Step 1: Understanding the Concept:

This question is based on the principle of superposition of electric fields. The net electric field at a point due to a system of charges is the vector sum of the electric fields at that point due to each individual charge.


Step 2: Detailed Explanation:

Let's analyze the electric field at point C due to the two charges at vertices A and B.

1. Electric Field due to charge at A (\(E_{CA}\)): A charge of -q is placed at A. Since the charge is negative, the electric field at point C will be attractive, meaning it will be directed from C towards A, i.e., along the line CA.

2. Electric Field due to charge at B (\(E_{CB}\)): Similarly, a charge of -q is placed at B. The electric field at C due to this charge will also be attractive and directed from C towards B, i.e., along the line CB.

3. Magnitude of Fields: Since ABC is an equilateral triangle, the distance AC is equal to the distance BC. As the magnitude of the charges at A and B is the same (q), the magnitudes of the electric fields will be equal: \(|E_{CA}| = |E_{CB}|\).

4. Resultant Field: The net electric field at C (\(E_{net}\)) is the vector sum of \(E_{CA}\) and \(E_{CB}\). Since the two vectors have equal magnitude, their resultant will lie along the angle bisector of the angle between them, which is \(\angle ACB\).

5. Geometry of Equilateral Triangle: In an equilateral triangle, the angle bisector of a vertex angle is also the median to the opposite side. Therefore, the angle bisector of \(\angle ACB\) is the line segment CM, where M is the midpoint of AB.

The resultant vector will point from C towards M.


Step 3: Final Answer:

The net electric field at C is the vector sum of the fields along CA and CB. Due to the symmetry of the equilateral triangle and equal charges, the resultant field points along the median CM.
Quick Tip: For symmetry problems involving geometric shapes like squares or equilateral triangles, always look for cancellations or resultant directions along axes of symmetry. This can save significant calculation time.

Step 1: Understanding the Concept:

This problem involves calculating the equivalent resistance of different combinations of resistors. The two basic combinations are series and parallel.

- Series Combination: \(R_{eq} = R_1 + R_2 + \dots\)

- Parallel Combination: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots\)


Step 2: Detailed Explanation:

Let's evaluate the equivalent resistance for each option given. We have three resistors, each with resistance R. The target equivalent resistance is \(\frac{2}{3}R\).


(A) All three resistors in series:

The equivalent resistance \(R_{eq}\) would be the sum of the individual resistances.
\[ R_{eq} = R + R + R = 3R \]
This is not equal to \(\frac{2}{3}R\).


(B) All three resistors in parallel:

The reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.
\[ \frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \] \[ \implies R_{eq} = \frac{R}{3} \]
This is not equal to \(\frac{2}{3}R\).


(C) Two resistors in series and then this combination in parallel with the third resistor:

First, find the resistance of the two resistors in series, let's call it \(R_s\).
\[ R_s = R + R = 2R \]
Now, this combination \(R_s\) is connected in parallel with the third resistor R.
\[ \frac{1}{R_{eq}} = \frac{1}{R_s} + \frac{1}{R} = \frac{1}{2R} + \frac{1}{R} = \frac{1 + 2}{2R} = \frac{3}{2R} \] \[ \implies R_{eq} = \frac{2R}{3} \]
This matches the required resistance.


(D) Two resistors in parallel and then this combination in series with the third resistor:

First, find the resistance of the two resistors in parallel, let's call it \(R_p\).
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_p = \frac{R}{2} \]
Now, this combination \(R_p\) is connected in series with the third resistor R.
\[ R_{eq} = R_p + R = \frac{R}{2} + R = \frac{3R}{2} \]
This is not equal to \(\frac{2}{3}R\).


Step 3: Final Answer:

The only configuration that yields an equivalent resistance of \(\frac{2}{3}R\) is connecting two resistors in series and then connecting that combination in parallel with the third resistor.
Quick Tip: In multiple-choice questions about resistor combinations, you can often eliminate options by estimation. For instance, connecting resistors in series always increases the total resistance above the largest individual resistance, while parallel connections always decrease it below the smallest.

Step 1: Understanding the Concept:

The magnetic field produced by a small current-carrying element is described by the Biot-Savart Law. This law relates the magnetic field to the magnitude, direction, length of the current element, and the distance to the point of observation.


Step 2: Key Formula or Approach:

The Biot-Savart Law in vector form is: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3} \]
where:

\(d\vec{B}\) is the magnetic field vector.
\(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7}\) T·m/A).
\(I\) is the current.
\(d\vec{l}\) is the length vector of the current element.
\(\vec{r}\) is the position vector from the current element to the point where the field is being calculated.
\(r\) is the magnitude of \(\vec{r}\).


Step 3: Detailed Explanation:

Given data:

Length of the segment, \(dl = 1 cm = 0.01 m\). Since it's in the x-direction, \(d\vec{l} = 0.01 \hat{i}\) m.
Current, \(I = 1\) A.
The element is at the origin (0, 0, 0).
The point P is at (1m, 1m, 0).

First, we find the position vector \(\vec{r}\) from the origin to point P.
\[ \vec{r} = (1-0)\hat{i} + (1-0)\hat{j} + (0-0)\hat{k} = 1\hat{i} + 1\hat{j} \]
The magnitude of \(\vec{r}\) is: \[ r = |\vec{r}| = \sqrt{1^2 + 1^2} = \sqrt{2} m \]
Next, we calculate the cross product \(d\vec{l} \times \vec{r}\): \[ d\vec{l} \times \vec{r} = (0.01 \hat{i}) \times (1\hat{i} + 1\hat{j}) \]
Using the properties of cross products (\(\hat{i} \times \hat{i} = 0\) and \(\hat{i} \times \hat{j} = \hat{k}\)): \[ d\vec{l} \times \vec{r} = (0.01 \times 1)(\hat{i} \times \hat{i}) + (0.01 \times 1)(\hat{i} \times \hat{j}) = 0 + 0.01 \hat{k} = 0.01 \hat{k} \]
Now, substitute these values into the Biot-Savart Law. Since the segment is small, we can approximate the field using \(\Delta\vec{B}\) instead of integrating. \[ \vec{B} \approx \frac{\mu_0}{4\pi} \frac{I (\Delta\vec{l} \times \vec{r})}{r^3} \] \[ \vec{B} = \frac{4\pi \times 10^{-7}}{4\pi} \frac{1 \times (0.01 \hat{k})}{(\sqrt{2})^3} \] \[ \vec{B} = 10^{-7} \frac{0.01 \hat{k}}{2\sqrt{2}} = \frac{10^{-9}}{2\sqrt{2}} \hat{k} T \]
The options provide the magnitude of the magnetic field, so we take the magnitude of our result: \[ |\vec{B}| = \frac{10^{-9}}{2\sqrt{2}} = \frac{1 \times 10 \times 10^{-10}}{2\sqrt{2}} = \frac{5}{\sqrt{2}} \times 10^{-10} T \]

Step 4: Final Answer:

The magnitude of the magnetic field is \(\frac{5.0}{\sqrt{2}} \times 10^{-10}\) T. This matches option (C).
Quick Tip: When using the Biot-Savart Law, pay close attention to the vector cross product. The direction of the resulting magnetic field is always perpendicular to both the current element vector (\(d\vec{l}\)) and the position vector (\(\vec{r}\)). The Right-Hand Rule is a quick way to determine this direction.

Step 1: Understanding the Concept:

This question asks for the standard formula for the magnetic field on the axial line of a magnetic dipole (like a short bar magnet). This formula is analogous to the electric field on the axis of an electric dipole.


Step 2: Key Formula or Approach:

The magnetic field (\(B_{axial}\)) at a point on the axis of a magnetic dipole with moment M, at a distance r from its center, is a standard result derived from the principles of magnetism.

The formula is: \[ B_{axial} = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \]
For comparison, the magnetic field on the equatorial line is: \[ B_{equatorial} = \frac{\mu_0}{4\pi} \frac{M}{r^3} \]
This shows that the axial field is twice as strong as the equatorial field at the same distance.


Step 3: Detailed Explanation:

Let's analyze the given options based on the standard formula.

(A) \(\frac{\mu_0}{4\pi} \frac{2M}{r^3}\): This is the correct formula for the magnetic field on the axial line of a small magnetic dipole.

(B) \(\frac{\mu_0}{4\pi} \frac{M}{r^3}\): This is the formula for the magnetic field on the equatorial line, not the axial line.

(C) \(\frac{\mu_0}{4\pi} \frac{M}{2r^3}\): This is dimensionally correct but numerically incorrect.

(D) \(\frac{\mu_0}{4\pi} \frac{2M}{r^2}\): This is incorrect. The magnetic field of a dipole falls off with the cube of the distance (\(1/r^3\)), not the square (\(1/r^2\)). An inverse square law applies to monopoles, which are not known to exist in magnetism.


Step 4: Final Answer:

The correct expression for the magnetic field along the axis of a small magnetic dipole is given by option (A).
Quick Tip: Remember the analogy between electrostatics and magnetism. An electric dipole's axial field is \( \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} \), and a magnetic dipole's axial field is \( \frac{\mu_0}{4\pi} \frac{2M}{r^3} \). The structure of the formulas is identical, which can help in memorization.

Step 1: Understanding the Concept:

This problem is an application of Faraday's Law of Induction and Lenz's Law. Lenz's Law states that the direction of the induced current in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes the change in the initial magnetic flux.


Step 2: Detailed Explanation:

Let's assume the permanent magnet has its North pole on the left (facing the coil) and the South pole on the right. The magnetic field lines emerge from the North pole.


Part 1: Entry of the Magnet

1. As the magnet enters the coil from the right, the North pole approaches the face Y of the coil.

2. The magnetic flux directed to the left through the coil increases.

3. According to Lenz's Law, the induced current will create a magnetic field to oppose this increase. It will create a magnetic field directed to the right.

4. For the coil to produce a magnetic field to the right, its face Y must become a North pole (to repel the incoming North pole).

5. Using the Right-Hand Grip Rule, if you curl your fingers in the direction of the current, your thumb points in the direction of the North pole. To make face Y a North pole, the current in the top wire (segment YZ) must flow from Z to Y.


Part 2: Exit of the Magnet

1. As the magnet exits the coil from the left, the North pole moves away from the face X of the coil.

2. The magnetic flux directed to the left through the coil is now decreasing.

3. According to Lenz's Law, the induced current will create a magnetic field to oppose this decrease. It will try to maintain the flux by creating a magnetic field directed to the left.

4. For the coil to produce a magnetic field to the left, its face X must become a North pole and its face Y must become a South pole.

5. Using the Right-Hand Grip Rule again, to make face X a North pole, the current in the top wire (segment YZ) must flow from Y to Z.


Step 3: Final Answer:

During entry, the current flows from Z to Y. During exit, the current flows from Y to Z. Therefore, the correct sequence is Z to Y and then Y to Z.
Quick Tip: A simple way to remember Lenz's Law is "the coil opposes the motion". When the magnet approaches, the coil repels it (like poles face each other). When the magnet recedes, the coil attracts it (opposite poles face each other). Use this to determine the required polarity of the coil face and then apply the right-hand rule to find the current direction.

Step 1: Understanding the Concept:

The relationship between the voltage (V) across an ideal inductor and the current (I) flowing through it is given by the formula involving the inductance (L) and the rate of change of current with respect to time (\(dI/dt\)).


Step 2: Key Formula or Approach:

The voltage across an inductor is given by: \[ V = L \frac{dI}{dt} \]
This means the voltage waveform is proportional to the slope (or derivative) of the current waveform.


Step 3: Detailed Explanation:

Let's analyze the given current (I) vs. time (t) graph in two parts:


1. Interval from \(t=0\) to \(t=T/2\):

- The graph of I vs. t is a straight line with a constant positive slope. The current increases linearly from zero to a maximum value.

- Since the slope \(\frac{dI}{dt}\) is a positive constant, the voltage \(V = L \frac{dI}{dt}\) must also be a positive constant during this interval.


2. Interval from \(t=T/2\) to \(t=T\):

- The graph of I vs. t is a straight line with a constant negative slope. The current decreases linearly from the maximum value back to zero.

- Since the slope \(\frac{dI}{dt}\) is a negative constant, the voltage \(V = L \frac{dI}{dt}\) must be a negative constant during this interval.


Matching with the options:

- The resulting voltage waveform is a constant positive value for the first half of the cycle and a constant negative value for the second half. This is a square or rectangular wave.

- Option (A) is a triangle, incorrect.

- Option (B) is a square wave, but it starts with a negative voltage, which is incorrect.

- Option (C) is a square wave that starts with a positive voltage and then becomes negative, which perfectly matches our analysis.

- Option (D) is a sinusoidal wave, incorrect.


Step 4: Final Answer:

The voltage waveform is the derivative of the triangular current waveform, resulting in a square wave that is positive for the first half-period and negative for the second. This corresponds to graph (C).
Quick Tip: Remember the relationship for basic circuit elements: Resistor: \(V = IR\) (V is proportional to I). Inductor: \(V = L \frac{dI}{dt}\) (V is proportional to the slope of I). Capacitor: \(I = C \frac{dV}{dt}\) (I is proportional to the slope of V). Knowing these helps to quickly determine the shape of one waveform from the other.

Step 1: Understanding the Concept:

This question deals with the principle of a transformer. An ideal transformer changes AC voltage levels according to the ratio of the number of turns in its primary and secondary coils.


Step 2: Key Formula or Approach:

The transformer equation relates the voltages and the number of turns in the primary and secondary coils: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
where:

\(V_p\) and \(V_s\) are the primary and secondary voltages, respectively.
\(N_p\) and \(N_s\) are the number of turns in the primary and secondary coils, respectively.


Step 3: Detailed Explanation:

We are given the following values:

Primary voltage, \(V_p = 200\) V.
Secondary voltage, \(V_s = 3000\) V.
Number of turns in the primary coil, \(N_p = 450\).

We need to find the number of turns in the secondary coil, \(N_s\).

Rearranging the transformer equation to solve for \(N_s\): \[ N_s = N_p \times \frac{V_s}{V_p} \]
Substituting the given values into the formula: \[ N_s = 450 \times \frac{3000 V}{200 V} \] \[ N_s = 450 \times 15 \]
Now, we perform the multiplication: \[ N_s = 6750 \]

Step 4: Final Answer:

The number of turns in the secondary coil is 6750. This is a step-up transformer as the secondary voltage is higher than the primary voltage, and consequently, \(N_s > N_p\).
Quick Tip: Before calculating, quickly identify if it's a step-up (voltage increases) or step-down (voltage decreases) transformer. For a step-up transformer, \(N_s\) must be greater than \(N_p\). For a step-down, \(N_s\) must be less than \(N_p\). This helps in eliminating incorrect options. Here, voltage goes from 200V to 3000V (step-up), so \(N_s\) must be greater than 450.

Step 1: Understanding the Concept:

This question tests the fundamental properties of the electrostatic field, specifically its conservative nature and the geometry of its field lines.


Step 2: Detailed Explanation:

Let's analyze the two properties mentioned:


1. Conservative Nature:

An electric field is said to be conservative if the work done by the field in moving a charge between two points is independent of the path taken. For an electrostatic field \(\vec{E}\), the line integral of \(\vec{E} \cdot d\vec{l}\) around any closed path is zero (\(\oint \vec{E} \cdot d\vec{l} = 0\)). This is a defining characteristic of a conservative field. The electric field produced by static (stationary) charges is always conservative.


2. Electric Field Lines:

Electric field lines are imaginary lines used to represent the direction and strength of an electric field. By convention:

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 they did, it would imply that the work done in a closed path is not zero, which would violate the conservative nature of the electrostatic field.


Evaluating the Options:

(A) conservative and field lines do not form closed loops: This statement is correct. Both properties accurately describe the electrostatic field.

(B) conservative and field lines form closed loops: This is incorrect because conservative fields cannot have closed-loop field lines.

(C) non-conservative and field lines do not form closed loops: This is incorrect because the electrostatic field is conservative.

(D) non-conservative and field lines form closed loops: This statement describes the induced electric field created by a changing magnetic field, not the field from static charges.


Step 3: Final Answer:

The electric field due to static charges is a conservative field, and its field lines start on positive charges and end on negative charges, never forming closed loops.
Quick Tip: Contrast the electrostatic field with the magnetic field. Magnetic field lines always form closed loops (as there are no magnetic monopoles), and the induced electric field (from changing magnetic flux) is non-conservative and also has closed-loop field lines. Keeping these distinctions clear is key.

Step 1: Understanding the Concept:

This problem relates the concepts of real depth, apparent depth, and the refractive index of a medium. The refractive index also connects the speed of light in a vacuum to its speed in the medium.


Step 2: Key Formula or Approach:

There are two key formulas needed:
1. The refractive index (\(\mu\)) of a medium in terms of real and apparent depth: \[ \mu = \frac{Real Depth}{Apparent Depth} \]
2. The refractive index (\(\mu\)) in terms of the speed of light: \[ \mu = \frac{Speed of light in vacuum (c)}{Speed of light in the liquid (v)} \]
The value of \(c\) is approximately \(3.0 \times 10^8 m/s\).


Step 3: Detailed Explanation:

Part 1: Calculate the refractive index of the liquid.

We are given:

Real Depth = 30.0 cm
Apparent Depth = 16.0 cm

Using the first formula: \[ \mu = \frac{30.0 cm}{16.0 cm} = \frac{30}{16} = \frac{15}{8} \]
So, the refractive index of the liquid is 1.875.


Part 2: Calculate the speed of light in the liquid.

Now we use the second formula, rearranging it to solve for v: \[ v = \frac{c}{\mu} \]
Substitute the values of c and \(\mu\): \[ v = \frac{3.0 \times 10^8 m/s}{15/8} \] \[ v = \frac{3.0 \times 10^8 \times 8}{15} \] \[ v = \frac{24.0 \times 10^8}{15} \]
To simplify the fraction \(\frac{24}{15}\), divide both by 3: \(\frac{8}{5}\). \[ v = \frac{8}{5} \times 10^8 = 1.6 \times 10^8 m/s \]

Step 4: Final Answer:

The speed of light in the liquid is \(1.6 \times 10^8 m s^{-1}\).
Quick Tip: Always remember that light slows down in a denser medium, so the speed of light in any transparent material will be less than \(c = 3.0 \times 10^8 m/s\). This allows you to immediately eliminate option (C) and any other option greater than c.

Step 1: Understanding the Concept:

This question combines the concepts of the hydrogen atomic spectrum and the photoelectric effect. For photoelectrons to be emitted, the energy of the incident photons must be greater than or equal to the work function of the metal surface (in this case, zinc).


Step 2: Key Formula or Approach:

1. Photoelectric Effect Condition: \(E_{photon} \ge \phi\), where \(\phi\) is the work function. The work function of Zinc (\(\phi_{Zn}\)) is approximately 4.3 eV.

2. Energy of Hydrogen Spectral Lines: The energy of a photon emitted when an electron transitions from a higher energy level \(n_2\) to a lower level \(n_1\) is given by the Rydberg formula in terms of energy: \[ E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) eV \]

Step 3: Detailed Explanation:

We need to find the energy range for each spectral series of hydrogen and compare it with the work function of zinc (\(\phi_{Zn} \approx 4.3 eV\)).


1. Lyman Series (\(n_1 = 1\)):

The transitions are from \(n_2 = 2, 3, 4, \dots\) to \(n_1 = 1\).
- Minimum energy (for \(n_2 = 2 \to n_1 = 1\)):
\[ E_{min} = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times \frac{3}{4} = 10.2 eV \]
- Maximum energy (for \(n_2 = \infty \to n_1 = 1\)):
\[ E_{max} = 13.6 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = 13.6 eV \]
The energy range for the Lyman series is [10.2 eV, 13.6 eV]. Since every photon in this series has an energy greater than 4.3 eV, all Lyman series lines will cause photoemission.


2. Balmer Series (\(n_1 = 2\)):

The transitions are from \(n_2 = 3, 4, 5, \dots\) to \(n_1 = 2\).
- Minimum energy (for \(n_2 = 3 \to n_1 = 2\)):
\[ E_{min} = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \times \frac{5}{36} \approx 1.89 eV \]
- Maximum energy (for \(n_2 = \infty \to n_1 = 2\)):
\[ E_{max} = 13.6 \left( \frac{1}{2^2} \right) = \frac{13.6}{4} = 3.4 eV \]
The energy range for the Balmer series is [1.89 eV, 3.4 eV]. All these energies are less than 4.3 eV. No photoemission will occur.


3. Paschen Series (\(n_1 = 3\)):

The transitions are from \(n_2 = 4, 5, 6, \dots\) to \(n_1 = 3\).
- Maximum energy (for \(n_2 = \infty \to n_1 = 3\)):
\[ E_{max} = 13.6 \left( \frac{1}{3^2} \right) = \frac{13.6}{9} \approx 1.51 eV \]
All energies in the Paschen series are less than 1.51 eV, which is well below 4.3 eV. No photoemission will occur.


Step 4: Final Answer:

Only the photons from the Lyman series have sufficient energy (\(>4.3\) eV) to eject photoelectrons from a zinc surface.
Quick Tip: Remember the energy hierarchy of the hydrogen series: Lyman (UV, highest energy), Balmer (Visible, medium energy), Paschen (Infrared, lower energy). For photoelectric effect questions, you often only need to calculate the minimum energy of the highest-energy series (Lyman) and the maximum energy of the next series (Balmer) to find the threshold.

Step 1: Understanding the Concept:

The energy of an electron in the \(n^{th}\) orbit of a hydrogen atom is quantized and is given by a specific formula related to the ground state energy.


Step 2: Key Formula or Approach:

The energy \(E_n\) of an electron in the orbit with principal quantum number \(n\) is given by: \[ E_n = \frac{E_1}{n^2} \]
where \(E_1\) is the energy of the electron in the ground state (\(n=1\)).


Step 3: Detailed Explanation:

We are given:

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 \(n\).

Using the formula: \[ -0.544 = \frac{-13.6}{n^2} \]
Rearranging the equation to solve for \(n^2\): \[ n^2 = \frac{-13.6}{-0.544} \] \[ n^2 = \frac{13.6}{0.544} = \frac{13600}{544} \]
To simplify the fraction, we can notice that \(136 \times 4 = 544\). \[ n^2 = \frac{136 \times 100}{136 \times 4} = \frac{100}{4} \] \[ n^2 = 25 \]
Taking the square root of both sides: \[ n = \sqrt{25} = 5 \]

Step 4: Final Answer:

The value of the principal quantum number \(n\) for the given energy level is 5.
Quick Tip: It's helpful to memorize the energies of the first few levels of the hydrogen atom: \(E_1 = -13.6\) eV, \(E_2 = -3.4\) eV, \(E_3 = -1.51\) eV, \(E_4 = -0.85\) eV, \(E_5 = -0.544\) eV. Recognizing these values can lead to the answer instantly without calculation.

Step 1: Understanding the Concept:

This question is about the V-I characteristics of a p-n junction diode. A diode exhibits very different resistance depending on the direction of the applied voltage (biasing).


Step 2: Detailed Explanation:

1. Forward Bias: When the p-end is connected to the positive terminal and the n-end to the negative terminal of a voltage source, the diode is forward-biased. The depletion region narrows, and the diode offers a very low resistance to the flow of current. It essentially acts like a closed circuit or a closed switch.

2. Reverse Bias: When the p-end is connected to the negative terminal and the n-end to the positive terminal, the diode is reverse-biased. The depletion region widens, and the diode offers a very high resistance, allowing almost no current to flow (ideally). It acts like an open circuit or an open switch.

3. Functionality: The question states that the resistance is high. This corresponds to the reverse-biased state where the diode blocks current. The ability of the diode to switch between a low-resistance state (ON) and a high-resistance state (OFF) based on the applied voltage is the fundamental principle of a switch.

Therefore, when its resistance is high (OFF state), it is part of its function as a switch.


Step 3: Final Answer:

The property of having a high resistance in one state and a low resistance in another allows a p-n junction diode to function as a switch.
Quick Tip: Think of a diode as a one-way street for current. Forward bias is the "go" direction (low resistance), and reverse bias is the "no entry" direction (high resistance). This binary on/off behavior is the essence of a digital switch.

Step 1: Understanding the Concept:

This question assesses the understanding of the depletion layer in a p-n junction and how its properties change with external conditions, specifically biasing.


Step 2: Detailed Explanation:

Analysis of Assertion (A):

The depletion layer is a region at the p-n junction that is depleted of free charge carriers (electrons and holes). Its thickness is not constant. When the diode is formed, a certain equilibrium thickness is established. However, applying an external voltage (biasing) alters this thickness. Under forward bias, the depletion layer becomes thinner. Under reverse bias, it becomes wider. Therefore, the assertion that the thickness is not fixed is true.


Analysis of Reason (R):

The reason states that the thickness depends on factors like biasing. This is the primary factor that controls the depletion layer width in an operating diode. The applied bias voltage either opposes or aids the built-in potential of the junction, causing the layer to shrink or expand. Other factors like doping concentration also determine the initial width, but biasing is the key dynamic factor. Hence, the reason is also true.


Relationship between Assertion and Reason:

The reason directly and correctly explains the assertion. The thickness of the depletion layer is not fixed \textit{because it depends on the applied bias. The reason provides the causal explanation for the statement made in the assertion.


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).
Quick Tip: Remember the effect of biasing on the depletion layer: Forward bias pushes charge carriers towards the junction, reducing the depletion width. Reverse bias pulls carriers away from the junction, increasing the depletion width.

Step 1: Understanding the Concept:

This question tests two key postulates of the Bohr model for the hydrogen atom: the quantization of angular momentum and the rules for electronic transitions.


Step 2: Detailed Explanation:

Analysis of Assertion (A):

According to Bohr's postulates:

The angular momentum (\(L_n\)) is quantized: \(L_n = n \frac{h}{2\pi}\). This implies \(L_n \propto n\).
The radius of the \(n^{th}\) orbit (\(r_n\)) is given by \(r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}\). This implies \(r_n \propto n^2\), or \(n \propto \sqrt{r_n}\).

By combining these two proportionalities, we can express \(L_n\) in terms of \(r_n\):
Since \(L_n \propto n\) and \(n \propto \sqrt{r_n}\), it follows that \(L_n \propto \sqrt{r_n}\).
Therefore, the Assertion (A) is true.


Analysis of Reason (R):

Bohr's model states that an electron jumps from a higher energy orbit to a lower energy orbit by emitting a photon. It does not restrict these jumps to only adjacent or "nearest" orbits. For example, an electron can jump from n=3 to n=1, or from n=4 to n=2, which are not nearest orbits. The spectral series (Lyman, Balmer, etc.) are based on transitions to a specific final orbit (\(n_f=1, 2, \dots\)) from any higher orbit (\(n_i > n_f\)). Therefore, the Reason (R) is false.


Step 3: Final Answer:

The Assertion (A) is true, but the Reason (R) is false.
Quick Tip: For Bohr model questions, keep the key dependencies straight: Radius \(r_n \propto n^2\), Velocity \(v_n \propto 1/n\), Energy \(E_n \propto 1/n^2\), and Angular Momentum \(L_n \propto n\). From these, you can derive any other relationship, like the one in the assertion.

Step 1: Understanding the Concept:

This question relates the phenomenon of diffraction to the properties (wavelength and frequency) of different types of electromagnetic waves.


Step 2: Detailed Explanation:

Analysis of Assertion (A):

Diffraction is the bending of waves as they pass around an obstacle or through an aperture. The effect is most significant when the wavelength (\(\lambda\)) of the wave is comparable to or larger than the size of the obstacle/aperture.

Let's compare the wavelengths of radio waves and infrared waves. In the electromagnetic spectrum, radio waves have the longest wavelengths (from meters to kilometers), while infrared waves have much shorter wavelengths (micrometers to millimeters).

Since \(\lambda_{radio} \gg \lambda_{infrared}\), radio waves will diffract much more significantly around everyday objects (like buildings, hills). This is why you can receive radio signals even when not in the line of sight of the transmitter. Therefore, the Assertion (A) is true.


Analysis of Reason (R):

The relationship between wave speed (c), frequency (f), and wavelength (\(\lambda\)) for electromagnetic waves is \(c = f\lambda\). Since c is constant in a vacuum, frequency and wavelength are inversely proportional (\(f \propto 1/\lambda\)).

As we established that radio waves have a longer wavelength than infrared waves, they must have a lower frequency. The Reason (R) states that radio waves have a greater frequency, which is incorrect. Therefore, the Reason (R) is false.


Step 3: Final Answer:

The Assertion (A) is true, but the Reason (R) is false.
Quick Tip: Remember the order of the electromagnetic spectrum by wavelength (longest to shortest): Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma Ray. Diffraction is more pronounced for waves with longer wavelengths.

Step 1: Understanding the Concept:

This question probes the definition of an ideal transformer and the relationship between voltage and current in a step-down transformer, based on the principle of conservation of energy.


Step 2: Detailed Explanation:

Analysis of Assertion (A):

An "ideal" transformer is a theoretical concept where there are no energy losses. This means the efficiency is 100%. The power input to the primary coil is exactly equal to the power output from the secondary coil (\(P_{in} = P_{out}\)). Since power is the rate of energy transfer, this implies that no electrical energy is lost (converted to heat, etc.). Therefore, the Assertion (A) is true.


Analysis of Reason (R):

A step-down transformer is designed to decrease the voltage, so the secondary voltage (\(V_s\)) is less than the primary voltage (\(V_p\)). From the principle of energy conservation for an ideal transformer: \[ P_{in} = P_{out} \] \[ V_p I_p = V_s I_s \] \[ \frac{I_s}{I_p} = \frac{V_p}{V_s} \]
Since for a step-down transformer \(V_p > V_s\), the ratio \(\frac{V_p}{V_s} > 1\). This means \(\frac{I_s}{I_p} > 1\), or \(I_s > I_p\). So, the voltage decreases while the current increases. The Reason (R) is true.


Relationship between Assertion and Reason:

While both statements are true, the Reason is not the fundamental explanation for the Assertion. The Assertion (energy is not lost) is true by the \textit{definition of an ideal transformer. The Reason (voltage decreases, current increases) is a \textit{consequence of this energy conservation principle applied to a step-down transformer. The lack of energy loss is the premise, not the conclusion of the reason. The actual reasons for no energy loss in an ideal model are the assumptions of no winding resistance, no flux leakage, no hysteresis loss, and no eddy currents. Therefore, Reason (R) is not the correct explanation for Assertion (A).


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
Quick Tip: When analyzing Assertion-Reason questions, ask "Is A true because of R?". In this case, energy conservation is the underlying principle for both. R is an outcome of A, not its cause. The cause of A is the 'ideal' nature of the transformer.

Step 1: Understanding the Concept:

This problem involves the division of current in a parallel circuit. The key principle is that current divides in the inverse ratio of the resistances of the parallel branches. The resistance of a wire is dependent on its length, material, and cross-sectional area.


Step 2: Key Formula or Approach:

1. Resistance of a wire: \(R = \rho \frac{l}{A}\), where \(\rho\) is resistivity, \(l\) is length, and \(A\) is the cross-sectional area.

2. Current division in parallel circuits: For two resistors \(R_1\) and \(R_2\) in parallel, the ratio of currents is \(\frac{I_1}{I_2} = \frac{R_2}{R_1}\).


Step 3: Detailed Explanation:

Let the two wires be denoted by 1 and 2.
We are given:

Same material \(\implies \rho_1 = \rho_2 = \rho\).
Same radius \(\implies A_1 = A_2 = A\).
Ratio of lengths: \(\frac{l_1}{l_2} = \frac{2}{3}\).
Total current: \(I_{total} = I_1 + I_2 = 15\) A.


First, let's find the ratio of their resistances. \[ R_1 = \rho \frac{l_1}{A} \quad and \quad R_2 = \rho \frac{l_2}{A} \]
The ratio of resistances is: \[ \frac{R_1}{R_2} = \frac{\rho l_1/A}{\rho l_2/A} = \frac{l_1}{l_2} = \frac{2}{3} \]
So, \(R_1 : R_2 = 2 : 3\).


When connected in parallel, the voltage (V) across both wires is the same. According to Ohm's law, \(V = I_1 R_1 = I_2 R_2\).
Therefore, the ratio of the currents is inversely proportional to the ratio of the resistances: \[ \frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{3}{2} \]
This means \(I_1 = \frac{3}{2} I_2\).


Now we use the total current information: \[ I_1 + I_2 = 15 A \]
Substitute \(I_1\) in terms of \(I_2\): \[ \frac{3}{2} I_2 + I_2 = 15 \] \[ \left(\frac{3}{2} + 1\right) I_2 = 15 \] \[ \frac{5}{2} I_2 = 15 \] \[ I_2 = 15 \times \frac{2}{5} = 6 A \]
Now find \(I_1\): \[ I_1 = 15 - I_2 = 15 - 6 = 9 A \]

Step 4: Final Answer:

The current through the wire of length \(l_1\) is 9 A, and the current through the wire of length \(l_2\) is 6 A.
Quick Tip: In parallel circuits, the current prefers the path of lower resistance. Since resistance is proportional to length here, the shorter wire will have less resistance and hence will draw more current. This can be a quick check for your answer.

Step 1: Understanding the Concept:

This problem involves analyzing a DC circuit in a steady state. In the steady state, a capacitor acts as an open circuit, meaning no current flows through the branch containing the capacitor. We can then use Kirchhoff's Voltage Law (KVL) to find currents and potential differences in the remaining parts of the circuit.


Step 2: Key Formula or Approach:

1. Steady State Condition: Current through the capacitor branch is zero (\(I_C = 0\)).
2. Kirchhoff's Voltage Law (KVL): The algebraic sum of potential changes around any closed loop is zero.


Step 3: Detailed Explanation:

At steady state, no current flows through the capacitor C. The circuit simplifies to a single outer loop containing the cells with e.m.f. V (top) and 2V (bottom), and resistors R and 2R.


Let's apply KVL to this outer loop. The 2V cell and the V cell are connected in opposition. The net e.m.f. driving the current is \(E_{net} = 2V - V = V\). The total resistance in the loop is \(R_{total} = R + 2R = 3R\).
The current \(I\) in the loop will be driven by the stronger cell (2V), so it will flow in a clockwise direction. \[ I = \frac{E_{net}}{R_{total}} = \frac{V}{3R} \]

(i) Potential difference between P and Q (\(V_{PQ} = V_P - V_Q\)):

We can find the potential difference by moving from P to Q along any path. Let's choose the bottom path through the 2V cell and 2R resistor. Let's assume the potential at P is \(V_P\).
Moving from P towards the 2V cell, we cross it from the negative to the positive terminal, so the potential increases by 2V. Then, moving through the 2R resistor in the direction of current \(I\), there is a potential drop of \(I(2R)\). \[ V_Q = V_P + 2V - I(2R) \]
Substitute the value of \(I\): \[ V_Q = V_P + 2V - \left(\frac{V}{3R}\right)(2R) = V_P + 2V - \frac{2V}{3} = V_P + \frac{4V}{3} \]
Rearranging for \(V_P - V_Q\): \[ V_P - V_Q = -\frac{4V}{3} \]
The potential difference is the magnitude, so \(|V_P - V_Q| = \frac{4V}{3}\).


(ii) Potential difference across capacitor C (\(V_C\)):

Since no current flows through the middle branch, we can find the potential difference across the capacitor by applying KVL to the loop containing P, the middle cell V, the capacitor C, and Q.
Let's traverse from P to Q through this middle branch.
From P, we cross the cell V from its positive to its negative terminal (potential drops by V). Let the point between this cell and the capacitor be M. The potential at M is \(V_M = V_P - V\).
The potential difference across the capacitor is \(V_C = V_M - V_Q\). \[ V_C = (V_P - V) - V_Q = (V_P - V_Q) - V \]
Substitute the value of \(V_P - V_Q\) we found: \[ V_C = -\frac{4V}{3} - V = -\frac{4V + 3V}{3} = -\frac{7V}{3} \]
The magnitude of the potential difference across the capacitor is \(|V_C| = \frac{7V}{3}\).


Step 4: Final Answer:

(i) The potential difference between P and Q is \(\frac{4V}{3}\).
(ii) The potential difference across the capacitor C is \(\frac{7V}{3}\).
Quick Tip: In steady-state DC analysis, the first step is always to redraw the circuit by treating capacitors as open circuits (breaks in the wire) and inductors as short circuits (plain wires). This greatly simplifies the circuit for applying Kirchhoff's laws.

Step 1: Understanding the Concept:

This problem deals with the interference pattern in Young's Double-Slit Experiment (YDSE) and how it shifts when a transparent sheet is introduced in the path of one of the beams. The sheet introduces an additional optical path difference, which changes the condition for interference at any given point on the screen.


Step 2: Key Formula or Approach:

1. Condition for Dark Fringes: Path difference \(\Delta x = (m - \frac{1}{2})\lambda\), where m = 1, 2, 3, ...
2. Condition for Bright Fringes: Path difference \(\Delta x = m\lambda\), where m = 0, 1, 2, 3, ...
3. Optical Path Difference due to a Sheet: A sheet of thickness t and refractive index n introduces an extra optical path of \((n-1)t\).


Step 3: Detailed Explanation:

Let the point on the screen be P.

Initial Situation (without the sheet):

The 6th dark fringe is observed at point P. The path difference at P is therefore: \[ \Delta x_{initial} = \left(6 - \frac{1}{2}\right)\lambda = 5.5\lambda \]

Final Situation (with the sheet):

A transparent sheet is introduced in one of the paths. This adds an optical path difference of \((n-1)t\). The new total path difference at the same point P is: \[ \Delta x_{final} = \Delta x_{initial} + (n-1)t \]
At this point P, the 8th bright fringe is now observed. The condition for the 8th bright fringe is: \[ \Delta x_{final} = 8\lambda \]

Equating and Solving for t:

We can now equate the two expressions for \(\Delta x_{final}\): \[ \Delta x_{initial} + (n-1)t = 8\lambda \]
Substitute the value of \(\Delta x_{initial}\): \[ 5.5\lambda + (n-1)t = 8\lambda \]
Now, solve for the thickness \(t\): \[ (n-1)t = 8\lambda - 5.5\lambda \] \[ (n-1)t = 2.5\lambda \] \[ t = \frac{2.5\lambda}{n-1} \]
This can also be written as: \[ t = \frac{5\lambda}{2(n-1)} \]

Step 4: Final Answer:

The relation for the thickness t is \(t = \frac{2.5\lambda}{n-1}\).
Quick Tip: The shift of the fringe pattern due to a thin sheet can be thought of as a change in the "effective" path length. The number of fringes shifted is given by \(\frac{(n-1)t}{\lambda}\). In this case, the shift is from the 5.5th fringe position to the 8th fringe position, a shift of \(8 - 5.5 = 2.5\) fringes. So, \(\frac{(n-1)t}{\lambda} = 2.5\), which directly gives the answer.

Step 1: Understanding the Concept:

This is a problem on the combination of lenses. The image formed by the first lens acts as the object for the second lens. We will apply the lens formula sequentially for each lens to find the final image position and its characteristics.


Step 2: Key Formula or Approach:

The lens formula is: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\), where \(u\) is the object distance, \(v\) is the image distance, and \(f\) is the focal length. The magnification is \(m = \frac{v}{u}\). We will use the Cartesian sign convention.


Step 3: Detailed Explanation:

Given:

Focal length of concave lens A, \(f_A = -8.0\) cm.
Focal length of concave lens B, \(f_B = -8.0\) cm.
Distance of object P from lens A, \(u_A = -4.0\) cm.
Separation between lenses, \(d = 16\) cm.


Image Formation by Lens A:

Using the lens formula for lens A: \[ \frac{1}{v_A} - \frac{1}{u_A} = \frac{1}{f_A} \] \[ \frac{1}{v_A} - \frac{1}{-4.0} = \frac{1}{-8.0} \] \[ \frac{1}{v_A} + \frac{1}{4} = -\frac{1}{8} \] \[ \frac{1}{v_A} = -\frac{1}{8} - \frac{1}{4} = \frac{-1 - 2}{8} = -\frac{3}{8} \] \[ v_A = -\frac{8}{3} \approx -2.67 cm \]
The first image (I1) is formed 2.67 cm to the left of lens A. Since \(v_A\) is negative, the image is virtual.


Image I1 as Object for Lens B:

This virtual image I1 now acts as a virtual object for lens B. The distance of I1 from lens B is: \[ u_B = - (d + |v_A|) = - \left(16 + \frac{8}{3}\right) = - \left(\frac{48 + 8}{3}\right) = -\frac{56}{3} cm \]

Image Formation by Lens B:

Using the lens formula for lens B: \[ \frac{1}{v_B} - \frac{1}{u_B} = \frac{1}{f_B} \] \[ \frac{1}{v_B} - \frac{1}{-56/3} = \frac{1}{-8} \] \[ \frac{1}{v_B} + \frac{3}{56} = -\frac{1}{8} \] \[ \frac{1}{v_B} = -\frac{1}{8} - \frac{3}{56} = \frac{-7 - 3}{56} = -\frac{10}{56} = -\frac{5}{28} \] \[ v_B = -\frac{28}{5} = -5.6 cm \]
The final image (I2) is formed at 5.6 cm to the left of lens B.


Nature of the Final Image:

Since \(v_B\) is negative, the final image is virtual.
To determine if it's erect or inverted, and its size, we find the total magnification \(M = m_A \times m_B\). \[ m_A = \frac{v_A}{u_A} = \frac{-8/3}{-4} = +\frac{2}{3} \] \[ m_B = \frac{v_B}{u_B} = \frac{-28/5}{-56/3} = \frac{28}{5} \times \frac{3}{56} = \frac{3}{10} = +0.3 \] \[ M = \left(+\frac{2}{3}\right) \times \left(+\frac{3}{10}\right) = +\frac{2}{10} = +0.2 \]
Since the total magnification M is positive, the final image is erect with respect to the original object.
Since \(|M| = 0.2 < 1\), the final image is diminished.


Step 4: Final Answer:

The final image is formed 5.6 cm to the left of lens B. Its nature is virtual, erect, and diminished.
Quick Tip: Always be careful with sign conventions, especially when the image from the first element acts as an object for the second. Drawing a rough ray diagram can help visualize the situation and prevent errors in calculating the object distance for the second lens.

Step 1: Understanding the Concept:

This problem applies Einstein's photoelectric equation, which describes the energy balance when a photon strikes a metal surface and ejects an electron. The photon's energy is used to overcome the metal's work function and to provide kinetic energy to the emitted electron.


Step 2: Key Formula or Approach:

1. Photon Energy: \(E = \frac{hc}{\lambda}\)
2. Einstein's Photoelectric Equation: \(K_{max} = E - \phi\)
3. Kinetic Energy: \(K_{max} = \frac{1}{2} m_e v_{max}^2\)
(Constants: \(h \approx 6.63 \times 10^{-34}\) J·s, \(c = 3 \times 10^8\) m/s, \(m_e \approx 9.1 \times 10^{-31}\) kg)


Step 3: Detailed Explanation:

Given:

Wavelength \(\lambda = 400 nm = 400 \times 10^{-9} m = 4 \times 10^{-7} m\).
Work function \(\phi = 3.0 \times 10^{-19} J\).


1. Calculate the energy of the incident photon (E): \[ E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34} J·s) \times (3 \times 10^8 m/s)}{4 \times 10^{-7} m} \] \[ E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} = 4.9725 \times 10^{-19} J \]

2. Calculate the maximum kinetic energy of the photoelectrons (\(K_{max}\)): \[ K_{max} = E - \phi = (4.9725 \times 10^{-19} J) - (3.0 \times 10^{-19} J) \] \[ K_{max} = 1.9725 \times 10^{-19} J \]

3. Calculate the speed of the fastest photoelectrons (\(v_{max}\)): \[ K_{max} = \frac{1}{2} m_e v_{max}^2 \] \[ v_{max} = \sqrt{\frac{2 K_{max}}{m_e}} = \sqrt{\frac{2 \times (1.9725 \times 10^{-19} J)}{9.1 \times 10^{-31} kg}} \] \[ v_{max} = \sqrt{\frac{3.945 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{0.4335 \times 10^{12}} \] \[ v_{max} = \sqrt{43.35 \times 10^{10}} = (\sqrt{43.35}) \times 10^5 m/s \] \[ v_{max} \approx 6.58 \times 10^5 m/s \]

Step 4: Final Answer:

The speed of the fastest photoelectrons emitted is approximately \(6.58 \times 10^5\) m/s.
Quick Tip: For quick calculations, remember the energy-wavelength product \(hc \approx 1240\) eV·nm. You can calculate energy in eV, convert the work function to eV (\(\phi (J) / 1.6 \times 10^{-19}\)), find \(K_{max}\) in eV, and then convert back to Joules to find the speed. This often simplifies the numbers involved.

Step 1: Understanding the Concept:

This is a simple DC circuit analysis problem involving a real diode. A forward-biased silicon diode, when conducting, has a nearly constant voltage drop across it, known as the threshold or knee voltage. We can use Kirchhoff's Voltage Law (KVL) to analyze the circuit.


Step 2: Key Formula or Approach:

1. Diode Model: Assume a constant voltage drop of \(V_d = 0.7\) V across the forward-biased silicon diode.
2. Ohm's Law: Voltage drop across the resistor is \(V_R = IR\).
3. KVL: The sum of voltage drops across the components must equal the source voltage: \(V = V_R + V_d\).


Step 3: Detailed Explanation:

Given:

Diode threshold voltage, \(V_d = 0.7\) V.
Series resistance, \(R = 1000 \, \Omega\).
Circuit current, \(I = 15 mA = 15 \times 10^{-3} A\).


The circuit consists of the battery (V), the resistor (R), and the diode connected in series. Since a current is flowing, the diode must be forward-biased.


First, calculate the voltage drop across the resistor (\(V_R\)) using Ohm's Law: \[ V_R = I \times R \] \[ V_R = (15 \times 10^{-3} A) \times (1000 \, \Omega) = 15 V \]

Now, apply KVL to the series circuit. The total voltage supplied by the battery (V) is the sum of the voltage drops across the resistor and the diode. \[ V = V_R + V_d \] \[ V = 15 V + 0.7 V \] \[ V = 15.7 V \]

Step 4: Final Answer:

The value of the battery voltage V is 15.7 V.
Quick Tip: For circuit problems with diodes, the first step is to determine if the diode is forward or reverse biased. If the circuit allows current flow, it's forward biased. Then, you can replace the ideal diode with its equivalent model, which is often a simple battery of voltage \(V_d\) (0.7V for Si, 0.3V for Ge) opposing the current flow.

Step 1: Understanding the Concept:

This question asks for the derivation of the formula for the terminal voltage of a real cell (one with internal resistance) when it is connected to an external load. The terminal voltage is the potential difference across the external resistor, which is less than the cell's e.m.f. due to the voltage drop across the internal resistance.


Step 2: Key Formula or Approach:

1. Ohm's Law for the complete circuit: The total current is the total e.m.f. divided by the total resistance.
2. Terminal Voltage: The potential difference across the external resistance.


Step 3: Detailed Explanation:

Consider a cell with e.m.f. E and internal resistance r connected to an external resistor R.


1. The total resistance in the series circuit is the sum of the external and internal resistances: \[ R_{total} = R + r \]

2. The current (I) flowing through the circuit is given by Ohm's law applied to the entire circuit: \[ I = \frac{E_{total}}{R_{total}} = \frac{E}{R+r} \]

3. The voltmeter measures the potential drop (V) across the external resistor R. According to Ohm's law applied to the external resistor: \[ V = I \times R \]

4. Now, substitute the expression for the current I from step 2 into the equation from step 3: \[ V = \left( \frac{E}{R+r} \right) R \]
This gives the relationship: \[ V = \frac{ER}{R+r} \]

This is the required relationship between the terminal voltage V, e.m.f. E, external resistance R, and internal resistance r.


Step 4: Final Answer:

The relationship between V, E, R, and r is \(V = \frac{ER}{R+r}\).
Quick Tip: Another useful form of this relationship is \(V = E - Ir\). This highlights that the terminal voltage V is always less than the EMF E by an amount \(Ir\), which is the 'lost volts' across the internal resistance. This form is often easier to use when the current is known.

Step 1: Understanding the Concept:

This question requires understanding the relationship between the terminal voltage (V) of a cell and the current (I) it delivers. This relationship can be represented graphically, and the key parameters of the cell (e.m.f. and internal resistance) can be extracted from this graph.


Step 2: Key Formula or Approach:

The governing equation is the terminal voltage formula: \[ V = E - Ir \]
This equation is in the form of a linear equation \(y = c + mx\).


Step 3: Detailed Explanation:

1. Drawing the Graph:

The equation is \(V = E - rI\). We want to plot V (on the y-axis) as a function of I (on the x-axis).
Comparing this to the standard equation of a straight line, \(y = mx + c\):

\(y = V\)
\(x = I\)
The slope \(m = -r\)
The y-intercept \(c = E\)

This means the graph of V versus I is a straight line with a negative slope (-r) and a y-intercept of E.

When \(I = 0\) (open circuit), \(V = E\). This is the starting point on the V-axis.
As the current I increases, the term \(Ir\) increases, and the terminal voltage V decreases linearly.
When \(V = 0\) (short circuit), \(0 = E - Ir \implies I = E/r\). This is the intercept on the I-axis.

The graph is as follows:

% A placeholder for a graphical representation
[A downward sloping straight line is drawn on a V vs. I axes. The line intersects the V-axis at a point labelled 'E'. The line intersects the I-axis at a point labelled 'E/r'.]


2. Determining E and r from the Graph:


To find the e.m.f. (E): The e.m.f. is the terminal voltage when no current is drawn from the cell (\(I=0\)). From the graph, this corresponds to the intercept on the V-axis (the y-intercept). So, by extending the plotted line to intersect the V-axis, we can read the value of E.

To find the internal resistance (r): The internal resistance is related to the slope of the line. The slope of the V-I graph is given by:
\[ Slope = m = \frac{\Delta V{\Delta I} \]
From our equation \(V = E - rI\), we see that the slope \(m = -r\). Therefore, the internal resistance is the negative of the slope of the graph.
\[ r = -(Slope) = -\frac{\Delta V}{\Delta I} \]
We can calculate this by picking two points \((I_1, V_1)\) and \((I_2, V_2)\) on the line and using the formula \(r = -\frac{V_2 - V_1}{I_2 - I_1}\).


Step 4: Final Answer:

The graph of V vs. I is a straight line with a negative slope. The e.m.f. (E) is determined from the V-axis intercept of the graph. The internal resistance (r) is determined by calculating the negative of the slope of the graph.
Quick Tip: This experiment is a standard practical in physics. Remember that in a real experiment, you would obtain several data points of (I, V) by changing the external resistance R, plot them, and then draw a "line of best fit". The E and r are then determined from this best-fit line to minimize experimental errors.

Step 1: Understanding the Concept:

The problem describes a charged particle moving with constant velocity, which implies that the net force on it is zero. The particle is subject to both an electric force and a magnetic force (from the current-carrying wire). For the net force to be zero, these two forces must be equal in magnitude and opposite in direction. This setup is characteristic of a velocity selector.


Step 2: Key Formula or Approach:

1. Magnetic Field due to a long straight wire: \(B = \frac{\mu_0 I}{2\pi r}\). Direction is given by the Right-Hand Thumb Rule.

2. Electric Force: \(\vec{F}_e = q\vec{E}\).

3. Magnetic Lorentz Force: \(\vec{F}_m = q(\vec{v} \times \vec{B})\).

4. Condition for constant velocity: \(\vec{F}_{net} = \vec{F}_e + \vec{F}_m = 0\).


Step 3: Detailed Explanation:

Let's set up a coordinate system. Let the conductor lie along the x-axis, and the particle move parallel to it at \(y=d\).

Current \(\vec{I}\) is in the \(-\hat{i}\) direction.
Velocity of the particle \(\vec{v} = -v_0 \hat{i}\).
Particle has a negative charge, let's say \(q\).


(i) Diagram and Direction of Fields:

Magnetic Field (\(\vec{B}\)): Using the Right-Hand Thumb Rule for the wire with current in the \(-\hat{i}\) direction, the magnetic field at a point \(y=d\) (above the wire) is directed out of the page, i.e., in the \(+\hat{k}\) direction. Its magnitude is \(B = \frac{\mu_0 I}{2\pi d}\).
Magnetic Force (\(\vec{F}_m\)):
\[ \vec{F}_m = q(\vec{v} \times \vec{B}) = q ((-v_0 \hat{i}) \times (\frac{\mu_0 I}{2\pi d} \hat{k})) = q (-\frac{v_0 \mu_0 I}{2\pi d}) (\hat{i} \times \hat{k}) \]
Since \(\hat{i} \times \hat{k} = -\hat{j}\),
\[ \vec{F}_m = q (\frac{v_0 \mu_0 I}{2\pi d} \hat{j}) \]
Since the particle is negatively charged (\(q < 0\)), the direction of \(\vec{F}_m\) is opposite to \(\hat{j}\), i.e., in the \(-\hat{j}\) direction (downwards, towards the wire).
Electric Force (\(\vec{F}_e\)): For the net force to be zero, \(\vec{F}_e\) must be opposite to \(\vec{F}_m\). So, \(\vec{F}_e\) must be in the \(+\hat{j}\) direction (upwards, away from the wire).
Electric Field (\(\vec{E}\)): The electric force is \(\vec{F}_e = q\vec{E}\). Since \(q\) is negative and \(\vec{F}_e\) is in the \(+\hat{j}\) direction, the electric field \(\vec{E}\) must be in the opposite direction, i.e., in the \(-\hat{j}\) direction (downwards).

% Placeholder for diagram. A proper diagram would show the wire on the x-axis, current to the left. The particle at y=d moving left. B field pointing out of the page. Fm pointing down, Fe pointing up, and E pointing down.

(ii) Forces acting on the particle:
The two forces acting on the charged particle are:
1. The electric force \(\vec{F}_e = q\vec{E}\), directed upwards.
2. The magnetic Lorentz force \(\vec{F}_m = q(\vec{v} \times \vec{B})\), directed downwards.

(iii) Finding the value of \(v_0\):
For the net force to be zero, the magnitudes of the electric and magnetic forces must be equal. \[ |\vec{F}_e| = |\vec{F}_m| \] \[ |q|E = |q|v_0 B \] \[ E = v_0 B \]
Substitute the expression for the magnetic field B: \[ E = v_0 \left( \frac{\mu_0 I}{2\pi d} \right) \]
Solving for \(v_0\): \[ v_0 = \frac{E}{\frac{\mu_0 I}{2\pi d}} = \frac{2\pi d E}{\mu_0 I} \]

Step 4: Final Answer:

The value of the velocity is \(v_0 = \frac{2\pi d E}{\mu_0 I}\).
Quick Tip: This scenario is a classic example of a velocity filter. When the electric and magnetic forces balance, only particles with a specific velocity (\(v = E/B\)) pass through undeflected. Remember that the directions of \(\vec{E}\), \(\vec{B}\), and \(\vec{v}\) must be mutually perpendicular for this simple relation to hold.

Step 1: Understanding the Concept:

This problem involves the principle of superposition for magnetic fields. The net magnetic field at a point due to multiple current sources is the vector sum of the magnetic fields produced by each individual source.


Step 2: Key Formula or Approach:

The magnetic field \(\vec{B}\) at a perpendicular distance \(r\) from an infinitely long straight wire carrying current \(I\) is given by the magnitude \(B = \frac{\mu_0 I}{2\pi r}\). The direction is determined by the Right-Hand Thumb Rule.


Step 3: Detailed Explanation:

Let's define the coordinate system: XX' axis is the x-axis, and YY' axis is the y-axis. The point P has coordinates (X, Y).

1. Magnetic Field due to conductor along X-axis (\(\vec{B}_1\)):

Current \(I_1\) flows along the -X axis (in the \(-\hat{i}\) direction).
The point P(X, Y) is at a perpendicular distance \(r_1 = Y\) from this conductor.
Using the Right-Hand Thumb Rule (pointing thumb in the direction of \(I_1\), i.e., \(-\hat{i}\)), the magnetic field at a point with a positive Y coordinate will curl out of the page.
Therefore, the direction of \(\vec{B}_1\) is along the \(+\hat{k}\) axis.
The magnitude is \(B_1 = \frac{\mu_0 I_1}{2\pi Y}\).
In vector form: \(\vec{B}_1 = \frac{\mu_0 I_1}{2\pi Y} \hat{k}\).


2. Magnetic Field due to conductor along Y-axis (\(\vec{B}_2\)):

Current \(I_2\) flows along the -Y axis (in the \(-\hat{j}\) direction).
The point P(X, Y) is at a perpendicular distance \(r_2 = X\) from this conductor.
Using the Right-Hand Thumb Rule (pointing thumb in the direction of \(I_2\), i.e., \(-\hat{j}\)), the magnetic field at a point with a positive X coordinate will curl into the page.
Therefore, the direction of \(\vec{B}_2\) is along the \(-\hat{k}\) axis.
The magnitude is \(B_2 = \frac{\mu_0 I_2}{2\pi X}\).
In vector form: \(\vec{B}_2 = -\frac{\mu_0 I_2}{2\pi X} \hat{k}\).


3. Net Magnetic Field (\(\vec{B}_{net}\)):
The net magnetic field is the vector sum of \(\vec{B}_1\) and \(\vec{B}_2\). \[ \vec{B}_{net} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 I_1}{2\pi Y} \hat{k} - \frac{\mu_0 I_2}{2\pi X} \hat{k} \] \[ \vec{B}_{net} = \left( \frac{\mu_0 I_1}{2\pi Y} - \frac{\mu_0 I_2}{2\pi X} \right) \hat{k} \] \[ \vec{B}_{net} = \frac{\mu_0}{2\pi} \left( \frac{I_1}{Y} - \frac{I_2}{X} \right) \hat{k} \]

Magnitude and Direction:

Magnitude: \( B_{net} = \left| \frac{\mu_0}{2\pi} \left( \frac{I_1}{Y} - \frac{I_2}{X} \right) \right| \).
Direction: The direction is along the z-axis (perpendicular to the plane of the wires). It is along \(+\hat{k}\) (out of the page) if \(\frac{I_1}{Y} > \frac{I_2}{X}\), and along \(-\hat{k}\) (into the page) if \(\frac{I_1}{Y} < \frac{I_2}{X}\). If they are equal, the net magnetic field is zero.


Step 4: Final Answer:

The net magnetic field at point P(X, Y) is \(\vec{B}_{net} = \frac{\mu_0}{2\pi} \left( \frac{I_1}{Y} - \frac{I_2}{X} \right) \hat{k}\).
Quick Tip: When dealing with multiple magnetic fields, always treat them as vectors. Determine the direction of each field component first using the appropriate rule (like the Right-Hand Rule), then perform the vector addition. Don't just add the magnitudes unless you are certain the fields are collinear.

Step 1: Understanding the Concept:

Part (a) tests the fundamental principle of electromagnetic induction. Part (b) tests the knowledge of the construction and working principle of an AC generator.


Step 2: Detailed Explanation:


(a) Lenz's Law:
Lenz's law states that the direction of the induced electromotive force (e.m.f.) and hence the induced current in a closed conducting loop is always such that it opposes the change in magnetic flux that is responsible for producing it. This law is a direct consequence of the principle of conservation of energy.


(b) Analysis of the Figure:
The figure shows a rotating coil in a magnetic field, connected to an external circuit via a system of rings and brushes, which is the basic setup of an electric generator.

(i) Identify the machine:

The machine is an AC Generator (or alternator/dynamo). It converts mechanical energy into electrical energy in the form of alternating current.


(ii) Name the parts:


P : These are Slip Rings. They rotate with the coil and provide a continuous connection to the external circuit.
Q : These are Brushes (typically carbon brushes). They are stationary and press against the slip rings to conduct the induced current from the coil to the external circuit.
R : This is the Armature Coil (or rotor coil). It is a coil of insulated wire wound on a soft iron core that rotates within the magnetic field.


(iii) Give the polarities of the magnetic poles:

The large C-shaped structure is a permanent magnet (or an electromagnet) that provides the magnetic field. For a magnetic field to exist across the coil, one pole must be North and the other must be South. Conventionally, one can label the top pole as North Pole (N) and the bottom pole as South Pole (S), so that the magnetic field lines are directed downwards.


(iv) Two ways of increasing the output voltage:

The magnitude of the induced e.m.f. (voltage) in a generator is given by \(\mathcal{E} = NBA\omega \sin(\omega t)\). The peak output voltage is \(\mathcal{E}_{max} = NBA\omega\). To increase this voltage, we can:

Increase the speed of rotation (\(\omega\)): Rotating the coil faster increases the rate of change of magnetic flux, thereby inducing a larger e.m.f.
Increase the number of turns (N) in the armature coil: A coil with more turns will have a greater total magnetic flux linkage, leading to a higher induced voltage.

Other methods include increasing the magnetic field strength (B) or increasing the area of the coil (A).
Quick Tip: Remember the key difference between an AC and a DC generator lies in the commutator. An AC generator uses two separate slip rings (like P in the diagram), while a DC generator uses a single split-ring commutator to reverse the current direction every half rotation.

Step 1: Understanding the Concept:

This question covers the fundamental properties of electromagnetic (EM) waves, including their transverse nature, the electromagnetic spectrum, and the applications of different types of EM waves.


Step 2: Detailed Explanation:


(a) Direction of Magnetic Field \(\vec{B}\):

For an electromagnetic wave, the electric field vector \(\vec{E}\), the magnetic field vector \(\vec{B}\), and the direction of wave propagation (\(\vec{k}\)) are mutually perpendicular. Their orientation follows a right-hand rule, where the direction of propagation is given by the direction of the vector cross product \(\vec{E} \times \vec{B}\).

Let's define a coordinate system:

Direction of propagation (North) = \(+\hat{x}\)
Direction of \(\vec{E}\) oscillation (Up and Down) = \(\pm\hat{z}\)

We need to find the direction of \(\vec{B}\) such that \(\vec{E} \times \vec{B}\) points in the \(+\hat{x}\) direction.

When \(\vec{E}\) is Up (\(+\hat{z}\)), we need \( (+\hat{z}) \times \vec{B}_{dir} = +\hat{x} \). According to the vector cross product rules, \(\hat{z} \times \hat{y} = -\hat{x}\) and \(\hat{z} \times (-\hat{y}) = +\hat{x}\). So, \(\vec{B}\) must be in the \(-\hat{y}\) direction.
When \(\vec{E}\) is Down (\(-\hat{z}\)), we need \( (-\hat{z}) \times \vec{B}_{dir} = +\hat{x} \). This requires \(\vec{B}\) to be in the \(+\hat{y}\) direction.

If we define East as \(+\hat{y}\) and West as \(-\hat{y}\), the magnetic field \(\vec{B}\) oscillates in the East-West direction.


(b) Wavelength Comparison:

The portion of the electromagnetic spectrum detectable by human eyes is called visible light. The order of EM waves by increasing wavelength is: Gamma rays, X-rays, Ultraviolet, Visible light, Infrared, Microwaves, Radio waves.
Therefore, both radio waves and microwaves have wavelengths that are longer than the wavelengths of visible light.


(c) Main Uses:

(i) Infrared waves:

Remote Controls: Used in remote controls for televisions, air conditioners, and other electronic devices to transmit signals.
Thermal Imaging: Infrared cameras (thermography) are used for night vision, building insulation analysis, and medical diagnostics by detecting heat signatures.
Physiotherapy: Infrared lamps are used to provide warmth for muscle pain relief.

(ii) Gamma rays:

Medical Treatment (Radiotherapy): High-energy gamma rays are used to destroy cancerous cells and tumors.
Sterilization: They are used to sterilize medical equipment and preserve food by killing harmful bacteria and microorganisms.
Industrial Radiography: Used to inspect metal castings and welds for internal flaws and cracks. Quick Tip: To easily remember the direction rule for EM waves, use the phrase "Eat Big Carrots" for \(\vec{E}\), \(\vec{B}\), and \(\vec{c}\) (direction of propagation). Align your fingers for \(\vec{E}\), curl them towards \(\vec{B}\), and your thumb will point in the direction of propagation \(\vec{c}\).

(a) Effect on the width of the beam:

When a parallel beam of light enters a denser medium (like water from air) obliquely, it refracts and bends towards the normal. According to Snell's law, \(n_1 \sin i = n_2 \sin r\). Since \(n_2 > n_1\), the angle of refraction \(r\) is less than the angle of incidence \(i\).

Let the width of the incident beam be \(w_i\). The projection of this width on the water surface is \(d = \frac{w_i}{\cos i}\). The refracted beam emerges from this same projected width \(d\). The width of the refracted beam, \(w_r\), is related to \(d\) by \(w_r = d \cos r\).

Substituting for \(d\), we get: \[ w_r = \left(\frac{w_i}{\cos i}\right) \cos r = w_i \frac{\cos r}{\cos i} \]
Since \(i > r\), it follows that \(\cos i < \cos r\). Therefore, the factor \(\frac{\cos r}{\cos i} > 1\).
This means \(w_r > w_i\). The width of the beam increases after it enters the water.


(b) A straw appears bent in water:

Explanation: The apparent bending of a straw is due to the phenomenon of refraction of light. Light rays traveling from the submerged part of the straw pass from a denser medium (water) into a rarer medium (air) before reaching the observer's eye. As the rays cross the water-air interface, they bend away from the normal. The eye perceives the object's position by tracing these refracted rays back in a straight line. These virtual rays intersect at a point that is shallower than the actual position of the object. Consequently, each point on the submerged portion of the straw appears to be raised. This makes the submerged part look shorter and creates the illusion that the straw is bent at the surface of the water.

Ray Diagram:
% A placeholder for a diagram showing:
% 1. A water-air interface.
% 2. A straight straw passing through the interface.
% 3. A point P on the submerged part of the straw.
% 4. Two light rays originating from P, moving towards the surface.
% 5. At the surface, the rays refract (bend away from the normal).
% 6. An eye is shown receiving these refracted rays.
% 7. The refracted rays are extended backward (as dotted lines) to intersect at a point P', which is vertically above P.
% 8. The apparent position of the straw is shown as a bent object, with the submerged part originating from P'.



(c) Signal transmission through an optical fibre:

Explanation: Transmission of optical signals through an optical fibre is based on the principle of Total Internal Reflection (TIR).
An optical fibre is a thin strand of high-quality glass or plastic consisting of two main parts:

Core: The inner part with a higher refractive index (\(n_{core}\)).
Cladding: The outer layer with a lower refractive index (\(n_{cladding}\)).

When a light signal (ray) enters the core at one end at a suitable angle, it travels and strikes the interface between the core and the cladding. If the angle of incidence (\(i\)) at this interface is greater than the critical angle (\(i_c\)), the light does not refract into the cladding but is completely reflected back into the core. The critical angle is defined by \(\sin(i_c) = n_{cladding}/n_{core}\). The light signal thus propagates along the length of the fibre by undergoing a series of successive total internal reflections, with almost no loss of intensity, even if the fibre is bent.

Diagram:
% A placeholder for a diagram showing:
% 1. A cross-section of an optical fibre with the core and cladding labelled, showing n_core > n_cladding.
% 2. A light ray entering the core.
% 3. The ray striking the core-cladding interface at an angle i > i_c.
% 4. The ray undergoing multiple total internal reflections as it travels down the length of the fibre. Quick Tip: For total internal reflection to occur, two conditions are essential: (1) Light must travel from a denser medium to a rarer medium. (2) The angle of incidence in the denser medium must be greater than the critical angle for the pair of media.

Step 1: Understanding the Concept:

This question asks for a graphical representation and interpretation of the binding energy per nucleon (BE/A) as a function of the mass number (A). This curve is fundamental to understanding nuclear stability and the energy released in nuclear reactions.


Step 2: Detailed Explanation:

Variation of Binding Energy per Nucleon with Mass Number:

The binding energy per nucleon is a measure of the stability of a nucleus. A higher BE/A value indicates a more stable nucleus. The graph is plotted with the mass number (A) on the x-axis and the binding energy per nucleon (BE/A in MeV) on the y-axis.

% Placeholder for the graph. The graph should show:
% 1. A rapid increase in BE/A for low mass numbers (A < 30).
% 2. A broad peak around A = 56 (Iron, Fe), with a maximum value of about 8.8 MeV.
% 3. A slow, gradual decrease in BE/A for high mass numbers (A > 60).
% 4. Some sharp peaks for light, stable nuclei like Helium-4, Carbon-12, and Oxygen-16.

Key features of the curve:

For very light nuclei (\(A < 30\)), the BE/A increases sharply with increasing A.
The curve has a broad maximum for nuclei with mass numbers in the range of \(30 < A < 170\). The peak of the curve is at A = 56 for Iron (\(^{56}Fe\)), which is one of the most stable nuclei.
For heavy nuclei (\(A > 170\)), the BE/A slowly decreases with increasing A.


Significance of the Binding Energy Curve:

The binding energy curve provides crucial insights into nuclear processes:

Nuclear Stability: It shows that nuclei with intermediate mass numbers (around A=56) are the most stable. Nuclei that are very light or very heavy are less stable.
Nuclear Fission: The curve explains why energy is released in nuclear fission. A heavy, less stable nucleus (e.g., Uranium, A > 200) with a lower BE/A splits into two lighter, more stable nuclei with higher BE/A. The increase in binding energy per nucleon is released as a large amount of energy.
Nuclear Fusion: The curve also explains why energy is released in nuclear fusion. Two very light, less stable nuclei (e.g., hydrogen isotopes) with low BE/A fuse together to form a heavier, more stable nucleus (e.g., Helium) with a significantly higher BE/A. This increase in binding energy is released as energy. Quick Tip: Remember the general shape: a steep rise, a broad peak around iron, and a slow fall. This shape is the key to explaining both fission (heavy nuclei splitting) and fusion (light nuclei joining), as both processes move towards the more stable peak of the curve.

Step 1: Understanding the Concept:

This question describes a nuclear process where the products are more stable (higher BE/A) than the reactants (lower BE/A). This is the fundamental principle behind energy-releasing nuclear reactions.


Step 2: Detailed Explanation:


(i) Type of nuclear reaction:

The process described is when lighter nuclei with lower binding energy per nucleon combine to form a heavier nucleus with a higher binding energy per nucleon. This type of reaction is called Nuclear Fusion. For example, two deuterium nuclei (\(^{2}H\)) can fuse to form a helium nucleus (\(^{4}He\)), moving up the binding energy curve and releasing energy.


(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, \(E_b = (\Delta m)c^2\). The mass defect is the difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus. A higher binding energy implies a larger mass defect.

Since the final nucleus has a higher binding energy than the initial nuclei, its mass defect is greater. This means that more mass has been converted into energy. Therefore, the total mass of the product nucleus is less than the total mass of the initial reactant nuclei. The total mass decreases during the reaction.


(iii) Energy requirement/production:

Since the total binding energy of the system increases (\(BE_{final} > BE_{initial}\)), this excess energy must be released to conserve energy. The energy released (Q-value) is given by: \[ Q = BE_{final} - BE_{initial} \]
As the product has more binding energy, \(Q > 0\), which signifies that the process produces energy. The decrease in mass is converted into this released energy.
Quick Tip: Think of binding energy as "negative energy". A more tightly bound (more stable) system has a lower total energy. To go from a less bound state to a more bound state, the system must release the energy difference.

Step 1: Understanding the Concept:

Extrinsic semiconductors are created by doping an intrinsic semiconductor (like Si or Ge) with impurity atoms. This doping process intentionally increases the concentration of one type of charge carrier (either electrons or holes) over the other.


Step 2: Detailed Explanation:

The charge carriers in a semiconductor are free electrons and holes. In an extrinsic semiconductor, their concentrations are unequal.

1. n-type Semiconductor:

Formation: Created by doping a pure semiconductor with pentavalent impurities (e.g., Phosphorus, Arsenic). These impurity atoms have five valence electrons.
Majority Carriers: Four valence electrons of the impurity form covalent bonds with the semiconductor atoms, and the fifth electron is loosely bound and easily becomes a free electron for conduction. This results in a large number of free electrons. Therefore, in an n-type semiconductor, the majority charge carriers are electrons.
Minority Carriers: Holes are also present due to the thermal breaking of covalent bonds, but their concentration is much smaller than the electron concentration. Therefore, the minority charge carriers are holes.


2. p-type Semiconductor:

Formation: Created by doping a pure semiconductor with trivalent impurities (e.g., Boron, Aluminum). These impurity atoms have three valence electrons.
Majority Carriers: The three valence electrons of the impurity form covalent bonds, but this leaves a vacancy or "hole" in the fourth bond. This hole can accept an electron from a neighboring atom, effectively making the hole a mobile positive charge carrier. Doping creates a large number of such holes. Therefore, in a p-type semiconductor, the majority charge carriers are holes.
Minority Carriers: Free electrons are also present due to thermal agitation, but their concentration is far less than the hole concentration. Therefore, the minority charge carriers are electrons. Quick Tip: A simple mnemonic: \textbf{n}-type has an excess of \textbf{n}egative electrons. \textbf{p}-type has an excess of \textbf{p}ositive holes.

Step 1: Understanding the Concept:

Forward biasing a p-n junction involves applying an external voltage that opposes the built-in potential barrier of the junction. This allows a significant current to flow across the junction.


Step 2: Detailed Explanation:

A p-n junction is forward biased by connecting the positive terminal of an external battery to the p-side and the negative terminal to the n-side.

Movement of Charge Carriers:

Effect on Depletion Region: The applied external voltage (\(V_{ext}\)) opposes the internal barrier potential (\(V_b\)). This reduces the effective potential barrier across the junction to (\(V_b - V_{ext}\)) and narrows the width of the depletion region.
Movement of Majority Carriers:

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 Current: With the potential barrier lowered, these majority carriers have enough energy to diffuse across the now-narrowed depletion region. Holes from the p-side diffuse into the n-side, and electrons from the n-side diffuse into the p-side.
Recombination: Once the carriers cross the junction, they become minority carriers (holes in the n-side, electrons in the p-side). They recombine with the majority carriers near the junction. For every electron-hole recombination, a covalent bond is completed, and an electron is supplied by the negative terminal of the battery to the n-side, travels to the p-side, and enters the positive terminal.
Conduction: This continuous flow of majority carriers across the junction and their subsequent recombination constitutes a large forward current, known as the diffusion current. The magnitude of this current is typically in milliamperes (mA) and increases exponentially with the applied forward voltage.

In summary, the forward current is primarily due to the diffusion of majority charge carriers across the junction, which is made possible by the reduction of the potential barrier.
Quick Tip: Remember: Forward Bias -> Potential Barrier Decreases -> Depletion Region Narrows -> Majority Carrier Diffusion -> Large Forward Current.

Step 1: Understanding the Concept:

The dynamic resistance (or AC resistance) of a diode is the resistance it offers to a small change in voltage. It is defined as the reciprocal of the slope of the I-V characteristic curve at a specific operating point.


Step 2: Key Formula or Approach:

The dynamic resistance \(r_d\) is given by: \[ r_d = \frac{\Delta V}{\Delta I} \]
where \(\Delta V\) is a small change in voltage around the operating point and \(\Delta I\) is the corresponding small change in current.


Step 3: Detailed Explanation:

We need to estimate the dynamic resistance at the operating point \(V = -0.6\) V.
This voltage is in the reverse bias region of the diode characteristic.
Let's examine the I-V graph in the vicinity of \(V = -0.6\) V.

The graph shows that in the reverse bias region (from V=0 to beyond V=-1.2 V), the current \(I\) is extremely small and is practically zero on the scale of the graph (which is in mA).
The curve is essentially a horizontal line lying on the V-axis (\(I=0\)).
To calculate the dynamic resistance, we look at the slope of the curve. The slope is \(\frac{\Delta I}{\Delta V}\).
Since the curve is a horizontal line in this region, the change in current \(\Delta I\) for any small change in voltage \(\Delta V\) is zero.
\[ \Delta I \approx 0 \]
Therefore, the slope is \(\frac{\Delta I}{\Delta V} \approx 0\).
The dynamic resistance is the reciprocal of the slope:
\[ r_d = \frac{\Delta V}{\Delta I} = \frac{1}{slope} = \frac{1}{0} \to \infty \]

The dynamic resistance of the diode in the reverse bias region is extremely high. Based on the provided graph, the current is effectively zero, making the dynamic resistance practically infinite.


Step 4: Final Answer:

The dynamic resistance of the diode at V = -0.6 V is extremely high, approaching infinity, as the current in this reverse bias region is essentially zero and does not change with voltage.
Quick Tip: For an ideal diode, the forward resistance is zero and the reverse resistance is infinite. For a real diode, the forward dynamic resistance is small (tens of ohms), and the reverse dynamic resistance is very large (megohms or more). Looking at the slope of the I-V curve is the key: a steep slope (forward bias) means low resistance, and a flat slope (reverse bias) means high resistance.

Step 1: Understanding the Concept:

The question asks how the electric field (E) between the plates of a parallel plate capacitor changes when both the separation between the plates (d) and the potential difference across them (V) are altered.


Step 2: Key Formula or Approach:

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

Step 3: Detailed Explanation:

Let the initial conditions be:

Initial electric field = \(E_{initial} = E\)
Initial plate separation = \(d_{initial} = d\)
Initial potential difference = \(V_{initial} = V\)

From the formula, we have the initial relationship: \[ E = \frac{V}{d} \]

Now, the conditions are changed as follows:

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

Let the new electric field be \(E_{new}\). Using the same formula for the new conditions: \[ E_{new} = \frac{V_{new}}{d_{new}} \]
Substitute the new values of V and d: \[ E_{new} = \frac{(V/2)}{(2d)} = \frac{V}{4d} \]
Now, we can express \(E_{new}\) in terms of the initial electric field \(E\). Since \(E = \frac{V}{d}\), we can substitute this into our expression for \(E_{new}\): \[ E_{new} = \frac{1}{4} \left( \frac{V}{d} \right) = \frac{1}{4} E \]

Step 4: Final Answer:

The new value of the electric field between the plates will be \(\frac{E}{4}\).
Quick Tip: When analyzing how a quantity changes, write down the formula, identify the variables that are changing, and see how they affect the result. In this case, \(E \propto V\) and \(E \propto 1/d\). So, halving V halves E, and doubling d also halves E. The combined effect is \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).

Step 1: Understanding the Concept:

The question asks for the relationship between the potential difference (V) and the plate separation (d) of a parallel plate capacitor, given that the electric field (E) between the plates is kept 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: \[ V = E \cdot d \]

Step 3: Detailed Explanation:

We are given that the electric field E is to be maintained constant.
Let's analyze the equation \(V = E \cdot d\).

Since E is a constant, the equation is of the form \(V = (constant) \times d\).
This is a linear relationship between V and d.
Comparing this to the equation of a straight line, \(y = mx + c\), we have:

\(y = V\) (plotted on the vertical axis)
\(x = d\) (plotted on the horizontal axis)
The slope \(m = E\) (which is a positive constant)
The y-intercept \(c = 0\) (since if \(d=0\), then \(V=0\))

Therefore, the graph of V versus d should be a straight line that passes through the origin and has a positive slope equal to the constant electric field E.

Looking at the options:

(A) shows a parabolic curve. Incorrect.
(B) shows a straight line passing through the origin with a positive slope. This correctly represents the relationship \(V \propto d\). Correct.
(C) shows a hyperbolic curve, which would represent an inverse relationship. Incorrect.
(D) shows a triangular graph. Incorrect.


Step 4: Final Answer:

To maintain a constant electric field E, the potential difference V must be directly proportional to the separation d. This relationship is represented by a straight line passing through the origin with a positive slope.
Quick Tip: The formula \(E = V/d\) is fundamental for parallel plate capacitors. Remember that if one quantity is constant, the relationship between the other two is simplified. Constant E means \(V \propto d\). Constant V means \(E \propto 1/d\). Constant d means \(E \propto V\).

Step 1: Understanding the Concept:

This question asks about the nature of the electric field inside and outside an ideal parallel plate capacitor. The line MN extends from a point M inside the capacitor to a point N outside the capacitor.


Step 2: Detailed Explanation:


Inside the Capacitor (between plates P and Q): For an ideal parallel plate capacitor, the electric field in the region between the plates is uniform and constant. This means its magnitude E is the same at all points between the plates. The direction of the field is from the positive plate (Q) to the negative plate (P).
Outside the Capacitor (region beyond Q): For an ideal, infinitely large parallel plate capacitor, the electric field outside the plates is zero. The fields from the positive and negative plates cancel each other out in the exterior region.

Now let's trace the magnitude of the electric field along the line MN:

From point M (between the plates) to the edge of plate Q, the electric field E has a constant, non-zero magnitude.
As we cross the boundary of plate Q and move towards point N (outside the plates), the electric field magnitude abruptly drops to zero (ideally).

This behavior is best represented by a graph that shows a constant value for E inside the plates and a value of zero outside.

Analyzing the options:

(A) shows a sinusoidal variation, which is incorrect.
(B) shows a constant, non-zero value of E between the plates (from M to the edge of Q) and then drops to zero for the region outside (towards N). This correctly represents the ideal electric field.
(C) shows a triangular variation, which is incorrect.
(D) shows a trapezoidal variation, which might represent the field with fringing effects, but the ideal model is a rectangular pulse. This graph is the best representation among the given choices.


Step 3: Final Answer:

The electric field between the plates of a parallel plate capacitor is constant, and it is zero outside. The graph that correctly depicts this variation along the line MN is a rectangular pulse.
Quick Tip: For ideal capacitors, always assume the electric field is uniform and confined strictly between the plates, and zero everywhere else. Real capacitors have "fringing fields" at the edges, but this is usually ignored in introductory problems unless specified.

Step 1: Understanding the Concept:

This problem deals with calculating the total potential difference across a system of multiple parallel plates with different electric fields between them. The total potential difference is the sum of the potential differences across each region.


Step 2: Key Formula or Approach:

The potential difference (\(V\)) between two points in a uniform electric field (\(\vec{E}\)) is given by the line integral \(V = -\int \vec{E} \cdot d\vec{l}\). For a path parallel to the field over a distance d, the magnitude of the potential difference is \(|\Delta V| = E \cdot d\). The total potential difference across multiple regions is the algebraic sum of the potential differences across each region.


Step 3: Detailed Explanation:

Let the three plates be Plate 1, Plate 2, and Plate 3, arranged from bottom to top (or first to third).

Let the potential of Plate 1 be \(V_1\).
Let the potential of Plate 2 be \(V_2\).
Let the potential of Plate 3 be \(V_3\).

We are given:

The electric field between Plate 1 and Plate 2 is \(\vec{E}_1\). The distance is d.
The electric field between Plate 2 and Plate 3 is \(\vec{E}_2\). The distance is d.

The potential difference between Plate 2 and Plate 1 is: \[ V_2 - V_1 = E_1 \cdot d \]
(Assuming the fields are directed from plate 1 to 2, and 2 to 3, etc. We are concerned with the magnitude of the potential difference here, so let's use scalar values and sum them up).
The potential difference between Plate 3 and Plate 2 is: \[ V_3 - V_2 = E_2 \cdot d \]
We need to find the potential difference between the third plate and the first plate, which is \(V_3 - V_1\).
We can write this as: \[ V_3 - V_1 = (V_3 - V_2) + (V_2 - V_1) \]
Substituting the expressions for the potential differences across each region: \[ V_3 - V_1 = (E_2 \cdot d) + (E_1 \cdot d) \] \[ V_3 - V_1 = (E_1 + E_2) \cdot d \]
If we treat the electric fields and displacement as vectors, let the displacement vector from plate 1 to 3 be \(2\vec{d}\), where \(\vec{d}\) points from one plate to the next. The total potential difference is: \[ V_{31} = (V_3 - V_2) + (V_2 - V_1) = (\vec{E}_2 \cdot \vec{d}) + (\vec{E}_1 \cdot \vec{d}) = (\vec{E}_1 + \vec{E}_2) \cdot \vec{d} \]
The options use scalar notation 'd', implying we are considering the magnitudes along the direction perpendicular to plates. Therefore, the total potential difference is \( (E_1 + E_2)d \). However, the options are written in vector dot product form, so (A) is the most appropriate representation.


Step 4: Final Answer:

The total potential difference is the sum of the potential differences across the two gaps, which is \((\vec{E}_1 + \vec{E}_2) \cdot \vec{d}\). The option provided seems to have a typo using \(d\) instead of \(\vec{d}\), but (A) is the correct expression.
Quick Tip: Potential is a scalar quantity, so potential differences add up algebraically. When moving through different regions of electric fields, you can find the total potential difference by simply summing the potential differences (\(E \cdot d\)) for each region along the path.

Step 1: Understanding the Concept:

This question asks about the effect of inserting a dielectric material into a parallel plate capacitor on its capacitance. A dielectric is an insulating material that increases the capacitance.


Step 2: Key Formula or Approach:

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

Step 3: Detailed Explanation:

Let the initial capacitance (with air or vacuum) be C. \[ C = \frac{\epsilon_0 A}{d} \]
Now, a dielectric material of constant K is introduced, filling the entire space between the plates. The new capacitance, let's call it \(C_{new}\), is: \[ C_{new} = \frac{K \epsilon_0 A}{d} \]
We can see the relationship between \(C_{new}\) and the initial capacitance C by factoring out the expression for C: \[ C_{new} = K \left( \frac{\epsilon_0 A}{d} \right) \] \[ C_{new} = K \cdot C \]
The new capacitance becomes K times the original capacitance.


Step 4: Final Answer:

The new value of the capacitance becomes CK.
Quick Tip: A dielectric material always increases the capacitance of a capacitor. The dielectric constant K is the factor by which the capacitance increases. Since K is always greater than 1 for any material (K=1 for vacuum), the new capacitance will always be larger than the original.

Step 1: Understanding the Concept:

This question asks about the relationship between the \(V_s\) vs \(\nu\) graphs for different metals. The key is to analyze the equation given and identify which parameters are universal constants and which are material-dependent.


Step 2: Key Formula or Approach:

The photoelectric equation relating stopping potential \(V_s\) and frequency \(\nu\) is: \[ eV_s = h\nu - h\nu_0 \]
Rearranging this to get \(V_s\) as a function of \(\nu\): \[ V_s = \left(\frac{h}{e}\right)\nu - \frac{h\nu_0}{e} \]

Step 3: Detailed Explanation:

This equation is in the form of a straight line, \(y = mx + c\), where:

\(y = V_s\)
\(x = \nu\)
The slope \(m = \frac{h}{e}\)
The y-intercept \(c = -\frac{h\nu_0}{e}\)

Let's analyze the components:

Slope (m): The slope is the ratio of Planck's constant (h) to the elementary charge (e). Both h and e are fundamental physical constants, independent of the material of the metal surface. Therefore, the slope of the \(V_s\) vs \(\nu\) graph is the same for all metals.
Intercepts: The threshold frequency \(\nu_0\) (and the work function \(\phi = h\nu_0\)) is a characteristic property of the specific metal. Different metals have different work functions and hence different threshold frequencies. This means the x-intercept (\(\nu_0\)) and the y-intercept (\(-\frac{h\nu_0}{e}\)) will be different for different metals.

Since the graphs for two different metals will have the same slope (\(h/e\)) but different intercepts, they will be parallel straight lines.


Step 4: Final Answer:

The straight line graphs obtained for two different metals are parallel to each other.
Quick Tip: In the photoelectric effect graph of \(V_s\) vs \(\nu\), the slope is a universal constant (\(h/e\)), while the intercepts depend on the material's work function. This is a key experimental verification of Einstein's photoelectric theory.

Step 1: Understanding the Concept:

The question asks to find an expression for Planck's constant (h) in terms of the slope (m) of the \(V_s\) vs \(\nu\) graph and the elementary charge (e).


Step 2: Key Formula or Approach:

The equation for the graph is: \[ V_s = \left(\frac{h}{e}\right)\nu - \frac{h\nu_0}{e} \]
The slope of this line is \(m = \frac{rise}{run} = \frac{\Delta V_s}{\Delta \nu}\).


Step 3: Detailed Explanation:

From the equation \(V_s = (\frac{h}{e})\nu - \frac{h\nu_0}{e}\), we can directly identify the slope by comparing it to \(y = mx + c\).
The slope of the \(V_s\) versus \(\nu\) graph is: \[ slope = m = \frac{h}{e} \]
The question asks for the value of Planck's constant, h. We can rearrange this equation to solve for h: \[ h = m \cdot e \]
The provided solution seems to have a typo. Based on the standard formula, Planck's constant is the product of the slope of the graph and the elementary charge. The option (C) `me` correctly represents this product. The options provided in the OCR might be slightly different from the intended ones. Assuming `m` represents the slope and `e` the charge, `me` is the correct expression for Planck's constant.


Step 4: Final Answer:

The value of Planck's constant is given by the product of the slope of the graph (m) and the elementary charge (e), which is \(h=me\).
Quick Tip: This is a very important result. The experimental determination of the slope of the \(V_s\) vs \(\nu\) graph was one of the first and most accurate ways to measure Planck's constant, h.

Step 1: Understanding the Concept:

This question asks for the x-intercept and y-intercept of the \(V_s\) vs \(\nu\) graph based on the photoelectric equation.


Step 2: Key Formula or Approach:

The governing equation is: \[ V_s = \frac{h}{e}\nu - \frac{h\nu_0}{e} \]

The \(\nu\)-axis intercept (x-intercept) is the value of \(\nu\) when \(V_s = 0\).
The \(V_s\)-axis intercept (y-intercept) is the value of \(V_s\) when \(\nu = 0\).


Step 3: Detailed Explanation:

1. Intercept on \(\nu\)-axis (x-intercept):

Set \(V_s = 0\) 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 frequency axis is the threshold frequency, \(\nu_0\).


2. Intercept on \(V_s\)-axis (y-intercept):

Set \(\nu = 0\) in the equation: \[ V_s = \frac{h}{e}(0) - \frac{h\nu_0}{e} \] \[ V_s = -\frac{h\nu_0}{e} \]
So, the intercept on the stopping potential axis is \(-\frac{h\nu_0}{e}\).


Comparing with options:
The intercepts are \(\nu_0\) (on the \(\nu\)-axis) and \(-\frac{h\nu_0}{e}\) (on the \(V_s\)-axis) respectively.
Let's re-examine the options:
(A) \(\nu_0, \frac{h\nu_0}{e}\) - The second term has the wrong sign.
(B) \(\nu_0, h\nu_0\) - The second term is incorrect.
(C) \(\frac{h\nu_0}{e}, \nu_0\) - The order is reversed and the sign is wrong.
(D) \(h\nu_0, \nu_0\) - The first term is incorrect.

There appears to be an error in all the options, as the y-intercept must be negative. Option (A) is the closest if we are asked to consider only the magnitude of the intercept. Assuming the question asks for the intercepts as (\(\nu\)-intercept, magnitude of \(V_s\)-intercept), then (A) would be the intended answer. However, strictly speaking, none are correct. The provided solution is likely (A) with the assumption of magnitude.


Step 4: Final Answer:

The intercept on the \(\nu\)-axis is \(\nu_0\), and the intercept on the \(V_s\)-axis is \(-\frac{h\nu_0}{e}\). None of the options provided are technically correct due to the missing negative sign on the \(V_s\) intercept. Option (A) is the closest if magnitude is considered.
Quick Tip: Always calculate intercepts carefully. The x-intercept is found by setting y=0, and the y-intercept by setting x=0. Be mindful of the signs, as they have physical meaning. The negative y-intercept here signifies that for frequencies below the threshold, a stopping potential is not needed (as no electrons are emitted).

Step 1: Understanding the Concept:

This question asks about the relationship between a photon's wavelength (\(\lambda\)), its wave number (\(k\)), and its frequency (\(\nu\)).


Step 2: Key Formula or Approach:

1. Wave Number (\(k\)): The wave number is the reciprocal of the wavelength: \(k = \frac{1}{\lambda}\). It represents the number of wavelengths per unit distance.
2. Frequency (\(\nu\)): The frequency is related to the wavelength by the speed of light (c): \(c = \nu \lambda\), which implies \(\nu = \frac{c}{\lambda}\).


Step 3: Detailed Explanation:

Let the initial wavelength be \(\lambda_{initial} = \lambda\).
The wavelength is doubled, so the new wavelength is \(\lambda_{new} = 2\lambda\).

1. Change in Wave Number:

Initial wave number: \(k_{initial} = \frac{1}{\lambda}\)
New wave number: \(k_{new} = \frac{1}{\lambda_{new}} = \frac{1}{2\lambda}\)

To find how many times the wave number has become, we take the ratio: \[ \frac{k_{new}}{k_{initial}} = \frac{1/(2\lambda)}{1/\lambda} = \frac{1}{2\lambda} \times \lambda = \frac{1}{2} \]
So, the wave number becomes \(\frac{1}{2}\) times its original value.

2. Change in Frequency:

Initial frequency: \(\nu_{initial} = \frac{c}{\lambda}\)
New frequency: \(\nu_{new} = \frac{c}{\lambda_{new}} = \frac{c}{2\lambda}\)

To find how many times the frequency has become, we take the ratio: \[ \frac{\nu_{new}}{\nu_{initial}} = \frac{c/(2\lambda)}{c/\lambda} = \frac{c}{2\lambda} \times \frac{\lambda}{c} = \frac{1}{2} \]
So, the frequency also becomes \(\frac{1}{2}\) times its original value.

Step 4: Final Answer:

The wave number becomes \(1/2\) times and the frequency becomes \(1/2\) times their respective initial values. The correct pair is (\(\frac{1}{2}, \frac{1}{2}\)).
Quick Tip: Remember that both wave number and frequency are inversely proportional to the wavelength. Therefore, if the wavelength is multiplied by a factor 'x', both the wave number and frequency will be multiplied by a factor '1/x'.

Step 1: Understanding the Concept:

This question relates the momentum of a photon to its wavelength using the de Broglie wavelength formula. For a photon, this relationship is fundamental.


Step 2: Key Formula or Approach:

The de Broglie relation gives the wavelength (\(\lambda\)) of a particle in terms of its momentum (p) and Planck's constant (h): \[ \lambda = \frac{h}{p} \]
(Constants: Planck's constant \(h \approx 6.63 \times 10^{-34}\) J·s)


Step 3: Detailed Explanation:

We are given:

Momentum of the photon, \(p = 5.0 \times 10^{-29}\) kg·m/s.

We need to find the wavelength \(\lambda\).
Using the formula: \[ \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} J·s}{5.0 \times 10^{-29} kg·m/s} \] \[ \lambda = \frac{6.63}{5.0} \times 10^{-34 - (-29)} m \] \[ \lambda = 1.326 \times 10^{-5} m \]
The options are given in micrometers (\(\mu\)m). We need to convert our answer.
Since \(1 \, \mum = 10^{-6}\) m, we can write: \[ \lambda = 1.326 \times 10 \times 10^{-6} m = 13.26 \times 10^{-6} m \] \[ \lambda = 13.26 \, \mum \]
This value is closest to option (D).


Step 4: Final Answer:

The wavelength of the photon is approximately 13.3 \(\mu\)m.
Quick Tip: Always check the units in the final step. Physics problems in exams often require a conversion to match the units given in the options (e.g., from meters to micrometers or nanometers).

Step 1: Understanding the Concept:

This problem deals with electrostatic potential and charge distribution on conductors. When two conductors are connected by a wire, they form a single conductor and must come to the same potential. Charge will redistribute itself accordingly.


Step 2: Key Formula or Approach:

1. Potential of a sphere: The potential of a conducting sphere of radius 'a' and charge 'Q' is \(V = \frac{kQ}{a}\), where \(k = \frac{1}{4\pi\epsilon_0}\).
2. Charge Redistribution: When conductors are connected, charge flows until the potential is uniform throughout the connected system.
3. Properties of Conductors: In electrostatic equilibrium, all charge on a conductor resides on its outer surface.


Step 3: Detailed Explanation:

Initial State:

Sphere A has radius r and is charged to a potential V. Its charge, let's call it q, can be found from the potential formula: \(V = \frac{kq}{r} \implies q = \frac{Vr}{k}\).
Shell B is initially uncharged (assumed, as not stated otherwise).


Final State (after connecting with a wire):

When sphere A and shell B are connected by a wire, they become a single conducting system.
According to the properties of conductors, in electrostatic equilibrium, charge must reside on the outermost surface of the conductor. In this system, the outer surface is the surface of shell B (radius R).
Therefore, the entire initial charge \(q\) from sphere A will flow through the wire and redistribute itself onto the outer surface of shell B. The final charge on sphere A becomes zero, and the final charge on shell B becomes \(q\).
Since the two are connected and form a single conductor, they must be at the same final potential, \(V_{final}\). This potential will be the potential of the outer shell B.
The potential of a spherical shell of radius R with charge q on its surface is given by:
\[ V_{shell} = \frac{kq}{R} \]
The potential inside this shell (and thus on the surface of sphere A) is constant and equal to the potential on the shell's surface.
So, the final potential of both sphere A and shell B is:
\[ V_{final} = \frac{kq}{R} \]

We can also express this in terms of the initial potential V of sphere A. Since \(q = \frac{Vr}{k}\): \[ V_{final} = \frac{k}{R} \left( \frac{Vr}{k} \right) = V \frac{r}{R} \]

Step 4: Final Answer:

After connecting, the entire charge moves to the outer shell B. The final potential of both the sphere and the shell becomes uniform and is equal to the potential of the outer shell, which is \(V_{final} = \frac{kq}{R}\) or \(V_{final} = V\frac{r}{R}\).
Quick Tip: A key principle of electrostatics: For a system of concentric conducting shells, if they are connected, the entire charge will always move to the outermost shell. The entire system then acts as a single conductor with a potential determined by the total charge and the radius of the outermost shell.

Step 1: Understanding the Concept:

This question has two parts. The first asks for the properties of equipotential surfaces. The second part involves calculating the potential at a point in a uniform electric field, given the potential at another point.


Step 2: Key Formula or Approach:

1. Equipotential Surfaces: Surfaces where the electric potential is constant.
2. Potential Difference in a Uniform Field: The relationship between electric field \(\vec{E}\) and potential V is \(\vec{E} = -\nabla V\). For a uniform field, the potential difference between two points A and B is given by \(\Delta V = V_B - V_A = -\vec{E} \cdot \Delta\vec{r}\), where \(\Delta\vec{r}\) is the displacement vector from A to B.


Step 3: Detailed Explanation:

Part 1: Characteristics of Equipotential Surfaces

No Work Done: No work is done in moving a 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 (\(W = q\Delta V = 0\)).
Perpendicular to Electric Field: The electric field lines are always perpendicular to the equipotential surfaces at every point. If they were not, there would be a component of the electric field along the surface, which would imply a potential difference along the surface, contradicting the definition.
No Intersection: Two different equipotential surfaces can never intersect. If they did, there would be two different values of electric potential at the point of intersection, which is not possible.


Part 2: Calculating Potential
We are given:

Uniform electric field \(\vec{E} = 50 \, \hat{i} \, N/C\) (along the +x axis).
Potential at the origin (Point A), \(V_A = V(0,0) = 220\) V.
We need to find the potential at Point B (4m, 3m), \(V_B = V(4,3)\).

The displacement vector from A to B is: \[ \Delta\vec{r} = \vec{r}_B - \vec{r}_A = (4\hat{i} + 3\hat{j}) - (0\hat{i} + 0\hat{j}) = 4\hat{i} + 3\hat{j} \]
Now, use the formula for potential difference: \[ V_B - V_A = -\vec{E} \cdot \Delta\vec{r} \] \[ V_B - 220 = -(50\hat{i}) \cdot (4\hat{i} + 3\hat{j}) \]
Calculate the dot product: \[ (50\hat{i}) \cdot (4\hat{i} + 3\hat{j}) = (50 \times 4) + (0 \times 3) = 200 \]
So, the equation becomes: \[ V_B - 220 = -200 \] \[ V_B = 220 - 200 = 20 V \]

Step 4: 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 potential only changes with the x-coordinate. The equipotential surfaces are planes parallel to the y-z plane. Therefore, the potential at (4m, 3m) is the same as the potential at (4m, 0m) or any point on the plane x=4. The change in potential is simply \(-E_x \Delta x\).

Step 1: Understanding the Concept:

This question asks for the topological distinction between open and closed surfaces, which is important in vector calculus and physics, particularly in the context of flux. It also asks to represent the area vector for a spherical surface.


Step 2: Detailed Explanation:

Difference between Open and Closed Surface:

Open Surface: An open surface is a surface that has a boundary or an edge. It does not enclose a volume. Examples include a flat sheet of paper, a hemisphere without its base, or a cylindrical surface without its top and bottom caps.
Closed Surface: A closed surface is a surface that has no boundary or edge and completely encloses a finite volume. You cannot get from the inside to the outside without crossing the surface. Examples include the surface of a sphere, a cube, or a complete cylinder with its top and bottom caps.

The main physical distinction arises when calculating flux. For an open surface, the direction of the area vector can be chosen, but for a closed surface, the area vector is conventionally defined to point outwards from the enclosed volume.


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

The elementary surface area vector \(d\vec{S}\) (also written as \(d\vec{A}\)) for any surface is a vector that has:

Magnitude: Equal to the infinitesimal area \(dS\).
Direction: Perpendicular to the surface at that point.

For a closed surface like a sphere, the direction is conventionally taken as the outward normal.

Diagram:
% A placeholder for a diagram showing:
% 1. A sphere S with its center at the origin.
% 2. A small patch of area dS on the surface of the sphere.
% 3. A vector dS originating from this patch, pointing radially outward, perpendicular to the tangent plane at that point.
The vector \(d\vec{S}\) for a spherical surface is always directed radially outward from the center of the sphere.
Quick Tip: The distinction between open and closed surfaces is crucial for applying integral theorems like Gauss's Law (which applies only to closed surfaces) and Stokes' Theorem (which relates a line integral over the boundary of an open surface to a surface integral over that surface).

Step 1: Understanding the Concept:

This question asks for the definition of electric flux and the role and properties of a Gaussian surface, a key concept in applying Gauss's Law.


Step 2: Detailed Explanation:

Definition of Electric Flux:

Electric flux (\(\Phi_E\)) through a surface is a measure of the total number of electric field lines passing normally through that surface. Mathematically, for a uniform electric field \(\vec{E}\) passing through a small planar area \(\Delta\vec{S}\), the flux is \(\Delta\Phi_E = \vec{E} \cdot \Delta\vec{S}\). For a general surface in a non-uniform field, the total electric flux is given by the surface integral of the electric field over the entire surface: \[ \Phi_E = \int_S \vec{E} \cdot d\vec{S} \]

Significance of a Gaussian Surface:

A Gaussian surface is an imaginary closed surface used in conjunction with Gauss's Law to calculate the electric field. Its significance lies in its utility:

It allows for the simplification of electric field calculations for symmetric charge distributions (e.g., spherical, cylindrical, planar).
By choosing a Gaussian surface on which the magnitude of the electric field is constant and its direction is either parallel or perpendicular to the surface normal, the flux integral \(\int \vec{E} \cdot d\vec{S}\) simplifies to \(E \cdot A\), making it easy to solve for E.
Gauss's Law, \(\Phi_E = \oint \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\epsilon_0}\), relates the flux through this closed surface to the net charge enclosed within it.


Why an external charge does not contribute to the total flux:

Consider a charge \(q_{out}\) located outside a closed Gaussian surface S.
The electric field lines originating from (or terminating on) this external charge will enter the Gaussian surface at some points and must exit the surface at other points, as the charge itself is not inside.

The flux entering the surface is considered negative (because \(\vec{E}\) and the outward normal \(d\vec{S}\) are in opposite directions, so \(\vec{E} \cdot d\vec{S}\) is negative).
The flux exiting the surface is considered positive (because \(\vec{E}\) and the outward normal \(d\vec{S}\) are in the same general direction, so \(\vec{E} \cdot d\vec{S}\) is positive).

For any closed surface, the number of field lines from an external charge that enter the surface is exactly equal to the number of field lines that leave the surface. Therefore, the total net flux through the closed surface due to the external charge is zero. The positive (outgoing) flux cancels out the negative (incoming) flux. This is the mathematical basis for Gauss's law only considering the enclosed charge.
Quick Tip: Gauss's Law is powerful but only useful for calculation when the charge distribution has high symmetry. The key is to choose a Gaussian surface that matches the symmetry of the problem.

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 through any closed surface is proportional to the total electric charge enclosed within that surface.


Step 2: Key Formula or Approach:

Gauss's Law: \(\Phi_E = \oint \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\epsilon_0}\)


Step 3: Detailed Explanation:

Let's apply Gauss's Law to each of the spherical shells.

1. Flux through the inner shell S\(_1\):

The flux through S\(_1\), denoted as \(\Phi_1\), depends only on the net charge enclosed within S\(_1\).
The charges inside S\(_1\) are \(q_1, q_2, and q_3\).
The total charge enclosed by S\(_1\) is: \[ q_{enc,1} = q_1 + q_2 + q_3 = (-3 \muC) + (-2 \muC) + (9 \muC) = 4 \muC \]
According to Gauss's Law, the flux through S\(_1\) is: \[ \Phi_1 = \frac{q_{enc,1}}{\epsilon_0} = \frac{4 \muC}{\epsilon_0} \]

2. Flux through the outer shell S\(_2\):

The flux through S\(_2\), denoted as \(\Phi_2\), depends on all the charges enclosed within S\(_2\).
The charges inside S\(_2\) include all the charges inside S\(_1\) (\(q_1, q_2, q_3\)) plus the charge Q placed between the shells.
The total charge enclosed by S\(_2\) is: \[ q_{enc,2} = q_1 + q_2 + q_3 + Q = q_{enc,1} + Q = (4 \muC) + Q \]
According to Gauss's Law, the flux through S\(_2\) is: \[ \Phi_2 = \frac{q_{enc,2}}{\epsilon_0} = \frac{(4 \muC) + Q}{\epsilon_0} \]

3. Relating the fluxes and solving for Q:

We are given the condition that the flux through S\(_2\) is four times the flux through S\(_1\): \[ \Phi_2 = 4 \Phi_1 \]
Substitute the expressions for the fluxes: \[ \frac{(4 \muC) + Q}{\epsilon_0} = 4 \left( \frac{4 \muC}{\epsilon_0} \right) \]
The \(\epsilon_0\) term cancels out from both sides: \[ (4 \muC) + Q = 4 \times (4 \muC) \] \[ 4 \muC + Q = 16 \muC \]
Now, solve for Q: \[ Q = 16 \muC - 4 \muC = 12 \muC \]

Step 4: Final Answer:

The value of the charge Q is 12 \(\mu\)C.
Quick Tip: When applying Gauss's Law, the most important step is to correctly identify the total charge enclosed by the imaginary Gaussian surface. Charges outside the surface contribute to the electric field but not to the net flux through the surface.

Step 1: Understanding the Concept:

This question covers the fundamental principles of magnetic forces. It asks for the origin of the force on a wire in a B-field, the derivation of the force between two parallel wires, and the resulting definition of the Ampere.


Step 2: Detailed Explanation:

Source of the Force:

A current in a conductor consists of moving charge carriers (usually electrons). When this conductor is placed in an external magnetic field, each of these moving charge carriers experiences a magnetic Lorentz force, given by \(\vec{F} = q(\vec{v} \times \vec{B})\). The total force on the conductor is the vector sum of the individual Lorentz forces acting on all the charge carriers within it. This net force is what we observe as the force on the current-carrying conductor.

Expression for Force between two Parallel Conductors:

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

Field produced by Wire 1 at Wire 2: Wire 1 produces a magnetic field (\(\vec{B}_1\)) 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 the currents are upwards, the direction of \(\vec{B}_1\) at Wire 2 is into the page.
Force on Wire 2: Wire 2, carrying current \(I_2\), is now situated in the magnetic field \(\vec{B}_1\). The force on a length L of Wire 2 is given by the formula \(\vec{F} = I(\vec{L} \times \vec{B})\).
The magnitude of the force on length L of Wire 2 is:
\[ F_2 = I_2 L B_1 \sin(90^\circ) = I_2 L B_1 \]
Substitute the expression for \(B_1\):
\[ F_2 = I_2 L \left( \frac{\mu_0 I_1}{2\pi d} \right) = \frac{\mu_0 I_1 I_2 L}{2\pi d} \]
Force per unit length: The force per unit length (\(f\)) on Wire 2 is:
\[ f = \frac{F_2}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]
Direction of Force: Using the Fleming's Left-Hand Rule (or vector cross product), with current \(\vec{I}_2\) upwards and field \(\vec{B}_1\) into the page, the force \(\vec{F}_2\) is directed towards Wire 1 (attractive).

By Newton's third law, Wire 1 experiences an equal and opposite force, also attractive. If the currents were in opposite directions, the force would be repulsive.

Definition of 'Ampere':

The SI unit of current, the ampere, is defined based on the force formula derived above.
One ampere is defined as that constant current which, if maintained in two long, straight parallel conductors of negligible circular cross-section, and placed one meter apart in vacuum, would produce between these conductors a force equal to \(2 \times 10^{-7}\) newton per meter of length.

We can see this from the formula:
If \(I_1 = I_2 = 1\) A and \(d = 1\) m, the force per unit length is: \[ f = \frac{\mu_0 (1)(1)}{2\pi (1)} = \frac{4\pi \times 10^{-7}}{2\pi} = 2 \times 10^{-7} N/m \] Quick Tip: Remember: "Parallel currents attract, anti-parallel currents repel." This can be a quick way to determine the direction of the force without using the full vector rules every time. The definition of the Ampere is a cornerstone of electromagnetism and is directly tied to the value of the permeability of free space, \(\mu_0\).

Step 1: Understanding the Concept:

This question asks for the work done by the magnetic Lorentz force. Work is done by a force only if there is a component of the force along the direction of displacement.


Step 2: Key Formula or Approach:

1. Magnetic Lorentz Force: The force \(\vec{F}_m\) on a charge q moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is given by \(\vec{F}_m = q(\vec{v} \times \vec{B})\).
2. Work Done: The work done (W) by a force \(\vec{F}\) for a small displacement \(d\vec{l}\) is \(dW = \vec{F} \cdot d\vec{l}\).


Step 3: Detailed Explanation:

From the formula for the magnetic Lorentz force, \(\vec{F}_m = q(\vec{v} \times \vec{B})\), we know that the force vector \(\vec{F}_m\) is the result of a cross product between the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\).

A fundamental property of the vector cross product is that the resulting vector is always perpendicular to both of the original vectors. Therefore, the magnetic force \(\vec{F}_m\) is always perpendicular to the velocity \(\vec{v}\) of the charged particle.

The instantaneous displacement of the particle, \(d\vec{l}\), is in the direction of its instantaneous velocity \(\vec{v}\).
So, we can say that the force \(\vec{F}_m\) is always perpendicular to the displacement \(d\vec{l}\).

The work done by this force is given by: \[ dW = \vec{F}_m \cdot d\vec{l} = |\vec{F}_m| |d\vec{l}| \cos\theta \]
where \(\theta\) is the angle between the force and the displacement. Since \(\vec{F}_m \perp d\vec{l}\), the angle \(\theta = 90^\circ\). \[ dW = |\vec{F}_m| |d\vec{l}| \cos(90^\circ) = |\vec{F}_m| |d\vec{l}| \times 0 = 0 \]
Since the work done for any small displacement is zero, the total work done by the magnetic force over any path is also zero.


Step 4: Final Answer:

The magnetic force is always perpendicular to the velocity of the charge, and thus to its displacement. Therefore, the work done by the magnetic force on the charge is always zero.
Quick Tip: The magnetic force can change the direction of a charged particle's motion, but it cannot change its speed or kinetic energy. This is a direct consequence of the fact that it does no work on the particle.

Step 1: Understanding the Concept:

The path of a charged particle in a uniform magnetic field depends on the angle between its velocity vector and the magnetic field vector. A helical path is a combination of circular and linear motion.


Step 2: Detailed Explanation:

For a charged particle to follow a helical (or spiral) path in a uniform magnetic field \(\vec{B}\), its initial velocity vector \(\vec{v}\) must have components both parallel and perpendicular to the magnetic field. This occurs when the angle \(\theta\) between \(\vec{v}\) and \(\vec{B}\) is not \(0^\circ\), \(90^\circ\), or \(180^\circ\).

Let's resolve the velocity vector \(\vec{v}\) into two components:

Parallel component (\(v_{\parallel}\)): This component is parallel to the magnetic field \(\vec{B}\). Its magnitude is \(v_{\parallel} = v \cos\theta\). The magnetic force due to this component is \(F = q(v_{\parallel} B \sin(0^\circ)) = 0\). Since there is no force related to this component, the particle continues to move along the direction of the magnetic field with a constant velocity \(v_{\parallel}\). This constitutes the linear part of the motion.
Perpendicular component (\(v_{\perp}\)): This component is perpendicular to the magnetic field \(\vec{B}\). Its magnitude is \(v_{\perp} = v \sin\theta\). The magnetic force due to this component is \(F = q(v_{\perp} B \sin(90^\circ)) = qv_{\perp}B\). This force is always perpendicular to both \(v_{\perp}\) and \(\vec{B}\), and it acts as a centripetal force, causing the particle to execute uniform circular motion in a plane perpendicular to the magnetic field.


Resulting Motion:
The superposition of these two motions—a constant linear motion along the field lines and a uniform circular motion in a plane perpendicular to the field lines—results in a three-dimensional path called a helix. The particle spirals around the magnetic field lines.

Necessary Conditions Summary:

A uniform magnetic field must be present.
The particle must be charged.
The initial velocity vector \(\vec{v}\) of the particle must be directed at an angle \(\theta\) to the magnetic field \(\vec{B}\) such that \(0^\circ < \theta < 90^\circ\) or \(90^\circ < \theta < 180^\circ\). Quick Tip: Remember the special cases: If \(\theta=0^\circ\) or \(180^\circ\), the path is a straight line (no force). If \(\theta=90^\circ\), the path is a perfect circle. For any other angle, the path is a helix.

Step 1: Understanding the Concept:

This question asks for the justification of treating a loop of current as a magnetic dipole. The key is the similarity between the magnetic field produced by the loop and the field of a standard magnetic dipole (like a short bar magnet).


Step 2: Detailed Explanation:

A magnetic dipole is a system of two equal and opposite magnetic poles separated by a small distance. A classic example is a small bar magnet. It produces a characteristic magnetic field pattern.

A current-carrying loop also produces a magnetic field. We can explain why it is considered a magnetic dipole as follows:

Magnetic Field Pattern: The magnetic field lines produced by a circular current loop are very similar to those of a bar magnet. The lines emerge from one face of the loop, loop around, and enter the other face. This creates two distinct magnetic poles.
Existence of Poles: One face of the loop acts as a North pole, and the other acts as a South pole. The polarity can be determined by the "Clock Rule":

If an observer looking at a face of the loop sees the current flowing in a counter-clockwise direction, that face behaves as a North pole.
If the observer sees the current flowing in a clockwise direction, that face behaves as a South pole.

Magnetic Dipole Moment: Just like a bar magnet, the current loop can be characterized by a vector quantity called the magnetic dipole moment, \(\vec{M}\). The magnitude of this moment is given by \(M = IA\), where I is the current and A is the area of the loop. Its direction is perpendicular to the plane of the loop.
Behavior in External Field: When placed in an external magnetic field, a current loop experiences a torque (\(\vec{\tau} = \vec{M} \times \vec{B}\)) that tends to align its magnetic moment with the external field, exactly like a bar magnet or a compass needle.

Because a current loop produces a dipole magnetic field and behaves like a bar magnet in an external field, it is considered a magnetic dipole.
Quick Tip: The equivalence between a current loop and a magnetic dipole is a fundamental concept. It forms the basis for understanding the magnetic properties of materials, as the magnetism of atoms can be modeled as arising from the orbital and spin motions of electrons, which are effectively tiny current loops.

Step 1: Understanding the Concept:

The magnetic dipole moment is a vector quantity that quantifies the strength and orientation of a magnet or any object that produces a magnetic field. For a current-carrying coil, it depends on the current, the area of the coil, and the number of turns.


Step 2: Derivation / Relation:

The magnetic dipole moment \(\vec{M}\) of a single planar loop of wire carrying a current \(I\) is defined as: \[ \vec{M} = I\vec{A} \]
where:

\(I\) is the magnitude of the steady current flowing in the loop.
\(\vec{A}\) is the area vector of the loop. The magnitude of \(\vec{A}\) is the area enclosed by the loop, and its direction is perpendicular to the plane of the loop.

If we have a coil that consists of N closely wound turns, each carrying the same current I and having the same area A, the total magnetic dipole moment is the vector sum of the individual moments of each turn. Since all turns are wound in the same direction, their magnetic moment vectors add up.
The relation for a coil with N turns is therefore: \[ \vec{M} = NI\vec{A} \]
The SI unit for magnetic dipole moment is ampere-meter squared (A·m\(^2\)).


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

The direction of the magnetic dipole moment vector \(\vec{M}\) (and thus the area vector \(\vec{A}\)) is determined by the Right-Hand Thumb Rule (or Right-Hand Curl Rule).
Rule: Curl the fingers of your right hand in the direction of the current flow around the coil. Your extended thumb will then point in the direction of the magnetic dipole moment \(\vec{M}\).
This direction is always normal (perpendicular) to the plane of the coil.
Quick Tip: The direction of the magnetic moment \(\vec{M}\) is the same as the direction of the magnetic field produced by the coil at its center. It points from the South pole face to the North pole face of the coil.

Step 1: Understanding the Concept:

This question explores the behavior of a magnetic dipole (a current coil) in a uniform external magnetic field, focusing on force, equilibrium, and magnetic flux.


Step 2: Detailed Explanation:

1. Net Force on the Coil:

In a uniform magnetic field \(\vec{B}_{ext}\), the forces on opposite sides of the current loop are equal in magnitude and opposite in direction. For example, in a rectangular loop, the force on one side is cancelled by the force on the opposite parallel side. The vector sum of the forces on all segments of the loop is zero.
Therefore, the net force acting on a current carrying coil in a uniform magnetic field is zero.


2. Orientation for Stable Equilibrium:

While the net force is zero, the coil experiences a torque given by \(\vec{\tau} = \vec{M} \times \vec{B}_{ext}\), where \(\vec{M}\) is the magnetic dipole moment of the coil. The potential energy of the dipole in the field is \(U = -\vec{M} \cdot \vec{B}_{ext}\).
Stable equilibrium is achieved when the system is at its lowest potential energy. \[ U = -MB_{ext}\cos\theta \]
The potential energy U is minimum when \(\cos\theta\) is maximum, which occurs when \(\cos\theta = 1\), i.e., when \(\theta = 0^\circ\).
An angle of \(\theta = 0^\circ\) means that the magnetic dipole moment vector \(\vec{M}\) is aligned and parallel to the external magnetic field vector \(\vec{B}_{ext}\). In this orientation, the torque \(\vec{\tau} = \vec{M} \times \vec{B}_{ext} = MB_{ext}\sin(0^\circ)\) is also zero.
So, the orientation for stable equilibrium is when \(\vec{M}\) is parallel to \(\vec{B}_{ext}\).


3. Flux in Stable Equilibrium:

The total magnetic field is the sum of the external field and the field produced by the loop itself: \(\vec{B}_{total} = \vec{B}_{ext} + \vec{B}_{loop}\).
The flux through the coil is \(\Phi = \int \vec{B}_{total} \cdot d\vec{A}\).

The field produced by the loop, \(\vec{B}_{loop}\), is in the same direction as its magnetic moment \(\vec{M}\).
In the stable equilibrium orientation, we found that \(\vec{M}\) is parallel to \(\vec{B}_{ext}\).
This implies that in stable equilibrium, \(\vec{B}_{loop}\) is also parallel to \(\vec{B}_{ext}\). The two fields are in the same direction.
The total field \(\vec{B}_{total}\) is therefore the sum of the magnitudes of the two fields, and it points in the same direction as both.
The area vector \(\vec{A}\) is also in the same direction as \(\vec{M}\) (and thus \(\vec{B}_{ext}\) and \(\vec{B}_{loop}\)).
The flux is \(\Phi = B_{total} A \cos(0^\circ) = (B_{ext} + B_{loop})A\).

Since in this orientation both constituent fields (\(\vec{B}_{ext}\) and \(\vec{B}_{loop}\)) are in the same direction and parallel to the area vector \(\vec{A}\) of the coil, their contributions to the flux add up constructively, resulting in the maximum possible flux through the coil.
Quick Tip: Think of a compass needle in the Earth's magnetic field. It aligns itself with the field (North pole pointing North) - this is its stable equilibrium position. A current loop behaves exactly the same way, aligning its magnetic moment \(\vec{M}\) with the external \(\vec{B}\) field.

Step 1: Understanding the Concept:

The problem involves image formation of an object of finite length placed along the principal axis of a mirror. A real and enlarged image is formed by a concave mirror when the object is placed between the center of curvature (C) and the focus (F). The condition "image just touches the pencil" implies that one end of the object and one end of the image coincide. This happens only at the center of curvature, where an object's image is formed at the same location.


Step 2: Key Formula or Approach:

1. Mirror Formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
2. Center of Curvature: \(R = 2f\). An object placed at \(u=-2f\) forms an image at \(v=-2f\).
3. Longitudinal Magnification (\(m_L\)): \(m_L = \frac{length of image}{length of object} = -\frac{v_2-v_1}{u_2-u_1}\). For a small object, \(m_L \approx -m^2\), where \(m\) is the lateral magnification.


Step 3: Detailed Explanation:

Since the image is real and enlarged, the mirror must be concave. Let its focal length be \(-f\). The center of curvature C is at a distance of \(2f\) from the pole.
Let one end of the pencil (end 1) be placed at the center of curvature C.

Object distance for end 1: \(u_1 = -2f\).
Using the mirror formula, the image distance is \(v_1 = -2f\). This image end coincides with the object end, satisfying the "just touches" condition.

The pencil has a length of \(L = f/4\) and is placed along the axis.
The other end of the pencil (end 2) must be closer to the focus.

Object distance for end 2: \(u_2 = u_1 + L = -2f + f/4 = -7f/4\).

Now, find the image position for end 2 (\(v_2\)): \[ \frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{-f} \] \[ \frac{1}{v_2} + \frac{1}{-7f/4} = \frac{1}{-f} \implies \frac{1}{v_2} - \frac{4}{7f} = -\frac{1}{f} \] \[ \frac{1}{v_2} = \frac{4}{7f} - \frac{1}{f} = \frac{4 - 7}{7f} = -\frac{3}{7f} \] \[ v_2 = -\frac{7f}{3} \]
The image of the pencil is formed between \(v_1 = -2f\) and \(v_2 = -7f/3\).
Length of the image = \(|v_2 - v_1| = |-\frac{7f}{3} - (-2f)| = |-\frac{7f}{3} + \frac{6f}{3}| = |-\frac{f}{3}| = \frac{f}{3}\).
Length of the object = \(f/4\).
The question asks for "the magnification produced". This is ambiguous.

Longitudinal Magnification: This measures how the length is magnified.
\[ m_L = \frac{Length of image}{Length of object} = \frac{f/3}{f/4} = \frac{4}{3} \]
Lateral Magnification: This is different for each point on the pencil. The magnitude of lateral magnification for end 2 is:
\[ |m_2| = \left|-\frac{v_2}{u_2}\right| = \left|-\frac{-7f/3}{-7f/4}\right| = \frac{4}{3} \]

Both interpretations lead to a value of 4/3. Since the object has length along the axis, longitudinal magnification is the most appropriate answer.

Step 4: Final Answer:

The longitudinal magnification produced by the mirror is \(\frac{4}{3}\).
Quick Tip: For an object placed along the principal axis, remember that the magnification is not uniform. The part of the object closer to the focus will be magnified more. The "touches the pencil" clue is a strong hint that one end of the object is at the center of curvature.

Step 1: Understanding the Concept:

This problem involves applying the prism formula, which relates the angle of incidence (i), angle of emergence (e), angle of prism (A), and angle of deviation (\(\delta\)), as well as Snell's law to find the refractive index.


Step 2: Key Formula or Approach:

1. Prism Formula: \(\delta = i + e - A\)
2. Relation for prism angle: \(A = r_1 + r_2\)
3. Snell's Law: \(n_1 \sin\theta_1 = n_2 \sin\theta_2\)


Step 3: Detailed Explanation:

We are given:

Angle of incidence, \(i = 45^\circ\)
Angle of deviation, \(\delta = 15^\circ\)
Angle of prism, \(A = 30^\circ\)

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

We use the prism formula to find the angle of emergence, e. \[ \delta = i + e - A \] \[ 15^\circ = 45^\circ + e - 30^\circ \] \[ 15^\circ = 15^\circ + e \] \[ e = 0^\circ \]
The angle of emergence (e) is the angle between the emergent ray and the normal to the emergent face (AC). An angle of emergence of \(0^\circ\) means that the emergent ray coincides with the normal. Therefore, the emergent ray is normal to the face AC.


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

Since the emergent ray is normal to the face AC, the angle of emergence \(e = 0^\circ\).
Inside the prism, the angle of refraction at the second face (AC) is \(r_2\). From Snell's law at this face, \(n \sin r_2 = 1 \sin e\). \[ n \sin r_2 = 1 \sin(0^\circ) = 0 \]
Since \(n \neq 0\), we must have \(\sin r_2 = 0\), which implies \(r_2 = 0^\circ\).

Now, we use the relation for the angle of the prism: \[ A = r_1 + r_2 \] \[ 30^\circ = r_1 + 0^\circ \] \[ r_1 = 30^\circ \]
Finally, we apply Snell's law at the first face (AB), where the ray enters the prism from air (\(n_1=1\)) into the material (\(n_2=n\)). \[ n_1 \sin i = n_2 \sin r_1 \] \[ 1 \cdot \sin(45^\circ) = n \cdot \sin(30^\circ) \] \[ \frac{1}{\sqrt{2}} = n \cdot \frac{1}{2} \]
Solving for n: \[ n = \frac{2}{\sqrt{2}} = \sqrt{2} \]

Step 4: Final Answer:

The refractive index of the material of the prism is \(\sqrt{2}\) (\(\approx 1.414\)).
Quick Tip: Normal emergence (\(e=0\)) or normal incidence (\(i=0\)) are common special cases in prism problems that greatly simplify the calculations. If \(e=0\), then \(r_2=0\) and \(A=r_1\). If \(i=0\), then \(r_1=0\) and \(A=r_2\).

Step 1: Understanding the Concept:

This problem involves Young's Double Slit Experiment (YDSE). Part (1) requires calculating the position of a specific bright fringe for a single wavelength. Part (2) involves finding the condition for the constructive interference maxima of two different wavelengths to overlap.


Step 2: Key Formula or Approach:

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


Step 3: Detailed Explanation:

Given data:

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

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

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

(2) Least distance for coincidence of bright fringes:

For the bright fringes to coincide, the position of the \(n_1\)-th bright fringe for \(\lambda_1\) must be the same as the position of the \(n_2\)-th bright fringe for \(\lambda_2\). \[ y_{coincide} = \frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \]
This simplifies to: \[ n_1 \lambda_1 = n_2 \lambda_2 \] \[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{480 nm}{600 nm} = \frac{48}{60} = \frac{4}{5} \]
We need the smallest non-zero integer values for \(n_1\) and \(n_2\) that satisfy this ratio. The least values are \(n_1 = 4\) and \(n_2 = 5\).
This means the 4th bright fringe of the 600 nm light coincides with the 5th bright fringe of the 480 nm light.
Now, we calculate the distance of this point from the central maximum using either wavelength: \[ y_{coincide} = \frac{n_1 \lambda_1 D}{d} = \frac{4 \times (600 \times 10^{-9} m) \times (1.0 m)}{1.0 \times 10^{-3} m} \] \[ y_{coincide} = 2400 \times 10^{-6} m = 2.4 \times 10^{-3} m = 2.4 mm \]

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 bright fringes coincide is 2.4 mm.
Quick Tip: For coincidence problems, the ratio of the orders of the fringes is the inverse of the ratio of the wavelengths: \(n_1/n_2 = \lambda_2/\lambda_1\). Find the simplest integer ratio to get the first point of coincidence.

Step 1: Understanding the Concept:

This question is about the single-slit diffraction phenomenon. Part (1) deals with the characteristics of the intensity pattern, while part (2) asks for a comparison of diffraction effects for light and sound.


Step 2: Detailed Explanation:

(1) Single-Slit Diffraction Pattern:

Intensity Variation Graph:
When monochromatic light passes through a narrow single slit, it diffracts, creating a characteristic pattern of bright and dark fringes on a screen. The graph of intensity (I) versus the angle of diffraction (\(\theta\)) shows:

A very bright and wide central maximum at \(\theta = 0\).
A series of much dimmer and narrower secondary maxima on either side of the central maximum.
The intensity of the secondary maxima decreases rapidly as we move away from the center. The first secondary maximum has an intensity of less than 5% of the central maximum.
Points of zero intensity, called minima, are located between the maxima.

% Placeholder for the graph of I vs. theta, showing a large central peak and smaller, decaying side peaks.

Expression for Zero Intensity (Minima):

The locations of the minima (zero intensity) in a single-slit diffraction pattern are given by the condition: \[ a \sin\theta = n\lambda \]
where:

\(a\) is the width of the slit.
\(\theta\) is the angle of diffraction.
\(\lambda\) is the wavelength of the light.
\(n\) is any non-zero integer (\(n = \pm 1, \pm 2, \pm 3, \dots\)). Note that \(n=0\) corresponds to the central maximum, not a minimum.


(2) Difference between Diffraction of Light and Sound:

The primary difference between the diffraction of light waves and sound waves in everyday experience stems from their vastly different wavelengths.

Condition for Diffraction: The phenomenon of diffraction is most prominent when the wavelength (\(\lambda\)) of the wave is comparable to or larger than the size of the obstacle or aperture (d), i.e., \(\lambda \gtrsim d\).
Wavelengths:

Sound Waves: Have long wavelengths, typically ranging from a few centimeters to several meters.
Light Waves: Have extremely short wavelengths, typically in the range of 400 to 700 nanometers (\(4 \times 10^{-7}\) to \(7 \times 10^{-7}\) m).

Observed Effect:

Because the wavelength of sound is comparable to the size of everyday objects (like doorways, corners, people), sound waves diffract readily around them. This is why we can hear sounds from around a corner even when we cannot see the source.
The wavelength of light is much smaller than everyday objects. Therefore, light travels in approximately straight lines and does not noticeably bend around large obstacles; it casts sharp shadows. To observe significant diffraction of light, a very narrow aperture or small obstacle with a size on the order of micrometers is required. Quick Tip: A simple rule of thumb: Longer wavelength = More diffraction. This explains why AM radio waves (long \(\lambda\)) can be received over hills, while FM radio and TV signals (shorter \(\lambda\)) require a clearer line of sight.