Step 1: Understanding Electric Flux.

Electric flux is the product of the electric field \( E \) and the area \( A \) through which the field lines pass, expressed as \( \Phi_E = E \times A \).

The unit of the electric field is \( N/C \) (Newton per Coulomb), and the unit of area is \( m^2 \). Therefore, the unit of electric flux becomes \( N \cdot m^2 \cdot C^{-1} \), which is also written as \( Nm^2C^{-1} \).


Step 2: Conclusion.

Thus, the correct unit for electric flux is \( Nm^2C^{-1} \), corresponding to option (ii). Quick Tip: The unit of electric flux is \( Nm^2C^{-1} \), as it is the product of electric field and area.

Step 1: Sensitivity of the Galvanometer.

The sensitivity of a galvanometer is directly related to both the number of turns in the coil and the area of the coil. By increasing either the number of turns or the area, the sensitivity of the galvanometer can be enhanced.


Step 2: Conclusion.

Hence, the sensitivity of a moving coil galvanometer can be increased by increasing the area of the coil. Therefore, the correct answer is option (iii). Quick Tip: Increasing the area of the coil increases the sensitivity of the moving coil galvanometer.

Step 1: Resistance of an Ideal p-n Junction.

For an ideal p-n junction in forward bias, the resistance is zero, allowing current to flow freely. In reverse bias, the resistance is infinite since no current passes through.


Step 2: Conclusion.

Therefore, the resistance of an ideal p-n junction is zero under forward bias and infinite under reverse bias. The correct answer is option (iv). Quick Tip: An ideal p-n junction has zero resistance in forward bias and infinite resistance in reverse bias.

Step 1: Matter Waves and Their Properties.

According to de Broglie, matter waves are associated with particles, and their wavelength is given by \( \lambda = \frac{h}{p} \), where \( h \) represents Planck's constant and \( p \) is the particle's momentum. The wavelength of matter waves is independent of the particle's charge and is solely dependent on its momentum.


Step 2: Conclusion.

Thus, the correct conclusion is that the wavelength of matter waves does not depend on the particle's charge. Therefore, the correct answer is option (iii). Quick Tip: Matter waves' wavelength is dependent on the momentum of the particle and not its charge.

Step 1: Angular Momentum of Electron.

The angular momentum \( L \) of an electron in a Bohr orbit is quantized and can be expressed as: \[ L = n \hbar \]
where \( n \) is the principal quantum number and \( \hbar \) is the reduced Planck's constant, given by \( \hbar = \frac{h}{2\pi} \).

For the second orbit (\( n = 2 \)) of the hydrogen atom, the angular momentum is: \[ L = 2 \times \frac{h}{2\pi} \]
Substituting \( h = 6.626 \times 10^{-34} \, J \cdot s \), we get: \[ L = 2 \times \frac{6.626 \times 10^{-34}}{2\pi} = 2.1 \times 10^{-34} \, J \cdot s \]

Step 2: Conclusion.

Thus, the angular momentum of the electron in the second orbit is \( 2.1 \times 10^{-34} \, J \cdot s \), which corresponds to option (iii). Quick Tip: The angular momentum of an electron in a Bohr orbit is quantized and given by \( L = n \hbar \), where \( n \) is the principal quantum number.

Step 1: Understanding the Focal Length and Focus of a Lens.

The focal length \( f \) of a convex lens is the distance from the lens to its focus. The first and second foci of the lens are located symmetrically on either side of the lens. Here, the given focal length of the lens is 10 cm.

Step 2: Object Placement and Image Formation.

The object is placed at a distance of 5 cm from the first focus. The total distance between the object and the lens is the sum of the distance from the first focus to the lens and the distance from the lens to the object: \[ Distance from lens to object = 5 \, cm + 10 \, cm = 15 \, cm \]

Using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. Since the object distance \( u = -15 \, cm \) (negative because the object is on the same side as the incoming light), and \( f = 10 \, cm \), we can substitute into the formula: \[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-15} \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{15} \]
Solving for \( v \): \[ \frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \] \[ v = 30 \, cm \]

Step 3: Conclusion.

Thus, the distance of the image from the second focus is 15 cm, corresponding to option (ii).
Quick Tip: For a thin convex lens, the total distance between the object and the lens is the sum of the distances from the first focus to the lens and from the lens to the object.

Step 1: Symbol of NOR Gate.

The symbol of a NOR gate is a combination of an OR gate with a NOT gate. It has two or more inputs and one output, and the output is true only when all inputs are false.


Truth Table of NOR Gate:


\begin{tabular{|c|c|c|
\hline
Input A & Input B & Output (A NOR B)

\hline
0 & 0 & 1

0 & 1 & 0

1 & 0 & 0

1 & 1 & 0

\hline
\end{tabular Quick Tip: In a NOR gate, the output is true (1) only when all inputs are false (0).

Step 1: Photon Momentum Formula.

The momentum \( p \) of a photon is related to its energy and wavelength. The formula is: \[ p = \frac{E}{c} = \frac{h}{\lambda} \]
where:
- \( p \) is the momentum of the photon,
- \( E \) is the energy of the photon,
- \( c \) is the speed of light,
- \( h \) is Planck's constant,
- \( \lambda \) is the wavelength of the photon. Quick Tip: The momentum of a photon is inversely proportional to its wavelength.

Step 1: Chain Reaction in Nuclear Bombs.

In nuclear bombs, the chain reaction is uncontrolled, and the fission of atoms releases a huge amount of energy in a very short time, resulting in an explosion.

Step 2: Chain Reaction in Nuclear Reactors.

In nuclear reactors, the chain reaction is controlled. The fission of atoms is moderated by control rods that absorb neutrons, ensuring a steady release of energy for power generation. Quick Tip: In nuclear bombs, the chain reaction is uncontrolled, whereas in nuclear reactors, it is carefully controlled to ensure safety and efficiency.

Step 1: Relationship Between Electric and Magnetic Fields.

For an electromagnetic wave, the electric field \( E \) and magnetic field \( B \) are related by the following equation: \[ B = \frac{E}{c} \]
where \( c \) is the speed of light. The magnetic field oscillates perpendicular to both the electric field and the direction of propagation.

Step 2: Equation for the Magnetic Field.

Given \( E_y = 60 \sin(500x - 1.5 \times 10^{11}t) \), we can write the magnetic field equation as: \[ B_z = \frac{60}{3 \times 10^8} \sin(500x - 1.5 \times 10^{11}t) \]
Simplifying this gives: \[ B_z = 2 \times 10^{-7} \sin(500x - 1.5 \times 10^{11}t) \, T \] Quick Tip: The magnetic field in an electromagnetic wave is always perpendicular to the electric field and the direction of wave propagation.

Step 1: Phase Difference in an Inductive Circuit.

In a pure inductive alternating circuit, the current lags the voltage by \( 90^\circ \). The voltage leads the current. Quick Tip: In a pure inductive circuit, the current lags the voltage by \( 90^\circ \).

Step 1: Total Resistance in the Circuit.

The total resistance \( R_{total} \) is the sum of the internal resistance of the cell and the external resistance: \[ R_{total} = 0.1 \, \Omega + 2.9 \, \Omega = 3 \, \Omega \]

Step 2: Current in the Circuit.

Using Ohm's law, the current \( I \) in the circuit is given by: \[ I = \frac{V}{R_{total}} = \frac{1.5 \, V}{3 \, \Omega} = 0.5 \, A \]

Step 3: Potential Difference Across the External Resistance.

The potential difference across the external resistance is given by: \[ V_{external} = I \times R_{external} = 0.5 \, A \times 2.9 \, \Omega = 1.45 \, V \]

Step 4: Conclusion.

The potential difference at the terminals of the cell is: \[ V_{terminals} = 1.5 \, V - (0.5 \, A \times 0.1 \, \Omega) = 1.5 \, V - 0.05 \, V = 1.45 \, V \] Quick Tip: The potential difference at the terminals of the cell is calculated by subtracting the voltage drop across the internal resistance from the e.m.f. of the cell.

Step 1: Definition of Binding Energy.

The binding energy of a nucleus is the energy required to disassemble a nucleus into its constituent protons and neutrons. It is a measure of the stability of the nucleus. The binding energy is related to the mass defect, which is the difference in mass between the nucleus and its constituent particles.

Step 2: Formula for Binding Energy.

The binding energy \( E_b \) is given by the relation: \[ E_b = \Delta m c^2 \]
where:
- \( \Delta m \) is the mass defect,
- \( c \) is the speed of light (\( 3 \times 10^8 \, m/s \)).

Step 3: Conversion of Mass Defect to Energy.

Given mass defect \( \Delta m = 10^{-6} \, kg \), we can calculate the binding energy in joules: \[ E_b = 10^{-6} \times (3 \times 10^8)^2 = 9 \times 10^{10} \, J \]

Step 4: Conversion to Electron Volts.

To convert joules to electron volts, use the conversion factor \( 1 \, J = 6.242 \times 10^{12} \, eV \): \[ E_b = 9 \times 10^{10} \, J \times 6.242 \times 10^{12} \, eV/J = 5.6178 \times 10^{23} \, eV \] Quick Tip: The binding energy of a nucleus is the energy required to break it into individual nucleons, and it is related to the mass defect by \( E_b = \Delta m c^2 \).

Step 1: Ampere's Circuital Law.

Ampere's circuital law states that the line integral of the magnetic field \( B \) around a closed loop is proportional to the total current \( I_{enc} \) passing through the loop. The law is given by: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( I_{enc} \) is the enclosed current.

Step 2: Magnetic Field due to a Straight Conductor.

Consider a long straight conductor carrying a current \( I \). We choose a circular path of radius \( r \) around the wire. Since the magnetic field due to a straight conductor is radially symmetric, the magnetic field at a distance \( r \) from the wire is tangential to the circle.

The line integral of the magnetic field around the circular path is: \[ B \times 2\pi r = \mu_0 I \]

Solving for \( B \), we get the magnetic field at a distance \( r \) from the conductor: \[ B = \frac{\mu_0 I}{2\pi r} \] Quick Tip: For a long straight conductor carrying current \( I \), the magnetic field at a distance \( r \) from the conductor is given by \( B = \frac{\mu_0 I}{2\pi r} \).

Step 1: Isotopic Nuclei.

Isotopic nuclei are nuclei that have the same number of protons (atomic number) but different numbers of neutrons, leading to different mass numbers. These nuclei belong to the same element but have different isotopes.

Step 2: Isobaric Nuclei.

Isobaric nuclei are nuclei that have the same mass number (total number of protons and neutrons) but different numbers of protons. These nuclei belong to different elements.

Step 3: Difference between Isotopic and Isobaric Nuclei.

- Isotopic nuclei have the same atomic number but different mass numbers.
- Isobaric nuclei have the same mass number but different atomic numbers. Quick Tip: Isotopic nuclei have the same atomic number, while isobaric nuclei have the same mass number.

Step 1: Photoelectric Equation.

The energy of the incident photon is given by: \[ E_{photon} = \frac{hc}{\lambda} \]
where:
- \( h = 6.626 \times 10^{-34} \, J \cdot s \) is Planck’s constant,
- \( c = 3 \times 10^8 \, m/s \) is the speed of light,
- \( \lambda = 5000 \, \overset{\small o}{A} = 5000 \times 10^{-10} \, m \) is the wavelength of light.
\[ E_{photon} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} = 3.976 \times 10^{-19} \, J \]

Step 2: Maximum Kinetic Energy of Photoelectrons.

The maximum kinetic energy of the emitted photoelectrons is given by: \[ K_{max} = E_{photon} - \phi \]
where \( \phi = 2.14 \, eV = 2.14 \times 1.602 \times 10^{-19} \, J = 3.429 \times 10^{-19} \, J \) is the work function.
\[ K_{max} = 3.976 \times 10^{-19} - 3.429 \times 10^{-19} = 5.47 \times 10^{-20} \, J \]

Step 3: Maximum Velocity of Photoelectrons.

The maximum kinetic energy of the photoelectrons is related to their velocity by the equation: \[ K_{max} = \frac{1}{2} m v_{max}^2 \]
where \( m = 9.11 \times 10^{-31} \, kg \) is the mass of the electron. Solving for \( v_{max} \): \[ v_{max} = \sqrt{\frac{2 K_{max}}{m}} = \sqrt{\frac{2 \times 5.47 \times 10^{-20}}{9.11 \times 10^{-31}}} = 3.57 \times 10^5 \, m/s \] Quick Tip: The maximum kinetic energy of emitted photoelectrons is the difference between the energy of the incident photon and the work function of the metal.

Step 1: Electric Dipole and Dipole Moment.

An electric dipole consists of two equal and opposite charges separated by a small distance. The dipole moment \( \vec{p} \) is given by: \[ \vec{p} = q \times \vec{d} \]
where:
- \( q \) is the charge,
- \( \vec{d} \) is the displacement vector between the charges.

Step 2: Potential Energy of the Dipole.

The potential energy \( U \) of a dipole in a uniform electric field \( E \) is given by: \[ U = - \vec{p} \cdot \vec{E} \]
where \( \vec{p} \) is the dipole moment and \( \vec{E} \) is the electric field.

For \( \theta = 30^\circ \), \[ U = - p E \cos(\theta) \]
Substitute the values: \[ U = - (2 \times 10^{-8} \, C \cdot m) \times (2 \times 10^5 \, V/m) \times \cos(30^\circ) \] \[ U = - 2 \times 10^{-8} \times 2 \times 10^5 \times \frac{\sqrt{3}}{2} \] \[ U = - 1.732 \times 10^{-3} \, J \]

Step 3: Moment of the Couple.

The moment \( \tau \) of the couple acting on the dipole is given by: \[ \tau = p E \sin(\theta) \]
Substitute the values: \[ \tau = (2 \times 10^{-8} \, C \cdot m) \times (2 \times 10^5 \, V/m) \times \sin(30^\circ) \] \[ \tau = 2 \times 10^{-8} \times 2 \times 10^5 \times \frac{1}{2} \] \[ \tau = 2 \times 10^{-3} \, N \cdot m \] Quick Tip: The potential energy of an electric dipole in a uniform electric field is given by \( U = - p E \cos(\theta) \), and the moment of the couple is \( \tau = p E \sin(\theta) \).

Step 1: Lens Maker's Formula.

The lens maker’s formula relates the focal length \( f \) of a thin lens to the refractive index of the lens material \( n \) and the radii of curvature of the two surfaces \( R_1 \) and \( R_2 \). The formula is given by: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \( n \) is the refractive index of the lens material,
- \( R_1 \) is the radius of curvature of the first lens surface,
- \( R_2 \) is the radius of curvature of the second lens surface.

Step 2: Effect of Refractive Index.

The focal length \( f \) is directly proportional to the refractive index \( n \). As the refractive index increases, the focal length of the lens decreases, making the lens more powerful (shorter focal length).

Step 3: Effect of Radius of Curvature.

The focal length \( f \) is also directly proportional to the radii of curvature \( R_1 \) and \( R_2 \). If the radii of curvature increase, the focal length increases, making the lens less powerful (longer focal length). Quick Tip: A higher refractive index and larger radii of curvature result in a longer focal length, whereas a smaller refractive index and smaller radii lead to a shorter focal length.

Step 1: Coefficient of Self-Induction.

The coefficient of self-induction \( L \) of a coil is a measure of the coil’s ability to induce an electromotive force (e.m.f.) in itself when the current passing through it changes. It is defined by the relation: \[ L = \frac{Induced e.m.f.}{Rate of change of current} \]
where \( L \) is the self-inductance of the coil.

Step 2: Coefficient of Mutual Induction.

The coefficient of mutual induction \( M \) of two coils is a measure of the ability of one coil to induce an e.m.f. in the other coil when the current in the first coil changes. It is defined by: \[ M = \frac{Induced e.m.f. in coil 2}{Rate of change of current in coil 1} \]

Step 3: Induced e.m.f. in the Coil.

The induced e.m.f. in a coil due to a change in current is given by Faraday's law of induction: \[ \mathcal{E} = - L \frac{\Delta I}{\Delta t} \]
where:
- \( L = 5 \, H \) is the self-inductance,
- \( \Delta I = 5 \, A \) (current decreases from 5 A to 0),
- \( \Delta t = 0.1 \, s \).

Substituting the values: \[ \mathcal{E} = - 5 \times \frac{5}{0.1} = - 250 \, V \]

Thus, the induced e.m.f. is 250 V (negative sign indicates the direction of the induced e.m.f. opposing the change in current). Quick Tip: The induced e.m.f. in a coil is proportional to the rate of change of current and the self-inductance of the coil, as given by Faraday’s law of induction.

Step 1: Formula for the Capacitance of a Parallel Plate Capacitor.

The capacitance \( C \) of a parallel plate capacitor is given by the formula: \[ C = \frac{\epsilon_0 A}{d} \]
where:
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, C^2 / N \cdot m^2 \)),
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.

Step 2: Formula for the Capacitance of a Charged Sphere.

The capacitance of a charged sphere of radius \( r \) is given by: \[ C = 4 \pi \epsilon_0 r \]
where \( r = 1 \, m \).

Step 3: Setting Capacitances Equal.

We are given that the capacitance of the parallel plate capacitor is equal to the capacitance of the charged sphere. So, we set the capacitance formulas equal: \[ \frac{\epsilon_0 A}{d} = 4 \pi \epsilon_0 r \]

Step 4: Solving for Distance.

The area \( A \) of the plates of the parallel plate capacitor is given by \( A = \pi R^2 \), where \( R = 3 \times 10^{-2} \, m \) is the radius of the plates. Substituting into the equation: \[ \frac{\epsilon_0 \pi R^2}{d} = 4 \pi \epsilon_0 r \]
Simplifying and solving for \( d \): \[ \frac{R^2}{d} = 4r \] \[ d = \frac{R^2}{4r} \]
Substituting the given values: \[ d = \frac{(3 \times 10^{-2})^2}{4 \times 1} = \frac{9 \times 10^{-4}}{4} = 2.25 \times 10^{-4} \, m \]

Step 5: Conclusion.

Thus, the distance between the plates of the capacitor is \( 2.25 \times 10^{-4} \, m \). Quick Tip: The capacitance of a parallel plate capacitor depends on the area of the plates and the distance between them. Increasing the area or reducing the distance increases the capacitance.

Step 1: Relationship Between Temperature and Resistance.

The resistance \( R \) of a conductor at a temperature \( T \) is related to its resistance at a reference temperature \( T_0 \) by: \[ R = R_0 (1 + \alpha (T - T_0)) \]
where:
- \( R_0 \) is the resistance at the reference temperature \( T_0 \),
- \( \alpha \) is the temperature coefficient of resistance,
- \( T \) is the temperature.

Step 2: Relationship Between Temperature and Resistivity.

The resistivity \( \rho \) of a conductor also depends on temperature, and it is given by: \[ \rho = \rho_0 (1 + \alpha (T - T_0)) \]
where:
- \( \rho_0 \) is the resistivity at the reference temperature \( T_0 \).

Step 3: Temperature Coefficient of Resistance.

The temperature coefficient of resistance \( \alpha \) is a constant that indicates how much the resistance changes with temperature. It is defined as: \[ \alpha = \frac{1}{R_0} \frac{dR}{dT} \]
Its unit is \( °C^{-1} \). Quick Tip: The resistance and resistivity of most materials increase with temperature. The temperature coefficient of resistance determines the rate of this change.

Step 1: Photoelectric Effect.

The photoelectric effect refers to the phenomenon where electrons are ejected from a material (usually a metal) when it is exposed to light or other electromagnetic radiation of sufficient frequency. The energy of the incident photons is transferred to the electrons, causing them to be ejected from the surface of the material.

Step 2: Energy of Incident Photon.

The energy of a photon is given by: \[ E_{photon} = h f \]
where:
- \( h = 6.626 \times 10^{-34} \, J \cdot s \) is Planck’s constant,
- \( f \) is the frequency of the incident light.

For the incident light frequency \( f = 8.2 \times 10^{14} \, Hz \), the energy is: \[ E_{photon} = (6.626 \times 10^{-34}) \times (8.2 \times 10^{14}) = 5.43 \times 10^{-19} \, J \]

Step 3: Work Function of the Surface.

The work function \( \phi \) is the minimum energy required to release an electron from the metal surface. The threshold frequency \( f_0 \) is the minimum frequency required to eject an electron. The work function is related to the threshold frequency by: \[ \phi = h f_0 \]
Given \( f_0 = 3.3 \times 10^{14} \, Hz \), the work function is: \[ \phi = (6.626 \times 10^{-34}) \times (3.3 \times 10^{14}) = 2.19 \times 10^{-19} \, J \]

Step 4: Maximum Kinetic Energy of the Emitted Electrons.

The maximum kinetic energy \( K_{max} \) of the emitted photoelectrons is given by: \[ K_{max} = E_{photon} - \phi \]
Substituting the values: \[ K_{max} = 5.43 \times 10^{-19} \, J - 2.19 \times 10^{-19} \, J = 3.24 \times 10^{-19} \, J \]

Step 5: Stopping Potential.

The stopping potential \( V_s \) is related to the maximum kinetic energy by: \[ K_{max} = e V_s \]
where \( e = 1.6 \times 10^{-19} \, C \) is the charge of an electron. Solving for \( V_s \): \[ V_s = \frac{K_{max}}{e} = \frac{3.24 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.025 \, V \] Quick Tip: The stopping potential is the potential required to stop the most energetic photoelectrons, and it is directly related to the maximum kinetic energy of the emitted electrons.

Step 1: Photoelectric Effect.

The photoelectric effect refers to the emission of electrons from a metal surface when it is illuminated by light of sufficient frequency. The energy of the incident photon is absorbed by the electron, and if the energy is enough, the electron is ejected.

Step 2: Energy of Incident Photon.

The energy of a photon is given by: \[ E_{photon} = h f \]
where:
- \( h = 6.626 \times 10^{-34} \, J \cdot s \) is Planck's constant,
- \( f \) is the frequency of the light.

For \( f = 8.2 \times 10^{14} \, Hz \), we get: \[ E_{photon} = (6.626 \times 10^{-34}) \times (8.2 \times 10^{14}) = 5.43 \times 10^{-19} \, J \]

Step 3: Work Function and Stopping Potential.

The work function \( \phi \) is the minimum energy required to remove an electron from the metal surface. It is related to the threshold frequency \( f_0 \) by: \[ \phi = h f_0 \]
Substituting the threshold frequency \( f_0 = 3.3 \times 10^{14} \, Hz \): \[ \phi = (6.626 \times 10^{-34}) \times (3.3 \times 10^{14}) = 2.19 \times 10^{-19} \, J \]

The maximum kinetic energy of the emitted photoelectron is: \[ K_{max} = E_{photon} - \phi = 5.43 \times 10^{-19} - 2.19 \times 10^{-19} = 3.24 \times 10^{-19} \, J \]

The stopping potential \( V_s \) is related to the maximum kinetic energy by: \[ K_{max} = e V_s \]
where \( e = 1.6 \times 10^{-19} \, C \) is the charge of an electron. Solving for \( V_s \): \[ V_s = \frac{K_{max}}{e} = \frac{3.24 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.025 \, V \] Quick Tip: The stopping potential is the potential required to stop the most energetic photoelectrons, and it is directly related to the maximum kinetic energy of the emitted electrons.

Step 1: Gauss's Theorem of Electrostatics.

Gauss's theorem states that the electric flux \( \Phi_E \) through a closed surface is proportional to the charge \( Q_{enc} \) enclosed within the surface. Mathematically, it is expressed as: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \]
where:
- \( \vec{E} \) is the electric field,
- \( d\vec{A} \) is the differential area vector on the closed surface,
- \( \epsilon_0 \) is the permittivity of free space.

Step 2: Equipotential Surface.

An equipotential surface is a surface on which the potential is constant at every point. No work is required to move a charge along an equipotential surface because the potential difference between any two points on the surface is zero. The electric field is always perpendicular to the equipotential surfaces.

Step 3: Electric Field Intensity at Points P and Q.

Given two insulator plates with equal charge density, the electric field between them is uniform. For an infinite plane sheet with charge density \( \sigma \), the electric field \( E \) at a distance \( r \) from the plate is given by: \[ E = \frac{\sigma}{2\epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

For points \( P \) and \( Q \), if they lie between the two plates, the electric field at both points will be the same in magnitude and direction, assuming the plates are infinitely large. The field will be directed from the positive plate towards the negative plate. Quick Tip: The electric field due to a uniformly charged infinite plane sheet is constant and does not depend on the distance from the sheet.

Step 1: Principle of Potentiometer.

A potentiometer is an instrument used to measure the potential difference (voltage) across two points in a circuit. The working principle is based on the fact that the potential difference is directly proportional to the length of the wire through which the current flows. A uniform wire of known resistance is used, and the voltage drop along the wire is measured using a galvanometer.

The principle is: \[ V = \frac{I R}{L} \times L \]
where \( V \) is the potential difference, \( I \) is the current through the wire, \( R \) is the resistance of the wire, and \( L \) is the length of the wire.

Step 2: Why Potentiometer is Superior to a Voltmeter.

- A voltmeter measures the potential difference directly by drawing current, which can alter the voltage in the circuit.
- A potentiometer, on the other hand, does not draw current from the circuit under test, so it does not alter the potential difference, providing more accurate measurements.

Step 3: Comparing e.m.f. of Two Cells Using a Potentiometer.

To compare the e.m.f. of two cells using a potentiometer, we use the following method:
1. Connect the two cells in separate circuits with the same current through the potentiometer.
2. Adjust the sliding contact to measure the potential drop across each cell.
3. The e.m.f. of the cells can be compared by the lengths of the wire that correspond to the potential difference for each cell.

Step 4: Circuit Diagram.

The circuit diagram for comparing the e.m.f. of two cells using a potentiometer is shown below:


% Replace with your image file Quick Tip: A potentiometer is a precise device for measuring voltage without drawing current from the circuit, unlike a voltmeter which can alter the potential.

Step 1: Huygens' Principle.

Huygens' principle states that every point on a wavefront of light can be considered as a source of secondary wavelets. The new position of the wavefront at any subsequent time is the envelope of these secondary wavelets. This principle helps explain phenomena such as reflection, refraction, and diffraction.

Step 2: Success of Huygens' Principle.

Huygens' principle successfully explains the propagation of light waves in terms of wavefronts and their movement over time. It is particularly useful in understanding diffraction and interference patterns.

Step 3: Failure of Huygens' Principle.

Huygens' principle fails when applied to the behavior of light in the presence of phenomena that require a more detailed understanding of the nature of light, such as the quantum theory of light or the photoelectric effect.

Step 4: Wavefront of Light from a Convex Lens.

When a point source of light is placed at the focus of a convex lens, the light emerging from the lens will form parallel rays. The wavefront of light emerging from the lens is a plane wavefront, as the lens focuses the light into parallel rays.


% Replace with your image file Quick Tip: Huygens' principle is used to explain the propagation of light waves, and it works successfully in most optical phenomena except in cases involving quantum mechanics.

Step 1: Meaning of Interference.

Interference is the phenomenon that occurs when two or more waves overlap and combine to form a resultant wave. This can lead to constructive interference (when the waves add up) or destructive interference (when the waves cancel out).

Step 2: Deriving the Expression for Intensity.

The total displacement \( y \) due to the interference of two waves \( y_1 \) and \( y_2 \) is given by: \[ y = y_1 + y_2 = a_1 \sin(\omega t) + a_2 \sin(\omega t + \phi) \]
Using the trigonometric identity: \[ \sin(A) + \sin(B) = 2 \sin\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right) \]
we get: \[ y = 2 \sin\left( \omega t + \frac{\phi}{2} \right) \cos\left( \frac{\phi}{2} \right) \]
The intensity \( I \) of the wave is proportional to the square of the amplitude, so: \[ I \propto y^2 = 4 a_1 a_2 \cos^2\left( \frac{\phi}{2} \right) \]

Step 3: Maximum and Minimum Intensity.

- For maximum intensity, \( \cos^2\left( \frac{\phi}{2} \right) = 1 \), so the maximum intensity \( I_{max} \) is: \[ I_{max} \propto 4 a_1 a_2 \]
- For minimum intensity, \( \cos^2\left( \frac{\phi}{2} \right) = 0 \), so the minimum intensity \( I_{min} \) is: \[ I_{min} \propto 0 \]

Step 4: Ratio of Maximum and Minimum Intensities.

The ratio of maximum to minimum intensities is: \[ \frac{I_{max}}{I_{min}} = \frac{4 a_1 a_2}{0} = \infty \]
Thus, the intensity can vary from maximum to zero, depending on the phase difference \( \phi \). Quick Tip: The intensity in interference phenomena depends on the phase difference between the waves. Maximum intensity occurs when the waves are in phase, and minimum intensity occurs when the waves are out of phase.

Step 1: Light Emitting Diodes (LEDs).

A Light Emitting Diode (LED) is a semiconductor device that emits light when an electric current passes through it. LEDs are made of materials that allow current to flow in only one direction and produce light by electroluminescence.

Step 2: Principle and Working of LED.

When a forward voltage is applied to an LED, electrons recombine with holes in the semiconductor material. As the electrons drop to lower energy levels, they release energy in the form of photons, producing light. The color of the light depends on the energy band gap of the semiconductor material.

Step 3: Advantages Over Traditional Filament Lamps.

LEDs are more energy efficient than traditional filament lamps because they produce less heat and convert more electrical energy into light. They also have a longer lifespan and are more durable. Unlike filament lamps, LEDs do not burn out or break easily. Moreover, LEDs are smaller in size, which allows for more compact lighting solutions. Quick Tip: LEDs are energy-efficient, have a long lifespan, and emit less heat compared to traditional filament lamps, making them ideal for modern lighting applications.

Step 1: Ray Diagram for an Astronomical Telescope.

In an astronomical telescope, the objective lens forms a real and inverted image of a distant object at its focal point. The eyepiece is then used to view the image at infinity, resulting in a final image that is virtual, erect, and at infinity.


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Step 2: Magnification of the Telescope.

The magnification \( M \) of an astronomical telescope when the final image is formed at infinity is given by: \[ M = \frac{Height of image}{Height of object} = \frac{f_{objective}}{f_{eyepiece}} \]
where \( f_{objective} \) is the focal length of the objective and \( f_{eyepiece} \) is the focal length of the eyepiece.

Step 3: Height of the Image.

We are given the height of the building (object) as 100 m and the distance as 2 km (which is much greater than the focal length of the telescope, so the object is considered to be at infinity for practical purposes). The magnification is then: \[ M = \frac{f_{objective}}{f_{eyepiece}} \]
Using the given focal length of the objective \( f_{objective} = 150 \, cm = 1.5 \, m \), and assuming the eyepiece has a much shorter focal length, the magnification can be calculated. For simplicity, assume that the magnification is large enough to give an easily observable image.

The height of the image will be: \[ Height of image = M \times Height of object \]

Given the magnification, you would substitute the known values to get the height of the image. Quick Tip: The magnification of an astronomical telescope is determined by the ratio of the focal lengths of the objective and the eyepiece, and it gives the size of the image relative to the object.

Step 1: Biot-Savart Law.

The Biot-Savart law gives the expression for the magnetic field produced by a small current element. It is given by: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I \, d\vec{l} \times \hat{r}}{r^2} \]
where:
- \( d\vec{B} \) is the infinitesimal magnetic field produced by the current element,
- \( I \) is the current through the conductor,
- \( d\vec{l} \) is the infinitesimal length vector of the current element,
- \( \hat{r} \) is the unit vector from the current element to the point of observation,
- \( r \) is the distance from the current element to the point of observation,
- \( \mu_0 \) is the permeability of free space.

Step 2: Magnetic Field Due to a Current Carrying Circular Loop.

The magnetic field at the center of a circular loop of radius \( r \), carrying a current \( I \), is given by the formula: \[ B = \frac{\mu_0 I}{2r} \]
This expression is derived using the Biot-Savart law by integrating over the entire loop.

Step 3: Magnetic Moment of the Current Loop.

The magnetic moment \( \mu \) of a current loop is given by: \[ \mu = I \cdot A \]
where:
- \( I \) is the current,
- \( A \) is the area of the loop.

For a circular loop, the area \( A = \pi r^2 \), so the magnetic moment becomes: \[ \mu = I \cdot \pi r^2 \] Quick Tip: The magnetic field at the center of a circular current loop is \( B = \frac{\mu_0 I}{2r} \), and the magnetic moment is \( \mu = I \cdot \pi r^2 \).

Step 1: Relationship Between Root-Mean-Square (rms) Value and Peak Value.

The root-mean-square (rms) value \( V_{rms} \) of an alternating voltage is related to the peak value \( V_{max} \) by: \[ V_{rms} = \frac{V_{max}}{\sqrt{2}} \]
This relation holds for both current and voltage in an AC circuit.

Step 2: Inductive Reactance.

The inductive reactance \( X_L \) is given by the formula: \[ X_L = 2 \pi f L \]
where:
- \( f \) is the frequency of the AC supply,
- \( L \) is the inductance of the coil.

Given that \( L = 8.1 \, mH = 8.1 \times 10^{-3} \, H \), we can calculate \( X_L \) if the frequency \( f \) is provided.

Step 3: Potential Difference Across the Resistance.

The total impedance \( Z \) in an R-L circuit is given by: \[ Z = \sqrt{R^2 + X_L^2} \]
where:
- \( R = 10 \, \Omega \) is the resistance,
- \( X_L \) is the inductive reactance.

The potential difference across the resistance is given by: \[ V_R = I \cdot R \]
where \( I \) is the current in the circuit, which can be calculated using the rms voltage and the total impedance: \[ I = \frac{V_{rms}}{Z} \]

Substitute the values to find the potential difference across the resistance. Quick Tip: The root-mean-square value of an alternating voltage is \( \frac{V_{max}}{\sqrt{2}} \), and the total impedance in an R-L circuit is \( Z = \sqrt{R^2 + X_L^2} \).