy Level Diagram


The diagram shows the various energy levels for hydrogen, and the transitions between these levels produce the spectral lines observed in the emission and absorption spectra of hydrogen. Quick Tip: In Bohr’s model, the emission or absorption of light occurs only when an electron transitions between discrete energy levels. The energy difference between these levels determines the frequency of the emitted or absorbed radiation.

The atomic mass unit (u), also known as the unified atomic mass unit, is defined as one twelfth of the mass of an unbound neutral atom of carbon-12. In other words:
\[ 1 \, u = \frac{1}{12} \times mass of one carbon-12 atom \]

The atomic mass unit is approximately equal to:
\[ 1 \, u = 1.66053906660 \times 10^{-27} \, kg \]

It is a standard unit of mass used to express atomic and molecular weights.

% Part (b) Calculation of Separation Energy
(b) Calculation of Separation Energy:

First, calculate the mass defect (\(\Delta m\)):
\[ \Delta m = \left( m_{H} + m_{n} \right) - m(D) \]

Substitute the given values:
\[ \Delta m = (1.007825 \, u + 1.008665 \, u) - 2.014102 \, u \]
\[ \Delta m = 2.016490 \, u - 2.014102 \, u = 0.002388 \, u \]

Next, convert the mass defect from atomic mass units to kilograms. 1 u is approximately \(1.660539 \times 10^{-27}\) kg:
\[ \Delta m = 0.002388 \, u \times 1.660539 \times 10^{-27} \, kg/u = 3.965 \times 10^{-30} \, kg \]

Now, calculate the energy using \(E = \Delta m \cdot c^2\), where \(c = 3 \times 10^8 \, m/s\):
\[ E = 3.965 \times 10^{-30} \, kg \times (3 \times 10^8 \, m/s)^2 \]
\[ E = 3.965 \times 10^{-30} \, kg \times 9 \times 10^{16} \, m^2/s^2 = 3.5685 \times 10^{-13} \, J \]

To express the energy in MeV (1 MeV = \(1.60218 \times 10^{-13}\) J):
\[ E = \frac{3.5685 \times 10^{-13} \, J}{1.60218 \times 10^{-13} \, J/MeV} \approx 2.23 \, MeV \]

Therefore, the energy required to separate a deuteron into a proton and a neutron is approximately \boxed{2.23 \, \text{MeV. Quick Tip: Understanding the concept of mass defect and binding energy is crucial for explaining why nuclei are stable and the energy processes involved in nuclear reactions.

% Circuit Diagrams for V-I Characteristics
Circuit Diagrams:

To obtain the V-I characteristics of a p-n junction diode, two setups are required:

1. Forward Biasing: Connect the positive terminal of a battery to the p-type and the negative to the n-type. Include a variable resistor to change the voltage and an ammeter to measure the current.

2. Reverse Biasing: Connect the negative terminal of the battery to the p-type and the positive to the n-type. Similarly, include a variable resistor and an ammeter.


% Features of V-I Characteristics
Features of V-I Characteristics:

(i) Forward Biasing:

In forward biasing, the diode conducts current easily after surpassing the threshold voltage (typically around 0.7V for silicon diodes). The current increases exponentially with an increase in voltage.





(ii) Reverse Biasing:

In reverse biasing, the diode does not conduct until a critical reverse breakdown voltage is reached. Under normal reverse bias conditions, the current is very small (leakage current) and nearly constant despite changes in voltage.



Quick Tip: Always ensure the diode is not subjected to the reverse breakdown voltage in typical applications, unless it is designed to handle such conditions (e.g., Zener diodes).

% Energy Band Diagrams Explanation
Energy Band Diagrams:


(i) Insulator:
Insulators have a very large band gap (usually greater than 6 eV) between the valence band and the conduction band. This large gap makes it extremely difficult for electrons to gain enough energy to move from the valence band to the conduction band, hence very low conductivity.




(ii) Semiconductor:
Semiconductors have a smaller band gap (about 1 eV for silicon). At absolute zero, they behave like insulators, but at higher temperatures, some electrons gain enough energy to jump to the conduction band, allowing the material to conduct electricity.





(iii) Conductor:
In conductors, the valence band and conduction band overlap, or the conduction band is partially filled with electrons. This allows electrons to move freely within the conduction band, resulting in high electrical conductivity.



Quick Tip: Understanding energy band diagrams is crucial for grasping how materials behave under different electrical conditions and temperatures, influencing their application in electronics.

% Electric Field Calculations
Given that the electric field \(E\) between two plates is calculated as: \[ E = \frac{\Delta V}{d} \]
where \(\Delta V\) is the potential difference and \(d\) is the distance between the plates (2.0 cm = 0.02 m in this case).

% Analyzing each pair for Electric Field

Pair I: \( \Delta V = -70V - (-50V) = -20V \)
\[ E_1 = \frac{20}{0.02} = 1000 \, V/m \]
Pair II: \( \Delta V = 150V - (-50V) = 200V \)
\[ E_2 = \frac{200}{0.02} = 10000 \, V/m \]
Pair III: \( \Delta V = 200V - (-20V) = 220V \)
\[ E_3 = \frac{220}{0.02} = 11000 \, V/m \]
Pair IV: \( \Delta V = -100V - (-400V) = 300V \)
\[ E_4 = \frac{300}{0.02} = 15000 \, V/m \]


% Conclusion
Answer: Comparing the magnitudes, \(E_4 > E_3 > E_2 > E_1\).
The correct option is (C) \(E_4 > E_3 > E_2 > E_1\). Quick Tip: Always remember, the electric field strength is proportional to the voltage difference divided by the separation distance. Greater voltage differences over the same distance result in stronger fields.

Given:

Electric potential of the left plate: \(-70 \, V\)
Electric potential of the right plate: \(-50 \, V\)
Distance between the plates: \(2.0 \, cm = 0.02 \, m\)
Charge-to-mass ratio of the electron: \(\frac{e}{m} = 1.76 \times 10^{11} \, C/kg\)


Step-by-Step Solution:

1. Determine the Electric Field (\(E\)):
The electric field between the plates is uniform and can be calculated using the potential difference (\(\Delta V\)) and the distance (\(d\)) between the plates:
\[ E = \frac{\Delta V}{d} \]
Here, \(\Delta V = V_{right} - V_{left} = -50 \, V - (-70 \, V) = 20 \, V\).

So,
\[ E = \frac{20 \, V}{0.02 \, m} = 1000 \, V/m \]

2. Calculate the Force on the Electron:
The force (\(F\)) on the electron due to the electric field is:
\[ F = eE \]
where \(e\) is the charge of the electron.

3. Determine the Acceleration (\(a\)):
Using Newton's second law, the acceleration of the electron is:
\[ a = \frac{F}{m} = \frac{eE}{m} \]
Given \(\frac{e}{m} = 1.76 \times 10^{11} \, C/kg\), we have:
\[ a = (1.76 \times 10^{11}) \times 1000 = 1.76 \times 10^{14} \, m/s^2 \]

4. Use the Kinematic Equation:
The electron comes to rest at the left plate, so its final velocity (\(v_f\)) is 0. The initial velocity (\(v_i\)) can be found using the kinematic equation:
\[ v_f^2 = v_i^2 + 2a d \]
Since \(v_f = 0\) and the electron travels the distance \(d = 0.02 \, m\):
\[ 0 = v_i^2 + 2 \times (1.76 \times 10^{14}) \times 0.02 \]
Solving for \(v_i\):
\[ v_i^2 = -2 \times (1.76 \times 10^{14}) \times 0.02 \]
\[ v_i^2 = 7.04 \times 10^{12} \]
\[ v_i = \sqrt{7.04 \times 10^{12}} \approx 2.65 \times 10^6 \, m/s \]

Final Answer:
The speed with which the electron was projected is approximately: \[ \boxed{2.6 \times 10^6 \, m/s} \] Quick Tip: When projecting an electron or any charged particle in an electric field, remember that kinetic energy is converted into electrical potential energy, and vice versa. The greater the potential difference, the faster the initial speed required to just reach the opposite plate.

The central peak of the diffraction pattern corresponds to the main lobe of the intensity distribution due to the diffraction effect. The number of interference peaks within the central diffraction peak is determined by the ratio of the width of the central diffraction peak to the fringe separation.

The diffraction angle for the first minimum is given by: \[ \sin \theta = \frac{\lambda}{d} \]
where \( \lambda = 450 \, nm \) (wavelength of the monochromatic light) and \( d = 6 \, \mum \) (distance between the slits).

The interference fringes fall within the diffraction envelope, and the number of peaks of the interference fringes within the central diffraction peak is 6.

Thus, the number of interference fringes within the central peak is 6. Quick Tip: Remember that the number of interference fringes within the central diffraction maximum depends on the relative widths of the interference and diffraction patterns. The calculation often involves approximations unless exact dimensions are provided.

When the slit width is doubled, the diffraction envelope becomes narrower because the angular position of the first diffraction minimum is inversely proportional to the slit width.

The number of interference fringes within the central diffraction peak will remain the same because the fringe separation is determined by the distance between the slits, not the slit width. Hence, the number of peaks of interference fringes within the central diffraction peak is 3.

Thus, the number of peaks of interference is 3. Quick Tip: When the width of the slit is increased, the diffraction effects become more pronounced, reducing the width of the central maximum and the effective space for interference fringes within it.

Step 1: Recalculate the angular width of the central maximum with the new wavelength.

With a wavelength of 680 nm, the angular width of the central maximum in the diffraction pattern is recalculated using the formula:
\[ \theta \approx \frac{\lambda}{a} \]
where \(\lambda\) is now 680 nm, and \(a\) remains as 2 m. Thus: \[ \theta \approx \frac{680 \times 10^{-9}}{2} = 340 \times 10^{-9} \, radians \]

Step 2: Determine the separation of the interference fringes with the new wavelength.

The fringe separation in the double-slit interference pattern, given by:
\[ \Delta y = \frac{\lambda L}{d} \]
will be recalculated with the new \(\lambda = 680 \times 10^{-9}\) m. Assuming \(d = 6\) m, we get:
\[ \Delta y \approx \frac{680 \times 10^{-9} L}{6} \]

Step 3: Calculate the number of peaks within the central maximum with the new wavelength.

The total width of the central diffraction peak, \(2\theta\), and the number of interference fringes fitting within this width, are recalculated:
\[ Number of peaks = \frac{2\theta L}{\Delta y} = \frac{2 \times 340 \times 10^{-9} L}{\frac{680 \times 10^{-9} L}{6}} = 6 \]
Given the larger wavelength, the angular width of the central maximum is wider, allowing more interference fringes to fit within. Quick Tip: Using a longer wavelength not only alters the diffraction pattern's angular width but also impacts the density and number of interference fringes within it.

Calculate the angle for the first minimum in the diffraction pattern.

The position of the first minimum in a single-slit diffraction pattern is given by the condition:
\[ a \sin(\theta) = \lambda \]
where \(a\) is the width of the slit, and \(\lambda\) is the wavelength of light used. For the initial setup with \(\lambda = 450\) nm and \(a = 2\) m:
\[ 2 \sin(\theta) = 450 \times 10^{-9} \Rightarrow \sin(\theta) = 225 \times 10^{-9} \]
To find the angle \(\theta\), we calculate:
\[ \theta = \sin^{-1}(225 \times 10^{-9}) \approx \sin^{-1}(0.225) \quad (approximation based on the correct magnitude) \] Quick Tip: When calculating angles for diffraction minima, ensure that dimensions are correctly scaled between the slit size and the wavelength. Large slit sizes or small wavelengths can result in very small angle values, which may require precise calculation or approximation techniques.

Solution

Given Data:

Wavelength of light, \(\lambda = 450 nm = 450 \times 10^{-9} m\)
Distance between the slits, \(d = 6 \mum = 6 \times 10^{-6} m\)
Distance from the slits to the screen, \(L = \frac{4}{3} m\)
Width of each slit, \(a = 2 \mum = 2 \times 10^{-6} m\)


Interference Fringe Spacing:

The spacing between bright fringes (interference maxima) on the screen is given by:
\[ \Delta y = \frac{\lambda L}{d} \]
Plugging in the values:
\[ \Delta y = \frac{450 \times 10^{-9} \times \frac{4}{3}}{6 \times 10^{-6}} = \frac{600 \times 10^{-9}}{6 \times 10^{-6}} = 0.1 m \]

Number of Fringes on 1 Meter of Screen:

To find the number of bright fringes in 1 meter of the screen:
\[ Number of fringes = \frac{1 m}{\Delta y} = \frac{1}{0.1} = 10 \]

Conclusion:
Therefore, the number of bright fringes formed due to interference on 1 meter of the screen is \boxed{10. Quick Tip: The theoretical number of fringes can be very high, but practical visibility and resolution limitations typically reduce the number that can be distinctly observed and counted on the screen.

The capacitance \( C \) of a parallel plate capacitor without any dielectric between its plates is given by: \[ C_0 = \frac{\epsilon_0 A}{d} \]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the separation between the plates.

When a dielectric medium with dielectric constant \( K \) is introduced between the plates, the capacitance increases by a factor of \( K \), because the dielectric reduces the effective electric field within the capacitor while maintaining the same charge. Thus, the new capacitance \( C \) is given by: \[ C = KC_0 = K \frac{\epsilon_0 A}{d} \]
This expression shows that the capacitance of a parallel plate capacitor is directly proportional to the dielectric constant of the medium between the plates, the permittivity of free space, and the area of the plates, and inversely proportional to the distance between the plates. Quick Tip: When inserting a dielectric material between the plates of a capacitor, the dielectric constant (\(K\)) of the material not only increases the capacitance but also affects the capacitor's ability to store electrical energy without breaking down. This is because the dielectric material reduces the electric field within the capacitor, allowing it to store more charge at the same voltage, or maintain the same charge at a lower voltage.

Given:

Charge \( Q = 6 \, \muC = 6 \times 10^{-6} \, C \)
Radius of the sphere \( R = 0.2 \, m \)


(i) Potential at the Surface of the Sphere

The electric potential \( V \) at the surface of a charged sphere is given by: \[ V = \frac{kQ}{R} \]
where \( k = \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \, N m^2/C^2 \).

Substituting the given values: \[ V = \frac{9 \times 10^9 \times 6 \times 10^{-6}}{0.2} = \frac{54 \times 10^3}{0.2} = 270 \times 10^3 \, V \] \[ V = \boxed{2.7 \times 10^5 \, V} \]

(ii) Potential at the Center of the Sphere

For a hollow metallic sphere, the potential inside the sphere (including at the center) is the same as the potential at the surface. Therefore: \[ V_{center} = V_{surface} = \boxed{2.7 \times 10^5 \, V} \] Quick Tip: Remember that the potential inside a charged conductor is uniform and equal to the potential at its surface, regardless of the shape of the conductor.

Step 1: Electric field inside the shell.
According to the properties of conductors in electrostatic equilibrium, the electric field inside a conducting shell is zero. This is because the charges reside on the surface and symmetrical distribution of charge ensures no net electric field points inside the shell.

Step 2: Electric field outside the shell.
For points outside the spherical shell, the shell can be considered as a point charge at the center for the purpose of calculating the electric field. The electric field \( E \) at a distance \( x \) from the center (where \( x > r \)) is given by Coulomb's Law: \[ E = \frac{1}{4\pi \epsilon_0} \frac{Q}{x^2} \]
where \( \epsilon_0 \) is the permittivity of free space. This formula indicates that the electric field behaves as if all the charge \( Q \) were concentrated at the center of the sphere. Quick Tip: Remember that the double electric field strength near a conducting plate compared to a nonconducting sheet is a result of the conductive property, which causes charges to redistribute and maximize the field on the exposed side. This principle is fundamental in designing effective electromagnetic shields and capacitors, where surface charge distribution plays a critical role in performance.

Step 1:
Consider a uniform surface charge density \( \sigma \) on both a conducting plate and a nonconducting sheet.

For a nonconducting sheet, the electric field at a point near the surface can be derived from Gauss's law.

Using a Gaussian pillbox with a small area \( A \) around the surface, we apply Gauss’s law:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \]
Since the sheet is nonconducting, the charge is only on one side of the sheet.

The total enclosed charge is \( \sigma A \).

The electric flux through the pillbox is:
\[ E \cdot A + E \cdot A = \frac{\sigma A}{\epsilon_0} \] \[ 2E = \frac{\sigma}{\epsilon_0} \quad \Rightarrow \quad E = \frac{\sigma}{2 \epsilon_0} \]

So, the electric field due to a nonconducting sheet with charge density \( \sigma \) is:
\[ E_{non-conducting} = \frac{\sigma}{2 \epsilon_0} \]

For a conducting plate, the situation is different because charges on a conductor move freely and spread out evenly. The field due to a conducting plate is calculated similarly using Gauss’s law. For a conducting plate, the charge distributes evenly on both sides of the plate, so the electric field is the sum of the fields from both sides. Each side contributes \( \frac{\sigma}{2 \epsilon_0} \), and thus the total electric field is:
\[ E_{conducting} = \frac{\sigma}{\epsilon_0} \]

Step 2:
We can now compare the electric fields for both cases:
For a nonconducting sheet, \( E_{non-conducting} = \frac{\sigma}{2 \epsilon_0} \).

For a conducting plate, \( E_{conducting} = \frac{\sigma}{\epsilon_0} \).

Therefore, the electric field for a conducting plate is twice that for a nonconducting sheet.

\[ E_{conducting} = 2 \cdot E_{non-conducting} \]

Thus, the electric field is twice in the case of a conducting plate compared to a nonconducting sheet. Quick Tip: Understanding the difference in electric fields between conducting and nonconducting materials is crucial in applications involving shielding and charge distribution.

N/A

Step 1: Calculate the total resistance needed for the initial voltmeter range \( 0 - V \).

The total resistance \( R_t \) for the voltmeter range \( 0 - V \) is given by:
\[ R_t = R + G \]
where \( R \) is the series resistance used to extend the range to \( V \) volts.


Step 2: Find the resistance required for the range \( 0 - V/2 \).

For a range of \( V/2 \), the total resistance needs to be higher to allow only half the voltage across the galvanometer for the same current. The new resistance \( R' \) can be calculated using the relation:
\[ V/2 = I(R' + G) \quad where \quad V = I(R + G) \]
From the first equation:
\[ I = \frac{V}{R + G} \]
Substitute \( I \) in the equation for \( V/2 \):
\[ \frac{V}{2} = \frac{V}{R + G}(R' + G) \Rightarrow 2(R' + G) = R + G \]
Solving for \( R' \) gives:
\[ R' = \frac{R + G}{2} - G = \frac{R - G}{2} \] Quick Tip: When modifying the range of a voltmeter by changing its series resistance, it’s crucial to ensure that the new resistance provides the correct voltage division to achieve the desired scale without exceeding the galvanometer's maximum current.

The induced emf (\( \mathcal{E} \)) in the coil is given by Faraday's law of induction, which states that: \[ \mathcal{E} = - \frac{d\phi}{dt} \]
where \( \phi \) is the magnetic flux. Substituting the given expression for \( \phi \), we first find the derivative of \( \phi \) with respect to time \( t \).

The given flux is: \[ \phi = 2.0 t^3 + 5.0 t^2 + 6.0 t \, mWb \]

Now, differentiate \( \phi \) with respect to \( t \): \[ \frac{d\phi}{dt} = 3 \times 2.0 t^2 + 2 \times 5.0 t + 6.0 = 6.0 t^2 + 10.0 t + 6.0 \]

At \( t = 2 \, s \), substitute the value of \( t \) into the derivative: \[ \frac{d\phi}{dt} = 6.0 \times (2)^2 + 10.0 \times (2) + 6.0 = 6.0 \times 4 + 10.0 \times 2 + 6.0 = 24 + 20 + 6 = 50.0 \, mWb/s \]

Thus, the induced emf is: \[ \mathcal{E} = - 50.0 \, mV \]

Now, using Ohm's law, the induced current \( I \) through the coil is given by: \[ I = \frac{\mathcal{E}}{R} \]
where \( R = 5 \, \Omega \) is the resistance of the coil. Substituting the values: \[ I = \frac{50.0 \times 10^{-3}}{5.0} = 10.0 \times 10^{-3} = 0.01 \, A = 10 \, mA \]

Thus, the magnitude of the induced current at \( t = 2 \, s \) is \( 10 \, mA \). Quick Tip: When calculating induced current due to changing magnetic flux, always check the units of magnetic flux and convert them to standard units (webers) if necessary to ensure the correct calculation of induced emf using Faraday's law.

At any instant of time, the magnetic flux \( \phi \) through the coil is given by: \[ \phi = N B A \cos(\theta) = N B A \cos(\omega t) \]
where:
- \( N \) is the number of turns,
- \( B \) is the magnetic field strength,
- \( A \) is the area of the coil,
- \( \omega \) is the angular velocity,
- \( t \) is the time,
- \( \theta = \omega t \) is the angle between the plane of the coil and the magnetic field at time \( t \).

Now, using Faraday's Law of Induction, the induced emf \( \mathcal{E} \) is the negative rate of change of magnetic flux: \[ \mathcal{E} = -\frac{d\phi}{dt} \]
Substitute the expression for flux: \[ \mathcal{E} = -\frac{d}{dt} \left( N B A \cos(\omega t) \right) \]
Taking the derivative of \( \cos(\omega t) \): \[ \mathcal{E} = - N B A \frac{d}{dt} \left( \cos(\omega t) \right) \] \[ \mathcal{E} = - N B A (-\omega \sin(\omega t)) \]
Simplifying: \[ \mathcal{E} = N B A \omega \sin(\omega t) \]

Thus, the induced emf at any instant of time is:
\[ \mathcal{E} = - N B A \omega \sin(\omega t) \] Quick Tip: The maximum emf is induced when the plane of the coil is perpendicular to the magnetic field, i.e., when \(\sin(\omega t) = 1\). This occurs twice during each complete rotation, once in each half-cycle.

Given:

Radius of \( L_1 \), \( r_1 = 1 \, cm = 0.01 \, m \)
Radius of \( L_2 \), \( r_2 = 100 \, cm = 1 \, m \)
\( \pi^2 = 10 \)
Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, H/m \)


The mutual inductance \( M \) between two coaxial circular loops is given by: \[ M = \frac{\mu_0 \pi r_1^2 r_2^2}{2(r_1^2 + r_2^2)^{3/2}} \]

Substitute the given values: \[ M = \frac{(4\pi \times 10^{-7}) \pi (0.01)^2 (1)^2}{2((0.01)^2 + (1)^2)^{3/2}} \]

Simplify the expression: \[ M = \frac{4\pi^2 \times 10^{-7} \times 10^{-4}}{2(0.0001 + 1)^{3/2}} \]

Calculate the denominator: \[ (0.0001 + 1)^{3/2} = (1.0001)^{3/2} \approx 1^{3/2} = 1 \]

Substitute \( \pi^2 = 10 \): \[ M = \frac{4 \times 10 \times 10^{-7} \times 10^{-4}}{2 \times 1} = \frac{4 \times 10^{-10}}{2} = 2 \times 10^{-10} \, H \]

Therefore, the mutual inductance of the loops is: \[ \boxed{2 \times 10^{-10} \, H} \] Quick Tip: Note that \(\pi^2 = 10\) significantly simplifies calculations in electromagnetic problems involving circular dimensions, especially in theoretical physics or academic exercises where approximate values are acceptable for learning concepts.

Step 1: Trace the path of the ray of light through the prism.
When a ray of light enters a prism with refractive index greater than that of the surrounding medium (typically air), it bends towards the normal at the first interface (at angle of incidence \(i\)), passes through the prism, and bends away from the normal at the second interface (at angle of exit \(e\)). The prism has an apex angle \(A\).

Step 2: Derive the formula for the angle of deviation (\(\delta\)).
The angle of deviation (\(\delta\)) is the angle by which the light ray deviates from its original direction after passing through the prism. It can be expressed in terms of the angle of incidence (\(i\)), the angle of exit (\(e\)), and the prism's apex angle (\(A\)) as follows: \[ \delta = i + e - A \]
This relationship arises because the external deviation is equal to the sum of the angles of incidence and emergence minus the angle of the prism.

Step 3: Graph the variation of \(\delta\) with \(i\).



The relationship between \(\delta\) and \(i\) is typically non-linear, showing that \(\delta\) decreases with an increase in \(i\) up to a minimum value (at the minimum deviation condition) and then increases. The graph of \(\delta\) versus \(i\) will have a "U" shape, indicating the minimum deviation occurs when the light ray passes symmetrically through the prism. Quick Tip: When plotting the graph of angle of deviation versus angle of incidence, note that the minimum angle of deviation occurs when the light path through the prism is symmetrical. This principle is used in optical instruments to achieve precise angular measurements.

Given:

Angle of incidence on the glass box: \(45^\circ\)
The emergent ray passes along the face AB, indicating that the angle of refraction at the glass-liquid interface is \(90^\circ\).


1. First Surface (Air-Glass Interface):
\[ \frac{\sin 45^\circ}{\sin \theta} = \mu \]
Given \(\sin 45^\circ = \frac{1}{\sqrt{2}}\):
\[ \frac{1}{\sqrt{2}} = \mu \sin \theta \]

2. Second Surface (Glass-Liquid Interface):
\[ \frac{\sin(90^\circ - \theta)}{\sin 90^\circ} = \frac{1}{\mu} \]
Since \(\sin(90^\circ - \theta) = \cos \theta\) and \(\sin 90^\circ = 1\):
\[ \cos \theta = \frac{1}{\mu} \]

3. Combining the Equations:
\[ \frac{1}{\sqrt{2} \sin \theta} = \frac{1}{\cos \theta} \]
Simplifying:
\[ \cos \theta = \sqrt{2} \sin \theta \]
\[ \tan \theta = \frac{1}{\sqrt{2}} \]

4. Finding \(\sin \theta\):
From the triangle GEF:
\[ \sin \theta = \frac{1}{\sqrt{3}} \]

5. Calculating the Refractive Index (\(\mu\)):
\[ \mu = \frac{1}{\sqrt{2} \sin \theta} = \frac{1}{\sqrt{2} \times \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}} \]

Therefore, the refractive index of the liquid is: \[ \boxed{\sqrt{\frac{3}{2}}} \] Quick Tip: The critical angle condition and refractive index relationship are key to understanding the behavior of light in different mediums, especially in designing optical instruments and understanding phenomena like total internal reflection.

Step 1: Express the resultant displacement.
When two waves superpose, the resultant displacement \(y\) is the sum of the individual displacements: \[ y = y_1 + y_2 = a \cos(\omega t) + a \cos(\omega t + \phi) \]
Using the trigonometric identity for the sum of cosines: \[ y = 2a \cos\left(\frac{\phi}{2}\right) \cos\left(\omega t + \frac{\phi}{2}\right) \]

Step 2: Determine the resultant intensity.
Intensity is proportional to the square of the amplitude of the wave. The amplitude of the resultant wave is \(2a \cos\left(\frac{\phi}{2}\right)\), so the intensity \(I\) is given by: \[ I = k \left(2a \cos\left(\frac{\phi}{2}\right)\right)^2 = 4ka^2 \cos^2\left(\frac{\phi}{2}\right) \]
where \(k\) is a constant of proportionality. Quick Tip: Remember that the maximum intensity occurs when \(\phi = 0\) (in phase) and the minimum intensity (possibly zero) occurs when \(\phi = \pi\) (out of phase), demonstrating constructive and destructive interference respectively.

N/A Quick Tip: In Young's double slit experiment, slight changes in the path difference lead to significant changes in intensity due to the sensitive dependence on the cosine squared relationship.