Step 1: Nature of Electromagnetic Waves.

Electromagnetic waves are generated by accelerating charges. A charge that is stationary or moving with constant velocity does not emit electromagnetic waves. The essential factor for wave production is the acceleration of the charge.


Step 2: Conclusion.

Hence, the correct answer is (iii) an accelerating charge.
Quick Tip: Electromagnetic waves are generated only when charges accelerate or decelerate.

Step 1: Planck's Constant Dimensions.

Planck's constant, \( h \), represents action, which has dimensions of energy multiplied by time. Since energy has the dimensions of force multiplied by distance, Planck's constant therefore has the dimensions of force multiplied by time.


Step 2: Conclusion.

Therefore, the correct answer is (i) force × time.
Quick Tip: The dimension of Planck's constant is force × time, which corresponds to the dimensions of action in quantum mechanics.

Step 1: Parallel Connection of Batteries.

When two ideal batteries with identical emf and internal resistances are connected in parallel, the combined emf remains the same. However, the internal resistances combine in parallel, resulting in a smaller equivalent internal resistance compared to each individual internal resistance.


Step 2: Analysis of options.

- (i) The difference in emf is incorrect for parallel batteries.

- (ii) The sum of emf is not applicable for parallel batteries.

- (iii) Correct: the combined emf stays the same, but the equivalent internal resistance is less than each individual internal resistance.

- (iv) Incorrect, since the internal resistance decreases in a parallel configuration, not increases.


Step 3: Conclusion.

Thus, the correct answer is (iii) the equivalent emf \( E \) is \( E_1 = E_2 \), but the equivalent internal resistance \( r < r_1 \) and \( r' < r_2 \).
Quick Tip: When connecting batteries in parallel, the emf remains the same, but the internal resistance is reduced.

Step 1: Effect of Impurity on Semiconductor.

When an impurity is introduced into an intrinsic semiconductor, it increases the number of free charge carriers (electrons or holes), which in turn enhances the material's conductivity.


Step 2: Conclusion.

Therefore, doping improves the conductivity of the semiconductor, and the correct answer is (ii) increases.
Quick Tip: Doping a semiconductor adds impurities, which increases the number of charge carriers and thus the conductivity.

Step 1: Current Relationships in a Transistor.

In a transistor, the emitter current \( i_e \) is the sum of the collector current \( i_c \) and the base current \( i_b \). Therefore, the collector current \( i_c \) is slightly less than the emitter current \( i_e \), as a small portion of the current flows through the base.


Step 2: Conclusion.

Thus, the correct answer is (i) \( i_c \) is slightly smaller than \( i_e \).
Quick Tip: In a transistor, the emitter current is the largest, and the collector current is slightly smaller than the emitter current.

Step 1: Voltmeter Resistance.

A voltmeter is used to measure the potential difference across a component and is connected in parallel with it. To ensure that it does not draw excessive current, the voltmeter must have a very high resistance.


Step 2: Ammeter Resistance.

In contrast, an ammeter is placed in series with the circuit to measure the current. It should have a minimal resistance to prevent any significant change in the current flowing through the circuit.


Step 3: Conclusion.

Therefore, the correct answer is (ii) A voltmeter should have large resistance.
Quick Tip: A voltmeter needs to have high resistance to avoid drawing current from the circuit, while an ammeter requires low resistance to ensure minimal impact on current flow.

Step 1: Understanding Ionising Potential.

The ionising potential refers to the energy required to remove an electron from the atom and bring it to the zero energy level (ionised state). It is equal to the magnitude of the energy of the atom in its ground state.


Step 2: Formula.

The ionising potential is given by: \[ Ionising potential = \left| E_{ground state} \right| \]
Where \( E_{ground state} \) is the energy of the atom in its ground state.


Step 3: Given Data.

It is given that the energy of the atom in its ground state is \( E_{ground state} = -54.4 \, eV \).


Step 4: Calculation.

Substituting the given value: \[ Ionising potential = \left| -54.4 \, eV \right| = 54.4 \, eV \]

Final Answer:

The ionising potential of the atom is \( \boxed{54.4 \, eV} \). Quick Tip: Ionising potential is the energy required to ionise an atom, which is equal to the magnitude of the energy of the ground state.

Step 1: Understanding the Lens Maker's Formula.

For a concave lens, the lens maker's formula is given by: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where:
- \( f \) is the focal length,
- \( n = 1.5 \) is the refractive index,
- \( R_1 = 30 \, cm \) is the radius of curvature of the first surface,
- \( R_2 = -30 \, cm \) is the radius of curvature of the second surface (negative for concave).

Step 2: Calculation.

Substituting the values into the formula: \[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{30} - \frac{1}{-30} \right) = 0.5 \times \left( \frac{2}{30} \right) \] \[ \frac{1}{f} = \frac{1}{30} \Rightarrow f = -30 \, cm \]

Final Answer:

The focal length of the concave lens is \( \boxed{-30 \, cm} \). Quick Tip: The focal length of a concave lens is always negative, and the lens maker's formula can be used to determine it based on the radii and refractive index.

Step 1: Definition of Mass Number.

The mass number \( A \) of a nucleus is the total number of protons and neutrons in the nucleus. It is given by the formula: \[ A = Z + N \]
Where:
- \( Z \) is the atomic number (number of protons),
- \( N \) is the number of neutrons.

Step 2: Definition of Atomic Number.

The atomic number \( Z \) of an element is the number of protons in the nucleus of the atom, which determines the element’s identity.

Step 3: Difference Between Mass Number and Atomic Number.

The atomic number represents the number of protons in an atom, while the mass number is the sum of the number of protons and neutrons in the nucleus.

Final Answer:

- The mass number \( A \) is the sum of protons and neutrons.

- The atomic number \( Z \) is the number of protons in the nucleus. Quick Tip: Mass number is the sum of protons and neutrons, while atomic number is only the number of protons in the nucleus.

Step 1: Understanding Nuclear Fusion.

Nuclear fusion is the process in which two light atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy. The energy released in fusion is described by Einstein’s mass-energy equivalence principle.

Step 2: Equation for Energy Released in Fusion.

The equation for the energy released in fusion is: \[ E = mc^2 \]
Where:
- \( E \) is the energy released,
- \( m \) is the mass defect (the difference in mass before and after fusion),
- \( c \) is the speed of light.

Final Answer:

The energy released in nuclear fusion is given by \( E = mc^2 \). Quick Tip: In nuclear fusion, the mass defect is converted into energy, as described by the equation \( E = mc^2 \).

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:
- \( C = 10 \, \muF = 10 \times 10^{-6} \, F \) is the capacitance,
- \( V = 2 \, V \) is the potential difference.

Step 2: Calculation.

Substituting the values into the formula: \[ E = \frac{1}{2} \times 10 \times 10^{-6} \times (2)^2 = \frac{1}{2} \times 10 \times 10^{-6} \times 4 = 20 \times 10^{-6} \, J = 0.01 \, J \]

Final Answer:

The energy stored in the capacitor is \( \boxed{0.01 \, J} \). Quick Tip: The energy stored in a capacitor is proportional to the square of the potential difference across it.

Step 1: Substitute the values of \( A \) and \( B \) into the Boolean expression.

The Boolean expression is: \[ Y = AB + BA' \]
Substituting the values \( A = 1 \) and \( B = 1 \): \[ Y = (1 \times 1) + (1 \times 0) = 1 + 0 = 1 \]

Final Answer:

The value of \( Y \) is \( \boxed{1} \). Quick Tip: In Boolean algebra, the complement of 1 is 0, and the complement of 0 is 1.

Step 1: Understanding the Units of N/C.

The unit N/C is the unit of electric field intensity. In terms of base units: \[ N = \frac{kg \cdot m}{s^2}, \quad C = coulomb \]
Thus, the unit N/C is: \[ \frac{kg \cdot m}{s^2 \cdot C} \]

Step 2: Understanding the Units of V/m.

The unit V/m is also the unit of electric field intensity. Voltage (V) is defined as: \[ V = \frac{J}{C} = \frac{kg \cdot m^2}{s^3 \cdot C} \]
Thus, the unit V/m is: \[ \frac{kg \cdot m^2}{s^3 \cdot C \cdot m} = \frac{kg \cdot m}{s^3 \cdot C} \]

Step 3: Comparison.

From both calculations, we can see that the units of N/C and V/m are the same. Hence, both N/C and V/m are the units of electric field intensity.

Final Answer:

Both N/C and V/m are the units of electric field intensity. Quick Tip: Electric field intensity can be expressed in both N/C and V/m, as they represent the same physical quantity.

Step 1: Formula for Work Done in Moving a Charge in an Electric Field.

The work done \( W \) in moving a charge \( q \) through a potential difference \( \Delta V \) is given by: \[ W = q \Delta V \]
Where:
- \( W = -10 \, J \) (since the kinetic energy decreases, the work done is negative),
- \( \Delta V = V_{final} - V_{initial} = 250 \, V - 200 \, V = 50 \, V \).

Step 2: Substituting Known Values.

Substitute the given values into the equation: \[ -10 = q \times 50 \]

Step 3: Solving for the Charge.

Solving for \( q \): \[ q = \frac{-10}{50} = -0.2 \, C \]

Final Answer:

The charge on the particle is \( \boxed{-0.2 \, C} \). Quick Tip: The work done in moving a charge in an electric field is equal to the change in the kinetic energy of the charge, and is related to the potential difference.

Step 1: Difficulty 1 – Stability of the Atom.

In Rutherford’s model, the electrons orbit the nucleus in circular paths. According to classical electromagnetism, an electron in motion should radiate energy in the form of electromagnetic waves. This would cause the electron to lose energy, spiral inward, and eventually crash into the nucleus. Hence, the atom should not be stable, but it is. This is a major difficulty of Rutherford’s model.

Step 2: Difficulty 2 – Discrete Line Spectra.

Rutherford’s model could not explain the observed discrete line spectra of atoms. According to the model, electrons would emit a continuous spectrum as they lose energy, but experimentally, only specific wavelengths (discrete lines) were observed in the atomic spectra. This discrepancy led to the development of a new model by Niels Bohr.

Final Answer:

1. Stability of the atom (electrons should spiral into the nucleus).

2. Failure to explain the discrete line spectra of atoms. Quick Tip: Rutherford’s model laid the foundation for modern atomic theory, but it was unable to explain the stability of atoms and the spectral lines observed in experiments.

Step 1: Definition of Work Function.

The work function (\( \phi \)) is the minimum energy required to remove an electron from the surface of a material (usually a metal) to a point where the electron is free from the material. This is the energy needed to overcome the attractive forces binding the electron to the material.

Step 2: Unit of Work Function.

The work function is measured in joules (J). Since energy is expressed in joules, the unit of the work function is also joules.

Final Answer:

The work function is the minimum energy required to free an electron from a material, and its unit is \( \boxed{Joules (J)} \). Quick Tip: The work function is an important concept in the photoelectric effect and is specific to the material from which electrons are emitted.

Step 1: Expression for Refractive Index.

The refractive index \( n \) of a medium is the ratio of the velocity of light in a vacuum (or air) to the velocity of light in that medium. For a liquid relative to air, the refractive index is given by: \[ n_{liquid/air} = \frac{c}{v} \]
Where:
- \( c \) is the speed of light in air,
- \( v \) is the speed of light in the liquid.

Step 2: Apparent Depth Formula.

When an object is placed in a liquid, its apparent depth is different from the actual depth due to the refraction of light. The formula for the apparent depth \( d_{apparent} \) is given by: \[ d_{apparent} = \frac{d}{n} \]
Where:
- \( d \) is the actual depth,
- \( n \) is the refractive index of the liquid relative to air.

Final Answer:

The refractive index of the liquid relative to air is \( n_{liquid/air} = \frac{c}{v} \), and the formula for the apparent depth of an object placed in a liquid is \( d_{apparent} = \frac{d}{n} \). Quick Tip: The refractive index determines how light slows down when passing through different mediums, affecting the apparent depth of objects.

Step 1: Use the Photoelectric Equation.

The energy of the incident photons can be related to the work function and stopping potential using the photoelectric equation: \[ E_{photon} = \phi + eV_s \]
Where:
- \( E_{photon} = \frac{hc}{\lambda} \) is the energy of the photon,
- \( \phi = 4.0 \, eV \) is the work function,
- \( eV_s = 2.5 \, eV \) is the stopping potential,
- \( h = 6.626 \times 10^{-34} \, J·s \) is Planck's constant,
- \( c = 3 \times 10^8 \, m/s \) is the speed of light,
- \( \lambda \) is the wavelength of light.

Step 2: Rearranging the Equation.

Substitute \( E_{photon} \) in the equation: \[ \frac{hc}{\lambda} = \phi + eV_s \] \[ \frac{hc}{\lambda} = 4.0 + 2.5 = 6.5 \, eV \]

Step 3: Convert eV to Joules.

1 eV = \( 1.6 \times 10^{-19} \, J \), so: \[ 6.5 \, eV = 6.5 \times 1.6 \times 10^{-19} \, J = 1.04 \times 10^{-18} \, J \]

Step 4: Solving for Wavelength \( \lambda \).

Now, solve for \( \lambda \): \[ \lambda = \frac{hc}{1.04 \times 10^{-18}} \]
Substitute the values of \( h \) and \( c \): \[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.04 \times 10^{-18}} = 1.91 \times 10^{-7} \, m = 191 \, nm \]

Final Answer:

The wavelength of light is \( \boxed{191 \, nm} \). Quick Tip: The wavelength of light required for the photoelectric effect can be determined by using the photoelectric equation, considering the work function and stopping potential.

Step 1: Formula for Magnetic Field at the Centre of a Circular Loop.

The magnetic field \( B \) at the centre of a current-carrying circular loop is given by Ampere’s Law: \[ B = \frac{\mu_0 I}{2R} \]
Where:
- \( B \) is the magnetic field,
- \( \mu_0 = 4\pi \times 10^{-7} \, N/A^2 \) is the permeability of free space,
- \( I \) is the current flowing through the loop,
- \( R \) is the radius of the circular loop.

Step 2: Derivation.

Using the Biot-Savart law for a current-carrying element, the magnetic field produced by a small segment of current is integrated over the entire loop. The result of this integration gives the above formula for the magnetic field at the centre of the loop.

Final Answer:

The formula for the magnetic field at the centre of a current-carrying circular loop is: \[ B = \frac{\mu_0 I}{2R} \] Quick Tip: The magnetic field at the centre of a current-carrying loop depends on the current and the radius of the loop. It is directly proportional to the current and inversely proportional to the radius.

Step 1: Formula for Refractive Index in Terms of Angle of Minimum Deviation.

The refractive index \( n \) of a prism can be found using the formula: \[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Where:
- \( A = 60^\circ \) is the angle of the prism,
- \( D = 30^\circ \) is the angle of minimum deviation.

Step 2: Substituting the Given Values.

Substituting the values of \( A \) and \( D \) into the formula: \[ n = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} \]

Step 3: Calculation.

Using known values: \[ \sin(45^\circ) = \frac{\sqrt{2}}{2}, \quad \sin(30^\circ) = \frac{1}{2} \] \[ n = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2} \approx 1.414 \]

Final Answer:

The refractive index of the prism material is \( \boxed{1.414} \). Quick Tip: The refractive index of a prism material can be found using the angle of the prism and the angle of minimum deviation in the formula.

Step 1: Understanding the Problem.

We are given a right-angled triangle \( \triangle ABC \) with:
- \( AB = 3 \, m \),
- \( BC = 4 \, m \),
- The velocity of the wire is \( v = 2.0 \, m/s \),
- The magnetic field is \( B = 0.5 \, T \),
- The magnetic field is perpendicular to the plane of the wire.

We need to find the induced emf in the three segments: \( BC \), \( AC \), and \( AB \).

Step 2: Formula for Induced EMF.

The induced emf in a moving conductor of length \( L \) moving with a velocity \( v \) in a magnetic field \( B \) is given by the formula: \[ \epsilon = B L v \]
Where:
- \( \epsilon \) is the induced emf,
- \( B \) is the magnetic field,
- \( L \) is the length of the conductor,
- \( v \) is the velocity of the conductor.

Step 3: Calculating Induced EMF in Segment BC.

For the segment \( BC \), the length is \( L_{BC} = 4 \, m \).

Substitute the given values into the formula: \[ \epsilon_{BC} = B \times L_{BC} \times v = 0.5 \times 4 \times 2 = 4 \, V \]

Step 4: Calculating Induced EMF in Segment AC.

For the segment \( AC \), the length is \( L_{AC} = 5 \, m \) (using the Pythagoras theorem, \( AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2 + 4^2} = 5 \, m \)).

Substitute the given values into the formula: \[ \epsilon_{AC} = B \times L_{AC} \times v = 0.5 \times 5 \times 2 = 5 \, V \]

Step 5: Calculating Induced EMF in Segment AB.

For the segment \( AB \), the length is \( L_{AB} = 3 \, m \).

Substitute the given values into the formula: \[ \epsilon_{AB} = B \times L_{AB} \times v = 0.5 \times 3 \times 2 = 3 \, V \]

Final Answer:

The induced emf in the segments are:
- \( \epsilon_{BC} = 4 \, V \),
- \( \epsilon_{AC} = 5 \, V \),
- \( \epsilon_{AB} = 3 \, V \). Quick Tip: The induced emf in a conductor moving through a magnetic field depends on the length of the conductor, its velocity, and the strength of the magnetic field.

Step 1: Use the lens formula for convex lens.

The lens formula is given by: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance (30 cm),
- \( u \) is the object distance (-15 cm, since the object is on the opposite side of the light source).

For the convex lens: \[ \frac{1}{f_{convex}} = \frac{1}{30} - \frac{1}{(-15)} \] \[ \frac{1}{f_{convex}} = \frac{1}{30} + \frac{1}{15} = \frac{1}{10} \]
So, \( f_{convex} = 10 \, cm \).

Step 2: Use the lens formula for concave lens.

When a concave lens is placed in contact with the convex lens, the total focal length of the system changes. The distance of the image shifts further by 30 cm. So the new image distance is 30 cm + 30 cm = 60 cm.

The total effective focal length \( f_{total} \) of two lenses in contact is given by: \[ \frac{1}{f_{total}} = \frac{1}{f_{convex}} + \frac{1}{f_{concave}} \]
Let the focal length of the concave lens be \( f_{concave} \).

For the new image distance: \[ \frac{1}{f_{total}} = \frac{1}{60} \quad and \quad \frac{1}{f_{convex}} = \frac{1}{10} \] \[ \frac{1}{60} = \frac{1}{10} + \frac{1}{f_{concave}} \] \[ \frac{1}{f_{concave}} = \frac{1}{60} - \frac{1}{10} = \frac{1}{60} - \frac{6}{60} = -\frac{5}{60} \]
Thus, \( f_{concave} = -12 \, cm \).

Final Answer:

The focal length of the convex lens is \( \boxed{10 \, cm} \), and the focal length of the concave lens is \( \boxed{-12 \, cm} \). Quick Tip: When two lenses are placed in contact, the total focal length of the system is the reciprocal sum of the focal lengths of the individual lenses.

Step 1: Apply Kirchhoff’s Rules.

We are given a circuit with resistors of \( 10 \, \Omega \), \( 3 \, \Omega \), and \( 6 \, \Omega \), and a voltage source of \( 3 \, V \) and \( 5 \, V \).

We will apply Kirchhoff’s Voltage Law (KVL) to the loop and solve for the current using Ohm’s Law: \[ I = \frac{V_{total}}{R_{total}} \]
Where \( V_{total} = 3 \, V + 5 \, V \) and \( R_{total} = 10 \, \Omega + 3 \, \Omega + 6 \, \Omega \).

After solving the equation, we get: \[ I = \frac{8}{19} \approx 0.42 \, A \]

Final Answer:

The current through the 10Ω resistor is \( \boxed{0.42 \, A} \). Quick Tip: To find the current in a circuit, always apply Kirchhoff’s Laws and Ohm’s Law. Simplify the circuit step by step if necessary.

Step 1: Formation of Depletion Layer.

At the p-n junction, when a p-type semiconductor is joined with an n-type semiconductor, free electrons from the n-region diffuse into the p-region and recombine with holes, creating a region depleted of charge carriers called the depletion layer.

Step 2: Reverse Bias and Avalanche Breakdown.

When the diode is reverse biased, the depletion layer widens. If the reverse voltage is increased to a critical value (known as the breakdown voltage), avalanche breakdown occurs, causing a large current to flow through the diode.

% (Here, you would draw the characteristic curve showing reverse bias and the avalanche breakdown.)

Final Answer:

The depletion layer forms due to the recombination of free electrons and holes at the p-n junction, and avalanche breakdown occurs when the reverse bias exceeds a certain threshold. Quick Tip: Avalanche breakdown in a reverse-biased diode occurs when the reverse voltage exceeds the breakdown voltage, causing a large increase in current.

Step 1: Magnetic Force on Electron.

The magnetic force on the electron is given by: \[ F = e v B \]
Where:
- \( e = 1.6 \times 10^{-19} \, C \) is the charge of the electron,
- \( v = 2.0 \times 10^7 \, m/s \) is the speed of the electron,
- \( B = \frac{E}{c} \) is the magnetic field, where \( E \) is the electric field.

Step 2: Maximum Magnetic Force.

The maximum magnetic force occurs when the electric field is at its peak: \[ F = e v \times \frac{E}{c} \]
Substitute the values: \[ F = 1.6 \times 10^{-19} \times 2.0 \times 10^7 \times \frac{300}{3 \times 10^8} = 3.2 \times 10^{-15} \, N \]

Final Answer:

The maximum magnetic force acting on the electron is \( \boxed{3.2 \times 10^{-15} \, N} \). Quick Tip: The magnetic force on an electron moving in an electric field depends on its speed, the strength of the electric field, and the speed of light.

Step 1: Energy of the Photon.

The energy of the emitted photon is given by: \[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 6.626 \times 10^{-34} \, J·s \),
- \( c = 3 \times 10^8 \, m/s \),
- \( \lambda = 1025 \times 10^{-10} \, m \).

Substitute the values: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1025 \times 10^{-10}} = 1.94 \times 10^{-18} \, J \]

Step 2: Transition and Quantum Numbers.

The energy of the photon corresponds to the energy difference between two energy levels of the hydrogen atom. Using the Rydberg formula: \[ E = -13.6 \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
Solve for \( n_1 \) and \( n_2 \) (the quantum numbers).

Final Answer:

The transition involves quantum numbers \( n_1 = 2 \) and \( n_2 = 3 \). Quick Tip: The wavelength of emitted radiation can be used to determine the quantum numbers of the states involved in the transition using the Rydberg formula.

Step 1: Definition of Capacitance.

The capacitance \( C \) of a capacitor is the ratio of the charge \( Q \) stored on one plate to the potential difference \( V \) between the plates. It is given by the formula: \[ C = \frac{Q}{V} \]
Where:
- \( C \) is the capacitance,
- \( Q \) is the charge on the capacitor,
- \( V \) is the potential difference across the capacitor.

Step 2: Analyze the Circuit.

Given the circuit:
- The capacitors are in a combination of series and parallel.
- Capacitors in series have an equivalent capacitance \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \).
- Capacitors in parallel have an equivalent capacitance \( C_{eq} = C_1 + C_2 \).

The circuit consists of capacitors \( 6 \, \muF, 3 \, \muF, 12 \, \muF, 4 \, \muF \) and a resistor \( 10 \, \Omega \) connected to a \( 20 \, V \) power supply.

Step 3: Find the Equivalent Capacitance of the Capacitors.

- The capacitors \( 6 \, \muF \) and \( 3 \, \muF \) are in series: \[ \frac{1}{C_1} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2} \quad \Rightarrow C_1 = 2 \, \muF \]

- Now, the \( 2 \, \muF \) capacitor is in parallel with the \( 12 \, \muF \) capacitor: \[ C_2 = 2 + 12 = 14 \, \muF \]

- The \( 14 \, \muF \) capacitor is in series with the \( 4 \, \muF \) capacitor: \[ \frac{1}{C_{eq}} = \frac{1}{14} + \frac{1}{4} = \frac{1}{\frac{14 \times 4}{14 + 4}} = \frac{1}{\frac{56}{18}} \quad \Rightarrow C_{eq} = 3.21 \, \muF \]

Step 4: Find the Total Charge in the Circuit.

Using the formula \( Q = C_{eq} V \), where \( V = 20 \, V \), we can find the total charge: \[ Q = 3.21 \times 20 = 64.2 \, \muC \]

Step 5: Find the Charge on Each Capacitor.

- The charge on each capacitor is the same in series combination, so the charge on \( 6 \, \muF \) and \( 3 \, \muF \) capacitors will be the same, \( Q_1 = 64.2 \, \muC \).
- The charge on the parallel combination of \( 2 \, \muF \) and \( 12 \, \muF \) will be the same, \( Q_2 = 64.2 \, \muC \).

Final Answer:

The charge on each capacitor is \( \boxed{64.2 \, \muC} \). Quick Tip: To find the charge on each capacitor, first find the equivalent capacitance of the entire circuit, and then use the formula \( Q = C_{eq} \times V \).

Step 1: Gauss’ Theorem.

Gauss’ theorem states that the total electric flux \( \Phi_E \) through a closed surface is equal to the charge enclosed \( Q_{enc} \) divided by the permittivity of free space \( \varepsilon_0 \): \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0} \]

Step 2: Electric Field Due to a Thin Charged Wire.

Consider an infinitely long, straight charged wire with linear charge density \( \lambda \) (charge per unit length). To find the electric field at a distance \( r \) from the wire, we use a cylindrical Gaussian surface with radius \( r \) and length \( L \).

The electric flux through the surface is given by: \[ \Phi_E = E \cdot 2\pi r L \]
Since the charge enclosed is \( Q_{enc} = \lambda L \), applying Gauss' law: \[ E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0} \]
Solving for \( E \): \[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \]

Final Answer:

The intensity of the electric field at a distance \( r \) from a thin charged wire of infinite length is: \[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} \] Quick Tip: The electric field due to an infinitely long charged wire decreases with the distance from the wire, and it is inversely proportional to the distance from the wire.

Step 1: Definition of Magnetic Moment.

Magnetic moment (\( \mu \)) is the measure of the strength and orientation of a magnetic source. For a moving charge, it is given by the formula: \[ \mu = I \cdot A \]
Where:
- \( I \) is the current,
- \( A \) is the area enclosed by the moving charge.

For an electron moving in a circular path, we can express the magnetic moment as: \[ \mu = m \cdot v \cdot r \]
Where:
- \( m \) is the mass of the electron (\( m_e = 9.11 \times 10^{-31} \, kg \)),
- \( v \) is the velocity of the electron (\( 2.0 \times 10^7 \, ms^{-1} \)),
- \( r \) is the radius of the circular orbit (\( 0.3 \, Å = 0.3 \times 10^{-10} \, m \)).

Step 2: Calculation of Magnetic Moment.

Substitute the values into the formula for magnetic moment: \[ \mu = 9.11 \times 10^{-31} \times 2.0 \times 10^7 \times 0.3 \times 10^{-10} \] \[ \mu = 5.46 \times 10^{-34} \, A·m^2 \]

Step 3: Unit of Magnetic Moment.

The SI unit of magnetic moment is \( A·m^2 \) (ampere square meter).

Final Answer:

The magnetic moment of the electron is \( \boxed{5.46 \times 10^{-34} \, A·m^2} \). Quick Tip: The magnetic moment for a moving charge in a circular orbit is given by \( \mu = m \cdot v \cdot r \), where \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of the orbit.

Step 1: Magnetic Force on the Electron.

The magnetic force \( F \) acting on the electron is given by the formula: \[ F = qvB \sin\theta \]
Where:
- \( q = 1.6 \times 10^{-19} \, C \) is the charge of the electron,
- \( v = 4 \times 10^5 \, ms^{-1} \) is the velocity of the electron,
- \( B = 0.3 \, T \) is the magnetic field,
- \( \theta = 60^\circ \) is the angle between the velocity and the magnetic field.

Step 2: Centripetal Force and Radius.

For circular motion, the magnetic force provides the centripetal force, so: \[ qvB \sin\theta = \frac{mv^2}{r} \]
Where \( r \) is the radius of the helical path and \( m = 9.11 \times 10^{-31} \, kg \) is the mass of the electron.

Substitute the known values: \[ 1.6 \times 10^{-19} \times 4 \times 10^5 \times 0.3 \times \sin(60^\circ) = \frac{9.11 \times 10^{-31} \times (4 \times 10^5)^2}{r} \] \[ r = \frac{9.11 \times 10^{-31} \times (4 \times 10^5)^2}{1.6 \times 10^{-19} \times 4 \times 10^5 \times 0.3 \times \sin(60^\circ)} \]
Solving for \( r \), we get: \[ r \approx 2.32 \times 10^{-2} \, m \]

Step 3: Pitch of the Helical Path.

The pitch of the helical path is the distance the electron moves along the direction of the magnetic field during one complete revolution. The pitch \( p \) is given by: \[ p = v \cos\theta \times T \]
Where \( T \) is the time period of the electron’s circular motion.

The time period \( T \) is: \[ T = \frac{2\pi r}{v \sin\theta} \]
Substitute the values: \[ p = v \cos\theta \times \frac{2\pi r}{v \sin\theta} \] \[ p = \frac{2\pi r \cos\theta}{\sin\theta} \]
Substituting the known values: \[ p = \frac{2\pi \times 2.32 \times 10^{-2} \times \cos(60^\circ)}{\sin(60^\circ)} \approx 1.5 \times 10^{-2} \, m \]

Final Answer:

The radius of the helical path is \( \boxed{2.32 \times 10^{-2} \, m} \) and the pitch of the electron’s path is \( \boxed{1.5 \times 10^{-2} \, m} \). Quick Tip: The radius of the helical path of a charged particle moving in a magnetic field depends on the velocity, magnetic field strength, and the angle between the velocity and the magnetic field.

Step 1: Definition of Impedance.

Impedance (\( Z \)) is the total opposition that a circuit presents to the flow of alternating current. It is the combination of resistance (\( R \)) and reactance (\( X \)) in the circuit. Impedance is a complex quantity and is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Where:
- \( R \) is the resistance,
- \( X_L \) is the inductive reactance,
- \( X_C \) is the capacitive reactance.

The unit of impedance is \( \Omega \) (Ohms), the same as resistance.

Step 2: Given Data.

From the given circuit:
- \( R = 30 \, \Omega \) (Resistance),
- \( X_L = 60 \, \Omega \) (Inductive reactance),
- \( X_C = 20 \, \Omega \) (Capacitive reactance),
- \( V = 200 \sqrt{2} \sin(\omega t) \, V \) (Voltage).

Step 3: Calculation of Impedance.

First, calculate the net reactance \( X = X_L - X_C \): \[ X = 60 - 20 = 40 \, \Omega \]
Now, calculate the impedance using the formula: \[ Z = \sqrt{R^2 + X^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega \]

Step 4: Finding the Ammeter and Voltmeter Readings.

The amplitude of the voltage is \( V_{max} = 200 \sqrt{2} \, V \). The RMS value of the voltage is: \[ V_{rms} = \frac{V_{max}}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 \, V \]

The current in the circuit can be calculated using Ohm’s law: \[ I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{50} = 4 \, A \]

Final Answer:

The impedance of the circuit is \( Z = 50 \, \Omega \), the reading of the ammeter is \( \boxed{4 \, A} \), and the reading of the voltmeter is \( \boxed{200 \, V} \). Quick Tip: Impedance in an AC circuit is similar to resistance in a DC circuit but includes both resistive and reactive components (inductive and capacitive reactances).

Step 1: Formula for Fringe Width.

In Young’s double slit experiment, the fringe width (\( \beta \)) is given by: \[ \beta = \frac{\lambda D}{d} \]
Where:
- \( \lambda = 6000 \, Å = 6000 \times 10^{-10} \, m \) is the wavelength of light,
- \( D = 1.5 \, m \) is the distance of the screen from the slits,
- \( d = 2 \, mm = 2 \times 10^{-3} \, m \) is the distance between the two slits.

Substitute the values into the formula: \[ \beta = \frac{6000 \times 10^{-10} \times 1.5}{2 \times 10^{-3}} = \frac{9000 \times 10^{-10}}{2 \times 10^{-3}} = 4.5 \times 10^{-4} \, m = 0.45 \, mm \]

Step 2: Distance of the Third Dark Fringe.

The position of the \( n \)-th dark fringe is given by: \[ y_n = (n - \frac{1}{2}) \beta \]
For the third dark fringe (\( n = 3 \)): \[ y_3 = \left( 3 - \frac{1}{2} \right) \times 0.45 = 2.5 \times 0.45 = 1.125 \, mm \]

Final Answer:

The fringe width is \( \boxed{0.45 \, mm} \), and the distance of the third dark fringe from the centre is \( \boxed{1.125 \, mm} \). Quick Tip: The fringe width is directly proportional to the wavelength of light and the distance to the screen, and inversely proportional to the distance between the slits.

Step 1: Definition of Amplification.

Amplification refers to the process of increasing the amplitude of a signal. In the context of electronic circuits, it is the process by which the amplitude of an electrical signal (voltage or current) is increased without altering its other characteristics, such as frequency.

Step 2: Given Data.

- Increase in collector current, \( \Delta I_C = 1 \, mA = 10^{-3} \, A \),
- Increase in base current, \( \Delta I_B = 5 \, \muA = 5 \times 10^{-6} \, A \).

Step 3: Current Gain \( \alpha \) and \( \beta \).

In a common emitter amplifier, the current gain \( \beta \) and \( \alpha \) are related as: \[ \beta = \frac{\Delta I_C}{\Delta I_B} \]
Substitute the given values: \[ \beta = \frac{1 \times 10^{-3}}{5 \times 10^{-6}} = 200 \]

The current gain \( \alpha \) is related to \( \beta \) by the formula: \[ \alpha = \frac{\beta}{\beta + 1} \]
Substitute the value of \( \beta \): \[ \alpha = \frac{200}{200 + 1} = \frac{200}{201} \approx 0.995 \]

Final Answer:

The current gain is \( \boxed{\alpha = 0.995} \) and \( \boxed{\beta = 200} \). Quick Tip: The current gain \( \beta \) is the ratio of the change in collector current to the change in base current, while \( \alpha \) is the ratio of the change in collector current to the total current.

Step 1: Definition of Oscillator.

An oscillator is a circuit that generates a continuous periodic waveform (usually sine or square) without requiring an external input signal. The circuit generates its own signal, often referred to as the "oscillations."

Step 2: Working of a Transistor as an Oscillator.

In a transistor oscillator circuit, feedback is used to produce an oscillating signal. A common configuration for a transistor oscillator is the Colpitts oscillator. This type of oscillator uses an LC tank circuit for frequency determination. The feedback network controls the oscillation, ensuring that the transistor amplifies the signal continuously.

The general working of a transistor oscillator:
1. The transistor is used to amplify the signal.
2. The LC network provides feedback from the output to the input.
3. The feedback loop sustains oscillations, and the circuit generates a stable waveform.

% (Diagram of a transistor oscillator like a Colpitts oscillator can be drawn here.)

Step 3: Circuit Diagram.

Here’s a simplified circuit diagram for a common emitter oscillator using a transistor:
\[ \begin{array}{c} Collector \ \longrightarrow L \ \longrightarrow Resistor \ \longrightarrow Vcc
\ \quad \quad \ \quad \downarrow
\quad Base \longleftarrow Capacitor \longrightarrow Emitter
\end{array} \]

In this diagram, the transistor amplifies the signal, while the LC circuit (inductor and capacitor) sets the frequency of oscillation.

Final Answer:

An oscillator is a circuit that generates a continuous periodic signal. A transistor oscillator uses feedback to sustain oscillations and typically uses an LC tank circuit to determine the frequency. Quick Tip: Transistor oscillators are widely used in generating radio frequencies and other high-frequency signals in communication systems.