Step 1: Understanding the Concept:

This question is based on the concept of electrostatic shielding. A conductor, when placed in an external electric field, rearranges its free charges in such a way that the net electric field inside the conductor becomes zero. This phenomenon is known as electrostatic shielding. A hollow conducting sphere acts as a perfect electrostatic shield for the region inside it.


Step 2: Detailed Explanation:

Initially, two identical point charges at vertices A and B produce a net electric field of magnitude E at the third vertex P.

When a hollow conducting sphere is placed at point P, it is now situated in this external electric field E.

The free electrons within the conducting sphere will move under the influence of this external field. They will accumulate on the side of the sphere that is opposite to the direction of the external field, leaving a net positive charge on the other side.

This separation of charges creates an induced electric field, \(E_{induced}\), inside the conductor, which is in the opposite direction to the external field E.

The rearrangement of charges continues until the induced field exactly cancels the external field at every point inside the conductor.
\[ \vec{E}_{net} = \vec{E}_{external} + \vec{E}_{induced} = 0 \]
Therefore, the net electric field at any point inside the hollow conducting sphere, including its center (the original point P), becomes zero.


Step 3: Final Answer:

The magnitude of the electric field at point P, which is now inside the hollow conducting sphere, becomes zero due to electrostatic shielding.
Quick Tip: Remember that the electric field inside any conductor (hollow or solid) in electrostatic equilibrium is always zero. The conductor shields its interior from any external static electric fields.

Step 1: Understanding the Concept:

This problem involves calculating the terminal potential difference of a battery under two conditions: open circuit (no current flowing) and closed circuit (current flowing). The terminal potential difference differs from the electromotive force (e.m.f.) when there is an internal resistance and current is drawn from the battery.


Step 2: Key Formula or Approach:

The terminal potential difference (V) of a battery is given by: \[ V = \mathcal{E} - Ir \]
where \(\mathcal{E}\) is the e.m.f., \(I\) is the current flowing through the circuit, and \(r\) is the internal resistance.

When the circuit is open, \(I = 0\).

When the circuit is closed, the current is given by Ohm's law for the entire circuit: \[ I = \frac{\mathcal{E}}{R + r} \]
where R is the external resistance.


Step 3: Detailed Explanation:

Case 1: Key is open

When the key is open, the circuit is incomplete, and no current flows. So, \(I = 0\).

The potential difference across the terminals is: \[ V_{open} = \mathcal{E} - (0)r = \mathcal{E} \]
Given \(\mathcal{E} = 12\) V, so \(V_{open} = 12\) V.


Case 2: Key is closed

When the key is closed, current flows through the circuit.
Given: \(\mathcal{E} = 12\) V, \(r = 0.5\) \(\Omega\), \(R = 9.5\) \(\Omega\).

The total resistance of the circuit is \(R_{total} = R + r = 9.5 + 0.5 = 10.0\) \(\Omega\).

The current in the circuit is: \[ I = \frac{\mathcal{E}}{R + r} = \frac{12 V}{10.0 \Omega} = 1.2 A \]
The potential difference across the terminals of the battery is: \[ V_{closed} = \mathcal{E} - Ir = 12 - (1.2)(0.5) = 12 - 0.6 = 11.4 V \]

Ratio Calculation:

The required ratio is \(\frac{V_{open}}{V_{closed}}\).
\[ Ratio = \frac{12}{11.4} = \frac{120}{114} = \frac{20 \times 6}{19 \times 6} = \frac{20}{19} \approx 1.0526 \]

Step 4: Final Answer:

The calculated ratio is approximately 1.05. Comparing with the options, the correct answer is 1.05.
Quick Tip: The terminal voltage of a battery is equal to its e.m.f. only when no current is drawn from it (open circuit). When current is drawn, the terminal voltage is always less than the e.m.f. due to the potential drop across the internal resistance.

Step 1: Understanding the Concept:

The relationship between the voltage (V) across a pure inductor and the current (I) flowing through it is fundamental to understanding AC circuits. The voltage across an inductor is directly proportional to the rate of change of current with respect to time.


Step 2: Key Formula or Approach:

The voltage across an inductor is given by the formula: \[ V = L \frac{dI}{dt} \]
where L is the inductance (a positive constant). This formula implies that the voltage waveform is the time derivative of the current waveform, scaled by the inductance L. The term \(\frac{dI}{dt}\) represents the slope of the I-t graph.


Step 3: Detailed Explanation:

Let's analyze the given I-t graph in two parts:

Part 1: From \(t = 0\) to \(t = T/2\)

In this interval, the graph of current I versus time t is a straight line with a constant positive slope. Let the peak current be \(I_0\).

The slope is \(\frac{\Delta I}{\Delta t} = \frac{I_0 - (-I_0)}{T/2 - 0} = \frac{2I_0}{T/2} = \frac{4I_0}{T}\) (This calculation is not needed, just the sign of the slope).
Since the slope \(\frac{dI}{dt}\) is a positive constant, the voltage \(V = L \frac{dI}{dt}\) must be a positive constant during this interval.


Part 2: From \(t = T/2\) to \(t = T\)

In this interval, the graph of current I versus time t is a straight line with a constant negative slope.

The slope is \(\frac{\Delta I}{\Delta t} = \frac{-I_0 - I_0}{T - T/2} = \frac{-2I_0}{T/2} = -\frac{4I_0}{T}\).
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.


Conclusion:

The voltage V should be a constant positive value from t=0 to t=T/2, and then it should abruptly switch to a constant negative value from t=T/2 to t=T. This behavior is represented by a square wave.

Looking at the options, graph (A) correctly depicts this behavior: a constant positive voltage for the first half-cycle and a constant negative voltage for the second half-cycle.


Step 4: Final Answer:

The correct representation of the voltage waveform is graph (A), which shows a square wave that is positive when the current slope is positive and negative when the current slope is negative.
Quick Tip: For AC circuits with inductors or capacitors, remember the relationship between voltage and current. For an inductor, \(V = L \frac{dI}{dt}\) (voltage leads current). For a capacitor, \(I = C \frac{dV}{dt}\) (current leads voltage). This means you can find one waveform by looking at the slope of the other.

Step 1: Understanding the Concept:

This question tests the fundamental property of diamagnetic materials when placed in an external magnetic field. Materials are classified as diamagnetic, paramagnetic, or ferromagnetic based on their response to magnetic fields. Diamagnetism is a property of all materials and is a weak form of magnetism that is induced by a change in the orbital motion of electrons.


Step 2: Detailed Explanation:

According to Lenz's law, when a diamagnetic substance is placed in a magnetic field, a magnetic moment is induced in it in the opposite direction to the applied field. This induced magnetism opposes the external field.

As a result, a diamagnetic substance is feebly repelled by a magnet.

The force experienced by a magnetic material in a non-uniform magnetic field is directed towards the region of weaker magnetic field if the material is diamagnetic, and towards the region of stronger magnetic field if it is paramagnetic.

The magnetic field of a bar magnet is strongest near its poles (both north and south) and weaker further away.

Therefore, when a diamagnetic substance is brought near either the north pole or the south pole of a bar magnet, it will be pushed away from the region of the strong field (the pole) towards a region of a weaker field.

This means the diamagnetic substance will be repelled by the north pole and also repelled by the south pole.


Step 3: Final Answer:

A diamagnetic substance is repelled by both poles of a magnet. Hence, it is repelled by the north pole as well as by the south pole.
Quick Tip: A simple way to remember the behavior of magnetic materials: \textbf{Diamagnetic:} Di- means 'against' or 'opposite'. They are repelled by magnets. \textbf{Paramagnetic:} 'Para' is like 'parallel' or 'alongside'. They are weakly attracted to magnets. \textbf{Ferromagnetic:} 'Ferro' relates to iron. They are strongly attracted to magnets.

Step 1: Understanding the Concept:

This problem requires the formulas for the self-inductance of a long solenoid and the mutual inductance of two long coaxial solenoids. Self-inductance relates the magnetic flux through a coil to the current in that same coil. Mutual inductance relates the magnetic flux through one coil to the current in a second, nearby coil.


Step 2: Key Formula or Approach:

Let the length of the solenoids be \(l\).

The self-inductance (L) of a long solenoid is given by: \[ L = \mu_0 n^2 A l \]
where \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(A\) is the cross-sectional area.


The mutual inductance (M) of two coaxial solenoids is given by: \[ M = \mu_0 n_1 n_2 A_{inner} l \]
where \(n_1\) and \(n_2\) are the turns per unit length of the two solenoids, and \(A_{inner}\) is the cross-sectional area of the inner solenoid, because the magnetic flux from the outer solenoid only links with the inner solenoid through the area of the inner one.


Step 3: Detailed Explanation:

Let solenoid 1 be the inner solenoid and solenoid 2 be the outer solenoid.
Given: Radius \(r_1\), turns per unit length \(n_1\).
Given: Radius \(r_2\), turns per unit length \(n_2\).
The cross-sectional area of the inner solenoid is \(A_1 = \pi r_1^2\).


Self-inductance of the inner solenoid (\(L_1\)):

Using the formula for self-inductance: \[ L_1 = \mu_0 n_1^2 A_1 l = \mu_0 n_1^2 (\pi r_1^2) l \]

Mutual inductance of the two solenoids (M):

The mutual inductance depends on the flux from one solenoid linking the other. The flux is confined within the area of the smaller (inner) solenoid.
Using the formula for mutual inductance: \[ M = \mu_0 n_1 n_2 A_1 l = \mu_0 n_1 n_2 (\pi r_1^2) l \]

Ratio Calculation:

We need to find the ratio \(\frac{L_1}{M}\).
\[ \frac{L_1}{M} = \frac{\mu_0 n_1^2 (\pi r_1^2) l}{\mu_0 n_1 n_2 (\pi r_1^2) l} \]
Canceling the common terms (\(\mu_0, n_1, \pi, r_1^2, l\)) from the numerator and denominator: \[ \frac{L_1}{M} = \frac{n_1}{n_2} \]

Step 4: Final Answer:

The ratio of the self-inductance of the inner solenoid to their mutual inductance is \(n_1/n_2\).
Quick Tip: For mutual inductance problems involving coaxial solenoids, remember that the effective area for flux linkage is always the area of the inner solenoid, regardless of which solenoid is carrying the current.

Step 1: Understanding the Concept:

This question requires the application of the Biot-Savart Law to find the magnetic field produced by a small current-carrying element. Since the point of observation (1m, 1m, 0) is very far from the current element (1 cm length) compared to its size, we can approximate the segment as a point-like current element located at the origin.


Step 2: Key Formula or Approach:

The Biot-Savart Law for the magnetic field \(d\vec{B}\) due to a current element \(I d\vec{l}\) at a position \(\vec{r}\) from the element is: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3} \]
where \(\mu_0 = 4\pi \times 10^{-7}\) T·m/A.


Step 3: Detailed Explanation:

Given values:
Current, \(I = 1\) A.

Length of the segment, \(\Delta l = 1\) cm = 0.01 m.

The segment is in the x-direction, so the current element vector is \(\vec{\Delta l} = 0.01 \hat{i}\) m.

The point P where the field is to be calculated is at (1m, 1m, 0). The position vector from the origin to point P is \(\vec{r} = 1\hat{i} + 1\hat{j} + 0\hat{k}\) m.

The magnitude of the position vector is \(r = |\vec{r}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}\) m.


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

Now, substitute these values into the Biot-Savart Law. We use \(\vec{B} \approx \vec{\Delta B}\) because the segment is small. \[ \vec{B} = \frac{\mu_0}{4\pi} \frac{I (\vec{\Delta l} \times \vec{r})}{r^3} \] \[ \vec{B} = (10^{-7}) \frac{(1 A) (0.01 \hat{k} m^2)}{(\sqrt{2} m)^3} \] \[ \vec{B} = \frac{10^{-7} \times 10^{-2}}{2\sqrt{2}} \hat{k} = \frac{10^{-9}}{2\sqrt{2}} \hat{k} T \]

To match the options, let's manipulate the expression for the magnitude: \[ B = \frac{10^{-9}}{2\sqrt{2}} = \frac{1 \times 10 \times 10^{-10}}{2\sqrt{2}} = \frac{5 \times 10^{-10}}{\sqrt{2}} T \]

The direction of the magnetic field is along the positive z-axis (\(+\hat{k}\)). The options provide scalar values, with some being negative. Since our result is positive, we choose the positive option with the correct magnitude.


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 vector form of the Biot-Savart Law, \(\frac{I (d\vec{l} \times \vec{r})}{r^3}\), be careful with the exponent of r in the denominator; it's \(r^3\). If you use the unit vector form, \(\frac{I (d\vec{l} \times \hat{r})}{r^2}\), it's \(r^2\). Both are equivalent. The vector form often simplifies calculations involving coordinate systems.

Step 1: Understanding the Concept:

The principle of an AC generator is based on electromagnetic induction. When a coil of wire rotates in a uniform magnetic field, the magnetic flux through the coil changes with time. According to Faraday's law of induction, this change in flux induces an electromotive force (e.m.f.) in the coil. The maximum e.m.f. depends on the number of turns, magnetic field strength, coil area, and the angular velocity of rotation.


Step 2: Key Formula or Approach:

The e.m.f. induced in a rotating coil at any time \(t\) is given by: \[ \mathcal{E}(t) = NBA\omega \sin(\omega t) \]
The maximum e.m.f. (\(\mathcal{E}_{max}\)) occurs when \(\sin(\omega t) = 1\): \[ \mathcal{E}_{max} = NBA\omega \]
where:
N = number of turns in the coil
B = magnetic field strength
A = area of the coil \(\omega\) = angular velocity of rotation.
The angular velocity \(\omega\) is related to the frequency of rotation \(f\) by \(\omega = 2\pi f\).


Step 3: Detailed Explanation:

First, let's identify the given values:
Number of turns, \(N = 100\).

Area of the coil, \(A = 0.1\) m\(^2\).

Magnetic field strength, \(B = 0.02\) T.

The coil rotates at "half a rotation per second". This is the frequency of rotation, \(f\).

Frequency, \(f = 0.5\) rotations per second (rps) or 0.5 Hz.


Next, calculate the angular velocity \(\omega\): \[ \omega = 2\pi f = 2\pi (0.5) = \pi rad/s \]

Now, we can calculate the maximum e.m.f. using the formula \(\mathcal{E}_{max} = NBA\omega\): \[ \mathcal{E}_{max} = (100) \times (0.02 T) \times (0.1 m^2) \times (\pi rad/s) \] \[ \mathcal{E}_{max} = (100 \times 0.02 \times 0.1) \pi V \] \[ \mathcal{E}_{max} = (2 \times 0.1) \pi V \] \[ \mathcal{E}_{max} = 0.2\pi V \]

To get a numerical value, we use the approximation \(\pi \approx 3.14159\): \[ \mathcal{E}_{max} = 0.2 \times 3.14159 \approx 0.6283 V \]

Step 4: Final Answer:

The calculated maximum e.m.f. is approximately 0.628 V. This value is closest to option (C) 0.63 V.
Quick Tip: Pay close attention to the units of rotation speed. Sometimes it's given in rotations per minute (rpm), sometimes in rotations per second (rps or Hz), and sometimes as angular velocity (rad/s). Always convert to rad/s for the formula \(\mathcal{E}_{max} = NBA\omega\). (1 rps = 2\(\pi\) rad/s).

Step 1: Understanding the Concept:

This question combines concepts from the atomic structure of hydrogen (Bohr model) and the photoelectric effect. For photoelectric emission to occur, the energy of the incident photons must be greater than or equal to the work function (\(\phi\)) of the metal surface. We need to compare the energies of photons emitted from different spectral series of hydrogen with the work function of zinc.


Step 2: Key Formula or Approach:

Condition for photoelectric effect: \(E_{photon} \ge \phi\).

Energy of a photon emitted during a transition in a hydrogen atom from a higher energy level \(n_f\) to a lower energy level \(n_i\) is: \[ E = -13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) eV = 13.6 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) eV \]
The work function of zinc is approximately \(\phi_{Zn} = 4.31\) eV.


Step 3: Detailed Explanation:

We need to check which spectral series of hydrogen can produce photons with energy \(E \ge 4.31\) eV.


Lyman Series (transitions to \(n_i = 1\)):

The final state is the ground state. The initial state can be \(n_f = 2, 3, 4, ...\).
The minimum energy in the Lyman series corresponds to the transition from \(n_f = 2\) to \(n_i = 1\). \[ E_{min, Lyman} = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times \frac{3}{4} = 10.2 eV \]
Since the minimum energy of a Lyman series photon (10.2 eV) is greater than the work function of zinc (4.31 eV), all lines in the Lyman series can cause photoelectric emission.


Balmer Series (transitions to \(n_i = 2\)):

The final state is the first excited state. The initial state can be \(n_f = 3, 4, 5, ...\).
The maximum energy in the Balmer series corresponds to the transition from \(n_f = \infty\) to \(n_i = 2\) (the series limit). \[ E_{max, Balmer} = 13.6 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = 13.6 \left( \frac{1}{4} - 0 \right) = \frac{13.6}{4} = 3.4 eV \]
Since the maximum energy of a Balmer series photon (3.4 eV) is less than the work function of zinc (4.31 eV), no lines from the Balmer series can cause photoelectric emission.


Paschen Series (transitions to \(n_i = 3\)):

The final state is the second excited state. The initial state can be \(n_f = 4, 5, 6, ...\).
The maximum energy in the Paschen series corresponds to the transition from \(n_f = \infty\) to \(n_i = 3\). \[ E_{max, Paschen} = 13.6 \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = 13.6 \left( \frac{1}{9} - 0 \right) = \frac{13.6}{9} \approx 1.51 eV \]
The maximum energy of a Paschen series photon (1.51 eV) is also less than the work function of zinc.


Step 4: Final Answer:

Only the photons from the Lyman series have sufficient energy to overcome the work function of zinc and cause photoelectric emission.
Quick Tip: Remember the energy ranges for the first few hydrogen series: Lyman (UV): \(E > 10.2\) eV Balmer (Visible): \(1.89 eV < E < 3.4\) eV Paschen (Infrared): \(0.66 eV < E < 1.51\) eV Knowing these approximate ranges can help you quickly solve problems involving the photoelectric effect.

Step 1: Understanding the Concept:

This question tests the understanding of how the properties of optical devices, specifically a spherical mirror, are affected by the surrounding medium. It's important to distinguish between mirrors and lenses in this context.


Step 2: Key Formula or Approach:

The focal length (\(f\)) of a spherical mirror is related to its radius of curvature (\(R\)) by the mirror formula: \[ f = \frac{R}{2} \]
The working principle of a mirror is based on the law of reflection (\(angle of incidence = angle of reflection\)). This law is independent of the refractive index of the medium in which the light travels.


Step 3: Detailed Explanation:

The focal length of a spherical mirror is a geometric property, determined solely by its radius of curvature. The derivation of \(f = R/2\) depends only on the geometry of the mirror and the laws of reflection.

The laws of reflection state that the angle of incidence equals the angle of reflection, and the incident ray, the reflected ray, and the normal all lie in the same plane. These laws hold true irrespective of the medium surrounding the mirror.

Unlike a lens, whose focal length depends on the refractive indices of the lens material and the surrounding medium (as described by the Lens Maker's Formula), a mirror's focal length is constant.

Therefore, immersing the concave mirror in a liquid, regardless of its refractive index, will not change its radius of curvature and hence will not change its focal length.


Step 4: Final Answer:

The focal length of the concave mirror will remain unchanged when immersed in the liquid. Thus, the new focal length will still be f.
Quick Tip: This is a classic conceptual question. Remember the key difference: \textbf{Mirror's focal length (\(f = R/2\))}: Depends only on geometry (R). It is \textbf{independent} of the surrounding medium. \textbf{Lens's focal length (Lens Maker's Formula)}: Depends on the refractive index of the lens material and the surrounding medium. It \textbf{changes} with the medium.

Step 1: Understanding the Concept:

This question asks about two fundamental properties of the electric field produced by static charges (an electrostatic field): its conservative nature and the geometry of its field lines.


Step 2: Detailed Explanation:

Conservative Nature:

An electric field \(\vec{E}\) is said to be conservative if the work done by the field in moving a test charge from one point to another is independent of the path taken. Mathematically, this is equivalent to the line integral of the electric field around any closed path being zero: \[ \oint \vec{E} \cdot d\vec{l} = 0 \]
The electric field created by static charges satisfies this condition. The work done depends only on the initial and final positions, which is why we can define a unique electric potential at every point in space. Therefore, the electrostatic field is a conservative field.


Field Lines:

Electric field lines are used to visualize the electric field. They have specific properties:

They originate from positive charges and terminate on negative charges (or extend to infinity if there is a net charge).
They never form closed loops. If a field line were to form a closed loop, moving a positive test charge along this loop would result in non-zero net work done by the field (since the force \(\vec{F} = q\vec{E}\) would always have a component along the path). This would violate the conservative nature of the field (\(\oint \vec{E} \cdot d\vec{l} \neq 0\)).

Since the electrostatic field is conservative, its field lines cannot form closed loops.


Evaluating the Options:

(A) conservative and field lines do not form closed loops. - This statement is correct on both counts.

(B) conservative and field lines form closed loops. - Incorrect. Conservative fields cannot have closed-loop field lines.

(C) non-conservative and field lines do not form closed loops. - Incorrect. The electrostatic field is conservative.

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


Step 3: Final Answer:

The correct statement is that the electric field due to static charges is conservative and its field lines do not form closed loops.
Quick Tip: Contrast the electrostatic field with the magnetic field and the induced electric field: \textbf{Electrostatic Field}: Conservative, lines start on + and end on -, no closed loops. \textbf{Magnetic Field}: Non-conservative, lines always form closed loops (no magnetic monopoles). \textbf{Induced Electric Field (from changing B-field)}: Non-conservative, lines form closed loops.

Step 1: Understanding the Concept:

This question relates to the V-I characteristics of a p-n junction diode. A diode exhibits very different resistance depending on the direction of the applied voltage, a property known as biasing.


Step 2: Detailed Explanation:

A p-n junction diode has two modes of operation based on the polarity of the voltage applied across it:


1. Forward Bias:

When the p-side is connected to a higher potential and the n-side to a lower potential, the diode is forward-biased. The width of the depletion region decreases, and a large current can flow through the junction. In this state, the diode offers very low resistance. It behaves like a closed switch (or a conductor).


2. Reverse Bias:

When the p-side is connected to a lower potential and the n-side to a higher potential, the diode is reverse-biased. The width of the depletion region increases, and only a very small leakage current can flow. In this state, the diode offers very high resistance. It behaves like an open switch (or an insulator).


The question states that the measured resistance is high. This corresponds to the reverse-biased condition. In this state, the diode effectively blocks the flow of current. This "on/off" or "low resistance/high resistance" behavior is the fundamental characteristic of a switch.


Step 3: Final Answer:

When its resistance is high, the p-n junction diode is acting as an open switch, blocking current flow. Therefore, it can be considered to be acting as a switch.
Quick Tip: Think of an ideal diode as a perfect electrical switch: \textbf{Forward Bias} \(\rightarrow\) Low resistance \(\rightarrow\) Closed switch \(\rightarrow\) Current ON. \textbf{Reverse Bias} \(\rightarrow\) High resistance \(\rightarrow\) Open switch \(\rightarrow\) Current OFF. This switching property is the basis for its use in rectifiers and digital logic circuits.

Step 1: Understanding the Concept:

According to the Bohr model for the hydrogen atom, the energy of an electron is quantized and depends on the principal quantum number, n. The energy of the electron in the nth orbit is inversely proportional to the square of n.


Step 2: Key Formula or Approach:

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


Step 3: Detailed Explanation:

We are given the following values:
Energy in the ground state, \(E_1 = -13.6\) eV.

Energy in the nth orbit, \(E_n = -0.544\) eV.


We need to find the value of the principal quantum number, n.

Rearranging the formula, we get: \[ n^2 = \frac{E_1}{E_n} \]
Substituting the given values: \[ n^2 = \frac{-13.6}{-0.544} \]
The negative signs cancel out: \[ n^2 = \frac{13.6}{0.544} \]
To simplify the division, we can multiply the numerator and denominator by 1000: \[ n^2 = \frac{13600}{544} \]
Now, let's perform the division. We can notice that \(136 \times 4 = 544\). \[ n^2 = \frac{136 \times 100}{136 \times 4} = \frac{100}{4} = 25 \]
Taking the square root of both sides: \[ n = \sqrt{25} = 5 \]

Step 4: Final Answer:

The value of the quantum number n corresponding to the energy level -0.544 eV 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 \(E_2 = -13.6/4 = -3.4\) eV \(E_3 = -13.6/9 \approx -1.51\) eV \(E_4 = -13.6/16 = -0.85\) eV \(E_5 = -13.6/25 = -0.544\) eV Recognizing these values can save you calculation time in an exam.

Step 1: Understanding the Concept:

This question assesses the understanding of diffraction of electromagnetic waves and the properties of the electromagnetic spectrum (wavelength and frequency). Diffraction is the bending of waves as they pass an obstacle or go through an aperture. The extent of diffraction is significant when the wavelength of the wave is comparable to or larger than the size of the obstacle.


Step 2: Detailed Explanation:

Analyzing Assertion (A):

The assertion states that radio waves show more diffraction than infrared waves.

The order of wavelengths (\(\lambda\)) in the electromagnetic spectrum is:
Radio waves > Microwaves > Infrared > Visible > Ultraviolet > X-rays > Gamma rays.

Since radio waves have a much larger wavelength (\(\lambda_{radio} >> \lambda_{infrared}\)), they will diffract much more significantly around everyday objects (like buildings, hills, etc.) compared to infrared waves. Therefore, Assertion (A) is true.


Analyzing Reason (R):

The reason states that radio waves have a greater frequency than infrared waves.

The relationship between the speed of light (\(c\)), frequency (\(f\)), and wavelength (\(\lambda\)) is \(c = f\lambda\). This implies that frequency is inversely proportional to wavelength (\(f \propto 1/\lambda\)).

Since radio waves have a larger wavelength than infrared waves, they must have a lower frequency. \[ \lambda_{radio} > \lambda_{infrared} \implies f_{radio} < f_{infrared} \]
Therefore, the statement in Reason (R) is false.


Step 3: Final Answer:

Assertion (A) is true, but Reason (R) is false. Based on the given codes, the correct option is (C).
Quick Tip: Remember the acronym for the EM spectrum in order of increasing frequency (and decreasing wavelength): "Raging Martians Invaded Venus Using X-ray Guns" (Radio, Microwaves, Infrared, Visible, UV, X-ray, Gamma). This helps to quickly compare wavelengths and frequencies.

Step 1: Understanding the Concept:

This question deals with the principles of an ideal transformer. An ideal transformer is a theoretical device that has 100% efficiency, meaning there are no energy losses. The question asks to relate this property to the voltage and current transformation in a step-down transformer.


Step 2: Detailed Explanation:

Analyzing Assertion (A):

An ideal transformer is defined as one with no energy losses. The sources of energy loss in a real transformer include resistive heating in the windings (copper loss), eddy currents in the core, and hysteresis loss. By definition, in an ideal transformer, these losses are zero. This means that the input power is equal to the output power (\(P_{in} = P_{out}\)). Since power is the rate of energy transfer, this implies that electrical energy is conserved (not lost). Therefore, Assertion (A) is true.


Analyzing Reason (R):

A step-down transformer is designed to decrease the voltage. In an ideal transformer, since power is conserved (\(P_{in} = P_{out} \implies V_p I_p = V_s I_s\)), if the secondary voltage (\(V_s\)) is decreased relative to the primary voltage (\(V_p\)), the secondary current (\(I_s\)) must increase proportionally to keep the product constant. Thus, in a step-down transformer, voltage decreases and current increases. Therefore, Reason (R) is also true.


Analyzing the link between A and R:

The Assertion (A) states a fundamental principle (energy conservation in an ideal transformer). The Reason (R) describes a consequence of this principle for a specific type of transformer (step-down). While the fact that current increases when voltage decreases is consistent with energy conservation, it is not the reason why energy is conserved. The reason for the conservation of energy is the "ideal" nature of the transformer itself – the absence of resistive losses, eddy currents, etc. The relationship in R is a result of the principle in A, not its cause. Therefore, R is not the correct explanation for A.


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). Based on the given codes, the correct option is (B).
Quick Tip: In Assertion-Reason questions, always ask "Is R the fundamental reason for A?". A consequence or an example of a principle is usually not considered its correct explanation. The explanation should be the underlying cause or definition.

Step 1: Understanding the Concept:

This question tests two aspects of the Bohr model for the hydrogen atom: the quantization of angular momentum and the rules for electronic transitions (jumps).


Step 2: Detailed Explanation:

Analyzing Assertion (A):

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

From Bohr's second postulate, the angular momentum is quantized: \[ L_n = \frac{nh}{2\pi} \]
This shows that \(L_n \propto n\).

The radius of the nth orbit in the Bohr model is given by: \[ r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2} \]
This shows that \(r_n \propto n^2\).

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

Now, substitute this into the angular momentum relation: \[ L_n \propto n \implies L_n \propto \sqrt{r_n} \]
So, the angular momentum of an electron is proportional to the square root of its orbit radius. Therefore, Assertion (A) is true.


Analyzing Reason (R):

The reason states that an electron can jump to its nearest orbits only.

Bohr's third postulate describes electronic transitions. An electron can jump from a higher energy orbit (\(n_f\)) to any lower energy orbit (\(n_i\)) by emitting a photon. It can also jump from a lower orbit to any higher orbit by absorbing a photon of the correct energy. The transitions are not restricted to adjacent or "nearest" orbits. For example, the Lyman series involves transitions from \(n=2, 3, 4, ...\) all the way down to \(n=1\). A jump from n=3 to n=1 is not to the nearest orbit. Therefore, Reason (R) is false.


Step 3: Final Answer:

Assertion (A) is true, but Reason (R) is false. Based on the given codes, the correct option is (C).
Quick Tip: For Bohr model dependencies on the principal quantum number (n), remember these key proportionalities: Radius \(r_n \propto n^2\) Velocity \(v_n \propto 1/n\) Energy \(E_n \propto -1/n^2\) Angular Momentum \(L_n \propto n\) Using these, you can derive other relationships, like the one between \(L_n\) and \(r_n\).

Step 1: Understanding the Concept:

This question pertains to the properties of the depletion layer in a p-n junction diode. The depletion layer is a region around the junction that is depleted of free charge carriers. Its width is a crucial parameter that determines the diode's electrical characteristics.


Step 2: Detailed Explanation:

Analyzing Assertion (A):

The depletion layer is formed due to the diffusion of electrons from the n-side to the p-side and holes from the p-side to the n-side, leaving behind immobile ionized donor and acceptor atoms. The width of this layer is not constant. It changes based on the external voltage (biasing) applied across the diode and the doping concentration of the semiconductor materials. Therefore, the statement that the thickness of the depletion layer is not fixed is true.


Analyzing Reason (R):

The reason states that the thickness depends on factors like biasing. This is correct.

Forward Biasing: When a positive voltage is applied to the p-side and a negative voltage to the n-side, the applied field opposes the internal field of the depletion region. This causes the depletion layer to narrow.
Reverse Biasing: When a negative voltage is applied to the p-side and a positive voltage to the n-side, the applied field reinforces the internal field. This pushes charge carriers further away from the junction, causing the depletion layer to widen.

Since biasing directly controls the width of the depletion layer, the Reason (R) is true.


Analyzing the link between A and R:

The Assertion (A) states that the thickness is not fixed. The Reason (R) provides the primary mechanism (biasing) that causes this thickness to change. Therefore, the Reason (R) is the correct explanation for the Assertion (A).


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).
Quick Tip: Remember the effect of biasing on the depletion layer: \textbf{Forward Bias} \(\rightarrow\) Pushes carriers towards junction \(\rightarrow\) \textbf{Narrows} the depletion layer. \textbf{Reverse Bias} \(\rightarrow\) Pulls carriers away from junction \(\rightarrow\) \textbf{Widens} the depletion layer.

Step 1: Understanding the Concept:

This problem involves a simple series circuit containing a battery, a resistor, and a forward-biased silicon diode. The threshold voltage (or forward voltage drop) of the diode is the voltage that must be overcome for it to conduct significantly. We can use Kirchhoff's Voltage Law (KVL) to analyze the circuit.


Step 2: Key Formula or Approach:

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


Step 3: Detailed Explanation:

Given values:

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

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

Step 4: Final Answer:

The value of the battery voltage V is 15.7 V.
Quick Tip: When analyzing circuits with diodes, treat a forward-biased ideal diode as a short circuit (0 V drop). For a practical diode, treat it as a small battery with a voltage equal to its threshold voltage (e.g., 0.7 V for silicon), opposing the current flow.

Step 1: Understanding the Concept:

Huygens' principle states that every point on a primary wavefront serves as the source of secondary wavelets that spread out in all directions with the speed of the wave in that medium. The new wavefront at a later time is the tangential surface (envelope) to all these secondary wavelets. We can use this principle to explain the phenomenon of refraction and derive Snell's law.


Step 2: Derivation using Huygens' Principle:

Consider a plane wavefront AB incident at an angle \(i\) on a plane interface XY, separating two media, medium 1 (rarer) and medium 2 (denser). Let the speed of light in medium 1 be \(v_1\) and in medium 2 be \(v_2\), where \(v_1 > v_2\).


Diagram Description:

XY is the plane interface.
AB is the incident wavefront, perpendicular to the incident rays.
CE is the refracted wavefront, perpendicular to the refracted rays.
The angle of incidence \(i\) is the angle between the incident wavefront AB and the interface XY.
The angle of refraction \(r\) is the angle between the refracted wavefront CE and the interface XY.

Let the wavefront first touch the interface at point A. The disturbance at point B on the wavefront will travel to point C on the interface in a time interval \(\Delta t\). \[ BC = v_1 \Delta t \]
According to Huygens' principle, point A acts as a source of secondary wavelets. In the same time interval \(\Delta t\), the secondary wavelet from A will travel into medium 2, covering a distance AE. \[ AE = v_2 \Delta t \]
The new refracted wavefront is the tangent plane CE drawn from point C to the spherical wavelet originating from A with radius AE.


Step 3: Proof of Snell's Law:

Now, consider the two right-angled triangles \(\triangle ABC\) and \(\triangle AEC\).

In \(\triangle ABC\): \[ \sin i = \frac{opposite}{hypotenuse} = \frac{BC}{AC} = \frac{v_1 \Delta t}{AC} \quad \ldots (1) \]
In \(\triangle AEC\): \[ \sin r = \frac{opposite}{hypotenuse} = \frac{AE}{AC} = \frac{v_2 \Delta t}{AC} \quad \ldots (2) \]
Now, divide equation (1) by equation (2): \[ \frac{\sin i}{\sin r} = \frac{(v_1 \Delta t) / AC}{(v_2 \Delta t) / AC} = \frac{v_1}{v_2} \]
The refractive index (\(n\)) of a medium is defined as the ratio of the speed of light in vacuum (\(c\)) to the speed of light in the medium (\(v\)), i.e., \(n = c/v\).
So, \(v_1 = c/n_1\) and \(v_2 = c/n_2\), where \(n_1\) and \(n_2\) are the refractive indices of medium 1 and medium 2, respectively.

Substituting these into our ratio: \[ \frac{\sin i}{\sin r} = \frac{c/n_1}{c/n_2} = \frac{n_2}{n_1} \]
Rearranging the terms, we get: \[ n_1 \sin i = n_2 \sin r \]
This is the mathematical expression for Snell's law of refraction.


Step 4: Final Answer:

Thus, Huygens' principle successfully explains the refraction of light and provides a proof for Snell's law.
Quick Tip: When drawing the diagram for this proof, remember that the wavelength changes upon refraction (\(\lambda_2 = \lambda_1 \frac{v_2}{v_1}\)). If light enters a denser medium (\(v_2 < v_1\)), the wavelength decreases, and the refracted wavefronts will be closer together than the incident wavefronts.

Step 1: Understanding the Concept:

This problem involves a combination of two lenses. The image formed by the first lens (A) serves as the object for the second lens (B). We will apply the lens formula sequentially for each lens to find the final image position and nature. All positions are measured from the origin of the optical bench.


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 sign convention is crucial: distances are measured from the optical center of the lens. Distances in the direction of incident light are positive, and those opposite are negative.


Step 3: Detailed Explanation:

For Lens A:

Position of Lens A: \(x_A = 50.0\) cm.
Focal length of Lens A: \(f_A = +10.0\) cm (convex lens).
Position of the object: \(x_{obj} = 20.0\) cm.

Object distance for Lens A, \(u_A\), is the distance from the object to Lens A. \[ u_A = x_{obj} - x_A = 20.0 - 50.0 = -30.0 cm \]
Using the lens formula to find the image distance \(v_A\): \[ \frac{1}{v_A} - \frac{1}{-30.0} = \frac{1}{10.0} \] \[ \frac{1}{v_A} = \frac{1}{10.0} - \frac{1}{30.0} = \frac{3 - 1}{30.0} = \frac{2}{30.0} = \frac{1}{15.0} \] \[ v_A = +15.0 cm \]
The image (let's call it I\(_1\)) is formed 15.0 cm to the right of Lens A. Its position on the optical bench is: \[ x_{I_1} = x_A + v_A = 50.0 + 15.0 = 65.0 cm \]

For Lens B:

Position of Lens B: \(x_B = 70.0\) cm.
Focal length of Lens B: \(f_B = +10.0\) cm (convex lens).
The image I\(_1\) from Lens A acts as the object for Lens B.

Object distance for Lens B, \(u_B\), is the distance from I\(_1\) to Lens B. \[ u_B = x_{I_1} - x_B = 65.0 - 70.0 = -5.0 cm \]
Using the lens formula to find the final image distance \(v_B\): \[ \frac{1}{v_B} - \frac{1}{-5.0} = \frac{1}{10.0} \] \[ \frac{1}{v_B} = \frac{1}{10.0} - \frac{1}{5.0} = \frac{1 - 2}{10.0} = -\frac{1}{10.0} \] \[ v_B = -10.0 cm \]
The final image is formed 10.0 cm to the left of Lens B. Its final position on the optical bench is: \[ x_{final} = x_B + v_B = 70.0 + (-10.0) = 60.0 cm \]

Nature of the Final Image:
Total magnification \(m = m_A \times m_B\). \[ m_A = \frac{v_A}{u_A} = \frac{+15.0}{-30.0} = -0.5 \] \[ m_B = \frac{v_B}{u_B} = \frac{-10.0}{-5.0} = +2 \] \[ m = (-0.5) \times (+2) = -1 \]
Since the final magnification is negative, the image is inverted with respect to the original object. Since the image distance \(v_B\) is negative, the final image is virtual.


Step 4: Final Answer:

The final image is located at the 60.0 cm mark on the optical bench. The nature of the image is virtual and inverted.
Quick Tip: When dealing with multiple lenses on an optical bench, it's often easier to work with absolute coordinates first to find the relative distances (\(u\) and \(v\)) for each lens. Be meticulous with the sign convention at each step.

Step 1: Understanding the Concept:

This problem is based on Einstein's photoelectric effect. When a photon with sufficient energy strikes a metal surface, it can eject an electron. The maximum kinetic energy of the ejected electron (\(K_{max}\)) is equal to the photon's energy minus the work function (\(\phi\)) of the metal.


Step 2: Key Formula or Approach:

Einstein's photoelectric equation is: \[ K_{max} = E_{photon} - \phi \]
The kinetic energy is related to the maximum speed (\(v_{max}\)) of the electron by: \[ K_{max} = \frac{1}{2} m v_{max}^2 \]
where \(m\) is the mass of the electron.


Step 3: Detailed Explanation:

Given values:

Work function of the metal, \(\phi = 2.0\) eV.
Energy of photons in the first case, \(E_1 = 2.5\) eV.
Energy of photons in the second case, \(E_2 = 4.5\) eV.


Case 1:
Maximum kinetic energy of the emitted electrons: \[ K_1 = E_1 - \phi = 2.5 eV - 2.0 eV = 0.5 eV \]
So, \(\frac{1}{2} m v_1^2 = 0.5 eV\), where \(v_1\) is the maximum speed in the first case.


Case 2:
Maximum kinetic energy of the emitted electrons: \[ K_2 = E_2 - \phi = 4.5 eV - 2.0 eV = 2.5 eV \]
So, \(\frac{1}{2} m v_2^2 = 2.5 eV\), where \(v_2\) is the maximum speed in the second case.


Finding the Ratio of Speeds:
Now, we find the ratio of the kinetic energies: \[ \frac{K_1}{K_2} = \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{v_1^2}{v_2^2} \]
Using the calculated values of K\(_1\) and K\(_2\): \[ \frac{v_1^2}{v_2^2} = \frac{0.5 eV}{2.5 eV} = \frac{1}{5} \]
Taking the square root of both sides to find the ratio of the speeds: \[ \frac{v_1}{v_2} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

Step 4: Final Answer:

The ratio of the maximum speed of the electrons in the two cases, \(v_1:v_2\), is \(1:\sqrt{5}\).
Quick Tip: In problems asking for ratios of speeds in the photoelectric effect, there is no need to convert the energy from electron-volts (eV) to Joules, as the conversion factor will cancel out in the ratio.

Step 1: Understanding the Concept:

This problem involves the division of current in a parallel circuit. When resistors are connected in parallel, the current divides among them, with more current flowing through the path of lower resistance. The current is inversely proportional to the resistance.


Step 2: Key Formula or Approach:

The resistance \(R\) of a wire is given by \(R = \rho \frac{l}{A}\), where \(\rho\) is the resistivity, \(l\) is the length, and \(A\) is the cross-sectional area.
For resistors in parallel, the voltage drop across them is the same: \(V = I_1 R_1 = I_2 R_2\).
This implies the ratio of currents is the inverse of the ratio of resistances: \(\frac{I_1}{I_2} = \frac{R_2}{R_1}\).


Step 3: Detailed Explanation:

Let the two wires be Wire 1 and Wire 2.
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: \(I = I_1 + I_2 = 15\) A.

First, find the ratio of their resistances: \[ \frac{R_1}{R_2} = \frac{\rho l_1 / A}{\rho l_2 / A} = \frac{l_1}{l_2} = \frac{2}{3} \]
Since the wires are connected in parallel, the current divides inversely to the resistance: \[ \frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{3}{2} \]
This means \(I_1 = \frac{3}{2} I_2\).

Now, use the total current equation: \[ I_1 + I_2 = 15 \]
Substitute the expression for \(I_1\): \[ \frac{3}{2} I_2 + I_2 = 15 \] \[ \left(\frac{3}{2} + 1\right) I_2 = 15 \] \[ \frac{5}{2} I_2 = 15 \] \[ I_2 = 15 \times \frac{2}{5} = 6 A \]
Now find \(I_1\): \[ I_1 = 15 - I_2 = 15 - 6 = 9 A \]

Step 4: Final Answer:

The current through the first wire (shorter length, lower resistance) is \(I_1 = 9\) A.
The current through the second wire (longer length, higher resistance) is \(I_2 = 6\) A.
Quick Tip: For current division in two parallel branches, you can use the formulas directly: \(I_1 = I_{total} \left( \frac{R_2}{R_1 + R_2} \right)\) and \(I_2 = I_{total} \left( \frac{R_1}{R_1 + R_2} \right)\). Here, the ratio R1:R2 is 2:3, so \(I_1 = 15 \times (\frac{3}{2+3}) = 9\) A and \(I_2 = 15 \times (\frac{2}{2+3}) = 6\) A.

Step 1: Understanding the Concept:

In a DC circuit, after a long time (in the steady state), a capacitor becomes fully charged and acts as an open circuit. This means that no current flows through the branch containing the capacitor. We can then analyze the remaining part of the circuit using Kirchhoff's laws.


Step 2: Key Formula or Approach:

We will use Kirchhoff's Voltage Law (KVL) to find the current in the circuit loops and then determine the potential difference between the required points.
KVL: The algebraic sum of changes in potential around any closed loop is zero.


Step 3: Detailed Explanation:

Analyzing the circuit in steady state:
In the steady state, no current flows through the middle branch containing the capacitor C. The circuit effectively becomes a single outer loop with cells V and 2V, and resistors R and 2R.


Finding the current in the outer loop:
Let the current flowing clockwise in the outer loop be \(I\). The cell 2V and cell V are in opposition. The net e.m.f. driving the current is \(2V - V = V\). The total resistance in the loop is \(R + 2R = 3R\).
Applying KVL to the outer loop (starting from the bottom-left and moving clockwise): \[ +2V - I(2R) - V - I(R) = 0 \] \[ V - 3IR = 0 \] \[ I = \frac{V}{3R} \]
The positive result confirms the current flows clockwise.


(i) Potential difference between P and Q (\(V_{PQ} = V_P - V_Q\)):
We can find this by calculating the potential change along any path from Q to P.
Let's take the top path (from Q, through R and cell V, to P): \[ V_P = V_Q + I R + V \]
(Potential increases by IR as we move against the current, and increases by V as we move from the negative to the positive terminal of the cell).
Substituting the value of \(I\): \[ V_{PQ} = V_P - V_Q = V + IR = V + \left(\frac{V}{3R}\right)R = V + \frac{V}{3} = \frac{4V}{3} \]
Alternatively, using the bottom path (from Q, through 2R and cell 2V, to P): \[ V_P = V_Q + I(2R) - 2V \]
(Potential increases by I(2R) as we move against the current, and decreases by 2V as we move from the positive to the negative terminal of the cell). \[ V_{PQ} = V_P - V_Q = 2IR - 2V = 2\left(\frac{V}{3R}\right)R - 2V = \frac{2V}{3} - 2V = -\frac{4V}{3} \]
Wait, there is a sign mistake in the second path. Let's re-examine. Path from P to Q via the bottom: \(V_Q = V_P + 2V - I(2R)\). Then \(V_P - V_Q = I(2R) - 2V\). The previous attempt was correct. Let's recheck the KVL signs again.
Let's start from P and go clockwise around the outer loop: \(V_P\). After top cell: \(V_P - V\). After R: \(V_P - V - IR\). This is \(V_Q\). So \(V_Q = V_P - V - IR\). Therefore \(V_P - V_Q = V + IR = 4V/3\). This calculation is correct. Let's recheck the bottom path from P to Q. Start at P. After 2V cell (from - to +): \(V_P + 2V\). After 2R resistor (in direction of current): \(V_P + 2V - I(2R)\). This is \(V_Q\). So \(V_Q = V_P + 2V - 2IR\). \(V_P - V_Q = 2IR - 2V = 2(V/3R)R - 2V = 2V/3 - 2V = -4V/3\). There must be a mistake in the diagram interpretation. Ah, the + and - of the V cell in the top path are drawn. Let's assume standard cell notation. Let's trust the first path calculation. Let's retry the second one. Path P -> bottom -> Q. \(V_{P} \xrightarrow{across 2V} V_P - 2V \xrightarrow{across 2R} V_P - 2V + I(2R) = V_Q\). Then \(V_P - V_Q = 2V - 2IR = 2V - 2(V/3) = 4V/3\). Okay, now both paths give the same result. The potential at P is higher than at Q.

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


(ii) Potential difference across capacitor C (\(V_C\)):
The potential difference across the capacitor is the potential difference across the middle branch.
Let's apply KVL to the loop containing P, the top cell V, the capacitor C, and Q.
Since no current flows, we only consider the e.m.f. and the capacitor voltage. Let the potential on the capacitor plate connected to the top wire be \(V_C\).
Path from P to Q through the middle branch: \[ V_P - V - V_C = V_Q \]
(Moving from P, potential drops by V across the cell, and drops by \(V_C\) across the capacitor). \[ V_P - V_Q = V + V_C \]
We already found \(V_P - V_Q = \frac{4V}{3}\). \[ \frac{4V}{3} = V + V_C \] \[ V_C = \frac{4V}{3} - V = \frac{V}{3} \]

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 \(\frac{V}{3}\).
Quick Tip: For complex circuits, always simplify the problem first. Recognizing that a capacitor acts as an open circuit in steady-state DC is the key. When finding potential differences, calculate it using two different paths to cross-check your answer and catch any sign errors.

Step 1: Definition of Resistivity:

Resistivity (or specific electrical resistance) is a fundamental property of a material that quantifies how strongly it resists the flow of electric current. It is defined as the resistance offered by a conductor of the material having a unit length and a unit cross-sectional area.

If \(R\) is the resistance of a conductor of length \(l\) and area of cross-section \(A\), then its resistivity \(\rho\) is given by: \[ \rho = R \frac{A}{l} \]
The SI unit of resistivity is the ohm-meter (\(\Omega \cdot m\)).


Step 2: Dependence on Temperature:

The resistivity of a conductor is not constant but depends on its temperature. For most metallic conductors, resistivity increases with an increase in temperature. This is because as the temperature rises, the ions of the conductor vibrate with greater amplitude about their mean positions. This increases the frequency of collisions between the free electrons (charge carriers) and the ions. The increased collision rate obstructs the flow of electrons, thereby increasing the resistivity.

The relationship between resistivity and temperature for a conductor, over a limited range of temperatures, can be approximated by the linear relation: \[ \rho_T = \rho_0 [1 + \alpha(T - T_0)] \]
where:

\(\rho_T\) is the resistivity at temperature T.
\(\rho_0\) is the resistivity at a reference temperature T\(_0\) (often 0\(^\circ\)C or 20\(^\circ\)C).
\(\alpha\) is the temperature coefficient of resistivity, which is positive for conductors.


Step 3: Plot for Copper:

The plot of resistivity (\(\rho\)) versus temperature (T in Kelvin) for copper (a typical conductor) is shown below.

% A simple description of the plot is provided here as drawing is not possible.
%
(A plot showing Resistivity on the y-axis and Temperature on the x-axis. The graph is a nearly straight line with a positive slope, starting from a small positive value on the y-axis at T=0K, indicating residual resistivity. The line curves slightly upwards at higher temperatures.)

The graph shows that resistivity increases almost linearly with temperature for a wide range, but it is not perfectly linear. At very low temperatures, the resistivity approaches a small constant value, known as residual resistivity, which is due to imperfections and impurities in the crystal lattice.
Quick Tip: Remember the key difference: Resistance is a property of an object (\(R = \rho l/A\)), while resistivity is an intrinsic property of the material itself. For conductors, \(\rho\) increases with T; for semiconductors, \(\rho\) decreases with T.

Step 1: Understanding the Concept:

Every real battery has an internal resistance (\(r\)), which causes a voltage drop within the battery when current is flowing. The voltage available to the external circuit is called the terminal voltage (\(V_T\)). The relationship between e.m.f. (\(\mathcal{E}\)), terminal voltage (\(V_T\)), current (\(I\)), and internal resistance (\(r\)) is key to this justification.


Step 2: Key Formula or Approach:

The terminal voltage of a battery supplying current \(I\) is given by: \[ V_T = \mathcal{E} - Ir \]
The current \(I\) delivered to an external resistance \(R\) is: \[ I = \frac{\mathcal{E}}{R + r} \]

Step 3: Justification:

A battery is designed to supply electrical energy to an external circuit. To draw a high current (\(I\)) from the battery, the total resistance of the circuit (\(R+r\)) should be as low as possible.

Furthermore, when a high current is drawn, the term \(Ir\) (the "lost volts" or internal voltage drop) becomes significant. \[ Lost Volts = Ir \]
If the internal resistance \(r\) is high, a large current \(I\) will cause a large internal voltage drop \(Ir\). This has two major negative effects:

Reduced Terminal Voltage: The terminal voltage \(V_T = \mathcal{E} - Ir\) will be significantly lower than the battery's e.m.f. This means less voltage is available for the external device.
Power Dissipation: A large amount of power is wasted as heat inside the battery (\(P_{loss} = I^2r\)). This reduces the efficiency of the battery and can cause it to overheat.

Therefore, for a battery to be able to supply a high current effectively, its internal resistance \(r\) must be very low. A low \(r\) minimizes the internal voltage drop and power loss, ensuring that the terminal voltage remains close to the e.m.f. and that most of the power is delivered to the external load. For example, car batteries, which need to provide hundreds of amperes to start the engine, are designed to have extremely low internal resistance.
Quick Tip: Think of internal resistance as a small resistor in series with the ideal voltage source inside the battery. To get maximum current out, you want this internal resistor to be as small as possible.

Step 1: Understanding the Concept:

This statement relates to the safety aspects of high voltage power sources. While low internal resistance is desirable for power delivery, it can be extremely dangerous in high voltage sources due to the risk of short circuits.


Step 2: Key Formula or Approach:

The short-circuit current (\(I_{sc}\)) is the maximum current a battery can deliver. It occurs when the external resistance \(R\) is nearly zero. The formula is: \[ I_{sc} = \frac{\mathcal{E}}{r} \]
The power dissipated as heat during a short circuit is \(P = I_{sc}^2 r = \mathcal{E}^2 / r\).


Step 3: Justification:

A high voltage battery (one with a large e.m.f. \(\mathcal{E}\)) is inherently dangerous. If such a battery were to have a very low internal resistance \(r\), an accidental short circuit (e.g., dropping a metal tool across its terminals) would result in an extremely large current flow. \[ I_{sc} = \frac{High \mathcal{E}}{Low r} = Very Large Current \]
This massive current can have catastrophic consequences:

Extreme Heat Generation: The power dissipated within the battery (\(P = \mathcal{E}^2 / r\)) would be enormous, causing rapid and dangerous overheating. This can lead to the battery catching fire or exploding.
Severe Electric Shock Hazard: The high current can cause fatal electric shocks if it passes through a person's body.
Damage to Equipment: The high current can melt wires and destroy any connected components.

To mitigate these risks, high voltage sources used in laboratories and other settings are often designed with a large internal resistance. This intentionally limits the maximum current that can be drawn from the source, even under short-circuit conditions, to a safer, more manageable level. Therefore, having a large internal resistance in a high voltage battery is a crucial safety feature.
Quick Tip: For batteries, there is a trade-off between power delivery efficiency and safety. Low internal resistance is good for high-power applications (like starting a car), while high internal resistance is a necessary safety feature for high-voltage sources.

Step 1: Understanding the Concept:

This question involves the refraction of a beam of light, which is a bundle of parallel rays. When light enters an optically denser medium (like water) from a rarer medium (like air), it bends towards the normal. We need to analyze how this bending affects the perpendicular distance between the outermost rays of the beam, which defines its width.


Step 2: Detailed Explanation:

Let a parallel beam of light of width \(w\) be incident on the surface of water at an angle of incidence \(i\). Let the angle of refraction be \(r\). Since light travels from air (rarer) to water (denser), it bends towards the normal, so \(r < i\).


Consider the wavefront of the incident beam. Let it be represented by the line segment AC, such that its width is \(w = AC\). The rays are perpendicular to this wavefront. The width of the beam is the perpendicular distance between the extreme rays.


Let's consider two points A and B on the interface where the two extreme rays of the beam strike the water surface. The distance between these points is \(d\). From the geometry of the incident beam, the width \(w\) is related to \(d\) by \(w = d \cos i\). \[ d = \frac{w}{\cos i} \]
After entering the water, the two extreme rays are now parallel to each other but travel in a new direction, making an angle \(r\) with the normal. The width of the refracted beam, \(w'\), is the perpendicular distance between these new rays. This new width is related to the distance \(d\) by \(w' = d \cos r\). \[ w' = d \cos r \]
Substituting the expression for \(d\): \[ w' = \left( \frac{w}{\cos i} \right) \cos r = w \frac{\cos r}{\cos i} \]
According to Snell's law, \(n_1 \sin i = n_2 \sin r\). Since water is denser than air, \(n_2 > n_1\), which implies \(\sin i > \sin r\). For angles between 0 and 90 degrees, if \(\sin i > \sin r\), then it must be that \(i > r\).

Also, for angles between 0 and 90 degrees, the cosine function is a decreasing function. Therefore, if \(i > r\), then \(\cos i < \cos r\).

This means the ratio \(\frac{\cos r}{\cos i}\) is greater than 1.

Therefore, \(w' = w \times (a number > 1)\), which means \(w' > w\).


Step 3: Final Answer:

When the parallel beam of light enters the water surface obliquely, it bends towards the normal, and its width increases.
Quick Tip: A good way to visualize this is to think of a column of soldiers marching from pavement (fast medium) onto sand (slow medium) at an angle. The soldier who hits the sand first slows down, causing the entire line of soldiers (the wavefront) to pivot and bunch up less, effectively widening the column's perpendicular width.

Step 1: Understanding the Concept:

The apparent bending of a straw or any object partially submerged in a liquid is an optical illusion caused by the refraction of light. Light rays traveling from the submerged portion of the straw bend as they pass from the denser medium (water) to the rarer medium (air) before reaching the observer's eye.


Step 2: Ray Diagram and Explanation:

Diagram:

% A simple description of the plot is provided here as drawing is not possible.
%
(A diagram showing a beaker of water with a straw partly submerged. An observer's eye is shown outside the water. Two light rays are drawn originating from a point P at the bottom of the straw. As these rays travel from water to air, they bend away from the normal at the water's surface. The observer's eye sees these refracted rays. When these refracted rays are traced backward as straight lines, they appear to originate from a point P', which is at a shallower depth than P. This makes the submerged part of the straw, OP, appear as OP', and the whole straw appears bent at the surface.)


Explanation:


Consider a point P on the part of the straw that is submerged in water. Light rays travel from this point P in all directions.
The rays of light that travel from point P (in water, a denser medium) to the observer's eye (in air, a rarer medium) undergo refraction at the water-air interface.
According to the laws of refraction, when light passes from a denser to a rarer medium, it bends away from the normal.
The human eye perceives the position of an object by assuming that light travels in straight lines. Therefore, the brain traces the refracted rays backward in a straight line.
These backward-extended rays appear to intersect at a point P', which is vertically above the actual point P. P' is the virtual image of point P.
Since this happens for every point on the submerged portion of the straw, the entire submerged part appears to be raised.
The portion of the straw that is in the air is seen directly without refraction.
As a result, the submerged part appears shallower than it actually is, while the part in the air appears at its true position. This difference in apparent depth causes the straw to look bent at the point where it enters the water. Quick Tip: Remember the concept of apparent depth. The apparent depth (\(d_{app}\)) is less than the real depth (\(d_{real}\)). The relationship is \(d_{app} = d_{real} / n\), where \(n\) is the refractive index of the denser medium. This is why the bottom of a swimming pool looks shallower than it is.

Step 1: Understanding the Concept:

An optical fibre is a device that transmits light signals over long distances with very little loss of energy. The working principle behind an optical fibre is Total Internal Reflection (TIR).


Structure of an Optical Fibre:
An optical fibre primarily consists of two parts made of transparent materials like glass or plastic:

Core: The inner cylindrical part through which the light signal travels. It has a higher refractive index (\(n_{core}\)).
Cladding: The outer layer that surrounds the core. It has a lower refractive index (\(n_{cladding}\)). So, \(n_{core} > n_{cladding}\).

This structure is essential for TIR to occur.


Step 2: Diagram and Explanation of Transmission:

Diagram:

% A simple description of the plot is provided here as drawing is not possible.
%
(A diagram showing the cross-section of an optical fibre with the inner core (refractive index n1) and outer cladding (refractive index n2), where n1 > n2. A light ray is shown entering the core from one end at a specific angle. The ray travels and strikes the core-cladding interface at an angle of incidence 'i'. The normal to the interface is shown. Since 'i' is greater than the critical angle 'c' (i > c), the ray undergoes total internal reflection and is reflected back into the core. This process is shown repeating multiple times, guiding the light ray along a zigzag path through the fibre to the other end.)


Explanation of Transmission:

A light signal (e.g., from a laser) is launched into the core of the optical fibre at one end. The angle of entry is controlled so that the light ray strikes the core-cladding interface at a suitable angle.
The ray travels through the core and strikes the boundary between the core and the cladding.
Since the ray is traveling from a denser medium (core, \(n_{core}\)) to a rarer medium (cladding, \(n_{cladding}\)), total internal reflection is possible.
TIR occurs if the angle of incidence (\(i\)) at the core-cladding interface is greater than the critical angle (\(i_c\)) for the two media. The critical angle is given by \(\sin(i_c) = \frac{n_{cladding}}{n_{core}}\).
If this condition (\(i > i_c\)) is met, the light is completely reflected back into the core with almost no loss of intensity.
This reflected ray then travels to the opposite side of the core and strikes the interface again. The process of total internal reflection is repeated.
Through a series of successive total internal reflections, the light signal is guided along the fibre, even if the fibre is bent, and emerges at the other end with minimal attenuation. Quick Tip: The two essential conditions for Total Internal Reflection (TIR) are: 1. Light must travel from an optically denser medium to an optically rarer medium. 2. The angle of incidence in the denser medium must be greater than the critical angle. These are the core principles behind optical fibres, mirages, and the sparkling of diamonds.

Step 1: Differentiation:

\begin{tabular{|p{3cm|p{6.5cm|p{6.5cm|
\hline
Aspect & Peak Value (\(I_0\) or \(I_{peak}\)) & Root Mean Square (RMS) Value (\(I_{rms}\))

\hline
Definition & It is the maximum instantaneous value or amplitude of the alternating current during a complete cycle. & It is the effective value of the AC. It is the value of steady DC that produces the same heating effect in a resistor over a given time as the AC does.

\hline
Representation & Represents the highest point the AC waveform reaches. & Represents the effective DC equivalent for power calculations.

\hline
Measurement & Can be directly observed on an oscilloscope. & Measured by standard AC ammeters and voltmeters (hot-wire instruments).

\hline
Usage & Important for determining insulation requirements and the maximum voltage/current a component must withstand. & Used in all power-related calculations, such as \(P = I_{rms}^2 R\) and \(P = V_{rms} I_{rms}\).

\hline
\end{tabular


Step 2: Derivation of RMS Value:

The RMS value of an alternating current is found by taking the square root of the mean (average) of the square of the current over one complete cycle.

Let the sinusoidal alternating current be represented by: \[ I(t) = I_0 \sin(\omega t) \]
where \(I_0\) is the peak value.


1. Square the current: \[ I^2(t) = I_0^2 \sin^2(\omega t) \]

2. Find the mean (average) value of \(I^2(t)\) over one cycle (from \(t=0\) to \(t=T\)):
The mean value \(\langle I^2 \rangle\) is given by: \[ \langle I^2 \rangle = \frac{\int_0^T I^2(t) dt}{\int_0^T dt} = \frac{1}{T} \int_0^T I_0^2 \sin^2(\omega t) dt \]
We use the trigonometric identity \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\). \[ \langle I^2 \rangle = \frac{I_0^2}{T} \int_0^T \frac{1 - \cos(2\omega t)}{2} dt \] \[ \langle I^2 \rangle = \frac{I_0^2}{2T} \left[ \int_0^T dt - \int_0^T \cos(2\omega t) dt \right] \] \[ \langle I^2 \rangle = \frac{I_0^2}{2T} \left[ t - \frac{\sin(2\omega t)}{2\omega} \right]_0^T \]
Now, we substitute the limits. Since \(\omega = 2\pi/T\), the term \(2\omega = 4\pi/T\). \[ \langle I^2 \rangle = \frac{I_0^2}{2T} \left[ (T - 0) - \left( \frac{\sin(4\pi T/T) - \sin(0)}{4\pi/T} \right) \right] \] \[ \langle I^2 \rangle = \frac{I_0^2}{2T} \left[ T - \left( \frac{\sin(4\pi) - 0}{4\pi/T} \right) \right] \]
Since \(\sin(4\pi) = 0\), the second part of the integral evaluates to zero. \[ \langle I^2 \rangle = \frac{I_0^2}{2T} [T - 0] = \frac{I_0^2}{2} \]

3. Take the square root of the mean:
The RMS value is the square root of \(\langle I^2 \rangle\). \[ I_{rms} = \sqrt{\langle I^2 \rangle} = \sqrt{\frac{I_0^2}{2}} \] \[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
This is the expression for the root mean square value of a sinusoidal alternating current in terms of its peak value. Numerically, \(I_{rms} \approx 0.707 I_0\).
Quick Tip: Unless specified otherwise, when you see values for AC voltage or current (e.g., "120 V AC outlet"), these are almost always the RMS values. This is because RMS values are used for power calculations, making them the most practical measure for everyday use.

Step 1: Fundamental Principle:

An electromagnetic (EM) wave is produced by an accelerated electric charge. This is the fundamental source of all electromagnetic radiation.


Step 2: Explanation:

Let's consider the fields produced by a charge in different states of motion:

Stationary Charge: A charge at rest produces only a static electric field (\(\vec{E}\)) in the space around it. This field does not change with time.
Charge Moving with Constant Velocity: A charge moving with a constant velocity constitutes a steady electric current. This produces both a static electric field and a static magnetic field (\(\vec{B}\)). The fields are constant in time at any given point in space and do not propagate outwards.
Accelerated Charge: When a charge accelerates (i.e., its velocity changes in magnitude, direction, or both), both the electric and magnetic fields it produces change with time and space. According to Maxwell's equations, a changing electric field generates a changing magnetic field, and a changing magnetic field generates a changing electric field.

This mutual generation of time-varying electric and magnetic fields creates a disturbance that propagates outwards from the accelerating charge. This propagating disturbance of coupled, oscillating electric and magnetic fields is an electromagnetic wave. The electric field (\(\vec{E}\)), magnetic field (\(\vec{B}\)), and the direction of wave propagation (\(\vec{v}\)) are mutually perpendicular to each other. A common way to produce EM waves is by making a charge oscillate, as in an LC oscillator circuit connected to an antenna. The continuous acceleration of the charge in the antenna radiates energy in the form of electromagnetic waves.
Quick Tip: A simple mnemonic is: Charge at rest \(\rightarrow\) Electric field only. Charge in uniform motion \(\rightarrow\) Electric + Magnetic fields. Charge in accelerated motion \(\rightarrow\) Radiates EM waves.

Step 1: Understanding the Concept:

In an electromagnetic wave, the electric field vector (\(\vec{E}\)), the magnetic field vector (\(\vec{B}\)), and the direction of wave propagation (\(\vec{v}\)) are always mutually perpendicular. Their relative orientation is given by the direction of the Poynting vector, \(\vec{S} \propto \vec{E} \times \vec{B}\), which points in the direction of wave propagation.


Step 2: Applying the Right-Hand Rule:

We can use the vector cross product rule (or a right-hand rule) to determine the direction of \(\vec{B}\). The relationship is \(direction(\vec{v}) = direction(\vec{E} \times \vec{B})\).


Step 3: Defining a Coordinate System:

Let's set up a standard geographical coordinate system:

East direction as the +x axis.
North direction as the +y axis.
Vertically Upward direction as the +z axis.

From this, it follows that:

West is the -x axis.
South is the -y axis.
Downward is the -z axis.


Step 4: Determining the Direction of \(\vec{B}\):

We are given:

Direction of propagation (\(\vec{v}\)): Vertically upward \(\rightarrow\) \(+\hat{k}\) (in the +z direction).
Direction of electric field (\(\vec{E}\)): West \(\rightarrow\) \(-\hat{i}\) (in the -x direction).

We need to find the direction of the magnetic field, let's call it \(\hat{b}\), such that: \[ \hat{k} = (-\hat{i}) \times \hat{b} \]
We know the cyclic properties of the vector cross product for unit vectors: \(\hat{i} \times \hat{j} = \hat{k}\).
Let's test the possible directions for \(\hat{b}\).
If we try \(\hat{b} = \hat{j}\) (North), we get \((-\hat{i}) \times \hat{j} = -(\hat{i} \times \hat{j}) = -\hat{k}\). This is incorrect, as we need \(+\hat{k}\).

If we try \(\hat{b} = -\hat{j}\) (South), we get \((-\hat{i}) \times (-\hat{j}) = (\hat{i} \times \hat{j}) = +\hat{k}\). This is the correct direction.


Final Answer: The magnetic field vector must point in the \(-\hat{j}\) direction, which corresponds to the South direction.
Quick Tip: Use the right-hand rule for \(\vec{E} \times \vec{B}\). Point your fingers in the direction of \(\vec{E}\) (West). You want your thumb to point in the direction of propagation (Up). To achieve this, you must curl your fingers towards the direction of \(\vec{B}\). This forces your palm (and the direction you curl towards) to face South. So, \(\vec{B}\) is towards the South.

Step 1: Understanding the Concept:

The electromagnetic spectrum is a continuous range of wavelengths and frequencies. The boundaries between different types of radiation (like radio waves and gamma rays) are not sharply defined, so this question asks for an estimate based on typical values.


Step 2: Estimating the Wavelengths:

We need to recall the approximate ranges for radio waves and gamma waves.


Radio Waves: This is the longest wavelength part of the spectrum. It includes waves from thousands of kilometers down to about 1 millimeter. The "shortest" radio waves are often considered to be in the microwave region.
\[ \lambda_{radio, shortest} \approx 10^{-3} meters (1 mm) \]
Gamma Waves (\(\gamma\)-rays): This is the shortest wavelength, highest energy part of the spectrum. The wavelengths are typically smaller than the size of an atom. The "longest" gamma-ray wavelengths overlap with the shortest X-ray wavelengths. There is no strict upper limit, but a typical value for the longest gamma rays is around 10 picometers.
\[ \lambda_{gamma, longest} \approx 10^{-11} to 10^{-12} meters \]

Let's use \(\lambda_{gamma, longest} \approx 10^{-12}\) m for our estimation.


Step 3: Calculating the Ratio:

The required ratio is: \[ Ratio = \frac{\lambda_{radio, shortest}}{\lambda_{gamma, longest}} \] \[ Ratio \approx \frac{10^{-3} m}{10^{-12} m} = 10^{-3 - (-12)} = 10^9 \]
If we use the value \(\lambda_{gamma, longest} \approx 10^{-11}\) m: \[ Ratio \approx \frac{10^{-3} m}{10^{-11} m} = 10^8 \]
Since this is an estimation, an order-of-magnitude answer is appropriate.


Step 4: Final Answer:

The estimated ratio of the shortest wavelength of radio waves to the longest wavelength of gamma waves is in the range of \(10^8\) to \(10^9\).
Quick Tip: Memorizing the order of the electromagnetic spectrum is crucial: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma Ray (in order of decreasing wavelength / increasing frequency). Knowing the approximate power-of-ten for the wavelengths (e.g., visible \(\sim 10^{-7}\) m, X-rays \(\sim 10^{-10}\) m) is very helpful for estimation problems.

Step 1: Understanding the Concept:

The problem states that a charged particle moves with a constant velocity. According to Newton's first law, this implies that the net force acting on the particle is zero. The particle is subject to two forces: an electric force due to the uniform electric field and a magnetic force due to its motion in the magnetic field created by the current-carrying wire. For the net force to be zero, these two forces must be equal in magnitude and opposite in direction.


(i) Diagram of Fields and Forces:
Let's set up a coordinate system where the conductor lies along the x-axis. The particle is moving parallel to it at a distance \(d\). Let's assume the particle is in the xy-plane, at \(y=d\).

Current (I): In the -x direction (\(-\hat{i}\)).
Magnetic Field (\(\vec{B}\)): Produced by the wire. Using the right-hand thumb rule (pointing thumb in the direction of current, -x), our fingers curl into the page above the wire. So, at the particle's position (\(y=d\)), the magnetic field \(\vec{B}\) is in the -z direction (\(-\hat{k}\)).
Magnetic Force (\(\vec{F}_B\)): We will calculate this in the next part. It will be directed away from the wire (+y direction).
Electric Field (\(\vec{E}\)): For the net force to be zero, the electric force must oppose the magnetic force. This means \(\vec{F}_E\) must be towards the wire (-y direction). Since the particle is negatively charged (\(q = -|q|\)), and \(\vec{F}_E = q\vec{E}\), the electric field \(\vec{E}\) must be in the +y direction.


% A simple description of the plot is provided here as drawing is not possible.
(A diagram showing the x-y plane. The wire is on the x-axis with current I pointing to the left (-\(\hat{i}\)). A negative charge (-q) is shown at y=d, moving to the left with velocity \(\vec{v}\). The magnetic field \(\vec{B}\) at this point is shown with a cross, indicating it's directed into the page (-\(\hat{k}\)). The resulting magnetic force \(\vec{F}_B\) is shown pointing upward (+\(\hat{j}\)). The electric force \(\vec{F}_E\) is shown pointing downward (-\(\hat{j}\)), and the electric field \(\vec{E}\) is shown pointing upward (+\(\hat{j}\)).)


(ii) Various Forces Acting:
There are two forces acting on the charged particle:

Electric Force (\(\vec{F}_E\)): Due to the uniform electric field \(\vec{E}\). It is given by \(\vec{F}_E = q\vec{E}\). Since the charge is negative, this force is opposite to the direction of \(\vec{E}\).
Magnetic Force (Lorentz Force) (\(\vec{F}_B\)): Due to the motion of the charged particle in the magnetic field \(\vec{B}\) of the wire. It is given by \(\vec{F}_B = q(\vec{v} \times \vec{B})\).


(iii) Finding the value of \(v_0\):
First, let's find the expression for the magnetic force.

Charge: \(q\) (we will use q, and remember it's negative).
Velocity: \(\vec{v} = -v_0 \hat{i}\).
Magnetic field at distance d from the wire: \(\vec{B} = \frac{\mu_0 I}{2\pi d} (-\hat{k})\).

The magnetic force is: \[ \vec{F}_B = q(\vec{v} \times \vec{B}) = q \left( (-v_0 \hat{i}) \times \left(-\frac{\mu_0 I}{2\pi d} \hat{k}\right) \right) \] \[ \vec{F}_B = q \frac{\mu_0 I v_0}{2\pi d} (\hat{i} \times \hat{k}) \]
Since \(\hat{i} \times \hat{k} = -\hat{j}\): \[ \vec{F}_B = -q \frac{\mu_0 I v_0}{2\pi d} \hat{j} \]
Since \(q\) is negative (\(q=-|q|\)), \(\vec{F}_B = |q| \frac{\mu_0 I v_0}{2\pi d} \hat{j}\). This confirms the magnetic force is in the +y direction (away from the wire).

The electric force is \(\vec{F}_E = q\vec{E}\).

For constant velocity, the net force is zero: \[ \vec{F}_{net} = \vec{F}_E + \vec{F}_B = 0 \implies \vec{F}_E = -\vec{F}_B \] \[ q\vec{E} = - \left( -q \frac{\mu_0 I v_0}{2\pi d} \hat{j} \right) = q \frac{\mu_0 I v_0}{2\pi d} \hat{j} \]
Cancelling \(q\) from both sides: \[ \vec{E} = \frac{\mu_0 I v_0}{2\pi d} \hat{j} \]
This shows the electric field must be uniform and directed along the +y axis. Its magnitude is given by \(E\). \[ E = |\vec{E}| = \frac{\mu_0 I v_0}{2\pi d} \]
Now, we solve for \(v_0\): \[ v_0 = \frac{2\pi d E}{\mu_0 I} \] Quick Tip: This problem is a classic example of a "velocity selector" condition. Whenever a charged particle moves with constant velocity through regions with both electric and magnetic fields, it means the net force is zero. Setting \(\vec{F}_E + \vec{F}_B = 0\) is the key to solving such problems.

Step 1: Understanding the Concept:

This problem requires the application of the principle of superposition for magnetic fields. The net magnetic field at point P will be the vector sum of the magnetic fields produced by each of the two current-carrying conductors individually. The magnetic field due to a long straight conductor is given by Ampere's circuital law.


Step 2: Key Formula or Approach:

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


Step 3: Detailed Explanation:

Let's analyze the contribution from each wire at point P(X, Y).


1. Magnetic Field due to wire along XX' (current \(I_1\)):

Current \(I_1\) is in the -X direction (\(-\hat{i}\)).
The point P(X, Y) is at a perpendicular distance \(r_1 = Y\) from the X-axis.
The magnitude of the magnetic field \(\vec{B}_1\) is:
\[ B_1 = \frac{\mu_0 I_1}{2\pi Y} \]
Direction of \(\vec{B}_1\): Using the right-hand thumb rule, point your thumb in the direction of the current (\(-X\)). At a point P(X,Y) with Y > 0 (above the wire), your fingers curl out of the page. So, the direction is along the +Z axis.
\[ \vec{B}_1 = \frac{\mu_0 I_1}{2\pi Y} \hat{k} \]


2. Magnetic Field due to wire along YY' (current \(I_2\)):

Current \(I_2\) is in the -Y direction (\(-\hat{j}\)).
The point P(X, Y) is at a perpendicular distance \(r_2 = X\) from the Y-axis.
The magnitude of the magnetic field \(\vec{B}_2\) is:
\[ B_2 = \frac{\mu_0 I_2}{2\pi X} \]
Direction of \(\vec{B}_2\): Using the right-hand thumb rule, point your thumb in the direction of the current (\(-Y\)). At a point P(X,Y) with X > 0 (to the right of the wire), your fingers curl into the page. So, the direction is along the -Z axis.
\[ \vec{B}_2 = -\frac{\mu_0 I_2}{2\pi X} \hat{k} \]


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

Magnitude of the Net Field:
The magnitude is the absolute value of the component along \(\hat{k}\). \[ B_{net} = \left| \frac{\mu_0}{2\pi} \left( \frac{I_1}{Y} - \frac{I_2}{X} \right) \right| \]
The question seems to have a typo and might have asked for perpendicular currents resulting in perpendicular fields. Let's assume a scenario where the fields are perpendicular for a more general magnitude formula. If one wire was along X and the other along Z, the fields at P(X,Y) would be perpendicular. As the question is stated, both fields are along the Z-axis.

Let's re-read the question carefully. It asks for magnitude and direction. What if the wires are not on the axes, but kept along lines parallel to the axes? No, "kept along XX' and YY' axes". Okay, the vector addition is correct.

Let's reconsider the problem as it might be intended, where the fields are perpendicular. This happens if the wires are in the same plane. Let's assume wire 1 is on the x-axis, current along -x. Wire 2 is on the y-axis, current along -y. Point P is (X,Y). Field from wire 1 at P is \(\vec{B}_1 = (\mu_0 I_1 / 2\pi Y) \hat{k}\). Field from wire 2 at P is \(\vec{B}_2 = (-\mu_0 I_2 / 2\pi X) \hat{k}\). The vector sum is correct. The question might have intended for the point P to be in the XZ plane, for instance, which would create perpendicular fields.

However, based on the literal interpretation, the result is: \[ \vec{B}_{net} = \frac{\mu_0}{2\pi} \left( \frac{I_1}{Y} - \frac{I_2}{X} \right) \hat{k} \]
Magnitude: \( B_{net} = \frac{\mu_0}{2\pi} \left| \frac{I_1}{Y} - \frac{I_2}{X} \right| \).
Direction: Along the +Z axis (out of page) if \(I_1/Y > I_2/X\), and along the -Z axis (into page) if \(I_1/Y < I_2/X\).

There might be a misunderstanding of the question. Let's assume the question meant one wire along X axis and one wire along Y axis, but the point P(X, Y) is in the XY plane.
Field from wire on X axis (current I1 in -i dir): at (X,Y), the perpendicular distance is Y. Direction by RHR is +k. \(B_1 = \frac{\mu_0 I_1}{2 \pi Y} \hat{k}\)
Field from wire on Y axis (current I2 in -j dir): at (X,Y), the perpendicular distance is X. Direction by RHR is -k. \(B_2 = \frac{-\mu_0 I_2}{2 \pi X} \hat{k}\)
The net field is indeed the sum of these two.

Let's consider the possibility that the question intends for the fields to be perpendicular, which is a more standard superposition problem. This would happen if, for example, the wires were perpendicular but not intersecting. Given the ambiguity, we'll stick to the literal interpretation. The final answer is as derived.
Final expression seems correct. Let's write the final answer.

Step 4: Final Answer:

The net magnetic field vector at point P(X, Y) is: \[ \vec{B}_{net} = \frac{\mu_0}{2\pi} \left( \frac{I_1}{Y} - \frac{I_2}{X} \right) \hat{k} \]
The magnitude is \( B_{net} = \frac{\mu_0}{2\pi} \left| \frac{I_1}{Y} - \frac{I_2}{X} \right| \).

The direction is along the +Z axis (out of the page) if \(I_1/Y > I_2/X\), and along the -Z axis (into the page) if \(I_2/X > I_1/Y\). If \(I_1/Y = I_2/X\), the net magnetic field is zero.
Quick Tip: When using the superposition principle, always calculate the magnetic field vector (magnitude and direction) from each source separately. Then, perform a vector addition to find the net field. Be very careful with directions using the Right-Hand Rule.

Step 1: Understanding the Concept:

An extrinsic semiconductor is a pure (intrinsic) semiconductor that has been intentionally doped with impurity atoms to increase its conductivity. This doping process creates an excess of either free electrons or holes, which become the primary charge carriers.


Step 2: Types of Extrinsic Semiconductors and Charge Carriers:

There are two types of extrinsic semiconductors:


1. n-type Semiconductor:

Doping: A pure semiconductor (like Silicon or Germanium, from Group IV) is doped with a pentavalent impurity (an element from Group V, like Phosphorus, Arsenic, or Antimony).
Mechanism: The pentavalent impurity atom forms covalent bonds with four neighboring semiconductor atoms. Its fifth valence electron is loosely bound and can easily move into the conduction band, becoming a free electron. These impurity atoms are called "donor" atoms.
Charge Carriers: The doping process creates a large number of free electrons. Therefore, in an n-type semiconductor:

Majority carriers are electrons.
Minority carriers are holes (which are created due to thermal energy breaking some covalent bonds).



2. p-type Semiconductor:

Doping: A pure semiconductor is doped with a trivalent impurity (an element from Group III, like Boron, Aluminum, or Gallium).
Mechanism: The trivalent impurity atom can only form three covalent bonds with its neighbors, leaving a vacancy or a "hole" in the fourth bond. This hole can accept an electron from a neighboring atom, causing the hole to move. These impurity atoms are called "acceptor" atoms.
Charge Carriers: The doping process creates a large number of holes. Therefore, in a p-type semiconductor:

Majority carriers are holes.
Minority carriers are electrons (which are created due to thermal energy). Quick Tip: A simple mnemonic to remember: \textbf{n-type}: Doped with pe\textbf{n}tavalent atoms, has excess \textbf{n}egative charge carriers (electrons). \textbf{p-type}: Doped with trivalent atoms, has excess \textbf{p}ositive charge carriers (holes).

Step 1: Understanding Forward Bias:

A p-n junction is forward-biased when the p-side of the junction is connected to the positive terminal of an external voltage source (a battery), and the n-side is connected to the negative terminal.


Step 2: Effect of Forward Bias on the Depletion Region:

The external applied voltage creates an electric field that opposes the internal barrier electric field of the depletion region. As a result, the effective potential barrier across the junction is lowered (\(V_B' = V_B - V_{applied}\)). This causes the width of the depletion region to decrease.


Step 3: Movement of Charge Carriers:

With the potential barrier significantly reduced, the charge carriers have enough energy to cross the junction.


Movement of Majority Carriers:

The positive terminal of the battery repels the majority carriers in the p-side (holes). These holes are pushed towards the junction.
The negative terminal of the battery repels the majority carriers in the n-side (electrons). These electrons are pushed towards the junction.
Since the barrier is low, the holes from the p-side can now easily diffuse across the junction into the n-side. Similarly, the electrons from the n-side can easily diffuse across the junction into the p-side.
This movement of majority carriers across the junction constitutes the diffusion current, which is the primary component of the forward current and is typically large (in the order of milliamperes).

Carrier Recombination:

Once the electrons from the n-side cross into the p-side, they become minority carriers there and recombine with the holes near the junction.
Similarly, when holes from the p-side cross into the n-side, they become minority carriers and recombine with electrons near the junction.
For every electron-hole recombination, a covalent bond is broken in the p-region near the positive terminal, releasing an electron that flows into the terminal. Simultaneously, an electron from the negative terminal enters the n-region to replenish the electron that was lost to recombination. This maintains a continuous flow of current in the external circuit.


In summary, the forward current in a p-n junction is predominantly due to the diffusion of majority charge carriers (holes from p to n, and electrons from n to p) across the junction, which is made possible by the reduction of the potential barrier.
Quick Tip: Remember the key effects of forward bias: Connects P to Positive, N to Negative. Opposes the barrier potential. Narrows the depletion layer. Allows majority carrier diffusion. Results in low resistance and high current.

Step 1: Understanding Dynamic Resistance:

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. \[ r_d = \frac{\Delta V}{\Delta I} = \left( \frac{dI}{dV} \right)^{-1} \]
It represents how the current changes in response to a small change in the applied voltage.


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

The point V = -0.6 V lies in the reverse bias region of the diode's characteristic curve. Looking at the graph:

The x-axis represents voltage (V) in volts.
The y-axis represents current (I) in milliamperes (mA).

In the entire reverse bias region shown (from 0 V to -1.2 V and beyond), the I-V curve is a flat horizontal line lying on the voltage axis. This indicates that the current \(I\) is approximately zero and, more importantly, it is not changing as the reverse voltage changes.


Step 3: Estimating the Dynamic Resistance:

To estimate the dynamic resistance at V = -0.6 V, we need to find the slope \(\frac{\Delta I}{\Delta V}\) around this point.
Let's pick a small interval around -0.6 V, for example, from V\(_1\) = -0.8 V to V\(_2\) = -0.4 V.

Change in Voltage, \(\Delta V = V_2 - V_1 = -0.4 - (-0.8) = +0.4\) V.
From the graph, the current at both V\(_1\) and V\(_2\) is approximately 0 mA.
Change in Current, \(\Delta I = I_2 - I_1 = 0 - 0 = 0\) mA.

Now, we calculate the dynamic resistance: \[ r_d = \frac{\Delta V}{\Delta I} = \frac{0.4 V}{0 A} \rightarrow \infty \]
Since the slope of the graph \(\frac{\Delta I}{\Delta V}\) is zero in the reverse bias region, its reciprocal, the dynamic resistance, is infinitely high.


Step 4: Final Answer:

The dynamic resistance of the diode at V = -0.6 V is extremely high. Based on the provided graph, the slope is zero, so the dynamic resistance is considered to be infinite. In a real diode, there would be a tiny leakage current, leading to a very large but finite resistance.
Quick Tip: Remember the general resistance properties of a diode from its I-V curve: \textbf{Forward Bias Region:} The curve is steep (\(\Delta I / \Delta V\) is large), so the dynamic resistance is \textbf{low}. \textbf{Reverse Bias Region:} The curve is flat (\(\Delta I / \Delta V\) is near zero), so the dynamic resistance is \textbf{very high}.

Step 1: The Binding Energy Curve:

The binding energy per nucleon (BE/A) is a measure of the stability of an atomic nucleus. It is the average energy required to remove one nucleon (a proton or a neutron) from the nucleus. The curve is a plot of this value against the mass number (A) for various nuclei.


Graph Description:

% A simple description of the plot is provided here as drawing is not possible.
%
(A graph with Binding Energy per Nucleon (in MeV) on the y-axis and Mass Number (A) on the x-axis. The curve starts at a low value for light nuclei (like Deuterium), rises sharply, and reaches a broad maximum around A = 56 (Iron), with a peak value of about 8.8 MeV. After the peak, the curve slowly and gradually decreases for heavier nuclei, dropping to about 7.6 MeV for Uranium.)


Key Features of the Curve:

For very light nuclei (A < 20), the BE/A is small and increases rapidly with A. There are some peaks for highly stable light nuclei like \(^4\)He, \(^{12}\)C, and \(^{16}\)O.
The curve has a broad maximum for nuclei with mass numbers in the range A = 40 to 120. The peak is at A = 56 (Iron, \(^{56}\)Fe), which is the most stable nucleus.
For heavy nuclei (A > 120), the BE/A slowly decreases as A increases.


Step 2: Significance of the Binding Energy Curve:

The shape of the binding energy curve is profoundly significant as it explains the release of energy in nuclear reactions:

Explains Nuclear Stability: The higher the binding energy per nucleon, the more stable the nucleus is. The curve shows that iron-56 is the most stable element. Elements to the left or right of the peak are less stable.
Explains Nuclear Fission: The curve shows that heavy nuclei (like Uranium, A > 230) have a lower BE/A than nuclei in the middle of the curve. If a heavy nucleus splits into two or more lighter nuclei (fission), the daughter nuclei will be located higher up on the curve. This means the total binding energy of the products is greater than that of the original nucleus. This increase in binding energy is released as a large amount of energy, according to Einstein's mass-energy equivalence, \(E = \Delta m c^2\).
Explains Nuclear Fusion: The curve also shows that very light nuclei (like Hydrogen isotopes) have a very low BE/A. If two light nuclei combine (fuse) to form a heavier nucleus, the resulting nucleus will be higher up on the curve (more stable). Again, the increase in binding energy is released as a tremendous amount of energy. This is the process that powers the Sun and other stars.

In essence, the curve shows that there are two ways to release nuclear energy: by splitting very heavy nuclei (fission) or by fusing very light nuclei (fusion), both of which are processes that move towards the peak of stability around Iron.
Quick Tip: Remember the main take-away from the BE curve: "Nature seeks stability". Both fission and fusion are processes where less stable nuclei transform into more stable nuclei, releasing the excess binding energy. The peak of the curve at Iron (A=56) is the key reference point for stability.

Step 1: Analyzing the Process:

The question describes a process where two nuclei, which are less stable (lower binding energy per nucleon), combine to form a single nucleus that is more stable (higher binding energy per nucleon). This process involves moving up the binding energy curve from the left side towards the peak.


(i) Type of Nuclear Reaction:
This process, where two or more light nuclei combine to form a single heavier nucleus, is known as nuclear fusion. For example: \[ ^2_1H + ^3_1H \rightarrow ^4_2He + ^1_0n + Energy \]
Here, Deuterium and Tritium (low BE/A) fuse to form Helium (higher BE/A).


(ii) Change in Total Mass:
Binding energy is the energy equivalent of the "mass defect" (\(\Delta m\)) of a nucleus, given by Einstein's 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.
The process is: Reactants \(\rightarrow\) Product.
Total Binding Energy of Product > Total Binding Energy of Reactants.
Since \(E_B \propto \Delta m\), this means:
Mass Defect of Product > Mass Defect of Reactants.
The mass of a nucleus is \(M_{nucleus} = (mass of constituents) - (mass defect)\).
Therefore, a larger mass defect means a smaller final nuclear mass.
So, the total mass of the product nucleus is less than the sum of the masses of the initial nuclei. The total mass of the system decreases. This "lost" mass is converted into the released energy.


(iii) Energy Requirement or Production:
Since the product nucleus has a higher binding energy per nucleon, it is in a more stable, lower energy state compared to the initial nuclei. To move from a higher energy state to a lower energy state, the system must release the excess energy.
The energy released (\(Q\)) is equal to the increase in the total binding energy: \[ Q = (Total BE of products) - (Total BE of reactants) > 0 \]
Therefore, the process produces energy (it is an exothermic reaction).
Quick Tip: Remember the relationship: \textbf{Higher Binding Energy \(\iff\) Larger Mass Defect \(\iff\) Lower Total Mass \(\iff\) Greater Stability.} In any spontaneous nuclear reaction (fission or fusion) that releases energy, the total mass of the products is always less than the total mass of the reactants.

Step 1: Understanding the Concept:

The question asks about the relationship between the graphs of stopping potential (\(V_0\)) versus frequency (\(\nu\)) for two different metals. This relationship is governed by Einstein's photoelectric equation.


Step 2: Key Formula or Approach:

Einstein's photoelectric equation is given as \(K_{max} = h\nu - \phi\), where \(\phi = h\nu_0\) is the work function of the metal.
We also know that \(K_{max} = eV_0\).
Combining these, we get: \[ eV_0 = h\nu - h\nu_0 \]
Rearranging this to express \(V_0\) as a function of \(\nu\), we get the equation of a straight line: \[ V_0 = \left(\frac{h}{e}\right)\nu - \frac{h\nu_0}{e} \]
This equation is in the form \(y = mx + c\), where:

\(y = V_0\) (the stopping potential)
\(x = \nu\) (the frequency)
\(m = \frac{h}{e}\) (the slope of the line)
\(c = -\frac{h\nu_0}{e}\) (the y-intercept)


Step 3: Detailed Explanation:

Let's analyze the components of the straight-line equation for two different metals.

Slope (\(m = h/e\)): The slope of the \(V_0\) vs. \(\nu\) graph is the ratio of Planck's constant (\(h\)) to the elementary charge (\(e\)). Both \(h\) and \(e\) are fundamental physical constants, and their values do not depend on the material of the metal. Therefore, the slope of the graph will be the same for all metals.
Intercepts: The work function (\(\phi = h\nu_0\)) is a characteristic property of a metal and is different for different metals.

The y-intercept is \(c = -\phi/e\). Since \(\phi\) is different for different metals, the y-intercepts will be different.
The x-intercept (where \(V_0 = 0\)) is the threshold frequency, \(\nu_0\). Since the work function is different, the threshold frequency \(\nu_0\) will also be different for different metals.


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


Step 4: Final Answer:

The straight-line graphs of stopping potential versus frequency for two different metals are parallel to each other.
Quick Tip: In the \(V_0\) vs. \(\nu\) plot, the slope (\(h/e\)) is a universal constant. The intercepts (\(\nu_0\) on the x-axis and \(-\phi/e\) on the y-axis) depend on the specific metal. This makes the graphs for different metals a set of parallel lines.

Step 1: Understanding the Concept:

The question asks to find the value of Planck's constant (\(h\)) in terms of the slope (\(m\)) of the given graph and the elementary charge (\(e\)).


Step 2: Key Formula or Approach:

From the previous analysis, the equation for the graph of stopping potential (\(V_0\)) versus frequency (\(\nu\)) is: \[ V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi}{e} \]
This is a straight line of the form \(y = (slope)x + (intercept)\).


Step 3: Detailed Explanation:

By comparing the photoelectric equation with the standard equation of a straight line, \(y = mx + c\), we can identify the slope.

Dependent variable \(y = V_0\).
Independent variable \(x = \nu\).
Slope \(m_{graph} = \frac{h}{e}\).

The problem states that the slope of the line is \(m\). Therefore, we have: \[ m = \frac{h}{e} \]
We need to find an expression for Planck's constant, \(h\). We can rearrange the equation by multiplying both sides by \(e\): \[ h = m \times e \]
or simply \(h = me\).


Step 4: Final Answer:

The value of Planck's constant is given by the product of the slope of the graph and the elementary charge, which is \(me\).
Quick Tip: Millikan's oil drop experiment determined the value of \(e\). By experimentally plotting the \(V_0\) vs. \(\nu\) graph for a metal and measuring its slope \(m\), he was able to use the relation \(h = me\) to make the first accurate determination of Planck's constant, \(h\).

Step 1: Understanding the Concept:

The intercepts of a graph are the points where the line crosses the coordinate axes. We need to find the intercepts for the \(V_0\) vs. \(\nu\) graph using the photoelectric equation.


Step 2: Key Formula or Approach:

The equation of the graph is: \[ V_0 = \frac{h}{e}\nu - \frac{h\nu_0}{e} \]

Step 3: Detailed Explanation:

1. Intercept on the \(\nu\)-axis (x-intercept):
The intercept on the frequency axis (\(\nu\)-axis) occurs when the stopping potential \(V_0 = 0\).
Setting \(V_0 = 0\) in the equation: \[ 0 = \frac{h}{e}\nu - \frac{h\nu_0}{e} \] \[ \frac{h}{e}\nu = \frac{h\nu_0}{e} \] \[ \nu = \nu_0 \]
So, the intercept on the \(\nu\)-axis is the threshold frequency, \(\nu_0\).


2. Intercept on the \(V_0\)-axis (y-intercept):
The intercept on the potential axis (\(V_0\)-axis) occurs when the frequency \(\nu = 0\).
Setting \(\nu = 0\) in the equation: \[ V_0 = \frac{h}{e}(0) - \frac{h\nu_0}{e} \] \[ V_0 = -\frac{h\nu_0}{e} \]
Since the work function is \(\phi = h\nu_0\), the y-intercept can also be written as \(-\phi/e\).


Step 4: Final Answer:

The intercept on the \(\nu\)-axis is \(\nu_0\), and the intercept on the \(V_0\)-axis is \(-\frac{h\nu_0}{e}\). The question asks for the intercepts respectively, so the answer is \(\nu_0\) and \(-h\nu_0/e\). This corresponds to option (A).
Quick Tip: The intercepts of the photoelectric graph have direct physical meaning: \textbf{x-intercept (\(\nu_0\))}: The minimum frequency of light required to cause photoemission. \textbf{y-intercept (\(-\phi/e\))}: Proportional to the work function, which is the minimum energy required to liberate an electron from the metal.

Step 1: Understanding the Concept:

This question asks about the relationship between a photon's wavelength (\(\lambda\)), its wave number (\(\bar{\nu}\) or \(k\)), and its frequency (\(f\) or \(\nu\)). These properties are all interrelated for an electromagnetic wave.


Step 2: Key Formula or Approach:

The key relationships are:

Wave Number (\(\bar{\nu}\)): It is defined as the reciprocal of the wavelength.
\[ \bar{\nu} = \frac{1}{\lambda} \]
Frequency (\(f\)): It is related to wavelength and the speed of light (\(c\)) by the wave equation.
\[ c = f \lambda \implies f = \frac{c}{\lambda} \]


Step 3: Detailed Explanation:

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

Change in Wave Number:
Let the initial and final wave numbers be \(\bar{\nu}_1\) and \(\bar{\nu}_2\). \[ \bar{\nu}_1 = \frac{1}{\lambda_1} \] \[ \bar{\nu}_2 = \frac{1}{\lambda_2} = \frac{1}{2\lambda_1} = \frac{1}{2} \bar{\nu}_1 \]
So, the wave number becomes 1/2 of its original value.


Change in Frequency:
Let the initial and final frequencies be \(f_1\) and \(f_2\). \[ f_1 = \frac{c}{\lambda_1} \] \[ f_2 = \frac{c}{\lambda_2} = \frac{c}{2\lambda_1} = \frac{1}{2} \left(\frac{c}{\lambda_1}\right) = \frac{1}{2} f_1 \]
So, the frequency also becomes 1/2 of its original value.


Step 4: Final Answer:

When the wavelength is doubled, both the wave number and the frequency become half of their initial values. The respective factors are 1/2 and 1/2.
Quick Tip: Remember that for any wave, wavelength (\(\lambda\)) is inversely proportional to both frequency (\(f\)) and wave number (\(\bar{\nu}\)). So, if wavelength increases by a factor of X, both frequency and wave number will decrease by a factor of X (i.e., become 1/X times).

Step 1: Understanding the Concept:

This problem uses the de Broglie relation, which connects the momentum of a particle (or photon) to its wavelength. For a photon, this relationship is a fundamental aspect of its wave-particle duality. The phrase "Ignoring relativistic effects" is a bit misleading here since photons are inherently relativistic, but the de Broglie formula \(p=h/\lambda\) is the correct one to use.


Step 2: Key Formula or Approach:

The momentum (\(p\)) of a photon is related to its wavelength (\(\lambda\)) by the de Broglie equation: \[ p = \frac{h}{\lambda} \]
where \(h\) is Planck's constant (\(h \approx 6.626 \times 10^{-34}\) J·s).
We can rearrange this formula to solve for the wavelength: \[ \lambda = \frac{h}{p} \]

Step 3: Detailed Explanation:

Given values:

Momentum of the photon, \(p = 5.0 \times 10^{-29}\) kg·m/s.
Planck's constant, \(h = 6.626 \times 10^{-34}\) J·s.

Now, substitute these values into the formula for wavelength: \[ \lambda = \frac{6.626 \times 10^{-34}}{5.0 \times 10^{-29}} \]
First, calculate the numerical part: \[ \frac{6.626}{5.0} \approx 1.3252 \]
Next, calculate the power of ten: \[ \frac{10^{-34}}{10^{-29}} = 10^{-34 - (-29)} = 10^{-5} \]
So, the wavelength is: \[ \lambda \approx 1.3252 \times 10^{-5} m \]
The options are given in micrometers (\(\mu\)m), where \(1 \mum = 10^{-6}\) m. We need to convert our answer to this unit. \[ \lambda = 1.3252 \times 10^{-5} m = (1.3252 \times 10) \times 10^{-6} m = 13.252 \times 10^{-6} m \] \[ \lambda \approx 13.3 \mum \]

Step 4: Final Answer:

The wavelength of the photon is approximately 13.3 \(\mu\)m. This corresponds to option (D).
Quick Tip: When performing calculations with scientific notation, separate the numerical parts and the powers of ten. This reduces the chance of making errors. Also, be mindful of the units in the final options and perform the necessary conversions.

Step 1: Understanding the Concept:

For a parallel plate capacitor, the electric field between the plates is assumed to be uniform. Its magnitude depends on the potential difference across the plates and the distance separating them.


Step 2: Key Formula or Approach:

The magnitude of the uniform electric field (\(E\)) between the plates of a parallel plate capacitor is related to the potential difference (\(V\)) and the separation distance (\(d\)) by the formula: \[ E = \frac{V}{d} \]

Step 3: Detailed Explanation:

Let the initial conditions be \(E_1, V_1,\) and \(d_1\).
We are given that the initial electric field is \(E_1 = E\).
So, \(E = \frac{V_1}{d_1}\).


Now, the conditions are changed:

The separation is doubled: \(d_2 = 2d_1\).
The potential difference is reduced to half: \(V_2 = \frac{1}{2}V_1\).

Let the new electric field be \(E_2\). Using the formula: \[ E_2 = \frac{V_2}{d_2} \]
Substitute the new values of V and d in terms of the initial values: \[ E_2 = \frac{\frac{1}{2}V_1}{2d_1} = \frac{1}{4} \frac{V_1}{d_1} \]
Since we know that \(E = \frac{V_1}{d_1}\), we can substitute this into the expression for \(E_2\): \[ E_2 = \frac{1}{4} E \]
The new value of the electric field is E/4.


Step 4: Final Answer:

The new value of the electric field between the plates will be E/4.
Quick Tip: This type of problem tests proportional reasoning. Since \(E \propto V\) and \(E \propto 1/d\), you can find the new value by multiplying the original value by the factors of change. Here, the factor for V is 1/2 and the factor for d is 2 (so the factor for 1/d is 1/2). The total change is \(E_{new} = E_{old} \times (\frac{1}{2}) \times (\frac{1}{2}) = E/4\).

Step 1: Understanding the Concept:

The problem requires finding the relationship between the potential difference (\(V\)) and the plate separation (\(d\)) under the condition that the electric field (\(E\)) between the plates remains constant.


Step 2: Key Formula or Approach:

The relationship between electric field, potential difference, and separation for a parallel plate capacitor is: \[ E = \frac{V}{d} \]

Step 3: Detailed Explanation:

We are given the condition that the electric field \(E\) must be constant. Let's represent this constant value by \(k\). \[ k = \frac{V}{d} \]
To find how \(V\) depends on \(d\), we can rearrange the equation: \[ V = k \times d \]
This equation, \(V = kd\), is in the form of a straight line, \(y = mx\), where:

\(y = V\) (the dependent variable)
\(x = d\) (the independent variable)
\(m = k = E\) (the slope, which is a positive constant)

This means that the graph of potential difference \(V\) versus separation \(d\) is a straight line passing through the origin (since if \(d=0\), \(V=0\)) with a positive slope equal to the magnitude of the constant electric field.

Let's examine the given graphs:

(A) shows V decreasing as d increases (inverse relationship).
(B) shows V increasing linearly with d, starting from the origin. This matches our derived relationship \(V = Ed\).
(C) shows a non-linear relationship (V \(\propto \sqrt{d}\) type curve).
(D) shows a non-linear relationship (V \(\propto d^2\) type curve).


Step 4: Final Answer:

The correct graph is (B), which shows a direct linear proportionality between V and d, starting from the origin.
Quick Tip: Whenever asked to find the relationship between two variables from a formula, try to rearrange the formula into the form \(y=f(x)\) and identify the type of function (linear, quadratic, inverse, etc.). This will directly tell you the shape of the graph.

Step 1: Understanding the Concept:

The question asks about the nature of the electric field between the plates of an ideal parallel plate capacitor. For such a capacitor, the electric field in the region between the plates is considered to be uniform, and it is zero outside the plates (ignoring fringing effects).


Step 2: Key Property of Parallel Plate Capacitors:

The electric field between two large, closely spaced, parallel conducting plates with equal and opposite charges is uniform and constant in magnitude and direction, except for small regions near the edges (this is called the "fringing effect"). The field lines are parallel and equally spaced, pointing from the positive plate to the negative plate.


Step 3: Detailed Explanation:

In the given figure:

Plate Q is at a positive potential and Plate P is at a negative potential. The electric field will be directed from Q to P.
The line MN is drawn perpendicular to the plates, representing the path along which we are observing the electric field strength E.
As one moves from M (on plate P) to N (on plate Q), one is moving through the region between the plates.

Since the electric field between the plates of a parallel plate capacitor is uniform, its magnitude \(E\) should be constant at all points along the line MN. The graph of E versus position along MN should therefore be a horizontal line, indicating a constant non-zero value.

Let's examine the graphs:

(A) and (B) show the electric field varying (increasing or decreasing) linearly, which is incorrect.
(C) shows the electric field E having a constant, non-zero value at all points between M and N. This correctly represents a uniform electric field.
(D) shows the electric field varying in a non-linear way, which is incorrect.


Step 4: Final Answer:

The correct graph is (C), which depicts a constant electric field magnitude E along the line MN between the capacitor plates.
Quick Tip: For an ideal parallel plate capacitor, remember these key points: Electric field \textbf{inside} is \textbf{uniform} and \textbf{constant} (\(E = \sigma/\epsilon_0\)). Electric field \textbf{outside} is (ideally) \textbf{zero}. The potential varies linearly with distance (\(V(x) = Ex\)), but the field E itself is constant.

Step 1: Understanding the Concept:

This problem involves calculating the total potential difference across a system of multiple parallel plates by adding the potential differences across each section. The potential difference across a region with a uniform electric field is the product of the field strength and the distance.


Step 2: Key Formula or Approach:

The potential difference (\(\Delta V\)) between two points in a uniform electric field \(E\) is given by \(\Delta V = E \cdot d\), where \(d\) is the distance between the points along the direction of the field. The total potential difference across multiple regions is the algebraic sum of the potential differences across each region. \[ V_{total} = \Delta V_1 + \Delta V_2 + \ldots \]

Step 3: Detailed Explanation:

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

Plate 1 is at the bottom.
Plate 2 is in the middle.
Plate 3 is at the top.

The separation between adjacent plates is \(d\).

The electric field between the first pair (Plate 1 and Plate 2) is \(E_1\).
The electric field between the second pair (Plate 2 and Plate 3) is \(E_2\).

We need to find the potential difference between the third and the first plate, which is \(V_3 - V_1\).
We can write this as the sum of the potential differences across the two gaps: \[ V_3 - V_1 = (V_3 - V_2) + (V_2 - V_1) \]
Now, let's find the potential difference for each gap. Assuming the electric fields are directed upwards (from 1 to 3).
The potential difference between Plate 2 and Plate 1 is: \[ V_2 - V_1 = E_1 \cdot d \]
(Assuming potential increases in the direction of E, this seems reversed. Let's use \(V = - \int E \cdot dl\). Let's assume Plate 1 is at \(z=0\), Plate 2 at \(z=d\), and Plate 3 at \(z=2d\)).
Then \(V_2 - V_1 = P.D. across first gap\). The magnitude is \(|V_2-V_1| = E_1 d\).
The potential difference between Plate 3 and Plate 2 is: \[ V_3 - V_2 = P.D. across second gap\). The magnitude is \(|V_3-V_2| = E_2 d\). The total potential difference between Plate 3 and Plate 1 is the sum of the magnitudes of the potential differences across each section, assuming the fields are in the same direction. \[ |V_3 - V_1| = |V_3 - V_2| + |V_2 - V_1| \] \[ |V_3 - V_1| = E_2 d + E_1 d = (E_1 + E_2)d \]
The question asks for the potential difference, which implies a scalar magnitude, and the options are all scalar magnitudes. So we assume the fields \(E_1\) and \(E_2\) are magnitudes and are in the same direction. Therefore, the total potential difference is the sum of the individual potential differences.


Step 4: Final Answer:

The potential difference between the third and the first plate is the sum of the potential differences across the two gaps, which is \((E_1+E_2)d\).
Quick Tip: For potential differences across multiple capacitors in series, the total voltage is the sum of the individual voltages: \(V_{total} = V_1 + V_2 + \dots\). This problem is a direct application of that principle, where each gap between plates acts like a capacitor.

Step 1: Understanding the Concept:

When a dielectric material is inserted between the plates of a capacitor, it increases the capacitor's ability to store charge for a given voltage. This results in an increase in its capacitance. The factor by which the capacitance increases is the dielectric constant, K.


Step 2: Key Formula or Approach:

The capacitance of a parallel plate capacitor with a vacuum (or air, approximately) between the plates is given by: \[ C_{air} = \frac{\epsilon_0 A}{d} \]
where \(\epsilon_0\) is the permittivity of free space, A is the area of the plates, and d is the separation.

When a dielectric material with dielectric constant K is completely filled between the plates, the permittivity of the medium becomes \(\epsilon = K \epsilon_0\). The new capacitance is: \[ C_{dielectric} = \frac{\epsilon A}{d} = \frac{K \epsilon_0 A}{d} \]

Step 3: Detailed Explanation:

We are given that the initial capacitance (presumably with air or vacuum as the medium) is C. \[ C = C_{air} = \frac{\epsilon_0 A}{d} \]
A dielectric material of constant K is then filled in the capacitor. The new capacitance, let's call it \(C'\), is: \[ C' = C_{dielectric} = \frac{K \epsilon_0 A}{d} \]
We can see the relationship between \(C'\) and \(C\) by substituting the expression for C: \[ C' = K \left( \frac{\epsilon_0 A}{d} \right) = K \cdot C \]
The new value of the capacitance becomes K times the original capacitance.


Step 4: Final Answer:

The new value of its capacitance becomes KC (or CK).
Quick Tip: The dielectric constant K is always greater than or equal to 1 (K=1 for vacuum). Therefore, introducing a dielectric material always increases (or keeps the same for vacuum) the capacitance of a capacitor. This is a fundamental concept in capacitance.

Step 1: Understanding the Concept:

The path of light through a prism is governed by the relation between the angles of incidence (\( i \)), emergence (\( e \)), prism angle (\( A \)), and deviation (\( \delta \)). Snell's law relates these angles to the refractive index of the material.


Step 2: Key Formula or Approach:

1. Prism relation: \( \delta = i + e - A \)

2. Internal angles: \( A = r_1 + r_2 \)

3. Snell's Law: \( \mu = \frac{\sin i}{\sin r_1} = \frac{\sin e}{\sin r_2} \)


Step 3: Detailed Explanation:

Given: \( i = 45^\circ \), \( \delta = 15^\circ \), \( A = 30^\circ \).

Substitute values into the prism relation to find the angle of emergence \( e \):
\[ 15^\circ = 45^\circ + e - 30^\circ \]
\[ 15^\circ = 15^\circ + e \implies e = 0^\circ \]

Since the angle of emergence \( e = 0^\circ \) with the normal, the emergent ray is normal to the face AC. This satisfies the first part of the question.

At the face AC, since \( e = 0^\circ \), by Snell's law, the internal angle \( r_2 = 0^\circ \).

Using the relation \( A = r_1 + r_2 \):
\[ 30^\circ = r_1 + 0^\circ \implies r_1 = 30^\circ \]

Now, apply Snell's Law at the first face AB:
\[ \mu = \frac{\sin i}{\sin r_1} = \frac{\sin 45^\circ}{\sin 30^\circ} \]
\[ \mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414 \]


Step 4: Final Answer:

The emergent ray is normal to face AC because \( e = 0^\circ \). The refractive index of the prism material is \( \sqrt{2} \).
Quick Tip: "Normal emergence" always means the angle of emergence \( e = 0^\circ \) and the internal angle of refraction at that second surface \( r_2 = 0^\circ \). In such cases, the prism angle \( A \) equals the angle of refraction \( r_1 \).

Step 1: Understanding the Concept:

In Young's Double Slit Experiment (YDSE), the position of bright fringes from the central maximum is determined by the wavelength of light and the geometric setup of the slits and screen.


Step 2: Key Formula or Approach:

The distance of the \( n^{th} \) bright fringe is given by: \[ y_n = \frac{n \lambda D}{d} \]


Step 3: Detailed Explanation:

Given data:

Wavelength \( \lambda = 600 nm = 600 \times 10^{-9} m \)

Distance to screen \( D = 1.0 m \)

Slit separation \( d = 1.0 mm = 1.0 \times 10^{-3} m \)

Order of the fringe \( n = 3 \)

Substituting the values:
\[ y_3 = \frac{3 \times 600 \times 10^{-9} \times 1.0}{1.0 \times 10^{-3}} \]
\[ y_3 = \frac{1800 \times 10^{-9}}{10^{-3}} = 1800 \times 10^{-6} m \]
\[ y_3 = 1.8 \times 10^{-3} m = 1.8 mm \]


Step 4: Final Answer:

The distance of the third bright fringe from the central maximum is 1.8 mm.
Quick Tip: Always convert all physical quantities to SI units (meters) before calculation. For example, \( 1 nm = 10^{-9} m \) and \( 1 mm = 10^{-3} m \).

Step 1: Understanding the Concept:

When bright fringes of two different wavelengths coincide, their positions from the central maximum must be equal. This occurs when the path difference for both is an integral multiple of their respective wavelengths.


Step 2: Key Formula or Approach:

Let the \( n_1^{th} \) bright fringe of wavelength \( \lambda_1 \) coincide with the \( n_2^{th} \) bright fringe of wavelength \( \lambda_2 \).
\[ n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d} \implies n_1 \lambda_1 = n_2 \lambda_2 \]


Step 3: Detailed Explanation:

Given: \( \lambda_1 = 600 nm \), \( \lambda_2 = 480 nm \).

Setting up the condition:
\[ n_1 \times 600 = n_2 \times 480 \]
\[ \frac{n_1}{n_2} = \frac{480}{600} = \frac{4}{5} \]

The smallest integers satisfying this ratio are \( n_1 = 4 \) and \( n_2 = 5 \).

This means the \( 4^{th} \) bright fringe of 600 nm coincides with the \( 5^{th} \) bright fringe of 480 nm.

The least distance \( y \) is:
\[ y = \frac{n_1 \lambda_1 D}{d} = \frac{4 \times 600 \times 10^{-9} \times 1.0}{1.0 \times 10^{-3}} \]
\[ y = 2400 \times 10^{-6} m = 2.4 \times 10^{-3} m = 2.4 mm \]


Step 4: Final Answer:

The least distance from the central maximum where the bright fringes coincide is 2.4 mm.
Quick Tip: To find where fringes coincide, simply calculate the ratio of the wavelengths. The least distance corresponds to the smallest integers that represent that ratio. \( n_1/n_2 = \lambda_2/\lambda_1 \).

Step 1: Intensity Variation Graph:

In a single-slit diffraction pattern, the intensity of light is not uniform. There is a central bright fringe (central maximum) which is the brightest and widest. On either side of the central maximum, there are alternating dark fringes (minima) and secondary bright fringes (secondary maxima) of decreasing intensity.


Graph:

% A simple description of the plot is provided here as drawing is not possible.
(A graph with Intensity on the y-axis and angle \(\theta\) on the x-axis, symmetric about \(\theta=0\). At \(\theta=0\), there is a large central peak (intensity \(I_0\)). The intensity drops to zero at \(\pm\lambda/a\), \(\pm 2\lambda/a\), etc. Between these zeros, there are much smaller secondary peaks whose intensity rapidly decreases as \(\theta\) increases.)

The intensity distribution is given by the function: \[ I(\theta) = I_0 \left( \frac{\sin \beta}{\beta} \right)^2 \quad where \quad \beta = \frac{\pi a \sin \theta}{\lambda} \]
Here, \(I_0\) is the intensity of the central maximum, \(a\) is the slit width, \(\lambda\) is the wavelength of light, and \(\theta\) is the angle of diffraction.


Step 2: Condition for Zero Intensity (Minima):

The locations of zero intensity (dark fringes or minima) occur when the numerator of the intensity expression is zero, but the denominator is not. This happens when \(\sin \beta = 0\), but \(\beta \neq 0\).
The condition \(\sin \beta = 0\) is satisfied when: \[ \beta = \pm n\pi, \quad where n = 1, 2, 3, \ldots \]
(The case n=0 is excluded because it corresponds to \(\beta=0\), which is the location of the central maximum).

Substituting the expression for \(\beta\): \[ \frac{\pi a \sin \theta}{\lambda} = n\pi \]
Cancelling \(\pi\) from both sides, we get the expression for the angle \(\theta_n\) corresponding to the n-th minimum: \[ a \sin \theta_n = n\lambda \]
For small angles, \(\sin \theta_n \approx \theta_n\), so the angular positions are approximately: \[ \theta_n \approx \frac{n\lambda}{a} \] Quick Tip: The key difference between interference and diffraction patterns is the intensity. In a double-slit interference pattern, all bright fringes are (ideally) of the same intensity. In a single-slit diffraction pattern, the central maximum is overwhelmingly bright, and the secondary maxima are very faint and decrease in intensity quickly.

The phenomenon of diffraction is common to all waves, but its manifestation differs significantly between light and sound waves primarily due to their vast difference in wavelength.

The main differences are:

Wavelength Scale: The most crucial difference lies in the wavelength.

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

Observability in Daily Life: Diffraction effects are most prominent when the size of the obstacle or aperture is comparable to the wavelength of the wave (\(\lambda \approx a\)).

Sound Waves: Since the wavelength of sound is comparable to the size of everyday objects like doors, corners, and people, diffraction of sound is a common experience. We can easily hear someone talking from around a corner because the sound waves bend around it.
Light Waves: The wavelength of light is minuscule compared to everyday objects. Therefore, to observe the diffraction of light, we need obstacles or apertures that are extremely small (on the order of micrometers), such as a very narrow slit or the edge of a razor blade. In daily life, light appears to travel in straight lines, and its diffraction is not easily noticeable.

Experimental Setup: To study diffraction:

Sound Waves: No special apparatus is needed; diffraction can be observed with large-scale objects.
Light Waves: Requires a carefully designed experiment with components like a monochromatic light source and precisely engineered narrow slits or gratings. Quick Tip: The core idea is scale. Sound waves bend around buildings; light waves require something as fine as a hair to show noticeable bending. The condition for significant diffraction is \(\lambda \ge a\).

Step 1: Understanding the Concept:

When two conductors are connected by a conducting wire, they form a single equipotential body. Charge will redistribute itself until the electric potential is the same everywhere on the surface of both conductors. Since the inner sphere is enclosed by the outer shell, all charge from the inner sphere will flow to the outer surface of the outer shell.


Step 2: Key Formula or Approach:

1. The potential of a single conducting sphere of radius \(r_0\) and charge \(q\) is \(V = \frac{1}{4\pi\epsilon_0} \frac{q}{r_0}\).
2. When conductors are connected, their final potentials are equal: \(V_{A, final} = V_{B, final} = V_f\).
3. The entire charge on an inner conductor moves to the outer surface of the enclosing conductor when they are connected.


Step 3: Detailed Explanation:

Initial State:
The small conducting sphere A has radius \(r\) and is charged to a potential \(V\). The potential of sphere A is solely due to its own charge, \(q_A\), since it is inside the shell B (which we assume is initially uncharged and does not contribute to the potential of A relative to infinity in this simple setup). \[ V = \frac{1}{4\pi\epsilon_0} \frac{q_A}{r} \]
From this, we can find the initial charge on sphere A: \[ q_A = (4\pi\epsilon_0 r) V \]

Final State:
When sphere A and shell B are connected by a wire, they become a single conductor. The charge \(q_A\) will redistribute. Since charges on a conductor reside on its outermost surface to be in electrostatic equilibrium, the entire charge \(q_A\) will move from sphere A to the outer surface of shell B.

Final charge on sphere A: \(q'_{A} = 0\).
Final charge on shell B: \(q'_{B} = q_A\).

The entire system is now at a single, uniform final potential, \(V_f\). This potential is determined by the charge \(q_A\) residing on the outer shell of radius \(R\). The potential of a spherical shell with charge \(q_A\) on its surface is constant at all points on and inside the shell. \[ V_f = \frac{1}{4\pi\epsilon_0} \frac{q_A}{R} \]
Now, substitute the expression for \(q_A\) from the initial state: \[ V_f = \frac{1}{4\pi\epsilon_0 R} \left( (4\pi\epsilon_0 r) V \right) \]
The term \(4\pi\epsilon_0\) cancels out: \[ V_f = \frac{rV}{R} \]

Step 4: Final Answer:

After connecting, the charge flows to the outer shell, and the whole system comes to a new common potential \(V_f\). Both sphere A and shell B will be at the final potential \(V_f = \frac{Vr}{R}\).
Quick Tip: This is the principle behind the Van de Graaff generator. Charge is delivered to an inner sphere and then transferred to the outer shell upon contact. Since all charge moves to the outer shell, the inner sphere can be "recharged" repeatedly, accumulating a very large charge and potential on the outer shell.

Part 1: Characteristics of Equipotential Surfaces

An equipotential surface is a surface on which the electric potential is the same at every point. Two key characteristics are:

No work is done in moving a test charge on an equipotential surface. The work done is \(W = q \Delta V\). Since \(\Delta V = 0\) for any two points on the surface, the work done \(W\) is zero.
The electric field is always perpendicular to the equipotential surface at every point. If the field had a component along the surface, it would exert a force and do work on a charge moving along the surface, which contradicts the definition.


Part 2: Calculating the Potential

Step 1: Understanding the Concept:

The potential difference between two points in a uniform electric field is related by the dot product of the electric field vector and the displacement vector between the two points.


Step 2: Key Formula or Approach:

The relationship is given by \(V_B - V_A = - \vec{E} \cdot \vec{r}_{AB}\), where \(\vec{r}_{AB}\) is the displacement vector from point A to point B.


Step 3: Detailed Explanation:

Given:

Uniform electric field: \(\vec{E} = 50 \hat{i}\) N/C (since it's along the +x axis).
Point A (origin): \(\vec{r}_A = 0\hat{i} + 0\hat{j}\).
Potential at A: \(V_A = 220\) V.
Point B: \(\vec{r}_B = 4\hat{i} + 3\hat{j}\).

We need to find the potential at B, \(V_B\).

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

Step 4: Final Answer:

The potential at the point (4m, 3m) is 20 V.
Quick Tip: For a uniform electric field \(\vec{E}\), the equipotential surfaces are planes perpendicular to \(\vec{E}\). In this case, \(\vec{E}\) is along the x-axis, so the equipotential surfaces are planes of constant x (the yz-planes). The potential only changes as you move along the x-direction. The displacement along y (3m) does not contribute to the change in potential. The change is simply \(-E \Delta x = -50 \times 4 = -200\) V.

The key difference between an open surface and a closed surface lies in whether they enclose a volume and have a boundary.


Open Surface:

An open surface is a surface that does not enclose a finite volume.
It has a distinct edge or boundary.
Examples include a flat disc, a sheet of paper, a hemisphere, or a cylindrical surface without its top and bottom caps.
The concept of "inside" and "outside" is not uniquely defined for an open surface in three-dimensional space.

Closed Surface:

A closed surface is a surface that completely encloses a finite volume, separating the space into a distinct "inside" region and an "outside" region.
It has no edges or boundaries. You cannot get from the inside to the outside without crossing the surface.
Examples include the surface of a sphere, a cube, a torus (donut shape), or a cylinder with its top and bottom caps included.
Closed surfaces are fundamental to Gauss's Law in electromagnetism, which relates the electric flux through a closed surface to the net charge enclosed within it. Quick Tip: A simple analogy: a deflated balloon is an open surface (with an opening). An inflated, tied-off balloon is a closed surface. It separates the air inside from the air outside.

1. Elementary Surface Vector \(d\vec{S}\):

For a small, elementary area patch \(dS\) on a spherical surface, the area vector \(d\vec{S}\) is a vector that has:

Magnitude: Equal to the area \(dS\).
Direction: Perpendicular (normal) to the surface at that point and directed radially outwards.


% A simple description of the plot is provided here as drawing is not possible.
(A diagram showing a sphere. A small patch of area dS is highlighted on its surface. A vector dS is drawn originating from this patch, pointing radially outwards, perpendicular to the tangent plane at that point.)


2. Electric Flux (\(\phi_E\)):

Electric flux is a measure of the flow of the electric field through a given surface. It quantifies the number of electric field lines crossing the surface. For a uniform electric field \(\vec{E}\) passing through a plane area \(\vec{A}\), the flux is \(\phi_E = \vec{E} \cdot \vec{A} = EA \cos\theta\). For a general surface and a non-uniform field, it is defined by the surface integral: \[ \phi_E = \int_S \vec{E} \cdot d\vec{S} \]

3. Significance of a Gaussian Surface:

A Gaussian surface is an imaginary closed surface used in the context of Gauss's Law (\(\phi_E = q_{enc}/\epsilon_0\)). Its significance lies in being a mathematical tool that simplifies the calculation of electric fields for symmetric charge distributions. By choosing a Gaussian surface that has the same symmetry as the charge distribution (e.g., a sphere for a point charge, a cylinder for a line charge), the integral \(\int \vec{E} \cdot d\vec{S}\) becomes very easy to evaluate, allowing for a straightforward calculation of \(\vec{E}\).


4. Why a Charge Outside a Gaussian Surface Contributes No Net Flux:

A charge \(q_{out}\) located outside a closed Gaussian surface produces an electric field that passes through the surface. However, its net contribution to the flux is zero. This is because:

The electric field lines from \(q_{out}\) that enter the closed surface at one point must necessarily exit the surface at another point.
At the point of entry, the angle between \(\vec{E}\) and the outward normal \(d\vec{S}\) is obtuse (\(> 90^\circ\)), so the dot product \(\vec{E} \cdot d\vec{S}\) is negative. This is considered an inward (negative) flux.
At the point of exit, the angle between \(\vec{E}\) and the outward normal \(d\vec{S}\) is acute (\(< 90^\circ\)), so the dot product \(\vec{E} \cdot d\vec{S}\) is positive. This is considered an outward (positive) flux.
For any external charge, the total number of field lines entering the surface is exactly equal to the total number of field lines leaving it. Therefore, the total negative flux perfectly cancels the total positive flux.
The net flux through the closed surface due to the external charge is zero. Quick Tip: Gauss's Law is a powerful statement about what's inside a closed surface. It tells us that to find the net flux, we can completely ignore all charges outside the "imaginary box" we've drawn.

Step 1: Understanding the Concept:

This problem requires the application of Gauss's Law for electric flux. Gauss's Law states that the total electric flux (\(\phi_E\)) through any closed surface (a Gaussian surface) is directly proportional to the total net electric charge (\(q_{enclosed}\)) inside that surface.


Step 2: Key Formula or Approach:

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


Step 3: Detailed Explanation:

Flux through surface S\(_1\) (\(\phi_1\)):
The surface S\(_1\) encloses the charges \(q_1, q_2,\) 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 electric flux through S\(_1\) is: \[ \phi_1 = \frac{q_{enc, 1}}{\epsilon_0} = \frac{4 \muC}{\epsilon_0} \]

Flux through surface S\(_2\) (\(\phi_2\)):
The surface S\(_2\) encloses everything that S\(_1\) encloses, plus the additional charge Q which is placed between the two shells. Therefore, 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 electric flux through S\(_2\) is: \[ \phi_2 = \frac{q_{enc, 2}}{\epsilon_0} = \frac{4 \muC + Q}{\epsilon_0} \]

Using the given relation:
The problem states 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 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 to nested surfaces, remember that the outer surface encloses all the charge that the inner surface does, plus any charge in the region between them.

1. Source of Force:

The force on a current-carrying conductor in a magnetic field arises from the collective Lorentz force acting on the individual moving charge carriers (electrons) within the conductor. A current is simply the flow of these charges. When a charge \(q\) moves with a velocity \(\vec{v}\) in a magnetic field \(\vec{B}\), it experiences a magnetic force \(\vec{F} = q(\vec{v} \times \vec{B})\). The total force on the conductor is the vector sum of these individual forces on all the charge carriers constituting the current.


2. Expression for Force Between Two Parallel Conductors:

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

Field produced by wire 1: Wire 1 produces a magnetic field (\(\vec{B}_1\)) at the location of wire 2. The magnitude of this field is given by Ampere's law:
\[ B_1 = \frac{\mu_0 I_1}{2\pi d} \]
Using the right-hand thumb rule, the direction of \(\vec{B}_1\) at wire 2 is perpendicular to wire 2 (e.g., into the page).
Force on wire 2: Now, wire 2, carrying current \(I_2\), is in the magnetic field \(\vec{B}_1\). The magnetic force (\(\vec{F}_2\)) on a length \(L\) of wire 2 is given by \(\vec{F} = I(\vec{L} \times \vec{B})\). Since \(\vec{L}\) and \(\vec{B}_1\) are perpendicular, the magnitude is:
\[ F_2 = I_2 L B_1 \]
Substitute 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:
\[ f = \frac{F_2}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]

By Newton's third law, wire 1 experiences an equal and opposite force. Using Fleming's left-hand rule, it can be shown that if the currents are in the same direction, the wires attract each other. If they are in opposite directions, they repel.


3. Definition of 'Ampere':

The SI unit of current, the ampere, is defined using the expression derived above. Consider the force per unit length between the two wires: \(f = \frac{\mu_0 I_1 I_2}{2\pi d}\).
Let's set the conditions as:

The currents are equal: \(I_1 = I_2 = 1\) A.
The separation is \(d = 1\) m.
The value of \(\mu_0\) is defined as \(4\pi \times 10^{-7}\) T·m/A.

Under these conditions, the force per unit length is: \[ f = \frac{(4\pi \times 10^{-7}) (1)(1)}{2\pi (1)} = 2 \times 10^{-7} N/m \]
Definition: One ampere is defined as that constant current which, if maintained in two straight, parallel conductors of infinite length and negligible circular cross-section, placed one metre apart in vacuum, would produce a force between them of \(2 \times 10^{-7}\) newtons per metre of length.
Quick Tip: A simple way to remember the interaction: "Parallel currents attract, anti-parallel currents repel." This can be easily verified using the combination of the right-hand thumb rule (for the field) and Fleming's left-hand rule (for the force).

Step 1: Understanding the Concept:

Work is done by a force when it causes a displacement in its own direction. We need to analyze the direction of the magnetic force relative to the direction of motion (displacement) of the charge.


Step 2: Key Formula or Approach:

1. The magnetic force (Lorentz force) on a charge \(q\) moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is:
\[ \vec{F}_B = q(\vec{v} \times \vec{B}) \]
2. The work done (\(dW\)) by a force \(\vec{F}\) over an infinitesimal displacement \(d\vec{s}\) is:
\[ dW = \vec{F} \cdot d\vec{s} \]
3. The instantaneous power (\(P\)) delivered by the force is \(P = \vec{F} \cdot \vec{v}\).


Step 3: Detailed Explanation:

From the definition of the vector cross product, the magnetic force vector \(\vec{F}_B\) is always perpendicular to the plane containing the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\).
This means that \(\vec{F}_B\) is always perpendicular to the velocity \(\vec{v}\) of the charged particle. \[ \vec{F}_B \perp \vec{v} \]
The infinitesimal displacement of the particle is always in the direction of its instantaneous velocity, \(d\vec{s} = \vec{v} dt\).

Therefore, the magnetic force \(\vec{F}_B\) is always perpendicular to the displacement \(d\vec{s}\).

Let's calculate the work done: \[ dW = \vec{F}_B \cdot d\vec{s} = |\vec{F}_B| |d\vec{s}| \cos(90^\circ) \]
Since \(\cos(90^\circ) = 0\), the work done over any infinitesimal displacement is zero. \[ dW = 0 \]
The total work done over any path is the integral of \(dW\), which will also be zero.
Alternatively, the power delivered by the magnetic force is: \[ P = \frac{dW}{dt} = \vec{F}_B \cdot \vec{v} = 0 \]
Since the power delivered is always zero, no work is done by the magnetic force. This implies that the magnetic force cannot change the kinetic energy or the speed of the charged particle; it can only change its direction of motion.


Step 4: Final Answer:

The work done by the magnetic force on the charge is always zero.
Quick Tip: A key consequence of the magnetic force doing no work is that a particle's kinetic energy is conserved when moving through a magnetic field. Any change in the particle's energy must be due to an electric field.

Step 1: Understanding the Concept:

A helical path is a three-dimensional spiral. For a charged particle in a uniform magnetic field to follow such a path, its motion must be a combination of circular motion in one plane and linear motion perpendicular to that plane.


Step 2: Necessary Conditions:

The necessary condition for a helical trajectory is that the initial velocity vector \(\vec{v}\) of the charged particle must be at an angle \(\theta\) to the uniform magnetic field \(\vec{B}\), where \(\theta\) is not \(0^\circ, 90^\circ,\) or \(180^\circ\).


Step 3: Explanation of the Motion:

When this condition is met, we can resolve the velocity vector \(\vec{v}\) into two components:

A component parallel to the magnetic field (\(v_{\parallel}\)):
\[ v_{\parallel} = v \cos\theta \]
The magnetic force on this component of velocity 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 motion is the linear drift that defines the axis of the helix.
A component perpendicular to the magnetic field (\(v_{\perp}\)):
\[ v_{\perp} = v \sin\theta \]
This component of velocity experiences a magnetic force of magnitude \(F = q(v_{\perp} B \sin(90^\circ)) = q v_{\perp} B\). This force is always perpendicular to both \(v_{\perp}\) and \(\vec{B}\). It acts as a centripetal force, causing the particle to execute a uniform circular motion in the plane perpendicular to the magnetic field.

Resultant Motion:
The trajectory of the particle is the superposition of these two motions:

A circular motion in a plane perpendicular to \(\vec{B}\).
A constant linear motion along the direction of \(\vec{B}\).

The combination of moving forward while simultaneously going in a circle results in a helical path. The radius of the helix is determined by \(v_{\perp}\), and the pitch (the distance between successive turns) is determined by \(v_{\parallel}\) and the time period of the circular motion.
Quick Tip: Think about the special cases: If \(\theta=0^\circ\) or \(180^\circ\), \(v_{\perp}=0\). The particle moves in a straight line parallel to \(\vec{B}\) (no force). If \(\theta=90^\circ\), \(v_{\parallel}=0\). The particle moves in a perfect circle. For any other angle, the path is a helix.

A current-carrying loop is considered a magnetic dipole because its magnetic field and its behavior in an external magnetic field are analogous to that of a short bar magnet, which is a classic example of a magnetic dipole. The explanation rests on two main points:


1. The Magnetic Field Produced by the Loop:

A current flowing through a loop generates a magnetic field. Using the right-hand curl rule (curling the fingers of your right hand in the direction of the current), your thumb points in the direction of the magnetic field inside the loop.
This means that one face of the loop acts as a magnetic North pole (where field lines emerge) and the opposite face acts as a magnetic South pole (where field lines enter).
The pattern of the magnetic field lines produced by the current loop, especially at distances far from the loop, is identical to the magnetic field pattern of a short bar magnet or an electric dipole. This structural similarity in the field is a key reason for the analogy.


2. The Behavior of the Loop in an External Magnetic Field:

When a current loop is placed in a uniform external magnetic field, it does not experience a net force, but it does experience a net torque (unless its magnetic moment is aligned with the field).
This torque tends to rotate the loop and align its axis with the direction of the external magnetic field. The torque is given by \(\vec{\tau} = \vec{M} \times \vec{B}\), where \(\vec{M}\) is the magnetic dipole moment of the loop.
This behavior is exactly the same as that of a bar magnet (a magnetic dipole) placed in a magnetic field. The bar magnet also experiences a torque that tries to align its North-South axis with the external field.

Because a current loop both creates a dipole-like magnetic field and responds to an external field by experiencing a torque just like a bar magnet, it is valid and useful to consider it as a magnetic dipole.
Quick Tip: The concept of a magnetic dipole moment, \(\vec{M}\), is what unifies the description of current loops, bar magnets, and even elementary particles like electrons. It's the fundamental quantity used to describe the magnetic properties of an object.

1. Relation for Magnetic Dipole Moment:

The magnetic dipole moment is a vector quantity that represents the strength and orientation of a magnetic dipole. For a current-carrying coil, its magnitude is directly proportional to the current, the area enclosed by the coil, and the number of turns in the coil.


Let \(I\) be the steady current flowing through the coil.
Let \(A\) be the area of the planar loop of the coil.
Let \(N\) be the number of turns in the coil.

The magnitude of the magnetic dipole moment (\(M\)) is given by the relation: \[ M = NIA \]
The SI unit for magnetic dipole moment is ampere-meter squared (A·m²).


2. Direction of \(\vec{M}\):

The magnetic dipole moment \(\vec{M}\) is a vector. Its direction is perpendicular to the plane of the current coil and is determined by the Right-Hand Curl Rule.

Rule: If you curl the fingers of your right hand in the direction of the current flowing in the coil, your extended thumb will point in the direction of the magnetic dipole moment vector \(\vec{M}\).

This direction is also the same as the direction of the magnetic field produced by the coil at its center. It corresponds to the direction from the South pole to the North pole of the equivalent bar magnet.

Therefore, the full vector relation can be written as: \[ \vec{M} = NIA\hat{n} \]
where \(\hat{n}\) is the unit vector normal to the plane of the coil, with its direction given by the right-hand rule.
Quick Tip: Don't confuse the direction of the magnetic moment \(\vec{M}\) with the direction of the magnetic field \(\vec{B}\) everywhere. The thumb rule gives the direction of \(\vec{M}\) and the direction of \(\vec{B}\) along the axis inside the loop. Outside the loop, the \(\vec{B}\) field lines loop around from the North to the South pole.

1. Net Force on the Coil:

In a uniform external magnetic field, the net force on a closed current-carrying coil is always zero. This is because the forces on opposite segments of the loop are equal in magnitude and opposite in direction, so their vector sum is zero. (Note: In a non-uniform field, the net force may be non-zero).


2. Orientation for Stable Equilibrium:

The coil experiences a torque given by \(\vec{\tau} = \vec{M} \times \vec{B}\), where \(\vec{M}\) is the magnetic dipole moment of the coil and \(\vec{B}\) is the external magnetic field. The potential energy of the coil in the field is \(U = - \vec{M} \cdot \vec{B} = -MB\cos\theta\).

For equilibrium, the net torque must be zero. This occurs when \(\sin\theta = 0\), i.e., when \(\theta = 0^\circ\) or \(\theta = 180^\circ\).

Stable Equilibrium: Occurs when the potential energy \(U\) is at a minimum. This happens when \(\cos\theta\) is maximum, i.e., \(\cos\theta = 1\), which corresponds to \(\theta = 0^\circ\). In this orientation, the magnetic dipole moment vector \(\vec{M}\) is parallel to the external magnetic field vector \(\vec{B}\).
Unstable Equilibrium: Occurs when the potential energy is maximum, which is at \(\theta = 180^\circ\) (\(\vec{M}\) is anti-parallel to \(\vec{B}\)).

So, the orientation for stable equilibrium is when the plane of the coil is perpendicular to the magnetic field, such that its magnetic moment vector \(\vec{M}\) is aligned with \(\vec{B}\).


3. Flux in Stable Equilibrium Orientation:

In the stable equilibrium orientation, \(\vec{M}\) is parallel to the external field \(\vec{B}_{ext}\).

The magnetic field produced by the loop itself, \(\vec{B}_{loop}\), is in the same direction as its magnetic moment \(\vec{M}\) (along the axis of the loop).

Therefore, in stable equilibrium, \(\vec{B}_{loop}\) and \(\vec{B}_{ext}\) are parallel and point in the same direction at points inside the loop.

The total magnetic field through the coil is the superposition of the external field and the loop's own field: \[ \vec{B}_{total} = \vec{B}_{ext} + \vec{B}_{loop} \]
The magnetic flux through the coil is given by \(\Phi = \int \vec{B}_{total} \cdot d\vec{A}\).
The area vector \(d\vec{A}\) is defined to be in the same direction as the magnetic moment \(\vec{M}\).

In stable equilibrium:

\(\vec{B}_{ext}\) is parallel to \(d\vec{A}\).
\(\vec{B}_{loop}\) is parallel to \(d\vec{A}\).

Thus, \(\vec{B}_{total}\) is parallel to \(d\vec{A}\), and the dot product \(\vec{B}_{total} \cdot d\vec{A}\) becomes the simple product of magnitudes \(B_{total} dA\). The flux is: \[ \Phi = \int (B_{ext} + B_{loop}) dA = (B_{ext} + B_{loop})A \]
This is the maximum possible value for the flux because in any other orientation, the angle between the fields and the area vector would be non-zero, introducing a \(\cos\theta\) factor that would reduce the value of the flux. Therefore, the flux of the total field through the coil is maximum in the stable equilibrium orientation.
Quick Tip: Stable equilibrium for any dipole (electric or magnetic) in an external field always corresponds to the orientation of minimum potential energy, which occurs when the dipole moment vector aligns with the external field vector.