Step 1: Formula for Electric Potential Energy.

The formula for the electric potential energy of two point charges is given by: \[ U = \frac{{k \cdot q_1 \cdot q_2}}{{r}} \]
where \( k = 9 \times 10^9 \, Nm^2/C^2 \) is Coulomb's constant, \( q_1 = q_2 = 1 \, \mu C = 1 \times 10^{-6} \, C \), and \( r = 1 \, m \).


Step 2: Substituting the values.
\[ U = \frac{{9 \times 10^9 \cdot (1 \times 10^{-6})^2}}{{1}} = \frac{{9 \times 10^9 \cdot 10^{-12}}}{{1}} = 9 \times 10^{-3} \, joule \]

Conclusion: The electric potential energy of the system is \( 9 \times 10^{-3} \) joule.
Quick Tip: The electric potential energy between two charges is directly proportional to the product of their charges and inversely proportional to the distance between them.

Step 1: Definition of capacitance.

Capacitance is the ability of a system to store charge per unit potential difference. The unit of capacitance is the farad (F), which is defined as: \[ 1 \, F = \frac{{1 \, coulomb}}{{1 \, volt}} \]

Step 2: Understanding the units.

The unit of capacitance is \(\frac{{coulomb}}{{volt}}\), which is choice (D).


Step 3: Elimination of options.

- (A) coulomb: Incorrect, unit of charge, not capacitance.

- (B) ampere: Incorrect, unit of current.

- (C) volt: Incorrect, unit of potential difference.

- (D) coulomb/volt: Correct, unit of capacitance.


Conclusion: The unit of capacitance is coulomb/volt.
Quick Tip: Capacitance is measured in farads (F), and it is the charge stored per unit voltage across the capacitor: \( 1 \, F = 1 \, \frac{{C}}{{V}} \).

Step 1: Understanding capacitance.

The capacity (or capacitance) of a condenser depends on the geometry and separation of the plates. Specifically, capacitance is directly proportional to the area of the plates and inversely proportional to the distance between them.


Step 2: Impact of charge.

While the charge on the plates affects the energy stored in the condenser, it does not affect its capacitance. The capacitance is determined by the properties of the material and geometry, not the charge.


Step 3: Elimination.

- (A) shape of plates: The shape of the plates does affect the capacitance.

- (B) size of plates: Larger plates increase capacitance.

- (C) charge on plates: Correct, the capacitance does not depend on the charge.

- (D) distance between plates: Increasing the distance reduces the capacitance.


Step 4: Conclusion.

Thus, the capacity of a condenser does not depend on the charge on the plates.
Quick Tip: Capacitance depends on the area of the plates and the distance between them, but not on the charge.

Step 1: Electric field of a conductor.

For a charged conductor, the electric field just outside the surface is perpendicular to the surface. This is a result of Gauss's law, which states that the electric field at the surface of a conductor must be normal (perpendicular) to the surface in electrostatic equilibrium.


Step 2: Elimination.

- (A) parallel to the surface: Incorrect, as the field is not parallel to the surface.

- (B) perpendicular to the surface: Correct, the electric field is always perpendicular to the surface of a conductor.

- (C) at 45° angle to the surface: Incorrect, not the case for conductors.

- (D) zero: Incorrect, there is always an electric field on the surface of a charged conductor.


Step 3: Conclusion.

Thus, the electric field on the outer surface of a charged conductor is perpendicular to the surface.
Quick Tip: For a charged conductor, the electric field at the surface is always perpendicular to the surface, as per Gauss's law.

Step 1: Behavior of electric field inside a conductor.

When a conductor is placed in an external electric field, free charges inside the conductor move in such a way that they cancel out the external field within the conductor. This results in the electric field inside the conductor becoming zero.


Step 2: Explanation of each option.

- (A) zero: Correct. Inside a conductor, the electric field is always zero in electrostatic equilibrium.

- (B) equal to the external field: Incorrect. The internal field is canceled out by the movement of free charges.

- (C) twice the external field: Incorrect. This does not hold for conductors.

- (D) half the external field: Incorrect. The field inside the conductor is zero, not a fraction of the external field.


Step 3: Conclusion.

The electric field inside the conductor is zero.
Quick Tip: Inside a conductor in electrostatic equilibrium, the electric field is always zero due to the rearrangement of charges.

Step 1: Electric potential energy between charges.

The potential energy between two point charges is given by the formula: \[ U = \frac{{k \cdot q_1 \cdot q_2}}{{r}} \]
where \(k\) is the Coulomb constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them.


Step 2: Analyzing the sign of potential energy.

- When two point charges have opposite signs (one positive and one negative), their potential energy is always negative because the product of \(q_1 \cdot q_2\) will be negative. This indicates an attractive interaction between the charges.


Step 3: Explanation of each option.

- (A) The potential energy is always negative: Correct. For opposite charges, the potential energy is negative due to the attractive force.

- (B) The potential energy is always positive: Incorrect. This happens only for like charges.

- (C) The potential energy can be either positive or negative: Incorrect for opposite charges, where the energy is always negative.

- (D) The potential energy is zero: Incorrect. The potential energy is not zero for opposite charges unless the charges are at infinite distance.


Step 4: Conclusion.

For two point charges of opposite signs, the potential energy is always negative.
Quick Tip: Opposite charges = attractive interaction = negative potential energy. Same sign charges = repulsive interaction = positive potential energy.

Step 1: Gauss’s Law.

Gauss’s law states that the total electric flux through a closed surface is directly proportional to the total charge enclosed within the surface. It is mathematically expressed as: \[ \Phi_E = \frac{Q_{enc}}{\epsilon_0} \]
where \(\Phi_E\) is the electric flux, \(Q_{enc}\) is the charge enclosed, and \(\epsilon_0\) is the permittivity of free space.


Step 2: Analyze options.

- (A) Proportional to the charge enclosed: Correct. Gauss's law directly relates the electric flux to the enclosed charge.

- (B) Inversely proportional to the charge enclosed: Incorrect. The flux increases with the charge enclosed, not inversely.

- (C) Zero: Incorrect. The flux is not necessarily zero unless the charge enclosed is zero.

- (D) Proportional to the square of the charge enclosed: Incorrect. There is no square relation in Gauss’s law; the flux is directly proportional to the charge.


Step 3: Conclusion.

Gauss's law indicates that the electric flux is proportional to the charge enclosed.
Quick Tip: Gauss’s law links electric flux and charge enclosed. No square or inverse relations exist in its basic form.

Step 1: Definition of Kilowatt-hour.

A kilowatt-hour (kWh) is a unit of energy. It represents the amount of energy consumed when a device with a power rating of 1 kilowatt operates for 1 hour. The energy is calculated as: \[ Energy = Power \times Time
Energy (kWh) = 1 \, kW \times 1 \, hr \]
Thus, 1 kWh = 1000 watts × 3600 seconds = 3.6 × 10^6 joules.


Step 2: Analyze options.

- (A) Energy: Correct. Kilowatt-hour is a unit of energy.

- (B) Power: Incorrect. Power is measured in watts or kilowatts, not kilowatt-hours.

- (C) Torque: Incorrect. Torque is a rotational force and is measured in newton-meters.

- (D) Force: Incorrect. Force is measured in newtons, not kilowatt-hours.


Step 3: Conclusion.

Kilowatt-hour is a unit of energy.
Quick Tip: Remember: kWh = energy, not power. Power is measured in watts (W).

Step 1: Understanding the principle.

The principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. Lenz’s law is directly based on this principle, as it states that the direction of an induced current in a closed loop will oppose the change that caused it, thus conserving energy.


Step 2: Explanation of the laws.

- (A) Ampere’s law: Describes the relationship between a current-carrying conductor and the magnetic field it creates, but not energy conservation.

- (B) Faraday’s law of electrolysis: Deals with the relationship between voltage and the rate of change of the magnetic field. While important in electromagnetism, it’s not directly related to energy conservation.

- (C) Lenz’s law: Directly related to energy conservation, as it ensures that the induced current opposes the change in the magnetic field, conserving energy.


Step 3: Conclusion.

Thus, Lenz’s law is based on the principle of energy conservation.

\[ \boxed{Correct Answer: Lenz's law} \] Quick Tip: Remember: Lenz’s law is a direct consequence of the conservation of energy — it ensures that the induced current always opposes the change in the magnetic flux.

Step 1: What is shunting?

Shunting an ammeter involves connecting a low resistance in parallel with the ammeter to allow most of the current to bypass the meter. This prevents the ammeter from being damaged by high currents.


Step 2: Effect on the measurement limit.

When the ammeter is shunted, its range of measurement is increased, meaning it can measure higher currents than it could without the shunt. The shunt bypasses most of the current, allowing a small current to pass through the meter and increasing its range.


Step 3: Conclusion.

Thus, when an ammeter is shunted, its measurement limit increases.

\[ \boxed{Correct Answer: Increases} \] Quick Tip: Shunting an ammeter increases its range, enabling it to measure higher currents without damage.

Step 1: Understanding permeability.

Permeability (\(\mu\)) is a physical constant that measures the ability of a material to support the formation of a magnetic field within itself. The dimensional formula for permeability is derived from the relationship between the magnetic field, the electric field, and the electric current.


Step 2: Derivation.

The permeability in the MKS system is represented as: \[ \mu = \frac{{Force}}{{Current^2 \cdot Length}} \]
Since force is measured in newtons (\( [MLT^{-2}] \)), current in amperes (\( [A] \)), and length in meters (\( [L] \)), the dimensional formula for permeability is: \[ \mu = [MLT^{-2}A^{-2}] \]

Conclusion: The dimensional formula of permeability is \( [MLT^{-2}A^{-2}] \).
Quick Tip: The permeability constant \( \mu_0 \) is used in electromagnetism and has the dimensional formula \( [MLT^{-2}A^{-2}] \).

Step 1: Understanding the quantity.

The quantity \( \frac{{\mu_0}}{{\epsilon_0}} \) is known as the characteristic impedance of free space, where \( \mu_0 \) is the permeability of free space and \( \epsilon_0 \) is the permittivity of free space. The value of \( \frac{{\mu_0}}{{\epsilon_0}} \) is used in the calculation of the speed of light in vacuum and is related to the unit of resistance, which is ohm.


Step 2: Relation to other units.

The unit of \( \frac{{\mu_0}}{{\epsilon_0}} \) is derived from the formula \( \mu_0 = 4\pi \times 10^{-7} \, H/m \) and \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/N·m^2 \). Combining these gives the unit as ohm (\( \Omega \)), the unit of electrical resistance.


Conclusion: The unit of \( \frac{{\mu_0}}{{\epsilon_0}} \) is ohm.
Quick Tip: The characteristic impedance of free space, \( \frac{{\mu_0}}{{\epsilon_0}} \), has the unit of ohms, which is the unit of electrical resistance.

Step 1: Sensitivity of a galvanometer.

The sensitivity of a moving coil galvanometer is given by \( S = \frac{N}{I} \), where \( N \) is the number of turns, and \( I \) is the current.


Step 2: Increasing number of turns.

When the number of turns (\( N \)) increases, the coil generates a stronger magnetic field, which increases the deflection for the same current. Hence, the sensitivity increases.


Step 3: Elimination.

- (A) increases: Correct, as sensitivity is directly proportional to the number of turns.

- (B) decreases: Incorrect, sensitivity increases with more turns.

- (C) remains unchanged: Incorrect, as increasing turns does affect sensitivity.

- (D) may increase or may decrease: Incorrect, it always increases.


Step 4: Conclusion.

Thus, increasing the number of turns in a galvanometer increases its sensitivity.
Quick Tip: In a moving coil galvanometer, increasing the number of turns results in increased sensitivity due to stronger induced magnetic fields.

Step 1: Understanding neutral points.

Neutral points are the locations where the magnetic field produced by the Earth (\( B_H \)) and the magnetic field produced by the magnet (\( B \)) cancel each other out. At these points, the net magnetic field is zero.


Step 2: Relationship between \( B \) and \( B_H \).

At neutral points, the magnetic field \( B \) created by the magnet is exactly equal to the horizontal component of the Earth's magnetic field \( B_H \). This results in the cancellation of the two fields.


Step 3: Elimination.

- (A) \( B > B_H \): Incorrect, the fields cancel out, so \( B \) cannot be greater.

- (B) \( B < B_H \): Incorrect, again the fields cancel out, so \( B \) cannot be less.

- (C) \( B = B_H \): Correct, at neutral points, the fields are equal and cancel each other.

- (D) \( B = 0 \): Incorrect, as the magnetic field is not zero at the neutral points.


Step 4: Conclusion.

Thus, at neutral points, \( B = B_H \).
Quick Tip: At neutral points, the magnetic fields of the magnet and the Earth are equal in magnitude but opposite in direction, resulting in no net magnetic field.

Step 1: Formula for Root Mean Square (RMS) value.

The relationship between the peak value (\(I_0\)) and the root mean square (RMS) value (\(I_{rms}\)) for alternating current is given by: \[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
where \(I_0 = 10 \, A\) is the peak current.


Step 2: Calculate the RMS value.

Substitute the peak value into the formula: \[ I_{rms} = \frac{10}{\sqrt{2}} \approx 7.07 \, A \]


Step 3: Analyze options.

- (A) 5 A: Incorrect. This is not the RMS value.

- (B) 7.07 A: Correct. This is the RMS value for a peak value of 10 A.

- (C) 10 A: Incorrect. The peak value is given, not the RMS value.

- (D) 14.14 A: Incorrect. This is the value for the peak value itself.


Step 4: Conclusion.

The RMS value of the current is 7.07 A.
Quick Tip: RMS value = Peak value / \(\sqrt{2}\) for sinusoidal AC.

Step 1: Power factor in an AC circuit.

The power factor (\( PF \)) in an alternating current (AC) circuit is the cosine of the phase angle (\(\theta\)) between the current and voltage. For purely inductive circuits, the phase angle between current and voltage is \(90^\circ\).


Step 2: Power factor for purely inductive circuits.

The power factor in a purely inductive circuit is given by: \[ PF = \cos(90^\circ) = 0 \]


Step 3: Analyze options.

- (A) 0: Correct. In a purely inductive circuit, the power factor is zero because current lags the voltage by 90° and no real power is transferred.

- (B) 1: Incorrect. A power factor of 1 happens only in a purely resistive circuit.

- (C) 0.5: Incorrect. This value does not apply to purely inductive circuits.

- (D) infinity: Incorrect. This is not a valid power factor value in any circuit.


Step 4: Conclusion.

In a purely inductive circuit, the power factor is 0.
Quick Tip: In purely inductive circuits, the current lags the voltage by 90°, leading to a power factor of zero.

Step 1: Understanding the phase difference in a capacitor.

In an alternating current (a.c.) circuit containing only a capacitor, the current leads the voltage by 90°. This is because the voltage across the capacitor lags the current due to the charging and discharging of the capacitor.


Step 2: Conclusion.

Thus, the phase difference between the current and voltage in a purely capacitive circuit is 90°.

\[ \boxed{90^\circ} \] Quick Tip: In a capacitor, current leads voltage by 90°, while in an inductor, voltage leads current by 90°.

Step 1: Formula for resonance frequency in an L-C circuit.

The resonance frequency of an L-C circuit is given by the formula: \[ f = \frac{1}{2\pi \sqrt{LC}} \]
where \( L \) is the inductance and \( C \) is the capacitance. This frequency occurs when the inductive reactance equals the capacitive reactance.


Step 2: Conclusion.

Thus, the correct expression for the resonance frequency is \( \frac{1}{2\pi} \sqrt{\frac{1}{LC}} \).

\[ \boxed{\frac{1}{2\pi} \sqrt{\frac{1}{LC}}} \] Quick Tip: Resonance frequency is crucial in L-C circuits for efficient energy transfer, and it is inversely proportional to the square root of the product of L and C.

Step 1: Understanding transformers.

In an ideal transformer, the voltage and current are related through the turns ratio. For a step-up transformer, the voltage in the secondary coil is higher than that in the primary coil.


Step 2: Current and voltage relation.

According to the law of conservation of energy and the transformer equation, when the voltage increases in the secondary coil, the current decreases. This inverse relationship ensures that the power transferred remains constant.


Step 3: Conclusion.

Thus, in a step-up transformer, the current in the secondary coil is less than the current in the primary coil.

\[ \boxed{Correct Answer: Less} \] Quick Tip: In a step-up transformer, voltage increases while current decreases. This is an inverse relationship governed by the power conservation principle.

Step 1: Understanding convex mirrors.

A convex mirror always forms a virtual, diminished image that is upright and located behind the mirror.


Step 2: Position of image in convex mirror.

For a convex mirror, the image is always formed between the pole and the focus, regardless of the object’s position. The rays diverge, and the image is formed behind the mirror at a point closer to the pole than the focus.


Step 3: Conclusion.

Thus, the image formed by a convex mirror is always located between the pole and the focus.

\[ \boxed{Correct Answer: In between pole and focus} \] Quick Tip: Convex mirrors always produce virtual images that are formed between the pole and the focus.

Step 1: Formula for critical angle.

The critical angle (\( \theta_c \)) is related to the refractive indices of the two media by the formula: \[ \sin(\theta_c) = \frac{{n_2}}{{n_1}} \]
where \( n_1 \) is the refractive index of the medium and \( n_2 \) is the refractive index of vacuum (\( n_2 = 1 \)). The refractive index \( n_1 \) is related to the velocity of light in the medium by: \[ n_1 = \frac{{c}}{{v}} \]
where \( c = 3 \times 10^8 \, m/sec \) is the speed of light in vacuum and \( v \) is the velocity of light in the medium.


Step 2: Using given information.

We know the critical angle is \( 30^\circ \), so \[ \sin(30^\circ) = \frac{1}{2} = \frac{1}{\frac{c}{v}} \Rightarrow v = 1.5 \times 10^8 \, m/sec \]

Conclusion: The velocity of light in the medium is \( 1.5 \times 10^8 \, m/sec \).
Quick Tip: Critical angle \( \theta_c \) and velocity of light in the medium are inversely related through refractive indices.

Step 1: Coulomb's law.

Coulomb’s law for the force between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is: \[ F = k \cdot \frac{{q_1 q_2}}{{r^2}} \]
where \( k \) is Coulomb’s constant.


Step 2: Applying changes.

When the distance is halved, \( r \rightarrow \frac{r}{2} \). So the force increases by a factor of \( 4 \) (since \( F \propto \frac{1}{r^2} \)). Also, when one charge is halved, \( q_1 \rightarrow \frac{q_1}{2} \), so the force decreases by a factor of \( 2 \).


Step 3: Final force.

Thus, the final force will be: \[ F' = 4F \times \frac{1}{2} = 2F \]

Conclusion: The force will become double the original force.
Quick Tip: Force between two charges increases with decreasing distance and is directly proportional to the product of charges.

Step 1: Energy of the electron.

The kinetic energy gained by the electron when accelerated through a potential difference \( V \) is given by: \[ K.E. = eV \]
where \( e \) is the charge of the electron.


Step 2: Relationship with speed.

The kinetic energy of the electron is also given by: \[ K.E. = \frac{1}{2}mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its velocity.


Step 3: Equating the two expressions.

Equating the kinetic energy expressions, we get: \[ eV = \frac{1}{2}mv^2 \]
Solving for \( v \), we get: \[ v = \sqrt{\frac{2eV}{m}} \]
Thus, the speed is proportional to \( \sqrt{V} \).


Step 4: Elimination.

- (A) proportional to \( V \): Incorrect, the speed is not directly proportional to \( V \).

- (B) proportional to \( \sqrt{V} \): Correct, as derived above.

- (C) proportional to \( \frac{1}{V} \): Incorrect.

- (D) proportional to \( V^2 \): Incorrect.


Step 5: Conclusion.

Hence, the speed of the electron is proportional to \( \sqrt{V} \).
Quick Tip: For an electron accelerated through a potential difference, its speed is proportional to the square root of the potential difference.

Step 1: Specific charge of electron.

The specific charge of an electron is the ratio of its charge to its mass. Mathematically, it is given by: \[ Specific charge = \frac{e}{m} \]
where \( e = 1.6 \times 10^{-19} \, C \) is the charge of an electron, and \( m = 9.11 \times 10^{-31} \, kg \) is the mass of the electron.


Step 2: Calculation.

Substituting the values: \[ \frac{1.6 \times 10^{-19}}{9.11 \times 10^{-31}} = 1.8 \times 10^{11} \, C/kg \]

Step 3: Elimination.

- (A) \( 1.8 \times 10^{-19} \, C/kg \): Incorrect.

- (B) \( 1.67 \times 10^{-19} \, C/kg \): Incorrect.

- (C) \( 1.8 \times 10^{11} \, C/kg \): Correct.

- (D) \( 6.67 \times 10^{11} \, C/kg \): Incorrect.


Step 4: Conclusion.

Therefore, the specific charge of an electron is \( 1.8 \times 10^{11} \, C/kg \).
Quick Tip: The specific charge of an electron is a constant \( \frac{e}{m} = 1.8 \times 10^{11} \, C/kg \).

Step 1: Parallel combination of capacitors.

In a parallel combination of capacitors, the potential difference across each capacitor remains the same. This is a fundamental property of parallel circuits: each component shares the same voltage.


Step 2: Explanation of each option.

- (A) Charge: Incorrect. The charge on each capacitor is different in a parallel combination.

- (B) Energy: Incorrect. The energy stored in each capacitor will differ based on their capacitance.

- (C) Potential difference: Correct. The potential difference is the same for all capacitors in a parallel combination.

- (D) Capacity: Incorrect. The capacitance of each capacitor in the parallel combination may differ, but the potential difference is the same.


Step 3: Conclusion.

In a parallel combination of capacitors, the potential difference remains the same for each condenser.
Quick Tip: In parallel circuits, all components experience the same potential difference.

Step 1: Coulomb's Law.

The repulsive force between two charges is given by Coulomb’s law: \[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \]
where \( F \) is the force, \( q_1 \) and \( q_2 \) are the magnitudes of the two charges, and \( r \) is the distance between them. The constant \( \epsilon_0 \) is the permittivity of free space.


Step 2: Effect of dielectric material.

When a dielectric material is inserted between two positive charges, the permittivity of the medium increases, and as a result, the electrostatic force between the charges decreases. The dielectric constant reduces the effective force.


Step 3: Explanation of each option.

- (A) Increase: Incorrect. The force will not increase; it decreases due to the dielectric.

- (B) Decrease: Correct. The dielectric reduces the force between the charges.

- (C) Remain same: Incorrect. The presence of a dielectric changes the electrostatic force.

- (D) Become zero: Incorrect. The force does not become zero; it decreases.


Step 4: Conclusion.

Inserting a dielectric material decreases the repulsive force between the charges.
Quick Tip: Dielectric materials reduce the effective force between charges by increasing the permittivity of the medium.

Step 1: Formula for electric dipole moment.

The electric dipole moment (\( p \)) is given by the formula: \[ p = q \times d \]
where:
- \( q \) is the charge
- \( d \) is the separation distance between the charges.

Step 2: Given values.

Given:
- \( q = 0.2 \, \mu C = 0.2 \times 10^{-6} \, C \)
- \( d = 3.0 \, cm = 0.03 \, m \)

Substitute the values into the formula: \[ p = (0.2 \times 10^{-6} \, C) \times (0.03 \, m) = 6.0 \times 10^{-8} \, C m \]

Step 3: Conclusion.

Thus, the electric dipole moment is \( 6.0 \times 10^{-8} \, C m \).

\[ \boxed{6.0 \times 10^{-8} \, C m} \] Quick Tip: Electric dipole moment = charge \(\times\) separation distance. Always convert units properly, like microcoulombs to coulombs and cm to meters.

Step 1: Concept of electric flux.

Electric flux through a closed surface is given by Gauss's law: \[ \Phi_E = \frac{q_{enc}}{\epsilon_0} \]
where:
- \( q_{enc} \) is the charge enclosed by the surface,
- \( \epsilon_0 \) is the permittivity of free space.

Step 2: Electric flux from dipoles.

For an arrangement of electric dipoles, the net charge enclosed inside a closed surface is zero because the dipoles have equal and opposite charges that cancel each other out.


Step 3: Conclusion.

Thus, the electric flux coming out from the closed surface will be zero.

\[ \boxed{0} \] Quick Tip: In Gauss's law, if the enclosed charge is zero (as in the case of dipoles), the electric flux is also zero.

Step 1: Understanding the superposition principle.

When two waves of equal amplitude and wavelength superimpose, the resultant amplitude is determined by the phase difference between the waves.


Step 2: Amplitude and phase difference.

The amplitude of the resultant wave will be maximum when the two waves are in phase, i.e., when the phase difference between them is zero. This results in constructive interference.


Step 3: Conclusion.

Thus, the amplitude will be maximum when the phase difference is zero.

\[ \boxed{Correct Answer: Zero} \] Quick Tip: Constructive interference (maximum amplitude) occurs when the phase difference between two waves is zero.

Step 1: Understanding polarization.

Polarized light refers to light in which the vibrations occur in a single plane. The plane of polarization is the direction in which the electric field vector of the light oscillates.


Step 2: Angle between vibration and polarization plane.

In polarized light, the plane of vibration and the plane of polarization are at a right angle, i.e., \(90^\circ\) apart. This means that the light wave’s electric field oscillates in a direction perpendicular to the plane of polarization.


Step 3: Conclusion.

Thus, the angle between the plane of vibration and the plane of polarization is \(90^\circ\).

\[ \boxed{Correct Answer: 90°} \] Quick Tip: In polarized light, the angle between the plane of vibration and the plane of polarization is always \(90^\circ\).

Step 1: Understanding electromagnetic waves.

Electromagnetic waves include visible light, radio waves, X-rays, gamma rays, and infrared rays. They all travel at the speed of light in a vacuum and can propagate through a vacuum.


Step 2: Elimination of options.

- (A) Alpha rays: Alpha rays are not electromagnetic waves. They are a form of particle radiation consisting of helium nuclei.

- (B) Gamma rays: Gamma rays are electromagnetic waves with high energy.

- (C) Infrared rays: Infrared rays are electromagnetic waves that are used for heat and communication.

- (D) X-rays: X-rays are also electromagnetic waves.


Conclusion: Alpha rays are not an electromagnetic wave.
Quick Tip: Electromagnetic waves are characterized by oscillating electric and magnetic fields, and include visible light, gamma rays, X-rays, and infrared rays.

Step 1: Speed of electromagnetic waves.

The speed of an electromagnetic wave in a material medium is given by the formula: \[ v = \frac{c}{\sqrt{\epsilon_r \mu_r}} \]
where \( c \) is the speed of light in a vacuum, \( \epsilon_r \) is the relative permittivity, and \( \mu_r \) is the relative permeability of the medium. The wave’s speed depends on the medium’s properties, not its frequency, wavelength, or intensity.


Step 2: Elimination of options.

- (A) Upon its wavelength: The speed of light is related to its wavelength and frequency through the equation \( v = \lambda f \). Wavelength is inversely proportional to frequency in a given medium.

- (B) Upon its frequency: The frequency of an electromagnetic wave is related to its speed in a particular medium.

- (C) Upon its intensity: The intensity of the wave does not affect the speed of light in the medium. Intensity is related to the amplitude of the wave, not its velocity.

- (D) Upon its permittivity: The speed of light depends on the permittivity of the medium.


Conclusion: The speed of the electromagnetic wave does not depend on its intensity.
Quick Tip: The speed of an electromagnetic wave in a medium depends on its permittivity and permeability, but not on its intensity.

Step 1: Photoelectric effect overview.

The photoelectric effect occurs when light strikes a material and ejects electrons from it. The current produced depends on the number of ejected electrons, which is influenced by the intensity of the incident light.


Step 2: Effect of intensity.

When the intensity of the incident light increases, more photons hit the material, leading to the emission of more electrons. This increases the photoelectric current.


Step 3: Elimination.

- (A) increases: Correct, the photoelectric current increases with intensity.

- (B) decreases: Incorrect, as the current increases with intensity.

- (C) remains unchanged: Incorrect, as intensity affects the current.

- (D) first increases then remains constant: Incorrect, current continuously increases with intensity.


Step 4: Conclusion.

Thus, the photoelectric current increases with the intensity of the incident light.
Quick Tip: In the photoelectric effect, the current increases with the intensity of the incident light, as more photons contribute to ejection of electrons.

Step 1: de Broglie wavelength.

According to de Broglie’s hypothesis, every moving particle has a wave associated with it. The wavelength \( \lambda \) of the de Broglie wave is given by: \[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.


Step 2: Dependence of wavelength.

The momentum \( p \) is related to the mass \( m \) and velocity \( v \) of the particle: \[ p = mv \]
Thus, the de Broglie wavelength depends on the mass and velocity of the particle, but not on its charge.


Step 3: Elimination.

- (A) mass: The wavelength depends on mass, as momentum depends on mass.

- (B) charge: Correct, charge does not affect the de Broglie wavelength.

- (C) velocity: The wavelength depends on velocity, as momentum is velocity-dependent.

- (D) momentum: The wavelength depends on momentum, as per de Broglie's relation.


Step 4: Conclusion.

Thus, the de Broglie wavelength does not depend on the charge of the particle.
Quick Tip: de Broglie wavelength depends on the momentum of the particle, which is a product of its mass and velocity, but not on its charge.

Step 1: Kinetic mass of photon.

The kinetic mass \(m\) of a photon can be expressed in terms of its energy \(E\) and speed of light \(c\). According to Einstein’s equation for energy and mass relation: \[ E = mc^2 \]
For a photon, its energy is also given by \(E = hv\), where \(h\) is Planck’s constant and \(v\) is the frequency of the photon. Combining these: \[ hv = mc^2 \]
Solving for \(m\): \[ m = \frac{hv}{c^2} \]


Step 2: Analyze options.

- (A) \(\frac{hv}{c}\): Incorrect. This does not correctly relate the energy and mass of a photon.

- (B) \(\frac{hv}{c^2}\): Correct. This matches the correct formula for the kinetic mass of a photon.

- (C) \(\frac{hc}{v}\): Incorrect. This is not the correct formula for the kinetic mass.

- (D) \(\frac{c^2}{hv}\): Incorrect. This does not match the formula.


Step 3: Conclusion.

The correct formula for the kinetic mass of a photon is \(\frac{hv}{c^2}\).
Quick Tip: The kinetic mass of a photon is given by \(m = \frac{hv}{c^2}\), where \(h\) is Planck’s constant and \(v\) is the frequency.

Step 1: Formula for focal length of two lenses in contact.

When two lenses of focal lengths \(f_1\) and \(f_2\) are in contact, the focal length \(f\) of the combination is given by: \[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]
For two lenses with equal focal lengths \(f_1 = f_2 = f\), this becomes: \[ \frac{1}{f} = \frac{1}{f} + \frac{1}{f} \]
Solving for \(f\): \[ \frac{1}{f} = \frac{2}{f} \quad \Rightarrow \quad f_{combination} = \frac{f}{2} \]


Step 2: Analyze options.

- (A) \(f\): Incorrect. This would be the case for two lenses in parallel, not in contact.

- (B) \(2f\): Incorrect. The focal length of the combination is not twice the focal length of one lens.

- (C) \(\frac{f}{2}\): Correct. The focal length of two lenses in contact is half the focal length of one lens.

- (D) \(3f\): Incorrect. This is not the correct relationship for two lenses in contact.


Step 3: Conclusion.

The focal length of the combination of two converging lenses in contact is \(\frac{f}{2}\).
Quick Tip: For two lenses in contact, the focal length of the combination is given by \(f = \frac{f}{2}\) if both lenses have equal focal lengths.

Step 1: Formula for focal length and power.

The relationship between power (\(P\)) and focal length (\(f\)) is given by the formula: \[ P = \frac{1}{f} \quad (in meters) \]
where \( P \) is the power of the lens in dioptres and \( f \) is the focal length in meters.

Step 2: Given values.

We are given that the power of the convex lens is 2 dioptres, so: \[ P = 2 \, dioptres \]
Using the formula, we can find the focal length: \[ f = \frac{1}{P} = \frac{1}{2} \, m = 0.5 \, m = 50 \, cm \]

Step 3: Conclusion.

Thus, the focal length of the lens is 50 cm.

\[ \boxed{50 \, cm} \] Quick Tip: The focal length is the reciprocal of the power, i.e., \( f = \frac{1}{P} \), where \( P \) is in dioptres.

Step 1: Understanding chromatic aberration.

Chromatic aberration occurs due to the dispersion of light into its constituent colors (spectrum). This happens because different colors (wavelengths) of light refract at different angles when passing through a lens, leading to color fringing.

Step 2: Solution to chromatic aberration.

An achromatic combination (made by combining a convex lens and a concave lens of different materials) is designed to minimize chromatic aberration. This combination corrects the color distortion by canceling out the effects of dispersion.


Step 3: Conclusion.

Thus, an achromatic combination is used to reduce chromatic aberration in lenses.

\[ \boxed{Achromatic combination} \] Quick Tip: Achromatic lenses are designed by combining different types of lenses (concave and convex) to correct chromatic aberration, especially in telescopes and microscopes.

Step 1: Refractive index and focal length.

The focal length \( f \) of a lens is related to its refractive index \( n \) and the refractive index of the surrounding medium. When a lens is immersed in a liquid with the same refractive index as the lens, the refraction at the surfaces of the lens becomes nullified. This leads to the lens losing its focusing ability.


Step 2: Conclusion.

When the refractive index of the surrounding medium matches that of the lens material, the focal length of the lens becomes infinity.

\[ \boxed{Correct Answer: Become infinity} \] Quick Tip: When the refractive index of the surrounding medium equals the refractive index of the lens material, the lens loses its power to focus, leading to a focal length of infinity.

Step 1: Understanding the phenomenon.

When light falls on a soap bubble, it undergoes interference due to the thin film of the bubble. The light waves reflecting off the two surfaces of the thin film interfere with each other, resulting in the appearance of bright and dark bands of color.


Step 2: Explanation of interference.

Interference occurs when two light waves of the same frequency superimpose, leading to constructive and destructive interference. This is the reason why soap bubbles show colors depending on the thickness of the bubble film.


Step 3: Conclusion.

Therefore, the soap bubble appears coloured due to the phenomenon of interference.

\[ \boxed{Correct Answer: Interference} \] Quick Tip: Soap bubbles show colours because of interference between light waves reflecting off the two surfaces of the bubble's thin film.

Step 1: Understanding polarization.

Polarization refers to the orientation of oscillations in the direction perpendicular to the direction of the wave. Light waves, radio waves, and X-rays are transverse waves, meaning their oscillations can be polarized. However, sound waves are longitudinal waves, and they cannot be polarized.


Step 2: Elimination of options.

- (A) Sound waves: Sound waves are longitudinal waves and cannot be polarized.

- (B) Light waves: Light waves are transverse waves and can be polarized.

- (C) Radio waves: Radio waves are also transverse waves and can be polarized.

- (D) X-rays: X-rays are electromagnetic waves and can be polarized.


Conclusion: Sound waves cannot be polarized.
Quick Tip: Only transverse waves, such as light, radio, and X-rays, can be polarized. Longitudinal waves like sound cannot be polarized.

Step 1: Understanding the power of a lens.

The power \( P \) of a lens is given by the formula: \[ P = \frac{1}{f} \, (in meters) \]
where \( f \) is the focal length of the lens. A positive lens (convex lens) is used to correct hypermetropia (farsightedness), where the image focuses behind the retina, requiring a converging lens to bring the focus to the retina.


Step 2: Identifying the defect.

- (A) Myopia: Myopia (nearsightedness) is corrected with a diverging (concave) lens, not a convex lens.

- (B) Hypermetropia: Hypermetropia is corrected with a positive (convex) lens, as the person cannot focus on nearby objects.

- (C) Presbyopia: Presbyopia is a condition related to aging, where the eye loses the ability to focus on nearby objects, often requiring reading glasses.

- (D) Astigmatism: Astigmatism requires cylindrical lenses to correct the distortion of images caused by the uneven curvature of the eye's surface.


Conclusion: The defect is hypermetropia, corrected with a +2D convex lens.
Quick Tip: A positive power lens (convex) is used to correct hypermetropia (farsightedness), while myopia is corrected with a negative power lens (concave).

Step 1: Type of semiconductor.

When boron, a trivalent element, is added as an impurity to silicon, it creates a "hole" or absence of an electron in the silicon lattice. These "holes" are positive charge carriers, leading to the formation of a p-type semiconductor.


Step 2: Explanation of p-type semiconductor.

In p-type semiconductors, the majority charge carriers are holes, which are created by the absence of electrons. As a result, the electrical conductivity is facilitated by the movement of holes.


Step 3: Elimination.

- (A) n-type semiconductor: Incorrect, n-type semiconductors are formed by adding pentavalent impurities (e.g., phosphorus).

- (B) p-type semiconductor: Correct, as boron creates "holes" that act as majority charge carriers.

- (C) n-type conductor: Incorrect, this refers to a conductor with excess electrons.

- (D) p-type conductor: Incorrect, semiconductors are being discussed, not general conductors.


Step 4: Conclusion.

Thus, when boron is added as an impurity in silicon, it forms a p-type semiconductor.
Quick Tip: In p-type semiconductors, the impurity atom (like boron) creates "holes" that serve as the majority charge carriers.

Step 1: Type of semiconductor.

In n-type semiconductors, the majority charge carriers are electrons, which are provided by the donor impurity atoms (usually pentavalent elements such as phosphorus). These extra electrons increase the conductivity of the material.


Step 2: Minority charge carriers.

The minority charge carriers in n-type semiconductors are holes. These are the empty spaces left when an electron moves, and they carry a positive charge. Since there are fewer holes than electrons, holes are the minority carriers.


Step 3: Elimination.

- (A) electrons: Incorrect, electrons are the majority charge carriers in n-type semiconductors.

- (B) holes: Correct, holes are the minority charge carriers in n-type semiconductors.

- (C) electron and hole: Incorrect, only holes are the minority charge carriers.

- (D) none of these: Incorrect.


Step 4: Conclusion.

Thus, in n-type semiconductors, the minority charge carriers are holes.
Quick Tip: In n-type semiconductors, electrons are the majority carriers and holes are the minority carriers.

Step 1: Understanding XOR gate.

The XOR (exclusive OR) gate is a digital logic gate that outputs true (1) if and only if the inputs are different. If both inputs are the same, the output is false (0).


Step 2: Explanation of each option.

- (A) OR gate: Incorrect. The OR gate outputs true if at least one of the inputs is true, regardless of whether they are the same or different.

- (B) AND gate: Incorrect. The AND gate outputs true only if both inputs are true.

- (C) NOT gate: Incorrect. The NOT gate inverts the input but does not depend on whether the inputs are the same or different.

- (D) XOR gate: Correct. The XOR gate outputs true only if the inputs are different.


Step 3: Conclusion.

The correct gate whose output is true only if the inputs are different is the XOR gate.
Quick Tip: An XOR gate outputs true when the inputs are different and false when they are the same.

Step 1: Understanding reverse biased diodes.

A diode is reverse biased when the positive terminal of the power supply is connected to the cathode and the negative terminal to the anode. This prevents current from flowing. In reverse bias, certain diodes exhibit specific characteristics.


Step 2: Explanation of each option.

- (A) Zener diode: Correct. A Zener diode is typically used in reverse bias to maintain a stable voltage across it. It is designed to break down at a certain reverse voltage, allowing current to flow.

- (B) LED: Incorrect. An LED (Light Emitting Diode) operates in forward bias, where current flows from anode to cathode, emitting light.

- (C) Photodiode: Correct. A photodiode operates in reverse bias and generates current when exposed to light.

- (D) both (A) and (C): Correct. Both Zener diodes and photodiodes operate in reverse bias.


Step 3: Conclusion.

Reverse biased diodes include both Zener diodes and photodiodes.
Quick Tip: Zener diodes and photodiodes are commonly used in reverse bias for their unique properties like voltage regulation and light detection.

Step 1: NAND gate logic.

A NAND gate is a NOT AND gate, meaning it outputs the opposite of an AND gate. The boolean expression for an AND gate is: \[ A . B = Y \]
The NOT operation is then applied to this output. Thus, the Boolean expression for the NAND gate is: \[ \overline{A . B} = Y \]

Step 2: Conclusion.

Hence, the correct Boolean expression for a NAND gate is \( \overline{A . B} \), which simplifies to \( A . B = Y \).

\[ \boxed{A . B = Y} \] Quick Tip: For NAND gates, the output is the complement of the AND gate. The Boolean expression is \( \overline{A . B} \).

Step 1: Understanding fibre optic communication.

Fibre optic communication relies on light signals transmitted through optical fibres. The medium used for transmitting data is electromagnetic waves, particularly light waves. These waves travel through the fibre core via total internal reflection.

Step 2: Conclusion.

Hence, the type of wave used in fibre optic communication is electromagnetic waves.

\[ \boxed{Electromagnetic waves} \] Quick Tip: Fibre optic communication uses light waves (electromagnetic waves) to carry data through optical fibres.

Step 1: Understanding TV transmission frequencies.

TV transmission utilizes electromagnetic waves in the radio frequency spectrum. The typical range for TV broadcast is between 30 MHz to 300 MHz. This range is commonly used for UHF and VHF channels in analog and digital TV transmission.


Step 2: Explanation of frequency ranges.

- (A) 30 - 300 Hz: This is a very low frequency range, used for signals such as audio frequencies, not for TV.

- (B) 30 - 300 kHz: This range is used for AM radio transmission, not for TV.

- (C) 30 - 300 MHz: This is the correct frequency range for TV transmission (UHF and VHF).

- (D) 30 - 300 GHz: This range is far higher than what is used for TV broadcasting, typically used for microwave transmission.


Step 3: Conclusion.

The correct frequency range for TV transmission is 30 - 300 MHz.

\[ \boxed{Correct Answer: 30 - 300 MHz} \] Quick Tip: TV transmission uses the VHF and UHF bands, which fall within the 30 MHz to 300 MHz frequency range.

Step 1: Understanding isotopes.

Isotopes are variants of the same element that have the same number of protons (atomic number) but differ in the number of neutrons. This difference in neutrons results in a different atomic mass but does not affect the chemical properties of the element.


Step 2: Explanation of options.

- (A) Number of protons: Isotopes of the same element have the same number of protons (this defines the element).

- (B) Number of neutrons: The key difference between isotopes of the same element is the number of neutrons.

- (C) Number of electrons: In isotopes, the number of electrons remains the same because they are neutral atoms.

- (D) Atomic number: The atomic number is the same for all isotopes of the same element, as it is determined by the number of protons.


Step 3: Conclusion.

The main difference between isotopes is the number of neutrons.

\[ \boxed{Correct Answer: Number of neutrons} \] Quick Tip: Isotopes of an element differ in the number of neutrons they contain, which affects their atomic mass but not their chemical behavior.

Step 1: Understanding types of radioactive decay.

Radioactive decay occurs when an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. Common types of radioactive decay are:

- (A) Alpha decay: Emission of an alpha particle (two protons and two neutrons).

- (B) Beta decay: Emission of an electron (beta-minus decay) or a positron (beta-plus decay).

- (C) Gamma decay: Emission of a high-energy photon (gamma ray).

- (D) Muon decay: Muons are unstable elementary particles that decay, but they are not a type of radioactive decay. Muons are produced by cosmic rays and other high-energy processes, not by the decay of unstable nuclei.


Conclusion: Muon decay is not a type of radioactive decay.
Quick Tip: Radioactive decay typically involves the emission of alpha, beta, or gamma radiation from unstable nuclei. Muon decay, however, is associated with high-energy particle physics, not radioactivity.

Step 1: Use the formula for the energy of a photon.

The energy of a photon is given by the formula: \[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.626 \times 10^{-34} \) Js (Planck’s constant),
- \( c = 3 \times 10^8 \) m/s (speed of light),
- \( \lambda = 500 \times 10^{-9} \) m (wavelength).


Step 2: Substituting the values.
\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 4 \times 10^{-19} \, J \]

Conclusion: The energy of the photon is \( 4 \times 10^{-19} \) J.
Quick Tip: The energy of a photon is inversely proportional to its wavelength. As the wavelength increases, the energy decreases.

Step 1: Hydrogen spectrum series.

The hydrogen spectrum consists of several series, each corresponding to transitions of electrons between energy levels. The visible portion of the hydrogen spectrum corresponds to the Balmer series.


Step 2: Series of hydrogen spectrum.

- Lyman series: Corresponds to transitions from higher levels to the first level and lies in the ultraviolet region.

- Balmer series: Correct, corresponds to transitions from higher levels to the second level and lies in the visible region.

- Paschen series: Corresponds to transitions from higher levels to the third level and lies in the infrared region.

- Brackett series: Corresponds to transitions from higher levels to the fourth level and lies in the infrared region.


Step 3: Conclusion.

Thus, the Balmer series lies in the visible portion of the hydrogen spectrum.
Quick Tip: The Balmer series is the only hydrogen spectrum series visible to the human eye, occurring in the visible light region.

Step 1: Chargeless particles.

Chargeless particles are particles that do not carry any electrical charge. A photon is the only particle in the list that is chargeless.


Step 2: Elimination.

- (A) α-particle: Incorrect, an alpha particle consists of two protons and two neutrons, thus it carries a positive charge.

- (B) β-particle: Incorrect, a beta particle is an electron, which carries a negative charge.

- (C) Proton: Incorrect, a proton carries a positive charge.

- (D) Photon: Correct, a photon is a particle of light and does not have any charge.


Step 3: Conclusion.

Thus, the photon is the chargeless particle.
Quick Tip: Photons are chargeless particles and are the quanta of electromagnetic radiation.

Step 1: Understanding X-rays.

X-rays are a form of electromagnetic radiation, much like light, but with much shorter wavelengths. They are produced when high-energy electrons collide with matter, releasing energy in the form of electromagnetic waves.


Step 2: Explanation of each option.

- (A) moving electron: Incorrect. X-rays are not composed of moving electrons; rather, they are electromagnetic waves produced by the interaction of electrons with matter.

- (B) moving positive ions: Incorrect. X-rays are not made of ions but rather electromagnetic radiation.

- (C) moving negative ion: Incorrect. Similar to option (B), X-rays are not generated by ions but by electromagnetic radiation.

- (D) electromagnetic waves: Correct. X-rays are a type of electromagnetic wave, just like visible light, radio waves, and gamma rays.


Step 3: Conclusion.

X-rays are electromagnetic waves.
Quick Tip: X-rays are high-energy electromagnetic radiation that can penetrate matter, which makes them useful in medical imaging and material analysis.

Step 1: Understanding the concept of AMU.

An atomic mass unit (amu) is a standard unit of mass that quantifies mass on an atomic or molecular scale. The energy equivalent of 1 amu can be calculated using Einstein's equation \(E = mc^2\). The energy equivalent of 1 amu is approximately 931 MeV (million electron volts).


Step 2: Explanation of each option.

- (A) 190 MeV: Incorrect. This value is not the standard energy equivalent for 1 amu.

- (B) 139 MeV: Incorrect. This value is not correct for the energy equivalent of 1 amu.

- (C) 913 MeV: Incorrect. While close, it is not the exact value.

- (D) 931 MeV: Correct. The energy equivalent of 1 amu is approximately 931 MeV, as calculated from the equation \(E = mc^2\).


Step 3: Conclusion.

The energy equivalent of 1 amu is 931 MeV.
Quick Tip: 1 amu is equivalent to 931 MeV. This conversion is useful when analyzing energy levels in nuclear reactions and particle physics.

Step 1: Formula for energy stored in a capacitor.

The energy (\( E \)) stored in a capacitor is given by the formula: \[ E = \frac{1}{2} C V^2 \]
where:
- \( C \) is the capacitance in farads (F),
- \( V \) is the voltage in volts (V).

Step 2: Given values.

- \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \),
- \( V = 100 \, V \).

Substituting these values into the formula: \[ E = \frac{1}{2} \times (100 \times 10^{-6}) \times (100)^2 = 0.5 \, joules \]

Step 3: Conclusion.

Thus, the energy stored in the capacitor is 0.5 joules.

\[ \boxed{0.5 \, joules} \] Quick Tip: The energy stored in a capacitor is proportional to the square of the voltage and the capacitance.

Step 1: Understanding the units.

The unit of electric field is defined as force per unit charge. In terms of SI units, it is expressed as: \[ Electric field = \frac{Force}{Charge} = \frac{Newtons}{Coulombs} = N/C \]
Alternatively, it can also be written in terms of voltage and distance: \[ Electric field = \frac{Voltage}{Distance} = \frac{V}{m} = volt/m \]

Step 2: Conclusion.

Thus, the unit of electric field is volt per metre (\(V/m\)), which is equivalent to volt-metre\(^{-1}\).

\[ \boxed{Electric field} \] Quick Tip: Electric field is defined as the force per unit charge and has units of volt per meter (V/m).

Step 1: Drift velocity and potential difference.

In electric conduction, the drift velocity \( v \) of free electrons in a conductor is directly proportional to the applied potential difference \( V \). This is a consequence of Ohm's law, which states that current \( I \) is proportional to the potential difference \( V \), and since current is related to drift velocity, the drift velocity follows the same relationship.


Step 2: Explanation of the relationship.

The drift velocity \( v \) is given by the formula \( v = \mu E \), where \( \mu \) is the mobility of the electrons and \( E \) is the electric field, which is related to the potential difference by \( E = \frac{V}{L} \), where \( L \) is the length of the conductor. Hence, \( v \) is proportional to \( V \).


Step 3: Conclusion.

Thus, the correct relationship is that drift velocity is proportional to the potential difference \( V \).

\[ \boxed{Correct Answer: Proportional to V} \] Quick Tip: Drift velocity in a conductor increases with the applied potential difference, following a direct proportionality as per Ohm's law.

Step 1: Understanding the graph of \( V \) and \( I \).

The graph between voltage \( V \) and current \( I \) for a conductor is a straight line, and its slope represents the resistance of the conductor. The slope is given by the tangent of the angle \( \theta \) the graph makes with the horizontal axis (which is voltage).


Step 2: Applying the formula.

From Ohm's law, \( V = IR \), where \( R \) is the resistance. If the graph between \( V \) and \( I \) is a straight line, the slope \( \frac{V}{I} = R \). The angle \( \theta \) between the line and the horizontal axis gives the tangent of the angle as \( \tan \theta = \frac{V}{I} \). Therefore, the resistance is given by \( \cot \theta \).


Step 3: Conclusion.

Thus, the resistance of the conductor is \( \cot \theta \), which is the inverse of the tangent of the angle.

\[ \boxed{Correct Answer: \cot \theta} \] Quick Tip: The slope of the \( V \)-\( I \) graph gives the resistance, which is \( \cot \theta \) when the angle \( \theta \) is measured with the horizontal axis.

Step 1: Understanding the formula for power in an electric circuit.

The power in an electric circuit can be calculated using the formula: \[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the resistor and \( R \) is the resistance of the circuit. This formula is derived from Ohm's law (\( V = IR \)) and the basic power formula (\( P = IV \)). By substituting \( I = \frac{V}{R} \), we get: \[ P = V \times \frac{V}{R} = \frac{V^2}{R} \]

Step 2: Conclusion.
Thus, the correct answer is \( \frac{V^2}{R} \).
Quick Tip: The power dissipated in an electrical circuit is inversely proportional to the resistance. As resistance increases, the power decreases for the same voltage.

Step 1: Understanding semiconductor behavior.

In a semiconductor, the resistance decreases as the temperature increases. This is because, with increased temperature, more charge carriers (electrons or holes) are released, which increases the electrical conductivity of the material. Therefore, the resistance of a semiconductor decreases with rising temperature.


Step 2: Conclusion.
The correct answer is that the resistance of a semiconductor decreases with rising temperature.
Quick Tip: Unlike metals, the resistance of semiconductors decreases with increasing temperature due to the increased movement of charge carriers.

Step 1: Formula for Resistance.

The formula for the resistance \( R \) of a conductor is given by:
\[ R = \rho \cdot \frac{l}{A} \]

Where:
- \( R \) is the resistance
- \( \rho \) is the resistivity
- \( l \) is the length of the conductor
- \( A \) is the cross-sectional area


Step 2: Elimination of options.

- (A) \( \rho \cdot \left( \frac{1}{A} \right) \): This is incorrect because the formula for resistance involves \( \frac{l}{A} \), not \( \frac{1}{A} \).

- (B) \( \rho \cdot \left( \frac{A}{l} \right) \): Incorrect, this would result in a reversed formula.

- (C) \( \frac{1}{\rho A} \): Incorrect, as the formula involves \( \frac{l}{A} \), not \( \frac{1}{A} \).

- (D) \( \frac{lA}{\rho} \): Incorrect, this expression does not correspond to the correct formula for resistance.


Step 3: Conclusion.

Therefore, the correct formula for resistance is \( R = \rho \cdot \frac{l}{A} \).
Quick Tip: Resistance is directly proportional to the length of the conductor and inversely proportional to its cross-sectional area.

Step 1: Definition of Relative Permeability.

The relative permeability (\( \mu_r \)) is the ratio of the permeability of a material to the permeability of free space (\( \mu_0 \)). It measures the ease with which a material can be magnetized.


Step 2: Characteristics of Ferromagnetic Substances.

Ferromagnetic substances, such as iron, cobalt, and nickel, have very high relative permeability (\( \mu_r \)), often much greater than 1. This allows them to be easily magnetized.


Step 3: Elimination of options.

- (A) \( \mu_r < 1 \): Incorrect, this is characteristic of diamagnetic materials, not ferromagnetic materials.

- (B) \( \mu_r = 1 \): Incorrect, this is characteristic of non-magnetic materials like air and vacuum.

- (C) \( \mu_r > 1 \): Incorrect, although ferromagnetic materials have a \( \mu_r > 1 \), the correct answer is \( \mu_r \gg 1 \) (much greater than 1).

- (D) \( \mu_r \gg 1 \): Correct, ferromagnetic materials have much greater relative permeability, indicating a high ability to be magnetized.


Step 4: Conclusion.

Therefore, the relative permeability of ferromagnetic substances is much greater than 1.
Quick Tip: Ferromagnetic materials have relative permeability values much higher than 1, indicating their strong magnetic properties.

Step 1: Understanding series circuit.

In a series circuit, the current that flows through all the resistors is the same because the current has only one path to follow. The total current flowing through the circuit is the same at every point in a series configuration.


Step 2: Explanation of each option.

- (A) same: Correct. In a series circuit, the current through each resistor is the same.

- (B) different: Incorrect. The current is the same throughout the series circuit.

- (C) zero: Incorrect. The current is not zero unless the circuit is open.

- (D) divided proportionally to the value of resistance: Incorrect. This statement is true for parallel circuits, not series circuits.


Step 3: Conclusion.

The current flowing through each resistor in a series circuit is the same.
Quick Tip: In a series circuit, the current is constant across all components, but the voltage is divided among the components.

Step 1: Understanding current density.

Current density is defined as the amount of current flowing per unit area. It is given by the formula: \[ J = \frac{I}{A} \]
where \( J \) is the current density, \( I \) is the current, and \( A \) is the cross-sectional area. The unit of current density is thus amperes per square meter (A/m²).


Step 2: Explanation of each option.

- (A) ampere (A): Incorrect. Ampere is the unit of current, not current density.

- (B) coulomb (C): Incorrect. Coulomb is the unit of charge, not current density.

- (C) ampere per square metre (A/m²): Correct. This is the correct unit for current density.

- (D) volt per metre (V/m): Incorrect. This is the unit of electric field, not current density.


Step 3: Conclusion.

The unit of current density is amperes per square meter (A/m²).
Quick Tip: Current density gives us an idea of how much current is flowing through a particular area of a conductor. Its unit is A/m².

Step 1: Understanding the relation between drift velocity and length of conductor.

The drift velocity (\( v_d \)) of electrons is given by the formula: \[ v_d = \frac{I}{nA e} \]
where:
- \( I \) is the current,
- \( n \) is the number of electrons per unit volume,
- \( A \) is the cross-sectional area of the conductor,
- \( e \) is the charge of an electron.

In the case of a conductor with a constant potential difference, the current remains the same because \( I = \frac{V}{R} \). However, the resistance of the conductor (\( R \)) is given by: \[ R = \rho \frac{L}{A} \]
where \( L \) is the length of the conductor and \( \rho \) is its resistivity. If the length \( L \) is doubled, the resistance doubles, which means the drift velocity is inversely proportional to the length of the conductor.

Step 2: Conclusion.

Thus, if the length of the conductor is doubled, the drift velocity of electrons will be halved.

\[ \boxed{be halved} \] Quick Tip: The drift velocity is inversely proportional to the length of the conductor, given that other factors remain constant.

Step 1: Understanding magnetic field in a solenoid.

A solenoid is a long wire wound in a cylindrical shape, and when current flows through it, it produces a magnetic field. The magnetic field lines inside the solenoid are parallel and uniform, aligning in the same direction as the axis of the solenoid.

Step 2: Conclusion.

Thus, the direction of the magnetic field inside a current-carrying solenoid is parallel to its axis.

\[ \boxed{parallel to axis} \] Quick Tip: The magnetic field inside a solenoid is uniform and parallel to its axis, while outside, it spreads out and behaves like that of a bar magnet.

Step 1: Understanding electromagnetic induction.

Electromagnetic induction refers to the process of generating an electromotive force (EMF) by changing the magnetic field within a conductor. This principle is applied in devices like electric generators.


Step 2: Explanation of options.

- (A) Voltmeter: A voltmeter measures the potential difference between two points but does not operate based on electromagnetic induction.

- (B) Electric motor: An electric motor uses magnetic fields to generate motion, but it does not work based on electromagnetic induction; instead, it works on the force created by current in a magnetic field.

- (C) Electric generator: This device generates electricity based on the principle of electromagnetic induction, as it converts mechanical energy into electrical energy by rotating a coil in a magnetic field.

- (D) Ammeter: An ammeter measures the current flowing through a circuit and does not operate based on electromagnetic induction.


Step 3: Conclusion.

An electric generator works based on electromagnetic induction, making it the correct choice.

\[ \boxed{Correct Answer: Electric generator} \] Quick Tip: An electric generator operates on the principle of electromagnetic induction, where a coil of wire rotates in a magnetic field to produce electricity.

Step 1: Understanding self-inductance.

The self-inductance of a solenoid is a property that measures its ability to induce a back emf in response to a change in current. It depends on factors such as the number of turns per unit length and the length of the solenoid.


Step 2: Explanation of options.

- (A) The current flowing through its medium: The current does not directly affect the self-inductance, although it influences the magnetic field.

- (B) The number of turns per unit length: The self-inductance of a solenoid increases with the number of turns per unit length. More turns result in a stronger magnetic field and higher inductance.

- (C) The length of the solenoid: The self-inductance of a solenoid is also affected by its length, as a longer solenoid has more space for the magnetic field to interact.

- (D) Both (B) and (C): The self-inductance of a solenoid depends on both the number of turns per unit length and the length of the solenoid.


Step 3: Conclusion.

The correct answer is (D) as the self-inductance of a solenoid depends on both the number of turns per unit length and its length.

\[ \boxed{Correct Answer: Both (B) and (C)} \] Quick Tip: Self-inductance increases with more turns per unit length and a longer solenoid, as it enhances the magnetic field produced.

The power of accommodation of the eye refers to its ability to adjust its focus to view objects at varying distances clearly. This is made possible by the lens of the eye, which changes its shape depending on the distance of the object being viewed. Accommodation allows the human eye to focus on objects ranging from a few centimeters (such as reading a book) to infinity (such as looking at distant objects).


This process is controlled by the ciliary muscles which are attached to the lens. The lens, being flexible, can change shape in response to these muscle movements. The power of accommodation involves the following steps:

1. For Near Vision (close objects):
When an object is closer to the eye, the ciliary muscles contract, reducing the tension in the \textit{zonular fibers (the ligaments that hold the lens in place). This allows the lens to become more spherical, increasing its curvature. The increased curvature helps the lens focus light from nearby objects onto the retina, ensuring a sharp image.


2. For Distant Vision (far objects):
When an object is far away, the ciliary muscles relax, increasing the tension in the zonular fibers, which in turn flattens the lens. This reduces the lens's curvature, allowing it to focus light from distant objects onto the retina, producing a sharp image.


The ability to accommodate is measured in \textit{diopters, and the closer the object, the greater the power required for accommodation. However, with age, this power diminishes, and individuals may experience difficulty focusing on near objects, a condition known as \textit{presbyopia. This is because, with age, the lens becomes less flexible, reducing its ability to change shape effectively.


In summary, the power of accommodation is a vital process that ensures clear vision at various distances by adjusting the shape of the eye’s lens through the ciliary muscles.
Quick Tip: \textbf{Power of accommodation = Ability of the eye to change the shape of its lens to focus on objects at varying distances. This power decreases with age (presbyopia).

Electrical resonance refers to the phenomenon that occurs in an electrical circuit when the frequency of the applied alternating current (AC) matches the natural frequency of the circuit, leading to the maximum transfer of energy between the circuit components. This is especially significant in circuits containing inductors (L) and capacitors (C), commonly known as LC circuits. Resonance can be observed in both series and parallel LC circuits, but the effects differ.


1. Resonance in Series LC Circuit:
In a series LC circuit, resonance occurs when the reactance of the inductor (\(X_L = \omega L\)) and the reactance of the capacitor (\(X_C = \frac{1}{\omega C}\)) are equal. Here, \(\omega = 2 \pi f\) is the angular frequency, where \(f\) is the frequency of the applied AC. At resonance, the inductive reactance (\(X_L\)) cancels out the capacitive reactance (\(X_C\)), making the total impedance of the circuit very low. As a result, the current in the circuit becomes maximum.


The resonance frequency \(f_0\) is given by the formula:
\[ f_0 = \frac{1}{2 \pi \sqrt{LC}} \]
where \(L\) is the inductance and \(C\) is the capacitance. At this frequency, the energy oscillates between the magnetic field of the inductor and the electric field of the capacitor.


2. Resonance in Parallel LC Circuit:
In a parallel LC circuit, resonance occurs when the impedance of the circuit becomes infinite, and the current through the circuit becomes minimal. The parallel LC circuit exhibits a different kind of behavior compared to the series circuit, but the concept of resonance remains the same—the total reactance of the circuit becomes zero at the resonance frequency.


3. Application of Electrical Resonance:
Electrical resonance is crucial in many applications, such as:

- Tuned circuits in radio receivers, where a circuit resonates at a particular frequency to select a desired signal.

- Filter circuits, where resonance can be used to select or block certain frequencies.

- Transformers and inductive coils in power systems use resonance to maximize energy efficiency.


In conclusion, electrical resonance occurs when the frequency of an applied AC voltage matches the natural frequency of a circuit, resulting in maximum energy transfer in series circuits and minimum impedance in parallel circuits. This phenomenon is fundamental to various electrical and electronic devices.
Quick Tip: \textbf{Electrical resonance} = Maximum energy transfer occurs when the applied frequency equals the circuit's natural frequency. In series circuits, this leads to maximum current; in parallel circuits, to minimum impedance.

In an alternating current (AC) circuit, the impedance \(Z\) is given by the relation: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Where:
- \(R\) is the resistance of the circuit,
- \(X_L\) is the inductive reactance,
- \(X_C\) is the capacitive reactance.

Here, we are given that:
- \(R = 8 \, \Omega\),
- \(X_L = X_C = 6 \, \Omega\).

Since \(X_L = X_C\), their difference is zero. Therefore, the impedance simplifies to: \[ Z = \sqrt{R^2} = R \]
Substituting the given value of \(R\): \[ Z = 8 \, \Omega \]

Now, the root mean square (RMS) value of the voltage is given by the formula: \[ V_{RMS} = \frac{V_{applied}}{\sqrt{2}} \]
Where \(V_{applied} = 220 \, V\). So, \[ V_{RMS} = \frac{220}{\sqrt{2}} \approx 155.56 \, V \]

Thus, the impedance of the circuit is \(8 \, \Omega\), and the RMS value of the voltage is approximately \(155.56 \, V\).
Quick Tip: Impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). If \(X_L = X_C\), impedance equals the resistance \(R\). RMS voltage \(V_{RMS} = \frac{V_{applied}}{\sqrt{2}}\).

The refractive index (\(n\)) of a medium is a measure of how much the speed of light is reduced inside that medium compared to the speed of light in a vacuum. It is given by the equation: \[ n = \frac{c}{v} \]
Where:
- \(c\) is the speed of light in vacuum (\(3 \times 10^8 \, m/s\)),
- \(v\) is the speed of light in the medium.

The refractive index is not constant for all wavelengths of light. It depends on the wavelength of the light passing through the medium, a phenomenon known as dispersion. Typically, the refractive index decreases as the wavelength of light increases, which is why shorter wavelengths (like violet light) bend more than longer wavelengths (like red light) when passing through a prism or any refractive medium.


The relationship between refractive index and wavelength can be described as follows:
- For shorter wavelengths (like blue or violet light), the refractive index is higher. This is because the interaction between light and the particles in the medium causes more bending of shorter wavelengths.
- For longer wavelengths (like red light), the refractive index is lower, and the light bends less.


Thus, the refractive index (\(n\)) is inversely related to the wavelength of light (\(\lambda\)) in most materials. This is summarized in the equation for dispersion: \[ n(\lambda) = n_0 \left(1 + \frac{A{\lambda^2}\right) \]
Where \(n_0\) is the refractive index at a reference wavelength, and \(A\) is a constant that depends on the material.


In summary, as the wavelength of light increases, the refractive index generally decreases, which results in less bending of longer wavelengths in the medium.
Quick Tip: Refractive index (\(n\)) decreases with increasing wavelength (\(\lambda\)) in most media. Shorter wavelengths (blue/violet) bend more than longer wavelengths (red).

The Curie temperature (denoted as \(T_C\)) is a critical temperature at which a ferromagnetic material undergoes a transition to a paramagnetic state. Below the Curie temperature, the material exhibits strong magnetic properties due to the alignment of its atomic magnetic dipoles. However, when the temperature exceeds the Curie point, thermal energy disrupts this alignment, causing the material to lose its ferromagnetic properties and become paramagnetic.


Ferromagnetic materials, such as iron, cobalt, and nickel, are characterized by their ability to retain magnetization even after the external magnetic field is removed. This property is a result of the alignment of individual magnetic moments within the material. However, as the temperature increases, the thermal energy becomes sufficient to overcome the alignment of these magnetic moments, resulting in the loss of ferromagnetic behavior.


The Curie temperature is unique to each material. For example, the Curie temperature of iron is around \(1043 \, K\), which means that at temperatures above this, iron will no longer exhibit ferromagnetic behavior and will behave as a paramagnetic substance.


In summary, the Curie temperature marks the boundary between the ferromagnetic and paramagnetic states, and it plays a crucial role in the study of magnetism and materials science.
Quick Tip: Curie temperature (\(T_C\)) = Temperature at which ferromagnetic material loses its magnetic properties and becomes paramagnetic.

Paramagnetic substances are materials that are weakly attracted by an external magnetic field and do not retain magnetic properties once the external field is removed. The two key properties of paramagnetic substances are:


1. Weak Attraction to Magnetic Fields:
Paramagnetic substances are weakly attracted to external magnetic fields. This occurs because, in these materials, the magnetic moments of individual atoms or molecules do not completely cancel out. However, the attraction is much weaker than that in ferromagnetic materials. The degree of magnetization is proportional to the applied magnetic field, following the relation \(M = \chi H\), where \(\chi\) is the magnetic susceptibility (which is positive for paramagnetic substances).


2. No Retention of Magnetization:
Unlike ferromagnetic materials, paramagnetic substances do not retain magnetization once the external magnetic field is removed. In these materials, the magnetic moments tend to align with the magnetic field when applied, but random thermal motion disrupts this alignment once the field is removed. This lack of residual magnetization is a key distinguishing feature of paramagnetism.


Other characteristics of paramagnetic substances include their temperature-dependent behavior. As the temperature increases, the thermal agitation of atoms or molecules increases, leading to a decrease in the alignment of the magnetic moments and, consequently, a reduction in the magnetization.


In summary, the two key properties of paramagnetic substances are their weak attraction to magnetic fields and the absence of permanent magnetization.
Quick Tip: Paramagnetic substances = Weakly attracted by magnetic fields, and they do not retain magnetization after the field is removed.

To convert a decimal number into binary, we divide the number by 2 and record the remainders until we get a quotient of 0. The binary equivalent is the sequence of remainders read from bottom to top.


1. Decimal number 21 to binary:

- \( 21 \div 2 = 10 \) remainder \( 1 \)

- \( 10 \div 2 = 5 \) remainder \( 0 \)

- \( 5 \div 2 = 2 \) remainder \( 1 \)

- \( 2 \div 2 = 1 \) remainder \( 0 \)

- \( 1 \div 2 = 0 \) remainder \( 1 \)


Reading the remainders from bottom to top, we get the binary equivalent of 21 as:
\[ 21 (decimal) = \textbf{10101 (binary)} \]

2. Decimal number 43 to binary:

- \( 43 \div 2 = 21 \) remainder \( 1 \)

- \( 21 \div 2 = 10 \) remainder \( 1 \)

- \( 10 \div 2 = 5 \) remainder \( 0 \)

- \( 5 \div 2 = 2 \) remainder \( 1 \)

- \( 2 \div 2 = 1 \) remainder \( 0 \)

- \( 1 \div 2 = 0 \) remainder \( 1 \)


Reading the remainders from bottom to top, we get the binary equivalent of 43 as:
\[ 43 (decimal) = \textbf{101011 (binary)} \]

Thus, the binary equivalents are:
- 21 (decimal) = 10101 (binary)
- 43 (decimal) = 101011 (binary)
Quick Tip: To convert decimal to binary, divide the number by 2 and record the remainders. The binary equivalent is the sequence of remainders from bottom to top.

Nuclear fission and nuclear fusion are two types of nuclear reactions that release energy. However, they differ significantly in their mechanisms and the materials involved.


1. Nuclear Fission:

- In nuclear fission, a heavy atomic nucleus (such as Uranium-235 or Plutonium-239) splits into two smaller nuclei, releasing a large amount of energy.

- It occurs when a neutron strikes a heavy nucleus, causing it to become unstable and break apart. The process also releases additional neutrons, which can trigger further fission reactions (a chain reaction).

- The energy released is in the form of kinetic energy of the fission fragments, gamma rays, and the energy carried by neutrons.

- Example: Nuclear reactors and atomic bombs use fission reactions.

- By-products: Fission produces radioactive by-products, which can be hazardous and require careful disposal.


2. Nuclear Fusion:

- In nuclear fusion, two light atomic nuclei, typically isotopes of hydrogen (such as deuterium and tritium), combine to form a heavier nucleus, releasing a vast amount of energy.

- Fusion reactions occur naturally in stars, including the Sun, where hydrogen nuclei fuse to form helium under extremely high pressure and temperature.

- Fusion requires extremely high temperatures (millions of degrees) to overcome the electrostatic repulsion between positively charged nuclei.

- Example: Hydrogen bombs and the Sun’s core produce energy through fusion.

- By-products: Fusion produces much less radioactive waste than fission and is considered a cleaner energy source. However, it is still very challenging to achieve controlled fusion for practical energy generation on Earth.


Key Differences:
- Fission splits a heavy nucleus into two lighter ones, while fusion combines two light nuclei into a heavier one.
- Fission is currently used in nuclear reactors and weapons, while fusion occurs naturally in stars and is a potential future energy source.
- Fission produces radioactive by-products, while fusion produces significantly less radioactive waste.

In summary, both fission and fusion release vast amounts of energy, but fusion holds the potential for cleaner and more sustainable energy, though it is much more difficult to achieve on Earth.
Quick Tip: Fission = Splitting of a heavy nucleus into lighter ones, used in reactors. Fusion = Combining of light nuclei to form a heavier one, happens naturally in stars and holds potential for clean energy.

Total internal reflection is a phenomenon that occurs when a light ray traveling from a denser medium to a rarer medium strikes the boundary at an angle greater than the critical angle. The necessary conditions for total internal reflection are:


1. The light must travel from a denser medium to a rarer medium:
For total internal reflection to occur, the light must be traveling from a medium with a higher refractive index (denser medium) to a medium with a lower refractive index (rarer medium). A typical example is light moving from water (denser) to air (rarer).


2. The angle of incidence must be greater than the critical angle:
The critical angle (\( \theta_c \)) is the minimum angle of incidence at which total internal reflection occurs. When the angle of incidence exceeds this critical angle, the light is entirely reflected within the denser medium and does not refract into the rarer medium. The critical angle is given by the formula:
\[ \theta_c = \sin^{-1} \left( \frac{n_2}{n_1} \right) \]
Where \(n_1\) is the refractive index of the denser medium and \(n_2\) is the refractive index of the rarer medium.


3. The refractive index of the denser medium must be greater than the refractive index of the rarer medium:
For total internal reflection to occur, the refractive index of the first medium (denser medium) must be greater than that of the second medium (rarer medium). This ensures that light is not refracted out but instead is entirely reflected back into the denser medium.


In summary, for total internal reflection to take place, light must travel from a denser to a rarer medium, the angle of incidence must be greater than the critical angle, and the refractive index of the denser medium must be greater than that of the rarer medium.
Quick Tip: Total internal reflection occurs when the angle of incidence exceeds the critical angle while light moves from a denser to a rarer medium.

Fleming's Left Hand Rule is used to determine the direction of motion in a motor. It helps us understand the relationship between the magnetic field, the current, and the force experienced by a conductor placed in the magnetic field. According to Fleming's Left Hand Rule, if you hold your left hand with the thumb, index finger, and middle finger mutually perpendicular to each other, then:


1. Thumb represents the direction of the motion (force) of the conductor.

2. Index Finger represents the direction of the magnetic field (from north to south).

3. Middle Finger represents the direction of the current flowing through the conductor.


In other words, if the magnetic field is in the direction of the index finger and the current flows in the direction of the middle finger, then the thumb will point in the direction of the force (motion) on the conductor. This rule is applicable to electric motors where the conductor experiences a force when placed in a magnetic field.


Fleming's Left Hand Rule is based on the principle of Lorentz force, which states that a charged particle experiences a force when moving through a magnetic field. In the case of a current-carrying conductor, the charges (electrons) moving through the conductor interact with the magnetic field, producing a force that causes the conductor to move.
Quick Tip: Fleming's Left Hand Rule = Thumb = Motion (Force), Index = Magnetic Field, Middle = Current. Use it for electric motors!

Interference and diffraction are both wave phenomena that involve the interaction of light waves, but they have distinct differences. The two main differences are:


1. Nature of the Phenomenon:
- Interference occurs when two or more waves meet and combine, resulting in constructive or destructive interference. The waves interfere with each other, creating regions of high and low intensities depending on their phase relationship.
- Diffraction, on the other hand, refers to the bending of light waves around obstacles or through narrow openings. It occurs when light encounters an obstacle or slit that is comparable in size to its wavelength, causing the light to spread out and form diffraction patterns.

2. Occurrence:
- Interference is typically observed when two coherent sources of light are involved. Coherent sources have a constant phase relationship, and interference is a result of the superposition of these two sources.

- Diffraction occurs due to a single light source passing through an opening or around an obstacle. It doesn't require two sources and is instead caused by the nature of the wave and the size of the obstacle or aperture.


In summary, interference involves the interaction of two or more coherent waves, while diffraction is the spreading of a single wave around obstacles or through slits.
Quick Tip: Interference = Superposition of waves from two coherent sources. Diffraction = Spreading of light around obstacles or slits.

Steel and soft iron are both ferromagnetic materials, but they exhibit distinct differences in their magnetic properties. The key differences are:


1. Magnetic Retention:

- Steel has a high retentivity, meaning it retains its magnetization even after the external magnetic field is removed. This property makes steel suitable for permanent magnets.

- Soft iron, on the other hand, has a low retentivity, meaning it loses its magnetization quickly when the external magnetic field is removed. This makes soft iron ideal for use in electromagnets, where quick magnetization and demagnetization are required.


2. Magnetic Susceptibility:

- Steel generally has a lower magnetic susceptibility compared to soft iron. This means that steel is not as easily magnetized as soft iron.

- Soft iron has a higher magnetic susceptibility, making it more easily magnetized and demagnetized than steel. It responds more readily to external magnetic fields.


In conclusion, steel is used for permanent magnets due to its high retentivity, while soft iron is used for electromagnets due to its ability to be easily magnetized and demagnetized.
Quick Tip: Steel = High retentivity, used for permanent magnets. Soft iron = Low retentivity, used for electromagnets.

The core of a transformer is made from a magnetic material that is highly efficient in conducting magnetic flux. The two characteristic properties of the material suitable for making the core of a transformer are:


1. High Magnetic Permeability:

The material used for the core should have high magnetic permeability (\(\mu\)), meaning it can easily allow magnetic lines of force to pass through. This ensures that the transformer operates efficiently by reducing energy losses. High magnetic permeability results in a strong magnetic field with less effort and less energy consumption. Materials such as silicon steel are often used for transformer cores because of their high permeability.


2. Low Hysteresis Loss:

The material should have low hysteresis loss, meaning that it should not lose a significant amount of energy when the magnetic field is reversed. Hysteresis loss is the energy lost as heat when the magnetic domains within the material realign with the alternating magnetic field. A material with low hysteresis loss ensures that the transformer operates efficiently, with minimal energy dissipation. Again, silicon steel is commonly used for transformer cores because of its low hysteresis loss.


In summary, a good transformer core material should have high magnetic permeability and low hysteresis loss to maximize energy efficiency.
Quick Tip: Transformer core material = High magnetic permeability + Low hysteresis loss. Common material: Silicon steel.

The de Broglie wavelength (\(\lambda\)) of a particle is given by the equation: \[ \lambda = \frac{h}{p} \]
Where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, J·s\)),
- \(p\) is the momentum of the particle.

Since both the proton and the electron have the same kinetic energy (\(K.E.\)), we can equate their kinetic energies. The kinetic energy is related to momentum as: \[ K.E. = \frac{p^2}{2m} \]
Where \(m\) is the mass of the particle and \(p\) is its momentum. Rearranging this equation to solve for momentum, we get: \[ p = \sqrt{2mK.E.} \]

Now, since the proton and the electron have the same kinetic energy, their momentum will depend on their masses. The mass of the proton (\(m_p\)) is much greater than the mass of the electron (\(m_e\)), so the momentum of the proton will be greater than that of the electron. This means that the de Broglie wavelength will be inversely proportional to the momentum.


Thus, for the same kinetic energy, the electron, having a smaller mass, will have a smaller momentum and, therefore, a larger de Broglie wavelength compared to the proton.


In summary, the electron has a greater de Broglie wavelength because it has a smaller mass and therefore smaller momentum for the same kinetic energy.
Quick Tip: For the same kinetic energy, the particle with the smaller mass has a greater de Broglie wavelength.

The volume density of charge (also known as charge density) is the amount of electric charge per unit volume at a specific point in space. It is denoted by \(\rho\) and is a measure of how much charge is present in a given volume of space. The formula for the volume density of charge is: \[ \rho = \frac{Q}{V} \]
Where:
- \(Q\) is the total charge in the volume,

- \(V\) is the volume in which the charge is distributed.


The SI unit of charge density is coulombs per cubic meter (\(C/m^3\)). It expresses the amount of charge (in coulombs) per unit volume (in cubic meters).


In summary, the volume density of charge is the charge per unit volume, and its SI unit is \(C/m^3\).
Quick Tip: Volume density of charge = \(\frac{Charge}{Volume}\). SI unit = \(C/m^3\).

The magnetic moment (\(M\)) is a vector quantity that represents the strength and direction of a magnetic source, such as a magnetic dipole. It is defined as the product of the current (\(I\)) flowing through a loop and the area (\(A\)) of the loop: \[ M = I \cdot A \]
Where:
- \(I\) is the current in amperes (A),
- \(A\) is the area of the loop in square meters (\(m^2\)).

The magnetic moment determines how a magnetic object will interact with an external magnetic field. It tends to align with the field, and the torque experienced by the magnetic moment in the field is given by: \[ \tau = M \cdot B \cdot \sin(\theta) \]
Where:
- \(\tau\) is the torque,

- \(B\) is the magnetic field strength,

- \(\theta\) is the angle between the magnetic moment and the magnetic field.


The SI unit of magnetic moment is \(Ampere-square meter (A·m^2)\). It is also sometimes expressed in joules per tesla (J/T), as torque (joules) per unit magnetic field strength (tesla).


In summary, the magnetic moment describes the strength and orientation of a magnetic source, and its SI unit is \(A·m^2\).
Quick Tip: Magnetic moment \(M = I \cdot A\), where \(I\) is current and \(A\) is area. SI unit = \(A·m^2\).

Beta rays are high-energy, high-speed electrons or positrons that are emitted from the nucleus of a radioactive atom during the process of beta decay. The two main properties of beta rays are:


1. Nature of the Particles:
Beta rays consist of fast-moving electrons (\(\beta^-\)) or positrons (\(\beta^+\)). The negatively charged particles (\(\beta^-\)) are electrons, while the positively charged particles (\(\beta^+\)) are the antimatter counterparts of electrons, known as positrons. Both types of beta particles are part of the decay process in radioactive isotopes.


2. Penetrating Power:
Beta rays have a greater penetrating power compared to alpha rays but are less penetrating than gamma rays. They can pass through paper or thin materials, but can be stopped by materials such as plastic, aluminum, or a few millimeters of water. Beta radiation is capable of ionizing atoms and molecules in its path, which is one of the reasons it can be harmful to living tissues.


In summary, beta rays are composed of electrons or positrons and have moderate penetrating power compared to other types of radiation.
Quick Tip: Beta rays = Fast electrons (\(\beta^-\)) or positrons (\(\beta^+\)) with moderate penetrating power.

Bohr's stable orbit refers to the concept proposed by Niels Bohr in 1913 to explain the stability of electrons in atoms. According to classical electromagnetism, electrons should spiral into the nucleus due to radiation of energy as they accelerate in orbit. However, Bohr introduced the idea of quantized orbits to solve this problem. The key points of Bohr's theory of stable orbits are:


1. Quantization of Angular Momentum:
Bohr proposed that electrons in an atom can only occupy certain discrete orbits around the nucleus. The electron's angular momentum (\(L\)) in these orbits is quantized and must satisfy the condition:
\[ L = n \hbar \]
Where \(n\) is a positive integer (the principal quantum number), and \(\hbar\) is the reduced Planck's constant. This quantization of angular momentum prevents the electron from spiraling into the nucleus.


2. No Radiation in Stable Orbits:
According to Bohr's model, electrons in stable orbits do not emit radiation, despite being accelerated in circular motion. This was a significant departure from classical electrodynamics, where moving charged particles emit radiation. The electron only emits radiation when it transitions from one orbit to another, leading to the emission or absorption of specific energy levels corresponding to the difference between orbits.


Thus, Bohr's stable orbits are those in which the electron has quantized angular momentum, and it remains stable without emitting radiation. These orbits explain the discrete spectral lines observed in atomic spectra.
Quick Tip: Bohr's stable orbit = Electron's angular momentum is quantized and it does not emit radiation in these orbits.

The work done in rotating an electric dipole in a uniform electric field can be calculated using the formula: \[ W = -\vec{p} \cdot (\vec{E}) \cdot (\cos \theta_2 - \cos \theta_1) \]
Where:
- \(\vec{p}\) is the dipole moment,

- \(\vec{E}\) is the electric field,

- \(\theta_1\) is the initial angle,

- \(\theta_2\) is the final angle.


In this case, the dipole is initially aligned parallel to the electric field, so \(\theta_1 = 0^\circ\), and after rotating it through \(180^\circ\), \(\theta_2 = 180^\circ\). The dipole moment and electric field are parallel in the initial position, and the final position is antiparallel.


Substituting into the formula: \[ W = -pE \left[ \cos 180^\circ - \cos 0^\circ \right] \] \[ W = -pE \left[ -1 - 1 \right] = 2pE \]

Thus, the work done in rotating the dipole through an angle of \(180^\circ\) is: \[ W = 2pE \] Quick Tip: Work done in rotating a dipole = \(W = 2pE\) for a \(180^\circ\) rotation.

In communication systems, different modes of propagation are used to transmit signals from one point to another. Two main modes of propagation are:


1. Ground Wave Propagation:

- In this mode, radio waves travel along the surface of the Earth. The waves follow the curvature of the Earth, making ground wave propagation ideal for long-range communication. Ground waves are typically used for AM radio broadcasting, where signals are transmitted at low frequencies and propagate along the ground. This method is especially useful for communications in the range of 30 Hz to 3 MHz.

- Example: AM radio signals use ground wave propagation.


2. Space Wave Propagation:

- In space wave propagation, radio waves travel through the atmosphere in a direct path from the transmitter to the receiver, usually over line-of-sight distances. This mode is used for high-frequency signals, including microwaves and satellite communication. Space wave propagation does not rely on the Earth’s surface but instead travels through the air, making it suitable for both terrestrial and satellite communication.

- Example: Satellite communication uses space wave propagation.


In summary, the two main modes of propagation in communication systems are ground wave propagation and space wave propagation, each suitable for different frequency ranges and communication needs.
Quick Tip: Ground wave = Travels along the Earth's surface (used in AM radio). Space wave = Direct line-of-sight propagation (used in satellite communication).

Gauss's Theorem states that the total electric flux (\(\Phi_E\)) passing through any closed surface is proportional to the total charge enclosed within the surface. Mathematically, it is expressed as: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \]
Where:
- \(\oint \vec{E} \cdot d\vec{A}\) is the surface integral of the electric field \(\vec{E}\) over the closed surface \(A\),

- \(Q_{enc}\) is the total charge enclosed within the surface,

- \(\epsilon_0\) is the permittivity of free space.


Proof:

The flux through a closed surface is the integral of the electric field over the surface area. If the charge distribution is symmetric, such as in the case of spherical symmetry, the electric field is normal to the surface at all points, and the magnitude is constant. Thus, Gauss's law simplifies to: \[ \Phi_E = E \cdot A \]
Where \(E\) is the magnitude of the electric field and \(A\) is the surface area. Using Gauss's law, we can calculate the electric field at any point outside a spherical charge distribution.


Electric Field Outside a Hollow Uniformly Charged Sphere:

For a uniformly charged spherical shell, we apply Gauss's theorem to find the electric field outside the sphere. Consider a spherical Gaussian surface of radius \(r\) greater than the radius of the sphere. Since the charge is uniformly distributed, by symmetry, the electric field at any point on the surface will be radially outward and have the same magnitude at all points.


By Gauss's law, we have: \[ \Phi_E = E \cdot A = \frac{Q_{enc}}{\epsilon_0} \]
Where \(A = 4\pi r^2\) is the surface area of the spherical Gaussian surface and \(Q_{enc}\) is the total charge on the sphere. Therefore, the electric field intensity \(E\) is given by: \[ E = \frac{Q_{enc}}{4 \pi \epsilon_0 r^2} \]
Thus, the electric field outside the hollow uniformly charged sphere behaves exactly as if all the charge were concentrated at the center of the sphere.
Quick Tip: Gauss's theorem: \(\Phi_E = \frac{Q_{enc}}{\epsilon_0}\). Electric field outside a uniformly charged sphere: \(E = \frac{Q_{enc}}{4 \pi \epsilon_0 r^2}\).

Kirchhoff's Laws:

1. Kirchhoff's Current Law (KCL):
This law states that the total current entering a junction is equal to the total current leaving the junction. Mathematically,
\[ \sum I_{in} = \sum I_{out} \]
This law is a consequence of the conservation of charge.


2. Kirchhoff's Voltage Law (KVL):
This law states that the sum of all voltages around any closed loop in a circuit is zero. Mathematically,
\[ \sum V_{across elements} = 0 \]
This law is a consequence of the conservation of energy in electrical circuits.


Balance of a Wheatstone Bridge:

The Wheatstone bridge consists of four resistors arranged in a diamond shape, with a battery providing a potential difference. The condition for balance occurs when the ratio of resistances in one arm equals the ratio in the opposite arm. Let the resistors in the four arms of the bridge be \(R_1, R_2, R_3, R_4\), with \(R_1\) and \(R_2\) forming one pair of opposite arms and \(R_3\) and \(R_4\) forming the other pair. A galvanometer is connected between the midpoints of the two arms. The balance condition for the bridge is: \[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]
In the balanced state, no current flows through the galvanometer. Using Kirchhoff's laws, we can derive this condition by applying KCL and KVL to the loops in the circuit. The condition for no current through the galvanometer is equivalent to the ratio of resistances being equal.


In summary, Kirchhoff's laws provide the basis for analyzing electrical circuits, and the balance condition for a Wheatstone bridge is given by \(\frac{R_1}{R_2} = \frac{R_3}{R_4}\).
Quick Tip: Kirchhoff’s Laws: KCL (sum of currents at junction) and KVL (sum of voltages around a loop = 0). Wheatstone bridge balance condition: \(\frac{R_1}{R_2} = \frac{R_3}{R_4}\).

Wavefront:

A wavefront is the surface over which an oscillation or wave has a constant phase. It represents all points in space where the wave has the same phase of oscillation at a given instant of time. Wavefronts are perpendicular to the direction of propagation of the wave. For example, in the case of light, spherical wavefronts are generated from a point source.


Secondary Wavelets:

Secondary wavelets are the waves that are produced at every point of a wavefront. According to Huygens' principle, each point on a wavefront acts as a secondary source of spherical wavelets that propagate out in the forward direction. The envelope of these wavelets forms the new wavefront at any later time.


Verification of the Law of Reflection Using Huygens' Wave Theory:

According to Huygens' principle, every point on a wavefront can be considered as the source of secondary wavelets. When a plane wavefront strikes a reflective surface, each point on the surface of the wavefront emits secondary wavelets. The new wavefront is formed by the envelope of these secondary wavelets.


In the case of reflection, consider a plane wavefront approaching a smooth reflective surface. The incident wavefront is at an angle to the surface. The secondary wavelets emanating from each point on the incident wavefront will form a new wavefront, which is reflected at an equal angle on the opposite side of the normal, following the rule that the angle of incidence is equal to the angle of reflection. Thus, Huygens' wave theory verifies the law of reflection: \[ \theta_i = \theta_r \]
Where:
- \(\theta_i\) is the angle of incidence,
- \(\theta_r\) is the angle of reflection.
Quick Tip: Huygens' principle: Each point on a wavefront is a secondary source of wavelets. The law of reflection: \(\theta_i = \theta_r\).

Equivalent Lens:

An equivalent lens is formed when two or more lenses are placed in contact with each other or at a finite distance. The focal length of the equivalent lens is determined by the combination of the focal lengths of the individual lenses.


Expression for Equivalent Focal Length:

For two lenses of focal lengths \(f_1\) and \(f_2\) kept at a distance \(d\), the formula for the equivalent focal length \(f_{eq}\) can be derived using the lens formula for each lens and the condition for combined focal length. The combined power \(P_{eq}\) of two lenses is the sum of their individual powers: \[ P_{eq} = P_1 + P_2 \]
Where: \[ P_1 = \frac{1}{f_1}, \quad P_2 = \frac{1}{f_2} \]
Thus: \[ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} \]
For lenses kept at a distance \(d\), the additional term due to the separation distance between the lenses must be included. The effective focal length is given by the relation: \[ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]
Therefore, the equivalent focal length of the two lenses placed at a distance \(d\) is: \[ f_{eq} = \left( \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \right)^{-1} \] Quick Tip: For two lenses at distance \(d\), the equivalent focal length is: \[ f_{eq} = \left( \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \right)^{-1} \]

Self-Inductance:

Self-inductance (\(L\)) is the property of a coil or solenoid that causes it to oppose any change in the current passing through it. When the current through a coil changes, a time-varying magnetic field is created, which induces an emf (electromotive force) in the coil, opposing the change in current. The self-inductance is defined as the ratio of the induced emf to the rate of change of current: \[ L = \frac{N \cdot \Phi_B}{I} \]
Where:
- \(L\) is the self-inductance,

- \(N\) is the number of turns of the coil,

- \(\Phi_B\) is the magnetic flux linked with the coil,

- \(I\) is the current passing through the coil.


Mutual Inductance:

Mutual inductance (\(M\)) is the property of two coils such that the change in current in one coil induces an emf in the other coil due to the time-varying magnetic field produced by the first coil. The mutual inductance is defined as the ratio of the induced emf in one coil to the rate of change of current in the other coil: \[ M = \frac{N_2 \cdot \Phi_{21}}{I_1} \]
Where:
- \(M\) is the mutual inductance,

- \(N_2\) is the number of turns in the second coil,

- \(\Phi_{21}\) is the magnetic flux through the second coil due to the current in the first coil,

- \(I_1\) is the current in the first coil.


Expression for Mutual Inductance of Two Coaxial Solenoids:

Consider two solenoids with lengths \(l_1\) and \(l_2\), radii \(r_1\) and \(r_2\), and number of turns \(N_1\) and \(N_2\), placed coaxially. The mutual inductance \(M\) between the two coils can be expressed as: \[ M = \frac{\mu_0 N_1 N_2 A}{l} \]
Where:
- \(\mu_0\) is the permeability of free space,

- \(N_1\) and \(N_2\) are the number of turns in the first and second solenoids,

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

- \(l\) is the length of the solenoids.


This expression holds when the solenoids are placed coaxially with their magnetic fields aligned.
Quick Tip: Self-inductance opposes current change in a coil, while mutual inductance measures the induced emf due to a current in another coil. For coaxial solenoids: \[ M = \frac{\mu_0 N_1 N_2 A}{l} \]

Photoelectric Effect:

The photoelectric effect is the phenomenon in which electrons are ejected from a material (usually metal) when light of sufficient frequency (or energy) strikes its surface. The emitted electrons are called photoelectrons. This effect was first observed by Heinrich Hertz in 1887, but it was Albert Einstein who provided the theoretical explanation in 1905, for which he was awarded the Nobel Prize in Physics.


Laws of Photoelectric Effect:

1. Emission of Electrons:
When light of suitable frequency (above a certain threshold frequency) is incident on a metal surface, it causes the emission of electrons from the surface. The emitted electrons are called photoelectrons.

2. Threshold Frequency:
There exists a minimum frequency of light, called the threshold frequency (\(f_0\)), below which no electrons are emitted, regardless of the intensity of light.

3. Effect of Light Intensity:
The number of emitted photoelectrons depends on the intensity of the light. A higher intensity of light leads to more photoelectrons being emitted, but the kinetic energy of the photoelectrons remains unchanged.

4. Effect of Light Frequency:
The kinetic energy of the emitted electrons depends on the frequency of the incident light. If the frequency is higher than the threshold frequency, the kinetic energy of the photoelectrons increases with the increase in frequency of the light.


Einstein's Explanation of the Photoelectric Effect:

Einstein proposed that light behaves as a stream of particles called photons, and each photon has energy given by: \[ E_{photon} = h f \]
Where:

- \(E_{photon}\) is the energy of the photon,

- \(h\) is Planck’s constant,

- \(f\) is the frequency of the light.


When a photon strikes the metal surface, it transfers its energy to an electron. If the photon has energy greater than the work function (\(W\)) of the metal, the electron absorbs this energy and is ejected from the surface. The kinetic energy of the emitted photoelectron is given by: \[ K.E. = h f - W \]
Where \(W\) is the work function, or the minimum energy required to eject an electron from the metal surface. Thus, Einstein’s equation explains the photoelectric effect by treating light as quantized packets of energy.
Quick Tip: Photoelectric effect: Emission of electrons when light strikes a metal surface. Einstein’s explanation: \(K.E. = h f - W\), where \(h\) is Planck’s constant, \(f\) is frequency, and \(W\) is work function.