Step 1: Understanding the Question:

The question asks for the new power consumed by a wire after it is stretched to twice its original length. The wire is connected to the same ideal battery, which means the voltage (V) across the wire remains constant.


Step 2: Key Formula or Approach:

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

2. When a wire is stretched, its volume remains constant. Volume \( V_{vol} = l \times A \).

3. Power consumed is given by \( P = \frac{V^2}{R} \), where V is the voltage.


Step 3: Detailed Explanation:

Let the initial length, area, and resistance be \( l_1, A_1, \) and \( R_1 \) respectively.

Initially, \( R_1 = R = \rho \frac{l_1}{A_1} \).

The wire is stretched to double its initial length, so the new length \( l_2 = 2l_1 \).

Since the volume remains constant, \( l_1 A_1 = l_2 A_2 \).
\( l_1 A_1 = (2l_1) A_2 \Rightarrow A_2 = \frac{A_1}{2} \).

The new resistance \( R_2 \) is:
\[ R_2 = \rho \frac{l_2}{A_2} = \rho \frac{2l_1}{A_1/2} = 4 \left( \rho \frac{l_1}{A_1} \right) = 4R_1 = 4R \]
The initial power consumed is \( P_1 = P = \frac{V^2}{R_1} = \frac{V^2}{R} \).

The new power consumed \( P_2 \) with the new resistance \( R_2 \) is:
\[ P_2 = \frac{V^2}{R_2} = \frac{V^2}{4R} = \frac{1}{4} \left( \frac{V^2}{R} \right) = \frac{P}{4} \]

Step 4: Final Answer:

The new power consumed will be \( \frac{P}{4} \).
Quick Tip: When a wire is stretched 'n' times its original length, its new resistance becomes \( n^2 \) times the original resistance. Consequently, if the voltage is constant, the new power becomes \( \frac{P}{n^2} \). Here n=2, so new power is \( \frac{P}{4} \).

Step 1: Understanding the Question:

A bar magnet is dropped through a copper ring. We need to determine its acceleration. The key feature is that the copper ring has a cut, making it an open circuit.


Step 2: Key Formula or Approach:

1. Faraday's Law of Induction: A changing magnetic flux through a loop induces an electromotive force (e.m.f.).

2. Lenz's Law: The direction of the induced current is such that it opposes the change in magnetic flux that produced it.

3. Ohm's Law: Current flows only in a closed circuit. \( I = \frac{V}{R} \).


Step 3: Detailed Explanation:

As the magnet falls towards the ring, the magnetic flux through the ring changes.

According to Faraday's law of induction, this changing flux induces an e.m.f. in the copper ring.

If the ring were a complete, closed loop, this induced e.m.f. would drive an induced current.

By Lenz's law, this induced current would create a magnetic field that opposes the falling magnet, exerting an upward repulsive force on it. This would make the net downward force less than the gravitational force (mg), and thus the acceleration would be less than g.

However, the ring in the question has a cut. This means the circuit is open.

Because the circuit is open, no induced current can flow through the ring, despite the presence of an induced e.m.f.

Without an induced current, there is no induced magnetic field to oppose the magnet's motion.

Therefore, the only force acting on the falling magnet is the force of gravity.

The acceleration of the magnet is thus equal to the acceleration due to gravity, g.


Step 4: Final Answer:

The acceleration of the falling magnet is g.
Quick Tip: Remember that Lenz's law opposition comes from the magnetic field produced by an \textbf{induced current}. If the circuit is open (like a ring with a cut), no current can flow, and therefore there is no opposing magnetic force. The object will fall freely under gravity.

Step 1: Understanding the Question:

We need to find the magnetic force on a 10 cm segment of a straight wire carrying a current of 2 A along the +x axis, placed in a given uniform magnetic field.


Step 2: Key Formula or Approach:

The magnetic force \( \vec{F} \) on a straight conductor of length vector \( \vec{L} \) carrying current \( I \) in a uniform magnetic field \( \vec{B} \) is given by the formula:
\[ \vec{F} = I (\vec{L} \times \vec{B}) \]
The vector cross product properties will be used: \( \hat{i} \times \hat{j} = \hat{k} \) and \( \hat{i} \times \hat{k} = -\hat{j} \).


Step 3: Detailed Explanation:

Given values:

Current, \( I = 2 \) A.

Length of the conductor, \( L = 10 \) cm \( = 0.1 \) m.

The current is in the +x direction, so the length vector is \( \vec{L} = 0.1\hat{i} \) m.

The magnetic field is \( \vec{B} = (0.6\hat{j} + 0.8\hat{k}) \) T.

Now, we calculate the cross product \( \vec{L} \times \vec{B} \):
\[ \vec{L} \times \vec{B} = (0.1\hat{i}) \times (0.6\hat{j} + 0.8\hat{k}) \] \[ \vec{L} \times \vec{B} = (0.1 \times 0.6)(\hat{i} \times \hat{j}) + (0.1 \times 0.8)(\hat{i} \times \hat{k}) \] \[ \vec{L} \times \vec{B} = 0.06(\hat{k}) + 0.08(-\hat{j}) \] \[ \vec{L} \times \vec{B} = -0.08\hat{j} + 0.06\hat{k} \]
Now, we find the force \( \vec{F} \):
\[ \vec{F} = I (\vec{L} \times \vec{B}) = 2 \times (-0.08\hat{j} + 0.06\hat{k}) \] \[ \vec{F} = -0.16\hat{j} + 0.12\hat{k} \]
The force acting on the conductor is \( (-0.16\hat{j} + 0.12\hat{k}) \) N.


Step 4: Final Answer:

The force acting on the conductor matches option (B).
Quick Tip: To quickly calculate the cross product \( \vec{A} \times \vec{B} \), you can use the determinant method. For this problem, \( \vec{L} = (0.1, 0, 0) \) and \( \vec{B} = (0, 0.6, 0.8) \). The determinant form makes calculations systematic and less prone to sign errors.

Step 1: Understanding the Question:

We are given the voltages across a resistor (\(V_R\)) and an inductor (\(V_L\)) in a series AC circuit. We need to find the total voltage of the AC source (\(V_S\)).


Step 2: Key Formula or Approach:

In a series R-L circuit, the voltage across the resistor (\(V_R\)) is in phase with the current, while the voltage across the inductor (\(V_L\)) leads the current by 90 degrees (\(\pi/2\) radians). Therefore, \(V_R\) and \(V_L\) are out of phase with each other by 90 degrees. The source voltage is the phasor sum of the individual voltages.

The magnitude of the source voltage is given by:
\[ V_S = \sqrt{V_R^2 + V_L^2} \]

Step 3: Detailed Explanation:

Given values:

Voltage across the resistor, \( V_R = 8 \) V.

Voltage across the inductor, \( V_L = 6 \) V.

Using the formula for the source voltage in a series R-L circuit:
\[ V_S = \sqrt{V_R^2 + V_L^2} \] \[ V_S = \sqrt{(8)^2 + (6)^2} \] \[ V_S = \sqrt{64 + 36} \] \[ V_S = \sqrt{100} \] \[ V_S = 10 V \]

Step 4: Final Answer:

The voltage of the source is 10 V.
Quick Tip: In series AC circuits, you cannot simply add the voltages arithmetically. You must use phasor addition (vector addition). For an R-L-C circuit, the formula is \( V_S = \sqrt{V_R^2 + (V_L - V_C)^2} \). Recognizing the 6-8-10 Pythagorean triplet can also save time in this specific problem.

Step 1: Understanding the Question:

A proton and an alpha particle enter a uniform magnetic field with the same velocity, perpendicular to the field. We need to find the ratio of the radii of their circular paths (proton radius to alpha particle radius).


Step 2: Key Formula or Approach:

When a charged particle moves perpendicular to a uniform magnetic field, it follows a circular path. The magnetic force provides the necessary centripetal force.

Magnetic force: \( F_B = qvB \).

Centripetal force: \( F_c = \frac{mv^2}{r} \).

Equating the two forces: \( qvB = \frac{mv^2}{r} \).

Solving for the radius \( r \): \( r = \frac{mv}{qB} \).


Step 3: Detailed Explanation:

Let's denote the properties of the proton with subscript 'p' and the alpha particle with subscript '\(\alpha\)'.

For the proton (p):

Mass: \( m_p \).

Charge: \( q_p = e \).

Radius of its path: \( r_p = \frac{m_p v}{q_p B} = \frac{m_p v}{eB} \).


For the alpha particle (\(\alpha\)):

An alpha particle is a helium nucleus, consisting of 2 protons and 2 neutrons.

Mass: \( m_\alpha \approx 4m_p \).

Charge: \( q_\alpha = 2e \).

Radius of its path: \( r_\alpha = \frac{m_\alpha v}{q_\alpha B} = \frac{(4m_p) v}{(2e)B} = 2 \frac{m_p v}{eB} \).


Ratio of the radii:

We need to find the ratio \( \frac{r_p}{r_\alpha} \).
\[ \frac{r_p}{r_\alpha} = \frac{\frac{m_p v}{eB}}{2 \frac{m_p v}{eB}} \] \[ \frac{r_p}{r_\alpha} = \frac{1}{2} \]

Step 4: Final Answer:

The ratio of the radii of their paths is \( \frac{1}{2} \).
Quick Tip: For problems comparing paths of different particles in a magnetic field, it's helpful to remember the basic properties:
Proton (\(p\)): mass \(m_p\), charge \(e\).
Deuteron (\(d\)): mass \(2m_p\), charge \(e\).
Alpha particle (\(\alpha\)): mass \(4m_p\), charge \(2e\).
The radius formula \( r = \frac{mv}{qB} \) shows that the radius is proportional to the mass-to-charge ratio (\(m/q\)).
For a proton, \(m/q \propto 1/1\). For an alpha particle, \(m/q \propto 4/2 = 2\). The ratio of radii is \(1/2\).

Step 1: Understanding the Question:

Two coherent waves of equal intensity \(I_0\) interfere. We need to find the average intensity over the entire interference pattern on the screen.


Step 2: Key Formula or Approach:

The intensity at a point in an interference pattern from two sources with intensities \(I_1\) and \(I_2\) and a phase difference \( \phi \) is:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
The average intensity is the average value of this expression over all possible phase differences. The principle of conservation of energy also states that the total energy is redistributed, but the average intensity is simply the sum of the individual intensities.


Step 3: Detailed Explanation:

Given \( I_1 = I_2 = I_0 \).

The resultant intensity at any point is:
\[ I = I_0 + I_0 + 2\sqrt{I_0 \cdot I_0} \cos(\phi) \] \[ I = 2I_0 + 2I_0 \cos(\phi) = 2I_0(1 + \cos(\phi)) \]
To find the average intensity (\(I_{avg}\)) on the screen, we need to average this expression over all points, which corresponds to averaging over a full cycle of phase difference \( \phi \) (from 0 to \( 2\pi \)).
\[ I_{avg} = \langle I \rangle = \langle 2I_0(1 + \cos(\phi)) \rangle \] \[ I_{avg} = 2I_0 \langle 1 + \cos(\phi) \rangle = 2I_0 ( \langle 1 \rangle + \langle \cos(\phi) \rangle ) \]
The average value of the cosine function over a full cycle is zero.
\[ \langle \cos(\phi) \rangle = \frac{1}{2\pi} \int_{0}^{2\pi} \cos(\phi) d\phi = \frac{1}{2\pi} [\sin(\phi)]_{0}^{2\pi} = 0 \]
Therefore,
\[ I_{avg} = 2I_0 (1 + 0) = 2I_0 \]
Alternatively, by the principle of conservation of energy, interference only redistributes the energy on the screen. The total power landing on the screen is the sum of the power from the two sources. Thus, the average intensity is the sum of the individual average intensities.
\[ I_{avg} = I_1 + I_2 = I_0 + I_0 = 2I_0 \]

Step 4: Final Answer:

The average intensity of light on the screen is \(2I_0\).
Quick Tip: Don't confuse average intensity with maximum or minimum intensity.
Maximum intensity (constructive interference, \( \cos(\phi) = 1 \)): \(I_{max} = 4I_0\).
Minimum intensity (destructive interference, \( \cos(\phi) = -1 \)): \(I_{min} = 0\).
Average intensity: \(I_{avg} = 2I_0\), which is simply the sum of the individual intensities. This holds true due to energy conservation.

Step 1: Understanding the Question:

We are asked to find the difference between the equivalent resistance of 'n' identical resistors (each with resistance R) when connected in series (\(R_s\)) and when connected in parallel (\(R_p\)).


Step 2: Key Formula or Approach:

1. For 'n' resistors of resistance R connected in series, the equivalent resistance is:

\( R_s = R + R + ... \) (n times) \( = nR \).

2. For 'n' resistors of resistance R connected in parallel, the equivalent resistance \(R_p\) is given by:

\( \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + ... \) (n times) \( = \frac{n}{R} \). This implies \( R_p = \frac{R}{n} \).


Step 3: Detailed Explanation:

First, calculate the equivalent resistance for the series combination, \(R_s\).
\[ R_s = nR \]
Next, calculate the equivalent resistance for the parallel combination, \(R_p\).
\[ R_p = \frac{R}{n} \]
Now, find the difference \( (R_s - R_p) \).
\[ R_s - R_p = nR - \frac{R}{n} \]
To combine the terms, find a common denominator, which is 'n'.
\[ R_s - R_p = \frac{n(nR) - R}{n} = \frac{n^2R - R}{n} \]
Factor out R from the numerator.
\[ R_s - R_p = \frac{R(n^2 - 1)}{n} = \left( \frac{n^2 - 1}{n} \right) R \]

Step 4: Final Answer:

The value of \( (R_s - R_p) \) is \( \left( \frac{n^2 - 1}{n} \right) R \). This matches option (C).
Quick Tip: Remember the basic rules for combining identical resistors:
\textbf{Series:} Multiply the individual resistance by the number of resistors (\(R_s = nR\)).
\textbf{Parallel:} Divide the individual resistance by the number of resistors (\(R_p = R/n\)).
These simple rules make calculations very fast.

Step 1: Understanding the Question:

The question asks how to modify a galvanometer to function as an ammeter.


Step 2: Key Formula or Approach:

An ammeter is a device used to measure current and must be connected in series with the component through which the current is to be measured. An ideal ammeter should have zero resistance so that it does not alter the current it is measuring.

A galvanometer is a sensitive device that gives a full-scale deflection for a small current (\(I_g\)). To measure a larger current (I), most of the current must be diverted away from the galvanometer.


Step 3: Detailed Explanation:

1. Purpose of an Ammeter: To measure the total current in a circuit branch. It is placed in series.

2. Ideal Property: An ideal ammeter has zero internal resistance. Therefore, a practical ammeter must have a very low resistance.

3. Galvanometer Limitation: A galvanometer has a certain resistance (\(R_g\)) and can only handle a small current (\(I_g\)).

4. Conversion: To convert a galvanometer into an ammeter that can measure a larger current I, we need to bypass the excess current (\(I - I_g\)) around the galvanometer. This is achieved by connecting a low-resistance path in parallel with the galvanometer. This low-resistance path is called a "shunt" resistance (S).

5. Why Parallel? A parallel connection provides an alternate path for the current to flow.

6. Why Small Resistance? To ensure that most of the large current I bypasses the sensitive galvanometer, the shunt resistance S must be much smaller than the galvanometer resistance \(R_g\). According to the current divider rule, more current flows through the path of lower resistance.


Therefore, connecting a small resistance in parallel (a shunt) converts a galvanometer into an ammeter. The overall resistance of the ammeter (\(R_A\)), which is the parallel combination of \(R_g\) and S, will be very low.


Step 4: Final Answer:

A galvanometer is converted into an ammeter by connecting a small resistance in parallel.
Quick Tip: Remember the contrast between converting to an ammeter and a voltmeter:
\textbf{Ammeter:} Connect a \textbf{small} resistance in \textbf{parallel} (Shunt).
\textbf{Voltmeter:} Connect a \textbf{large} resistance in \textbf{series}.
Think about their function: Ammeters are in series (need low R), Voltmeters are in parallel (need high R).

Step 1: Understanding the Question:

The question asks to compare the magnitudes of the nuclear force between different pairs of nucleons (protons and neutrons) inside a nucleus.


Step 2: Key Formula or Approach:

The key concept here is the properties of the strong nuclear force. One of its fundamental properties is charge independence.


Step 3: Detailed Explanation:

The strong nuclear force is the force that binds protons and neutrons together in the atomic nucleus. Experimental evidence from scattering experiments and the analysis of nuclear structure have revealed several key properties of this force:

1. It is the strongest force in nature (at subatomic distances).

2. It is short-ranged, acting only over distances of about 1-2 femtometers (\(10^{-15}\) m).

3. It is charge-independent. This means the force does not depend on the electric charge of the interacting nucleons. The nuclear force between two protons is the same as the force between two neutrons, which is also the same as the force between a proton and a neutron, assuming they are in the same quantum state.


Based on the property of charge independence:

The force between two protons (\(F_{pp}\)), the force between a proton and a neutron (\(F_{pn}\)), and the force between two neutrons (\(F_{nn}\)) are all considered equal.
\[ F_{pp} = F_{pn} = F_{nn} \]
It is important to note that this refers only to the \textit{strong nuclear force component. In reality, there is also a much weaker electrostatic (Coulomb) repulsive force between two protons, which is not present between neutrons or between a proton and a neutron. However, the question specifically asks about the "nuclear forces". In this context, the charge-independence property is the dominant principle.


Step 4: Final Answer:

The nuclear forces between the nucleon pairs are equal.
Quick Tip: A key property of the strong nuclear force is its charge independence. This is a fundamental concept in nuclear physics. Don't get confused by the electrostatic repulsion between protons; the question specifically asks about the nuclear force, which is treated as being the same for all nucleon pairs.

Step 1: Understanding the Question:

We need to calculate the de Broglie wavelength of an electron that has a kinetic energy of 5 eV.


Step 2: Key Formula or Approach:

The de Broglie wavelength (\(\lambda\)) of a particle is given by \( \lambda = \frac{h}{p} \), where h is Planck's constant and p is the momentum of the particle.

The kinetic energy (K) is related to momentum by \( K = \frac{p^2}{2m} \), so \( p = \sqrt{2mK} \).

Substituting this into the de Broglie equation gives:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]
For an electron accelerated through a potential V, its kinetic energy is K = eV. The formula can be simplified for electrons to:
\[ \lambda (in Angstroms, \AA) = \frac{12.27}{\sqrt{V (in Volts)}} \]

Step 3: Detailed Explanation:

The kinetic energy of the electron is given as K = 5 eV. This means the electron has been accelerated through a potential difference of V = 5 Volts.

We can use the shortcut formula for an electron's de Broglie wavelength.
\[ \lambda (\AA) = \frac{12.27}{\sqrt{V}} \]
Substitute V = 5 V into the formula:
\[ \lambda (\AA) = \frac{12.27}{\sqrt{5}} \]
We know that \( \sqrt{5} \approx 2.236 \).
\[ \lambda (\AA) \approx \frac{12.27}{2.236} \approx 5.487 \AA \]
Now, we need to convert this to nanometers (nm). Since 1 nm = 10 Å, we divide the result by 10.
\[ \lambda (nm) = \frac{5.487}{10} \approx 0.549 nm \]
This value is closest to 0.55 nm.


Step 4: Final Answer:

The de Broglie wavelength is approximately 0.55 nm.
Quick Tip: Memorizing the formula \( \lambda (\AA) = \frac{12.27}{\sqrt{V}} \) is extremely useful for quickly solving problems involving the de Broglie wavelength of electrons. This formula saves you from substituting the values of h, m, and e every time.

Step 1: Understanding the Question:

The question asks about the motion of a diamagnetic material when it is placed in a uniform magnetic field and is free to move.


Step 2: Key Formula or Approach:

The force on a magnetic dipole (like a piece of magnetic material) in a magnetic field \( \vec{B} \) is given by \( \vec{F} = \nabla(\vec{m} \cdot \vec{B}) \). In a uniform magnetic field, the field strength \( \vec{B} \) is constant in magnitude and direction, which means its spatial derivatives are zero.


Step 3: Detailed Explanation:

1. Diamagnetic Materials: These materials are weakly repelled by magnetic fields. When placed in an external magnetic field, they develop a magnetization that opposes the external field.

2. Force in a Non-Uniform Field: If the magnetic field were non-uniform, the diamagnetic material would experience a net force that pushes it from the region of stronger magnetic field to the region of weaker magnetic field. This is the basis of magnetic levitation of diamagnetic substances.

3. Force in a Uniform Field: The question explicitly states that the magnetic field is uniform. In a uniform field, the magnetic field strength is the same everywhere. There is no field gradient (\( \nabla B = 0 \)).

A piece of material placed in the field can be thought of as a collection of magnetic dipoles. A magnetic dipole experiences a force only if the field is non-uniform. In a uniform field, the forces on the north and south poles (or equivalent induced poles) are equal and opposite, resulting in zero net translational force. There might be a net torque that aligns the material, but this torque will also be zero for a diamagnetic material since its induced moment is antiparallel to the field.

Since there is no net force, and the object is initially at rest, it will not move at all.


Step 4: Final Answer:

A piece of diamagnetic material placed in a uniform magnetic field experiences no net force and therefore does not move.
Quick Tip: Pay close attention to the word \textbf{uniform} when dealing with forces on magnetic materials.
- \textbf{Uniform Field:} No net force on any magnetic material (paramagnetic, diamagnetic, or ferromagnetic). There can be a torque.
- \textbf{Non-Uniform Field:} Net force exists. Paramagnetic/Ferromagnetic materials are attracted to stronger fields. Diamagnetic materials are repelled from stronger fields.

Step 1: Understanding the Question:

We need to calculate the momentum of a photon given its frequency.


Step 2: Key Formula or Approach:

The energy of a photon (E) is given by two fundamental relations:

1. Planck's relation: \( E = hf \), where h is Planck's constant and f is the frequency.

2. Mass-energy equivalence for a massless particle (photon): \( E = pc \), where p is the momentum and c is the speed of light.

By equating these two expressions for energy, we can find the momentum.
\[ pc = hf \Rightarrow p = \frac{hf}{c} \]

Step 3: Detailed Explanation:

Given values:

Frequency, \( f = 6.0 \times 10^{14} \) Hz.

Planck's constant, \( h \approx 6.63 \times 10^{-34} \) J·s.

Speed of light, \( c \approx 3.0 \times 10^8 \) m/s.

Using the formula \( p = \frac{hf}{c} \):
\[ p = \frac{(6.63 \times 10^{-34} J·s) \times (6.0 \times 10^{14} Hz)}{3.0 \times 10^8 m/s} \] \[ p = \frac{6.63 \times 6.0}{3.0} \times 10^{-34 + 14 - 8} kg·m/s \] \[ p = (6.63 \times 2.0) \times 10^{-28} kg·m/s \] \[ p = 13.26 \times 10^{-28} kg·m/s \]
To express this in standard scientific notation:
\[ p = 1.326 \times 10^{-27} kg·m/s \]

Step 4: Final Answer:

The momentum of the photon is \( 1.326 \times 10^{-27} \) kg m/s. This matches option (B).
Quick Tip: The relation \( p = h/\lambda \) is often remembered, but \( p = hf/c \) is equally important and directly applicable when frequency is given. Always check your units and powers of ten carefully in such calculations. Simple arithmetic mistakes are common under exam pressure.

Step 1: Analyzing the Assertion (A):

The assertion states that a hole is an apparent free particle with an effective positive charge. In semiconductor physics, a "hole" is the absence of an electron in a filled valence band. When an electron moves to fill this vacancy, it leaves a new vacancy behind. This movement of the vacancy is equivalent to the movement of a positive charge carrier. Thus, a hole is treated as a mobile, quasi-particle with a positive charge equal in magnitude to the electronic charge. The assertion is correct.


Step 2: Analyzing the Reason (R):

The reason states that a hole is not necessarily a vacancy left behind by an electron in the valence band. This statement is incorrect. By definition, a hole in a semiconductor is precisely the vacancy created in the valence band when an electron is excited to a higher energy level (like the conduction band or an acceptor level). The entire concept of a hole is based on this vacancy. Therefore, the reason is false.


Step 3: Conclusion:

The Assertion (A) is a true statement describing the nature of a hole in a semiconductor.

The Reason (R) provides an incorrect definition of a hole.

Therefore, Assertion (A) is true, but Reason (R) is false.
Quick Tip: For Assertion-Reason questions in physics, first, evaluate the truthfulness of each statement independently.
\textbf{Assertion (A):} Is it a correct physical statement? Yes, holes are treated as positive charge carriers.
\textbf{Reason (R):} Is it a correct physical statement? No, a hole is by definition a vacancy in the valence band.
Once you determine one is true and one is false, the choice is straightforward (either C or the equivalent option if A is false and R is true).

Step 1: Analyzing the Assertion (A):

The assertion states that a reflecting telescope's image is free from chromatic aberration. Reflecting telescopes use mirrors (specifically, a large concave mirror as the objective) to collect and focus light. The law of reflection (angle of incidence equals angle of reflection) is independent of the wavelength (color) of light. Since all colors are reflected at the same angle, they all focus at the same point. Thus, there is no color separation, and the image is free from chromatic aberration. The assertion is true.


Step 2: Analyzing the Reason (R):

The reason states that chromatic aberration occurs only due to refraction of light through an optical medium. Chromatic aberration is the phenomenon where a lens fails to focus all colors to the same convergence point. This happens because the refractive index of the lens material (like glass) is a function of the wavelength of light (a phenomenon called dispersion). Different colors bend by different amounts upon refraction. Since this effect is tied to the wavelength-dependence of the refractive index, it is a characteristic of refraction, not reflection. The reason is true.


Step 3: Connecting Assertion and Reason:

The reason correctly explains why the assertion is true. A reflecting telescope is free from chromatic aberration precisely because it uses reflection instead of refraction. And, as the reason states, chromatic aberration is a problem exclusive to refraction-based optical systems (like refracting telescopes and lenses). Therefore, the reason is the correct explanation for the assertion.
Quick Tip: Remember the key difference between lenses and mirrors regarding aberrations:
- \textbf{Lenses (Refraction):} Suffer from both chromatic and spherical aberration.
- \textbf{Mirrors (Reflection):} Free from chromatic aberration but can suffer from spherical aberration (which is often corrected by using parabolic mirrors). This is a major advantage of reflecting telescopes over refracting ones.

Step 1: Analyzing the Assertion (A):

The assertion states that the binding energy per nucleon (BE/A) is nearly constant for nuclei with mass numbers (A) between 30 and 170. This is a well-known feature of the binding energy curve. The curve rises sharply for light nuclei, reaches a broad plateau around 8.5 MeV per nucleon for intermediate-mass nuclei (from roughly A=30 to A=170), and then slowly decreases for heavy nuclei. Therefore, the assertion is true.


Step 2: Analyzing the Reason (R):

The reason claims that nuclear forces are not short-range in this mass number range. This is fundamentally incorrect. The strong nuclear force is a short-range force, effective only over distances of a few femtometers. This property is true for all nuclei, regardless of their mass number. In fact, it is the short-range nature of the nuclear force that leads to the saturation property (a nucleon interacts only with its immediate neighbors), which in turn explains why the binding energy per nucleon is nearly constant for medium and heavy nuclei. Because the force is short-range, adding more nucleons doesn't significantly increase the binding energy of the existing ones, leading to the plateau in the BE/A curve. Thus, the reason is false.


Step 3: Conclusion:

The Assertion (A) is true, as it correctly describes the binding energy curve.

The Reason (R) is false, as the nuclear force is always a short-range force.

Therefore, Assertion (A) is true, but Reason (R) is false.
Quick Tip: The saturation and near-constancy of binding energy per nucleon are direct consequences of the \textbf{short-range} nature of the nuclear force. A nucleon only 'feels' the attraction of its nearest neighbors. This is a key concept to remember about nuclear forces and the structure of the nucleus.

Step 1: Analyzing the Assertion (A):

The assertion states that X-rays are produced when slow moving electrons are stopped by a metal target. This is incorrect. The production of X-rays (specifically Bremsstrahlung or "braking radiation") requires the rapid deceleration of high-energy (fast-moving) electrons. When these energetic electrons strike a target (typically a metal with a high atomic number to be more efficient), their kinetic energy is converted into electromagnetic radiation in the form of X-ray photons. The use of "slow moving" makes the assertion false.


Step 2: Analyzing the Reason (R):

The reason states that X-rays consist of low-energy photons. This is also incorrect. In the electromagnetic spectrum, X-rays are characterized by their high energy, high frequency, and short wavelength. Their typical energies range from hundreds to hundreds of thousands of electron volts (eV), placing them well above visible light and ultraviolet radiation in terms of energy. Therefore, the reason is false.


Step 3: Conclusion:

Both the Assertion (A) and the Reason (R) are incorrect statements.

Therefore, the correct option is (D).
Quick Tip: Remember the key conditions for X-ray production in an X-ray tube:
1. A source of electrons (heated filament).
2. A high accelerating voltage to produce \textbf{fast-moving, high-energy} electrons.
3. A metal target (anode), usually with a \textbf{high atomic number} and high melting point, to stop the electrons.
X-rays are \textbf{high-energy} photons.

Step 1: Relationship between V, E, I, and r

For a cell of emf E and internal resistance r, connected to an external resistance R, the current flowing in the circuit is \( I = \frac{E}{R+r} \).

The terminal voltage (V) across the cell is the potential difference across the external resistance R, so \( V = IR \).

Alternatively, the terminal voltage can be expressed as the emf minus the potential drop across the internal resistance:
\[ V = E - Ir \]
This equation shows a linear relationship between the terminal voltage V and the current I.


Step 2: Plotting the Graph

The equation \( V = -rI + E \) is in the form of a straight line equation \( y = mx + c \), where:

- y-axis represents Terminal Voltage (V).

- x-axis represents Current (I).

- The slope of the line, \( m = -r \).

- The y-intercept, \( c = E \).


The graph of V versus I is a straight line with a negative slope.

- When \( I = 0 \) (open circuit, \( R \rightarrow \infty \)), the terminal voltage is \( V = E \). This is the intercept on the V-axis.

- When \( V = 0 \) (short circuit, \( R = 0 \)), the current is maximum, \( I_{max} = \frac{E}{r} \). This is the intercept on the I-axis.


\begin{tikzpicture
\begin{axis[
axis lines=left,
xlabel={Current (I),
ylabel={Terminal Voltage (V),
xmin=0,
xmax=4,
ymin=0,
ymax=2,
xtick={0,
ytick={0,
extra y ticks={1.5,
extra y tick labels={E,
extra x ticks={3,
extra x tick labels={\( \frac{E}{r} \),
width=10cm,
height=7cm,
grid=major,
grid style={dashed, gray!30,
legend pos=outer north east
]

% The plot for V = E - Ir. Let's assume E=1.5V and r=0.5 Ohm for plotting.
% So V = 1.5 - 0.5*I.
% Y-intercept (I=0) is V=E=1.5.
% X-intercept (V=0) is I=E/r=1.5/0.5=3.
\addplot[
domain=0:3,
samples=100,
color=blue,
thick,
] {1.5 - 0.5*x;

% Annotate the slope
\draw[dashed, red] (axis cs:1,1) -- (axis cs:2,1) node[midway, above] {\(\Delta I\);
\draw[dashed, red] (axis cs:2,1) -- (axis cs:2,0.5) node[midway, right] {\(\Delta V\);
\node[red, anchor=west] at (axis cs:2.1, 0.75) {Slope \( = \frac{\Delta V}{\Delta I} = -r \);

\end{axis
\end{tikzpicture

Step 3: Determining E and r from the graph

1. To find the emf (E): The emf of the cell is equal to the terminal voltage when the current drawn from the cell is zero. This corresponds to the y-intercept of the V-I graph. By extending the plotted line to intersect the V-axis (where I=0), the value of the intercept gives the emf E.

\[ E = (V-intercept of the graph) \]
2. To find the internal resistance (r): The internal resistance is related to the slope of the graph. The slope of the V-I graph is \( \frac{\Delta V}{\Delta I} \). From the equation \( V = E - Ir \), we can see that the slope is equal to \(-r\).

\[ Slope = \frac{\Delta V}{\Delta I} = -r \]
Therefore, the internal resistance is the negative of the slope of the graph.

\[ r = -(Slope of the V-I graph) \] Quick Tip: The equation \( V = E - Ir \) is fundamental for circuits with real batteries. Always remember that the terminal voltage V is equal to the emf E only in an open circuit (I=0). When the cell supplies current, V is always less than E due to the internal voltage drop 'Ir'. When charging a cell, the equation becomes \( V = E + Ir \).

Step 1: Understanding the process in an n-type semiconductor

An n-type semiconductor is created by doping an intrinsic semiconductor (like Si or Ge) with pentavalent impurity atoms (like Phosphorus or Arsenic). These impurity atoms are called donors.

At room temperature, two processes occur simultaneously:

1. Generation: Due to thermal energy, some covalent bonds break, creating electron-hole pairs. This process generates an equal number of free electrons and holes.

2. Recombination: Free electrons and holes move randomly, and they can meet and recombine, annihilating each other. An electron from the conduction band falls into a hole in the valence band.


Step 2: Role of Donor Impurities

The key to understanding n-type semiconductors is the role of the donor atoms. Each pentavalent donor atom has five valence electrons. Four of these form covalent bonds with the neighboring semiconductor atoms. The fifth electron is very loosely bound to the donor atom.

At room temperature, the thermal energy is sufficient to easily free this fifth electron, making it a charge carrier in the conduction band. This process creates a free electron without creating a corresponding hole in thevalence band. The donor atom becomes a fixed positive ion.


Step 3: Explaining the Concentration Difference

Let \(n_e\) be the electron concentration and \(n_h\) be the hole concentration.

- In an n-type semiconductor, the total number of free electrons (\(n_e\)) comes from two sources: those generated from broken covalent bonds (thermal generation) and those contributed by the donor atoms.

- The total number of holes (\(n_h\)) comes only from one source: thermal generation.

- Since the doping concentration is usually significant, the number of electrons contributed by donor atoms is far greater than the number of electrons generated thermally.

- Although recombination is a continuous process where electrons and holes are annihilated, thermal generation is also a continuous process creating new pairs. At thermal equilibrium, the rate of recombination equals the rate of generation, leading to steady-state concentrations of electrons and holes.

- Because of the large number of electrons supplied by the donor atoms, the equilibrium concentration of electrons (\(n_e\)) is much higher than the equilibrium concentration of holes (\(n_h\)). The electrons are the majority charge carriers, and holes are the minority charge carriers.

- Therefore, despite the continuous recombination, the vast supply of electrons from donor atoms ensures that \(n_e \gg n_h\) is always maintained at room temperature.
Quick Tip: Think of it like a crowded room (the conduction band for electrons) and an almost empty room (the valence band for holes). In an n-type material, doping adds a huge number of people directly to the crowded room without taking anyone from the empty room. Even if people continuously move between rooms (recombination/generation), the crowded room will always have far more people.

Step 1: Understanding the Question

We need to calculate the force exerted by a laser beam on a surface. We are given the beam's power, wavelength, and the fact that it strikes normally and is perfectly reflected.


Step 2: Key Formula or Approach

Force is the rate of change of momentum (\( F = \frac{dp}{dt} \)).

The momentum of a single photon is \( p = \frac{h}{\lambda} \).

When a photon is perfectly reflected from a surface at normal incidence, its momentum changes from \( p \) to \( -p \). The change in momentum for one photon is \( \Delta p = p - (-p) = 2p \).

The force is the total change in momentum per second. This is equal to (Number of photons striking per second) \( \times \) (Change in momentum per photon).

A simpler approach is to relate force directly to power (P) for reflection:
\[ F = \frac{2P}{c} \]
where c is the speed of light. The factor of 2 accounts for the perfect reflection. For complete absorption, the formula would be \( F = \frac{P}{c} \).


Step 3: Detailed Calculation

Given values:

Power of the laser beam, \( P = 5 \) mW \( = 5 \times 10^{-3} \) W.

Wavelength, \( \lambda = 500 \) nm = \( 500 \times 10^{-9} \) m (This is not needed for the direct formula).

Area, A = 1 mm\(^2\) (This is also not needed, assuming the entire beam hits the area).

Speed of light, \( c = 3 \times 10^8 \) m/s.

The surface is perfectly reflecting, so we use the formula \( F = \frac{2P}{c} \).
\[ F = \frac{2 \times (5 \times 10^{-3} W)}{3 \times 10^8 m/s} \] \[ F = \frac{10 \times 10^{-3}}{3 \times 10^8} \] \[ F = \frac{10}{3} \times 10^{-3-8} \] \[ F = 3.33 \times 10^{-11} N \]

Step 4: Final Answer

The force exerted by the laser beam on the body is approximately \( 3.33 \times 10^{-11} \) N.
Quick Tip: Remember the formulas for radiation pressure and force. They depend on whether the surface is absorbing or reflecting.
- \textbf{Perfect Absorption:} Force \( F = P/c \), Pressure \( P_{rad} = I/c \) (where I is intensity).
- \textbf{Perfect Reflection (Normal Incidence):} Force \( F = 2P/c \), Pressure \( P_{rad} = 2I/c \).
The wavelength and area information can sometimes be distractors if the power is already given.

Step 1: Understanding the Condition of Minimum Deviation

When a prism is in the position of minimum deviation (\(\delta = \delta_m\)), the light ray passes symmetrically through it. This symmetry implies specific relationships between the angles.

The key conditions for minimum deviation are:

1. The angle of incidence (\(i\)) is equal to the angle of emergence (\(e\)). \( \implies i=e \)

2. The angle of refraction at the first face (\(r_1\)) is equal to the angle of refraction at the second face (\(r_2\)). \( \implies r_1=r_2=r \)


Step 2: Key Formulas for a Prism

The general relations for any prism are:

1. Angle of prism: \( A = r_1 + r_2 \)

2. Angle of deviation: \( \delta = (i + e) - A \)


Step 3: Applying Formulas for Minimum Deviation

We apply the conditions from Step 1 to the formulas in Step 2. The question uses \( \delta \) to denote the angle of minimum deviation.

(ii) To find the angle of refraction (r) on face AB:

Using the formula \( A = r_1 + r_2 \) and the condition \( r_1=r_2=r \), we get:
\[ A = r + r = 2r \]
Solving for r, we find the angle of refraction:
\[ r = \frac{A}{2} \]

(i) To find the angle of incidence (i):

Using the formula \( \delta = (i + e) - A \) and the condition \( i=e \), we get:
\[ \delta = (i + i) - A = 2i - A \]
Rearranging this equation to solve for the angle of incidence i:
\[ 2i = A + \delta \] \[ i = \frac{A + \delta}{2} \]

Step 4: Final Answer

(i) The angle of incidence is \( i = \frac{A + \delta}{2} \).

(ii) The angle of refraction on face AB is \( r = \frac{A}{2} \).
Quick Tip: The condition of minimum deviation is all about symmetry. Remembering that the ray passes symmetrically (\(i=e\) and \(r_1=r_2\)) is the key to deriving all related formulas, including the prism formula for refractive index: \( n = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \).

Step 1: Key Formula or Approach

In Young's double-slit experiment, if two coherent waves of equal intensity \( I_0 \) interfere, the resultant intensity \( I_R \) at a point is given by: \[ I_R = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( \phi \) is the phase difference between the waves. The phase difference is related to the path difference \( \Delta x \) by the formula: \[ \phi = \frac{2\pi}{\lambda} \Delta x \]

Step 2: Calculation for Case (i)

Path difference, \( \Delta x = \frac{\lambda}{3} \).

First, calculate the phase difference \( \phi \): \[ \phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3} radians \]
Now, substitute this into the intensity formula: \[ I_R = 4I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right) \]
We know that \( \cos\left(\frac{\pi}{3}\right) = \cos(60^\circ) = \frac{1}{2} \). \[ I_R = 4I_0 \left(\frac{1}{2}\right)^2 = 4I_0 \left(\frac{1}{4}\right) \] \[ I_R = I_0 \]

Step 3: Calculation for Case (ii)

Path difference, \( \Delta x = \frac{\lambda}{2} \).

First, calculate the phase difference \( \phi \): \[ \phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{2}\right) = \pi radians \]
Now, substitute this into the intensity formula: \[ I_R = 4I_0 \cos^2\left(\frac{\pi}{2}\right) \]
We know that \( \cos\left(\frac{\pi}{2}\right) = \cos(90^\circ) = 0 \). \[ I_R = 4I_0 (0)^2 \] \[ I_R = 0 \]
This corresponds to the condition for a dark fringe (minimum intensity).

Step 4: Final Answer

(i) For a path difference of \( \frac{\lambda}{3} \), the intensity is \( I_0 \).

(ii) For a path difference of \( \frac{\lambda}{2} \), the intensity is 0.
Quick Tip: Memorize the conditions for maxima and minima in terms of both path difference and phase difference.
- \textbf{Maxima (Bright Fringes):} Path difference \( \Delta x = n\lambda \), Phase difference \( \phi = 2n\pi \). Intensity \( I_{max} = 4I_0 \).
- \textbf{Minima (Dark Fringes):} Path difference \( \Delta x = (n + 1/2)\lambda \), Phase difference \( \phi = (2n+1)\pi \). Intensity \( I_{min} = 0 \).
Here, n = 0, 1, 2, ...

Step 1: Understanding the Question and Sign Convention

We need to find the image position and nature for an object placed in front of a single convex spherical refracting surface. We will use the Cartesian sign convention.
- Light travels from left to right.
- The pole (vertex) of the surface is the origin (0,0).
- Distances measured in the direction of incident light are positive.
- Distances measured opposite to the direction of incident light are negative.


Step 2: Key Formula or Approach

The formula for refraction at a single spherical surface is: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
where:
- \( n_1 \) = refractive index of the object medium.
- \( n_2 \) = refractive index of the image medium.
- \( u \) = object distance.
- \( v \) = image distance.
- \( R \) = radius of curvature of the surface.

Step 3: Applying the Formula

Given values based on the sign convention:
- The object is in air, so \( n_1 = 1 \).
- The surface is made of glass, so \( n_2 = 1.5 \).
- The object is 12 cm in front of the surface, so \( u = -12 \) cm.
- The surface is convex, and its center of curvature is to the right (in the glass), so \( R = +30 \) cm.

Substitute these values into the formula: \[ \frac{1.5}{v} - \frac{1}{-12} = \frac{1.5 - 1}{+30} \] \[ \frac{1.5}{v} + \frac{1}{12} = \frac{0.5}{30} \] \[ \frac{1.5}{v} + \frac{1}{12} = \frac{1}{60} \]
Now, solve for v: \[ \frac{1.5}{v} = \frac{1}{60} - \frac{1}{12} \]
Find a common denominator for the right side: \[ \frac{1.5}{v} = \frac{1 - 5}{60} = \frac{-4}{60} = \frac{-1}{15} \] \[ 1.5 = v \left( \frac{-1}{15} \right) \] \[ v = 1.5 \times (-15) \] \[ v = -22.5 cm \]

Step 4: Nature and Position of the Image

- Position: The image distance is \( v = -22.5 \) cm. The negative sign indicates that the image is formed on the same side as the object (in the air), at a distance of 22.5 cm from the pole of the surface.

- Nature: Since the image is formed in the medium from which the rays are originating (indicated by the negative sign for v), and the rays of light do not actually meet there but appear to diverge from that point, the image is virtual.


Final Answer: The image is formed at a distance of 22.5 cm from the surface, on the same side as the object. The image is virtual.
Quick Tip: Properly applying the sign convention is crucial for problems in geometric optics. For a single refracting surface: - If v is positive, the image is real and formed in the second medium. - If v is negative, the image is virtual and formed in the first medium. Always draw a rough diagram to help visualize the setup and apply the signs correctly.

Step 1: Concept of Electron Motion in a Conductor

In a metallic conductor, in the absence of an external electric field, the free conduction electrons are in a state of continuous random motion due to thermal energy.
They move in all possible directions with high thermal velocities, colliding with the fixed positive ions of the metal.
The average velocity of all these electrons over any period is zero, so there is no net flow of charge, and hence no current.


Step 2: Effect of an External Electric Field

When an external electric field is applied across the conductor, each free electron experiences an electrostatic force in the direction opposite to the field.
This force accelerates the electron.
However, this acceleration is short-lived as the electron soon collides with a positive ion, losing the velocity gained.
After the collision, it again accelerates. This process repeats.


Step 3: Definition of Drift Velocity

As a result of these repeated collisions and accelerations, the electrons acquire a small, constant average velocity, superimposed on their random thermal motion, in the direction opposite to the electric field.

Drift velocity (\(v_d\)) is defined as the average velocity with which free electrons in a conductor get drifted towards the positive end of the conductor under the influence of an externally applied electric field.
Quick Tip: The drift velocity of electrons is surprisingly small, typically on the order of \(10^{-4}\) m/s or a few millimeters per second.
This is much smaller than the random thermal velocities of electrons, which are around \(10^5\) to \(10^6\) m/s.
The large number of charge carriers is what allows for a significant current despite the low drift speed.

Step 1: Force and Acceleration on an Electron

When a potential difference V is applied across a conductor of length l, a uniform electric field \( E = \frac{V}{l} \) is set up inside it.

A free electron of charge -e in this field experiences a force:
\[ \vec{F} = -e\vec{E} \]

Due to this force, the electron experiences an acceleration \(\vec{a}\):
\[ \vec{a} = \frac{\vec{F}}{m} = -\frac{e\vec{E}}{m} \]

where m is the mass of the electron.


Step 2: Relating Drift Velocity to Relaxation Time

The relaxation time (\(\tau\)) is the average time interval between two successive collisions of an electron with the positive ions.

The drift velocity (\(v_d\)) is the average velocity gained by the electron during this time.
Using the equation of motion \(v = u + at\), for an average electron starting with zero average initial velocity (\(u_{avg}=0\)) and accelerating for time \(\tau\):
\[ \vec{v}_d = 0 + \vec{a}\tau = -\frac{e\vec{E}\tau}{m} \]

The magnitude of the drift velocity is \( v_d = \frac{eE\tau}{m} \).


Step 3: Deriving Current Density

The current density (\(j\)) is defined as the current per unit cross-sectional area, \(j = I/A\).

The relationship between current I and drift velocity \(v_d\) is given by \( I = n e A v_d \), where n is the number density of free electrons.

Therefore, the current density is:
\[ j = \frac{I}{A} = \frac{n e A v_d}{A} = n e v_d \]

Now, substitute the expression for the magnitude of drift velocity from Step 2:
\[ j = n e \left( \frac{eE\tau}{m} \right) \]
\[ j = \frac{ne^2E\tau}{m} \]

Since \( E = \frac{V}{l} \), we can also write this as:
\[ j = \frac{ne^2\tau}{m} \left( \frac{V}{l} \right) \]

This is the required formula for current density in terms of relaxation time \(\tau\).
Quick Tip: The expression \( j = \frac{ne^2\tau}{m} E \) is a form of Ohm's law.
By comparing it with the microscopic form of Ohm's law, \( \vec{j} = \sigma \vec{E} \), we can derive the expression for electrical conductivity (\(\sigma\)) as \( \sigma = \frac{ne^2\tau}{m} \).
This shows how conductivity depends on the material properties n, m, and \(\tau\).

Part 1: Lenz's Law

Statement: Lenz's law states that the direction of the induced electromotive force (emf) and hence the induced current in a closed circuit is always such that it opposes the change in magnetic flux that produces it.

This law is a direct consequence of the principle of conservation of energy.
To do work against the opposing force created by the induced current, mechanical or other forms of energy must be expended, and this is what gets converted into electrical energy.


Part 2: Derivation of Induced EMF in a Rotating Rod

Step 1: Setup

Consider a metallic rod MN of length L rotating with a constant angular velocity \( \omega \) about an axis passing through its end M.
The rotation occurs in a uniform magnetic field \( \vec{B} \) which is parallel to the axis of rotation (i.e., perpendicular to the plane of rotation).


Step 2: Motional EMF in a small element

Consider a small element of the rod of length \(dr\) at a distance \(r\) from the pivot point M.

The linear velocity of this element is perpendicular to its length and is given by \( v = r\omega \).

Since the velocity \( \vec{v} \), the magnetic field \( \vec{B} \), and the length element \( d\vec{r} \) are mutually perpendicular, the motional emf induced across this small element is:
\[ d\epsilon = B v dr = B (r\omega) dr \]


Step 3: Integrating to find the total EMF

The total emf induced between the ends of the rod M (at \(r=0\)) and N (at \(r=L\)) is the integral of \(d\epsilon\) over the entire length of the rod.
The emfs of all such elements add up as they are in series.
\[ \epsilon = \int_{0}^{L} d\epsilon = \int_{0}^{L} B\omega r dr \]

Since B and \( \omega \) are constant:
\[ \epsilon = B\omega \int_{0}^{L} r dr \]
\[ \epsilon = B\omega \left[ \frac{r^2}{2} \right]_{0}^{L} \]
\[ \epsilon = B\omega \left( \frac{L^2}{2} - \frac{0^2}{2} \right) \]
\[ \epsilon = \frac{1}{2}B\omega L^2 \]

This is the required expression for the emf induced between the ends of the rotating rod.
Quick Tip: The polarity of the induced emf can be found using the Lorentz force (\( \vec{F} = q(\vec{v} \times \vec{B}) \)).
The force on the free electrons in the rod will push them towards one end, making it negatively charged and the other end positively charged.
In this setup, the outer end N becomes positive and the center M becomes negative, creating the potential difference.

Part 1: Definition of Self-Inductance

Self-inductance is the property of a coil (or any electrical circuit) by virtue of which it opposes any change in the strength of the current flowing through it by inducing an electromotive force (emf) in itself.
This phenomenon is called self-induction.

The magnetic flux (\(\Phi_B\)) linked with the coil is directly proportional to the current (I) flowing through it.
\[ \Phi_B \propto I \quad or \quad \Phi_B = LI \]

where L is the constant of proportionality called the self-inductance or coefficient of self-induction of the coil.

Thus, self-inductance can be defined as the magnetic flux linked with the coil when a unit current flows through it.
Its SI unit is the Henry (H).

Alternatively, from Faraday's law, the induced emf is \( \epsilon = -\frac{d\Phi_B}{dt} = -L\frac{dI}{dt} \).
So, L can also be defined as the magnitude of the induced emf in the coil when the rate of change of current is unity.


Part 2: Derivation for a Long Solenoid

Step 1: Magnetic Field inside a Solenoid

Consider a long solenoid of length l, cross-sectional area A, and n turns per unit length.
The total number of turns is \( N = nl \).

When a current I flows through the solenoid, a uniform magnetic field is produced inside it, given by:
\[ B = \mu_0 n I \]

where \( \mu_0 \) is the permeability of free space.


Step 2: Magnetic Flux through the Solenoid

The magnetic flux linked with each turn of the solenoid is:
\[ \Phi_{turn} = B \times A = (\mu_0 n I)A \]

The total magnetic flux linked with the entire solenoid is the flux per turn multiplied by the total number of turns N:
\[ \Phi_{total} = N \times \Phi_{turn} = (nl) \times (\mu_0 n I A) \]
\[ \Phi_{total} = \mu_0 n^2 A l I \]


Step 3: Calculating Self-Inductance

From the definition of self-inductance, we have \( \Phi_{total} = LI \).

Comparing this with the expression derived in Step 2:
\[ LI = \mu_0 n^2 A l I \]

Cancelling I from both sides, we get the expression for the self-inductance of the long solenoid:
\[ L = \mu_0 n^2 A l \]
Quick Tip: The self-inductance of a solenoid depends on its geometry (number of turns, area, length) and the medium inside it.
If the core is filled with a material of relative permeability \( \mu_r \), the formula becomes \( L = \mu_0 \mu_r n^2 A l \).
This shows that inserting a ferromagnetic core can significantly increase the inductance.

Part 1: Ray Diagram

The path of the light ray through a parallel-sided glass slab is shown below.
The emergent ray is parallel to the incident ray but is displaced laterally.


\begin{tikzpicture[scale=1.5, >=stealth]
% Draw the glass slab
\draw[thick, fill=cyan!10] (0,0) rectangle (4, -2);
\node at (0.5, -1.8) {Glass Slab;

% Define coordinates
\coordinate (P) at (0.5, 0.5); % A point on the incident ray
\coordinate (Q) at (1.5, 0); % Point of incidence
\coordinate (R) at (2.7, -2); % Point of emergence

% Draw the incident ray
\draw[->, thick, red] (P) -- (Q);
\node[above left] at (P) {P;

% Draw the first normal
\draw[dashed] (Q) -- (1.5, -2.5) node[below] {N;

% Draw the refracted ray
% Snell's law: sin(i)/sin(r) = 1.5. Angle i is ~26.5 deg. sin(i)~0.447. sin(r)~0.298. r~17.3 deg.
% Using coordinates is easier.
\draw[->, thick, red] (Q) -- (R);
\node[above left, yshift=-2] at (Q) {Q;

% Draw the second normal
\draw[dashed] (R) -- (2.7, 0.5) node[above] {N';

% Calculate the emergent ray's direction (parallel to incident)
\coordinate (S) at (\((R) + (Q) - (P)\));
\draw[->, thick, red] (R) -- (S);
\node[below right] at (R) {R;
\node[below right] at (S) {S;

% Extend the incident ray
\coordinate (T_point) at (3.1, -2);
\draw[dashed, red!50] (Q) -- (T_point);

% Draw and label the lateral shift
\draw[<->, thick, blue] (R) -- (2.3, -1.333); % Manually found perpendicular point for visual
\node[blue, right] at (2.4, -1.6) {Lateral Shift (x);

% Draw and label the thickness
\draw[<->] (4.2, 0) -- (4.2, -2) node[midway, right] {Thickness (d);

% Label angles
\draw (1.2, 0) arc (180:243.5:0.4);
\node at (1, 0.1) {\(i\);
\draw (1.8, 0) arc (0:-54.5:0.4);
\node at (1.9, -0.2) {\(r\);
\draw (2.7-0.3, -2) arc(180:125.5:0.4);
\node at (2.3, -1.8) {\(r\);
\end{tikzpicture


In the diagram, PQ is the incident ray, QR is the refracted ray, and RS is the emergent ray.
The lateral shift is the perpendicular distance RT.


Part 2: Expression for Lateral Shift

In the right-angled triangle QTR, the angle \( \angle RQT \) is equal to \( (i-r) \).

From this triangle, we have:
\[ \sin(i-r) = \frac{RT}{QR} = \frac{x}{QR} \]
\[ \implies x = QR \sin(i-r) \quad \dots(1) \]

Now, consider the right-angled triangle QNR, where QN is the normal and QN = d (thickness of the slab).
\[ \cos(r) = \frac{QN}{QR} = \frac{d}{QR} \]
\[ \implies QR = \frac{d}{\cos(r)} \quad \dots(2) \]

Substituting the value of QR from equation (2) into equation (1):
\[ x = \frac{d}{\cos(r)} \sin(i-r) \]

This is the required expression for the lateral shift.


Part 3: Condition for Minimum Shift

The lateral shift 'x' depends on the thickness of the slab (d), the refractive index of the slab (which determines r for a given i), and the angle of incidence (i).

For the shift to be minimum, the term \( \frac{\sin(i-r)}{\cos(r)} \) should be minimum.
This occurs when the angle of incidence \(i\) is minimum.

The minimum possible value for the angle of incidence is \( i = 0^\circ \) (normal incidence).

When \( i = 0^\circ \), according to Snell's law (\( n_1 \sin i = n_2 \sin r \)), the angle of refraction is also \( r = 0^\circ \).

Substituting these values into the expression for lateral shift:
\[ x = \frac{d}{\cos(0^\circ)} \sin(0^\circ - 0^\circ) = \frac{d}{1} \sin(0^\circ) = d \times 0 = 0 \]

Thus, the lateral shift is minimum (zero) when the ray of light is incident normally on the glass slab.
Quick Tip: For a parallel-sided slab, the angle of emergence 'e' is always equal to the angle of incidence 'i'.
This means the emergent ray is always parallel to the incident ray, and the only effect is a lateral shift.
The shift increases with the angle of incidence and the thickness of the slab.

Part 1: Geiger-Marsden Experiment

The Geiger-Marsden experiment, also known as Rutherford's alpha-particle scattering experiment, was a pivotal experiment designed to investigate the internal structure of the atom.

Setup:

1. A narrow beam of high-energy alpha particles was obtained from a radioactive source (like Bismuth-214) placed inside a lead cavity.

2. This beam was directed at a very thin gold foil (approximately \(10^{-7}\) m thick).

3. The entire setup was enclosed in an evacuated chamber to prevent the alpha particles from being scattered by air molecules.

4. A rotatable detector, consisting of a zinc sulphide (ZnS) screen and a microscope, was used to observe the alpha particles scattered by the foil. Each alpha particle striking the screen produced a tiny, visible flash of light called a scintillation.

Procedure:

The number of alpha particles scattered at different angles (\(\theta\)) from their original path was counted by moving the detector around the gold foil.


Part 2: Graph of Observations

The experimental results were plotted as a graph of the number of scattered alpha particles, N(\(\theta\)), versus the scattering angle, \(\theta\).



\begin{tikzpicture
\begin{axis[
width=11cm,
height=7cm,
axis lines=left,
xlabel={Scattering angle, \(\theta\) (degrees),
ylabel style={rotate=90, anchor=center,
ylabel={Number of scattered particles, \(N(\theta)\),
ymode=log,
xlabel style={anchor=north,
ylabel style={rotate=90, anchor=south, yshift=-3.5em,
xmin=0, xmax=180,
ymin=1, ymax=1e7,
xtick={0, 30, 60, 90, 120, 150, 180,
ytick={1, 10, 100, 1000, 10000, 100000, 1000000,
log ticks with fixed point,
grid=major,
grid style={dashed, gray!30,
legend pos=south west,
]
% Plotting the Rutherford scattering formula
\addplot[
domain=5:180, % Start domain > 0 to avoid singularity
samples=150,
color=blue,
thick,
no marks,
] {1e7 / (sin(x/2))^4;
\addlegendentry{\(N(\theta) \propto \frac{1}{\sin^4(\theta/2)}\)
\end{axis
\end{tikzpicture


The key observations from the graph are:

1. The vast majority of alpha particles passed through the gold foil with little or no deflection (\(\theta\) is small).

2. A small fraction of alpha particles were deflected through moderate angles.

3. A very tiny number of alpha particles (about 1 in 8000) were deflected by large angles (\(\theta > 90^\circ\)), with some even bouncing back along their original path (\(\theta \approx 180^\circ\)).


Part 3: Discovery of the Nucleus

These observations were in stark contradiction with the then-prevalent Thomson's "plum pudding" model, which suggested that positive charge was spread throughout the atom. That model could not explain large-angle scattering.

Rutherford interpreted the results as follows:

1. Most of the Atom is Empty Space: Since most alpha particles passed straight through, it implied that they did not encounter anything substantial, meaning most of the atom must be empty.

2. Concentrated Positive Charge and Mass: The large-angle scattering of a few alpha particles could only be explained if they experienced an immense repulsive force. Such a strong force would only occur if the entire positive charge and most of the mass of the atom were concentrated in a very small, dense region at the center. An alpha particle that happened to travel very close to this dense core would be strongly repelled and deflected by a large angle.

3. The Nucleus: Rutherford named this small, dense, positively charged core the nucleus. This experiment provided the first direct experimental evidence for the nuclear model of the atom, where a tiny nucleus contains the positive charge and mass, with electrons orbiting it at a relatively large distance.
Quick Tip: Remember the three key observations and their direct implications:
1. Most \(\alpha\)-particles passed straight through \(\implies\) Most of the atom is empty space.
2. Some \(\alpha\)-particles were deflected by small angles \(\implies\) The center of the atom is positively charged.
3. A very few \(\alpha\)-particles were deflected by large angles (\(>90^\circ\)) \(\implies\) The positive charge and most of the atom's mass are concentrated in a tiny, dense nucleus.

Step 1: Understanding Q Value

The Q value of a nuclear reaction is the energy released or absorbed during the reaction.
It is calculated as the difference between the total mass of the reactants and the total mass of the products, multiplied by \(c^2\).
\[ Q = (Mass_{reactants} - Mass_{products})c^2 \]

If Q > 0, the reaction is exothermic (energy is released).

If Q < 0, the reaction is endothermic (energy is absorbed).


Step 2: Calculating Mass Difference (\(\Delta m\))

First, find the total mass of the reactants:

Mass of reactants = \( m({}^{12}C) + m({}^{12}C) = 12.000000 + 12.000000 = 24.000000 \) u.

Next, find the total mass of the products:

Mass of products = \( m({}^{20}Ne) + m({}^{4}He) = 19.992439 + 4.002603 = 23.995042 \) u.

Now, calculate the mass difference, \( \Delta m = Mass_{reactants} - Mass_{products} \):
\[ \Delta m = 24.000000 - 23.995042 = 0.004958 u \]


Step 3: Calculating Q Value

The energy equivalent of 1 u is given as 931 MeV. So, \( Q = \Delta m \times 931 \) MeV.
\[ Q = 0.004958 \times 931 MeV \]
\[ Q \approx 4.615898 MeV \]


Step 4: Determining the Nature of the Reaction

Since the Q value is positive (Q > 0), mass has been converted into energy, and energy is released during the reaction.

Therefore, the reaction is exothermic.
Quick Tip: A positive Q value means the products are more tightly bound (have less mass per nucleon) than the reactants.
This excess binding energy is released during the reaction.
Always be careful with the arithmetic, as the mass differences are very small.

Part (i): Torque on the loop

Step 1: Analyze the Magnetic Field and Forces

The straight long wire produces a magnetic field \( \vec{B} \) whose field lines are concentric circles around the wire.
In the plane of the rectangular loop, the magnetic field is directed perpendicularly into the plane of the paper.
The force on a current-carrying segment of the loop is given by \( \vec{F} = I(\vec{L} \times \vec{B}) \).
The forces on all four sides of the rectangular loop will lie in the plane of the loop itself.

Step 2: Condition for Torque

A torque is produced when forces create a turning effect.
For a current loop in a uniform magnetic field, the torque is \( \vec{\tau} = \vec{m} \times \vec{B} \), where \( \vec{m} \) is the magnetic dipole moment.
In this case, the magnetic field \( \vec{B} \) is perpendicular to the plane of the loop everywhere.
The magnetic moment vector \( \vec{m} \) is also perpendicular to the plane of the loop (its direction can be found by the right-hand curl rule).
This means that the angle between \( \vec{m} \) and \( \vec{B} \) is either 0 or 180 degrees.

Therefore, the torque \( \vec{\tau} = \vec{m} \times \vec{B} = mB \sin(0^\circ) = 0 \).

Alternatively, since all the forces on the sides of the loop lie within the plane of the loop, there are no forces that can cause the loop to rotate about an axis in its plane.

Answer (i): The torque acting on the loop is zero.


Part (ii): Net force on the loop

Let the sides of the loop be named AB (top), BC (right), CD (bottom), and DA (left). Let the current in the loop be \(I_{loop} = 1\) A (let's assume clockwise) and in the wire be \(I_{wire} = 2\) A.

Length of sides BC and DA = 5 cm = 0.05 m.
Length of sides AB and CD = 1 cm = 0.01 m.

Step 1: Forces on horizontal sides (AB and CD)

The forces on sides AB and CD are equal in magnitude and opposite in direction. They will cancel each other out.

Step 2: Force on vertical side DA (closer to the wire)

The distance of side DA from the wire is \(r_1 = 1\) cm = 0.01 m.
The magnetic field at this distance is \( B_1 = \frac{\mu_0 I_{wire}}{2\pi r_1} \).
The force on side DA is \( F_{DA} = I_{loop} L_{DA} B_1 \). Assuming clockwise current, the current in DA is downwards. Using the right-hand rule, the force is attractive (towards the wire).
\[ F_{DA} = \frac{\mu_0 I_{wire} I_{loop} L_{DA}}{2\pi r_1} = \] \[\frac{(4\pi \times 10^{-7}) \times 2 \times 1 \times 0.05}{2\pi \times 0.01} = 2 \times 10^{-7} \times \frac{2 \times 0.05}{0.01} = 2 \times 10^{-5} N (towards wire) \]

Step 3: Force on vertical side BC (farther from the wire)

The distance of side BC from the wire is \(r_2 = 1 + 1 = 2\) cm = 0.02 m.
The magnetic field at this distance is \( B_2 = \frac{\mu_0 I_{wire}}{2\pi r_2} \).
The force on side BC is \( F_{BC} = I_{loop} L_{BC} B_2 \). The current in BC is upwards. The force is repulsive (away from the wire).
\[ F_{BC} = \frac{\mu_0 I_{wire} I_{loop} L_{BC}}{2\pi r_2} = \frac{(4\pi \times 10^{-7}) \times 2 \times 1 \times 0.05}{2\pi \times 0.02} \] \[= 2 \times 10^{-7} \times \frac{2 \times 0.05}{0.02} = 1 \times 10^{-5} N (away from wire) \]

Step 4: Net Force

The net force is the vector sum of these forces. Since \(F_{DA}\) and \(F_{BC}\) are in opposite directions, the net force is the difference between their magnitudes.
\[ F_{net} = F_{DA} - F_{BC} = (2 \times 10^{-5} - 1 \times 10^{-5}) N = 1 \times 10^{-5} N \]

The direction is that of the larger force, which is the attractive force \(F_{DA}\).

Answer (ii): The magnitude of the net force is \( 1 \times 10^{-5} \) N, and its direction is towards the long straight wire (attractive).
Quick Tip: For a closed loop in a non-uniform magnetic field, there is generally a net force, even if the net torque is zero.
The net force arises because the forces on opposite sides of the loop do not cancel if the magnetic field strength is different at their locations.
In this configuration, the attractive force on the nearer side is always stronger than the repulsive force on the farther side, resulting in a net attractive force.

(i) In Radar:

EM Wave: Microwaves are used in radar systems for aircraft navigation, speed detection, and weather forecasting.

Wavelength Range: The wavelength of microwaves ranges from approximately 1 millimeter (mm) to 1 meter (m).


(ii) In Eye Surgery:

EM Wave: Ultraviolet (UV) waves are used in eye surgery, specifically in procedures like LASIK to reshape the cornea. Excimer lasers, which produce UV light, are commonly used.

Wavelength Range: The wavelength of UV waves ranges from approximately 10 nanometers (nm) to 400 nanometers (nm).


(iii) As Diagnostic Tool in Medicine:

EM Wave: X-rays are widely used as a diagnostic tool in medicine to image bones and internal organs (radiography, CT scans).

Wavelength Range: The wavelength of X-rays ranges from approximately 0.01 nanometers (nm) to 10 nanometers (nm).
Quick Tip: Remember the order of the electromagnetic spectrum by energy/frequency/wavelength.
A common mnemonic is: "Roman Men Invented Very Unusual X-ray Guns" (Radio, Microwaves, Infrared, Visible, Ultraviolet, X-rays, Gamma rays).
This order is from lowest frequency (longest wavelength) to highest frequency (shortest wavelength).
Knowing the order helps in recalling the properties and applications of each type of wave.

Step 1: Understanding Donor Impurities

Germanium (Ge) is a Group 14 element and is a tetravalent semiconductor (it has 4 valence electrons).
A donor impurity is an atom that, when added to a semiconductor, "donates" a free electron to the conduction band, creating an n-type semiconductor.
To do this, the impurity atom must have more valence electrons than the semiconductor atom.
Therefore, for Ge, a donor impurity must be a pentavalent element (from Group 15 of the periodic table).


Step 2: Analyzing the Options

(A) Boron (B) is in Group 13 (trivalent). It is an acceptor impurity.

(B) Antimony (Sb) is in Group 15 (pentavalent). It is a donor impurity.

(C) Aluminium (Al) is in Group 13 (trivalent). It is an acceptor impurity.

(D) Indium (In) is in Group 13 (trivalent). It is an acceptor impurity.


Step 3: Conclusion

Antimony is the only pentavalent element among the options and thus acts as a donor impurity for Germanium.
Quick Tip: A simple way to remember is the "P-A-N" rule for doping:
\textbf{P}entavalent impurities (like \textbf{A}rsenic, \textbf{A}ntimony, \textbf{P}hosphorus) create \textbf{N}-type semiconductors (donors).
\textbf{T}rivalent impurities (like \textbf{B}oron, \textbf{A}luminium, \textbf{G}allium, \textbf{I}ndium) create \textbf{P}-type semiconductors (acceptors).

Step 1: Understanding Ionization Energy of Dopants

When a pentavalent donor atom is introduced into a silicon (Si) lattice, its fifth valence electron is not involved in covalent bonding.
It is loosely bound to its parent positive ion by a weak electrostatic force.
The energy required to detach this electron from the donor atom and move it to the conduction band is called the ionization energy.
This energy is very small compared to the band gap energy of the semiconductor (which is about 1.1 eV for Si).


Step 2: Standard Values for Si and Ge

The ionization energy can be estimated using a model similar to the hydrogen atom, but modified by the dielectric constant of the semiconductor and the effective mass of the electron.
For practical purposes, these are standard experimental values that are useful to remember for competitive exams.

- For Silicon (Si), the ionization energy for common donor impurities (like Phosphorus) is about 0.05 eV.

- For Germanium (Ge), the ionization energy for common donor impurities is about 0.01 eV.


Step 3: Conclusion

The question asks for the ionization energy in Silicon (Si). Based on the standard value, the energy required is approximately 0.05 eV.
Quick Tip: Remember the approximate ionization energies for dopants in Si and Ge:
- Silicon (Si): \(\approx\) 0.05 eV
- Germanium (Ge): \(\approx\) 0.01 eV
These energies are small, which is why most donor/acceptor atoms are ionized at room temperature (\(kT \approx 0.025\) eV).

Step 1: Initial Process - Diffusion

When a p-type and an n-type semiconductor are joined, there is a high concentration of holes on the p-side and a high concentration of electrons on the n-side.
Due to this concentration gradient, majority carriers begin to diffuse across the junction: holes diffuse from the p-side to the n-side, and electrons diffuse from the n-side to the p-side.
This constitutes a diffusion current. Initially, this diffusion current is large.


Step 2: Formation of the Depletion Region

When an electron from the n-side diffuses to the p-side, it leaves behind a positively charged, immobile donor ion (\( Nd^+ \)) on the n-side.
When a hole from the p-side diffuses to the n-side, it leaves behind a negatively charged, immobile acceptor ion (\( Na^- \)) on the p-side.
This process creates a layer of uncovered immobile ions on both sides of the junction.
This region, devoid of mobile charge carriers, is called the depletion region or space-charge region.


Step 3: Analyzing the Charge Layers

The layer on the n-side of the junction consists of positive donor ions.
The layer on the p-side of the junction consists of negative acceptor ions.
Therefore, a layer of positive charge appears on the n-side, and a layer of negative charge appears on the p-side. This matches option (B).


Step 4: Analyzing Other Options

(A) is incorrect; the charge layers are reversed.

(C) is incorrect; electrons are majority carriers on the n-side and they move to the p-side initially.

(D) is incorrect; initially, the diffusion current is large due to the high concentration gradient, and the drift current is small. The drift current grows as the electric field in the depletion region builds up.
Quick Tip: A simple way to remember the charge layers in the depletion region is that each side loses its majority carrier and is left with the charge of the fixed ion core.
The \textbf{n}-side loses \textbf{n}egative electrons, leaving behind \textbf{p}ositive ions.
The \textbf{p}-side loses \textbf{p}ositive holes, leaving behind \textbf{n}egative ions.

Step 1: Understanding Reverse Bias

A p-n junction is reverse-biased when the positive terminal of the external battery is connected to the n-side and the negative terminal is connected to the p-side.
This applied field is in the same direction as the internal barrier field of the depletion region.


Step 2: Effect of Reverse Bias

- Depletion Region Width: The applied field aids the internal field, pulling the majority carriers (electrons on n-side, holes on p-side) further away from the junction. This widens the depletion region. So, (C) is incorrect.

- Current: The reverse bias opposes the flow of majority carriers, so the diffusion current becomes negligible. However, it supports the flow of minority carriers (electrons from p-side, holes from n-side) across the junction. This constitutes a small drift current, known as the reverse saturation current. This current is very small, typically of the order of microamperes (\(\mu A\)) or nanoamperes (nA), not milliamperes (mA). So, (A) is incorrect.

- Voltage Dependence of Current: The reverse saturation current is due to thermally generated minority carriers and is largely independent of the applied reverse voltage (until breakdown voltage is reached). So, (D) is incorrect.

- Potential Drop: The widened depletion region is almost devoid of mobile charge carriers and has a very high resistance compared to the p-type and n-type regions on either side. Therefore, when an external voltage is applied, almost the entire potential drop occurs across this high-resistance depletion region. So, (B) is correct.
Quick Tip: Think of the p-n junction as a resistor whose resistance depends on the bias.
In \textbf{forward bias}, the depletion region narrows, resistance is low, and current flows easily.
In \textbf{reverse bias}, the depletion region widens, resistance is extremely high, and very little current flows.
Almost all the applied voltage drops across the component with the highest resistance, which is the depletion region in reverse bias.

Step 1: Understanding AC Input

The input is an AC signal with a frequency of 50 Hz.
This means the signal completes 50 full cycles (one positive half-cycle and one negative half-cycle) in one second.
The time period of the input signal is \(T_{in} = \frac{1}{f_{in}} = \frac{1}{50}\) s.


Step 2: Action of a Full-Wave Rectifier

A full-wave rectifier converts both halves of the AC input cycle into a pulsating DC output.
During the positive half-cycle of the input, the rectifier produces a positive output pulse.
During the negative half-cycle of the input, the rectifier inverts it and produces another positive output pulse.


Step 3: Analyzing the Output Waveform and Frequency

For every one full cycle of the input AC signal, the output of the full-wave rectifier produces two positive pulses.
This means the output waveform repeats itself twice as fast as the input waveform.
The time period of the output signal (\(T_{out}\)) is half the time period of the input signal (\(T_{in}\)).
\[ T_{out} = \frac{T_{in}}{2} \]
Since frequency is the reciprocal of the time period (\( f = 1/T \)), the output frequency (\(f_{out}\)) is double the input frequency (\(f_{in}\)).
\[ f_{out} = \frac{1}{T_{out}} = \frac{1}{T_{in}/2} = 2f_{in} \]

Given \( f_{in} = 50 \) Hz.
\[ f_{out} = 2 \times 50 Hz = 100 Hz \]
Quick Tip: Remember the output frequencies for different rectifiers:
- \textbf{Half-Wave Rectifier:} It only passes one half of the AC cycle. The output frequency is the \textbf{same} as the input frequency. (\(f_{out} = f_{in}\)).
- \textbf{Full-Wave Rectifier:} It passes both halves of the AC cycle. The output frequency is \textbf{double} the input frequency. (\(f_{out} = 2f_{in}\)).

Step 1: Formula for Capacitance

The capacitance of a parallel plate capacitor filled with air is given by the formula:
\[ C = \frac{\epsilon_0 A}{d} \]

where \( \epsilon_0 \) is the permittivity of free space, A is the area of the plates, and d is the separation between them.


Step 2: Analyzing the Changes

The initial capacitance is given as \( C = \frac{\epsilon_0 A}{d} \).

The problem states two changes:

1. The distance between the plates is increased to \( d' = 2d \).

2. One of the plates is shifted. The diagram [Fig. (b)] shows the plates are still parallel and fully overlapping, just with a larger separation. The term "shifted" refers to moving one plate away from the other to increase the separation. The overlapping area A remains the same.


Step 3: Calculating the New Capacitance

The new capacitance, \( C' \), can be calculated using the new distance \( d' \):
\[ C' = \frac{\epsilon_0 A}{d'} \]

Substitute \( d' = 2d \):
\[ C' = \frac{\epsilon_0 A}{2d} \]

We can rewrite this in terms of the original capacitance C:
\[ C' = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) = \frac{C}{2} \]


Step 4: Final Answer

The capacitance of the new system is C/2.
Quick Tip: Capacitance is directly proportional to the plate area (A) and inversely proportional to the plate separation (d).
- Doubling the area doubles the capacitance.
- Doubling the separation halves the capacitance.
Always check what parameters are changing to quickly determine the effect on capacitance.

Step 1: Electric Fields

Let \( E_0 \) be the electric field between the capacitor plates without the dielectric. This field is produced by the charge density \( \sigma \) on the plates.
\[ E_0 = \frac{\sigma}{\epsilon_0} \]

When the dielectric slab is inserted, it gets polarized. This polarization creates an induced surface charge density \( -\sigma_p \) and \( +\sigma_p \) on the faces of the slab.
This induced charge produces an opposing electric field, \( E_p \), inside the dielectric.
\[ E_p = \frac{\sigma_p}{\epsilon_0} \]


Step 2: Net Field inside the Dielectric

The net electric field (E) inside the dielectric is the vector sum of the external field and the induced field. Since they are in opposite directions, the magnitude is:
\[ E = E_0 - E_p = \frac{\sigma}{\epsilon_0} - \frac{\sigma_p}{\epsilon_0} = \frac{\sigma - \sigma_p}{\epsilon_0} \]


Step 3: Definition of Dielectric Constant (K)

The dielectric constant K is defined as the factor by which the electric field inside a material is reduced compared to the external field.
\[ E = \frac{E_0}{K} \]

Substituting the expressions for E and \( E_0 \):
\[ \frac{\sigma - \sigma_p}{\epsilon_0} = \frac{(\sigma / \epsilon_0)}{K} \]
\[ \sigma - \sigma_p = \frac{\sigma}{K} \]


Step 4: Solving for K

Rearranging the equation to solve for K:
\[ K = \frac{\sigma}{\sigma - \sigma_p} \]
Quick Tip: The dielectric constant K is a measure of how much a dielectric material can reduce the electric field.
\(K=1\) for vacuum, and \(K > 1\) for all materials.
The formula \( K = \frac{\sigma}{\sigma - \sigma_p} \) is a key relationship between the free charge density (\(\sigma\)) and the induced polarization charge density (\(\sigma_p\)).

Step 1: Energy Stored in a Capacitor

The energy U stored in a capacitor of capacitance C, charged to a potential difference \( V_{cap} \), is given by:
\[ U = \frac{1}{2}CV_{cap}^2 \]


Step 2: Relating C and \(V_{cap}\) to E and V

For a parallel plate capacitor with plate area A and separation d:

- Capacitance is \( C = \frac{\epsilon_0 A}{d} \).

- The uniform electric field is \( E = \frac{V_{cap}}{d} \), so \( V_{cap} = Ed \).

- The volume of the space between the plates is \( V = A \times d \).


Step 3: Deriving the Expression

Substitute the expressions for C and \(V_{cap}\) into the energy formula:
\[ U = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) (Ed)^2 \]
\[ U = \frac{1}{2} \frac{\epsilon_0 A}{d} (E^2 d^2) \]
\[ U = \frac{1}{2} \epsilon_0 E^2 (A \times d) \]

Since the volume of the space is \( V = A \times d \), we get:
\[ U = \frac{1}{2}\epsilon_0 E^2 V \]


Step 4: Analyzing the Options

- Option (A) \( \frac{1}{2}\epsilon_0 E^2 \) represents the energy density (energy per unit volume), not the total energy.

- Option (C) matches our derived result for the total energy stored.
Quick Tip: It is very important to distinguish between total energy and energy density.
- \textbf{Energy Density (u):} Energy per unit volume. For an electric field, \( u_E = \frac{1}{2}\epsilon_0 E^2 \).
- \textbf{Total Energy (U):} Energy density multiplied by the volume over which the field exists. \( U = u_E \times V \).
This concept is fundamental in electromagnetism.

Step 1: Analyzing the Circuit

Let the voltage of the battery be \( V_{bat} \).

The capacitors A, B, and M are connected in series with each other.
This series combination (let's call it \(C_{ABM}\)) is connected in parallel with capacitor N.
The entire parallel combination is connected across the battery.


Step 2: Equivalent Capacitance of the Series Combination

The equivalent capacitance of the three capacitors A, B, and M in series is:
\[ \frac{1}{C_{ABM}} = \frac{1}{C_A} + \frac{1}{C_B} + \frac{1}{C_M} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \]
\[ C_{ABM} = \frac{C}{3} \]


Step 3: Calculating Charges Q and Q'

Since the series branch (ABM) and capacitor N are in parallel, the voltage across both is the same and equal to the battery voltage, \( V_{bat} \).

The charge Q' on capacitor N is:
\[ Q' = C_N \times V_{bat} = (2C) V_{bat} \]

The charge Q on capacitor A is the same as the charge on the equivalent capacitor \(C_{ABM}\) (since charge is the same for capacitors in series).
\[ Q = C_{ABM} \times V_{bat} = \left(\frac{C}{3}\right) V_{bat} \]


Step 4: Finding the Ratio \( \frac{Q}{Q'} \)

Now, we can find the ratio of the charges:
\[ \frac{Q}{Q'} = \frac{(C/3) V_{bat}}{(2C) V_{bat}} \]

The terms C and \(V_{bat}\) cancel out.
\[ \frac{Q}{Q'} = \frac{1/3}{2} = \frac{1}{6} \]
Quick Tip: Remember the key rules for combinations:
- \textbf{Series Capacitors:} Same charge (Q), voltages add, \( \frac{1}{C_{eq}} = \sum \frac{1}{C_i} \).
- \textbf{Parallel Capacitors:} Same voltage (V), charges add, \( C_{eq} = \sum C_i \).
Breaking down complex circuits into simpler series and parallel parts is the standard approach.

Step 1: Initial Capacitance

The initial capacitance of the air-filled parallel plate capacitor is:
\[ C_0 = \frac{\epsilon_0 A}{d} \]


Step 2: Capacitance with the Dielectric Slab

When the dielectric slab of thickness d/2 is inserted, the capacitor can be viewed as a series combination of two capacitors.

- Capacitor 1 (\(C_1\)): Filled with the dielectric of constant K, with thickness \(t = d/2\).

\[ C_1 = \frac{K \epsilon_0 A}{d/2} = \frac{2K \epsilon_0 A}{d} \]

- Capacitor 2 (\(C_2\)): The remaining part, which is air-filled, with thickness \( d - t = d - d/2 = d/2 \).

\[ C_2 = \frac{\epsilon_0 A}{d/2} = \frac{2 \epsilon_0 A}{d} \]


Step 3: Calculating the Equivalent Capacitance C

Since these two parts are in series, the equivalent capacitance C is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]
\[ \frac{1}{C} = \frac{1}{(2K \epsilon_0 A / d)} + \frac{1}{(2 \epsilon_0 A / d)} = \frac{d}{2K \epsilon_0 A} + \frac{d}{2 \epsilon_0 A} \]
\[ \frac{1}{C} = \frac{d}{2 \epsilon_0 A} \left( \frac{1}{K} + 1 \right) = \frac{d}{2 \epsilon_0 A} \left( \frac{1+K}{K} \right) \]

Inverting to find C:
\[ C = \frac{2 \epsilon_0 A}{d} \left( \frac{K}{K+1} \right) \]


Step 4: Finding the Ratio \( \frac{C}{C_0} \)

Now, we find the ratio of the new capacitance C to the original capacitance \(C_0\).
\[ \frac{C}{C_0} = \frac{\frac{2 \epsilon_0 A}{d} \left( \frac{K}{K+1} \right)}{\frac{\epsilon_0 A}{d}} \]

The term \( \frac{\epsilon_0 A}{d} \) cancels out.
\[ \frac{C}{C_0} = 2 \left( \frac{K}{K+1} \right) = \frac{2K}{K+1} \]
Quick Tip: When a dielectric slab of thickness 't' is inserted into a capacitor of separation 'd', the general formula for the new capacitance is \( C = \frac{\epsilon_0 A}{d - t + t/K} \).
Applying this to our case, \(t=d/2\):
\( C = \frac{\epsilon_0 A}{d - d/2 + (d/2)/K} = \frac{\epsilon_0 A}{d/2 + d/(2K)} = \frac{\epsilon_0 A}{(d/2)(1 + 1/K)} = \frac{2\epsilon_0 A}{d( (K+1)/K )} = C_0 \frac{2K}{K+1} \).
This is a very useful formula to remember.

Part 1: Ray Diagram

The ray diagram for a compound microscope in normal adjustment (final image at infinity) is as follows:


Objective Lens: An object AB is placed just beyond the focal point \(F_o\) of the objective lens. The objective forms a real, inverted, and magnified intermediate image A'B'.

Eyepiece: The eyepiece is positioned such that the intermediate image A'B' is formed exactly at its focal point \(F_e\).

Final Image: Since the object for the eyepiece (A'B') is at its focal point, the rays emerging from the eyepiece are parallel. These parallel rays enter the eye, which perceives the final image as being virtual, highly magnified, and formed at infinity.






Part 2: Expression for Total Magnification

The total magnifying power (M) of a compound microscope is the product of the linear magnification produced by the objective (\(m_o\)) and the angular magnification produced by the eyepiece (\(m_e\)).
\[ M = m_o \times m_e \]

Magnification by Objective (\(m_o\)):

The linear magnification by the objective lens is given by \( m_o = \frac{v_o}{u_o} \), where \(v_o\) is the image distance and \(u_o\) is the object distance from the objective. For high magnification, the object is placed very close to the first focal point \(F_o\), so \( u_o \approx -f_o \). The intermediate image is formed close to the eyepiece, so \( v_o \approx L \), where L is the tube length (distance between the second focal point of the objective and the first focal point of the eyepiece).

So, \( m_o \approx -\frac{L}{f_o} \). The negative sign indicates a real and inverted image.


Magnification by Eyepiece (\(m_e\)):

When the final image is formed at infinity, the eyepiece acts as a simple microscope in normal adjustment. Its angular magnification is given by:
\[ m_e = \frac{D}{f_e} \]

where D is the least distance of distinct vision (typically 25 cm) and \(f_e\) is the focal length of the eyepiece.


Total Magnification (M):

Combining the two expressions, the total magnification is:
\[ M = \left(-\frac{L}{f_o}\right) \left(\frac{D}{f_e}\right) \]

This is the required expression for the magnifying power of a compound microscope when the final image is at infinity.
Quick Tip: The tube length L is formally defined as the distance between the second focal point of the objective and the first focal point of the eyepiece.
The approximation \(m_o \approx -L/f_o\) is very useful for problems involving normal adjustment.
Remember that for the final image at the near point (D), the eyepiece magnification becomes \(m_e = (1 + D/f_e)\).

Step 1: Find the position of the intermediate image (\(v_o\))

We use the lens formula for the objective lens: \( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \).

Given:

Object distance, \( u_o = -1.5 \) cm (by sign convention).

Focal length of objective, \( f_o = +1.25 \) cm.

Substituting the values:
\[ \frac{1}{1.25} = \frac{1}{v_o} - \frac{1}{-1.5} \]
\[ \frac{1}{1.25} = \frac{1}{v_o} + \frac{1}{1.5} \]
\[ \frac{1}{v_o} = \frac{1}{1.25} - \frac{1}{1.5} = \frac{4}{5} - \frac{2}{3} \]
\[ \frac{1}{v_o} = \frac{12 - 10}{15} = \frac{2}{15} \]
\[ v_o = \frac{15}{2} = 7.5 cm \]

The intermediate image is formed 7.5 cm from the objective lens.


Step 2: Position the eyepiece

The problem states that the final image is formed at infinity.
This condition implies that the intermediate image formed by the objective must be located at the first focal point of the eyepiece.
Therefore, the distance of the intermediate image from the eyepiece, \( |u_e| \), must be equal to the focal length of the eyepiece, \(f_e\).

Given \( f_e = 5 \) cm, so \( |u_e| = 5 \) cm.


Step 3: Calculate the distance between the lenses

The distance between the objective lens and the eyepiece (L) is the sum of the distance of the intermediate image from the objective (\(v_o\)) and its distance from the eyepiece (\(|u_e|\)).
\[ L = v_o + |u_e| \]
\[ L = 7.5 cm + 5 cm = 12.5 cm \]


Final Answer: The distance between the objective and the eyepiece is 12.5 cm.
Quick Tip: For a microscope in normal adjustment (final image at infinity), the length of the tube is always given by \( L = v_o + f_e \).
First, use the lens formula on the objective to find \(v_o\). Then, simply add \(f_e\) to it.
This is a standard two-step process for such problems.

Step 1: Huygens' Principle and Setup

Huygens' principle states that every point on a wavefront is a source of secondary wavelets that spread out in all directions with the speed of the wave in that medium. The new wavefront at any later time is the forward envelope (common tangent) of these secondary wavelets.

Consider a plane wavefront AB incident at an angle 'i' on a plane interface XY separating two media, Medium 1 (air, speed \(v_1\)) and Medium 2 (glass, speed \(v_2\)), where \(v_1 > v_2\).





Step 2: Construction of the Refracted Wavefront

As the wavefront AB strikes the interface, the point A becomes a source of secondary wavelets in Medium 2.
Let the wavefront take time 't' to travel from B to C. The distance covered is \( BC = v_1 t \).
During this time 't', the secondary wavelets from A travel a distance \( AE = v_2 t \) into Medium 2.
To find the new refracted wavefront, we draw a sphere (a circle in 2D) of radius \( AE = v_2 t \) with A as the center.
The tangent plane CE drawn from point C to this sphere represents the refracted wavefront.


Step 3: Verification of Snell's Law

Let 'i' be the angle of incidence and 'r' be the angle of refraction. From the geometry of the figure:

In the right-angled triangle ABC:
\[ \sin i = \frac{BC}{AC} = \frac{v_1 t}{AC} \]

In the right-angled triangle AEC:
\[ \sin r = \frac{AE}{AC} = \frac{v_2 t}{AC} \]

Now, divide the first equation by the second:
\[ \frac{\sin i}{\sin r} = \frac{v_1 t / AC}{v_2 t / AC} = \frac{v_1}{v_2} \]

The refractive index of a medium is defined as \( n = \frac{c}{v} \), where c is the speed of light in vacuum.
So, \( v_1 = c/n_1 \) and \( v_2 = c/n_2 \).
Substituting these into the equation:
\[ \frac{\sin i}{\sin r} = \frac{c/n_1}{c/n_2} = \frac{n_2}{n_1} \]
\[ \implies n_1 \sin i = n_2 \sin r \]

This is the mathematical expression for Snell's law of refraction.
Quick Tip: The key to this derivation is the construction based on the time taken.
The distance travelled by the wavefront in the first medium (\(v_1 t\)) is the hypotenuse for the angle of incidence.
The distance travelled by the secondary wavelet in the second medium (\(v_2 t\)) is the hypotenuse for the angle of refraction.
The common base AC links the two triangles, allowing the ratio to be taken.

Step 1: Mirror Formula and Sign Convention

The mirror formula relates the object distance (u), image distance (v), and focal length (f) of a spherical mirror:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

We will use the Cartesian sign convention, where the pole of the mirror is the origin and the direction of incident light is taken as positive.


Step 2: Applying Sign Convention to a Convex Mirror

For a convex mirror:

- The object is a real object placed in front of the mirror. Therefore, the object distance 'u' is always negative (\(u < 0\)).

- The focal point (F) of a convex mirror is behind the mirror. Therefore, its focal length 'f' is always positive (\(f > 0\)).


Step 3: Deducing the Nature of Image Distance (v)

Rearrange the mirror formula to solve for \( \frac{1}{v} \):
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]

Now, substitute the sign conventions:
\[ \frac{1}{v} = \frac{1}{(+f)} - \frac{1}{(-|u|)} \]
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{|u|} \]

Since f and |u| are both positive magnitudes, their reciprocals \( \frac{1}{f} \) and \( \frac{1}{|u|} \) are also positive quantities.

The sum of two positive quantities is always positive. Therefore:
\[ \frac{1}{v} > 0 \]

This implies that the image distance 'v' must always be positive.


Step 4: Conclusion

According to the sign convention, a positive image distance for a mirror means that the image is formed behind the mirror.
An image formed behind the mirror cannot be formed by the actual intersection of reflected rays; it is formed by their apparent intersection.
Therefore, the image is always virtual.

Thus, a convex mirror always produces a virtual image of a real object placed in front of it.
Quick Tip: The sign of the focal length is the key differentiator for mirrors.
Convex Mirror \(\implies\) Diverging \(\implies\) Positive Focal Length (\(f > 0\)).
Concave Mirror \(\implies\) Converging \(\implies\) Negative Focal Length (\(f < 0\)).
A positive image distance (\(v > 0\)) for a mirror always means a virtual image, while a negative image distance (\(v < 0\)) means a real image.

Step 1: Formula for Work Done

The work done by an external agent in moving a charge q from point A to point B in an electric field \( \vec{E} \) is given by the line integral:
\[ W_{ext} = -q \int_{A}^{B} \vec{E} \cdot d\vec{l} \]

For a unit positive charge, q = +1 C. The expression becomes:
\[ W = - \int_{A}^{B} \vec{E} \cdot d\vec{l} \]


Step 2: Setting up the Integral

The given electric field is \( \vec{E} = 40x \hat{i} \).

The differential displacement vector is \( d\vec{l} = dx \hat{i} + dy \hat{j} \).

The dot product is \( \vec{E} \cdot d\vec{l} = (40x \hat{i}) \cdot (dx \hat{i} + dy \hat{j}) = 40x \, dx \).

The initial point is A = (0, 3m) and the final point is B = (5m, 0).

The integral for the work done is:
\[ W = - \int_{(0,3)}^{(5,0)} 40x \, dx \]


Step 3: Evaluating the Integral

Since the integrand depends only on x, the limits of integration for x are from \(x_A = 0\) to \(x_B = 5\). The path taken in the y-direction does not affect the work done because the force is purely in the x-direction.
\[ W = -40 \int_{0}^{5} x \, dx \]
\[ W = -40 \left[ \frac{x^2}{2} \right]_{0}^{5} \]
\[ W = -40 \left( \frac{5^2}{2} - \frac{0^2}{2} \right) \]
\[ W = -40 \left( \frac{25}{2} \right) \]
\[ W = -20 \times 25 = -500 J \]


Final Answer: The amount of work done is -500 J.
Quick Tip: The electric field \( \vec{E} = 40x \hat{i} \) is a conservative field.
This means the work done is independent of the path taken and depends only on the initial and final positions.
You can also solve this by finding the potential difference: \( V(x) = -\int E_x dx = -\int 40x dx = -20x^2 \).
\( W = q(V_B - V_A) = 1 \times (V(5) - V(0)) = (-20(5^2)) - (-20(0^2)) = -500 \) J.

Step 1: Distribute the Charge Q

Let the charges on the inner and outer spheres be \(q_1\) and \(q_2\) respectively.

Total charge: \( q_1 + q_2 = Q \quad \dots(1) \)

The surface charge densities (\(\sigma\)) are equal:
\[ \sigma = \frac{q_1}{Area_1} = \frac{q_2}{Area_2} \implies \frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2} \]
\[ q_2 = q_1 \frac{R^2}{r^2} \quad \dots(2) \]

Substitute (2) into (1):
\[ q_1 + q_1 \frac{R^2}{r^2} = Q \implies q_1 \left( 1 + \frac{R^2}{r^2} \right) = Q \implies q_1 = Q \frac{r^2}{r^2 + R^2} \]

And substituting back into (2):
\[ q_2 = \left( Q \frac{r^2}{r^2 + R^2} \right) \frac{R^2}{r^2} = Q \frac{R^2}{r^2 + R^2} \]


(I) Electric Field at the Common Centre

The electric field inside a uniformly charged hollow spherical shell is zero.

The common centre lies inside both the inner sphere and the outer sphere.

- Electric field at the centre due to the charge \(q_1\) on the inner sphere is \(E_1 = 0\).

- Electric field at the centre due to the charge \(q_2\) on the outer sphere is \(E_2 = 0\).

The total electric field at the common centre is the vector sum:
\[ \vec{E}_{centre} = \vec{E}_1 + \vec{E}_2 = 0 \]


(II) Potential at the Common Centre

The electric potential at any point inside a uniformly charged hollow spherical shell is constant and equal to the potential at its surface (\( V = \frac{1}{4\pi\epsilon_0} \frac{q}{radius} \)).

The total potential at the centre is the scalar sum of the potentials due to each sphere.

- Potential at the centre due to the inner sphere: \( V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r} \).

- Potential at the centre due to the outer sphere: \( V_2 = \frac{1}{4\pi\epsilon_0} \frac{q_2}{R} \).

Total potential:
\[ V_{centre} = V_1 + V_2 = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right) \]

Substitute the expressions for \(q_1\) and \(q_2\):
\[ V_{centre} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q r^2}{(r^2 + R^2)r} + \frac{Q R^2}{(r^2 + R^2)R} \right) \]
\[ V_{centre} = \frac{1}{4\pi\epsilon_0} \left( \frac{Q r}{r^2 + R^2} + \frac{Q R}{r^2 + R^2} \right) \]
\[ V_{centre} = \frac{Q(r+R)}{4\pi\epsilon_0(r^2+R^2)} \]
Quick Tip: Key results from Gauss's Law for spherical shells are crucial here:
- \textbf{E-field inside} a shell is always zero.
- \textbf{Potential inside} a shell is constant and equals the potential on its surface.
Remember that potential is a scalar, so you simply add the potentials from all sources algebraically.

Part 1: Derivation of Electric Field on Equatorial Plane

Consider an electric dipole consisting of charges -q and +q separated by a distance 2a. Let P be a point on the equatorial plane at a distance r from the centre of the dipole.

The distance of P from both charges is \( \sqrt{r^2 + a^2} \).

The magnitude of the electric field at P due to +q is \( E_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2} \).

The magnitude of the electric field at P due to -q is \( E_- = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2} \). So, \( E_+ = E_- \).

The vertical components of these fields (\(E_+ \sin\theta\) and \(E_- \sin\theta\)) are equal and opposite, so they cancel out.

The horizontal components (\(E_+ \cos\theta\) and \(E_- \cos\theta\)) add up.

The resultant field is \( E_{eq} = E_+ \cos\theta + E_- \cos\theta = 2E_+ \cos\theta \).

From the geometry, \( \cos\theta = \frac{a}{\sqrt{r^2 + a^2}} \).
\[ E_{eq} = 2 \left( \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2} \right) \left( \frac{a}{\sqrt{r^2 + a^2}} \right) = \frac{1}{4\pi\epsilon_0} \frac{2qa}{(r^2 + a^2)^{3/2}} \]

Since the dipole moment magnitude is \( p = q \times 2a \), the expression is:
\[ E_{eq} = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + a^2)^{3/2}} \]

Direction: The direction of the field is from P towards the left, which is opposite to the direction of the dipole moment vector \( \vec{p} \) (which points from -q to +q).

In vector form: \( \vec{E}_{eq} = - \frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{(r^2 + a^2)^{3/2}} \).


Part 2: Value of Electric Field at specific points

(I) At the centre of the dipole (r = 0):

Substitute r=0 into the expression:
\[ E_{centre} = \frac{1}{4\pi\epsilon_0} \frac{p}{(0^2 + a^2)^{3/2}} = \frac{1}{4\pi\epsilon_0} \frac{p}{(a^2)^{3/2}} = \frac{p}{4\pi\epsilon_0 a^3} \]


(II) At a point r \(>>\) a:

For a point far from the dipole, we can neglect \(a^2\) in comparison to \(r^2\). So, \( r^2 + a^2 \approx r^2 \).
\[ E_{far} \approx \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2)^{3/2}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \]

This shows the field of a dipole falls off as \(1/r^3\).
Quick Tip: Remember the key differences between axial and equatorial fields for a dipole:
- \textbf{Axial Field:} \( \vec{E}_{axial} \approx \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{r^3} \) (parallel to \( \vec{p} \)).
- \textbf{Equatorial Field:} \( \vec{E}_{eq} \approx - \frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3} \) (antiparallel to \( \vec{p} \)).
The axial field is twice as strong as the equatorial field for the same distance r.

Step 1: Analyzing the Electric Field and Flux

The electric field is non-uniform and directed only along the x-axis: \( \vec{E} = (10x + 5) \hat{i} \).

According to Gauss's law, the net electric flux through a closed surface is \( \Phi_{net} = \oint \vec{E} \cdot d\vec{A} \).

For the cube, flux will pass only through the faces perpendicular to the electric field (i.e., the faces perpendicular to the x-axis). For the four faces parallel to the x-axis (top, bottom, front, back), the area vector \( d\vec{A} \) is perpendicular to \( \vec{E} \), so \( \vec{E} \cdot d\vec{A} = 0 \). Their flux contribution is zero.

Let's assume the cube is placed with its back face (A1-A-B-Z plane) in the y-z plane at \(x=0\), and it extends to \(x=L\).


Step 2: Flux through the Left Face (at x=0)

The electric field at this face is \( \vec{E}_{left} = (10(0) + 5) \hat{i} = 5 \hat{i} \) N/C.

The area vector for the left face points outwards, i.e., in the negative x-direction: \( \vec{A}_{left} = -L^2 \hat{i} \).

The flux through the left face is:
\[ \Phi_{left} = \vec{E}_{left} \cdot \vec{A}_{left} = (5 \hat{i}) \cdot (-L^2 \hat{i}) = -5L^2 N m^2/C \]


Step 3: Flux through the Right Face (at x=L)

The electric field at this face is \( \vec{E}_{right} = (10L + 5) \hat{i} \) N/C.

The area vector for the right face points outwards, i.e., in the positive x-direction: \( \vec{A}_{right} = +L^2 \hat{i} \).

The flux through the right face is:
\[ \Phi_{right} = \vec{E}_{right} \cdot \vec{A}_{right} = ((10L + 5) \hat{i}) \cdot (L^2 \hat{i}) = (10L + 5)L^2 = (10L^3 + 5L^2) N m^2/C \]


Step 4: Calculating the Net Flux

The net flux through the cube is the sum of the fluxes through all faces.
\[ \Phi_{net} = \Phi_{left} + \Phi_{right} + \Phi_{other\_faces} \]
\[ \Phi_{net} = (-5L^2) + (10L^3 + 5L^2) + 0 \]
\[ \Phi_{net} = 10L^3 N m^2/C \]
Quick Tip: Alternatively, you can use the differential form of Gauss's Law: \( \Phi_{net} = \frac{Q_{enc}}{\epsilon_0} \).
The enclosed charge can be found from the volume charge density \( \rho = \epsilon_0 (\nabla \cdot \vec{E}) \).
Here, \( \nabla \cdot \vec{E} = \frac{\partial E_x}{\partial x} = \frac{\partial}{\partial x}(10x+5) = 10 \).
So, \( \rho = 10\epsilon_0 \).
The total enclosed charge is \( Q_{enc} = \rho \times Volume = (10\epsilon_0) \times L^3 \).
The net flux is \( \Phi_{net} = \frac{10\epsilon_0 L^3}{\epsilon_0} = 10L^3 \). This confirms the result.

Principle:

An AC generator works on the principle of electromagnetic induction. It states that whenever the magnetic flux linked with a coil changes, an electromotive force (emf) is induced in the coil. If the coil is part of a closed circuit, an induced current flows through it.


Labelled Diagram:





Working:


1. The armature coil ABCD is rotated in the uniform magnetic field provided by the permanent magnets.

2. As the coil rotates, the angle between the magnetic field vector and the area vector of the coil changes continuously. This causes the magnetic flux (\(\Phi = NBA \cos\theta\)) linked with the coil to change.

3. According to Faraday's law of induction, this change in flux induces an emf in the coil.

4. Consider one rotation: Initially, the coil is vertical. As it rotates, arm AB moves down and arm CD moves up. By applying Fleming's Right-Hand Rule, the induced current flows from A to B in arm AB and from C to D in arm CD. The current flows out through brush B1, through the external load, and back in through brush B2.

5. After half a rotation, the positions of the arms are interchanged. Arm AB now moves up, and arm CD moves down. The direction of the induced current in each arm reverses (now from B to A and D to C).

6. The current now flows out through brush B2 and back in through brush B1. The slip rings ensure that each brush remains in contact with its respective arm's connection, so the direction of current in the external circuit also reverses every half rotation.

7. This periodic reversal of current direction results in an alternating current (AC). The induced emf is sinusoidal in nature, given by \( \epsilon = \epsilon_0 \sin(\omega t) \), where \( \epsilon_0 = NBA\omega \) is the peak emf.
Quick Tip: The key difference between an AC generator and a DC generator lies in the connection to the external circuit.
- \textbf{AC Generator} uses \textbf{slip rings} which allow the current to reverse its direction in the external circuit every half rotation.
- \textbf{DC Generator} uses a \textbf{split-ring commutator} which reverses the connection every half rotation, ensuring the current in the external circuit always flows in the same direction.

Step 1: Extract Information and Calculate Reactances


From the source voltage equation \( v = 140 \sin(100\pi t) \):

Peak voltage \( V_0 = 140 \) V.

Angular frequency \( \omega = 100\pi \) rad/s.

Given: \( R = 400 \, \Omega \), \( L = \frac{5}{\pi} \, H \), \( C = \frac{50}{\pi} \times 10^{-6} \, F \).

Inductive Reactance: \( X_L = \omega L = (100\pi) \left(\frac{5}{\pi}\right) = 500 \, \Omega \).

Capacitive Reactance: \( X_C = \frac{1}{\omega C} = \frac{1}{(100\pi) \left(\frac{50}{\pi} \times 10^{-6}\right)} = \frac{1}{5000 \times 10^{-6}} = 200 \, \Omega \).


Step 2: Calculate Impedance and RMS Current


Impedance \( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{400^2 + (500 - 200)^2} \).
\( Z = \sqrt{400^2 + 300^2} = \sqrt{160000 + 90000} = \sqrt{250000} = 500 \, \Omega \).

Source RMS Voltage: \( V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{140}{\sqrt{2}} = 70\sqrt{2} \) V \( \approx 98.99 \) V.

RMS Current: \( I_{rms} = \frac{V_{rms}}{Z} = \frac{70\sqrt{2}}{500} = \frac{7\sqrt{2}}{50} \) A \( \approx 0.198 \) A.


Step 3: Calculate RMS Voltages across R, L, and C


RMS Voltage across Resistor:
\( V_{R, rms} = I_{rms} \times R = \frac{7\sqrt{2}}{50} \times 400 = 56\sqrt{2} \) V \( \approx 79.2 \) V.

RMS Voltage across Inductor:
\( V_{L, rms} = I_{rms} \times X_L = \frac{7\sqrt{2}}{50} \times 500 = 70\sqrt{2} \) V \( \approx 99.0 \) V.

RMS Voltage across Capacitor:
\( V_{C, rms} = I_{rms} \times X_C = \frac{7\sqrt{2}}{50} \times 200 = 28\sqrt{2} \) V \( \approx 39.6 \) V.


Step 4: Explanation

Algebraic sum of voltages = \( V_{R, rms} + V_{L, rms} + V_{C, rms} = 56\sqrt{2} + 70\sqrt{2} + 28\sqrt{2} = 154\sqrt{2} \) V \( \approx 217.8 \) V.

This sum (217.8 V) is clearly greater than the source RMS voltage (99.0 V).

Explanation: This occurs because the voltages across the resistor, inductor, and capacitor are not in phase with each other. In a series LCR circuit, the voltage across the inductor (\(V_L\)) leads the current by 90°, while the voltage across the capacitor (\(V_C\)) lags the current by 90°. This makes \(V_L\) and \(V_C\) 180° out of phase. The voltage across the resistor (\(V_R\)) is in phase with the current.

The total source voltage is the phasor sum (vector sum) of the individual voltages, not their simple algebraic sum. The phasor sum is given by:
\[ V_{rms} = \sqrt{V_{R,rms}^2 + (V_{L,rms} - V_{C,rms})^2} \]
\[ V_{rms} = \sqrt{(56\sqrt{2})^2 + (70\sqrt{2} - 28\sqrt{2})^2} = \sqrt{(56\sqrt{2})^2 + (42\sqrt{2})^2} \]
\[ V_{rms} = \sqrt{2(56^2 + 42^2)} = \sqrt{2(3136 + 1764)} = \sqrt{2(4900)} = 70\sqrt{2} V \]

This matches the source voltage. Because the individual voltages peak at different times, their simple sum can exceed the source voltage at any given instant.
Quick Tip: In an AC circuit, always treat voltages (and currents in parallel circuits) as phasors (vectors).
They must be added vectorially, taking their phase differences into account.
The algebraic sum of RMS voltages across components in a series AC circuit is only equal to the source voltage if all components are purely resistive.

Principle:

A transformer works on the principle of mutual induction. It states that if two coils are magnetically coupled, a changing current in one coil (the primary coil) induces an electromotive force (emf) in the other coil (the secondary coil) due to the change in magnetic flux linked with it.


Labelled Diagram of a Step-up Transformer:




Working of a Step-up Transformer:

1. An alternating voltage source is connected to the primary coil. This causes an alternating current to flow through it.

2. This alternating primary current produces a continuously changing magnetic flux in the laminated soft iron core. The soft iron core concentrates the magnetic field lines, ensuring that almost all the flux from the primary coil links with the secondary coil (ideal case).

3. According to Faraday's law of induction, this changing magnetic flux induces an alternating emf (and hence voltage) in the secondary coil.

4. Let \(\Phi\) be the magnetic flux linked with each turn of the coils. The induced emf in the primary and secondary coils are:

\( \epsilon_p = -N_p \frac{d\Phi}{dt} \) and \( \epsilon_s = -N_s \frac{d\Phi}{dt} \)

5. Assuming an ideal transformer with no flux leakage, the ratio of the secondary voltage (\(V_s\)) to the primary voltage (\(V_p\)) is equal to the ratio of the number of turns in the coils:

\[ \frac{V_s}{V_p} \approx \frac{\epsilon_s}{\epsilon_p} = \frac{N_s}{N_p} \]

6. For a step-up transformer, the number of turns in the secondary coil is greater than the number of turns in the primary coil (\(N_s > N_p\)).

7. Consequently, the output voltage is greater than the input voltage (\(V_s > V_p\)), i.e., the voltage is "stepped up".

8. For an ideal transformer, the output power equals the input power (\(P_{out} = P_{in}\)), which means \( V_s I_s = V_p I_p \). Since \(V_s > V_p\), it follows that \(I_s < I_p\). The current is stepped down.
Quick Tip: The key to a transformer's function is \textbf{changing} flux. This is why transformers only work with AC, not DC.
A constant DC current produces a constant magnetic flux, which does not induce any emf in the secondary coil (except at the moments of switching on or off).
The ratio \( k = N_s / N_p \) is called the transformation ratio. For step-up, \(k>1\); for step-down, \(k<1\).

There is an ambiguity in the problem statement. The "design" voltages (50V to 250V) suggest a transformation ratio, while the "operating" voltage source is given by an equation. We will assume the "design" gives the turns ratio, and the operation is with the given source.

Transformation Ratio \( k = \frac{N_s}{N_p} = \frac{V_{s, design}}{V_{p, design}} = \frac{250}{50} = 5 \).


(I) RMS value of input current (\(I_{i, rms}\))

Power drawn from the source, \( P_{in} = 200 \) W.

From the operating voltage source equation \( v_i = 20 \sin(100\pi t) \), the peak input voltage is \( V_{i, peak} = 20 \) V.

The RMS value of the operating input voltage is:
\[ V_{i, rms} = \frac{V_{i, peak}}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} V \]

The RMS input current is calculated from the power and RMS voltage:
\[ I_{i, rms} = \frac{P_{in}}{V_{i, rms}} = \frac{200}{10\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} A \]
\[ I_{i, rms} \approx 14.14 A \]


(II) Expression for instantaneous output voltage (\(v_o\))

The transformation ratio applies to peak voltages as well: \( \frac{V_{o, peak}}{V_{i, peak}} = k = 5 \).

Peak output voltage: \( V_{o, peak} = k \times V_{i, peak} = 5 \times 20 = 100 \) V.

The frequency and phase (assuming a resistive load) remain the same. The angular frequency is \( \omega = 100\pi \) rad/s.

The expression for the instantaneous output voltage is:
\[ v_o = V_{o, peak} \sin(\omega t) = 100 \sin(100\pi t) V \]


(III) Expression for instantaneous output current (\(i_o\))

For an ideal transformer, input power equals output power: \( P_{out} = P_{in} = 200 \) W.

The RMS output voltage is \( V_{o, rms} = \frac{V_{o, peak}}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 50\sqrt{2} \) V.

The RMS output current is:
\[ I_{o, rms} = \frac{P_{out}}{V_{o, rms}} = \frac{200}{50\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} A \]

The peak output current is:
\[ I_{o, peak} = I_{o, rms} \times \sqrt{2} = (2\sqrt{2}) \times \sqrt{2} = 4 A \]

Assuming the current is in phase with the voltage, the expression for the instantaneous output current is:
\[ i_o = I_{o, peak} \sin(\omega t) = 4 \sin(100\pi t) A \]
Quick Tip: For an ideal transformer, the following relations hold:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} = k \]
Power is conserved: \( P_{in} = P_{out} \).
When given conflicting information, state your assumption clearly. Here, using the design specs to find the turns ratio and then applying it to the given operating source is the most logical approach.