Step 1: Understanding the Concept:

The torque (\(\tau\)) experienced by an electric dipole in a uniform electric field is given by the cross product of the dipole moment vector (\(\vec{p}\)) and the electric field vector (\(\vec{E}\)). The magnitude of the torque depends on the magnitudes of the dipole moment and the electric field, as well as the angle between their vectors.


Step 2: Key Formula or Approach:

The formula for the torque on a dipole is given by:
\[ \vec{\tau} = \vec{p} \times \vec{E} \]
The magnitude of the torque is given by:
\[ \tau = |\vec{p}| |\vec{E}| \sin\theta \]
where \(\theta\) is the angle between the dipole moment vector \(\vec{p}\) and the electric field vector \(\vec{E}\).


Step 3: Detailed Explanation:

Given data:

Dipole moment, \(p = 1.0 \times 10^{-12}\) Cm. Since it lies along the x-axis, the vector is \(\vec{p} = (1.0 \times 10^{-12}) \hat{i}\) Cm.

Magnitude of the electric field, \(E = 2.0 \times 10^4\) NC\(^{-1}\).

The unit vector along the electric field is \(\hat{E} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\).

The electric field vector is \(\vec{E} = E \hat{E} = (2.0 \times 10^4) \left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right)\) NC\(^{-1}\).


Method 1: Using the cross product
\[ \vec{\tau} = \vec{p} \times \vec{E} \] \[ \vec{\tau} = \left(1.0 \times 10^{-12} \hat{i}\right) \times \left( (2.0 \times 10^4) \left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) \right) \] \[ \vec{\tau} = (1.0 \times 10^{-12}) (2.0 \times 10^4) \left[ \hat{i} \times \left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) \right] \] \[ \vec{\tau} = (2.0 \times 10^{-8}) \left[ \frac{\sqrt{3}}{2}(\hat{i} \times \hat{i}) + \frac{1}{2}(\hat{i} \times \hat{j}) \right] \]
We know that \(\hat{i} \times \hat{i} = 0\) and \(\hat{i} \times \hat{j} = \hat{k}\).
\[ \vec{\tau} = (2.0 \times 10^{-8}) \left[ 0 + \frac{1}{2}\hat{k} \right] = 1.0 \times 10^{-8} \hat{k} Nm \]
The magnitude of the torque is \(|\vec{\tau}| = 1.0 \times 10^{-8}\) Nm.


Method 2: Using the angle between vectors

The dipole moment vector is along the x-axis, so its unit vector is \(\hat{p} = \hat{i}\).

The electric field unit vector is \(\hat{E} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\).

The cosine of the angle \(\theta\) between \(\vec{p}\) and \(\vec{E}\) is given by the dot product of their unit vectors:
\[ \cos\theta = \hat{p} \cdot \hat{E} = (\hat{i}) \cdot \left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) = \frac{\sqrt{3}}{2} \]
This implies that \(\theta = 30^\circ\).

Now, we can find \(\sin\theta\):
\[ \sin\theta = \sin(30^\circ) = \frac{1}{2} \]
Using the magnitude formula for torque:
\[ \tau = p E \sin\theta \] \[ \tau = (1.0 \times 10^{-12}) \times (2.0 \times 10^4) \times \frac{1}{2} \] \[ \tau = 1.0 \times 10^{-8} Nm \]

Step 4: Final Answer:

Both methods yield the same result. The magnitude of the torque acting on the dipole is \(1.0 \times 10^{-8}\) Nm. This corresponds to option (B).
Quick Tip: When dealing with vector quantities like torque, remember the formula \(\vec{\tau} = \vec{p} \times \vec{E}\). If magnitudes and the angle are easier to find, use \(\tau = pE\sin\theta\). In this problem, finding the angle from the unit vectors is straightforward.

Step 1: Understanding the Concept:

This question relates to the mechanism of electric current flow in a conductor. The key is to distinguish between the speed of the charge carriers (electrons) and the speed at which the effect of the electric field propagates.


Step 2: Detailed Explanation:

Let's analyze each option:

(A) The filament is a conductor, which means it already has a vast number of free electrons. It is not initially empty. Also, the individual electrons from the power source take a very long time to travel to the bulb. This statement is incorrect.


(B) The age or resistance of the filament affects its brightness and lifespan, but it does not explain the instantaneous nature of the glow. A bulb glows instantaneously whether it is new or old. This statement is not the fundamental reason.


(C) When a switch is closed, an electric potential difference is applied across the conductor. This establishes an electric field throughout the entire circuit almost instantaneously (at a speed close to the speed of light). This electric field exerts a force (\(F = -eE\)) on every free electron already present in the filament. This force causes all the free electrons throughout the filament to start drifting simultaneously, constituting an electric current. The collisions of these drifting electrons with the lattice ions of the filament cause it to heat up and glow. This is the correct explanation.


(D) The free electrons themselves move with a very small average velocity called drift velocity, which is typically on the order of millimeters per second (mm/s). They do not travel at the speed of light. It is the electric field that propagates at nearly the speed of light. This statement is incorrect.


Step 3: Final Answer:

The correct reason for the instantaneous glow of the bulb is the near-instantaneous establishment of the electric field across the filament, which forces the free electrons already present in the filament to move, causing a current to flow immediately. Therefore, option (C) is the correct answer.
Quick Tip: A common misconception is that the electrons from the switch travel to the bulb. Remember the analogy of a pipe filled with marbles: if you push one marble in at one end, a marble almost instantly comes out the other end, even though each individual marble only moved a short distance. Similarly, the electric field "pushes" all the electrons in the wire at once.

Step 1: Understanding the Concept:

A charged particle moving in a magnetic field experiences a magnetic force, known as the Lorentz force. This force is always perpendicular to both the velocity of the particle and the magnetic field direction.


Step 2: Key Formula or Approach:

The magnetic force \(\vec{F}\) on a charge \(q\) moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is given by the Lorentz force equation:
\[ \vec{F} = q(\vec{v} \times \vec{B}) \]
The magnitude of the force is found by first calculating the cross product and then finding the magnitude of the resulting vector.


Step 3: Detailed Explanation:

Given data:

Charge of the particle = \(q\).

Velocity of the particle, \(\vec{v} = v_0\hat{i}\).

Magnetic field, \(\vec{B} = B_1\hat{j} + B_2\hat{k}\).


First, we calculate the cross product \(\vec{v} \times \vec{B}\):
\[ \vec{v} \times \vec{B} = (v_0\hat{i}) \times (B_1\hat{j} + B_2\hat{k}) \]
Using the distributive property of the cross product:
\[ \vec{v} \times \vec{B} = (v_0\hat{i} \times B_1\hat{j}) + (v_0\hat{i} \times B_2\hat{k}) \] \[ \vec{v} \times \vec{B} = v_0 B_1 (\hat{i} \times \hat{j}) + v_0 B_2 (\hat{i} \times \hat{k}) \]
We know the cyclic properties of unit vectors: \(\hat{i} \times \hat{j} = \hat{k}\) and \(\hat{i} \times \hat{k} = -\hat{j}\).

Substituting these into the equation:
\[ \vec{v} \times \vec{B} = v_0 B_1 (\hat{k}) + v_0 B_2 (-\hat{j}) = -v_0 B_2 \hat{j} + v_0 B_1 \hat{k} \]
Now, substitute this back into the Lorentz force equation:
\[ \vec{F} = q(\vec{v} \times \vec{B}) = q(-v_0 B_2 \hat{j} + v_0 B_1 \hat{k}) \]
The magnitude of the force vector \(\vec{F}\) is:
\[ |\vec{F}| = |q| \sqrt{(-v_0 B_2)^2 + (v_0 B_1)^2} \] \[ |\vec{F}| = q \sqrt{v_0^2 B_2^2 + v_0^2 B_1^2} \]
Factor out \(v_0^2\) from under the square root:
\[ |\vec{F}| = q \sqrt{v_0^2 (B_1^2 + B_2^2)} \] \[ |\vec{F}| = q v_0 \sqrt{B_1^2 + B_2^2} \]

Step 4: Final Answer:

The magnitude of the force experienced by the particle is \(qv_0\sqrt{B_1^2 + B_2^2}\). This matches option (C).
Quick Tip: To solve cross products quickly, use the determinant method for \( \vec{v} \times \vec{B} \): \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
v_x & v_y & v_z
B_x & B_y & B_z \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
v_0 & 0 & 0
0 & B_1 & B_2 \end{vmatrix} = \hat{i}(0) - \hat{j}(v_0 B_2) + \hat{k}(v_0 B_1) = -v_0 B_2 \hat{j} + v_0 B_1 \hat{k} \] This avoids errors with signs in unit vector cross products.

Step 1: Understanding the Concept:

A bar magnet (magnetic dipole) placed in a uniform magnetic field experiences a torque that tends to align it with the field. The magnitude of this torque depends on the magnetic moment of the magnet, the strength of the magnetic field, and the angle between the magnet's axis and the field lines.


Step 2: Key Formula or Approach:

The magnitude of the torque (\(\tau\)) on a bar magnet with magnetic moment \(m\) in a uniform magnetic field \(B\) is given by:
\[ \tau = mB\sin\theta \]
where \(\theta\) is the angle between the magnetic moment vector and the magnetic field vector.


Step 3: Detailed Explanation:

Initial State:

The bar magnet is initially at right angles to the magnetic field.

So, the initial angle is \(\theta_1 = 90^\circ\).

The initial torque (\(\tau_1\)) is:
\[ \tau_1 = mB\sin(\theta_1) = mB\sin(90^\circ) \]
Since \(\sin(90^\circ) = 1\), the initial torque is maximum:
\[ \tau_1 = mB \]

Final State:

The magnet is rotated until the torque becomes one-half of its initial value.

Let the final torque be \(\tau_2\) and the final angle be \(\theta_2\).

Given: \(\tau_2 = \frac{1}{2}\tau_1\).
\[ mB\sin(\theta_2) = \frac{1}{2}(mB) \] \[ \sin(\theta_2) = \frac{1}{2} \]
This gives the final angle \(\theta_2 = 30^\circ\) (assuming the angle decreases from 90\(^\circ\)).


Angle of Rotation:

The question asks for the angle *through which* the bar magnet is rotated. This is the difference between the initial and final angles.

Angle of rotation, \(\Delta\theta = \theta_1 - \theta_2\).
\[ \Delta\theta = 90^\circ - 30^\circ = 60^\circ \]

Step 4: Final Answer:

The angle through which the magnet is rotated is \(60^\circ\). This corresponds to option (C).
Quick Tip: Be careful with the wording. The question asks for the "angle through which it is rotated" (\(\Delta\theta\)), not the final angle (\(\theta_2\)). Always read the question carefully to ensure you are solving for the correct quantity.

Step 1: Understanding the Concept:

Magnetic susceptibility (\(\chi\)) is a dimensionless proportionality constant that indicates the degree of magnetization of a material in response to an applied magnetic field. The sign of \(\chi\) determines the type of magnetic material.


Step 2: Detailed Explanation:

Materials are classified based on their magnetic susceptibility as follows:

(A) Diamagnetic materials: These materials are weakly repelled by a magnetic field. When placed in an external magnetic field, they develop a magnetization in the direction opposite to the applied field. This property results in a small, negative magnetic susceptibility (\(\chi < 0\)). Examples include copper, gold, water, and bismuth.


(B) Paramagnetic materials: These materials are weakly attracted by a magnetic field. They have a small, positive magnetic susceptibility (\(\chi > 0\)). Their atoms have permanent magnetic dipole moments that are randomly oriented but tend to align with an external field. Examples include aluminum, platinum, and oxygen.


(C) Ferromagnetic materials: These materials are strongly attracted by a magnetic field and can be permanently magnetized. They have a large, positive magnetic susceptibility (\(\chi \gg 0\)). Examples include iron, nickel, and cobalt.


(D) Non-magnetic materials: This is not a formal classification in physics. All materials exhibit some form of magnetic behavior. The term is often used colloquially for materials with very weak magnetic responses (i.e., diamagnetic and paramagnetic materials). However, in the context of physics classifications, it is imprecise.


The question specifically asks for materials with negative magnetic susceptibility. Based on the definitions above, these are diamagnetic materials.


Step 3: Final Answer:

Materials with negative magnetic susceptibility are classified as diamagnetic. Therefore, option (A) is the correct answer.
Quick Tip: Remember the signs of magnetic susceptibility (\(\chi\)) for different materials: \textbf{Dia}magnetic: \(\chi\) is \textbf{negative} (think "di" for different/opposite direction). \textbf{Para}magnetic: \(\chi\) is small and \textbf{positive}. \textbf{Ferro}magnetic: \(\chi\) is large and \textbf{positive}. This simple sign convention is key to solving many conceptual questions on magnetism.

Step 1: Understanding the Concept:

This problem deals with the phenomenon of self-induction. According to Faraday's law of electromagnetic induction, a changing current in a coil induces an electromotive force (emf) in the coil itself. This induced emf opposes the change in current. The property of the coil that quantifies this opposition is called self-inductance (L).


Step 2: Key Formula or Approach:

The induced emf (\(\varepsilon\)) in a coil due to a change in current is given by the formula:
\[ \varepsilon = -L \frac{dI}{dt} \]
where \(L\) is the self-inductance, and \(\frac{dI}{dt}\) is the rate of change of current. For calculations involving magnitude, we can write:
\[ |\varepsilon| = L \left| \frac{\Delta I}{\Delta t} \right| \]

Step 3: Detailed Explanation:

Given data:

Initial current, \(I_1 = 8\) A.

Final current, \(I_2 = 6\) A.

Time interval, \(\Delta t = 4 ms = 4 \times 10^{-3} s\).

Induced emf, \(\varepsilon = 1.5\) V.


First, calculate the change in current (\(\Delta I\)):
\[ \Delta I = I_2 - I_1 = 6 A - 8 A = -2 A \]
Next, calculate the rate of change of current \(\left| \frac{\Delta I}{\Delta t} \right|\):
\[ \left| \frac{\Delta I}{\Delta t} \right| = \left| \frac{-2 A}{4 \times 10^{-3} s} \right| = \frac{2}{4 \times 10^{-3}} = 0.5 \times 10^3 A/s = 500 A/s \]
Now, rearrange the formula for self-inductance \(L\):
\[ L = \frac{|\varepsilon|}{\left| \frac{\Delta I}{\Delta t} \right|} \]
Substitute the given values:
\[ L = \frac{1.5 V}{500 A/s} = \frac{1.5}{500} H \] \[ L = 0.003 H \]
To express the answer in millihenries (mH), we multiply by 1000:
\[ L = 0.003 \times 1000 mH = 3 mH \]

Step 4: Final Answer:

The value of the self-inductance of the coil is 3 mH. This corresponds to option (C).
Quick Tip: Always be careful with units. In this problem, the time is given in milliseconds (ms) and must be converted to seconds (s) for calculations in SI units. Also, remember that the negative sign in the formula \(\varepsilon = -L \frac{dI}{dt}\) represents Lenz's law (opposition to the change), but for finding the magnitude of L, we use the absolute values.

Step 1: Understanding the Concept:

This question is about displacement current, a concept introduced by James Clerk Maxwell. Displacement current (\(I_d\)) is not a current of moving charges but is produced by a time-varying electric field. It has the same units as electric current and produces a magnetic field just as a conduction current does. In the space between capacitor plates, the changing electric field during charging or discharging constitutes a displacement current.


Step 2: Key Formula or Approach:

The displacement current \(I_d\) is related to the rate of change of electric flux \(\Phi_E\) by the formula:
\[ I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]
For a parallel plate capacitor with a uniform electric field \(E\) perpendicular to the plates of area \(A\), the electric flux is \(\Phi_E = E \cdot A\). Therefore, the formula becomes:
\[ I_d = \varepsilon_0 A \frac{dE}{dt} \]
where \(\varepsilon_0\) is the permittivity of free space (\(\approx 8.85 \times 10^{-12}\) F/m).


Step 3: Detailed Explanation:

Given data:

Area of each plate, \(A = 2.5 \times 10^{-3} m^2\).

Rate of change of electric field, \(\frac{dE}{dt} = 4 \times 10^6 Vm^{-1}s^{-1}\).

Permittivity of free space, \(\varepsilon_0 = 8.85 \times 10^{-12} C^2N^{-1}m^{-2}\).


Now, substitute these values into the formula for displacement current:
\[ I_d = \varepsilon_0 A \frac{dE}{dt} \] \[ I_d = (8.85 \times 10^{-12}) \times (2.5 \times 10^{-3}) \times (4 \times 10^6) \]
Let's simplify the calculation:
\[ I_d = 8.85 \times (2.5 \times 4) \times 10^{-12 - 3 + 6} \] \[ I_d = 8.85 \times 10 \times 10^{-9} \] \[ I_d = 8.85 \times 10^{-8} A \]

Step 4: Final Answer:

The displacement current between the plates of the capacitor is \(8.85 \times 10^{-8}\) A. This matches option (C).
Quick Tip: Remember the value of the permittivity of free space, \(\varepsilon_0 \approx 8.85 \times 10^{-12}\) SI units, as it is frequently used in electrostatics and electromagnetism problems. The concept of displacement current is crucial for understanding Maxwell's equations and the propagation of electromagnetic waves.

Step 1: Understanding the Concept:

A current-carrying conductor produces a magnetic field in the space around it. The pattern (shape and direction) of the magnetic field lines depends on the geometry of the conductor and the direction of the current. For a long, straight wire, the magnetic field lines form closed loops around the wire.


Step 2: Key Formula or Approach:

The direction of the magnetic field lines around a straight current-carrying wire is determined by the Right-Hand Thumb Rule. This rule states: If you point the thumb of your right hand in the direction of the current, your fingers will curl in the direction of the magnetic field lines.


Step 3: Detailed Explanation:

In this problem, the wire is held vertically, and the current is in the upward direction.

1. Apply the Right-Hand Thumb Rule: Point your right thumb upwards, parallel to the wire.

2. Curl your fingers around the wire. Your fingers will wrap around the wire in a counter-clockwise direction when viewed from above.

3. The path traced by your curling fingers represents the magnetic field lines. This path consists of circles centered on the wire.

4. Since the wire is vertical, these circles lie in horizontal planes.


Therefore, the shape of the magnetic field lines are concentric horizontal circles around the wire.

Let's analyze the other options:

(A) Radially outward lines describe the electric field of a positive line charge, not a magnetic field from a current.

(B) Parallel lines would imply the magnetic field is in the same direction as the current, which is incorrect.

(D) Helical lines are the trajectory of a charged particle moving in a uniform magnetic field at an angle, not the shape of the field lines themselves from a straight wire.


Step 4: Final Answer:

The correct description of the magnetic field lines is concentric horizontal circles around the wire. This corresponds to option (C).
Quick Tip: The Right-Hand Thumb Rule is fundamental for determining the direction of magnetic fields from currents. It's essential to visualize this rule correctly. For a straight wire, the field lines are circles. For a circular loop, the field line through the center is a straight line.

Step 1: Understanding the Concept:

Materials are classified based on their ability to conduct electricity. This property is determined by the availability of free charge carriers (usually electrons) to move through the material.


Conductors: Have a large number of free electrons and offer very low resistance to the flow of current. Metals are excellent conductors.
Insulators: Have very few free electrons and offer very high resistance to current flow.
Semiconductors: Have electrical properties between those of conductors and insulators. Their conductivity can be significantly changed by temperature or by adding impurities (doping).


Step 2: Detailed Explanation:

Let's analyze each of the given options:

(A) Sn (Tin): Tin is a metal. Metals are characterized by a 'sea' of delocalized electrons that are free to move throughout the metallic lattice. This makes them excellent electrical conductors at room temperature.


(B) Mica: Mica is a group of silicate minerals. It is a very good electrical insulator, meaning it does not conduct electricity well. It is often used as an insulating material in electrical components.


(C) Si (Silicon): Silicon is a classic example of a semiconductor. In its pure (intrinsic) form at room temperature, it has a limited number of free charge carriers and is a poor conductor compared to metals. Its conductivity increases with temperature.


(D) C (Carbon): Carbon exists in various forms (allotropes). Diamond is an excellent insulator. Graphite, another allotrope, is a conductor due to its layered structure with delocalized electrons. However, Tin (Sn) is a metal and is unambiguously classified as a conductor in general contexts. Given the choices, Sn is the best and most direct answer for an electrical conductor.


Step 3: Final Answer:

Among the given options, Tin (Sn) is a metal and therefore is the best example of an electrical conductor at room temperature. Option (A) is the correct choice.
Quick Tip: For classification questions, it's helpful to remember key examples for each category. \textbf{Conductors:} Metals (Copper, Silver, Gold, Aluminum, Tin). \textbf{Insulators:} Glass, Rubber, Plastic, Mica, Diamond. \textbf{Semiconductors:} Silicon (Si), Germanium (Ge). This basic knowledge is often tested in competitive exams.

Step 1: Understanding the Concept:

The magnification (\(m\)) produced by a spherical mirror provides two key pieces of information:

Sign of magnification: A negative sign (\(m < 0\)) indicates that the image is real and inverted. A positive sign (\(m > 0\)) indicates that the image is virtual and erect.
Magnitude of magnification: The absolute value \(|m|\) tells about the size of the image relative to the object. If \(|m| > 1\), the image is magnified. If \(|m| < 1\), the image is diminished. If \(|m| = 1\), the image is of the same size.


Step 2: Detailed Explanation:

Given magnification, \(m = -2.0\).


Analysis of the sign:

The magnification is negative (\(m = -2.0\)). This implies that the image formed is real and inverted.


Analysis of the magnitude:

The magnitude of the magnification is \(|m| = |-2.0| = 2.0\). Since \(|m| > 1\), the image is magnified (enlarged).


Identifying the mirror type:

We need to determine which type of spherical mirror can produce a real, inverted, and magnified image.

A convex mirror always forms a virtual, erect, and diminished image, regardless of the object's position. Its magnification is always positive and less than 1 (\(0 < m < 1\)).
A concave mirror can form different types of images depending on the object's position. It forms a real and inverted image when the object is placed beyond its focal point (F). Specifically, it forms a magnified, real, and inverted image when the object is placed between the center of curvature (C) and the focal point (F).

Since the image is real and magnified, the mirror must be a concave mirror.


Step 3: Final Answer:

The mirror used is concave, and the image formed is real. This corresponds to option (B).
Quick Tip: Remember the sign convention for magnification in mirrors: \textbf{N}egative is \textbf{R}eal and \textbf{I}nverted (\textbf{NRI}). \textbf{P}ositive is \textbf{V}irtual and \textbf{E}rect (\textbf{PVE}). A convex mirror can never produce a real image or a magnified image.

Step 1: Understanding the Concept:

This question tests the understanding of the wave-particle duality, a fundamental concept in modern physics. This principle states that all matter and radiation exhibit both wave-like and particle-like properties. The particle aspect of light is the photon, and the wave aspect of matter (like electrons) is described by the de Broglie wavelength.


Step 2: Detailed Explanation:

Let's analyze each statement:

(A) Photons of light show diffraction whereas electrons do not show diffraction.

This statement is incorrect. Light (photons) exhibits wave properties like diffraction. The Davisson-Germer experiment famously demonstrated that electrons also exhibit wave properties, including diffraction, confirming the de Broglie hypothesis.


(B) Electrons have momentum whereas photons do not have momentum.

This statement is incorrect. Both electrons and photons possess momentum. The momentum of an electron is given by \(p = mv\). The momentum of a photon is given by \(p = E/c = h/\lambda\), where \(E\) is its energy, \(c\) is the speed of light, \(h\) is Planck's constant, and \(\lambda\) is its wavelength.


(C) Photons of light and electrons both exhibit dual nature.

This statement is correct. This is the essence of wave-particle duality. Light behaves as a wave (showing interference, diffraction) and as a particle (photons in the photoelectric effect). Similarly, electrons behave as particles (having mass and charge) and as waves (showing diffraction).


(D) All electromagnetic radiations do not have photons.

This statement is incorrect. According to the quantum theory of light, all electromagnetic radiation is quantized. It consists of discrete packets of energy called photons. The energy of each photon is proportional to the frequency of the radiation (\(E = hf\)).


Step 3: Final Answer:

The only true statement is that both photons and electrons exhibit dual nature. This corresponds to option (C).
Quick Tip: Wave-particle duality is a cornerstone of quantum mechanics. Remember that not just photons and electrons, but all particles (protons, neutrons, atoms, etc.) have a de Broglie wavelength (\(\lambda = h/p\)) and thus exhibit wave properties, although these are only significant for particles with very small mass.

Step 1: Understanding the Concept:

This problem connects the concepts of light intensity, photon energy, and the electromagnetic spectrum. Intensity (\(I\)) of a light beam is the power delivered per unit area. It can also be defined as the total energy of photons passing through a unit area per unit time. The energy of a single photon (\(E\)) is determined by its frequency (\(f\)) or wavelength (\(\lambda\)).


Step 2: Key Formula or Approach:

The energy of a single photon is given by:
\[ E = hf = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant.

The intensity of a light beam can be expressed as:
\[ I = n \times E \]
where \(n\) is the number of photons incident per unit area per unit time.


Step 3: Detailed Explanation:

Comparing Photon Energies:

In the visible spectrum, red light has a longer wavelength than blue light (\(\lambda_{red} > \lambda_{blue}\)).

Since photon energy is inversely proportional to wavelength (\(E \propto 1/\lambda\)), the energy of a single blue photon is greater than the energy of a single red photon.
\[ E_{blue} > E_{red} \]
This makes statement (D) incorrect. Statement (C) is also incorrect as it states the opposite of the known fact about wavelengths.


Comparing Number of Photons for Equal Intensity:

We are given that the intensities of the two beams are equal:
\[ I_{red} = I_{blue} \]
Let \(n_{red}\) and \(n_{blue}\) be the number of photons per unit area per second for the red and blue beams, respectively.

Using the intensity formula:
\[ n_{red} \times E_{red} = n_{blue} \times E_{blue} \]
We can write the ratio of the number of photons:
\[ \frac{n_{red}}{n_{blue}} = \frac{E_{blue}}{E_{red}} \]
Since we established that \(E_{blue} > E_{red}\), the ratio \(\frac{E_{blue}}{E_{red}} > 1\).

Therefore, \(\frac{n_{red}}{n_{blue}} > 1\), which implies \(n_{red} > n_{blue}\).

This means that for the intensities to be equal, the red beam (composed of lower-energy photons) must have a greater number of photons than the blue beam (composed of higher-energy photons). This makes statement (B) correct and statement (A) incorrect.


Step 4: Final Answer:

The correct statement is that the red beam has more number of photons than the blue beam. This corresponds to option (B).
Quick Tip: Think of intensity as the total energy delivered. To deliver the same total energy, you need more low-energy "packets" (red photons) than high-energy "packets" (blue photons). This analogy can help you quickly solve problems comparing intensities of different colors of light.

Step 1: Understanding the Concept:

This question relates to the photoelectric effect as described by Einstein's photoelectric equation. It tests the understanding of why emitted photoelectrons have a range of kinetic energies, and the role of the work function.


Step 2: Key Formula or Approach:

Einstein's photoelectric equation is:
\[ K_{max} = h\nu - \phi_0 \]
where \(K_{max}\) is the maximum kinetic energy of the emitted photoelectrons, \(h\nu\) is the energy of the incident photon, and \(\phi_0\) is the work function of the metal.


Step 3: Detailed Explanation:

Analyzing the Assertion (A):

When monochromatic light (photons of a single energy \(h\nu\)) falls on a metal surface, the photons are absorbed by electrons.

An electron on the very surface of the metal, upon absorbing a photon, needs the minimum energy (\(\phi_0\)) to escape. It will be emitted with the maximum possible kinetic energy, \(K_{max}\).
An electron from deeper inside the metal, after absorbing a photon, loses some of its energy in collisions with other atoms on its way to the surface. Therefore, it will be emitted with a kinetic energy \(K\) that is less than \(K_{max}\).

This means the kinetic energies of the emitted photoelectrons will range from 0 to \(K_{max}\), and consequently, their speeds will range from zero to a maximum value. Thus, the Assertion (A) is true.


Analyzing the Reason (R):

The work function (\(\phi_0\)) is the minimum energy required to liberate an electron from the surface of a given metal. It is a characteristic property of the metal. For example, cesium has a low work function, while platinum has a high one. So, each metal has a definite work function. Thus, the Reason (R) is true.


Connecting Assertion and Reason:

The Reason (R) explains why there is a definite *maximum* kinetic energy (\(K_{max}\)) for a given metal and light frequency. However, it does not explain why there is a *range* of energies from zero up to this maximum. The range exists because electrons are emitted from different depths within the metal and suffer energy loss due to internal collisions. Therefore, while both statements are true, the Reason is not the correct explanation for the Assertion.


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, but Reason (R) does not correctly explain Assertion (A). This corresponds to option (B).
Quick Tip: In photoelectric effect questions, remember that the work function (\(\phi_0\)) sets the *threshold* and the *maximum* kinetic energy. The *range* of kinetic energies is due to electrons coming from different depths and losing energy in collisions.

Step 1: Understanding the Concept:

This question differentiates between the phenomena of interference and diffraction, both of which are characteristic of waves. Interference arises from the superposition of waves from two or more coherent sources, while diffraction is the bending of waves as they pass through an aperture or around an obstacle.


Step 2: Detailed Explanation:

Analyzing the Assertion (A):

Young's double-slit experiment demonstrates interference between waves from two slits. If one of these slits is closed, the setup is no longer a double-slit experiment but a single-slit experiment. Light passing through the single open slit will spread out due to diffraction. The pattern observed on the screen will be a characteristic single-slit diffraction pattern (a broad central maximum flanked by dimmer, narrower secondary maxima). Therefore, the Assertion (A) is true.


Analyzing the Reason (R):

The principle of interference requires the superposition of two or more waves. For a stable interference pattern, the sources of these waves must be coherent. Therefore, to observe interference, at least two waves are necessary. Thus, the Reason (R) is true.


Connecting Assertion and Reason:

The Assertion describes the result of having only one slit open (diffraction). The Reason states the condition for interference (at least two waves). While the Reason explains why an *interference* pattern is no longer observed when one slit is closed, it does not explain why a *diffraction* pattern is formed. The diffraction pattern arises from the interference of secondary wavelets originating from different points within the single open slit (Huygens' principle). So, both statements are correct facts from wave optics, but the Reason is not the explanation for the Assertion.


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A). This corresponds to option (B).
Quick Tip: Remember the key difference: Interference is the superposition of waves from a few (at least two) discrete coherent sources. Diffraction is the superposition of waves from a continuum of coherent sources (like the points across a single slit). In a double-slit experiment, both phenomena occur: diffraction at each slit and interference between the waves from the two slits.

Step 1: Understanding the Concept:

This question concerns the behavior of a series LCR circuit at resonance. Resonance is a special condition where the circuit's impedance is at its minimum, leading to maximum current.


Step 2: Key Formula or Approach:

The impedance of a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where \(X_L = \omega L\) is the inductive reactance and \(X_C = \frac{1}{\omega C}\) is the capacitive reactance.


Step 3: Detailed Explanation:

Analyzing the Assertion (A):

At resonance, the inductive reactance \(X_L\) equals the capacitive reactance \(X_C\). The term \((X_L - X_C)\) in the impedance formula becomes zero.
\[ Z = \sqrt{R^2 + (0)^2} = R \]
Since the impedance \(Z\) is equal to the resistance \(R\), the net reactance of the circuit is zero. This means the circuit behaves as if it contains only resistance. In this state, the voltage and current are in phase, which is the characteristic of a purely resistive circuit. Thus, the Assertion (A) is true.


Analyzing the Reason (R):

The condition for resonance in a series LCR circuit is \(X_L = X_C\). Substituting the expressions for reactance:
\[ \omega L = \frac{1}{\omega C} \]
Solving for the angular frequency \(\omega\):
\[ \omega^2 = \frac{1}{LC} \implies \omega = \frac{1}{\sqrt{LC}} \]
This specific frequency is called the resonant angular frequency. Thus, the Reason (R) is a true statement.


Connecting Assertion and Reason:

The Assertion states that the circuit becomes purely resistive at resonance. The Reason provides the mathematical condition for resonance (\(X_L = X_C\)). It is precisely this condition that causes the reactive components of the impedance to cancel out, making the total impedance purely resistive (\(Z=R\)). Therefore, the Reason (R) is the correct explanation for the Assertion (A).


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). This corresponds to option (A).
Quick Tip: For a series LCR circuit at resonance, remember: \(X_L = X_C\), Impedance \(Z\) is minimum (\(Z=R\)), current is maximum, and the phase difference between voltage and current is zero.

Step 1: Understanding the Concept:

This question deals with the electrical neutrality of doped semiconductors. It's a common point of confusion: although an n-type semiconductor has an abundance of negative charge *carriers* (electrons), the material as a whole is electrically neutral.


Step 2: Detailed Explanation:

Analyzing the Assertion (A):

An n-type semiconductor is formed by doping an intrinsic semiconductor (like silicon, which is neutral) with pentavalent impurity atoms (like phosphorus, which are also neutral). Since we are combining two neutral materials, the resulting material must also be electrically neutral. The total number of positive charges (protons in the nuclei of silicon and dopant atoms) equals the total number of negative charges (electrons). Therefore, an n-type semiconductor is not negatively charged. The Assertion (A) is true.


Analyzing the Reason (R):

When a neutral pentavalent impurity atom (e.g., phosphorus with 5 valence electrons) is added to a neutral silicon crystal (4 valence electrons), it forms four covalent bonds. Its fifth valence electron is loosely bound and is easily donated to the conduction band, becoming a free electron. By donating a negatively charged electron, the originally neutral impurity atom becomes a fixed positive ion (a donor ion) in the crystal lattice. Thus, the Reason (R) is a true statement describing the doping process.


Connecting Assertion and Reason:

The Reason explains the mechanism that ensures charge neutrality. For every free electron (negative charge) created by a donor atom, a corresponding positive ion is created and fixed in the lattice. The positive charge of the donor ions perfectly balances the negative charge of the excess free electrons they contribute. This explains *why* the n-type semiconductor as a whole remains electrically neutral. Therefore, the Reason (R) is the correct explanation for the Assertion (A).


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). This corresponds to option (A).
Quick Tip: Do not confuse charge carriers with net charge. An n-type semiconductor has many free electrons (negative carriers), and a p-type has many holes (positive carriers), but both materials are electrically neutral overall.

Step 1: Understanding the Concept:

This problem involves the properties of doped semiconductors. In an intrinsic (pure) semiconductor, the concentration of electrons (\(n_e\)) is equal to the concentration of holes (\(n_h\)), and this is called the intrinsic carrier concentration (\(n_i\)). Doping changes these concentrations but the mass-action law, \(n_e n_h = n_i^2\), still holds.


Step 2: Key Formula or Approach:

Mass-Action Law: \(n_e n_h = n_i^2\).

Classification:

If \(n_h > n_e\), it is a p-type semiconductor, created by adding acceptor (trivalent) impurities.
If \(n_e > n_h\), it is an n-type semiconductor, created by adding donor (pentavalent) impurities.


Step 3: Detailed Explanation:

Given data:

Intrinsic carrier concentration, \(n_i = 5 \times 10^8\) m\(^{-3}\).

After doping, hole concentration, \(n_h = 8 \times 10^{12}\) m\(^{-3}\).


Part (a):

(i) Type of dopant:

We compare the hole concentration in the doped semiconductor (\(n_h = 8 \times 10^{12}\) m\(^{-3}\)) with the intrinsic concentration (\(n_i = 5 \times 10^8\) m\(^{-3}\)).

Since \(n_h \gg n_i\), the concentration of holes has increased dramatically. This is achieved by adding impurities that create holes. Such impurities are trivalent atoms (e.g., Boron, Aluminium), which have one less valence electron than the semiconductor (e.g., Silicon). They accept an electron from a covalent bond, creating a hole. Therefore, the dopant is of the acceptor type (trivalent).

(ii) Type of extrinsic semiconductor:

A semiconductor in which holes are the majority charge carriers (\(n_h > n_e\)) is called a p-type semiconductor.


Part (b):

To find the electron concentration (\(n_e\)) in the extrinsic (doped) semiconductor, we use the mass-action law:
\[ n_e n_h = n_i^2 \]
Rearranging for \(n_e\):
\[ n_e = \frac{n_i^2}{n_h} \]
Substituting the given values:
\[ n_e = \frac{(5 \times 10^8 m^{-3})^2}{8 \times 10^{12} m^{-3}} \] \[ n_e = \frac{25 \times 10^{16}}{8 \times 10^{12}} m^{-3} \] \[ n_e = 3.125 \times 10^{(16-12)} m^{-3} \] \[ n_e = 3.125 \times 10^4 m^{-3} \]

Step 4: Final Answer:

(a) The dopant is trivalent (acceptor type), and the semiconductor formed is p-type.

(b) The electron concentration in the extrinsic semiconductor is \(3.125 \times 10^4\) m\(^{-3}\).
Quick Tip: The mass-action law \(n_e n_h = n_i^2\) is fundamental for doped semiconductors. It shows that if you increase the concentration of one type of carrier (e.g., holes in p-type), the concentration of the other type (electrons) must decrease to keep the product constant.

Step 1: Understanding the Concept:

This problem deals with Young's double-slit experiment (YDSE) using two different wavelengths of light simultaneously. We need to find the location where a bright fringe from one wavelength's pattern overlaps, or coincides with, a bright fringe from the other wavelength's pattern.


Step 2: Key Formula or Approach:

The position (\(y_n\)) of the \(n\)-th bright fringe from the central maximum in a YDSE is given by:
\[ y_n = \frac{n \lambda D}{d} \]
where \(n\) is the order of the fringe (\(n = 0, 1, 2, ...\)), \(\lambda\) is the wavelength of light, \(D\) is the distance to the screen, and \(d\) is the slit separation.

For coincidence, the position of a bright fringe for \(\lambda_1\) must be equal to the position of a bright fringe for \(\lambda_2\).


Step 3: Detailed Explanation:

Given data:

Slit separation, \(d = 1.5 mm = 1.5 \times 10^{-3} m\).

Screen distance, \(D = 1.5 m\).

First wavelength, \(\lambda_1 = 600 nm = 600 \times 10^{-9} m\).

Second wavelength, \(\lambda_2 = 400 nm = 400 \times 10^{-9} m\).


Let the \(n_1\)-th bright fringe of \(\lambda_1\) coincide with the \(n_2\)-th bright fringe of \(\lambda_2\) at a distance \(y\) from the central maximum.
\[ y = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \]
This simplifies to:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
We need to find the ratio of the integers \(n_1\) and \(n_2\):
\[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{400 \times 10^{-9}}{600 \times 10^{-9}} = \frac{400}{600} = \frac{2}{3} \]
We are looking for the *minimum distance* from the central maximum, which corresponds to the smallest non-zero integer values of \(n_1\) and \(n_2\) that satisfy this ratio.

From the ratio \(\frac{n_1}{n_2} = \frac{2}{3}\), the smallest integers are \(n_1 = 2\) and \(n_2 = 3\).

This means the 2nd order bright fringe of the 600 nm light coincides with the 3rd order bright fringe of the 400 nm light.


Now, we can calculate the distance \(y\) using the formula for either wavelength. Using \(\lambda_1\):
\[ y = \frac{n_1 \lambda_1 D}{d} \] \[ y = \frac{2 \times (600 \times 10^{-9} m) \times (1.5 m)}{1.5 \times 10^{-3} m} \]
The \(1.5\) in the numerator and denominator cancels out:
\[ y = \frac{2 \times 600 \times 10^{-9}}{10^{-3}} m \] \[ y = 1200 \times 10^{-6} m = 1.2 \times 10^{-3} m \]
Converting to millimeters:
\[ y = 1.2 mm \]

Step 4: Final Answer:

The minimum distance from the central maximum at which the bright fringes coincide is 1.2 mm.
Quick Tip: For fringe coincidence problems, the condition is always \(n_1 \lambda_1 = n_2 \lambda_2\). The ratio of the integers is the inverse of the ratio of the wavelengths: \(\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}\). Find the simplest integer ratio to get the first point of coincidence after the center.

Step 1: Understanding the Concept:

This problem requires the use of the Lens Maker's formula, which relates the focal length of a lens to its refractive index, the refractive index of the surrounding medium, and the radii of curvature of its two surfaces.


Step 2: Key Formula or Approach:

The Lens Maker's formula is given by:
\[ \frac{1}{f} = \left(\frac{n_{lens}}{n_{medium}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
where \(f\) is the focal length, \(n_{lens}\) is the refractive index of the lens material, \(n_{medium}\) is the refractive index of the surrounding medium, \(R_1\) is the radius of curvature of the first surface, and \(R_2\) is the radius of curvature of the second surface.


Step 3: Detailed Explanation:

Given data:

Refractive index of the lens, \(n_{lens} = 1.5\).

Refractive index of the liquid (medium), \(n_{medium} = 1.25\).

For a plano-convex lens, one surface is convex and the other is plane.

Radius of curvature of the convex surface, \(R_1 = +10\) cm (by sign convention, as it is convex towards the incident light).

Radius of curvature of the plane surface, \(R_2 = \infty\).


Substitute these values into the Lens Maker's formula:
\[ \frac{1}{f} = \left(\frac{1.5}{1.25} - 1\right) \left(\frac{1}{10} - \frac{1}{\infty}\right) \]
First, simplify the ratio of refractive indices:
\[ \frac{1.5}{1.25} = \frac{150}{125} = \frac{6 \times 25}{5 \times 25} = \frac{6}{5} = 1.2 \]
Now, substitute this back into the formula. Note that \(\frac{1}{\infty} = 0\).
\[ \frac{1}{f} = (1.2 - 1) \left(\frac{1}{10}\right) \] \[ \frac{1}{f} = 0.2 \times \frac{1}{10} \] \[ \frac{1}{f} = \frac{1}{5} \times \frac{1}{10} = \frac{1}{50} \]
Therefore, the focal length \(f\) is:
\[ f = 50 cm \]

Step 4: Final Answer:

The focal length of the plano-convex lens when immersed in the liquid is 50 cm.
Quick Tip: Always apply the sign convention for radii of curvature carefully. For a convex surface, R is positive if the center of curvature is on the side of the transmitted light. For a plane surface, R is infinite. The focal length of a converging lens increases when placed in a denser medium (but less dense than the lens itself).

Step 1: Understanding the Concept:

This problem deals with the emission spectrum of the hydrogen atom, specifically the Lyman series, which involves electron transitions to the ground state (\(n=1\)). The wavelength of the emitted radiation is related to the energy difference between the initial and final states. The minimum wavelength corresponds to the maximum energy transition, and the maximum wavelength corresponds to the minimum energy transition.


Step 2: Key Formula or Approach:

The Rydberg formula gives the reciprocal of the wavelength of the emitted photon:
\[ \frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \]
where \(R\) is the Rydberg constant, \(n_f\) is the principal quantum number of the final state, and \(n_i\) is the principal quantum number of the initial state.


Step 3: Detailed Explanation:

For transitions to the ground state, the final state is \(n_f = 1\). The initial state \(n_i\) can be any integer greater than 1, i.e., \(n_i = 2, 3, 4, ..., \infty\).


Maximum Wavelength (\(\lambda_{max}\)):

The wavelength is maximum when the energy difference is minimum. This occurs for the smallest possible jump, which is from the next higher energy level, \(n_i = 2\), to the ground state, \(n_f = 1\).
\[ \frac{1}{\lambda_{max}} = R \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \left(1 - \frac{1}{4}\right) = \frac{3R}{4} \] \[ \implies \lambda_{max} = \frac{4}{3R} \]

Minimum Wavelength (\(\lambda_{min}\)):

The wavelength is minimum when the energy difference is maximum. This occurs for the largest possible jump, which is from \(n_i = \infty\) to the ground state, \(n_f = 1\).
\[ \frac{1}{\lambda_{min}} = R \left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = R \left(1 - 0\right) = R \] \[ \implies \lambda_{min} = \frac{1}{R} \]

Ratio Calculation:

The required ratio is \(\frac{\lambda_{min}}{\lambda_{max}}\).
\[ \frac{\lambda_{min}}{\lambda_{max}} = \frac{1/R}{4/(3R)} = \frac{1}{R} \times \frac{3R}{4} = \frac{3}{4} \]

Step 4: Final Answer:

The ratio of minimum to maximum wavelength for radiations emitted for transitions to the ground state is 3/4.
Quick Tip: For any spectral series (Lyman, Balmer, etc.), the maximum wavelength (lowest energy) corresponds to the transition from the level just above the final level (\(n_f+1 \to n_f\)). The minimum wavelength (highest energy), known as the series limit, corresponds to the transition from infinity (\(\infty \to n_f\)).

Step 1: Understanding the Concept:

The brightness of an incandescent bulb is determined by the power it dissipates. Power (\(P\)) can be calculated using \(P = I^2 R\), where \(I\) is the current flowing through the bulb and \(R\) is its resistance. Since all three bulbs are identical, they have the same resistance, \(R\). Therefore, their brightness is directly proportional to the square of the current passing through them (\(P \propto I^2\)).


Step 2: Detailed Explanation:

Let the resistance of each identical bulb be \(R\) and the emf of the battery be \(V\).


(i) When key K is closed:

Bulbs P and Q are connected in parallel. Their equivalent resistance, \(R_{PQ}\), is:
\[ \frac{1}{R_{PQ}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_{PQ} = \frac{R}{2} \]
This parallel combination is in series with bulb S. The total resistance of the circuit, \(R_{total}\), is:
\[ R_{total} = R_S + R_{PQ} = R + \frac{R}{2} = \frac{3R}{2} \]
The total current from the battery, which flows through bulb S, is:
\[ I_S = \frac{V}{R_{total}} = \frac{V}{(3R/2)} = \frac{2V}{3R} \]
This current \(I_S\) splits equally between the identical bulbs P and Q.
\[ I_P = I_Q = \frac{I_S}{2} = \frac{1}{2} \left(\frac{2V}{3R}\right) = \frac{V}{3R} \]
Brightness Comparison:

Power of bulb S: \(P_S = I_S^2 R = \left(\frac{2V}{3R}\right)^2 R = \frac{4V^2}{9R}\)

Power of bulbs P and Q: \(P_P = I_P^2 R = \left(\frac{V}{3R}\right)^2 R = \frac{V^2}{9R}\) and \(P_Q = I_Q^2 R = \frac{V^2}{9R}\)

Comparing the powers: \(P_P = P_Q = \frac{1}{4} P_S\).

Justification: Bulb S is 4 times brighter than bulbs P and Q. Bulbs P and Q have equal brightness.


(ii) When key K is opened:

When the key K is opened, the circuit branch containing bulb P is broken, and no current flows through it (\(I_P = 0\)). Bulbs S and Q are now in series with each other.

The new total resistance of the circuit, \(R'_{total}\), is:
\[ R'_{total} = R_S + R_Q = R + R = 2R \]
The current flowing through the series combination (and thus through both S and Q) is:
\[ I'_S = I'_Q = \frac{V}{R'_{total}} = \frac{V}{2R} \]
Brightness Comparison:

Since the same current flows through both S and Q, they dissipate the same amount of power:
\[ P'_S = (I'_S)^2 R = \left(\frac{V}{2R}\right)^2 R = \frac{V^2}{4R} \] \[ P'_Q = (I'_Q)^2 R = \left(\frac{V}{2R}\right)^2 R = \frac{V^2}{4R} \]
Justification: Bulbs S and Q have the same brightness because they are in series and carry the same current. Bulb P is off.
Quick Tip: For circuits with identical components, comparing currents is the key to comparing power and brightness. Remember that current divides in parallel (inversely to resistance) and is the same for all components in series.

Step 1: Understanding the Concept:

This problem requires analyzing a DC circuit with parallel branches. The key is to identify the potential difference across the parallel components. When identical ideal cells are connected in parallel, the equivalent emf is the same as the emf of a single cell.


Step 2: Key Formula or Approach:

1. Find the equivalent emf of the parallel voltage sources.
2. Identify the parallel branches connected across this equivalent source.
3. Apply Ohm's Law (\(V = IR\)) to the specific branch containing the bulb to find the current through it.


Step 3: Detailed Explanation:

The circuit has three main parallel branches connected between two common points.

Branch 1 & 2 (Left): Two identical cells of emf 10 V each are connected in parallel. Assuming they are ideal cells, the equivalent emf of this combination is 10 V. This combination provides a constant potential difference of 10 V across the points it is connected to.


Branch 3 (Middle): This branch contains the bulb B with a resistance of \(R_B = 10 \, \Omega\).


Branch 4 (Right): This branch contains a \(20 \, \Omega\) resistor and a \(10 \, \Omega\) resistor connected in series.


Since the branch with the bulb B is connected in parallel to the combination of the cells, the potential difference across the bulb B is equal to the equivalent emf of the cells.

Potential difference across the bulb, \(V_B = 10\) V.

Resistance of the bulb, \(R_B = 10 \, \Omega\).


Now, we can find the current (\(I_B\)) flowing through the bulb using Ohm's Law:
\[ V_B = I_B \times R_B \] \[ 10 V = I_B \times 10 \, \Omega \]
Solving for \(I_B\):
\[ I_B = \frac{10 V}{10 \, \Omega} = 1 A \]

The current in the other resistor branch does not affect the current in the bulb's branch because they are in parallel across a fixed voltage source.


Step 4: Final Answer:

The current that flows through the bulb B is 1 A.
Quick Tip: When analyzing parallel circuits, remember that the voltage across each parallel branch is the same. If a branch is connected directly across an ideal voltage source, the current in that branch depends only on the voltage of the source and the resistance of that branch, independent of other parallel branches.

Step 1: Understanding the Concept:

Einstein explained the photoelectric effect by proposing that light energy is quantized in discrete packets called photons. The energy of each photon is \(E = h\nu\). An electron in a metal absorbs the entire energy of a single photon. A part of this energy is used to overcome the binding energy of the electron (the work function, \(\phi_0\)), and the rest appears as the kinetic energy of the electron.


Step 2: Key Formula or Approach:

Einstein's photoelectric equation is the cornerstone for explaining all observations:
\[ K_{max} = h\nu - \phi_0 \]
where \(K_{max}\) is the maximum kinetic energy of an emitted photoelectron, \(h\) is Planck's constant, \(\nu\) is the frequency of incident light, and \(\phi_0\) is the work function of the metal.


Step 3: Detailed Explanation:


(a) Existence of Threshold Frequency

For an electron to be emitted, its kinetic energy must be non-negative, i.e., \(K_{max} \geq 0\).

Using Einstein's equation:
\[ h\nu - \phi_0 \geq 0 \] \[ \implies h\nu \geq \phi_0 \]
This shows that the energy of the incident photon (\(h\nu\)) must be greater than or equal to the work function (\(\phi_0\)) for photoemission to occur.

If we define a minimum frequency, called the threshold frequency (\(\nu_0\)), such that \(h\nu_0 = \phi_0\), then the condition for emission becomes \(\nu \geq \nu_0\).

If the incident frequency \(\nu\) is less than this minimum value \(\nu_0\), no matter how high the intensity of light, no photoelectric emission will take place.


(b) Effect of Frequency and Intensity on Maximum Kinetic Energy

From Einstein's equation, \(K_{max} = h\nu - \phi_0\), it is clear that the maximum kinetic energy \(K_{max}\) is linearly dependent on the frequency \(\nu\) of the incident light. As \(\nu\) increases, \(K_{max}\) increases.

The intensity of light is related to the number of photons incident per unit area per unit time. Increasing the intensity means increasing the number of photons, which will result in the emission of more photoelectrons per second (a larger photoelectric current). However, the energy of each individual photon (\(h\nu\)) remains the same. Since an electron absorbs a single photon, the maximum kinetic energy it can gain is fixed by the photon's energy and the work function, and is therefore independent of the intensity of the light.


(c) Cut-off Voltage versus Frequency Graph

The cut-off voltage or stopping potential (\(V_0\)) is the potential required to stop the most energetic photoelectrons. Thus, the work done by this potential equals the maximum kinetic energy:
\[ K_{max} = eV_0 \]
where \(e\) is the charge of an electron.

Substituting this into Einstein's equation:
\[ eV_0 = h\nu - \phi_0 \]
Rearranging this equation to express \(V_0\) as a function of \(\nu\):
\[ V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \]
This equation is in the form of a straight line, \(y = mx + c\), where:


\(y = V_0\) (the cut-off voltage)
\(x = \nu\) (the frequency)
The slope \(m = \frac{h}{e}\)
The y-intercept \(c = -\frac{\phi_0}{e}\)

Therefore, the plot of cut-off voltage (\(V_0\)) versus frequency (\(\nu\)) is a straight line, and its slope is the universal constant \(\frac{h}{e}\).
Quick Tip: Remember that Einstein's equation links macroscopic observable quantities (\(K_{max}\), \(V_0\)) to the microscopic quantum nature of light (photon energy \(h\nu\)). This equation is central to solving almost any problem on the photoelectric effect.

Step 1: Understanding the Concept:

The motion of the charged particle is governed by the Lorentz force, which is the vector sum of the electric force and the magnetic force acting on it. We need to analyze the effect of each force on the particle's velocity.


Step 2: Key Formula or Approach:

The total Lorentz force is given by \(\vec{F} = \vec{F}_E + \vec{F}_B = q\vec{E} + q(\vec{v} \times \vec{B})\).

We analyze the components of motion parallel and perpendicular to the fields.


Step 3: Detailed Explanation:

Let's assume the magnetic field \(\vec{B}\) and electric field \(\vec{E}\) are along the z-axis, so \(\vec{B} = B\hat{k}\) and \(\vec{E} = E\hat{k}\). The initial velocity \(\vec{v}\) is perpendicular to \(\vec{B}\), so it lies in the x-y plane.


Effect of the Magnetic Force (\(\vec{F}_B\)):

The magnetic force is \(\vec{F}_B = q(\vec{v} \times \vec{B})\). Since \(\vec{v}\) is in the x-y plane and \(\vec{B}\) is along the z-axis, the force \(\vec{F}_B\) is always in the x-y plane and perpendicular to \(\vec{v}\). This magnetic force provides the necessary centripetal force for the particle to move in a circle in the x-y plane. The magnetic force does no work and does not change the speed of the particle in this plane. The radius of this circular path is \(r = \frac{mv}{qB}\).


Effect of the Electric Force (\(\vec{F}_E\)):

The electric force is \(\vec{F}_E = q\vec{E} = qE\hat{k}\). This force is constant and acts along the z-axis (parallel to \(\vec{B}\)). This force will cause a constant acceleration, \(a_z = \frac{F_E}{m} = \frac{qE}{m}\), in the z-direction. It does not affect the motion in the x-y plane.


Combined Motion:

The particle's motion is a superposition of two independent motions:

Uniform circular motion in the plane perpendicular to \(\vec{B}\) (the x-y plane).
Uniformly accelerated motion along the direction of \(\vec{E}\) and \(\vec{B}\) (the z-axis).

The combination of these two motions results in a helical path. However, because the particle is accelerating along the axis of the helix, the distance it travels along the axis in each revolution (the pitch of the helix) is not constant. The pitch continuously increases.

The shape is a helix with a variable (increasing) pitch.
Quick Tip: When analyzing motion in combined electric and magnetic fields, it's often easiest to resolve the velocity and forces into components parallel and perpendicular to the magnetic field. The magnetic force only affects the perpendicular components of velocity, causing rotation.

Step 1: Understanding the Concept:

For a long (ideal) solenoid, the magnetic field inside is strong, uniform, and directed along the axis. The field outside is negligible. The strength of the internal field depends on the number of turns per unit length and the current flowing through the wire.


Step 2: Key Formula or Approach:

The magnitude of the magnetic field inside a long solenoid is given by:
\[ B = \mu_0 n I \]
where \(\mu_0\) is the permeability of free space, \(I\) is the current, and \(n\) is the number of turns per unit length (\(n = N/L\), where \(N\) is the total number of turns and \(L\) is the length of the solenoid).

Substituting for \(n\), the formula can be written as:
\[ B = \mu_0 \frac{N}{L} I \]

Step 3: Detailed Explanation:


(i) The radius of the turns of the solenoid is increased:

Let's examine the formula \(B = \mu_0 \frac{N}{L} I\). The variables in this formula are the total number of turns (\(N\)), the length (\(L\)), and the current (\(I\)). The radius of the turns (\(r\)) does not appear in the expression for the magnetic field inside an ideal solenoid.

Conclusion: Therefore, increasing the radius of the turns will have no effect on the magnetic field inside the solenoid, assuming it remains a 'long' solenoid. The field remains unchanged.


(ii) The length of solenoid as well as the total number of its turns are doubled:

Let the initial parameters be \(N\) and \(L\), and the initial magnetic field be \(B = \mu_0 \frac{N}{L} I\).

The new parameters are \(N' = 2N\) and \(L' = 2L\).

The new magnetic field, \(B'\), will be:
\[ B' = \mu_0 \frac{N'}{L'} I \]
Substituting the new values:
\[ B' = \mu_0 \frac{2N}{2L} I = \mu_0 \frac{N}{L} I \]
We can see that \(B' = B\).

Conclusion: The magnetic field remains unchanged. This is because the number of turns per unit length (\(n = N/L\)) remains constant: \(n' = N'/L' = 2N/2L = N/L = n\). Since the field depends only on \(n\) and \(I\), it is unaffected.
Quick Tip: The key quantity for a solenoid's magnetic field is the density of turns, \(n = N/L\). Any change that leaves this ratio constant (like doubling both N and L) will not change the magnetic field strength inside. Also, remember that for an ideal solenoid, the field is independent of the radius.

Step 1: Understanding the Concept:

Magnetic field (\(\vec{B}\)) is a fundamental vector quantity that describes the magnetic influence in a region of space. Magnetic flux (\(\Phi_B\)), on the other hand, is a scalar quantity that measures the total amount of magnetism flowing through a particular surface or area.


Step 2: Detailed Explanation:

The key differences between magnetic field and magnetic flux are tabulated below:

\begin{tabular{|l|l|l|
\hline
Characteristic & Magnetic Field (\(\vec{B}\)) & Magnetic Flux (\(\Phi_B\))

\hline
Definition & The force experienced by a unit charge & The total number of magnetic field

& moving with unit velocity & lines passing normally through a

& perpendicular to the field. & given surface area.

\hline
Nature & It is a vector quantity, having both & It is a scalar quantity, having

& magnitude and direction. & only magnitude.

\hline
Property & It is a property of a point in space. & It is a property of an area or surface.

\hline
Formula & Defined by the Lorentz force, & Defined as the surface integral of

& \(\vec{F} = q(\vec{v} \times \vec{B})\). & the magnetic field, \(\Phi_B = \int \vec{B} \cdot d\vec{A}\).

\hline
SI Unit & Tesla (T) or Weber/m\(^2\) (Wb/m\(^2\)). & Weber (Wb) or Tesla-m\(^2\) (T·m\(^2\)).

\hline
\end{tabular
Quick Tip: An easy analogy is to think of rain. The 'magnetic field' is like the velocity (speed and direction) of the raindrops at a particular point. The 'magnetic flux' is like the total volume of water collected by a bucket (the area) in a certain time. The flux depends on both the rain's velocity and the bucket's area and orientation.

Step 1: Understanding the Concept:

This phenomenon is explained by Faraday's Law of Electromagnetic Induction and Lenz's Law. A change in the magnetic flux linked with a closed circuit induces an electromotive force (emf) and hence a current. The direction of this induced current opposes the change that causes it.


Step 2: Detailed Explanation:

1. Induction of Current (Faraday's Law):

A bar magnet produces a magnetic field. When the magnet is held near the coil, some of its magnetic field lines pass through the area of the coil. This is the magnetic flux linked with the coil.

When the magnet is moved towards the coil, the distance between them decreases. As a result, the strength of the magnetic field passing through the coil increases, and consequently, the magnetic flux linked with the coil increases.

According to Faraday's Law of Induction, whenever the magnetic flux through a closed coil changes, an emf is induced. Since the coil is a closed circuit, this induced emf drives an induced current through it.


2. Direction of Current (Lenz's Law):

Lenz's Law states that the direction of the induced current is such that it creates a magnetic field that opposes the change in flux.

Cause of Change: The south pole of the magnet is approaching the coil, causing an increase in magnetic flux directed into the coil (field lines emerge from North and enter South).
Opposition: To oppose this increase in flux, the coil must generate its own magnetic field in the opposite direction. This means the face of the coil nearest to the magnet must act as a repelling pole. Since the approaching pole is a South pole, the coil's face must also become a South pole to repel it.
Determining Current Direction: According to the Clock Face Rule (or Right-Hand Grip Rule), for a face of a coil to behave like a South pole, the current in it must flow in the clockwise direction when viewed from the side of the approaching magnet.


Conclusion:
A current is induced due to the change in magnetic flux as the magnet moves. The direction of the current in the coil will be clockwise when observed from the magnet's side.
Quick Tip: Remember "Lenz's Law is a law of protest". The induced effect always opposes its cause. If a magnet approaches, the coil repels it. If a magnet recedes, the coil attracts it. Use this to determine the required polarity of the coil face, and then use the clock rule (S-pole \(\rightarrow\) Clockwise, N-pole \(\rightarrow\) Anti-clockwise) to find the current direction.

Step 1: Understanding the Concept:

Electromagnetic (EM) waves are disturbances consisting of time-varying electric and magnetic fields that propagate through space. They are described by Maxwell's equations and encompass a wide spectrum from radio waves to gamma rays.


Step 2: Detailed Explanation:

Three key characteristics of electromagnetic waves are:

Transverse Nature: EM waves are transverse. This means that the oscillating electric field vector (\(\vec{E}\)) and the oscillating magnetic field vector (\(\vec{B}\)) are both perpendicular to the direction of wave propagation (\(\vec{c}\)). Furthermore, the electric and magnetic fields are also mutually perpendicular to each other, such that \(\vec{E} \perp \vec{B} \perp \vec{c}\).

Propagation in Vacuum: EM waves do not require any material medium for their propagation. They can travel through a vacuum. In a vacuum, all electromagnetic waves, regardless of their frequency, travel at the same constant speed, the speed of light, \(c \approx 3 \times 10^8\) m/s. The speed is given by the relation \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\), where \(\mu_0\) is the permeability and \(\epsilon_0\) is the permittivity of free space.

Energy and Momentum: EM waves transport energy and momentum from one region of space to another. The energy is shared equally between the electric and magnetic fields. The energy density of the wave is \(u = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2\). The rate of energy flow per unit area is given by the Poynting vector, \(\vec{S} = \frac{1}{\mu_0}(\vec{E} \times \vec{B})\). Quick Tip: A useful mnemonic to remember the relationship between E, B, and c is the order of the letters in "E-B-C". The direction of propagation (\(\vec{c}\)) is given by the direction of the cross product \(\vec{E} \times \vec{B}\).

Step 1: Understanding the Concept:

James Clerk Maxwell introduced the concept of displacement current to remove an inconsistency in Ampere's circuital law and to account for the propagation of electromagnetic waves. It is not a current due to the flow of charge but arises from a time-varying electric field.


Step 2: Detailed Explanation:


How Displacement Current is Produced:


During the process of charging a parallel-plate capacitor, charge accumulates on the plates. Let \(Q(t)\) be the charge on the plates at time \(t\).
This accumulation of charge creates an electric field \(E(t)\) in the region between the plates. Since the charge is changing with time (\(dQ/dt \neq 0\)), the electric field between the plates is also changing with time (\(dE/dt \neq 0\)).
The electric flux (\(\Phi_E\)) through any surface between the plates is proportional to the electric field (\(\Phi_E = EA\), for a uniform field). Therefore, a changing electric field results in a changing electric flux (\(d\Phi_E/dt \neq 0\)).
Maxwell proposed that this time-varying electric flux is equivalent to a current, which he named the displacement current (\(I_d\)). Its value is given by:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]

This displacement current produces a magnetic field in the same way that a conduction current (flow of charges) does.


Where Displacement Current Exists:

The displacement current exists exclusively in the region of space where the electric field is changing with time.

In the context of a charging capacitor:

The conduction current (\(I_c\)), which is the flow of electrons, exists in the connecting wires leading up to the capacitor plates.
The displacement current (\(I_d\)) exists in the insulating gap (e.g., vacuum or dielectric) between the capacitor plates.

Crucially, for a charging capacitor, the magnitude of the conduction current in the wires is exactly equal to the magnitude of the displacement current in the gap (\(I_c = I_d\)). This ensures that the principle of continuity of current is maintained throughout the entire circuit.
Quick Tip: Think of the displacement current as the "missing link" that makes the current continuous in a capacitor circuit. Current flows in the wires (conduction), and the "flow" continues across the gap not as moving charges, but as a changing electric field (displacement).

Step 1: Understanding the Concept:

This question asks for the definition of a wavefront and the application of this concept to the refraction of light through a convex lens, based on Huygens' principle.


Step 2: Detailed Explanation:

Definition of Wavefront:

A wavefront is defined as the continuous locus of all points in a medium that are vibrating in the same phase at a given instant. The direction of propagation of the wave is always perpendicular to the wavefront.


Refraction of a Plane Wavefront by a Convex Lens:

A convex lens is a converging lens. According to Huygens' principle, different parts of an incident wavefront travel through different thicknesses of the lens.

A plane wavefront is incident on the convex lens. This means the incident rays are parallel to the principal axis.
The central part of the wavefront travels through the thickest part of the lens, where it is slowed down the most.
The parts of the wavefront near the edges of the lens travel through thinner sections of glass and are slowed down less.
As a result, the central part of the wavefront lags behind the edges. The emergent wavefront is no longer plane but becomes curved.
Since a convex lens converges parallel rays to its principal focus, the emergent wavefront is a spherical wavefront that converges towards the focus.


Diagram:

% Placeholder for a diagram showing a plane wavefront becoming a converging spherical wavefront after passing through a convex lens.
\textit{Diagram showing a plane wavefront on the left of a convex lens, and a curved, converging wavefront on the right, shrinking towards the focal point F.

The shape of the refracted wavefront is spherical and converging.
Quick Tip: Remember that lenses and mirrors change the shape of wavefronts. A plane wavefront corresponds to parallel rays. A converging lens or a concave mirror will change a plane wavefront into a converging spherical wavefront.

Step 1: Understanding the Concept:

This question requires the derivation of Snell's law of refraction using Huygens' principle. This principle states that every point on a wavefront is a source of secondary spherical wavelets. The new wavefront at a later time is the forward envelope of these wavelets.


Step 2: Detailed Explanation and Derivation:

Consider a plane wavefront AB incident at an angle \(i\) on a plane surface XY separating a denser medium (Medium 1, with speed of light \(v_1\)) from a rarer medium (Medium 2, with speed of light \(v_2\)). Since Medium 2 is rarer, \(v_2 > v_1\).


Diagram:

% Placeholder for a diagram of refraction using Huygens' principle.
Diagram showing incident wavefront AB, refracted wavefront CE, interface XY. Ray at A strikes first. Ray at B travels to C. While B reaches C, the wavelet from A reaches E. Normals, angles i and r are marked.


Let the time taken for the wavefront to travel from B to C be \(t\). So, \(BC = v_1 t\).

During this time \(t\), the secondary wavelet from point A will travel a distance \(AE = v_2 t\) into the second medium.

To find the new refracted wavefront, we draw a sphere of radius \(AE\) centered at A. The tangent CE drawn from point C to this sphere represents the refracted wavefront. The angle of refraction is \(r\).


From the right-angled triangle \(\triangle ABC\):
\[ \sin i = \frac{BC{AC} \implies BC = AC \sin i \]
From the right-angled triangle \(\triangle AEC\):
\[ \sin r = \frac{AE}{AC} \implies AE = AC \sin r \]
Now, we can express the time \(t\) in two ways:
\[ t = \frac{BC}{v_1} = \frac{AC \sin i}{v_1} \] \[ t = \frac{AE}{v_2} = \frac{AC \sin r}{v_2} \]
Equating the two expressions for time \(t\):
\[ \frac{AC \sin i}{v_1} = \frac{AC \sin r}{v_2} \] \[ \frac{\sin i}{v_1} = \frac{\sin r}{v_2} \]
Rearranging the terms, we get:
\[ \frac{\sin i}{\sin r} = \frac{v_1}{v_2} \]
The refractive index of a medium is defined as \(n = \frac{c}{v}\), where \(c\) is the speed of light in vacuum.

So, \(v_1 = c/n_1\) and \(v_2 = c/n_2\). Substituting these:
\[ \frac{\sin i}{\sin r} = \frac{c/n_1}{c/n_2} = \frac{n_2}{n_1} \] \[ \implies n_1 \sin i = n_2 \sin r \]
This is Snell's law of refraction, which has been verified using Huygens' principle.
Quick Tip: The key to deriving laws of reflection or refraction using Huygens' principle is to consider the time taken for different parts of the wavefront to travel. The time taken for the disturbance to travel from point B to C on the interface is the same as the time for the secondary wavelet to travel from A to E in the second medium.

Step 1: Understanding the Concept:

Extrinsic semiconductors are created by doping an intrinsic semiconductor with specific impurities to increase the concentration of one type of charge carrier over the other. The more abundant carrier is called the majority carrier, and the less abundant one is the minority carrier.


Step 2: Detailed Explanation:


In p-type semiconductors:

Formation: Formed by doping an intrinsic semiconductor (like Si or Ge, Group 14) with a trivalent impurity (Group 13, e.g., Boron, Aluminium).
Mechanism: The trivalent atom has one less valence electron than needed for bonding. This creates a vacancy or a "hole".
Majority Carriers: The concentration of holes (\(n_h\)) is significantly increased due to doping, making holes the majority charge carriers.
Minority Carriers: The concentration of free electrons (\(n_e\)) is very low, arising mainly from thermal generation. Therefore, electrons are the minority charge carriers. (\(n_h \gg n_e\))


In n-type semiconductors:

Formation: Formed by doping an intrinsic semiconductor with a pentavalent impurity (Group 15, e.g., Phosphorus, Arsenic).
Mechanism: The pentavalent atom has one extra valence electron that is not involved in bonding and is easily donated to the conduction band.
Majority Carriers: The concentration of free electrons (\(n_e\)) is significantly increased due to doping, making electrons the majority charge carriers.
Minority Carriers: The concentration of holes (\(n_h\)) is very low. Therefore, holes are the minority charge carriers. (\(n_e \gg n_h\)) Quick Tip: A simple mnemonic: \textbf{p}-type has \textbf{p}ositive holes as majority carriers. \textbf{n}-type has \textbf{n}egative electrons as majority carriers.

Step 1: Understanding the Concept:

When a p-type and an n-type semiconductor are joined, a p-n junction is formed. Due to the differences in carrier concentrations on either side of the junction, two types of current mechanisms, diffusion and drift, are established, which eventually balance each other at equilibrium.


Step 2: Detailed Explanation:


1. Formation of Diffusion Current:

At the moment the junction is formed, there is a high concentration of holes on the p-side and a high concentration of electrons on the n-side.
Due to this concentration gradient, the majority carriers tend to move from a region of higher concentration to a region of lower concentration.
Holes from the p-side diffuse across the junction into the n-side, and electrons from the n-side diffuse across the junction into the p-side.
This movement of charge carriers due to the concentration gradient constitutes the diffusion current. The conventional direction of the diffusion current is from the p-side to the n-side (the same direction as the flow of holes).


2. Formation of Drift Current:

As holes diffuse from the p-side, they leave behind negatively charged acceptor ions. As electrons diffuse from the n-side, they leave behind positively charged donor ions.
This creates a layer of immobile positive and negative ions on either side of the junction. This region is devoid of free charge carriers and is called the depletion region.
The layer of fixed ions sets up an internal electric field (\(E_i\)) across the junction, directed from the positive n-side to the negative p-side. This field creates a potential barrier.
This electric field exerts a force on the minority charge carriers (electrons on the p-side and holes on the n-side) that happen to be near the junction.
The field sweeps these minority carriers across the junction: electrons from the p-side are moved to the n-side, and holes from the n-side are moved to the p-side.
This movement of charge carriers due to the electric field constitutes the drift current. The conventional direction of the drift current is from the n-side to the p-side, opposite to the diffusion current.


In an unbiased p-n junction, a dynamic equilibrium is reached where the magnitude of the diffusion current becomes equal to the magnitude of the drift current, resulting in zero net current across the junction.




\begin{quicktipbox
Remember the cause for each current: Diffusion is due to a Difference in concentration. Drift is due to the Driving electric field in the depletion region.
\end{quicktipbox Quick Tip: Remember the cause for each current: \textbf{Diffusion} is due to a \textbf{D}ifference in concentration. \textbf{Drift} is due to the \textbf{D}riving electric field in the depletion region.

Step 1: Understanding the Concept:

Resistivity is an intrinsic property of a material that measures its opposition to the flow of electric current. This derivation connects this macroscopic property (\(\rho\)) to the microscopic properties of the charge carriers (electrons) in the conductor, such as their number density, charge, mass, and relaxation time.


Step 2: Key Formula or Approach:

The derivation starts with the expression for the drift velocity of electrons under an external electric field and relates it to current density and Ohm's law.


Step 3: Detailed Derivation:

Consider a conductor with a number density of free electrons \(n\). Let an external electric field \(\vec{E}\) be applied across it.

Each free electron experiences an electric force \(\vec{F} = -e\vec{E}\), where \(e\) is the magnitude of the electron's charge.

This force produces an acceleration \(\vec{a} = \frac{\vec{F}}{m} = -\frac{e\vec{E}}{m}\), where \(m\) is the mass of the electron.

The electrons accelerate but frequently collide with the positive ions of the conductor. The average time interval between two successive collisions is called the relaxation time, \(\tau\).

The average velocity acquired by an electron between collisions, in the direction opposite to the field, is called the drift velocity (\(v_d\)). It is given by:
\[ \vec{v}_d = \vec{a}\tau = -\frac{e\vec{E}}{m}\tau \]
The magnitude of the drift velocity is \(v_d = \frac{eE}{m}\tau\).

The current density (\(J\)) in the conductor is related to the drift velocity by the equation:
\[ J = n e v_d \]
Substituting the expression for \(v_d\):
\[ J = n e \left(\frac{eE}{m}\tau\right) = \frac{ne^2\tau}{m}E \]
According to the microscopic form of Ohm's Law, the electric field \(E\) and current density \(J\) are related by the resistivity \(\rho\):
\[ E = \rho J \quad or \quad J = \frac{E}{\rho} \]
Comparing this with the expression we derived for \(J\):
\[ \frac{E}{\rho} = \left(\frac{ne^2\tau}{m}\right)E \] \[ \frac{1}{\rho} = \frac{ne^2\tau}{m} \]
Inverting this expression gives the resistivity \(\rho\):
\[ \rho = \frac{m}{ne^2\tau} \]
This is the required expression for resistivity.
Quick Tip: The derivation follows a logical flow: Force \(\rightarrow\) Acceleration \(\rightarrow\) Drift Velocity \(\rightarrow\) Current Density \(\rightarrow\) Ohm's Law \(\rightarrow\) Resistivity. Practice this sequence to master the derivation.

Step 1: Understanding the Concept:

The electric current \(I\) is the rate of flow of charge \(Q\), i.e., \(I = \frac{dQ}{dt}\). To find the total charge that has passed through a cross-section over a time interval, we need to integrate the current over that time interval: \(Q = \int I \, dt\). Graphically, this integral represents the area under the current-time (I-t) graph.


Step 2: Key Formula or Approach:

Charge \(Q\) = Area under the I-t graph.


Step 3: Detailed Explanation:

We need to calculate the area under the graph for the two intervals corresponding to \(Q_1\) and \(Q_2\).


Calculation of Charge \(Q_1\):

The charge \(Q_1\) corresponds to the time interval from \(t=1\) s to \(t=2\) s.

In this interval, the graph is a rectangle.

The height of the rectangle is the constant current, \(I = 2.0\) A.

The width of the rectangle is the time interval, \(\Delta t = 2 s - 1 s = 1\) s.

The area, which is the charge \(Q_1\), is:
\[ Q_1 = height \times width = (2.0 A) \times (1 s) = 2.0 C \]

Calculation of Charge \(Q_2\):

The charge \(Q_2\) corresponds to the time interval from \(t=4\) s to \(t=6\) s.

In this interval, the graph is a triangle.

The base of the triangle is the time interval, \(\Delta t = 6 s - 4 s = 2\) s.

The height of the triangle is the maximum current at \(t=6\) s, which is \(I = 2.0\) A.

The area, which is the charge \(Q_2\), is:
\[ Q_2 = \frac{1}{2} \times base \times height = \frac{1}{2} \times (2 s) \times (2.0 A) = 2.0 C \]

Comparison:

From our calculations, we find that \(Q_1 = 2.0\) C and \(Q_2 = 2.0\) C.

Therefore, the charges are equal: \(Q_1 = Q_2\).
Quick Tip: For any graph-based problem in physics, remember what the area under the curve and the slope of the curve represent. For an I-t graph, the area is charge (\(Q = \int I dt\)) and the slope is the rate of change of current (\(dI/dt\)).

Step 1: Understanding the Concept:

This question deals with a simple DC circuit containing a real battery (with internal resistance) and an external load. We need to find expressions for the current and terminal voltage and determine the conditions under which they are maximized.


Step 2: Key Formula or Approach:

Apply Ohm's law to the entire circuit. The total resistance is the sum of the external and internal resistances. The terminal voltage is the potential difference across the external resistance, which is also the emf minus the voltage drop across the internal resistance.


Step 3: Detailed Explanation:


(I) Current and Maximum Current:

The total resistance in the circuit is the series combination of the external resistance \(R\) and the internal resistance \(r\).

Total Resistance = \(R + r\).

According to Ohm's law for the complete circuit, the current \(I\) is given by:
\[ I = \frac{Total EMF}{Total Resistance} = \frac{E}{R+r} \]
To find the maximum possible current (\(I_{max}\)), we need to minimize the denominator \((R+r)\). Since \(E\) and \(r\) are constant for a given battery, the current is maximized when the external resistance \(R\) is minimized. The minimum possible value for resistance is \(R=0\) (a short circuit).
\[ I_{max} = \frac{E}{0+r} = \frac{E}{r} \]

(II) Terminal Voltage and Maximum Terminal Voltage:

The terminal voltage \(V\) is the potential difference across the terminals of the battery when it is supplying current. This is equal to the potential difference across the external resistor \(R\).
\[ V = I R \]
Substituting the expression for \(I\):
\[ V = \left(\frac{E}{R+r}\right)R = \frac{ER}{R+r} \]
Alternatively, it can be expressed as the EMF minus the 'lost volts' across the internal resistance: \(V = E - Ir\).

To find the maximum possible value of \(V\), we analyze the expression \(V = E - Ir\). \(V\) will be maximum when the term \(Ir\) is minimum. This occurs when the current \(I\) is minimum.
From \(I = \frac{E}{R+r}\), the current is minimum when the external resistance \(R\) is maximum, i.e., \(R \to \infty\) (an open circuit).
When \(R \to \infty\), the current \(I \to 0\).
The terminal voltage becomes:
\[ V_{max} = E - (0 \times r) = E \]
The maximum possible terminal voltage is the emf of the battery itself, which is achieved in an open circuit condition.
Quick Tip: Remember the difference between EMF and terminal voltage. EMF (\(E\)) is the maximum potential difference a source can provide (open circuit). Terminal voltage (\(V\)) is the actual potential difference provided when current is flowing (\(V = E-Ir\)) and is always less than or equal to the EMF.

Step 1: Understanding the Concept:

We are given two different states of the same circuit, with different external resistances and corresponding currents. Since the battery is the same, its emf \(E\) and internal resistance \(r\) are constant. We can set up two equations for the two states and solve them simultaneously to find \(r\).


Step 2: Key Formula or Approach:

Use the circuit equation \(E = I(R+r)\) for each of the two cases provided.


Step 3: Detailed Derivation:

From part (b)(i), we have the general relation between emf, current, external resistance, and internal resistance:
\[ E = I(R+r) \]
Case 1: When the external resistance is \(R_1\), the current is \(I_1\).
\[ E = I_1(R_1 + r) \quad \ldots(1) \]
Case 2: When the external resistance is \(R_2\), the current is \(I_2\).
\[ E = I_2(R_2 + r) \quad \ldots(2) \]
Since the emf \(E\) of the battery is the same in both cases, we can equate the right-hand sides of equations (1) and (2):
\[ I_1(R_1 + r) = I_2(R_2 + r) \]
Now, we need to solve this equation for the internal resistance \(r\). Expand the terms:
\[ I_1 R_1 + I_1 r = I_2 R_2 + I_2 r \]
Group the terms containing \(r\) on one side and the other terms on the opposite side:
\[ I_1 r - I_2 r = I_2 R_2 - I_1 R_1 \]
Factor out \(r\):
\[ r(I_1 - I_2) = I_2 R_2 - I_1 R_1 \]
Finally, divide to isolate \(r\):
\[ r = \frac{I_2 R_2 - I_1 R_1}{I_1 - I_2} \]
This is the required expression for the internal resistance of the battery.
Quick Tip: This method of using two different load conditions to find the internal parameters (\(E\) and \(r\)) of a source is a standard experimental and problem-solving technique. The key is to recognize that \(E\) and \(r\) are constants for the source.

Step 1: Understanding the Concept:

As stated in the passage, for the electron to maintain a stable circular orbit, the electrostatic force of attraction provided by the proton must be equal to the centripetal force required for the circular motion.


Step 2: Key Formula or Approach:

1. Electrostatic Force between proton (+e) and electron (-e): \(F_e = K \frac{|(+e)(-e)|}{r^2} = K\frac{e^2}{r^2}\).

2. Centripetal Force for an electron of mass m moving with speed v in a circle of radius r: \(F_c = \frac{mv^2}{r}\).

Equate these two forces: \(F_e = F_c\).


Step 3: Detailed Explanation:

Setting the magnitude of the electrostatic force equal to the centripetal force:
\[ \frac{mv^2}{r} = K\frac{e^2}{r^2} \]
We need to solve this equation for the speed, \(v\).

First, cancel one factor of \(r\) from both sides:
\[ mv^2 = K\frac{e^2}{r} \]
Next, divide by the mass \(m\):
\[ v^2 = \frac{Ke^2}{mr} \]
Finally, take the square root of both sides to find \(v\):
\[ v = \sqrt{\frac{Ke^2}{mr}} \]

Step 4: Final Answer:

The expression for the speed of the electron is \(\sqrt{\frac{Ke^2}{mr}}\), which corresponds to option (C).
Quick Tip: This fundamental relationship, equating the electrical force to the centripetal force, is the starting point for many derivations in the Bohr model. Mastering this step is crucial.

Step 1: Understanding the Concept:

The total energy (\(E\)) of the electron in its orbit is the sum of its kinetic energy (\(KE\)) and its electrostatic potential energy (\(PE\)).
\[ E = KE + PE \]

Step 2: Key Formula or Approach:

1. Kinetic Energy: \(KE = \frac{1}{2}mv^2\).

2. Potential Energy between a proton (+e) and an electron (-e): \(PE = K\frac{(+e)(-e)}{r} = -\frac{Ke^2}{r}\).


Step 3: Detailed Explanation:

Calculate Kinetic Energy (KE):

From the previous part (i), we derived the relation for stable orbits:
\[ mv^2 = \frac{Ke^2}{r} \]
The kinetic energy is \(KE = \frac{1}{2}mv^2\). Substituting the expression for \(mv^2\):
\[ KE = \frac{1}{2} \left(\frac{Ke^2}{r}\right) = \frac{Ke^2}{2r} \]

Calculate Potential Energy (PE):

The electrostatic potential energy of the electron-proton system is:
\[ PE = -\frac{Ke^2}{r} \]
The negative sign indicates that the force is attractive and that the electron is bound to the nucleus.


Calculate Total Energy (E):

Now, add the kinetic and potential energies:
\[ E = KE + PE = \frac{Ke^2}{2r} + \left(-\frac{Ke^2}{r}\right) \] \[ E = \frac{Ke^2}{2r} - \frac{2Ke^2}{2r} \] \[ E = -\frac{Ke^2}{2r} \]

Step 4: Final Answer:

The total energy of the atom is \(-\frac{Ke^2}{2r}\), which corresponds to option (B).
Quick Tip: For any particle in a circular orbit under an inverse-square force (like gravity or electrostatic force), there's a simple relationship between the energies: Total Energy (\(E\)) = -Kinetic Energy (\(KE\)) = \(\frac{1}{2}\) Potential Energy (\(PE\)). Knowing this can be a great shortcut. For instance, once you find \(KE = \frac{Ke^2}{2r}\), you immediately know \(E = -\frac{Ke^2}{2r}\).

Step 1: Understanding the Concept:

When an electron transitions to a lower energy state, a photon is emitted. The energy of this photon is equal to the decrease in the total energy of the electron. We also need to relate the change in total energy to the change in kinetic energy using the relationships derived in the previous part of the case study.


Step 2: Key Formula or Approach:

1. Energy of the emitted photon: \(E_{photon} = \frac{hc}{\lambda}\). To get the energy in eV, we can use the formula \(E (eV) = \frac{1240}{\lambda (nm)}\).

2. Change in total energy: \(\Delta E_{total} = E_{final} - E_{initial} = -E_{photon}\) (since energy is lost).

3. Relationship between Total Energy and Kinetic Energy from the Bohr model: \(E_{total} = -KE\). Therefore, \(\Delta E_{total} = -\Delta KE\).


Step 3: Detailed Explanation:

Given wavelength of the emitted photon, \(\lambda = 500\) nm.


Calculate the Change in Total Energy (\(\Delta E_{total}\)):

First, find the energy of the emitted photon in eV.
\[ E_{photon} = \frac{1240}{\lambda (nm)} = \frac{1240}{500} eV = 2.48 eV \]
Since the photon is emitted, the atom loses this amount of energy. The change in the total energy of the electron is therefore negative.
\[ \Delta E_{total} = -E_{photon} = -2.48 eV \]

Calculate the Change in Kinetic Energy (\(\Delta KE\)):

From the Bohr model, we know the relationship \(E_{total} = -KE\).

This implies that the change in total energy is the negative of the change in kinetic energy.
\[ \Delta E_{total} = \Delta(-KE) = -\Delta KE \]
Therefore,
\[ \Delta KE = -\Delta E_{total} \]
Substituting the value we found for \(\Delta E_{total}\):
\[ \Delta KE = -(-2.48 eV) = +2.48 eV \]
So, as the electron moves to a lower, more tightly bound orbit, its total energy decreases, but its kinetic energy (and speed) increases.


The two values are \(-2.48\) eV and \(+2.48\) eV, respectively.


Step 4: Final Answer:

The change in total energy is -2.48 eV and the change in kinetic energy is 2.48 eV. This corresponds to option (C).
Quick Tip: For any transition in a Bohr-like atom, remember that \(\Delta E_{total} = - \Delta KE = \frac{1}{2} \Delta PE\). This set of relationships is extremely useful. When an electron "falls" to a lower orbit, it loses total energy, loses potential energy, but gains kinetic energy (it moves faster).

Step 1: Understanding the Concept:

The frequency of revolution (\(f\)) is the number of orbits the electron completes per second. It is related to the orbital speed (\(v\)) and the radius of the orbit (\(r\)) by \(f = \frac{v}{2\pi r}\). We need to find how the speed and radius depend on the principal quantum number \(n\).


Step 2: Key Formula or Approach:

From the Bohr model postulates, we can derive the dependencies of radius and velocity on \(n\):

1. Radius of the n\(^{th}\) orbit: \(r_n \propto n^2\).

2. Velocity of the electron in the n\(^{th}\) orbit: \(v_n \propto \frac{1}{n}\).

3. Frequency of revolution: \(f_n = \frac{v_n}{2\pi r_n}\).


Step 3: Detailed Explanation:

Let's find the proportionality for the frequency \(f_n\).
\[ f_n \propto \frac{v_n}{r_n} \]
Substitute the known proportionalities for \(v_n\) and \(r_n\):
\[ f_n \propto \frac{(1/n)}{(n^2)} \] \[ f_n \propto \frac{1}{n \cdot n^2} \] \[ f_n \propto \frac{1}{n^3} \]
Therefore, the frequency of revolution of the electron is inversely proportional to the cube of the principal quantum number.


Step 4: Final Answer:

The frequency of revolution is proportional to \(\frac{1}{n^3}\), which corresponds to option (D).
Quick Tip: It is extremely helpful to memorize the key proportionalities in the Bohr model for the hydrogen atom: Radius \(r_n \propto n^2\) Velocity \(v_n \propto 1/n\) Energy \(E_n \propto 1/n^2\) Angular momentum \(L_n \propto n\) Frequency \(f_n \propto 1/n^3\) These can help you solve many multiple-choice questions quickly.

Step 1: Understanding the Concept:

We need to identify the principal quantum numbers (\(n\)) corresponding to the given energy levels for the hydrogen atom. Then, we can use the formula for the radius of the n\(^{th}\) Bohr orbit to find the initial and final radii and calculate the change.


Step 2: Key Formula or Approach:

1. Energy of the n\(^{th}\) state in a hydrogen atom: \(E_n = -\frac{13.6}{n^2}\) eV.

2. Radius of the n\(^{th}\) orbit in a hydrogen atom: \(r_n = r_1 \times n^2\), where \(r_1\) is the radius of the ground state (Bohr radius).


Step 3: Detailed Explanation:

Given data:

Initial energy state, \(E_i = -3.4\) eV.

Final state is the ground state, so \(n_f = 1\).

Radius of the ground state, \(r_1 = 0.53\) \AA.


Identify the initial quantum number (\(n_i\)):

Use the energy formula to find the \(n\) corresponding to -3.4 eV.
\[ E_i = -\frac{13.6}{n_i^2} \] \[ -3.4 = -\frac{13.6}{n_i^2} \] \[ n_i^2 = \frac{13.6}{3.4} = 4 \] \[ n_i = 2 \]
So, the electron makes a transition from the first excited state (\(n=2\)) to the ground state (\(n=1\)).


Calculate the initial and final radii:

Initial radius (for \(n_i = 2\)):
\[ r_i = r_1 \times n_i^2 = (0.53 \AA) \times (2^2) = 0.53 \times 4 = 2.12 \AA \]
Final radius (for \(n_f = 1\)), which is the ground state radius:
\[ r_f = r_1 = 0.53 \AA \]

Calculate the change in radius:

The change in the radius of the orbit is the difference between the initial and final radii.
\[ \Delta r = r_i - r_f = 2.12 \AA - 0.53 \AA = 1.59 \AA \]

Step 4: Final Answer:

The radius of the orbit changes by 1.59 \AA, which corresponds to option (C).
Quick Tip: Memorize the first few energy levels for the hydrogen atom: \(E_1 = -13.6\) eV, \(E_2 = -3.4\) eV, \(E_3 = -1.51\) eV, etc. This will allow you to quickly identify the quantum numbers involved in a transition, saving valuable time in exams.

Step 1: Understanding the Concept:

The given arrangement of three plates can be viewed as a combination of two separate capacitors. We need to identify how these capacitors are connected (in series or parallel) and then calculate the equivalent capacitance of the combination.


Step 2: Key Formula or Approach:

1. Capacitance of a single parallel plate capacitor with a dielectric: \(C = \frac{K \epsilon_0 A}{d}\).

2. Equivalent capacitance for parallel combination: \(C_{eq} = C_1 + C_2 + \dots\).

3. Equivalent capacitance for series combination: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots\).


Step 3: Detailed Explanation:

The system consists of two capacitors:

- Capacitor \(C_1\) is formed by plates P\(_1\) and P\(_2\).

- Capacitor \(C_2\) is formed by plates P\(_2\) and P\(_3\).


Let's analyze the connections:

- Plate P\(_2\) is connected to terminal A.

- Plates P\(_1\) and P\(_3\) are connected together to terminal B.

This means plate P\(_1\) of \(C_1\) is at potential B, and plate P\(_2\) of \(C_1\) is at potential A. So, \(C_1\) is connected across A and B.

Similarly, plate P\(_2\) of \(C_2\) is at potential A, and plate P\(_3\) of \(C_2\) is at potential B. So, \(C_2\) is also connected across A and B.

Since both capacitors are connected across the same two points (A and B), they are in a parallel combination.


Now, let's calculate the capacitance of each capacitor.

The plates are square with side 'L', so the area is \(A = L^2\). The distance between the plates is \(d\), and the dielectric constant is K.

Capacitance of \(C_1\) is \(C_1 = \frac{K \epsilon_0 L^2}{d}\).

Capacitance of \(C_2\) is \(C_2 = \frac{K \epsilon_0 L^2}{d}\).


For a parallel combination, the equivalent capacitance \(C_{eq}\) is the sum of the individual capacitances:
\[ C_{eq} = C_1 + C_2 \] \[ C_{eq} = \frac{K \epsilon_0 L^2}{d} + \frac{K \epsilon_0 L^2}{d} \] \[ C_{eq} = \frac{2 K \epsilon_0 L^2}{d} \]

Step 4: Final Answer:

The total capacitance of the system is \(\frac{2\epsilon_0 K L^2}{d}\), which corresponds to option (C).
Quick Tip: To analyze multi-plate capacitor systems, redraw the circuit. Assign potentials (like A and B) to the plates and see how the capacitors (formed by adjacent plates) are connected between these potentials. If they are connected across the same two points, they are in parallel.

Step 1: Understanding the Concept:

The charge on a capacitor plate is determined by its capacitance and the potential difference across it. In a parallel combination, the potential difference across each capacitor is the same. Plate P\(_1\) is one of the plates of capacitor \(C_1\).


Step 2: Key Formula or Approach:

The charge \(Q\) on a capacitor is given by the formula \(Q = CV\), where \(C\) is the capacitance and \(V\) is the potential difference.


Step 3: Detailed Explanation:

From the analysis in part (i), we know that the system consists of two capacitors, \(C_1\) and \(C_2\), connected in parallel across the potential difference V.

The potential difference across capacitor \(C_1\) (formed by plates P\(_1\) and P\(_2\)) is V.

The capacitance of \(C_1\) is:
\[ C_1 = \frac{K \epsilon_0 L^2}{d} \]
The charge on capacitor \(C_1\), denoted \(Q_1\), is the magnitude of the charge on each of its plates.
\[ Q_1 = C_1 V \]
Substituting the expression for \(C_1\):
\[ Q_1 = \left(\frac{K \epsilon_0 L^2}{d}\right) V = \frac{\epsilon_0 V K L^2}{d} \]
This \(Q_1\) is the magnitude of the charge on plate P\(_1\). Since point A (connected to P\(_2\)) is at a positive potential with respect to point B (connected to P\(_1\)), plate P\(_1\) will hold a negative charge of this magnitude, and the inner surface of plate P\(_2\) facing P\(_1\) will hold a positive charge of the same magnitude. The question asks for the charge on plate P\(_1\), which refers to this magnitude.


Step 4: Final Answer:

The charge on plate P\(_1\) is \(\frac{\epsilon_0 V K L^2}{d}\), which corresponds to option (B).
Quick Tip: The total charge stored in the system would be \(Q_{total} = C_{eq}V = (\frac{2\epsilon_0 K L^2}{d})V\). Option (C) represents this total charge. Be careful to read the question and identify whether it asks for the charge on a single plate or the total charge of the system.

Step 1: Understanding the Question

The question asks for the magnitude of the electric field in the region between two points, P\(_1\) and P\(_2\). Based on the context of parallel plate capacitors, it's reasonable to assume that P\(_1\) and P\(_2\) represent the two plates of a simple parallel plate capacitor.


Step 2: Key Formula or Approach

For a parallel plate capacitor, the electric field (E) between the plates is uniform (neglecting fringe effects) and is related to the potential difference (V) across the plates and the distance (d) between them by the formula:
\[ E = \frac{V}{d} \]


Step 3: Detailed Explanation

Let's assume:

- V is the potential difference applied across the plates P\(_1\) and P\(_2\).

- d is the separation distance between the plates P\(_1\) and P\(_2\).

The electric field in the region between the plates is defined as the potential gradient. For a uniform field, this relationship simplifies to the potential difference divided by the distance.

Therefore, the electric field E is given by \( E = \frac{V}{d} \).


Step 4: Final Answer

Comparing this result with the given options, we find that option (A) correctly represents the electric field.
Quick Tip: Remember the fundamental relationship between electric field and potential: \( E = -\frac{dV}{dr} \). For a uniform field between capacitor plates, this simplifies to \( E = \frac{\Delta V}{\Delta r} = \frac{V}{d} \). This formula is crucial for many problems involving capacitors.

Step 1: Understanding the Question

The question requires us to find the plate separation of a parallel plate air capacitor that has the same capacitance as a given "system". Based on the options involving a dielectric constant 'K', we can infer that the "system" is a capacitor of the same plate area L\(^2\) and separation 'd', but filled with a dielectric material of constant K.


Step 2: Key Formula or Approach

The capacitance (C) of a parallel plate capacitor is given by:

1. With a dielectric medium of constant K: \( C_{system} = \frac{K \epsilon_0 A}{d} \)

2. With air (or vacuum) as the medium: \( C_{air} = \frac{\epsilon_0 A}{x} \)

where A is the area of the plates, d is the separation in the first case, and x is the new separation we need to find.


Step 3: Detailed Explanation

Let A be the area of the plates, so A = L\(^2\).

The capacitance of the system (capacitor filled with dielectric K) is:
\[ C_{system} = \frac{K \epsilon_0 L^2}{d} \]

Let x be the separation of the equivalent air capacitor. Its capacitance is:
\[ C_{air} = \frac{\epsilon_0 L^2}{x} \]

According to the problem, the capacitances are equal:
\[ C_{air} = C_{system} \]
\[ \frac{\epsilon_0 L^2}{x} = \frac{K \epsilon_0 L^2}{d} \]

Canceling the common terms \( \epsilon_0 L^2 \) from both sides, we get:
\[ \frac{1}{x} = \frac{K}{d} \]

Solving for x, we find:
\[ x = \frac{d}{K} \]


Step 4: Final Answer

The required separation for the air capacitor is \( \frac{d}{K} \). This corresponds to option (A).
Quick Tip: The concept of "equivalent air thickness" is very useful. A dielectric slab of thickness 'd' and constant 'K' is equivalent to an air gap of thickness d/K in terms of its effect on capacitance when charges are constant. This is a shortcut for solving problems with composite dielectrics.

Step 1: Understanding the Question

The question describes a process: first, a capacitor (with plates A and B) is charged by a potential source. Then, the source is disconnected. Finally, the two plates are connected to each other with a conducting wire. We need to find the final net charge on the system.


Step 2: Detailed Explanation

1. Charging the Capacitor: When the potential source is connected, the plates A and B acquire equal and opposite charges. Let's say plate A gets a charge of +Q and plate B gets a charge of -Q.

2. Removing the Source: After the source is removed, the capacitor remains isolated. The charges +Q and -Q are trapped on the plates. The net charge on the isolated two-plate system is (+Q) + (-Q) = 0.

3. Connecting Plates with a Wire: When a conducting wire connects plates A and B, it provides a low-resistance path for the charge to flow. The excess electrons from the negatively charged plate (-Q) will flow through the wire to the positively charged plate (+Q), which has a deficit of electrons.

4. Discharging: This flow of charge continues until the potential difference between the plates becomes zero. At this point, both plates become electrically neutral. The final charge on plate A is 0, and the final charge on plate B is 0.

5. Final Net Charge: The net charge on the system is the sum of the charges on the plates, which is 0 + 0 = 0. This process is called discharging the capacitor.


Step 3: Final Answer

The net charge on the system becomes zero after the plates are connected by a conducting wire. This matches option (D).
Quick Tip: Connecting the two terminals of a charged capacitor with an ideal conductor is called "short-circuiting". This action always leads to the complete discharge of the capacitor, resulting in zero final charge and zero potential difference across its plates.

Step 1: Statement of Lenz's Law

Lenz's law states that the direction of the induced electromotive force (e.m.f.) and hence the induced current in a closed electrical circuit is always such that it opposes the change in magnetic flux that is responsible for producing it.


Step 2: Explanation as a Consequence of Conservation of Energy

Lenz's law is a direct manifestation of the principle of conservation of energy. This can be explained with the following example:

Consider pushing the north pole of a bar magnet towards a closed coil of wire.

1. As the magnet moves towards the coil, the magnetic flux through the coil increases.

2. According to Faraday's law of induction, this change in flux induces an e.m.f. and hence a current in the coil.

3. According to Lenz's law, the direction of this induced current will be such that it opposes the increase in flux. To do this, the face of the coil towards the magnet must become a north pole (since like poles repel).

4. This repulsive force opposes the motion of the magnet. Therefore, we must do mechanical work against this repulsive force to keep the magnet moving towards the coil.

5. This mechanical work done by the external agent is converted into electrical energy in the circuit, which then dissipates as heat (Joule heating). So, mechanical energy is converted into electrical energy, upholding the law of conservation of energy.


Hypothetical Contradiction:

If we assume the opposite, that the induced current's direction aids the change in flux, the face of the coil would become a south pole. This would create an attractive force on the approaching north pole of the magnet. The magnet would accelerate towards the coil without any external work being done. This acceleration would increase the rate of change of flux, which would induce more current, creating an even stronger attractive force. This sets up a perpetual motion machine where kinetic energy and electrical energy are continuously generated from nothing. This is a clear violation of the law of conservation of energy.

Therefore, the induced current must oppose the change in flux, as stated by Lenz's law, to be consistent with the principle of conservation of energy.
Quick Tip: A simple way to remember Lenz's law is to think that "nature opposes change". Whenever there is a change in magnetic flux, the induced effects will try to counteract that change. This principle is fundamental to understanding generators, motors, and transformers.

Step 1: Understanding the Concept:

This problem involves electromagnetic induction. As the conducting loop moves through the magnetic field, the magnetic flux linked with it changes. According to Faraday's Law of Induction, this change in flux induces an electromotive force (emf) and hence a current in the loop. The direction of the current is determined by Lenz's Law, which states that the induced current will flow in a direction that opposes the change in magnetic flux that produced it.


Step 2: Key Formula or Approach:

1. Magnetic Flux (\(\phi\)): \(\phi = \vec{B} \cdot \vec{A} = BA\cos\theta\). For this case, \(\phi = BA\), where A is the area of the loop inside the field.

2. Faraday's Law of Induction: Induced emf \(E = -\frac{d\phi}{dt}\).

3. Lenz's Law / Right-Hand Rule: To determine the direction of the induced current.


Step 3: Detailed Explanation:

The motion of the loop can be divided into three stages: entering the field, moving completely within the field, and leaving the field. Let the side of the square loop be \(s = l/2\). The width of the magnetic field region is \(l\).


(I) Directions of Induced Current:

1. When the loop enters the field:

The magnetic field \(\vec{B}\) is directed into the page.
As the loop enters, the area inside the field increases, so the magnetic flux directed into the page increases.
According to Lenz's law, the induced current must create a magnetic field that opposes this change, i.e., a field directed out of the page.
Using the right-hand grip rule, a current that produces a field out of the page must be in the anti-clockwise direction.


2. When the loop leaves the field:

As the loop leaves, the area inside the field decreases, so the magnetic flux directed into the page decreases.
According to Lenz's law, the induced current must create a magnetic field that opposes this change, i.e., a field directed into the page to reinforce the diminishing external field.
Using the right-hand grip rule, a current that produces a field into the page must be in the clockwise direction.


(II) Plots of Flux and EMF vs. Time:

Let's define the time intervals for the three stages of motion.

Stage 1 (Entering): Time taken to enter is \(t_1 = \frac{distance}{speed} = \frac{l/2}{v} = \frac{l}{2v}\). This occurs for \(0 \le t \le \frac{l}{2v}\).
Stage 2 (Inside): Time spent fully inside is \(t_{in} = \frac{l-l/2}{v} = \frac{l}{2v}\). This occurs for \(\frac{l}{2v} \le t \le \frac{l}{v}\).
Stage 3 (Leaving): Time taken to leave is \(t_3 = \frac{l/2}{v} = \frac{l}{2v}\). This occurs for \(\frac{l}{v} \le t \le \frac{3l}{2v}\).


Plot of Magnetic Flux (\(\phi\)) vs. Time (t):


For \(0 \le t \le \frac{l}{2v}\): The area inside the field increases linearly with time, \(A(t) = (\frac{l}{2})(vt)\). The flux increases linearly: \(\phi(t) = B A(t) = \frac{Blv}{2}t\). It goes from 0 to \(\phi_{max} = B(\frac{l}{2})^2 = \frac{Bl^2}{4}\).
For \(\frac{l}{2v} \le t \le \frac{l}{v}\): The loop is completely inside. The flux is constant at its maximum value, \(\phi(t) = \phi_{max} = \frac{Bl^2}{4}\).
For \(\frac{l}{v} \le t \le \frac{3l}{2v}\): The area inside the field decreases linearly. The flux decreases linearly from \(\phi_{max}\) to 0.

The graph is a trapezoid.


% Placeholder for Flux vs Time graph
Graph of \(\phi\) vs. t: Starts at (0,0), rises linearly to (\(l/2v, Bl^2/4\)), remains constant until (\(l/v, Bl^2/4\)), then decreases linearly to (\(3l/2v, 0\)).


Plot of Induced EMF (E) vs. Time (t):

The induced EMF is the negative of the slope of the \(\phi\)-t graph (\(E = -d\phi/dt\)).

For \(0 \le t \le \frac{l{2v}\): The slope of the flux graph is constant and positive: \(\frac{d\phi}{dt} = \frac{Blv}{2}\). So, the EMF is constant and negative: \(E = -\frac{Blv}{2}\).
For \(\frac{l}{2v} \le t \le \frac{l}{v}\): The slope of the flux graph is zero. So, the EMF is zero: \(E=0\).
For \(\frac{l}{v} \le t \le \frac{3l}{2v}\): The slope of the flux graph is constant and negative: \(\frac{d\phi}{dt} = -\frac{Blv}{2}\). So, the EMF is constant and positive: \(E = -(-\frac{Blv}{2}) = +\frac{Blv}{2}\).

The graph consists of two rectangular pulses.


% Placeholder for EMF vs Time graph
\textit{Graph of E vs. t: For \(0 < t < l/2v\), E is a constant \( -Blv/2 \). For \(l/2v < t < l/v\), E is 0. For \(l/v < t < 3l/2v\), E is a constant \( +Blv/2 \). Quick Tip: Motional EMF in a straight conductor of length \(L\) moving with velocity \(v\) perpendicular to a field \(B\) is given by \(E = BLv\). Here, when the loop enters or leaves, one vertical side of length \(l/2\) is cutting the flux lines, so the magnitude of the induced EMF is \(E = B(l/2)v\). This is a quick way to find the magnitude of the EMF.

Step 1: Understanding the Concept:

Alternating current (AC) varies sinusoidally with time. To characterize its magnitude, we use different measures like peak value and RMS value. The peak value represents the maximum instantaneous current, while the RMS value represents the effective or heating value of the current.


Step 2: Detailed Explanation:

Differentiation between Peak and RMS Values:

\begin{tabular{|l|l|l|
\hline
Aspect & Peak Value (\(I_0\)) & RMS Value (\(I_{rms}\))

\hline
Definition & It is the maximum instantaneous value or & It is the effective value of AC, defined as

& amplitude of the alternating current & the equivalent steady DC that produces

& during one complete cycle. & the same heating effect in the same

& & resistor over the same time.

\hline
Representation & Represents the highest magnitude the & Represents the average heating power.

& current reaches. & It is also called the "effective value" or

& & "virtual value".

\hline
Measurement & Can be directly observed on an & Standard AC voltmeters and ammeters

& oscilloscope. & are calibrated to read RMS values.

\hline
Calculation & It is the amplitude \(I_0\) in the equation & It is the Root Mean Square of the

& \(I(t) = I_0 \sin(\omega t)\). & instantaneous current over a full cycle.

\hline
\end{tabular


Relation between Peak and RMS Values:

For a sinusoidal alternating current given by the equation \(I(t) = I_0 \sin(\omega t)\), the RMS value (\(I_{rms}\)) is related to the peak value (\(I_0\)) by the following formula:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0 \]
This means the RMS value of a sinusoidal AC is approximately 70.7% of its peak value.


Derivation:
The mean square value of the current over one cycle (from \(t=0\) to \(t=T\)) is:
\[ \overline{I^2} = \frac{1}{T}\int_0^T I(t)^2 dt = \frac{1}{T}\int_0^T (I_0 \sin(\omega t))^2 dt \] \[ \overline{I^2} = \frac{I_0^2}{T}\int_0^T \sin^2(\omega t) dt \]
Using the trigonometric identity \(\sin^2(\theta) = \frac{1-\cos(2\theta)}{2}\):
\[ \overline{I^2} = \frac{I_0^2}{T}\int_0^T \frac{1 - \cos(2\omega t)}{2} dt = \frac{I_0^2}{2T} \left[ t - \frac{\sin(2\omega t)}{2\omega} \right]_0^T \]
Since the integral of the cosine term over a full period is zero, we get:
\[ \overline{I^2} = \frac{I_0^2}{2T} [T - 0] = \frac{I_0^2}{2} \]
The RMS value is the square root of this mean square value:
\[ I_{rms} = \sqrt{\overline{I^2}} = \sqrt{\frac{I_0^2}{2}} = \frac{I_0}{\sqrt{2}} \] Quick Tip: When you see an AC voltage or current rating, for example, "220 V AC" for household supply, this value is the RMS value, not the peak value. The peak voltage would be \(220 \times \sqrt{2} \approx 311\) V. The RMS value is used because it allows direct application of power formulas like \(P = I^2R\) in the same way as for DC.

Step 1: Understanding the Concept:

This question deals with the phase relationship between voltage and current in purely inductive and purely capacitive AC circuits. When these elements are combined in series, they can exhibit resonance, a condition where the reactances cancel each other out.


Step 2: Detailed Explanation:


(I) Identification of Elements X and Y


Element X: The voltage leads the current by \(\frac{\pi}{2}\) (or 90\(^\circ\)). This is the characteristic behavior of a pure inductor. In an inductor, the back emf opposes the change in current, causing the current to lag behind the voltage.

Phasor Diagram for Inductor (X):

The phasor for voltage (\(V_L\)) is rotated 90\(^\circ\) anti-clockwise with respect to the phasor for current (\(I\)).

% Placeholder for Inductor Phasor Diagram
Diagram shows the current phasor I along the positive x-axis and the voltage phasor \(V_L\) along the positive y-axis, indicating a +90\(^\circ\) phase lead.


Element Y: The voltage lags behind the current by \(\frac{\pi{2}\) (or 90\(^\circ\)). This is the characteristic behavior of a pure capacitor. In a capacitor, the current must flow to build up charge (and thus voltage), so the current leads the voltage.

Phasor Diagram for Capacitor (Y):

The phasor for voltage (\(V_C\)) is rotated 90\(^\circ\) clockwise with respect to the phasor for current (\(I\)).

% Placeholder for Capacitor Phasor Diagram
Diagram shows the current phasor I along the positive x-axis and the voltage phasor \(V_C\) along the negative y-axis, indicating a -90\(^\circ\) phase lag.


(II) Resonance in Series LC Circuit

When element X (Inductor L) and element Y (Capacitor C) are connected in series, it forms a series LC circuit.


Condition of Resonance:

Resonance in a series LCR circuit occurs when the net reactance of the circuit is zero. In an LC circuit, this happens when the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)).
\[ X_L = X_C \]

Expression for Resonant Frequency:

We know that \(X_L = \omega L\) and \(X_C = \frac{1{\omega C}\), where \(\omega = 2\pi\nu\) is the angular frequency.

The resonance condition becomes:
\[ \omega_r L = \frac{1}{\omega_r C} \]
where \(\omega_r\) is the resonant angular frequency.
\[ \omega_r^2 = \frac{1}{LC} \implies \omega_r = \frac{1}{\sqrt{LC}} \]
Substituting \(\omega_r = 2\pi\nu_r\), where \(\nu_r\) is the resonant frequency:
\[ 2\pi\nu_r = \frac{1}{\sqrt{LC}} \] \[ \nu_r = \frac{1}{2\pi\sqrt{LC}} \]

Impedance Value at Resonance:

The impedance of a series circuit is given by \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).

For an ideal LC circuit, the resistance \(R=0\). At resonance, \(X_L = X_C\), so the term \((X_L - X_C)\) is also zero.
\[ Z = \sqrt{0^2 + (0)^2} = 0 \]
The impedance of an ideal series LC circuit at resonance is zero.
Quick Tip: A useful mnemonic to remember the phase relationships is "ELI the ICE man". For an inductor (L), the EMF (E) or Voltage leads the Current (I) - ELI. For a capacitor (C), the Current (I) leads the EMF (E) or Voltage - ICE.

N/A

Step 1: Understanding the Concept:

This question deals with the phase relationship between voltage and current in purely inductive and purely capacitive AC circuits. When these elements are combined in series, they can exhibit resonance, a condition where the reactances cancel each other out.


Step 2: Detailed Explanation:


(I) Identification of Elements X and Y


Element X: The voltage leads the current by \(\frac{\pi}{2}\) (or 90\(^\circ\)). This is the characteristic behavior of a pure inductor. In an inductor, the back emf opposes the change in current, causing the current to lag behind the voltage.

Phasor Diagram for Inductor (X):

The phasor for voltage (\(V_L\)) is rotated 90\(^\circ\) anti-clockwise with respect to the phasor for current (\(I\)).

% Placeholder for Inductor Phasor Diagram
Diagram shows the current phasor I along the positive x-axis and the voltage phasor \(V_L\) along the positive y-axis, indicating a +90\(^\circ\) phase lead.


Element Y: The voltage lags behind the current by \(\frac{\pi{2}\) (or 90\(^\circ\)). This is the characteristic behavior of a pure capacitor. In a capacitor, the current must flow to build up charge (and thus voltage), so the current leads the voltage.

Phasor Diagram for Capacitor (Y):

The phasor for voltage (\(V_C\)) is rotated 90\(^\circ\) clockwise with respect to the phasor for current (\(I\)).

% Placeholder for Capacitor Phasor Diagram
Diagram shows the current phasor I along the positive x-axis and the voltage phasor \(V_C\) along the negative y-axis, indicating a -90\(^\circ\) phase lag.


(II) Resonance in Series LC Circuit

When element X (Inductor L) and element Y (Capacitor C) are connected in series, it forms a series LC circuit.


Condition of Resonance:

Resonance in a series LCR circuit occurs when the net reactance of the circuit is zero. In an LC circuit, this happens when the inductive reactance (\(X_L\)) equals the capacitive reactance (\(X_C\)).
\[ X_L = X_C \]

Expression for Resonant Frequency:

We know that \(X_L = \omega L\) and \(X_C = \frac{1{\omega C}\), where \(\omega = 2\pi\nu\) is the angular frequency.

The resonance condition becomes:
\[ \omega_r L = \frac{1}{\omega_r C} \]
where \(\omega_r\) is the resonant angular frequency.
\[ \omega_r^2 = \frac{1}{LC} \implies \omega_r = \frac{1}{\sqrt{LC}} \]
Substituting \(\omega_r = 2\pi\nu_r\), where \(\nu_r\) is the resonant frequency:
\[ 2\pi\nu_r = \frac{1}{\sqrt{LC}} \] \[ \nu_r = \frac{1}{2\pi\sqrt{LC}} \]

Impedance Value at Resonance:

The impedance of a series circuit is given by \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).

For an ideal LC circuit, the resistance \(R=0\). At resonance, \(X_L = X_C\), so the term \((X_L - X_C)\) is also zero.
\[ Z = \sqrt{0^2 + (0)^2} = 0 \]
The impedance of an ideal series LC circuit at resonance is zero.
Quick Tip: A useful mnemonic to remember the phase relationships is "ELI the ICE man". For an inductor (L), the EMF (E) or Voltage leads the Current (I) - ELI. For a capacitor (C), the Current (I) leads the EMF (E) or Voltage - ICE.

Step 1: Understanding the Concept:

This problem involves two scenarios. First, finding the image position for a single convex lens. Second, finding the focal length of a combination of lenses, and from that, the focal length of the unknown concave lens. We will use the thin lens formula for both cases.


Step 2: Key Formula or Approach:

1. Thin Lens Formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)

2. Combination of thin lenses in contact: \(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\)

Sign convention: Distances measured in the direction of light are positive, and opposite are negative. The object is placed to the left of the lens.


Step 3: Detailed Explanation:


Case 1: Convex lens only

Given:
Object distance, \(u = -30\) cm.

Focal length of convex lens, \(f_1 = +10\) cm.

Let the initial image position be \(v_1\). Using the lens formula:
\[ \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u} \] \[ \frac{1}{10} = \frac{1}{v_1} - \frac{1}{-30} = \frac{1}{v_1} + \frac{1}{30} \] \[ \frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30} = \frac{1}{15} \] \[ v_1 = +15 cm \]
The initial sharp image is formed 15 cm to the right of the lens.


Case 2: Combination of lenses

A concave lens (focal length \(f_2\)) is placed in contact with the convex lens.

The screen has to be moved by 45 cm from its initial position to get a sharp image. Since adding a diverging lens makes the system less converging, the final image will be formed further away.

New image distance, \(v_{final} = v_1 + 45 cm = 15 + 45 = 60\) cm. So, \(v_{final} = +60\) cm.

The object distance remains the same, \(u = -30\) cm.

Let \(F\) be the equivalent focal length of the combination. Using the lens formula for the combination:
\[ \frac{1}{F} = \frac{1}{v_{final}} - \frac{1}{u} \] \[ \frac{1}{F} = \frac{1}{60} - \frac{1}{-30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20} \] \[ F = +20 cm \]
Now, use the formula for the combination of lenses:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \] \[ \frac{1}{20} = \frac{1}{10} + \frac{1}{f_2} \] \[ \frac{1}{f_2} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = -\frac{1}{20} \] \[ f_2 = -20 cm \]
The focal length of the concave lens is -20 cm.
Quick Tip: When lenses are combined, first analyze the initial setup to find the first image position. Then, use this information to determine the final image position for the combined system. Finally, use the lens combination formula to find the unknown focal length.

Step 1: Understanding the Concept:

The angle of minimum deviation (\(\delta_m\)) is the smallest angle of deviation a prism can produce for a given wavelength of light. It occurs when the angle of incidence and emergence are equal, and the refracted ray inside the prism is parallel to its base.


Step 2: Key Formula or Approach:

1. For an equilateral prism, the angle of the prism, \(A = 60^\circ\).

2. The prism formula relates refractive index (\(n\)), prism angle (\(A\)), and angle of minimum deviation (\(\delta_m\)):
\[ n = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
3. At minimum deviation, the angle of incidence (\(i\)) is related to \(A\) and \(\delta_m\) by:
\[ i = \frac{A+\delta_m}{2} \]

Step 3: Detailed Explanation:

Given data:

Refractive index, \(n = \sqrt{3}\).

Angle of prism, \(A = 60^\circ\).


Calculate the angle of minimum deviation (\(\delta_m\)):

Substitute the known values into the prism formula:
\[ \sqrt{3} = \frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} \] \[ \sqrt{3} = \frac{\sin\left(30^\circ+\frac{\delta_m}{2}\right)}{\sin(30^\circ)} \]
Since \(\sin(30^\circ) = \frac{1}{2}\):
\[ \sqrt{3} = \frac{\sin\left(30^\circ+\frac{\delta_m}{2}\right)}{1/2} \] \[ \sin\left(30^\circ+\frac{\delta_m}{2}\right) = \frac{\sqrt{3}}{2} \]
We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\). Therefore:
\[ 30^\circ+\frac{\delta_m}{2} = 60^\circ \] \[ \frac{\delta_m}{2} = 60^\circ - 30^\circ = 30^\circ \] \[ \delta_m = 60^\circ \]

Calculate the angle of incidence (\(i\)):

Using the formula for the angle of incidence at minimum deviation:
\[ i = \frac{A+\delta_m}{2} = \frac{60^\circ+60^\circ}{2} = \frac{120^\circ}{2} = 60^\circ \] Quick Tip: A special case to remember: when the angle of minimum deviation is equal to the angle of the prism (\(\delta_m = A\)), the refractive index of the prism is given by \(n = 2\cos(A/2)\). In this problem, \(A=60^\circ\), \(\delta_m = 60^\circ\), so \(n = 2\cos(30^\circ) = 2(\frac{\sqrt{3}}{2}) = \sqrt{3}\), which matches the given information.

Step 1: Understanding the Concept:

This problem applies the formulas of Young's Double Slit Experiment (YDSE). We need to calculate the slit separation required to produce a given fringe width, and then find the position of the first dark fringe (minimum).


Step 2: Key Formula or Approach:

1. Fringe width: \(\beta = \frac{\lambda D}{d}\)

2. Position of the n-th minimum (dark fringe): \(y_{min, n} = \left(n - \frac{1}{2}\right) \frac{\lambda D}{d}\)

where \(\lambda\) is the wavelength, \(D\) is the screen distance, and \(d\) is the slit separation.


Step 3: Detailed Explanation:

Given data:

Wavelength, \(\lambda = 633 nm = 633 \times 10^{-9}\) m.

Screen distance, \(D = 5.0\) m.

Fringe width, \(\beta = 5 mm = 5 \times 10^{-3}\) m.


(I) Find the slit separation (d):

Rearrange the fringe width formula to solve for \(d\):
\[ d = \frac{\lambda D}{\beta} \]
Substitute the given values:
\[ d = \frac{(633 \times 10^{-9} m) \times (5.0 m)}{5 \times 10^{-3} m} \] \[ d = \frac{633 \times 5.0 \times 10^{-9}}{5 \times 10^{-3}} = 633 \times 10^{-9 - (-3)} = 633 \times 10^{-6} m \]
To express the answer in millimeters (mm):
\[ d = 0.633 \times 10^{-3} m = 0.633 mm \]

(II) Find the position of the first minimum:

The first minimum corresponds to \(n=1\) in the formula for dark fringes.
\[ y_{min, 1} = \left(1 - \frac{1}{2}\right) \frac{\lambda D}{d} = \frac{1}{2} \frac{\lambda D}{d} \]
We recognize that \(\frac{\lambda D}{d}\) is the fringe width \(\beta\).
\[ y_{min, 1} = \frac{1}{2} \beta \]
Substitute the given value of \(\beta\):
\[ y_{min, 1} = \frac{1}{2} \times (5 mm) = 2.5 mm \]
The first minimum will be 2.5 mm from the central maximum.
Quick Tip: In YDSE, the distance from the center to the first dark fringe is always exactly half the fringe width. The distance between consecutive bright or dark fringes is equal to the fringe width \(\beta\).

Step 1: Understanding the Concept:

This problem deals with single-slit diffraction. When light passes through a narrow slit, it diffracts, creating a pattern of bright and dark fringes on a screen. The central fringe (maximum) is the brightest and widest. The question asks for the width of this central maximum, which is defined as the distance between the first minima on either side of the center.


Step 2: Key Formula or Approach:

The position (\(y\)) of the n-th minimum in a single-slit diffraction pattern is given by the formula:
\[ y = \frac{n \lambda D}{a} \]
where \(n\) is an integer (\(\pm 1, \pm 2, \dots\)), \(\lambda\) is the wavelength of light, \(D\) is the distance from the slit to the screen, and \(a\) is the width of the slit.

The width of the central maximum is the distance between the first minimum on the positive side (\(n=1\)) and the first minimum on the negative side (\(n=-1\)).

Width \(W = y_1 - y_{-1} = \frac{1 \cdot \lambda D}{a} - \frac{-1 \cdot \lambda D}{a} = \frac{2\lambda D}{a}\).


Step 3: Detailed Explanation:

Given data:

Wavelength, \(\lambda = 650 nm = 650 \times 10^{-9}\) m.

Slit width, \(a = 0.6 mm = 0.6 \times 10^{-3}\) m.

Screen distance, \(D = 60 cm = 0.6\) m.


Using the formula for the width of the central maximum:
\[ W = \frac{2\lambda D}{a} \]
Substitute the given values into the formula:
\[ W = \frac{2 \times (650 \times 10^{-9} m) \times (0.6 m)}{0.6 \times 10^{-3} m} \]
The term 0.6 in the numerator and denominator cancels out.
\[ W = 2 \times 650 \times \frac{10^{-9}}{10^{-3}} m \] \[ W = 1300 \times 10^{-6} m = 1.3 \times 10^{-3} m \]
To express the answer in millimeters (mm):
\[ W = 1.3 mm \]

Step 4: Final Answer:

The distance between the first order minima on both sides of the central maximum is 1.3 mm.
Quick Tip: Be careful to distinguish between the formulas for interference (double-slit) and diffraction (single-slit). For single-slit diffraction, the width of the central maximum (\(2\lambda D/a\)) is twice the width of the other secondary maxima (\(\lambda D/a\)).

Step 1: Understanding the Concept:

This setup describes an electric dipole. We need to find the electric field at a point on the equatorial line (or perpendicular bisector) of the dipole. The net field is the vector sum of the fields produced by each individual charge.


Step 2: Key Formula or Approach:

1. Electric field due to a point charge \(Q\): \(\vec{E} = k\frac{Q}{r^2} \hat{r}\), where \(r\) is the distance and \(\hat{r}\) is the unit vector from the charge to the point.

2. Principle of Superposition: \(\vec{E}_{net} = \vec{E}_1 + \vec{E}_2\).


Step 3: Detailed Explanation:

Let the point be P(0, y). The charges are at A(-a, 0) and B(a, 0).
The distance from each charge to point P is \(r = AP = BP = \sqrt{(y-0)^2 + (0-(-a))^2} = \sqrt{y^2 + a^2}\).


The electric field at P due to \(-q\) at A is \(\vec{E}_A\). Its magnitude is \(E_A = \frac{k q}{r^2} = \frac{k q}{y^2+a^2}\). Its direction is from P towards A.

The electric field at P due to \(+q\) at B is \(\vec{E}_B\). Its magnitude is \(E_B = \frac{k q}{r^2} = \frac{k q}{y^2+a^2}\). Its direction is from B towards P.

So, \(|\vec{E}_A| = |\vec{E}_B| = E\).


Let \(\theta\) be the angle between the y-axis and the line connecting a charge to P.
By resolving the vectors into components:
- The y-components, \(E_A \cos\theta\) and \(-E_B \cos\theta\), are equal and opposite, so they cancel out.
- The x-components both point in the negative x-direction. \(E_{Ax} = -E_A \sin\theta\) and \(E_{Bx} = -E_B \sin\theta\).

The net electric field \(\vec{E}_{net}\) is along the negative x-axis.
\[ E_{net} = E_{Ax} + E_{Bx} = 2 E_A \sin\theta \]
From the geometry of the triangle, \(\sin\theta = \frac{a}{r} = \frac{a}{\sqrt{y^2+a^2}}\).

Substitute the expressions for \(E_A\) and \(\sin\theta\):
\[ E_{net} = 2 \left(\frac{k q}{y^2+a^2}\right) \left(\frac{a}{\sqrt{y^2+a^2}}\right) = \frac{2kqa}{(y^2+a^2)^{3/2}} \]
In vector form, since the direction is along \(\hat{-i}\):
\[ \vec{E}_{net} = -\frac{2kqa}{(y^2+a^2)^{3/2}} \hat{i} \]
This is the expression for the net electric field. The quantity \(p = 2qa\) is the magnitude of the electric dipole moment.


Electric field at a far off point (\(y \gg a\)):

In this case, \(a^2\) is negligible compared to \(y^2\) in the denominator.
\[ y^2 + a^2 \approx y^2 \]
So, the denominator becomes \((y^2)^{3/2} = y^3\).

The expression for the electric field simplifies to:
\[ \vec{E}_{net} \approx -\frac{2kqa}{y^3} \hat{i} = -\frac{kp}{y^3} \hat{i} \] Quick Tip: For an electric dipole, remember the two standard results for far-field approximation: On the axial line, the field magnitude is \(E \approx \frac{2kp}{r^3}\) (in the direction of \(\vec{p}\)). On the equatorial line, the field magnitude is \(E \approx \frac{kp}{r^3}\) (opposite to the direction of \(\vec{p}\)).

Step 1: Understanding the Concept:

The work done in moving a configuration of charges is equal to the change in the electrostatic potential energy of the system. We need to calculate the potential energy of the system in its initial configuration and its final configuration and find the difference.


Step 2: Key Formula or Approach:

1. Work Done, \(W = \Delta U = U_{final} - U_{initial}\).

2. Potential energy of a system of three charges: \(U = k\left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right)\).


Step 3: Detailed Explanation:

Given charges:
\(q_A = -2 nC = -2 \times 10^{-9}\) C
\(q_B = -1 nC = -1 \times 10^{-9}\) C
\(q_C = +5 nC = +5 \times 10^{-9}\) C

Coulomb's constant, \(k = 9 \times 10^9 Nm^2C^{-2}\).


Initial Configuration:

The charges are at the vertices A, B, C of an equilateral triangle of side \(s = 0.2\) m. The distance between each pair of charges is \(r_{initial} = 0.2\) m.

The initial potential energy \(U_{initial}\) is:
\[ U_{initial} = \frac{k}{r_{initial}} (q_A q_B + q_B q_C + q_C q_A) \] \[ q_A q_B = (-2 \times 10^{-9})(-1 \times 10^{-9}) = 2 \times 10^{-18} \] \[ q_B q_C = (-1 \times 10^{-9})(5 \times 10^{-9}) = -5 \times 10^{-18} \] \[ q_C q_A = (5 \times 10^{-9})(-2 \times 10^{-9}) = -10 \times 10^{-18} \] \[ Sum of products = (2 - 5 - 10) \times 10^{-18} = -13 \times 10^{-18} C^2 \] \[ U_{initial} = \frac{9 \times 10^9}{0.2} (-13 \times 10^{-18}) = (4.5 \times 10^{10})(-13 \times 10^{-18}) = -58.5 \times 10^{-8} J = -5.85 \times 10^{-7} J \]

Final Configuration:

The charges are moved to the midpoints A\(_1\), B\(_1\), C\(_1\). These midpoints form a smaller equilateral triangle whose side length is half the original side length.

The distance between each pair of charges is now \(r_{final} = s/2 = 0.2/2 = 0.1\) m.

The final potential energy \(U_{final}\) is:
\[ U_{final} = \frac{k}{r_{final}} (q_A q_B + q_B q_C + q_C q_A) \] \[ U_{final} = \frac{9 \times 10^9}{0.1} (-13 \times 10^{-18}) = (9 \times 10^{10})(-13 \times 10^{-18}) = -117 \times 10^{-8} J = -11.7 \times 10^{-7} J \]

Work Done:
\[ W = U_{final} - U_{initial} \] \[ W = (-11.7 \times 10^{-7}) - (-5.85 \times 10^{-7}) J \] \[ W = (-11.7 + 5.85) \times 10^{-7} J = -5.85 \times 10^{-7} J \] Quick Tip: Notice that \(U_{final} = \frac{k}{s/2} \times (sum) = 2 \times \frac{k}{s} \times (sum) = 2 U_{initial}\). So, the work done is \(W = U_{final} - U_{initial} = 2 U_{initial} - U_{initial} = U_{initial}\). You only need to calculate the initial potential energy. This shortcut applies when all dimensions are scaled by the same factor.

Step 1: Understanding the Concept:

Gauss's law is a fundamental law of electrostatics that relates the electric flux through a closed surface to the net charge enclosed by that surface. It can be used to derive Coulomb's law, and vice versa, showing their consistency. It is a powerful tool for calculating the electric field for charge distributions with high symmetry, like a spherical shell.


Step 2: Key Formula or Approach:

Gauss's Law: \(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}\)


Step 3: Detailed Explanation:


Consistency of Gauss's Law and Coulomb's Law:

Let's derive Coulomb's law from Gauss's law. Consider a point charge \(q\). To find the electric field at a distance \(r\) from this charge, we choose a spherical Gaussian surface of radius \(r\) concentric with the charge.

By symmetry, the electric field \(\vec{E}\) must be directed radially outward and have the same magnitude \(E\) at every point on the spherical surface.
The area vector \(d\vec{A}\) for any small area element on the sphere is also radially outward. Thus, \(\vec{E}\) is parallel to \(d\vec{A}\), and \(\vec{E} \cdot d\vec{A} = E \, dA\).
Applying Gauss's Law: \(\oint \vec{E} \cdot d\vec{A} = \oint E \, dA = E \oint dA = E(4\pi r^2)\).
The charge enclosed by the surface is \(q_{enc} = q\).
So, \(E(4\pi r^2) = \frac{q}{\epsilon_0}\), which gives \(E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\).
The force on a test charge \(q_0\) placed at this point is \(F = q_0 E = \frac{1}{4\pi\epsilon_0}\frac{q q_0}{r^2}\), which is Coulomb's Law. This demonstrates their consistency.


Electric Field of a Uniformly Charged Thin Spherical Shell:

Let the shell have radius \(r\) and a total charge \(Q\) uniformly distributed on its surface. We want to find the field at a distance \(y\) from its center.


(I) Outside the shell (\(y > r\)):

We construct a concentric spherical Gaussian surface of radius \(y\).

The total charge enclosed is \(q_{enc} = Q\).
By spherical symmetry, the electric field \(\vec{E}\) is radial and has a constant magnitude \(E\) on the Gaussian surface.
Applying Gauss's Law: \(E(4\pi y^2) = \frac{Q}{\epsilon_0}\).
Solving for \(E\): \[ E = \frac{1}{4\pi\epsilon_0} \frac{Q}{y^2} \]
For any point outside, the shell behaves as if its entire charge were concentrated at its center.


(II) Inside the shell (\(y < r\)):

We construct a concentric spherical Gaussian surface of radius \(y\).

Since the charge resides only on the surface of the shell, the charge enclosed by this Gaussian surface is \(q_{enc} = 0\).
Applying Gauss's Law: \(E(4\pi y^2) = \frac{0}{\epsilon_0} = 0\).
This implies: \[ E = 0 \]
The electric field is zero at all points inside a uniformly charged thin spherical shell. Quick Tip: The key to applying Gauss's law effectively is to choose a Gaussian surface that matches the symmetry of the charge distribution. For spheres, use concentric spheres; for cylinders/wires, use coaxial cylinders; for planes, use pillboxes.

Step 1: Understanding the Concept:

Electric potential is a scalar quantity. The total potential at a point due to a system of point charges is the algebraic sum of the potentials due to each individual charge. We are asked to find a charge \(q_2\) such that the sum of potentials from \(q_1\) and \(q_2\) is zero at a specified point.


Step 2: Key Formula or Approach:

1. Potential due to a point charge: \(V = k\frac{q}{r}\).

2. Principle of Superposition for Potential: \(V_{total} = V_1 + V_2\).


Step 3: Detailed Explanation:

Let the given charges and points be:

Charge \(q_1 = +2 nC = +2 \times 10^{-9}\) C at the origin O(0, 0, 0).

Unknown charge \(q_2\) is placed at point P\(_1\)(0, 0, -6 m).

The potential is to be zero at point P\(_2\)(0, 0, 2 m).


The condition is \(V_{total} at P_2 = 0\).
\[ V_1 + V_2 = 0 \]
Where \(V_1\) is the potential at P\(_2\) due to \(q_1\), and \(V_2\) is the potential at P\(_2\) due to \(q_2\).


Calculate \(V_1\):

The distance between \(q_1\) (at origin) and P\(_2\) is \(r_1 = 2\) m.
\[ V_1 = k\frac{q_1}{r_1} = (9 \times 10^9) \frac{2 \times 10^{-9}}{2} = 9 V \]

Calculate \(V_2\):

The distance between \(q_2\) (at P\(_1\)) and P\(_2\) is \(r_2 = |2 - (-6)| = 8\) m.
\[ V_2 = k\frac{q_2}{r_2} = (9 \times 10^9) \frac{q_2}{8} \]

Solve for \(q_2\):

Substitute \(V_1\) and \(V_2\) into the condition \(V_1 + V_2 = 0\).
\[ 9 + (9 \times 10^9) \frac{q_2}{8} = 0 \] \[ (9 \times 10^9) \frac{q_2}{8} = -9 \]
Cancel the 9 from both sides:
\[ \frac{10^9 q_2}{8} = -1 \] \[ q_2 = -8 \times 10^{-9} C \] \[ q_2 = -8 nC \]

Step 4: Final Answer:

The required charge is negative in type.

The magnitude of the charge is \(8 nC\).
Quick Tip: Remember that potential is a scalar, so you just need to do algebraic addition. Pay careful attention to the signs of the charges and the distances, which are always positive.