Step 1: Understanding the Concept:

The electric dipole moment (\(\vec{p}\)) is a vector quantity used to measure the separation of positive and negative electric charges in a system. The direction of the dipole moment vector is from the negative charge to the positive charge. The net dipole moment of a system of multiple dipoles is the vector sum of the individual dipole moments.


Step 2: Key Formula or Approach:

The dipole moment \(\vec{p}\) for a pair of charges +q and -q separated by a displacement vector \(\vec{d}\) is given by: \[ \vec{p} = q\vec{d} \]
where \(\vec{d}\) is the vector pointing from the negative charge to the positive charge.

The net dipole moment of the system (\(\vec{p}_{net}\)) is the vector sum of the individual dipole moments: \[ \vec{p}_{net} = \vec{p}_1 + \vec{p}_2 \]


Step 3: Detailed Explanation:

For dipole D₁:

The negative charge (-q) is at the origin (0, 0).

The positive charge (+q) is at (a, 0).

The displacement vector \(\vec{d}_1\) from the negative charge to the positive charge is: \[ \vec{d}_1 = (a - 0)\hat{i} + (0 - 0)\hat{j} = a\hat{i} \]
The dipole moment of D₁ is: \[ \vec{p}_1 = q\vec{d}_1 = qa\hat{i} \]


For dipole D₂:

The charges are at (0, a) and (0, 2a). Assuming the negative charge (-q) is at (0, a) and the positive charge (+q) is at (0, 2a).

The displacement vector \(\vec{d}_2\) from the negative charge to the positive charge is: \[ \vec{d}_2 = (0 - 0)\hat{i} + (2a - a)\hat{j} = a\hat{j} \]
The dipole moment of D₂ is: \[ \vec{p}_2 = q\vec{d}_2 = qa\hat{j} \]


Net Dipole Moment:

The net dipole moment of the system is the vector sum of \(\vec{p}_1\) and \(\vec{p}_2\): \[ \vec{p}_{net} = \vec{p}_1 + \vec{p}_2 = qa\hat{i} + qa\hat{j} \] \[ \vec{p}_{net} = qa(\hat{i} + \hat{j}) \]


Step 4: Final Answer:

The net dipole moment of the system is qa(\^i + \^j). This matches option (A). Note that option (B) in the provided paper is likely a typo and is identical to (A).
Quick Tip: Always remember that the dipole moment is a vector. The direction is crucial and is conventionally defined as pointing from the negative charge to the positive charge. When dealing with multiple dipoles, always perform vector addition to find the net dipole moment.

Step 1: Understanding the Concept:

The problem involves a simple DC circuit with an external resistor and a power source with internal resistance. An ideal voltmeter has infinite resistance and measures the potential difference across the component it is connected in parallel with. An ideal ammeter has zero resistance and measures the current flowing through the circuit in series. The readings must satisfy the relationships for the entire circuit, including the internal resistance of the source.


Step 2: Key Formula or Approach:

Let E be the EMF of the power source, r be its internal resistance, and R be the external resistance.

1. The current (I) in the circuit is given by Ohm's law for the entire circuit: \( I = \frac{E}{R + r} \).

2. The voltmeter reading (V) across the external resistor R is the terminal voltage, given by \( V = I \times R \).

3. The terminal voltage can also be expressed in terms of EMF: \( V = E - Ir \).


Step 3: Detailed Explanation:

Given values:

External Resistance, \( R = 24 \, \Omega \).

Internal Resistance, \( r = 3 \, \Omega \).


The voltmeter reading (V) and the ammeter reading (I) must be related by Ohm's law for the external resistor: \[ V = I \times R \implies V = 24I \]
We can check each option to see which pair satisfies this fundamental relationship.


Checking the options:

(A) V = 12.0 V, I = 2.0 A
\( V/I = 12.0 / 2.0 = 6 \, \Omega \neq 24 \, \Omega \). This pair is not possible.


(B) V = 2.0 V, I = 0.5 A
\( V/I = 2.0 / 0.5 = 4 \, \Omega \neq 24 \, \Omega \). This pair is not possible.


(C) V = 6.0 V, I = 2.0 A
\( V/I = 6.0 / 2.0 = 3 \, \Omega \neq 24 \, \Omega \). This pair is not possible.


(D) V = 12 V, I = 0.5 A
\( V/I = 12 / 0.5 = 24 \, \Omega \). This pair satisfies the condition \( V = 24I \).


Now, let's verify if this possible pair is consistent with the given internal resistance. We can find the required EMF (E) of the source for these readings.
Using the terminal voltage formula \( V = E - Ir \): \[ 12 = E - (0.5)(3) \] \[ 12 = E - 1.5 \] \[ E = 12 + 1.5 = 13.5 \, V \]
So, if the power source has an EMF of 13.5 V, the readings would indeed be 12 V and 0.5 A. Since such a source can exist, this pair of readings is possible.


Step 4: Final Answer:

The only pair of readings that satisfies Ohm's law for the external resistor and is consistent with the circuit parameters is 12 V and 0.5 A.
Quick Tip: For questions like this, the fastest way to eliminate incorrect options is to check if \(V = IR\) holds for the external resistor. Often, only one option will satisfy this, making it the correct answer. You can then perform a quick check with the internal resistance to be fully certain.

Step 1: Understanding the Concept:

This question asks about the fundamental properties of the electric field created by static (stationary) charges, also known as the electrostatic field. The two key properties are its conservative nature and the geometry of its field lines.


Step 2: Detailed Explanation:

Conservative Nature:

An electrostatic field is a conservative field. This means that the work done by the electric field in moving a test charge from one point to another is independent of the path taken. It only depends on the initial and final positions.

Mathematically, this is equivalent to saying that the line integral of the electric field around any closed path is zero: \[ \oint \vec{E} \cdot d\vec{l} = 0 \]
This property allows us to define a scalar quantity called electric potential.


Electric Field Lines:

Electric field lines are a visual representation of the electric field. For an electrostatic field:

They originate from positive charges and terminate on negative charges (or extend to infinity if there is a net charge).
They never form closed loops. If a field line were to form a closed loop, moving a charge along this loop would result in non-zero work done (\(\oint \vec{E} \cdot d\vec{l} \neq 0\)), which would contradict the conservative nature of the electrostatic field.

Therefore, the electric field due to static charges is both conservative and has field lines that do not form closed loops.


Step 3: Final Answer:

Based on the fundamental principles of electrostatics, the correct statement is that the electric field is conservative and its field lines do not form closed loops.
Quick Tip: It is helpful to contrast the electrostatic field with the magnetic field. Magnetic field lines always form closed loops, and the magnetic field (induced by changing currents) is non-conservative. Remembering this distinction can help you answer questions about both electric and magnetic fields correctly.

Step 1: Understanding the Concept:

Materials respond differently when placed in an external magnetic field. This response is categorized based on how they get magnetized. The question describes a material that is repelled by the magnetic field ("pushed out").


Step 2: Detailed Explanation:

Let's analyze the behavior of different types of magnetic materials:


Diamagnetic Materials: When placed in a magnetic field, they develop a feeble magnetization in the direction opposite to the applied field. Consequently, they are weakly repelled by magnets and tend to move from a region of a stronger magnetic field to a region of a weaker magnetic field. This is the behavior described as being "pushed out". Examples include bismuth, copper, water, and gold.

Paramagnetic Materials: These materials get feebly magnetized in the \textit{same direction as the applied magnetic field. They are weakly attracted to magnets and tend to move from a weaker to a stronger part of the field. Examples include aluminum, platinum, and oxygen.

Ferromagnetic Materials: These materials get strongly magnetized in the \textit{same direction as the applied field. They are strongly attracted to magnets. Examples include iron, nickel, and cobalt.

Non-magnetic is not a standard classification, as almost all materials exhibit some form of magnetic response.



Step 3: Final Answer:

Since the material is pushed out of the magnetic field, it is being repelled. This is the characteristic property of a diamagnetic material.
Quick Tip: A simple mnemonic: \textbf{Diamagnetic materials are "di-vergent" from the field (pushed out). \textbf{Para}magnetic materials are "parallel" to the field (pulled in). \textbf{Ferro}magnetic (from Ferrum, Latin for iron) materials are very strongly attracted.

Step 1: Understanding the Concept:

This problem is based on Faraday's Law of Electromagnetic Induction and Lenz's Law. When the magnetic flux through a coil changes, an electromotive force (EMF) and hence a current is induced in the coil. Lenz's Law states that the direction of the induced current is such that it creates a magnetic field that opposes the change in magnetic flux that produced it.


Step 2: Detailed Explanation:

1. Initial Magnetic Flux: The U-shaped magnet creates a magnetic field. The magnetic field lines emerge from the North (N) pole and enter the South (S) pole. The coil is wrapped around the arm of the S pole. Within the iron core of the magnet, the magnetic flux is directed downwards through the coil.


2. Change in Magnetic Flux: Soft iron is a ferromagnetic material with high magnetic permeability. When the soft iron rod X falls and approaches the poles of the magnet, it provides a low reluctance path for the magnetic field lines. This concentrates the magnetic field lines and strengthens the magnetic flux passing through the entire U-shaped core, including the part where the coil is wrapped. Therefore, as the rod falls towards the magnet, the downward magnetic flux through the coil increases.


3. Applying Lenz's Law: According to Lenz's Law, the induced current must create a magnetic field that opposes this increase in downward flux. To oppose the increase, the induced magnetic field (\(\vec{B_{induced}\)) must be directed upwards.


4. Determining Current Direction: We use the Right-Hand Grip Rule to find the direction of the current that produces an upward magnetic field inside the coil. If you curl the fingers of your right hand in the direction of the current around the coil, your thumb points in the direction of the magnetic field produced. To have the thumb point upwards, the current must flow from terminal Y to terminal Z. When viewed from above, this is an anticlockwise direction.


Step 3: Final Answer:

As the rod falls, the downward magnetic flux increases. To oppose this, an upward magnetic field is induced, which corresponds to an anticlockwise current in the coil.
Quick Tip: Lenz's Law is all about opposing change. Think of it as magnetic inertia. If flux \textbf{increases, the induced field is in the \textbf{opposite} direction to the original field. If flux \textbf{decreases}, the induced field is in the \textbf{same} direction as the original field to try and maintain it.

Step 1: Understanding the Concept:

The magnetic field produced by a small current-carrying element is described by the Biot-Savart Law. Since the length of the conductor segment (1 cm) is much smaller than the distance to the point of interest (\(\sqrt{1^2 + 1^2} = \sqrt{2} \, m \approx 141 \, cm\)), we can approximate the segment as a differential current element located at the origin.


Step 2: Key Formula or Approach:

The Biot-Savart Law gives the magnetic field \(d\vec{B}\) produced by a current element \(I d\vec{l}\) at a position \(\vec{r}\) away from the element: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3} \]
where \(\mu_0 = 4\pi \times 10^{-7} \, T\cdotm/A\) is the permeability of free space.


Step 3: Detailed Explanation:

Given values:

Current, \(I = 1 \, A\)
Length of the segment, \( \Delta l = 1 \, cm = 0.01 \, m \)
The segment is in the x-direction at the origin, so we can represent the current element vector as \(\Delta\vec{l} \approx 0.01 \hat{i} \, m\).
The point P is at (1m, 1m, 0). The position vector \(\vec{r}\) from the origin to P is \(\vec{r} = 1\hat{i} + 1\hat{j} \, m\).


First, calculate the magnitude of the position vector, \(r\): \[ r = |\vec{r}| = \sqrt{1^2 + 1^2} = \sqrt{2} \, m \]

Next, calculate the cross product \(\Delta\vec{l} \times \vec{r}\): \[ \Delta\vec{l} \times \vec{r} = (0.01 \hat{i}) \times (1\hat{i} + 1\hat{j}) \] \[ \Delta\vec{l} \times \vec{r} = (0.01 \times 1)(\hat{i} \times \hat{i}) + (0.01 \times 1)(\hat{i} \times \hat{j}) \]
Since \(\hat{i} \times \hat{i} = 0\) and \(\hat{i} \times \hat{j} = \hat{k}\), we have: \[ \Delta\vec{l} \times \vec{r} = 0 + 0.01 \hat{k} = 0.01 \hat{k} \, m^2 \]

Now, substitute all values into the Biot-Savart Law formula: \[ \vec{B} \approx \frac{\mu_0}{4\pi} \frac{I (\Delta\vec{l} \times \vec{r})}{r^3} \] \[ \vec{B} = (10^{-7} \, T\cdotm/A) \frac{(1 \, A)(0.01 \hat{k} \, m^2)}{(\sqrt{2} \, m)^3} \] \[ \vec{B} = 10^{-7} \frac{10^{-2} \hat{k}}{2\sqrt{2}} = \frac{10^{-9}}{2\sqrt{2}} \hat{k} \, T \]

To match the format of the options, we can rewrite this expression: \[ \vec{B} = \frac{1 \times 10 \times 10^{-10}}{2\sqrt{2}} \hat{k} \, T = \frac{10}{2\sqrt{2}} \times 10^{-10} \hat{k} \, T \] \[ \vec{B} = \frac{5}{\sqrt{2}} \times 10^{-10} \hat{k} \, T \]

Step 4: Final Answer:

The calculated magnetic field is \(\frac{5.0}{\sqrt{2}} \times 10^{-10} \hat{k} \, T\), which matches option (C).
Quick Tip: When applying the Biot-Savart law, be meticulous with the vector cross product. The direction of the resulting magnetic field is perpendicular to both the current element vector (\(d\vec{l}\)) and the position vector (\(\vec{r}\)), as determined by the right-hand rule.

Step 1: Understanding the Concept:

This question deals with a step-up transformer with a tapped secondary winding. A transformer works on the principle of mutual induction. For an ideal transformer, the ratio of the output (secondary) voltage to the input (primary) voltage is equal to the ratio of the number of turns in the secondary coil to the number of turns in the primary coil.


Step 2: Key Formula or Approach:

The transformer equation is: \[ \frac{V_S}{V_P} = \frac{N_S}{N_P} \]
where:

\(V_S\) is the secondary voltage (output).
\(V_P\) is the primary voltage (input).
\(N_S\) is the number of turns in the secondary coil.
\(N_P\) is the number of turns in the primary coil.



Step 3: Detailed Explanation:

Given values:

Input voltage, \(V_P = 20 \, V\).
Number of primary turns (between A and B), \(N_P = 20\).
Desired output voltage, \(V_S = 120 \, V\).


We need to find the number of secondary turns (\(N_S\)) required to produce the desired output voltage. Rearranging the transformer equation: \[ N_S = N_P \times \frac{V_S}{V_P} \] \[ N_S = 20 \times \frac{120}{20} \] \[ N_S = 20 \times 6 = 120 \, turns \]

Now, we must find which pair of terminals on the secondary coil corresponds to a total of 120 turns. The secondary coil is tapped, and the turns in each section are given:

Turns between P and Q = 20
Turns between Q and R = 40
Turns between R and S = 60

We need to find the combination of sections that sums to 120 turns.


Checking the options:

(A) P and Q: Total turns = \(N_{PQ} = 20\). (Incorrect)

(B) Q and S: Total turns = \(N_{QR} + N_{RS} = 40 + 60 = 100\). (Incorrect)

(C) P and R: Total turns = \(N_{PQ} + N_{QR} = 20 + 40 = 60\). (Incorrect)

(D) P and S: Total turns = \(N_{PQ} + N_{QR} + N_{RS} = 20 + 40 + 60 = 120\). (Correct)


Step 4: Final Answer:

To get an output of 120 V, we need 120 turns in the secondary coil. This is achieved by connecting the output across terminals P and S.
Quick Tip: In a transformer with a tapped secondary, the total number of turns between two taps is the sum of the turns of the individual sections between them. Always calculate the required number of turns first, then check the diagram to see how that number can be achieved.

Step 1: Understanding the Concept:

The relationship between the voltage (V) across an ideal inductor and the current (I) flowing through it is defined by the rate of change of the current. The voltage is directly proportional to the time derivative of the current.


Step 2: Key Formula or Approach:

The voltage across an inductor is given by the formula: \[ V(t) = L \frac{dI(t)}{dt} \]
where L is the inductance. This means the voltage waveform is the scaled derivative of the current waveform. We need to find the derivative (slope) of the given current graph at different time intervals.


Step 3: Detailed Explanation:

The given graph for current I(t) is a triangular wave. Let's analyze its slope in two parts of the cycle.


Interval 1: \(0 < t < T/2\)

In this interval, the graph of I versus t is a straight line rising from 0 to a maximum value, \(I_{max}\). A straight line has a constant slope.

The slope is: \[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{I_{max} - 0}{T/2 - 0} = \frac{2I_{max}}{T} \]
This is a positive constant value.

Therefore, the voltage in this interval is: \[ V = L \left( \frac{2I_{max}}{T} \right) = Positive Constant \]


Interval 2: \(T/2 < t < T\)

In this interval, the graph of I versus t is a straight line falling from \(I_{max}\) to 0. This is also a straight line with a constant slope.

The slope is: \[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{0 - I_{max}}{T - T/2} = \frac{-I_{max}}{T/2} = -\frac{2I_{max}}{T} \]
This is a negative constant value.

Therefore, the voltage in this interval is: \[ V = L \left( -\frac{2I_{max}}{T} \right) = Negative Constant \]


Conclusion:
The voltage V is a constant positive value for the first half of the cycle and a constant negative value for the second half. This describes a square wave. Graph (C) correctly depicts this behavior.


Step 4: Final Answer:

The derivative of a triangular wave is a square wave. Therefore, the voltage waveform across the inductor will be a square wave.
Quick Tip: Remember the relationship between voltage and current for basic components in AC circuits: \textbf{Inductor:} Voltage is the derivative of current (\(V = L \frac{dI}{dt}\)). \textbf{Capacitor:} Current is the derivative of voltage (\(I = C \frac{dV}{dt}\)). \textbf{Resistor:} Voltage is proportional to current (\(V = IR\)). Visualizing the graph of the derivative of a function is a key skill. The derivative of a line is a constant, and the derivative of a curve gives another curve (e.g., sine -> cosine).

Step 1: Understanding the Concept:

When a lens is silvered on one side, it acts as a mirror. A ray of light entering the system first refracts through the lens surface, then reflects from the silvered mirror surface, and finally refracts back out through the lens surface. The overall behavior is that of a mirror, and its equivalent focal length can be determined by considering the powers of the lens and the mirror.


Step 2: Key Formula or Approach:

The power of the equivalent mirror (\(P_{eq}\)) is given by the sum of the powers of the components, accounting for the path of light: \[ P_{eq} = P_{lens} + P_{mirror} + P_{lens} = 2P_{lens} + P_{mirror} \]
The equivalent focal length of the mirror is \(F = -1/P_{eq}\).

The power of a lens (\(P_{lens}\)) is given by the Lens Maker's formula: \(P_{lens} = \frac{1}{f_{lens}} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).

The power of a mirror (\(P_{mirror}\)) is \(P_{mirror} = -1/f_{mirror}\).


Step 3: Detailed Explanation:

First, let's find the power of the plano-convex lens.

For a plano-convex lens, one surface is curved with radius R and the other is plane with radius \(\infty\).
Let the curved surface have radius \(R_1 = R\) and the plane surface have radius \(R_2 = \infty\).

Using the Lens Maker's formula: \[ P_{lens} = (n-1)\left(\frac{1}{R} - \frac{1}{\infty}\right) = \frac{n-1}{R} \]
Next, let's find the power of the mirror. The plane face is silvered, so it acts as a plane mirror.
The focal length of a plane mirror is infinite (\(f_{mirror} = \infty\)).
Therefore, the power of the plane mirror is: \[ P_{mirror} = \frac{-1}{\infty} = 0 \]
Now, we can find the power of the equivalent mirror system: \[ P_{eq} = 2P_{lens} + P_{mirror} = 2 \left(\frac{n-1}{R}\right) + 0 = \frac{2(n-1)}{R} \]
The focal length of this equivalent mirror, \(F\), is given by \(F = 1/P_{eq}\) (considering magnitude for comparison with options). \[ F = \frac{1}{P_{eq}} = \frac{R}{2(n-1)} \]
Since the combination reflects light to a focus, it behaves as a concave mirror. The focal length is \( \frac{R}{2(n-1)} \).


Step 4: Final Answer:

The combination will behave as a concave mirror of focal length \(\frac{R}{2(n-1)}\).
Quick Tip: The formula for a lens silvered on one side is \(P_{eq} = 2P_{lens} + P_{mirror}\). Remember that light passes through the lens twice. For a plane silvered surface, \(P_{mirror} = 0\), simplifying the calculation. For a curved silvered surface, use \(P_{mirror} = -1/f_m = 2/R_{mirror}\).

Step 1: Understanding the Concept:

A p-n junction diode is a semiconductor device that allows current to flow easily in one direction (forward bias) but restricts current flow in the opposite direction (reverse bias). This behavior is characterized by its resistance.


Step 2: Detailed Explanation:

Forward Bias: When the p-side is connected to a higher potential than the n-side, the diode is forward-biased. The width of the depletion region decreases, and the diode offers a very low resistance to the current flow. Current flows easily. This corresponds to a "closed" or "ON" state.


Reverse Bias: When the n-side is connected to a higher potential than the p-side, the diode is reverse-biased. The width of the depletion region increases, and the diode offers a very high resistance. Only a very small leakage current flows. This high resistance state effectively blocks the current. This corresponds to an "open" or "OFF" state.


The question states that the measured resistance is high. This corresponds to the reverse-biased condition. The ability to switch between a low-resistance ("ON") state and a high-resistance ("OFF") state is the fundamental property of a switch. Therefore, when its resistance is high, it acts as an open switch.


Step 3: Final Answer:

Because a p-n junction can exhibit both very low and very high resistance depending on the biasing, its primary application in this context is as an electronic switch. The high resistance state represents the "open" or "off" position of the switch.
Quick Tip: Think of a diode's I-V characteristics. In the forward region, the slope (\(\Delta I / \Delta V\)) is large, meaning resistance (\(\Delta V / \Delta I\)) is low. In the reverse region, the slope is near zero, meaning resistance is very high. This on/off characteristic is the essence of a switch.

Step 1: Understanding the Concept:

The photoelectric effect occurs when a photon with sufficient energy strikes a metal surface, causing an electron to be ejected. The minimum energy required to eject an electron is called the work function (\(\phi\)) of the metal. For photoemission to occur, the energy of the incident photon (\(E_{photon}\)) must be greater than or equal to the work function (\(E_{photon} \geq \phi\)).


Step 2: Key Formula or Approach:

The energy of a photon emitted from a hydrogen atom during a transition from an initial state \(n_i\) to a final state \(n_f\) is given by: \[ E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \, eV \]
The work function of zinc is approximately \(\phi_{Zn} \approx 4.3 \, eV\). We need to find which spectral series of hydrogen produces photons with energy greater than 4.3 eV.


Step 3: Detailed Explanation:

Let's calculate the energy range for each spectral series:


Lyman Series (\(n_f = 1\)):

The transitions are from \(n_i = 2, 3, 4, ...\) to \(n_f = 1\).

The minimum energy (for \(n_i = 2 \to 1\)) is: \[ E_{min} = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \times \frac{3}{4} = 10.2 \, eV \]
The maximum energy (for \(n_i = \infty \to 1\)) is \(13.6 \, eV\).

Since all photon energies in the Lyman series are greater than 10.2 eV, and \(10.2 \, eV > 4.3 \, eV\), all lines of the Lyman series can cause photoemission from zinc.


Balmer Series (\(n_f = 2\)):

The transitions are from \(n_i = 3, 4, 5, ...\) to \(n_f = 2\).

The maximum energy (for \(n_i = \infty \to 2\)) is: \[ E_{max} = 13.6 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{13.6}{4} = 3.4 \, eV \]
Since the maximum energy in the Balmer series is 3.4 eV, which is less than the work function of zinc (4.3 eV), no line from the Balmer series can cause photoemission.


Paschen Series (\(n_f = 3\)):

The transitions are from \(n_i = 4, 5, 6, ...\) to \(n_f = 3\).

The maximum energy (for \(n_i = \infty \to 3\)) is: \[ E_{max} = 13.6 \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = \frac{13.6}{9} \approx 1.51 \, eV \]
This energy is also less than 4.3 eV, so no line from the Paschen series can cause photoemission.


Step 4: Final Answer:

Only the photons from the Lyman series have sufficient energy to overcome the work function of zinc and emit photoelectrons.
Quick Tip: Remember the energy levels of hydrogen: -13.6 eV, -3.4 eV, -1.51 eV, ... . Lyman series involves transitions to the -13.6 eV level, resulting in high-energy UV photons. Balmer involves transitions to -3.4 eV (visible light), and Paschen to -1.51 eV (infrared). For most metals, work functions are in the range of 2-5 eV, which typically requires UV light (like the Lyman series) for photoemission.

Step 1: Understanding the Concept:

According to the Bohr model for the hydrogen atom, the energy of an electron in a specific orbit (or energy level) is quantized and depends on the principal quantum number, n.


Step 2: Key Formula or Approach:

The energy \(E_n\) of an electron in the n-th orbit of a hydrogen atom is given by the formula: \[ E_n = \frac{E_1}{n^2} \]
where \(E_1\) is the energy of the ground state (n=1).


Step 3: Detailed Explanation:

We are given the following values:

Ground state energy, \(E_1 = -13.6 \, eV\)
Energy in the n-th orbit, \(E_n = -0.544 \, eV\)

We need to find the value of n. Substituting the given values into the formula: \[ -0.544 = \frac{-13.6}{n^2} \]
Now, we solve for \(n^2\): \[ n^2 = \frac{-13.6}{-0.544} = \frac{13.6}{0.544} \]
To simplify the fraction, we can multiply the numerator and denominator by 1000: \[ n^2 = \frac{13600}{544} \]
We can notice that \(544 = 4 \times 136\). \[ n^2 = \frac{136 \times 100}{136 \times 4} = \frac{100}{4} = 25 \]
Taking the square root of both sides: \[ n = \sqrt{25} = 5 \]
The principal quantum number n must be a positive integer, so n = 5.


Step 4: Final Answer:

The value of the quantum number n corresponding to the energy -0.544 eV is 5.
Quick Tip: It's very helpful to memorize the energies of the first few levels of the hydrogen atom: n=1: -13.6 eV n=2: -13.6 / 4 = -3.4 eV n=3: -13.6 / 9 = -1.51 eV n=4: -13.6 / 16 = -0.85 eV n=5: -13.6 / 25 = -0.544 eV Having these values in mind can allow you to answer such questions almost instantly.

Step 1: Analyzing the Assertion (A):

The assertion states that in an ideal step-down transformer, electrical energy is not lost. The term "ideal" in physics implies a perfect system with no dissipative forces or energy losses. For a transformer, this means there is no energy loss due to factors like resistance of windings (copper loss), eddy currents in the core, hysteresis loss, or flux leakage. In an ideal transformer, the power input to the primary coil equals the power output from the secondary coil (\(P_{in} = P_{out}\)). Since energy is the integral of power over time, conserved power implies conserved energy. Thus, Assertion (A) is true.


Step 2: Analyzing the Reason (R):

The reason states that in a step-down transformer, voltage decreases but the current increases. A step-down transformer is defined as one where the secondary voltage is less than the primary voltage (\(V_S < V_P\)). From the principle of energy conservation for an ideal transformer, \(V_P I_P = V_S I_S\). This can be rearranged to \(\frac{I_S}{I_P} = \frac{V_P}{V_S}\). Since \(V_P > V_S\), it follows that \(I_S > I_P\). This means that as the voltage is stepped down, the current is stepped up. Thus, Reason (R) is also a true statement.


Step 3: Evaluating if Reason (R) explains Assertion (A):

Assertion (A) is true by definition of an ideal transformer. The reason for no energy loss is the assumption of ideality (no resistance, no hysteresis, etc.). Reason (R) describes the relationship between voltage and current that is a \textit{consequence of energy being conserved. It explains \textit{how an ideal transformer works (trading voltage for current to keep power constant), but it does not explain \textit{why the energy is conserved in the first place. The 'why' is simply the "ideal" assumption. Therefore, Reason (R) is not the correct explanation for Assertion (A).


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true statements, but Reason (R) does not correctly explain Assertion (A). This corresponds to option (B).
Quick Tip: In Assertion-Reason questions, always check the "because" link. Read the statements as: "(Assertion) \textbf{because (Reason)". Here, "In an ideal transformer, energy is not lost \textbf{because} in a step-down transformer voltage decreases and current increases." This link is not logical. The correct reason would be "...because an ideal transformer has no energy dissipation mechanisms like winding resistance or eddy currents."

Step 1: Analyzing the Assertion (A):

Diffraction is the phenomenon of waves bending around obstacles. The effect of diffraction is more pronounced when the wavelength (\(\lambda\)) of the wave is comparable to or larger than the size of the obstacle. We need to compare the wavelengths of radio waves and infrared waves.

In the electromagnetic spectrum, the order of decreasing wavelength is:
Radio waves > Microwaves > Infrared > Visible light > Ultraviolet > X-rays > Gamma rays.

Since radio waves have a much longer wavelength than infrared waves, they will diffract more significantly around everyday objects (like buildings, hills, etc.). Therefore, Assertion (A) is true.


Step 2: Analyzing the Reason (R):

The reason states that radio waves have a greater frequency than infrared waves. The relationship between frequency (f), wavelength (\(\lambda\)), and the speed of light (c) is \(c = f\lambda\). Since c is constant for all electromagnetic waves in a vacuum, frequency is inversely proportional to wavelength (\(f \propto 1/\lambda\)).

As we established in Step 1, \(\lambda_{radio} > \lambda_{infrared}\).

Therefore, it must be that \(f_{radio} < f_{infrared}\).

The reason states the opposite, that radio waves have a greater frequency. Thus, Reason (R) is false.


Step 3: Final Answer:

Assertion (A) is true, but Reason (R) is false. This corresponds to option (C).
Quick Tip: Memorize the order of the electromagnetic spectrum, either by wavelength or by frequency. A common mnemonic for increasing frequency/decreasing wavelength is: "\textbf{R}ich \textbf{M}en \textbf{I}n \textbf{V}egas \textbf{U}se \textbf{X}-pensive \textbf{G}adgets" (Radio, Microwave, Infrared, Visible, UV, X-ray, Gamma). Also, remember that diffraction is more significant for longer wavelengths.

Step 1: Analyzing the Assertion (A):

The assertion states that the thickness of the depletion layer in a semiconductor diode is not fixed. The depletion layer (or depletion region) is a region around the p-n junction that is depleted of free charge carriers. Its width is influenced by the electric field across the junction. When an external voltage (biasing) is applied, this electric field changes, which in turn changes the width of the depletion layer. Therefore, its thickness is not fixed but variable. The assertion is true.


Step 2: Analyzing the Reason (R):

The reason states that the thickness of the depletion layer depends on factors like the biasing of the semiconductor. This is the primary factor controlling the width:

Forward Biasing: Applying a forward voltage opposes the built-in potential, reducing the net electric field and causing the depletion layer to narrow.

Reverse Biasing: Applying a reverse voltage aids the built-in potential, increasing the net electric field and causing the depletion layer to widen.


Other factors like the doping concentration of the p and n materials also affect the thickness. Thus, the reason is a true and accurate statement.


Step 3: Evaluating if Reason (R) explains Assertion (A):

The assertion says the thickness is not fixed. The reason explains precisely why it is not fixed: because it depends on external factors like biasing. The reason provides the correct scientific explanation for the assertion.


Step 4: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A). This corresponds to option (A).
Quick Tip: Remember the effect of biasing on the depletion layer: Forward Bias \(\rightarrow\) Narrower depletion layer \(\rightarrow\) Low resistance. Reverse Bias \(\rightarrow\) Wider depletion layer \(\rightarrow\) High resistance. This relationship is fundamental to how diodes and transistors function.

Step 1: Analyzing the Assertion (A):

Let's check the relationship between angular momentum (\(L_n\)) and radius (\(r_n\)) in the Bohr model.

According to Bohr's postulates:

Angular momentum is quantized: \(L_n = n \frac{h}{2\pi}\), where n is the principal quantum number. So, \(L_n \propto n\).

The radius of the n\(^{th}\) orbit is given by \(r_n = r_0 n^2\), where \(r_0\) is the Bohr radius. So, \(r_n \propto n^2\).


From the radius relation, we can write \(n \propto \sqrt{r_n}\).

Substituting this into the angular momentum relation, we get \(L_n \propto n \propto \sqrt{r_n}\).

Therefore, the assertion that angular momentum is proportional to the square root of the orbit radius is true.


Step 2: Analyzing the Reason (R):

The reason states that an electron can jump only to its nearest orbits. This is incorrect. Bohr's model allows for transitions (jumps) between \textit{any two allowed stationary orbits. For example, an electron can jump from n=3 to n=1 (part of the Lyman series) or from n=4 to n=2 (part of the Balmer series), not just from n=3 to n=2. The concept of selection rules (\(\Delta l = \pm 1\)) that restrict transitions exists in the more advanced quantum mechanical model, but the "nearest orbit only" rule is not a part of the Bohr model. Therefore, the reason is false.


Step 3: Final Answer:

Assertion (A) is true, but Reason (R) is false. This corresponds to option (C).
Quick Tip: For Bohr model questions, it's crucial to know the dependencies on the quantum number n: Radius \(r_n \propto n^2\).
Velocity \(v_n \propto 1/n\).
Angular Momentum \(L_n = m v_n r_n \propto n\).
Energy \(E_n \propto -1/n^2\).
By knowing these, you can derive any proportionality, like the one between L\(_n\) and r\(_n\).

Step 1: Understanding the Concept:

This problem involves a simple series DC circuit containing a battery, a resistor, and a silicon diode. When a diode is conducting (forward-biased), it maintains a nearly constant voltage drop across it, known as the threshold or cut-in voltage. We can apply Kirchhoff's Voltage Law (KVL) to the circuit to find the unknown battery voltage.


Step 2: Key Formula or Approach:

According to Kirchhoff's Voltage Law (KVL), the sum of the voltage rises (from the battery) must equal the sum of the voltage drops across all components in a closed loop. \[ V_{battery} = V_{resistor} + V_{diode} \]
The voltage drop across the resistor is given by Ohm's Law: \(V_{resistor} = I \times R\).


Step 3: Detailed Explanation:

Given values:

Threshold voltage of the diode, \(V_d = 0.7 \, V\).
Resistance, \(R = 1000 \, \Omega\).
Current in the circuit, \(I = 15 \, mA = 15 \times 10^{-3} \, A\).

First, calculate the voltage drop across the resistor (\(V_R\)): \[ V_R = I \times R = (15 \times 10^{-3} \, A) \times (1000 \, \Omega) = 15 \, V \]
Now, apply KVL to the series circuit. The total voltage V supplied by the battery is the sum of the voltage drop across the resistor and the voltage drop across the diode. \[ V = V_R + V_d \] \[ V = 15 \, V + 0.7 \, V = 15.7 \, V \]

Step 4: Final Answer:

The value of the battery voltage V is 15.7 V.
Quick Tip: When analyzing circuits with ideal or near-ideal diodes, treat a forward-biased diode as a small battery with a voltage equal to its threshold voltage (e.g., 0.7 V for Si, 0.3 V for Ge) that opposes the main current flow. This simplifies KVL calculations. A reverse-biased ideal diode is treated as an open circuit.

Step 1: Understanding the Concept:

In a Young's double-slit experiment (YDSE), interference fringes are formed on a screen. The angular width of a fringe is the angle subtended by the distance between two consecutive bright or dark fringes at the slits. For small angles, this is directly related to the wavelength of light and the distance between the slits.


Step 2: Key Formula or Approach:

The angular width (\(\theta\)) of a fringe is given by the formula: \[ \theta = \frac{\lambda}{d} \]
where \(\lambda\) is the wavelength of the light and \(d\) is the separation between the slits. Note that this formula requires the angle \(\theta\) to be in radians.


Step 3: Detailed Explanation:

Given values:

Angular width, \(\theta = 0.2^\circ\).
Wavelength, \(\lambda = 500 \, nm = 500 \times 10^{-9} \, m\).

First, we must convert the angular width from degrees to radians. \[ \theta (rad) = 0.2^\circ \times \frac{\pi}{180^\circ} \approx 0.00349 \, rad \]
Now, we rearrange the formula to solve for the slit separation, \(d\): \[ d = \frac{\lambda}{\theta} \]
Substitute the values into the formula: \[ d = \frac{500 \times 10^{-9} \, m}{0.00349 \, rad} \] \[ d \approx 143266 \times 10^{-9} \, m \] \[ d \approx 1.43 \times 10^{-4} \, m \]
This can also be expressed as 0.143 mm.


Step 4: Final Answer:

The separation of the two slits is approximately \(1.43 \times 10^{-4}\) m.
Quick Tip: Always be careful with units in physics problems. Angles in trigonometric formulas like \(\theta = \lambda/d\) must be in radians. Forgetting to convert from degrees to radians is a very common mistake in exams. Remember the conversion: \( radians = degrees \times \frac{\pi}{180} \).

Step 1: Understanding the Concept:

When a lens is placed in the path of a converging beam, the original point of convergence (O) acts as an object for the lens. Since the rays are converging towards a point on the right side of the lens, this object is a virtual object. We can use the lens formula to find the position of the new convergence point (O'), which will be the final image.


Step 2: Key Formula or Approach:

The lens formula relates the object distance (u), image distance (v), and focal length (f): \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Sign convention: We take the optical center of the lens as the origin. The direction of incident light is positive.

Focal length of a concave lens is negative.
The virtual object O is to the right of the lens, so its distance u is positive.



Step 3: Detailed Explanation:

Given values:

Focal length of the concave lens, \(f = -15 \, cm\).
The lens is placed 10 cm before point O. So, the distance to the virtual object is \(u = +10 \, cm\).

Substitute these values into the lens formula to find the image distance v: \[ \frac{1}{v} - \frac{1}{+10} = \frac{1}{-15} \] \[ \frac{1}{v} = \frac{1}{10} - \frac{1}{15} \]
To solve this, find a common denominator (which is 30): \[ \frac{1}{v} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \] \[ v = +30 \, cm \]
The positive sign for v indicates that the final image O' is formed 30 cm to the right of the lens.


Calculating the shift OO':

The original convergence point O was 10 cm to the right of the lens.
The new convergence point O' is 30 cm to the right of the lens.
The shift is the distance between O and O'. \[ Shift OO' = (Position of O') - (Position of O) = 30 \, cm - 10 \, cm = 20 \, cm \]
The shift is 20 cm, and since the final position is further from the lens, the shift is away from the lens (or to the right).


Step 4: Final Answer:

The magnitude of the shift OO' is 20 cm. The direction of the shift is away from the lens, in the direction of the light propagation.
Quick Tip: The concept of a virtual object is important. A virtual object is formed when converging rays are intercepted by an optical element before they can converge. The position of a virtual object is taken as positive according to the Cartesian sign convention.

Step 1: Understanding the Concept:

This problem applies the principles of the photoelectric effect. The work function (\(\phi\)) is the minimum energy required to remove an electron from a metal surface, and it corresponds to the threshold wavelength (\(\lambda_0\)). When a photon with energy greater than the work function strikes the metal, an electron is ejected with a maximum kinetic energy (\(K_{max}\)) equal to the excess energy.


Step 2: Key Formula or Approach:

We will use the following formulas:

Work function: \(\phi = \frac{hc}{\lambda_0}\)
Einstein's Photoelectric Equation: \(K_{max} = E_{photon} - \phi\), where \(E_{photon} = \frac{hc}{\lambda}\).

A very useful shortcut for calculations is to use the value \(hc \approx 1240 \, eV\cdotnm\). This allows direct conversion from wavelength in nm to energy in eV.


Step 3: Detailed Explanation:

Given values:

Threshold wavelength, \(\lambda_0 = 450 \, nm\).
Incident wavelength, \(\lambda = 250 \, nm\).


(i) Calculation of the work function (\(\phi\)):

Using the shortcut formula: \[ \phi (in eV) = \frac{1240 \, eV\cdotnm}{\lambda_0 (in nm)} \] \[ \phi = \frac{1240}{450} = \frac{124}{45} \approx 2.76 \, eV \]

(ii) Calculation of the maximum kinetic energy (\(K_{max}\)):

First, find the energy of the incident photon (\(E_{photon}\)): \[ E_{photon} (in eV) = \frac{1240 \, eV\cdotnm}{\lambda (in nm)} \] \[ E_{photon} = \frac{1240}{250} = \frac{124}{25} = 4.96 \, eV \]
Now, use the photoelectric equation to find \(K_{max}\): \[ K_{max} = E_{photon} - \phi \] \[ K_{max} = 4.96 \, eV - 2.76 \, eV = 2.20 \, eV \]

Step 4: Final Answer:

(i) The work function of the metal is approximately 2.76 eV.

(ii) The maximum energy of the ejected photoelectrons is 2.20 eV.
Quick Tip: Memorizing and using \(hc \approx 1240 \, eV\cdotnm\) is a huge time-saver in competitive exams for problems involving photons and the photoelectric effect. It avoids dealing with Planck's constant (h) and the speed of light (c) in SI units and the subsequent conversion to electron-volts.

Step 1: Understanding the Concept:

This problem involves the principle of current division in a parallel circuit. When resistors are connected in parallel, the current from the source divides among them. The current is inversely proportional to the resistance of each branch.


Step 2: Key Formula or Approach:

1. Resistance of a wire: \(R = \frac{\rho L}{A}\), where \(\rho\) is resistivity, L is length, and A is the cross-sectional area.
2. For parallel connection, the voltage (V) across both wires is the same.
3. Current division rule: The ratio of currents in two parallel branches is the inverse of the ratio of their resistances: \(\frac{I_1}{I_2} = \frac{R_2}{R_1}\).
4. Total current: \(I_{total} = I_1 + I_2\).


Step 3: Detailed Explanation:

Given information:

Same material (\(\rho_1 = \rho_2 = \rho\)).
Same radius (\(r_1 = r_2 \implies A_1 = A_2 = A\)).
Ratio of lengths: \(\frac{L_1}{L_2} = \frac{2}{3}\).
Total current: \(I_{total} = 15 \, A\).

First, find the ratio of their resistances: \[ \frac{R_1}{R_2} = \frac{\rho L_1 / A}{\rho L_2 / A} = \frac{L_1}{L_2} = \frac{2}{3} \]
Now, use the current division principle. Since voltage is the same (\(V = I_1 R_1 = I_2 R_2\)), the ratio of currents is: \[ \frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{3}{2} \]
This means \(I_1 = \frac{3}{2} I_2\).

We also know that the total current is 15 A: \[ I_1 + I_2 = 15 \]
Substitute the expression for \(I_1\) into the sum equation: \[ \left(\frac{3}{2} I_2\right) + I_2 = 15 \] \[ \frac{5}{2} I_2 = 15 \] \[ I_2 = 15 \times \frac{2}{5} = 6 \, A \]
Now, find \(I_1\): \[ I_1 = 15 - I_2 = 15 - 6 = 9 \, A \]
So, the current through the first wire (with length \(L_1\)) is 9 A, and the current through the second wire (with length \(L_2\)) is 6 A.


Step 4: Final Answer:

The currents through the wires are 9 A and 6 A.
Quick Tip: For current division in parallel circuits, remember that "current prefers the path of least resistance." The wire with lower resistance (the shorter wire in this case, \(R_1\)) will get a larger share of the total current.

Step 1: Understanding the Concept:

The problem involves a DC circuit with resistors, cells, and a capacitor. The key phrase is "in the steady state." In a DC circuit, after a long time (steady state), a capacitor gets fully charged and acts as an open circuit. This means no DC current can flow through the branch containing the capacitor.


Step 2: Key Formula or Approach:

1. In steady state, current through the capacitor branch is zero (\(I_C = 0\)).
2. Use Nodal Analysis (a form of Kirchhoff's Current Law, KCL) to find the potential at key nodes in the circuit. KCL states that the algebraic sum of currents entering a node is zero.
3. The potential difference across any component can be found by traversing a path between its terminals.


Step 3: Detailed Explanation:

The circuit shows three parallel branches connected between nodes P and Q.
Let's apply Nodal Analysis. Assume the potential at node Q is zero, i.e., \(V_Q = 0\). We need to find the potential at node P, \(V_P\). The potential difference between P and Q will then be \(V_{PQ} = V_P - V_Q = V_P\).


At node P, the sum of currents leaving the node through the three branches must be zero.
Let the current in the top branch be \(I_1\), middle branch be \(I_C\), and bottom branch be \(I_2\). \[ I_1 + I_C + I_2 = 0 \]
In steady state, \(I_C = 0\). So, \(I_1 + I_2 = 0\), or \(I_1 = -I_2\).


Let's write the expressions for \(I_1\) and \(I_2\) using Ohm's law for each branch. The potential on the other side of the cells and resistors is \(V_P\).
For the top branch: The branch contains a cell of EMF V and resistor R. The potential at P is \(V_P\). The potential at the positive terminal of the cell is higher. So, \(I_1 = \frac{V_P - V}{R}\).
For the bottom branch: The branch contains a cell of EMF 2V and resistor 2R. So, \(I_2 = \frac{V_P - 2V}{2R}\).

Now apply KCL at node P: \[ \frac{V_P - V}{R} + \frac{V_P - 2V}{2R} = 0 \]
Multiply the entire equation by 2R to eliminate the denominators: \[ 2(V_P - V) + (V_P - 2V) = 0 \] \[ 2V_P - 2V + V_P - 2V = 0 \] \[ 3V_P - 4V = 0 \] \[ 3V_P = 4V \implies V_P = \frac{4V}{3} \]

(i) Potential difference between P and Q:
\[ V_{PQ} = V_P - V_Q = \frac{4V}{3} - 0 = \frac{4V}{3} \]

(ii) Potential difference across capacitor C:

Now consider the middle branch containing cell V and capacitor C. The total potential difference across this branch is \(V_{PQ}\). Let the potential difference across the capacitor be \(V_C\). Traversing from P to Q through the middle branch: \[ V_P - V - V_C = V_Q \] \[ V_P - V_Q = V + V_C \] \[ V_{PQ} = V + V_C \]
We know \(V_{PQ} = \frac{4V}{3}\), so: \[ \frac{4V}{3} = V + V_C \] \[ V_C = \frac{4V}{3} - V = \frac{4V - 3V}{3} = \frac{V}{3} \]

Step 4: Final Answer:

(i) The potential difference between P and Q is \(\frac{4V}{3}\).

(ii) The potential difference across capacitor C is \(\frac{V}{3}\).
Quick Tip: Nodal analysis is a very powerful tool for solving complex circuits, especially those with multiple parallel branches. The key steps are: 1. Choose a reference node (usually ground, potential = 0). 2. Assign variables for the potential at other major nodes. 3. Apply KCL at each non-reference node, writing currents in terms of node potentials. 4. Solve the resulting system of linear equations.

Step 1: Understanding the Concept:

Electrical Conductivity (\(\sigma\)): Electrical conductivity is an intrinsic property of a material that measures its ability to conduct an electric current. It is the reciprocal of electrical resistivity (\(\rho\)). A material with high conductivity allows electric current to flow easily through it. \[ \sigma = \frac{1}{\rho} \]
The SI unit of conductivity is Siemens per meter (\(S/m\)) or \((\Omega \cdot m)^{-1}\).


Step 2: Derivation of the Expression:

Consider a conductor with length 'l' and cross-sectional area 'A'. Let 'n' be the number density of free electrons (number of free electrons per unit volume). When an external electric field 'E' is applied across the conductor, each free electron experiences an electrostatic force.


Force on an electron: The force on an electron of charge \(-e\) in an electric field \(\vec{E}\) is:
\[ \vec{F} = -e\vec{E} \]
Acceleration of the electron: Due to this force, the electron accelerates. If 'm' is the mass of the electron, the acceleration is:
\[ \vec{a} = \frac{\vec{F}}{m} = -\frac{e\vec{E}}{m} \]
Drift Velocity (\(v_d\)): As the electrons accelerate, they collide with the positive ions of the conductor. The average velocity with which the electrons drift towards the positive end of the conductor under the influence of the electric field is called drift velocity. It is given by:
\[ \vec{v}_d = \vec{a}\tau = -\frac{e\vec{E}}{m}\tau \]
where \(\tau\) is the relaxation time, the average time interval between two successive collisions. The magnitude of the drift velocity is \(v_d = \frac{eE}{m}\tau\).

Current and Current Density (J): The total current 'I' flowing through the conductor is related to the drift velocity by the equation:
\[ I = neAv_d \]
Current density is the current per unit area, \(J = I/A\).
\[ J = nev_d \]
Relating J and E: Substituting the expression for \(v_d\) into the equation for J:
\[ J = ne \left( \frac{eE}{m}\tau \right) = \left( \frac{ne^2\tau}{m} \right) E \]
Deriving Conductivity (\(\sigma\)): From the microscopic form of Ohm's Law, we know that \(J = \sigma E\). Comparing this with the expression we derived:
\[ \sigma E = \left( \frac{ne^2\tau}{m} \right) E \]
This gives the expression for electrical conductivity:
\[ \sigma = \frac{ne^2\tau}{m} \]


Step 3: Final Answer:

Electrical conductivity is defined as the reciprocal of electrical resistivity. Its expression in terms of number density (n) and relaxation time (\(\tau\)) is \(\sigma = \frac{ne^2\tau}{m}\).
Quick Tip: The derivation hinges on connecting the microscopic motion of electrons (\(v_d\)) to the macroscopic quantities (J and E). Remember the key relations: \(I = nAev_d\) and \(v_d = a\tau\). This logical chain is fundamental to understanding conduction in metals.

Step 1: Understanding the Concept:

Resistivity (\(\rho\)) is the reciprocal of conductivity (\(\sigma\)). Using the expression for conductivity derived in part (a), we can explain how resistivity changes with temperature for a conductor.


Step 2: Key Formula or Approach:

From part (a), the conductivity is \(\sigma = \frac{ne^2\tau}{m}\).

Therefore, the resistivity \(\rho\) is: \[ \rho = \frac{1}{\sigma} = \frac{m}{ne^2\tau} \]
For a given conductor (metal), the mass of an electron (m), the charge of an electron (e), and the number density of free electrons (n) are essentially constant and do not change significantly with temperature.

Therefore, the resistivity is inversely proportional to the relaxation time \(\tau\). \[ \rho \propto \frac{1}{\tau} \]

Step 3: Detailed Explanation:


Effect of Temperature: When the temperature of a conductor increases, the thermal energy of the atoms in the crystal lattice increases. This causes the positive ions (kernels) in the lattice to vibrate with a larger amplitude about their mean positions.

Effect on Collisions: The free electrons, while drifting through the conductor, collide with these vibrating ions. Since the ions are vibrating with a larger amplitude, the frequency of collisions between the electrons and the ions increases.

Effect on Relaxation Time (\(\tau\)): Relaxation time (\(\tau\)) is the average time between two successive collisions. As the frequency of collisions increases, the average time between collisions decreases. Thus, as temperature increases, \(\tau\) decreases.

Effect on Resistivity (\(\rho\)): Since resistivity is inversely proportional to the relaxation time (\(\rho \propto 1/\tau\)), a decrease in \(\tau\) will lead to an increase in \(\rho\).


Conclusion: For a conductor, as the temperature increases, the relaxation time of free electrons decreases, which in turn causes the electrical resistivity to increase.
Quick Tip: A simple way to remember: Hotter metal \(\rightarrow\) More vibrating atoms \(\rightarrow\) More obstacles for electrons \(\rightarrow\) More collisions \(\rightarrow\) Shorter time between collisions (\(\tau\downarrow\)) \(\rightarrow\) Higher resistivity (\(\rho\uparrow\)). This logic applies specifically to conductors.

Step 1: The Binding Energy Curve:

The binding energy per nucleon (BE/A) is a measure of the stability of an atomic nucleus. The curve is a plot of the binding energy per nucleon versus the mass number (A).

Description of the Curve:

The curve starts at a low value for light nuclei (like deuterium).
It rises steeply for low mass numbers, showing a series of peaks for nuclei like \(^4He\), \(^{12}C\), and \(^{16}O\), which are particularly stable.
The curve reaches a broad maximum (a plateau) for mass numbers in the range of A = 50 to 80. The peak of the curve is at A = 56 (Iron, Fe), with a BE/A of about 8.8 MeV, making it one of the most stable nuclei.
For mass numbers greater than about 80, the curve gradually decreases. Heavy nuclei like Uranium (A = 238) have a lower BE/A (about 7.6 MeV) compared to the nuclei in the middle of the curve.

(A sketch of the graph would typically be drawn here in an exam).


Step 2: Significance of the Binding Energy Curve:

The shape of the binding energy curve has profound implications for nuclear stability and energy release in nuclear reactions:

Nuclear Stability: A higher binding energy per nucleon corresponds to a more stable nucleus. The peak of the curve around A=56 indicates that nuclei in this mass range (like Iron and Nickel) are the most tightly bound and therefore the most stable.

Nuclear Fusion: For light nuclei with low mass numbers (A \(<\) 56), the BE/A is relatively low. If two or more light nuclei combine (fuse) to form a heavier nucleus that is further up the curve, the resulting nucleus will have a higher BE/A. This increase in binding energy is released as a large amount of energy. This is the principle behind the energy generation in stars and hydrogen bombs.

Nuclear Fission: For heavy nuclei with high mass numbers (A \(>\) 80), the BE/A is lower than that of nuclei in the middle range. If a heavy nucleus splits (fission) into two or more lighter nuclei, the products will have a higher total BE/A than the original nucleus. This difference in binding energy is released as energy. This is the principle behind nuclear power reactors and atomic bombs.

Nature of Nuclear Force: The initial steep rise shows that the nuclear force is strong and attractive. The plateau and subsequent slow decline indicate that the nuclear force is short-ranged and saturates; a nucleon only interacts with its immediate neighbors. Quick Tip: Remember the key idea: Nature favors states of higher binding energy. The BE/A curve is like a "valley of stability". Processes that move nuclei "uphill" towards the peak at Iron-56 (fusion for light nuclei, fission for heavy nuclei) will release energy.

Step 1: Analyzing the Process:

The problem describes a process where two lighter nuclei (implied by "lower binding energy per nucleon", which is characteristic of the left side of the BE curve) combine to form a heavier nucleus with a higher binding energy per nucleon.


(i) Type of nuclear reaction:

The process of combining two or more lighter nuclei to form a single, heavier nucleus is called nuclear fusion. This process moves the product nucleus up the binding energy curve towards the peak of stability.


(ii) Change in total mass:

Binding energy is the energy equivalent of the mass defect (\(\Delta m\)) according to Einstein's mass-energy equivalence relation, \(E_b = \Delta m c^2\). A higher binding energy implies a larger mass defect.

The product nucleus has a higher binding energy per nucleon, and thus a higher total binding energy, than the sum of the binding energies of the initial nuclei.
This increase in total binding energy means that the mass defect of the product is greater than the sum of the mass defects of the reactants.
Consequently, the rest mass of the product nucleus is less than the sum of the rest masses of the initial nuclei.
Therefore, the total mass of the nuclei decreases during the reaction.



(iii) Energy requirement:

Since the total binding energy of the system increases, the system moves to a more stable, lower energy state. The difference in binding energy (\(E_{b,final} - E_{b,initial}\)) is released during the reaction. Therefore, the process produces energy. The amount of energy produced is equal to the decrease in mass multiplied by \(c^2\).
Quick Tip: Binding Energy vs. Mass: Think of binding energy as "negative energy". A more tightly bound (more stable) system has a more negative potential energy, and therefore a lower total mass-energy. So, higher binding energy always means lower mass.

Step 1: Phasor Diagram for a Series LCR Circuit:

In a series circuit, the current \(I\) is the same through all components. Therefore, we use the current phasor as the reference, drawn along the positive x-axis.

Voltage across Resistor (\(V_R\)): The voltage across the resistor is in phase with the current. So, the phasor \(V_R\) is drawn along the same direction as the current phasor \(I\).
Voltage across Inductor (\(V_L\)): The voltage across the inductor leads the current by \(90^\circ\) or \(\pi/2\) radians. So, the phasor \(V_L\) is drawn along the positive y-axis.
Voltage across Capacitor (\(V_C\)): The voltage across the capacitor lags the current by \(90^\circ\) or \(\pi/2\) radians. So, the phasor \(V_C\) is drawn along the negative y-axis.

The total voltage \(V\) across the circuit is the vector sum of \(V_R\), \(V_L\), and \(V_C\). Assuming \(V_L > V_C\), the resultant of \(V_L\) and \(V_C\) is a phasor of magnitude \(V_L - V_C\) pointing along the positive y-axis.
(A sketch of the phasor diagram showing \(V_R\) on the x-axis, \(V_L-V_C\) on the y-axis, and V as the hypotenuse would be drawn here).


Step 2: Derivation of Impedance (Z):

From the phasor diagram, we can see a right-angled triangle formed by the phasors \(V_R\), \((V_L - V_C)\), and the resultant voltage \(V\). Using the Pythagorean theorem: \[ V^2 = V_R^2 + (V_L - V_C)^2 \]
The magnitudes of the individual voltages are given by: \[ V_R = IR \] \[ V_L = IX_L = I(\omega L) \] \[ V_C = IX_C = I\left(\frac{1}{\omega C}\right) \]
where \(X_L\) is the inductive reactance and \(X_C\) is the capacitive reactance.

Substituting these into the voltage equation: \[ V^2 = (IR)^2 + (IX_L - IX_C)^2 \] \[ V^2 = I^2 R^2 + I^2(X_L - X_C)^2 \] \[ V^2 = I^2 [R^2 + (X_L - X_C)^2] \]
Taking the square root of both sides: \[ V = I \sqrt{R^2 + (X_L - X_C)^2} \]
The impedance (Z) of the circuit is defined as the ratio of the total voltage to the total current, \(Z = V/I\). \[ Z = \frac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the expressions for \(X_L\) and \(X_C\): \[ Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]
This is the expression for the impedance of a series LCR circuit.
Quick Tip: A useful mnemonic for the phase relationship is "ELI the ICE man". In an inductor (L), Voltage (E) leads Current (I). In a capacitor (C), Current (I) leads Voltage (E). This helps in correctly drawing the phasor diagram.

Step 1: Concept of Resonance:

Resonance in a series LCR circuit is a special condition that occurs at a particular frequency of the AC source, called the resonant frequency. At this frequency, the inductive reactance (\(X_L\)) becomes equal to the capacitive reactance (\(X_C\)). \[ X_L = X_C \]
This condition has several important consequences for the behavior of the circuit.


Step 2: Characteristics of the Circuit at Resonance:


Minimum Impedance: The impedance of the circuit is given by \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). When \(X_L = X_C\), the reactive term \((X_L - X_C)\) becomes zero. The impedance becomes minimum and is equal to the resistance of the circuit.
\[ Z_{min} = \sqrt{R^2 + 0^2} = R \]
Maximum Current: Since the impedance is at its minimum value, the current flowing through the circuit becomes maximum for a given source voltage \(V\). This maximum current is given by:
\[ I_{max} = \frac{V}{Z_{min}} = \frac{V}{R} \]
The circuit is said to be purely resistive at resonance.
Phase Angle: The phase angle \(\phi\) between the voltage and current is given by \(\tan \phi = \frac{X_L - X_C}{R}\). At resonance, since \(X_L - X_C = 0\), we have \(\tan \phi = 0\), which means \(\phi = 0^\circ\). The total voltage and current are in the same phase.


Step 3: Expression for Resonant Frequency:

The resonant condition is \(X_L = X_C\).
We know that \(X_L = \omega L\) and \(X_C = \frac{1}{\omega C}\).
Let \(\omega_r\) be the angular resonant frequency. At resonance: \[ \omega_r L = \frac{1}{\omega_r C} \] \[ \omega_r^2 = \frac{1}{LC} \] \[ \omega_r = \frac{1}{\sqrt{LC}} \]
This is the expression for the angular resonant frequency.

The linear resonant frequency, \(f_r\), is related to \(\omega_r\) by \(\omega_r = 2\pi f_r\). \[ 2\pi f_r = \frac{1}{\sqrt{LC}} \] \[ f_r = \frac{1}{2\pi \sqrt{LC}} \] Quick Tip: Resonance is the key to tuning circuits, like in a radio receiver. By changing the capacitance (C) or inductance (L), you change the resonant frequency (\(f_r\)). When \(f_r\) matches the frequency of a desired radio station, the current from that station's signal is maximized, and you "tune in" to it.

Step 1: Understanding the Concept:

The intensity of any wave, including a light wave, is a measure of the power it carries per unit area. This intensity is directly related to the square of the amplitude of the wave.


Step 2: Key Formula or Approach:

The intensity (I) of a wave is proportional to the square of its amplitude (A). \[ I \propto A^2 \]
This can be written as an equation with a proportionality constant k: \[ I = kA^2 \]

Step 3: Detailed Explanation:

Let the initial amplitude be \(A_1\) and the initial intensity be \(I_1\). \[ I_1 = kA_1^2 \quad \quad ...(1) \]
According to the problem, the new amplitude (\(A_2\)) is 'n' times the original amplitude. \[ A_2 = nA_1 \]
The new intensity (\(I_2\)) is 'm' times the original intensity. \[ I_2 = mI_1 \]
Now, let's write the expression for the new intensity using the amplitude relationship: \[ I_2 = kA_2^2 \]
Substitute \(A_2 = nA_1\) into this equation: \[ I_2 = k(nA_1)^2 = k(n^2 A_1^2) = n^2 (kA_1^2) \quad \quad ...(2) \]
From equation (1), we know that \(kA_1^2 = I_1\). Substituting this into equation (2): \[ I_2 = n^2 I_1 \]
We are also given that \(I_2 = mI_1\). Comparing the two expressions for \(I_2\): \[ mI_1 = n^2 I_1 \]
Dividing both sides by \(I_1\) (assuming \(I_1 \neq 0\)), we get the relation between m and n: \[ m = n^2 \]

Step 4: Final Answer:

The relation between m and n is \(m = n^2\).
Quick Tip: This is a fundamental wave property. Always remember: Intensity \(\propto\) (Amplitude)\(^2\). If the amplitude is doubled (n=2), the intensity becomes four times (m=4). If the amplitude is tripled (n=3), the intensity becomes nine times (m=9), and so on.

Step 1: Understanding the Concept:

Light, composed of photons, carries momentum. When light strikes a surface, it exerts a pressure known as radiation pressure. This pressure arises from the transfer of momentum from the photons to the surface. The magnitude of the pressure depends on how the surface interacts with the light (absorbs, reflects, or transmits).


Step 2: Key Formula or Approach:

Pressure is force per unit area. Force is the rate of change of momentum (\(F = \frac{\Delta p}{\Delta t}\)).

For a perfectly absorbing surface, a photon of momentum \(p\) is absorbed, so its final momentum is 0. The change in momentum is \(\Delta p = p_{final} - p_{initial} = 0 - p = -p\). The momentum transferred to the surface is \(p\). The pressure exerted is \(P_{abs} = I/c\), where I is the intensity of light and c is the speed of light.
For a perfectly reflecting surface, a photon of momentum \(p\) is reflected, so its final momentum is \(-p\). The change in momentum is \(\Delta p = p_{final} - p_{initial} = -p - p = -2p\). The momentum transferred to the surface is \(2p\). The pressure exerted is \(P_{ref} = 2I/c\).


Step 3: Justification and Conclusion:

Let's analyze the three surfaces:

Black Surface: An ideal black surface absorbs all light incident upon it. Therefore, it experiences the pressure due to complete absorption. \(P_{black} = I/c\).
White Surface: An ideal white surface reflects all light incident upon it. It experiences the pressure due to complete reflection. \(P_{white} = 2I/c\).
Yellow Surface: A yellow surface is partially absorbing and partially reflecting. It absorbs certain wavelengths (like blue) and reflects others (like yellow, red, green). The total momentum transferred will be more than for a purely absorbing surface but less than for a purely reflecting surface. Therefore, \(I/c < P_{yellow} < 2I/c\).


(i) Maximum Pressure:

The greatest change in momentum occurs when the light is perfectly reflected, as the momentum vector of each photon is reversed. This results in the largest force and hence the maximum pressure.
Maximum pressure is exerted on the white surface.


(ii) Minimum Pressure:

The smallest change in momentum (for non-transmitted light) occurs when the light is completely absorbed. The photon's momentum is entirely transferred to the surface, but not doubled as in reflection.
Minimum pressure is exerted on the black surface.
Quick Tip: Think of it like throwing a ball against a wall. If the ball is made of putty and sticks to the wall (absorption), it transfers its momentum. If the ball is bouncy and rebounds with the same speed (reflection), it transfers twice its initial momentum. The bouncy ball exerts more force on the wall.

Step 1: Understanding the Concept:

Extrinsic semiconductors are created by doping an intrinsic (pure) semiconductor with a small amount of a suitable impurity. This process, called doping, significantly increases the number of free charge carriers of one type, making them the "majority" carriers, while the other type becomes the "minority". There are two types of extrinsic semiconductors.


Step 2: Detailed Explanation:

1. n-type Semiconductor:

Formation: Created by doping a pure semiconductor (like Si or Ge, from Group 14) with a pentavalent impurity (an element with 5 valence electrons, from Group 15, e.g., Phosphorus, Arsenic, Antimony).
Mechanism: Four of the five valence electrons of the impurity atom form covalent bonds with the four neighboring semiconductor atoms. The fifth electron is loosely bound and can easily become a free electron for conduction, even at room temperature. Each impurity atom thus "donates" one free electron.
Charge Carriers: The doping process creates a large number of free electrons. Thermally generated electron-hole pairs also exist, but the number of donated electrons far exceeds the number of thermally generated holes.
Majority Carriers: Electrons.
Minority Carriers: Holes.


2. p-type Semiconductor:

Formation: Created by doping a pure semiconductor with a trivalent impurity (an element with 3 valence electrons, from Group 13, e.g., Boron, Aluminum, Gallium).
Mechanism: The three valence electrons of the impurity atom form covalent bonds with three neighboring semiconductor atoms. This leaves a vacancy or a "hole" in the bond with the fourth neighbor. This hole can easily accept an electron from a nearby bond, effectively causing the hole to move. Each impurity atom thus "accepts" an electron, creating one mobile hole.
Charge Carriers: The doping process creates a large number of holes. While some electron-hole pairs are still generated thermally, the number of created holes is much larger than the number of thermally generated electrons.
Majority Carriers: Holes.
Minority Carriers: Electrons. Quick Tip: A simple mnemonic: \textbf{n}-type has an excess of \textbf{n}egative charge carriers (electrons). \textbf{p}-type has an excess of \textbf{p}ositive charge carriers (holes). The "other" carrier is always the minority.

Step 1: Understanding Forward Bias:

A p-n junction is forward biased when the positive terminal of an external voltage source (battery) is connected to the p-side and the negative terminal is connected to the n-side. This applied external electric field (\(E_{ext}\)) opposes the internal built-in electric field (\(E_{int}\)) across the depletion region. As a result, the net electric field is reduced, and the potential barrier height is lowered.


Step 2: Movement of Charge Carriers:

The lowering of the potential barrier allows a significant flow of charge carriers across the junction, resulting in a large forward current. The process occurs as follows:

Movement of Majority Carriers towards the Junction:

The positive terminal of the battery repels the majority carriers in the p-side (holes), pushing them towards the junction.
The negative terminal of the battery repels the majority carriers in the n-side (electrons), pushing them towards the junction.

Diffusion Across the Junction: With the potential barrier reduced, a large number of these majority carriers have sufficient kinetic energy to overcome the barrier and diffuse across the junction into the opposite region.

Holes from the p-side diffuse into the n-side.
Electrons from the n-side diffuse into the p-side.

Minority Carrier Injection: This process is called minority carrier injection because the majority carriers that cross the junction become minority carriers in the new region (e.g., holes injected into the n-side are now minority carriers there).
Recombination: Once injected, these minority carriers are in a region with a high concentration of majority carriers. They travel a short distance (the diffusion length) before they recombine. For example, an electron injected into the p-side recombines with a hole.
Current Flow in the External Circuit: To maintain the continuous flow, for every electron-hole recombination that occurs near the junction, an electron is supplied by the negative terminal of the battery to the n-side. Simultaneously, an electron is drawn out from the p-side (from a broken covalent bond) into the positive terminal of the battery, creating a new hole.

This continuous process of diffusion, injection, recombination, and charge supply from the battery constitutes the forward current. This current is primarily due to the diffusion of majority carriers and is typically in the milliampere (mA) range.
Quick Tip: In forward bias, think "FLOOD GATES OPEN". The external voltage lowers the barrier, allowing a flood of majority carriers to diffuse across the junction, resulting in a large current.

Step 1: Understanding the Concept:

Dynamic resistance (or AC resistance) of a diode, denoted by \(r_d\), is the resistance offered by the diode to a changing voltage. It is defined as the ratio of a small change in voltage (\(\Delta V\)) across the diode to the corresponding small change in current (\(\Delta I\)) through it. Mathematically, it is the reciprocal of the slope of the I-V characteristic curve at a specific operating point. \[ r_d = \frac{\Delta V}{\Delta I} \]

Step 2: Analyzing the Graph at V = -0.6 V:

The operating point given is V = -0.6 V. This voltage is negative, which means the diode is operating in the reverse bias region.

We need to examine the I-V curve in the reverse bias region around V = -0.6 V.
Looking at the provided graph:

The x-axis represents voltage (V) in volts.
The y-axis represents current (I) in milliamperes (mA).
In the entire reverse bias region shown (from V = 0 to V = -1.2 V), the graph is a horizontal line lying on the voltage axis.
This indicates that the current (I) is constant and equal to 0 mA for all these negative voltages.


Step 3: Estimation of Dynamic Resistance:

To estimate \(r_d\), we can pick two points on the curve around V = -0.6 V. Let's choose:

\(V_1 = -0.4 \, V\), at which \(I_1 = 0 \, mA\)
\(V_2 = -0.8 \, V\), at which \(I_2 = 0 \, mA\)

Now we calculate \(\Delta V\) and \(\Delta I\): \[ \Delta V = V_2 - V_1 = -0.8 - (-0.4) = -0.4 \, V \] \[ \Delta I = I_2 - I_1 = 0 \, mA - 0 \, mA = 0 \, A \]
Now, we calculate the dynamic resistance: \[ r_d = \frac{\Delta V}{\Delta I} = \frac{-0.4 \, V}{0 \, A} \to \infty \]
The slope of the curve (\(\Delta I / \Delta V\)) is zero in this region. The dynamic resistance, being the reciprocal of the slope, is infinitely large.


Step 4: Final Answer:

Based on the provided graph, the current in the reverse bias region is practically zero and does not change with voltage. Therefore, the change in current (\(\Delta I\)) is zero. This leads to a dynamic resistance (\(r_d = \Delta V / \Delta I\)) that is extremely high or effectively infinite.

(Note: In a real diode, there is a very small, non-zero reverse saturation current, which would result in a very high but finite resistance. However, based on the resolution of this graph, the resistance is considered infinite.)
Quick Tip: Dynamic resistance is the inverse of the slope of the I-V curve. \textbf{Forward Bias (steep slope):} Small \(\Delta V\) causes large \(\Delta I\). Slope is large \(\implies\) \(r_d\) is small. \textbf{Reverse Bias (flat slope):} Large \(\Delta V\) causes almost no \(\Delta I\). Slope is near-zero \(\implies\) \(r_d\) is very large.

Step 1: Understanding the Concept:

The key information is that the charged particle moves with a constant velocity. According to Newton's first law, this implies that the net force acting on the particle is zero. The particle is in a region with both an electric field and a magnetic field, so it will experience both an electric force and a magnetic force. For the net force to be zero, these two forces must be equal in magnitude and opposite in direction.


Step 2: Analysis of Forces and Fields:

Let the charge of the negatively charged particle be \(q = -e\).

Velocity: \(\vec{v} = -v_0 \hat{i}\).
Conductor: Along the x-axis, with current \(\vec{I}\) in the \(-\hat{i}\) direction.
Position of particle: At a distance d. Let's assume the particle is on the y-axis at the point (0, d, 0).


(i) Diagram and Directions of Fields:

Magnetic Field (\(\vec{B}\)): The magnetic field produced by the long straight conductor at a distance d (at point (0, d, 0)) can be found using the Right-Hand Thumb Rule. Pointing the thumb in the direction of the current (\(-\hat{i}\)), the fingers curl such that at a point above the wire (positive y-axis), the field points out of the page. So, the magnetic field is in the \(+\hat{k}\) direction. Its magnitude is \(B = \frac{\mu_0 I}{2\pi d}\).
\[ \vec{B} = \frac{\mu_0 I}{2\pi d} \hat{k} \]
Magnetic Force (\(\vec{F}_m\)): The magnetic force on the charge is \(\vec{F}_m = q(\vec{v} \times \vec{B})\).
\[ \vec{F}_m = (-e) \left( (-v_0 \hat{i}) \times \left(\frac{\mu_0 I}{2\pi d} \hat{k}\right) \right) = (-e) \left( -\frac{\mu_0 I v_0}{2\pi d} \right) (\hat{i} \times \hat{k}) \]
Since \(\hat{i} \times \hat{k} = -\hat{j}\),
\[ \vec{F}_m = (-e) \left( -\frac{\mu_0 I v_0}{2\pi d} \right) (-\hat{j}) = - \frac{e \mu_0 I v_0}{2\pi d} \hat{j} \]
The magnetic force is directed downwards, along the negative y-axis.
Electric Field (\(\vec{E}\)): For the net force to be zero, the electric force \(\vec{F}_e\) must be equal and opposite to \(\vec{F}_m\).
\[ \vec{F}_e = -\vec{F}_m = +\frac{e \mu_0 I v_0}{2\pi d} \hat{j} \]
The electric force must be directed upwards. The electric field is related by \(\vec{F}_e = q\vec{E} = (-e)\vec{E}\).
\[ (-e)\vec{E} = \frac{e \mu_0 I v_0}{2\pi d} \hat{j} \implies \vec{E} = -\frac{\mu_0 I v_0}{2\pi d} \hat{j} \]
The uniform electric field \(\vec{E}\) must be directed downwards, along the negative y-axis.

(The diagram would show the x-y axes, the wire on the x-axis with current to the left, the particle at (0,d) moving left, \(\vec{B}\) pointing out of the page, and \(\vec{E}\) pointing down).


(ii) Various forces acting on the particle:
The two forces acting on the charged particle are:

Electric Force (\(\vec{F}_e = q\vec{E}\)): An upward force exerted by the uniform electric field.
Magnetic Force (\(\vec{F}_m = q(\vec{v} \times \vec{B})\)): A downward force exerted by the magnetic field of the current-carrying conductor.


(iii) Finding the value of \(v_0\):
For constant velocity, the net force is zero: \(\vec{F}_e + \vec{F}_m = 0\), which means their magnitudes are equal: \(|\vec{F}_e| = |\vec{F}_m|\).
Let the magnitude of the electric field be E. \[ |\vec{F}_e| = |qE| = eE \] \[ |\vec{F}_m| = |q v_0 B \sin\theta| = e v_0 B \sin(90^\circ) = e v_0 B \]
Equating the magnitudes: \[ eE = e v_0 B \implies E = v_0 B \]
Substitute the magnitude of the magnetic field, \(B = \frac{\mu_0 I}{2\pi d}\): \[ E = v_0 \left( \frac{\mu_0 I}{2\pi d} \right) \]
Solving for \(v_0\): \[ v_0 = \frac{E}{\frac{\mu_0 I}{2\pi d}} = \frac{2\pi d E}{\mu_0 I} \] Quick Tip: This problem is an example of a velocity selector. When electric and magnetic forces balance, only particles with a specific velocity (\(v = E/B\)) can pass through undeflected. Remember that \(\vec{F}_e\) and \(\vec{F}_m\) must be anti-parallel for this to work.

Step 1: Understanding the Concept:

The net magnetic field at a point due to multiple current-carrying conductors is the vector sum of the magnetic fields produced by each conductor individually. This is the principle of superposition. The magnetic field from an infinitely long straight wire is given by the Ampere's Law or Biot-Savart Law.


Step 2: Key Formula or Approach:

The magnitude of the magnetic field (\(B\)) at a perpendicular distance \(r\) from an infinitely long straight wire carrying current \(I\) is: \[ B = \frac{\mu_0 I}{2\pi r} \]
The direction is found using the Right-Hand Thumb Rule.


Step 3: Detailed Explanation:

Let the coordinates of the point P be (x, y). We assume x > 0 and y > 0.


Magnetic Field due to Conductor 1 (along X-axis):

Current \(I_1\) is along the \(-X\) axis (\(-\hat{i}\) direction).
The point P(x, y) is at a perpendicular distance \(r_1 = y\) from this conductor.
Using the Right-Hand Thumb Rule, with the thumb pointing in the \(-\hat{i}\) direction, the magnetic field at a point with positive y-coordinate points into the page.
The direction is \(-\hat{k}\).
The magnetic field vector is: \(\vec{B}_1 = -\frac{\mu_0 I_1}{2\pi y} \hat{k}\).


Magnetic Field due to Conductor 2 (along Y-axis):

Current \(I_2\) is along the \(-Y\) axis (\(-\hat{j}\) direction).
The point P(x, y) is at a perpendicular distance \(r_2 = x\) from this conductor.
Using the Right-Hand Thumb Rule, with the thumb pointing in the \(-\hat{j}\) direction, the magnetic field at a point with positive x-coordinate points out of the page.
The direction is \(+\hat{k}\).
The magnetic field vector is: \(\vec{B}_2 = +\frac{\mu_0 I_2}{2\pi x} \hat{k}\).


Net Magnetic Field:
The net magnetic field \(\vec{B}_{net}\) at point P is the vector sum of \(\vec{B}_1\) and \(\vec{B}_2\). \[ \vec{B}_{net} = \vec{B}_1 + \vec{B}_2 = -\frac{\mu_0 I_1}{2\pi y} \hat{k} + \frac{\mu_0 I_2}{2\pi x} \hat{k} \] \[ \vec{B}_{net} = \left( \frac{\mu_0 I_2}{2\pi x} - \frac{\mu_0 I_1}{2\pi y} \right) \hat{k} \] \[ \vec{B}_{net} = \frac{\mu_0}{2\pi} \left( \frac{I_2}{x} - \frac{I_1}{y} \right) \hat{k} \]
(Using X, Y as in the question: \(\vec{B}_{net} = \frac{\mu_0}{2\pi} \left( \frac{I_2}{X} - \frac{I_1}{Y} \right) \hat{k}\))


Magnitude and Direction:

Magnitude: The magnitude of the net magnetic field is the absolute value of the z-component:
\[ B_{net} = \left| \frac{\mu_0}{2\pi} \left( \frac{I_2}{X} - \frac{I_1}{Y} \right) \right| \]
Direction: The direction is along the z-axis.

If \( \frac{I_2}{X} > \frac{I_1}{Y} \), the direction is along the positive z-axis (\(+\hat{k}\)), i.e., perpendicular to the XY plane and pointing outwards.
If \( \frac{I_2}{X} < \frac{I_1}{Y} \), the direction is along the negative z-axis (\(-\hat{k}\)), i.e., perpendicular to the XY plane and pointing inwards.
If \( \frac{I_2}{X} = \frac{I_1}{Y} \), the net magnetic field is zero. Quick Tip: When dealing with multiple wires, always treat the magnetic fields as vectors. Calculate the magnitude and determine the direction for each wire separately using the right-hand rule, then perform a vector addition. Be careful with signs and coordinate directions.

Step 1: Understanding the Concept:

When a beam of light travels from a rarer medium (like air) to a denser medium (like water) at an oblique angle, it refracts, bending towards the normal. We need to analyze how this bending affects the perpendicular distance between the rays in the beam, which defines the beam's width.


Step 2: Ray Diagram and Geometry:

Let's consider a parallel beam of light of width 'w' incident from air (\(n_1=1\)) to water (\(n_2=n>1\)).
Let the angle of incidence be \(i\) and the angle of refraction be \(r\).
Consider a wavefront AB of the incident beam, which is perpendicular to the rays. The width of the beam is \(w = AB\).
Let the rays at A and B strike the water surface at points A and C respectively. The distance AC lies along the water surface.
From the geometry of the incident beam, the angle between the wavefront AB and the surface AC is \(i\). So, we have a right-angled triangle, and we can write: \[ \cos(i) = \frac{AB}{AC} = \frac{w}{AC} \implies AC = \frac{w}{\cos(i)} \]
Now, consider the refracted beam. All points on the wavefront CD in the water are in the same phase. This wavefront is perpendicular to the refracted rays and makes an angle \(r\) with the surface AC.
The width of the refracted beam, \(w'\), is the perpendicular distance CD. From the new right-angled triangle, we have: \[ \cos(r) = \frac{CD}{AC} = \frac{w'}{AC} \implies w' = AC \cos(r) \]
Substitute the expression for AC: \[ w' = \left(\frac{w}{\cos(i)}\right) \cos(r) = w \frac{\cos(r)}{\cos(i)} \]

Step 3: Applying Snell's Law:

According to Snell's Law, \(n_1 \sin(i) = n_2 \sin(r)\), or \( \sin(i) = n \sin(r) \).
Since the light enters a denser medium, \(n > 1\), which implies \( \sin(i) > \sin(r) \).
For angles between 0 and 90 degrees, this means \( i > r \).
The cosine function is a decreasing function in the first quadrant (from 0 to 90 degrees). Therefore, if \( i > r \), then \( \cos(i) < \cos(r) \).
This implies that the ratio \( \frac{\cos(r)}{\cos(i)} > 1 \).


Step 4: Final Conclusion:

Since \(w' = w \frac{\cos(r)}{\cos(i)}\) and \( \frac{\cos(r)}{\cos(i)} > 1 \), it follows that \(w' > w\).
Therefore, the width of the beam increases when it enters the water obliquely from the air.
Quick Tip: A simple way to visualize this: as the beam bends towards the normal, the rays become "more upright", so the perpendicular distance between them must increase to accommodate the same cross-section along the interface.

Step 1: Ray Diagram:

(The diagram shows a beaker of water with a straight straw PQ partly submerged. P is the tip inside the water, and Q is the end in the air. Two light rays are drawn originating from the tip P. One ray travels towards the surface at an angle of incidence \(i\). At the water-air interface, it refracts and bends away from the normal, with an angle of refraction \(r > i\). A second ray from P is also drawn. The two refracted rays in the air are extended backwards, where they intersect at a point P', which is above P. The observer's eye is shown viewing these refracted rays. The apparent position of the submerged part is the line from P' to the surface, making the straw appear as P'Q.)


Step 2: Explanation:

The apparent bending of a straw partially dipped in water is a phenomenon caused by the refraction of light.

When we look at an object, we see it because light rays from the object travel to our eyes.
Consider the tip of the straw (P) that is underwater. Light rays originating from P travel from the water (a denser medium) into the air (a rarer medium) before reaching the observer's eye.
As these rays cross the water-air interface, they bend away from the normal, according to Snell's law (\(n_w \sin i = n_a \sin r\)). Since \(n_w > n_a\), the angle of refraction \(r\) is greater than the angle of incidence \(i\).
The human brain interprets the position of an object by assuming that light travels in straight lines. Therefore, our eye traces the refracted rays back in a straight line.
These back-projected rays appear to diverge from a point P', which is located at a shallower depth than the actual tip P. This point P' is the virtual image of the tip P.
The same phenomenon occurs for every point on the submerged portion of the straw. Each point appears to be raised.
The part of the straw that is in the air is seen directly without refraction. The combination of seeing the virtual image of the submerged part and the actual position of the part in the air makes the straw appear to be bent at the water's surface. Quick Tip: This is a classic example of "apparent depth". The apparent depth (\(d'\)) is always less than the real depth (\(d\)) when viewing from a rarer medium into a denser one. The relationship is \(d' = d/n\), where n is the refractive index of the denser medium.

Step 1: Structure and Principle:

An optical fibre is a thin, flexible strand of glass or plastic that acts as a waveguide for light. It consists of two main parts:

Core: The inner cylindrical part made of a material with a high refractive index (\(n_1\)). The light signal travels through the core.
Cladding: The outer layer that surrounds the core, made of a material with a slightly lower refractive index (\(n_2\)), such that \(n_1 > n_2\).

The transmission of light signals through the optical fibre is based on the principle of Total Internal Reflection (TIR).


Step 2: Diagram:

(The diagram shows a cross-section of an optical fibre with the inner core (\(n_1\)) and outer cladding (\(n_2\)). A light ray is shown entering the core from the left. After entering, it strikes the core-cladding interface at an angle of incidence \(i\). Since \(i\) is greater than the critical angle \(c\), the ray undergoes TIR and is reflected back into the core. The diagram shows the ray propagating down the fibre through a series of successive total internal reflections.)


Step 3: Explanation of Transmission:


For Total Internal Reflection (TIR) to occur at the core-cladding interface, two conditions must be met:

The light ray must be traveling from a denser medium (core, \(n_1\)) to a rarer medium (cladding, \(n_2\)).
The angle of incidence (\(i\)) at the interface must be greater than the critical angle (\(c\)), where the critical angle is defined by \(\sin(c) = \frac{n_2}{n_1}\).

The optical signal, which is a beam of light (often from a laser), is launched into one end of the fibre. The angle of entry is controlled such that after refraction at the air-core interface, the ray strikes the core-cladding boundary at an angle of incidence greater than the critical angle.
When the ray hits the boundary, it is not refracted into the cladding but is completely reflected back into the core.
This reflected ray then travels to the opposite side of the core, where it again strikes the boundary at an angle greater than the critical angle and undergoes another TIR.
This process of successive total internal reflections continues, guiding the light signal along the entire length of the fibre, even around bends.
Since the reflection is total, there is very little loss of light intensity as the signal propagates. This allows the signal to travel over very long distances with minimal attenuation, making optical fibres highly efficient for telecommunications. Quick Tip: The key to optical fibre communication is trapping light by TIR. The core must be denser than the cladding (\(n_1 > n_2\)) for this to be possible. The efficiency of this process is why optical fibres have replaced copper wires for high-speed data transmission.

Step 1: Understanding the Concept:

This question relates to the graph of stopping potential (\(V_0\)) versus frequency (\(\nu\)) for the photoelectric effect, as described in the case study. We need to determine how the graphs for two different metals would relate to each other.


Step 2: Key Formula or Approach:

The equation for the graph is \(V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}\), where \(\phi_0\) is the work function of the metal.

The slope of this graph is \(m = \frac{h}{e}\).

The x-intercept is the threshold frequency, \(\nu_0 = \phi_0/h\).


Step 3: Detailed Explanation:


Slope of the graph: The slope is \(\frac{h}{e}\), where 'h' is Planck's constant and 'e' is the charge of an electron. Both of these are fundamental physical constants. Their value does not change regardless of the material used. Therefore, the slope of the \(V_0\) vs \(\nu\) graph will be the same for all metals.
Parallel Lines: Since the slope is the same for both metals, their graphs must be parallel straight lines.
Intercepts: Different metals have different work functions (\(\phi_0\)). A metal with a higher work function will have a higher threshold frequency (\(\nu_0\)). This means the x-intercept will be different for the two metals. Consequently, the y-intercept (\(-\phi_0/e\)) will also be different.
Conclusion: Because the slopes are identical but the intercepts are different, the straight-line graphs for two different metals will be parallel to each other but will not coincide.


Step 4: Final Answer:

The graphs for two different metals are parallel to each other.
Quick Tip: The \(V_0\) vs \(\nu\) graph's slope is a universal constant (\(h/e\)). Changing the metal only changes the work function, which shifts the line horizontally (changing the \(\nu_0\) intercept) and vertically (changing the \(V_0\) intercept) without altering its slope.

Step 1: Understanding the Concept:

The question asks for the value of Planck's constant (h) in terms of the slope (m) of the \(V_0\) vs \(\nu\) graph and the elementary charge (e).


Step 2: Key Formula or Approach:

From the case study and the derivation in the previous question, we have the equation for the graph: \[ V_0 = \left(\frac{h}{e}\right)\nu - \frac{h\nu_0}{e} \]
The slope of this straight line is \(m = \frac{change in V_0}{change in \nu}\).


Step 3: Detailed Explanation:

By comparing the equation of the line with the standard form \(y = mx + c\), we identify the slope as: \[ m = \frac{h}{e} \]
The question asks for the value of Planck's constant, 'h'. We need to rearrange this equation to solve for h. \[ m = \frac{h}{e} \]
Multiply both sides by 'e': \[ me = h \]
So, Planck's constant 'h' is equal to the product of the slope of the graph 'm' and the elementary charge 'e'.


Step 4: Final Answer:

The value of Planck's constant is me.
Quick Tip: This is a direct application of understanding the photoelectric equation in the form of a linear graph. Simply identify the slope from \(V_0 = (h/e)\nu - \phi_0/e\) and solve for h.

Step 1: Understanding the Concept:

We need to find the points where the graph of \(V_0\) vs \(\nu\) intersects the horizontal (\(\nu\)-axis) and vertical (\(V_0\)-axis).


Step 2: Key Formula or Approach:

The equation of the graph is: \[ eV_0 = h\nu - h\nu_0 \quad or \quad V_0 = \frac{h}{e}\nu - \frac{h\nu_0}{e} \]
To find an intercept, we set the other variable to zero.

\(\nu\)-axis intercept: Set \(V_0 = 0\).
\(V_0\)-axis intercept: Set \(\nu = 0\).


Step 3: Detailed Explanation:

Intercept on the \(\nu\)-axis (x-intercept):

Set \(V_0 = 0\) in the equation. \[ 0 = \frac{h}{e}\nu - \frac{h\nu_0}{e} \] \[ \frac{h}{e}\nu = \frac{h\nu_0}{e} \] \[ \nu = \nu_0 \]
So, the intercept on the \(\nu\)-axis is the threshold frequency, \(\nu_0\).


Intercept on the \(V_0\)-axis (y-intercept):

Set \(\nu = 0\) in the equation. \[ V_0 = \frac{h}{e}(0) - \frac{h\nu_0}{e} \] \[ V_0 = -\frac{h\nu_0}{e} \]
This can also be written in terms of the work function \(\phi_0 = h\nu_0\) as \(V_0 = -\frac{\phi_0}{e}\).
So, the intercept on the \(V_0\)-axis is \(-\frac{h\nu_0}{e}\).


Step 4: Final Answer:

The intercepts on the \(\nu\)-axis and \(V_0\)-axis are \(\nu_0\) and \(-\frac{h\nu_0}{e}\) respectively. This corresponds to option (A).
Quick Tip: Finding intercepts is a basic algebraic skill. For any linear graph \(y=mx+c\), the x-intercept is where \(y=0\) (\(x=-c/m\)) and the y-intercept is where \(x=0\) (\(y=c\)). Apply this directly to the photoelectric equation.

Step 1: Understanding the Concept:

We need to find the relationship between wavelength (\(\lambda\)), wave number (\(\bar{\nu}\) or k), and frequency (\(\nu\)). Then we'll see how wave number and frequency change when the wavelength is doubled.


Step 2: Key Formula or Approach:

The key relationships are:

Wave Number (\(\bar{\nu}\)): The wave number is the reciprocal of the wavelength. It represents the number of waves per unit length.
\[ \bar{\nu} = \frac{1}{\lambda} \]
Frequency (\(\nu\)): The frequency is related to wavelength and the speed of light (c) by the wave equation.
\[ c = \nu \lambda \implies \nu = \frac{c}{\lambda} \]


Step 3: Detailed Explanation:

Let the initial wavelength be \(\lambda_1\). The new wavelength \(\lambda_2\) is doubled. \[ \lambda_2 = 2\lambda_1 \]

Change in Wave Number:

Initial wave number: \(\bar{\nu}_1 = \frac{1}{\lambda_1}\)

New wave number: \(\bar{\nu}_2 = \frac{1}{\lambda_2} = \frac{1}{2\lambda_1} = \frac{1}{2} \left(\frac{1}{\lambda_1}\right) = \frac{1}{2}\bar{\nu}_1\)

So, the wave number becomes \(\frac{1}{2}\) times the original value.


Change in Frequency:

Initial frequency: \(\nu_1 = \frac{c}{\lambda_1}\)

New frequency: \(\nu_2 = \frac{c}{\lambda_2} = \frac{c}{2\lambda_1} = \frac{1}{2} \left(\frac{c}{\lambda_1}\right) = \frac{1}{2}\nu_1\)

So, the frequency also becomes \(\frac{1}{2}\) times the original value.


Step 4: Final Answer:

The wave number becomes \(1/2\) times and the frequency becomes \(1/2\) times. The respective factors are \(1/2\) and \(1/2\). This corresponds to option (B).
Quick Tip: Both wave number and frequency are inversely proportional to wavelength. So, if you double the wavelength, you must halve both the wave number and the frequency.

Step 1: Understanding the Concept:

This question relates the momentum of a photon to its wavelength. The relationship is given by the de Broglie wavelength formula, which applies to photons as well. The note about ignoring relativistic effects is redundant for a photon, as its properties are inherently relativistic, but the formula remains the same.


Step 2: Key Formula or Approach:

The momentum (p) of a photon is related to its wavelength (\(\lambda\)) by the de Broglie relation: \[ p = \frac{h}{\lambda} \]
where 'h' is Planck's constant. We can rearrange this to find the wavelength: \[ \lambda = \frac{h}{p} \]

Step 3: Detailed Explanation:

Given values:

Momentum, \(p = 5.0 \times 10^{-29} \, kg \cdot m/s\).
Planck's constant, \(h \approx 6.626 \times 10^{-34} \, J \cdot s\) (or kg m\(^2\)/s).

Substitute these values into the formula for wavelength: \[ \lambda = \frac{6.626 \times 10^{-34} \, J \cdot s}{5.0 \times 10^{-29} \, kg \cdot m/s} \] \[ \lambda = \left(\frac{6.626}{5.0}\right) \times 10^{-34 - (-29)} \, m \] \[ \lambda = 1.3252 \times 10^{-5} \, m \]
The options are given in micrometers (\(\mu\)m), where \(1 \, \mum = 10^{-6} \, m\). To convert our answer to micrometers, we can write: \[ \lambda = 13.252 \times 10^{-6} \, m \] \[ \lambda \approx 13.3 \, \mum \]

Step 4: Final Answer:

The wavelength of the photon is approximately 13.3 \(\mu\)m. This corresponds to option (D).
Quick Tip: The de Broglie relation \( \lambda = h/p \) is one of the cornerstones of quantum mechanics, linking the wave nature (\(\lambda\)) and particle nature (p) of any entity. For photons, you can also derive it from \(E=pc\) and \(E=h\nu = hc/\lambda\), which gives \(pc = hc/\lambda \implies p = h/\lambda\).

Step 1: Understanding the Concept:

For a parallel plate capacitor, the electric field (E) between the plates is considered uniform (neglecting fringe effects). This uniform electric field is directly related to the potential difference (V) across the plates and the separation (d) between them.


Step 2: Key Formula or Approach:

The relationship between the uniform electric field (E), potential difference (V), and plate separation (d) is: \[ E = \frac{V}{d} \]

Step 3: Detailed Explanation:

Initial State:
Let the initial electric field be E, the initial potential difference be V, and the initial separation be d. \[ E = \frac{V}{d} \quad \quad ...(1) \]

Final State:
The problem states the following changes are made:

The separation is doubled: \(d' = 2d\).
The potential difference is reduced to half: \(V' = \frac{V}{2}\).

The new electric field, E', will be: \[ E' = \frac{V'}{d'} \]
Substitute the new values of V' and d' into this equation: \[ E' = \frac{(V/2)}{(2d)} = \frac{V}{4d} \]
Now, substitute the expression for the original field from equation (1), \(E = V/d\): \[ E' = \frac{1}{4} \left(\frac{V}{d}\right) = \frac{1}{4} E \] \[ E' = \frac{E}{4} \]

Step 4: Final Answer:

The new value of the electric field between the plates will be E/4.
Quick Tip: For questions involving changes to capacitor parameters, write down the initial formula, then write the new formula with the modified parameters (\(V', d'\), etc.). Finally, substitute the initial formula into the new one to find the relationship. This systematic approach prevents errors.

Step 1: Understanding the Concept:

The question asks for the relationship between the potential difference (V) and the plate separation (d) under the condition that the electric field (E) between the plates remains constant.


Step 2: Key Formula or Approach:

The relationship between the uniform electric field (E), potential difference (V), and plate separation (d) for a parallel plate capacitor is: \[ E = \frac{V}{d} \]
We can rearrange this formula to express V as a function of d.


Step 3: Detailed Explanation:

We are given that the electric field E must be maintained at a constant value. Let's rearrange the formula to solve for V: \[ V = E \times d \]
This equation shows the relationship between V and d. Since E is a constant, this equation is of the form: \[ y = mx + c \]
where:

\(y\) corresponds to the potential difference V.
\(x\) corresponds to the separation d.
The slope \(m\) corresponds to the constant electric field E.
The y-intercept \(c\) is zero (since if \(d=0\), \(V=0\)).

An equation of the form \(y=mx\) represents a straight line passing through the origin with a positive slope (assuming E is positive). Therefore, the graph of V versus d is a straight line with a positive slope E, starting from the origin.

Graph (D) shows a straight line with a positive slope passing through the origin, which correctly represents the relationship \(V = Ed\).


Step 4: Final Answer:

The graph correctly depicting the potential difference V as a function of separation d is a straight line with a positive slope starting from the origin.
Quick Tip: Whenever a question asks for the relationship between two physical quantities while a third is held constant, rearrange the relevant formula to match the standard linear equation form \(y = mx + c\). This will immediately tell you the shape of the graph (line, parabola, etc.), its slope, and its intercept.

Step 1: Understanding the Concept:

The question asks for the variation of the magnitude of the electric field (E) along a line MN that is drawn between the two plates of an ideal parallel plate capacitor.


Step 2: Key Properties of an Ideal Capacitor Field:

For an ideal parallel plate capacitor, the electric field in the region between the plates is assumed to be:

Uniform: The electric field has the same magnitude and direction at all points in the space between the plates.
Perpendicular to the plates: The electric field lines are straight, parallel, and point from the positive plate (Q) to the negative plate (P).

The electric field outside the plates is considered to be zero.
(This is an idealization; in reality, there are "fringing fields" near the edges of the plates where the field is non-uniform and bows outwards.)


Step 3: Detailed Explanation:

The line MN is located entirely within the region between the plates P and Q. According to the properties of an ideal capacitor, the electric field strength (E) should be constant at every point along this line.

As we move from point M to point N, we are moving parallel to the plates, but staying within the uniform field region.
The magnitude of the electric field does not depend on the position along the line MN. It remains constant, with a value of \(E = \sigma / \epsilon_0\), where \(\sigma\) is the surface charge density on the plates.
Therefore, a graph of E versus position along the line MN should be a horizontal line, indicating a constant non-zero value.

Graph (C) shows the magnitude of the electric field E as a constant value between M and N, which correctly represents the uniform field inside an ideal capacitor.
Graphs (A), (B), and (D) show a varying electric field, which would be incorrect for the region between the plates of an ideal capacitor.


Step 4: Final Answer:

The correct graph is the one showing a constant electric field strength along the line MN.
Quick Tip: Unless a question specifically mentions "fringe effects" or asks about the region near the edges of the plates, always assume the standard ideal model for a parallel plate capacitor: the electric field inside is uniform, and the electric field outside is zero.

Step 1: Understanding the Concept:

The potential difference between two points is the line integral of the electric field between those points. For a uniform electric field, this simplifies to \(V = - \vec{E} \cdot \vec{l}\), where \(\vec{l}\) is the displacement vector. When there are multiple regions with different uniform fields, the total potential difference is the sum of the potential differences across each region.


Step 2: Key Formula or Approach:

The potential difference between two points A and B is given by: \[ V_B - V_A = - \int_A^B \vec{E} \cdot d\vec{l} \]
If the electric field is uniform, \(V = - \vec{E} \cdot \vec{l}\) or \( \Delta V = \vec{E} \cdot \vec{d} \) if \(\Delta V\) is the potential drop in the direction of \(\vec{d}\).
The total potential difference is the sum of the potential differences across each segment: \(V_{total} = V_1 + V_2\).


Step 3: Detailed Explanation:

Let the three plates be Plate 1, Plate 2, and Plate 3, starting from the bottom.

Plate 1 is at position 0.
Plate 2 is at position \(\vec{d}\).
Plate 3 is at position \(2\vec{d}\).

We are given:

The electric field between Plate 1 and Plate 2 is \(\vec{E}_1\).
The electric field between Plate 2 and Plate 3 is \(\vec{E}_2\).
The displacement vector between neighboring plates is \(\vec{d}\). This vector points from a plate to the next one above it.

We want to find the potential difference between the third and the first plate, i.e., \(V_3 - V_1\).
We can find this by summing the potential differences: \[ V_3 - V_1 = (V_3 - V_2) + (V_2 - V_1) \]
Now, let's calculate each term.
The potential difference between Plate 2 and Plate 1 is: \[ V_2 - V_1 = - \int_{Plate 1}^{Plate 2} \vec{E}_1 \cdot d\vec{l} \]
Since the field is uniform and the path is a straight line of displacement \(\vec{d}\), this becomes: \[ V_2 - V_1 = - \vec{E}_1 \cdot \vec{d} \]
Similarly, the potential difference between Plate 3 and Plate 2 is: \[ V_3 - V_2 = - \int_{Plate 2}^{Plate 3} \vec{E}_2 \cdot d\vec{l} = - \vec{E}_2 \cdot \vec{d} \]
Now, add the two potential differences: \[ V_3 - V_1 = (- \vec{E}_2 \cdot \vec{d}) + (- \vec{E}_1 \cdot \vec{d}) \] \[ V_3 - V_1 = - (\vec{E}_1 \cdot \vec{d} + \vec{E}_2 \cdot \vec{d}) \] \[ V_3 - V_1 = - (\vec{E}_1 + \vec{E}_2) \cdot \vec{d} \]
The question asks for "the potential difference", which is often interpreted as the magnitude or the potential drop. If we consider the potential drop from plate 3 to plate 1, \(V_{13} = V_1 - V_3\), it would be: \[ V_1 - V_3 = (\vec{E}_1 + \vec{E}_2) \cdot \vec{d} \]
This matches option (A). The phrasing "potential difference between the third and the first plate" is slightly ambiguous, but given the options, it most likely refers to the potential drop \(V_{1}-V_{3}\) or the magnitude of the potential difference. The expression \((\vec{E}_1 + \vec{E}_2) \cdot \vec{d}\) represents this value.


Step 4: Final Answer:

The potential difference between the third and the first plate is \((\vec{E}_1 + \vec{E}_2) \cdot \vec{d}\).
Quick Tip: Remember the negative sign in the potential integral formula \(V_B - V_A = - \int \vec{E} \cdot d\vec{l}\). This means that potential decreases as you move in the direction of the electric field. Potential difference is a scalar, so vector dot products are essential for calculating it correctly.

Step 1: Understanding the Concept:

When a dielectric material is introduced between the plates of a capacitor, it reduces the effective electric field, which allows more charge to be stored for the same potential difference. This results in an increase in the capacitance.


Step 2: Key Formula or Approach:

The capacitance of a parallel plate capacitor with air or vacuum between the plates is given by: \[ C = \frac{\epsilon_0 A}{d} \]
When the space between the plates is completely filled with a dielectric material of dielectric constant K, the permittivity of the medium becomes \( \epsilon = K\epsilon_0 \). The new capacitance, C', is given by: \[ C' = \frac{K\epsilon_0 A}{d} \]

Step 3: Detailed Explanation:

We are given the initial capacitance as C. \[ C = \frac{\epsilon_0 A}{d} \]
The new capacitance, C', after filling the dielectric is: \[ C' = \frac{K\epsilon_0 A}{d} \]
We can see that the new expression is simply K times the original expression for C. \[ C' = K \left( \frac{\epsilon_0 A}{d} \right) \] \[ C' = KC \]
The capacitance increases by a factor of K.


Step 4: Final Answer:

The new value of the capacitance becomes CK.
Quick Tip: A simple rule to remember: inserting a dielectric with constant K always increases the capacitance by a factor of K (assuming the capacitor is completely filled). Capacitance is directly proportional to the dielectric constant.

Step 1: Source of the Force:

The source of the force on a current-carrying conductor in a magnetic field is the magnetic Lorentz force acting on the individual charge carriers (usually electrons) that constitute the current. An electric current is a flow of charges. When the conductor is placed in a magnetic field, each moving charge experiences a magnetic force given by \(\vec{F} = q(\vec{v} \times \vec{B})\). The net force on the conductor is the vector sum of the forces on all the moving charges within it.


Step 2: Expression for Force between Two Parallel Conductors:

Consider two long, straight, parallel conductors (Wire 1 and Wire 2) separated by a distance 'd' in vacuum. Let them carry steady currents \(I_1\) and \(I_2\) respectively in the same direction.

Magnetic Field of Wire 1 at Wire 2: Wire 1 produces a magnetic field (\(\vec{B}_1\)) at all points around it. At the location of Wire 2, the magnitude of this field is:
\[ B_1 = \frac{\mu_0 I_1}{2\pi d} \]
By the right-hand thumb rule, if \(I_1\) is upwards, \(\vec{B}_1\) at Wire 2 is directed into the page.
Force on Wire 2: Now, Wire 2, carrying current \(I_2\), is in the magnetic field \(\vec{B}_1\). The force (\(\vec{F}_2\)) on a length 'L' of Wire 2 is given by \(\vec{F}_2 = I_2(\vec{L} \times \vec{B}_1)\).
The magnitude of this force is:
\[ F_2 = I_2 L B_1 \sin(90^\circ) = I_2 L B_1 \]
(since \(\vec{L}\) is perpendicular to \(\vec{B}_1\)).
Substituting B\(_1\):
\[ F_2 = I_2 L \left( \frac{\mu_0 I_1}{2\pi d} \right) = \frac{\mu_0 I_1 I_2 L}{2\pi d} \]
The force per unit length on Wire 2 is:
\[ \frac{F_2}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]
By Fleming's Left-Hand Rule, this force is directed towards Wire 1, indicating an attractive force. If the currents were in opposite directions, the force would be repulsive.


Step 3: Definition of 'Ampere':

The SI unit of current, the ampere, is defined using the expression for the force per unit length derived above. \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]
Let's set \(I_1 = I_2 = 1 \, A\) and \(d = 1 \, m\). \[ \frac{F}{L} = \frac{\mu_0 (1)(1)}{2\pi (1)} = \frac{4\pi \times 10^{-7}}{2\pi} = 2 \times 10^{-7} \, N/m \]
Based on this, the ampere is defined as:

One ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to \(2 \times 10^{-7}\) newtons per meter of length.
Quick Tip: Remember the final expression for force per unit length: \(F/L = \mu_0 I_1 I_2 / (2\pi d)\). A key takeaway is that parallel currents attract, and anti-parallel currents repel.

Step 1: Understanding the Concept:

Work is done by a force when it has a component along the direction of displacement. We need to analyze the direction of the magnetic Lorentz force relative to the direction of motion (velocity) of the charged particle.


Step 2: Key Formula or Approach:

The magnetic Lorentz force on a charge q moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is given by: \[ \vec{F}_m = q(\vec{v} \times \vec{B}) \]
The work done (W) by a force \(\vec{F}\) over a small displacement \(d\vec{s}\) is \(dW = \vec{F} \cdot d\vec{s}\). The rate of doing work (power) is \(P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}\).


Step 3: Detailed Explanation:

By the mathematical definition of a cross product, the vector \(\vec{F}_m = q(\vec{v} \times \vec{B})\) is always perpendicular to both the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\). \[ \vec{F}_m \perp \vec{v} \]
The work done by the magnetic force is calculated by integrating the dot product of the force and the displacement vector \(d\vec{s}\). The displacement vector for a moving particle is always in the direction of its instantaneous velocity, so we can write \(d\vec{s} = \vec{v} dt\).
The rate at which work is done (the power delivered by the magnetic force) is: \[ P = \vec{F}_m \cdot \vec{v} \]
Since \(\vec{F}_m\) is always perpendicular to \(\vec{v}\), the angle between them is \(90^\circ\). The dot product is therefore: \[ \vec{F}_m \cdot \vec{v} = |\vec{F}_m| |\vec{v}| \cos(90^\circ) = 0 \]
Since the power delivered by the force is zero at all times, the total work done by the magnetic force over any period of time is also zero. \[ W = \int P \, dt = \int 0 \, dt = 0 \]

Step 4: Final Answer:

The work done by the magnetic force on the charge is always zero. The magnetic force can change the direction of the particle's velocity, but it cannot change its speed or its kinetic energy.
Quick Tip: A fundamental property to remember: magnetic forces do no work. They act as a deflecting force, changing the direction of motion but not the speed. This is why a magnetic field alone cannot be used to accelerate a charged particle from rest.

Step 1: Understanding the Concept:

A helical path is a three-dimensional spiral. It is the result of a combination of two simultaneous motions: a circular motion in a plane and a linear motion perpendicular to that plane. We need to find the conditions on the particle's initial velocity relative to the magnetic field that produce this combined motion.


Step 2: Necessary Conditions:

For a charged particle to follow a helical trajectory in a uniform magnetic field \(\vec{B}\), the following conditions are necessary:

The particle must be charged.
A uniform magnetic field must be present.
The initial velocity vector \(\vec{v}\) of the particle must be at an angle \(\theta\) to the direction of the magnetic field \(\vec{B}\), such that \(\theta \neq 0^\circ\), \(90^\circ\), or \(180^\circ\).


Step 3: Detailed Explanation of Motion:

When the velocity \(\vec{v}\) is at an angle \(\theta\) to \(\vec{B}\), we can resolve \(\vec{v}\) into two components:

Parallel Component (\(v_{\parallel}\)): This component is parallel to the magnetic field, with magnitude \(v_{\parallel} = v \cos\theta\). The magnetic force due to this component is \(F = q(v_{\parallel} \hat{B} \times B \hat{B}) = 0\), since the cross product of a vector with itself is zero. Therefore, this component of velocity remains constant, causing the particle to move with uniform linear motion along the magnetic field lines.
Perpendicular Component (\(v_{\perp}\)): This component is perpendicular to the magnetic field, with magnitude \(v_{\perp} = v \sin\theta\). This component experiences a magnetic force of magnitude \(F_m = q v_{\perp} B\). This force is always perpendicular to both \(v_{\perp}\) and \(\vec{B}\). It acts as a centripetal force, forcing the particle to execute a uniform circular motion in the plane perpendicular to the magnetic field.

Resulting Trajectory:

The superposition of these two motions—a constant velocity translation along the field direction and a uniform circular motion in the plane perpendicular to the field—results in a helical path. The particle spirals around the magnetic field lines. The radius of the helix is determined by the circular motion (\(r = mv_{\perp}/qB\)), and the pitch (the distance between successive turns) is determined by the linear motion (\(p = v_{\parallel} \times T\), where T is the period of the circular motion).
Quick Tip: Think of the velocity vector as having two "jobs". The part parallel to B is "invisible" to the magnetic force and just moves the particle along. The part perpendicular to B is what feels the force and gets turned into a circle. Putting them together gives a spiral (helix).

Step 1: Understanding the Concept:

A magnetic dipole is a system that has two opposite magnetic poles (a north and a south pole) separated by a distance. A bar magnet is a common example. We need to explain why a loop of wire carrying an electric current behaves like a magnetic dipole. This is done by showing that the magnetic field produced by the loop is similar to the field of a bar magnet.


Step 2: Explanation:


Magnetic Field of a Current Loop: A current flowing through a loop of wire generates a magnetic field. The direction of this field can be determined by the Right-Hand Thumb Rule. If you curl the fingers of your right hand in the direction of the current, your thumb points in the direction of the magnetic field inside the loop.
Formation of Magnetic Poles: The magnetic field lines emerge from one face of the loop and loop around to enter the other face.

The face of the loop from which the magnetic field lines emerge acts as the North pole. If you look at this face, the current will appear to be flowing in the anticlockwise direction.
The face of the loop into which the magnetic field lines enter acts as the South pole. If you look at this face, the current will appear to be flowing in the clockwise direction.

Since the current loop has two distinct faces acting as north and south poles, it behaves as a magnetic dipole.
Comparison with a Bar Magnet: The pattern of the magnetic field lines produced by a current loop is very similar to the pattern produced by a short bar magnet, especially at points far from the loop.
Mathematical Analogy: The magnetic field on the axis of a circular current loop of radius R and current I at a distance x from its center is given by \(B = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}\). For a point far from the loop (\(x \gg R\)), this simplifies to \(B \approx \frac{\mu_0 I R^2}{2x^3}\). If we define the magnetic dipole moment as \(M = I \cdot A = I(\pi R^2)\), this expression becomes \(B \approx \frac{\mu_0 (M/\pi)}{2x^3} = \frac{\mu_0}{4\pi} \frac{2M}{x^3}\). This expression is mathematically identical to the expression for the magnetic field on the axis of a short bar magnet.

Because a current loop creates a two-poled magnetic field and the mathematical description of its far-field is identical to that of a magnetic dipole, it is considered to be a magnetic dipole.
Quick Tip: Remember the "face rule" for poles: If you look at a loop and the current is \textbf{A}nti-\textbf{C}lockwise, that face is a \textbf{N}orth pole. If the current is \textbf{C}lockwise, that face is a \textbf{S}outh pole. This helps visualize the dipole nature.

Step 1: Understanding the Concept:

The magnetic dipole moment (\(\vec{M}\)) is a vector quantity that measures the strength and orientation of a magnetic dipole. For a current loop, it depends on the current flowing through it, the area enclosed by it, and the number of turns in the coil.


Step 2: Relation for Magnetic Dipole Moment:

The magnetic dipole moment of a current-carrying coil is defined as the product of the number of turns in the coil, the current flowing through it, and the area vector of the coil.
For a coil with \(N\) turns, carrying a steady current \(I\), and enclosing a planar area \(A\), the magnitude of the magnetic dipole moment is given by: \[ M = NIA \]
In vector form, the relation is: \[ \vec{M} = NI\vec{A} \]
where \(\vec{A}\) is the area vector.


Step 3: Direction of \(\vec{M}\):

The direction of the magnetic dipole moment vector \(\vec{M}\) is the same as the direction of the area vector \(\vec{A}\). The direction of the area vector is perpendicular to the plane of the coil and is determined by the Right-Hand Thumb Rule.
Rule: If you curl the fingers of your right hand in the direction of the current flowing in the coil, your outstretched thumb will point in the direction of the magnetic dipole moment vector \(\vec{M}\).
Quick Tip: The formula \(\vec{M} = NI\vec{A}\) is fundamental. Note its similarity to the definition of electric dipole moment \(\vec{p} = q\vec{d}\). For magnetism, "current times area" plays a role similar to "charge times separation" for electricity.

Step 1: Net Force on the Coil:

In a uniform magnetic field, the magnetic forces on opposite sides of a current-carrying coil are equal in magnitude and opposite in direction. For instance, in a rectangular loop, the force on one side is cancelled by the force on the opposite side. Therefore, the vector sum of all the forces acting on the coil is zero. \[ \vec{F}_{net} = 0 \]
The coil will not experience any net translational force, but it may experience a torque.


Step 2: Orientation for Stable Equilibrium:

The potential energy (U) of a magnetic dipole \(\vec{M}\) in an external magnetic field \(\vec{B}\) is given by: \[ U = -\vec{M} \cdot \vec{B} = -MB \cos\theta \]
where \(\theta\) is the angle between \(\vec{M}\) and \(\vec{B}\).
For stable equilibrium, the system must be in a state of minimum potential energy. The potential energy U is minimum when \(\cos\theta\) has its maximum value, which is +1.
This occurs when \(\theta = 0^\circ\).
Therefore, the orientation for stable equilibrium is when the magnetic dipole moment vector \(\vec{M}\) of the coil is aligned parallel to the external magnetic field vector \(\vec{B}\). In this orientation, the plane of the coil is perpendicular to the magnetic field.


Step 3: Flux of the Total Field in Stable Equilibrium:

The total magnetic field \(\vec{B}_{total}\) is the vector sum of the external field \(\vec{B}_{ext}\) and the field produced by the loop itself, \(\vec{B}_{loop}\). \[ \vec{B}_{total} = \vec{B}_{ext} + \vec{B}_{loop} \]
The total magnetic flux \(\Phi_{total}\) through the coil is the surface integral of the total field over the area of the coil: \[ \Phi_{total} = \int \vec{B}_{total} \cdot d\vec{A} = \int (\vec{B}_{ext} + \vec{B}_{loop}) \cdot d\vec{A} \]
Let's analyze the fields in the stable equilibrium orientation (\(\theta = 0^\circ\)):

Direction of \(\vec{B}_{ext}\): It is in a fixed direction.
Direction of \(\vec{B}_{loop}\): The magnetic field produced by the loop (\(\vec{B}_{loop}\)) is in the same direction as its magnetic moment \(\vec{M}\) (by the Right-Hand Rule).
Alignment: In stable equilibrium, \(\vec{M}\) is parallel to \(\vec{B}_{ext}\). This means that \(\vec{B}_{loop}\) is also parallel to \(\vec{B}_{ext}\) within the loop. The area vector \(\vec{A}\) is also in the same direction.
Flux Calculation: Since \(\vec{B}_{ext}\) and \(\vec{B}_{loop}\) are in the same direction and parallel to the area vector \(d\vec{A}\), the dot products become simple multiplications of magnitudes, and the fluxes add up constructively.
\[ \Phi_{ext} = \int B_{ext} dA = B_{ext}A \]
\[ \Phi_{loop} = \int B_{loop} dA \]
The total flux is \(\Phi_{total} = \Phi_{ext} + \Phi_{loop}\).

In any other orientation, the flux due to the external field would be \(\Phi_{ext} = B_{ext}A \cos\theta\), which is less than its maximum value. More importantly, the two fields \(\vec{B}_{ext}\) and \(\vec{B}_{loop}\) would not be perfectly aligned, leading to a smaller total field magnitude and thus a smaller total flux. Therefore, the total flux is maximum when the coil is in its stable equilibrium position.
Quick Tip: Equilibrium positions correspond to zero torque (\(\theta=0^\circ\) or \(180^\circ\)). Stable equilibrium is the position of minimum energy (\(U=-MB\), \(\theta=0^\circ\)). Unstable equilibrium is the position of maximum energy (\(U=+MB\), \(\theta=180^\circ\)).

Step 1: Understanding the Concept:

The image is real and enlarged, which is only possible with a concave mirror when the object is placed between the center of curvature (C) and the principal focus (F). The object (pencil) lies along the principal axis, so we are dealing with longitudinal magnification. The key condition is that the image "just touches" the pencil.


Step 2: Key Formula or Approach:

The mirror formula is \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\). For a concave mirror, f is negative.
Let the focal length be \(-f\). The formula becomes \(\frac{1}{v} + \frac{1}{u} = -\frac{1}{f}\).
The condition "image just touches the pencil" implies that one end of the image coincides with one end of the object. Since the object is between C and F, the image is formed beyond C and is further away from the mirror. This means the image of an object point cannot coincide with the same point (except at C). The only plausible configuration is that the image of one end of the pencil forms at the location of the other end. Since the image is formed further away, the image of the end closer to C (the far end of the pencil) must be formed at C itself. Let's test this.
The simplest point where an object and image can touch is the center of curvature, C (\(u=-2f, v=-2f\)). Let's assume one end of the pencil is at C.


Step 3: Detailed Explanation:

Let the far end of the pencil be at the center of curvature, C.

Object distance for the far end, \(u_{far} = -2f\).
The pencil has length \(L=f/4\) and lies along the principal axis. So, the near end is closer to the mirror.
Object distance for the near end, \(u_{near} = -2f + f/4 = -7f/4\).

This places the entire pencil between C (\(-2f\)) and F (\(-f\)), which is the correct region for a real, enlarged image.

Now, let's find the image positions for both ends:

Image of the far end (\(u_{far} = -2f\)): The image is formed at \(v_{far} = -2f\).
Image of the near end (\(u_{near} = -7f/4\)): Using the mirror formula:
\[ \frac{1}{v_{near}} + \frac{1}{-7f/4} = -\frac{1}{f} \]
\[ \frac{1}{v_{near}} = \frac{4}{7f} - \frac{1}{f} = \frac{4-7}{7f} = -\frac{3}{7f} \]
\[ v_{near} = -\frac{7f}{3} \]

The image of the pencil extends from \(v_{far} = -2f\) to \(v_{near} = -7f/3\).
The object (pencil) extends from \(u_{near} = -7f/4\) to \(u_{far} = -2f\).
The point \(-2f\) (the center of curvature) is common to both the object and the image. Thus, the image "just touches" the pencil at this point. This configuration satisfies all the given conditions.

The question asks for "the magnification". Since the object is extended along the axis, the lateral magnification is different for each point. A common interpretation is to ask for the longitudinal magnification, or the lateral magnification of the end not at C. Let's calculate the lateral magnification of the near end. \[ m_{near} = -\frac{v_{near}}{u_{near}} = -\frac{(-7f/3)}{(-7f/4)} = -\frac{4}{3} \]
The magnitude of this magnification is \(|m| = 4/3 \approx 1.33\). Since it's greater than 1, the image is enlarged. The negative sign indicates it's inverted (though for an object on the axis, "inverted" is not visually meaningful).

If longitudinal magnification (\(m_L\)) is asked: \[ m_L = \frac{Length of image}{Length of object} = \frac{|v_{near} - v_{far}|}{|u_{near} - u_{far}|} = \frac{|-7f/3 - (-2f)|}{|-7f/4 - (-2f)|} = \frac{|-f/3|}{|f/4|} = \frac{f/3}{f/4} = \frac{4}{3} \]
Both interpretations lead to the value 4/3.

Step 4: Final Answer:

The magnification produced is \(\frac{4}{3}\).
Quick Tip: For objects extended along the principal axis, the image length and magnification are found by calculating the image positions of the two endpoints separately using the mirror formula. The condition "image touches object" often simplifies the problem by placing one end at a special point like the center of curvature.

Step 1: Understanding the Concept:

This problem involves the refraction of light through a prism. We are given the angle of incidence, angle of deviation, and the prism angle. We need to find the angle of emergence to prove the condition of normal emergence and then use Snell's law to find the refractive index.


Step 2: Key Formula or Approach:

The key formulas for a prism are:

Prism angle relation: \(A = r_1 + r_2\)
Angle of deviation relation: \(\delta = (i + e) - A\)
Snell's Law at the two faces: \(n_1 \sin i = n_2 \sin r_1\) and \(n_2 \sin r_2 = n_1 \sin e\)

where \(i\) is the angle of incidence, \(e\) is the angle of emergence, \(r_1\) and \(r_2\) are the angles of refraction at the first and second faces respectively, \(A\) is the prism angle, \(\delta\) is the angle of deviation, and \(n_1, n_2\) are the refractive indices of the surrounding medium and the prism.


Step 3: Detailed Explanation:

Given values:

Angle of incidence, \(i = 45^\circ\)
Angle of deviation, \(\delta = 15^\circ\)
Angle of prism, \(A = 30^\circ\)
Surrounding medium is assumed to be air, so \(n_1 = 1\). Let prism refractive index be \(n_2 = n\).


Part 1: Show that the emergent ray is normal to face AC.

To show that the emergent ray is normal (perpendicular) to the face AC, we need to prove that the angle of emergence \(e = 0^\circ\).
Using the angle of deviation formula: \[ \delta = (i + e) - A \]
Substitute the known values: \[ 15^\circ = (45^\circ + e) - 30^\circ \] \[ 15^\circ = 15^\circ + e \] \[ e = 15^\circ - 15^\circ = 0^\circ \]
Since the angle of emergence \(e\) is \(0^\circ\), the emergent ray makes an angle of \(0^\circ\) with the normal to the face AC. This means the ray emerges along the normal, i.e., it is perpendicular to the face AC. This proves the statement.


Part 2: Find the refractive index (n) of the prism.

To find n, we apply Snell's Law at the first face (AB): \(1 \cdot \sin i = n \cdot \sin r_1\). We need to find the value of \(r_1\).
First, find \(r_2\). At the second face (AC), Snell's Law is \(n \sin r_2 = 1 \sin e\).
Since \(e = 0^\circ\), we have: \[ n \sin r_2 = \sin(0^\circ) = 0 \]
This implies \(\sin r_2 = 0\), which means \(r_2 = 0^\circ\).
Now, use the prism angle formula: \[ A = r_1 + r_2 \] \[ 30^\circ = r_1 + 0^\circ \] \[ r_1 = 30^\circ \]
Finally, apply Snell's Law at the first face (AB) with \(i = 45^\circ\) and \(r_1 = 30^\circ\): \[ 1 \cdot \sin(45^\circ) = n \cdot \sin(30^\circ) \] \[ \frac{1}{\sqrt{2}} = n \cdot \frac{1}{2} \]
Solving for n: \[ n = \frac{2}{\sqrt{2}} = \sqrt{2} \]
The refractive index of the material of the prism is \(\sqrt{2} \approx 1.414\).
Quick Tip: The condition of normal emergence (\(e=0\)) is a significant simplification in prism problems, as it immediately implies that \(r_2=0\). This allows you to find \(r_1\) directly from the prism angle \(A\), making the application of Snell's law straightforward.

Note: The slit separation is given as 1.0 nm, which is physically unrealistic for an optical experiment (it's smaller than an atom). A typical value is 1.0 mm. We will assume the value is a typo and proceed with d = 1.0 mm = \(1.0 \times 10^{-3\) m.

Step 1: Understanding the Concept:

This problem involves Young's Double-Slit Experiment (YDSE). We use the formula for the position of bright fringes. When two different wavelengths are used, their fringe patterns are superimposed. A point of coincidence is where a bright fringe from one pattern overlaps with a bright fringe from the other.


Step 2: Key Formula or Approach:

The distance (\(y_n\)) of the n-th bright fringe from the central maximum is given by: \[ y_n = \frac{n\lambda D}{d} \]
where n is an integer (0, 1, 2, ...), \(\lambda\) is the wavelength, D is the distance to the screen, and d is the slit separation.
For coincidence of bright fringes of two wavelengths \(\lambda_1\) and \(\lambda_2\), the position must be the same: \[ \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \implies n_1 \lambda_1 = n_2 \lambda_2 \]

Step 3: Detailed Explanation:

Given values:

\(\lambda_1 = 600 \, nm = 600 \times 10^{-9} \, m\)
\(\lambda_2 = 480 \, nm = 480 \times 10^{-9} \, m\)
\(D = 1.0 \, m\)
\(d = 1.0 \, mm = 1.0 \times 10^{-3} \, m\) (assumed value)


(1) Distance of the third bright fringe for \(\lambda_1 = 600\) nm:

Here, \(n = 3\) and \(\lambda = \lambda_1 = 600 \times 10^{-9}\) m. \[ y_3 = \frac{3 \times (600 \times 10^{-9} \, m) \times (1.0 \, m)}{1.0 \times 10^{-3} \, m} \] \[ y_3 = 1800 \times 10^{-6} \, m = 1.8 \times 10^{-3} \, m = 1.8 \, mm \]

(2) Least distance for coincidence:

We need to find the smallest integers \(n_1\) and \(n_2\) (other than zero) such that \(n_1 \lambda_1 = n_2 \lambda_2\). \[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{480 \, nm}{600 \, nm} = \frac{48}{60} = \frac{4}{5} \]
The least integer solution is \(n_1 = 4\) and \(n_2 = 5\). This means the 4th bright fringe of \(\lambda_1\) coincides with the 5th bright fringe of \(\lambda_2\).
Now, we calculate the distance (y) for this coincidence using either fringe:
Using \(\lambda_1\): \[ y = \frac{n_1 \lambda_1 D}{d} = \frac{4 \times (600 \times 10^{-9} \, m) \times (1.0 \, m)}{1.0 \times 10^{-3} \, m} \] \[ y = 2400 \times 10^{-6} \, m = 2.4 \times 10^{-3} \, m = 2.4 \, mm \]
(As a check, using \(\lambda_2\): \(y = \frac{5 \times (480 \times 10^{-9}) \times 1.0}{1.0 \times 10^{-3}} = 2400 \times 10^{-6} \, m = 2.4 \, mm\)).

Step 4: Final Answer:

(1) The distance of the third bright fringe for 600 nm is 1.8 mm.

(2) The least distance from the central maximum where the bright fringes coincide is 2.4 mm.
Quick Tip: For fringe coincidence problems, the key is the condition \(n_1 \lambda_1 = n_2 \lambda_2\). Find the simplest integer ratio \(n_1/n_2\) to find the first point of overlap (after the center).

(1) Single Slit Diffraction Pattern:

Intensity Graph:
The graph of intensity versus diffraction angle (\(\theta\)) has the following features:

A central bright maximum that is very intense and wide.
A series of secondary maxima on either side of the central one.
The intensity of the secondary maxima decreases rapidly as we move away from the center. The first secondary maximum has an intensity of only about 4.5% of the central maximum.
The width of the central maximum is twice the width of the secondary maxima.
Minima (points of zero intensity) are located between the maxima.

(A diagram showing a large central peak with much smaller, rapidly decaying side peaks would be drawn here).

Expression for Zero Intensity (Minima):

The condition for destructive interference, which leads to minima (zero intensity) in a single-slit diffraction pattern, is given by: \[ a \sin\theta = n\lambda \]
where:

'a' is the width of the slit.
'\(\theta\)' is the angle of diffraction.
'\(\lambda\)' is the wavelength of the light.
'n' is a non-zero integer (\(n = \pm 1, \pm 2, \pm 3, \ldots\)).

The value of the angle corresponding to these locations is: \[ \theta = \arcsin\left(\frac{n\lambda}{a}\right) \]

(2) Difference between Diffraction of Light and Sound:

The primary difference in the observable diffraction effects of light and sound in our daily lives stems from their vastly different wavelengths.

Condition for Diffraction: Significant diffraction (bending of waves) occurs only when the wavelength of the wave is comparable to or larger than the size of the obstacle or aperture (\(\lambda \gtrsim d\)).
Wavelengths:

Sound Waves: Have relatively long wavelengths, typically ranging from a few centimeters to several meters (e.g., for \(f = 340\) Hz, \(\lambda \approx 1\) m).
Light Waves: Have extremely short wavelengths, in the range of 400 nm to 700 nm (\(4 \times 10^{-7}\) m to \(7 \times 10^{-7}\) m).

Observable Effect:

Sound: The wavelength of sound is comparable to the size of everyday objects like doorways, corners of buildings, and furniture. As a result, sound waves readily diffract around these obstacles. This is why we can hear a person talking even if they are around a corner and not in our direct line of sight.
Light: The wavelength of light is thousands of times smaller than everyday objects. Therefore, light does not diffract noticeably around them and appears to travel in straight lines, casting sharp shadows. To observe the diffraction of light, we need very small obstacles or apertures, such as a narrow slit or a pinhole, with dimensions on the order of micrometers. Quick Tip: The rule of thumb for diffraction is: if the wavelength is about the same size as the opening, the wave will spread out significantly. If the wavelength is much smaller, it will pass through like a beam. This explains why we hear around corners but can't see around them.

Step 1: Understanding the Concept:

When two conductors are connected by a wire, they form a single conductor and thus become an equipotential body. Charge will redistribute itself until the potential is the same everywhere on both conductors. For a system of concentric spherical conductors, any charge placed inside will move to the outermost surface upon connection.


Step 2: Key Formula or Approach:

1. The potential of an isolated conducting sphere of radius 'r' and charge 'q' is \(V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}\).
2. When connected, the two spheres A and B act as a single conductor. The entire charge resides on the outer surface, which is the surface of shell B.
3. The potential of this combined conductor will be the potential of a sphere of radius R carrying the total charge.


Step 3: Detailed Explanation:

Initial State:
Sphere A of radius r has a potential V. Let's assume this is its potential in isolation. The charge on sphere A is \(q_A\). \[ V = \frac{1}{4\pi\epsilon_0} \frac{q_A}{r} \implies q_A = 4\pi\epsilon_0 r V \]
Shell B is initially uncharged (this is a reasonable assumption if not stated otherwise). The total charge of the system is \(q_{total} = q_A\).


Final State:
When sphere A and shell B are connected by a wire, charge flows until they reach a common potential, \(V_{final}\). Since A is inside B, the charge \(q_A\) will flow from the inner sphere to the outer shell. This is because charges on a conductor always reside on its outermost surface to minimize potential energy. The final charge distribution will be:

Final charge on sphere A, \(q'_A = 0\).
Final charge on shell B, \(q'_B = q_A\).

The entire system (A + wire + B) is now at a single potential, which is determined by the charge on the outer surface (radius R).
The final potential \(V_{final}\) is: \[ V_{final} = \frac{1}{4\pi\epsilon_0} \frac{q_{total}}{R} = \frac{1}{4\pi\epsilon_0} \frac{q_A}{R} \]
Now, substitute the expression for \(q_A\) from the initial state: \[ V_{final} = \frac{1}{4\pi\epsilon_0} \frac{4\pi\epsilon_0 r V}{R} \] \[ V_{final} = \frac{rV}{R} \]
Since they are connected, the final potential on sphere A is the same as the final potential on shell B. \[ V_{final, A} = V_{final, B} = \frac{rV}{R} \]

Step 4: Final Answer:

The final potential on both sphere A and shell B is the same and is equal to \(\frac{rV}{R}\).
Quick Tip: A key principle of electrostatics: when an inner conductor is connected to an enclosing outer conductor, all of the inner conductor's charge flows to the outer one. This principle is used in devices like the Van de Graaff generator.

Step 1: Characteristics of Equipotential Surfaces:

An equipotential surface is a surface on which the electric potential is the same at all points. Two of its key characteristics are:

No work is done in moving a test charge from one point to another on the same equipotential surface. This is because the potential difference between any two points on the surface is zero (\(W = q\Delta V = 0\)).
The electric field lines are always perpendicular to the equipotential surfaces at every point. If the field had a component along the surface, it would do work on a charge moving along the surface, which contradicts the definition.
(Another characteristic) Two equipotential surfaces can never intersect each other. If they did, there would be two different values of potential at the point of intersection, which is not possible.


Step 2: Potential Calculation:

Concept: The relationship between potential difference and a uniform electric field is given by \(V_B - V_A = -\vec{E} \cdot \vec{r}_{AB}\), where \(\vec{r}_{AB}\) is the displacement vector from point A to point B.

Given values:

Electric field, \(\vec{E} = 50 \, \hat{i} \, N/C\) (along +x axis).
Point A is the origin (0,0), with potential \(V_A = 220 \, V\).
Point B is the point P(4m, 3m).


Calculation:
The displacement vector from the origin (A) to the point P (B) is: \[ \vec{r}_{AB} = \vec{r}_B - \vec{r}_A = (4\hat{i} + 3\hat{j}) - (0\hat{i} + 0\hat{j}) = 4\hat{i} + 3\hat{j} \, m \]
Now, use the formula for potential difference: \[ V_P - V_{origin} = - \vec{E} \cdot \vec{r}_{AB} \] \[ V_P - 220 \, V = - (50 \, \hat{i}) \cdot (4\hat{i} + 3\hat{j}) \]
Calculate the dot product: \[ (50 \, \hat{i}) \cdot (4\hat{i} + 3\hat{j}) = (50 \times 4)(\hat{i} \cdot \hat{i}) + (50 \times 3)(\hat{i} \cdot \hat{j}) \]
Since \(\hat{i} \cdot \hat{i} = 1\) and \(\hat{i} \cdot \hat{j} = 0\): \[ (50 \, \hat{i}) \cdot (4\hat{i} + 3\hat{j}) = 200 \]
Substitute this back into the potential equation: \[ V_P - 220 = -200 \] \[ V_P = 220 - 200 = 20 \, V \]

Step 3: Final Answer:

The potential at the point (4m, 3m) is 20 V.
Quick Tip: In a uniform electric field pointing along the x-axis, the equipotential surfaces are planes parallel to the y-z plane. The potential only changes as you move along the x-direction. The change in potential is simply \( \Delta V = -E \Delta x \). In this problem, \(\Delta x = 4\) m, so \(\Delta V = -50 \times 4 = -200\) V. The final potential is \(220 - 200 = 20\) V.

Step 1: Difference between Open and Closed Surfaces:


Open Surface: An open surface is a surface that has a boundary or an edge. It does not enclose a volume. Imagine a leaf or a flat sheet of paper; they have edges. Other examples include a disc, a hemispherical bowl without its lid, or a cylindrical surface without its top and bottom caps.
Closed Surface: A closed surface is a surface that is continuous, has no boundary, and completely encloses a region of space (a volume). There is a distinct "inside" and "outside" to a closed surface. Examples include the surface of a sphere, a cube, or a complete cylinder with its top and bottom lids. Closed surfaces are fundamental to Gauss's Law in electrostatics.


Step 2: Elementary Surface Vector \(d\vec{S}\) for a Spherical Surface:

The elementary surface vector \(d\vec{S}\) represents an infinitesimally small patch of area on a surface. It is a vector quantity with two properties:

Magnitude: Its magnitude, \(dS = |d\vec{S}|\), is equal to the area of the infinitesimal patch.
Direction: Its direction is defined to be perpendicular (normal) to the surface at that point. For a closed surface like a sphere, the direction is conventionally chosen to point outward from the enclosed volume.


Diagram:
(A diagram of a sphere is drawn with its center at the origin. A small patch of area \(dS\) is shown on its surface. A vector labeled \(d\vec{S}\) is drawn starting from this patch, pointing radially outwards, away from the center of the sphere. The vector is shown to be perpendicular to the tangent plane at that point.)
Quick Tip: The concept of a "closed surface" is crucial for Gauss's Law, which relates the flux "out of" a volume to the charge "inside" that volume. An open surface doesn't have a clear inside/outside, so Gauss's Law in its integral form doesn't apply to it.

1. Definition of Electric Flux:

Electric flux (\(\Phi_E\)) is a measure of the flow of the electric field through a given surface. It quantifies the total number of electric field lines passing perpendicularly through that surface. For a uniform electric field \(\vec{E}\) passing through a planar area \(\vec{A}\), the flux is \(\Phi_E = \vec{E} \cdot \vec{A}\). More generally, for a non-uniform field and a curved surface, the electric flux is defined as the surface integral of the electric field over that surface: \[ \Phi_E = \int_S \vec{E} \cdot d\vec{S} \]

2. Significance of a Gaussian Surface:

A Gaussian surface is an imaginary closed surface chosen to evaluate the electric flux in a region of space. Its primary significance lies in its application with Gauss's Law (\(\Phi_E = q_{enc}/\epsilon_0\)). By choosing a Gaussian surface that has the same symmetry as the charge distribution (e.g., a sphere around a point charge, a cylinder around a line charge), the calculation of the electric field becomes vastly simplified. The surface integral for flux can often be reduced to a simple algebraic expression (\(E \times Area\)), allowing for an easy calculation of E. It is a powerful mathematical tool for calculating electric fields in situations with high symmetry.


3. Why an External Charge does not Contribute to Total Flux:

A charge placed outside a closed Gaussian surface does not contribute to the net electric flux through that surface. This is because:

The electric field lines from an external positive charge radiate outwards.
Any field line that enters the closed surface from the outside must also exit the surface at some other point.
At the point of entry, the angle between the electric field vector \(\vec{E\) and the outward area vector \(d\vec{S}\) is obtuse (greater than 90°), so the dot product \(\vec{E} \cdot d\vec{S}\) is negative. This is considered an inward (negative) flux.
At the point of exit, the angle between \(\vec{E}\) and \(d\vec{S}\) is acute (less than 90°), so the dot product \(\vec{E} \cdot d\vec{S}\) is positive. This is an outward (positive) flux.
For any field line from an external charge, the negative flux on entry is exactly cancelled by the positive flux on exit.
When integrated over the entire closed surface, the total flux due to the external charge is zero. This is a direct consequence of Gauss's Law, which states that the net flux depends only on the \textit{enclosed charge. Quick Tip: Think of flux as a "flow". For a closed surface with no sources or sinks inside (no enclosed charge), whatever flows in must flow out. The net flow (total flux) is zero. An enclosed charge acts as a source (positive charge) or sink (negative charge), creating a net outflow or inflow.

Step 1: Understanding the Concept:

This problem requires the application of Gauss's Law for electric flux. Gauss's Law states that the total electric flux (\(\Phi_E\)) through any closed surface (Gaussian surface) is equal to \(1/\epsilon_0\) times the net electric charge (\(q_{enclosed}\)) enclosed by the surface.


Step 2: Key Formula or Approach:

Gauss's Law: \[ \Phi_E = \oint \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\epsilon_0} \]
We will apply this law to both spherical shells, S\(_1\) and S\(_2\).


Step 3: Detailed Explanation:

Flux through surface S\(_1\):
The flux through S\(_1\), denoted as \(\Phi_1\), depends only on the total charge enclosed within S\(_1\). \[ q_{enclosed, S_1} = q_1 + q_2 + q_3 \] \[ q_{enclosed, S_1} = (-3 \, \mu C) + (-2 \, \mu C) + (9 \, \mu C) = 4 \, \mu C \]
According to Gauss's Law: \[ \Phi_1 = \frac{q_{enclosed, S_1}}{\epsilon_0} = \frac{4 \, \mu C}{\epsilon_0} \]

Flux through surface S\(_2\):
The flux through S\(_2\), denoted as \(\Phi_2\), depends on the total charge enclosed within S\(_2\). This includes all the charges inside S\(_1\) as well as the charge Q placed between the shells. \[ q_{enclosed, S_2} = (q_1 + q_2 + q_3) + Q = q_{enclosed, S_1} + Q \] \[ q_{enclosed, S_2} = (4 \, \mu C) + Q \]
According to Gauss's Law: \[ \Phi_2 = \frac{q_{enclosed, S_2}}{\epsilon_0} = \frac{(4 \, \mu C + Q)}{\epsilon_0} \]

Using the given relation:
We are given that the flux through S\(_2\) is four times the flux through S\(_1\). \[ \Phi_2 = 4 \Phi_1 \]
Substitute the expressions for \(\Phi_1\) and \(\Phi_2\): \[ \frac{(4 \, \mu C + Q)}{\epsilon_0} = 4 \times \left( \frac{4 \, \mu C}{\epsilon_0} \right) \]
The \(\epsilon_0\) term cancels out from both sides: \[ 4 \, \mu C + Q = 16 \, \mu C \]
Solve for Q: \[ Q = 16 \, \mu C - 4 \, \mu C = 12 \, \mu C \]

Step 4: Final Answer:

The value of charge Q is 12 \(\mu\)C.
Quick Tip: In problems involving concentric Gaussian surfaces, remember that the flux through an outer surface is determined by the total charge inside it, which includes all charges contained within any inner surfaces.