The statement describes the Zeroth Law of Thermodynamics. This law defines the concept of temperature. If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other. This means they are at the same temperature. Quick Tip: Remember the laws: {Zeroth} defines Temperature. {First} is about Conservation of Energy. {Second} is about Entropy (disorder).

Bernoulli's theorem is a statement of the principle of conservation of energy for a flowing fluid. It states that for an ideal fluid in steady flow, the sum of its pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant along a streamline. Therefore, it is the total energy of the fluid that is conserved. Quick Tip: Bernoulli's theorem = Conservation of Energy for fluids. The equation \( P + \frac{1}{2}\rho v^2 + \rho gh = constant \) represents pressure energy, kinetic energy, and potential energy.

A diathermanous substance is one that is transparent to thermal radiation (heat). An athermanous substance is opaque to thermal radiation. Wood, iron, and copper are athermanous; they absorb heat but do not transmit it as radiation. Glass is largely transparent to visible light but is also partially transparent to infrared radiation, making it the most diathermanous substance among the choices, though materials like rock salt are better examples. However, there seems to be a common misconception in the question as wood is typically athermanous. If the intended answer is 'wood' it might be based on a specific, less common context or an error in the question. Assuming standard physics definitions, glass would be the better answer. If we must choose from the options as presented, and there's a possibility of error, let's re-evaluate. A better interpretation might be what allows heat transfer. All listed options conduct heat, but transparency to heat radiation is key. Let's assume the question is flawed and selects the intended common-knowledge answer which might be (a) in some contexts. (Note: In standard physics, glass is the correct answer. This solution reflects the option provided). Let's proceed with wood as per some curriculum's interpretation. Wood is an insulator, it does not allow heat to pass through it easily. Substances that do not allow heat radiation to pass are athermanous. Substances that allow heat radiation to pass are diathermanous. The question is likely asking which is a poor conductor of heat, but uses the wrong term. In that case, wood is the poorest conductor. Let's stick to the definition: Wood is athermanous. Glass is diathermanous. Iron and copper are athermanous. The correct answer should be (C) Glass. Assuming there is an error in the provided answer key and selecting the scientifically accurate one. Let's re-answer assuming the question or options are tricky. Given the options, let's re-evaluate. Wood is opaque to IR, iron and copper are opaque, glass is largely transparent to visible and near-IR but opaque to far-IR. Perhaps the question is about transmission in general. Let's assume the provided answer (A) is correct and try to justify it. There is no standard physical justification for wood being diathermanous. It is a classic insulator and athermanous. The question is flawed. Let's provide the correct physics answer.

The correct answer based on physics principles is (C) glass. Glass transmits light and a significant portion of heat radiation. Wood, iron, and copper are opaque to thermal radiation (athermanous). Quick Tip: {Diathermanous} = Transmits heat radiation (like transparent glass). {Athermanous} = Opaque to heat radiation (like a wall or wood).

The electric potential (V) created by a point charge (q) at a distance (r) from the charge is given by the formula: \[ V = k \frac{q}{r} \]
where k is Coulomb's constant. From this formula, it is clear that the electric potential V is inversely proportional to the distance r, or directly proportional to \( \frac{1}{r} \). Quick Tip: Remember the difference: Electric {Potential} (\(V\)) is proportional to \( \frac{1}{r} \). Electric {Field} (\(E\)) is proportional to \( \frac{1}{r^2} \).

In a potentiometer experiment, we measure the emf (E) of the cell (open circuit) and the terminal voltage (V) when a known external resistance (R) is connected (closed circuit). The emf corresponds to a balancing length \( l_1 \), and the terminal voltage corresponds to a balancing length \( l_2 \). So, \( E \propto l_1 \) and \( V \propto l_2 \).
The relationship between emf, terminal voltage, and internal resistance (r) is \( E = I(R+r) \) and \( V = IR \).
Dividing the two equations gives \( \frac{E}{V} = \frac{R+r}{R} = 1 + \frac{r}{R} \).
Rearranging for r: \( \frac{r}{R} = \frac{E}{V} - 1 \), which gives \( r = R \left( \frac{E}{V} - 1 \right) \). Since \( E/V = l_1/l_2 \), this is also written as \( r = R \left( \frac{l_1}{l_2} - 1 \right) \). Quick Tip: The internal resistance causes a "voltage drop" inside the cell, so the terminal voltage V is always less than the emf E when current flows. The formula reflects this difference.

The de-Broglie wavelength (\(\lambda\)) is related to kinetic energy (K) and mass (m) by the equation: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \]
where h is Planck's constant. Since the kinetic energy K is the same for all particles, the wavelength \(\lambda\) is inversely proportional to the square root of the mass (\( \lambda \propto \frac{1}{\sqrt{m}} \)).
To have the longest wavelength, the particle must have the smallest mass.
Comparing the masses:
Mass of electron (\(m_e\)) < Mass of proton (\(m_p\)) < Mass of hydrogen atom (\(m_H \approx m_p\)) < Mass of \(\alpha\)-particle (\(m_\alpha \approx 4m_p\)).
The electron has the smallest mass, and therefore it will have the longest de-Broglie wavelength. Quick Tip: De-Broglie Wavelength: Lighter particles have longer wavelengths for the same energy. Think of it as smaller things having "wavier" properties.

Let's check the condition "both inputs are high (1)" for each gate:

AND gate: Output is 1 only if BOTH inputs are 1. (1 AND 1 = 1). This matches the condition.
XOR gate: Output is 1 if inputs are different. (1 XOR 1 = 0).
NAND gate: Output is the opposite of AND. (1 NAND 1 = 0).
NOR gate: Output is the opposite of OR. (1 OR 1 = 1, so 1 NOR 1 = 0).

The only gate that produces a high output (1) when both inputs are high is the AND gate. Quick Tip: {AND:} Output is high ONLY IF all inputs are high. {OR:} Output is high IF ANY input is high.

The power (P) in a rotational system is given by the product of torque (\(\tau\)) and angular speed (\(\omega\)). \[ P = \tau \times \omega \]
Given:
Torque, \(\tau = 2\) Nm
Angular speed, \(\omega = 2\pi\) rad/s
Substituting the values: \[ P = 2 \, Nm \times 2\pi \, rad/s = 4\pi \, W \]
The power rating of the ceiling fan is \(4\pi\) W. Quick Tip: Rotational power is the rotational analogue of linear power (\(P = F \times v\)). Just replace force (F) with torque (\(\tau\)) and linear velocity (v) with angular velocity (\(\omega\)).

When a string vibrates in 'n' loops, its length 'L' is equal to 'n' times half the wavelength (\(\lambda\)). \[ L = n \frac{\lambda}{2} \]
Given:
Length of the string, \(L = 2\) m
Number of loops, \(n = 2\)
Substituting the values: \[ 2 = 2 \times \frac{\lambda}{2} \] \[ \lambda = 2 \, m \]
The wavelength of the wave is 2 m.
The distance between a node and an adjacent antinode is always one-quarter of a wavelength (\(\frac{\lambda}{4}\)). \[ Distance = \frac{\lambda}{4} = \frac{2}{4} = 0.5 \, m \] Quick Tip: Key distances in a standing wave: Node to next Node = \(\lambda/2\) Antinode to next Antinode = \(\lambda/2\) Node to next Antinode = \(\lambda/4\)

For an ideal transformer, 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 secondary coil (\(N_s\)) to the number of turns in the primary coil (\(N_p\)). \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
Given:
Primary voltage, \(V_p = 220\) V
Secondary voltage, \(V_s = 880\) V
Number of turns in primary coil, \(N_p = 1000\)
Substituting the values: \[ \frac{880}{220} = \frac{N_s}{1000} \] \[ 4 = \frac{N_s}{1000} \] \[ N_s = 4 \times 1000 = 4000 \]
The number of turns in the secondary coil is 4000. Quick Tip: For a step-up transformer, the voltage increases, so the number of turns in the secondary coil must be greater than in the primary. For a step-down, it's the opposite. Here, voltage increased 4 times, so turns must also increase 4 times.

The surface tension of a liquid becomes zero at its critical temperature. At this temperature, the distinction between the liquid and gas phases disappears, and the intermolecular cohesive forces that cause surface tension become negligible. Quick Tip: The critical temperature is the point where a substance can no longer be liquefied by pressure alone. Think of it as the temperature where the liquid and gas states merge.

Self-inductance is the property of a coil by virtue of which it opposes any change in the strength of the electric current flowing through it by inducing an electromotive force (e.m.f.) in itself. It is numerically equal to the magnetic flux linked with the coil when a unit current flows through it. Its SI unit is the Henry (H). Quick Tip: Self-inductance is like electrical inertia. Just as inertia resists changes in motion, self-inductance resists changes in current.

The work done by the magnetic field is zero. The magnetic force on a moving charge (Lorentz force) is given by \( \vec{F} = q(\vec{v} \times \vec{B}) \). By the definition of the cross product, the force \( \vec{F} \) is always perpendicular to both the velocity \( \vec{v} \) and the magnetic field \( \vec{B} \). Since work done is \( W = \vec{F} \cdot \vec{d} \), and the displacement \( \vec{d} \) is in the direction of velocity, the force is always perpendicular to the displacement. Therefore, the dot product is zero, and no work is done. Quick Tip: A magnetic field can change the direction of a charged particle's motion (making it move in a circle), but it cannot change its speed or kinetic energy because it does no work on it.

A thermodynamic system is a well-defined macroscopic region of the universe that is selected for study. It consists of a specific quantity of matter with a distinct boundary (either real or imaginary) that separates it from its surroundings. A system can interact with its surroundings by exchanging energy (in the form of heat and work) and/or matter. Quick Tip: Think of it as putting an imaginary box around what you want to study. Everything inside the box is the "system," and everything outside is the "surroundings."

In a hysteresis loop, when the external magnetizing field (H) is reduced to zero, the remaining magnetic flux density (B) in the material is called Retentivity or Remanence. It represents the ability of a magnetic material to retain its magnetism after the external magnetizing field is removed. Quick Tip: {Retentivity} = {Retained} magnetism. It's the memory of the magnetic field that was applied.

The formula for the ideal angle of banking (\(\theta\)) for a vehicle moving at speed \(v\) on a curved road with radius \(r\) is: \[ \tan \theta = \frac{v^2}{rg} \]
where \(g\) is the acceleration due to gravity. This angle allows the vehicle to make the turn without relying on friction. Quick Tip: To go faster or turn tighter (smaller r), you need a steeper bank (larger \(\theta\)). This is why race tracks have very steep banks in the corners.

For a large (ideally infinite) charged plane sheet, the electric field intensity (E) at a point near it is uniform and given by Gauss's law. Assuming it's a non-conducting sheet: \[ E = \frac{\sigma}{2\epsilon_0} \]
where \(\sigma\) is the surface charge density and \(\epsilon_0\) is the permittivity of free space (\( 8.85 \times 10^{-12} \, C^2/N \cdot m^2 \)).
Given: \(\sigma = 8.85 \, \mu C/m^2 = 8.85 \times 10^{-6} \, C/m^2 \). \[ E = \frac{8.85 \times 10^{-6} \, C/m^2}{2 \times 8.85 \times 10^{-12} \, C^2/N \cdot m^2} \] \[ E = \frac{10^{-6}}{2 \times 10^{-12}} = 0.5 \times 10^6 \, N/C \]
The electric field intensity is \( 5 \times 10^5 \, N/C \). The distance from the mid-point does not affect the result for a large plane sheet. Quick Tip: The electric field from an infinite sheet is constant everywhere in space. For a conducting sheet, the formula would be \( E = \sigma / \epsilon_0 \), which is twice as strong.

The total translational kinetic energy of an ideal gas is given by the formula: \[ K.E. = \frac{3}{2} PV \]
At Standard Temperature and Pressure (S.T.P.):
Pressure, \( P = 1 \, atm \approx 1.013 \times 10^5 \, Pa \)
Volume, \( V = 1 \, litre = 10^{-3} \, m^3 \)
Substituting the values: \[ K.E. = \frac{3}{2} \times (1.013 \times 10^5 \, Pa) \times (10^{-3} \, m^3) \] \[ K.E. = 1.5 \times 1.013 \times 10^2 \, J \] \[ K.E. \approx 152 \, J \]
The kinetic energy of 1 litre of an ideal gas at S.T.P. is approximately 152 Joules. Quick Tip: The formula \( K.E. = \frac{3}{2} PV \) is a direct and quick way to find the total kinetic energy if you know the pressure and volume, avoiding the need to calculate the number of moles.

In the context of standing waves produced in a medium (like a string or an air column), harmonics and overtones describe the different frequencies at which the medium can naturally vibrate.


Fundamental Frequency (\(n_0\)): This is the lowest possible frequency at which a system can vibrate. It is also called the first harmonic.

Harmonics: Harmonics are all the integer multiples of the fundamental frequency. The series of frequencies is given by \(n_0, 2n_0, 3n_0, 4n_0, \ldots\). The first harmonic is the fundamental frequency itself, the second harmonic is \(2n_0\), the third is \(3n_0\), and so on. All possible vibrational modes are harmonics.

Overtones: Overtones are any frequencies that can be produced by the system that are higher than the fundamental frequency. The first overtone is the next possible frequency above the fundamental, the second overtone is the next frequency after that, and so on.


Relationship:

For a system where all harmonics are present (e.g., a string fixed at both ends, an open organ pipe), the first overtone corresponds to the second harmonic, the second overtone corresponds to the third harmonic, and the \(p^{th}\) overtone corresponds to the \((p+1)^{th}\) harmonic.
For a system where only odd harmonics are present (e.g., a closed organ pipe), the first overtone corresponds to the third harmonic, the second overtone corresponds to the fifth harmonic, etc. Quick Tip: {Harmonics} are a complete musical scale (1x, 2x, 3x... of the base note). {Overtones} are just the notes "over" the base one (the 2nd, 3rd, 4th... possible notes). They are often the same, but not always.

Newton's Law of Viscosity states that for a streamline flow of a fluid, the viscous force (\(F\)) acting between two adjacent layers is directly proportional to the area of the layers (\(A\)) and the velocity gradient (\(dv/dx\)) between them. \[ F \propto A \frac{dv}{dx} \]
Introducing a constant of proportionality, \(\eta\), we get: \[ F = -\eta A \frac{dv}{dx} \]
The negative sign indicates that the viscous force opposes the relative motion between the layers. The constant \(\eta\) is called the coefficient of viscosity.

Derivation of the Expression for \(\eta\):
From the formula above, we can rearrange the terms to solve for the coefficient of viscosity, \(\eta\). Considering only the magnitude of the force: \[ \eta = \frac{F}{A \frac{dv}{dx}} \]
This is the expression for the coefficient of viscosity.

\(F\) is the tangential viscous force.
\(A\) is the area of the layer.
\(\frac{dv}{dx}\) is the velocity gradient, which is the rate of change of velocity with distance perpendicular to the direction of flow.

The SI unit of the coefficient of viscosity is Pascal-second (Pa·s) or N·s/m². Quick Tip: Think of viscosity (\(\eta\)) as the "stickiness" or internal friction of a fluid. The formula shows it's the force required to slide layers past each other at a certain speed. High \(\eta\) means high force is needed (like for honey).

According to the Biot-Savart law, the magnetic induction (\(B\)) at the centre of a circular arc of wire with radius \(r\), carrying a current \(I\), and subtending an angle \(\theta\) (in radians) at the centre is given by: \[ B = \frac{\mu_0 I \theta}{4\pi r} \]
where \(\mu_0\) is the permeability of free space.

Magnetic Induction at the Centre of a Circular Loop:
To find the magnetic induction at the centre of a full circular loop, we consider the arc to be a complete circle.
For a complete circle, the angle \(\theta\) subtended at the centre is \(2\pi\) radians.
Substituting \(\theta = 2\pi\) into the formula for a circular arc: \[ B_{loop} = \frac{\mu_0 I (2\pi)}{4\pi r} \] \[ B_{loop} = \frac{\mu_0 I}{2r} \]
This is the formula for the magnetic induction at the centre of a current-carrying circular loop. Quick Tip: Start with the general arc formula \(B = \frac{\mu_0 I \theta}{4\pi r}\). For a full circle, the angle \(\theta\) is \(2\pi\). For a semi-circle, \(\theta\) is \(\pi\). This makes it easy to find the field for any part of a circle.

Statement: The law of conservation of angular momentum states that if the net external torque acting on a system is zero, then the total angular momentum of the system remains constant (conserved).

Proof:
The angular momentum (\(\vec{L}\)) of a system is related to its moment of inertia (\(I\)) and angular velocity (\(\vec{\omega}\)) by \(\vec{L} = I\vec{\omega}\).
The relationship between torque (\(\vec{\tau}\)) and angular momentum (\(\vec{L}\)) is that the net external torque is equal to the time rate of change of the angular momentum. \[ \vec{\tau}_{ext} = \frac{d\vec{L}}{dt} \]
According to the law, if the net external torque on the system is zero, then \(\vec{\tau}_{ext} = 0\).
Substituting this into the equation: \[ 0 = \frac{d\vec{L}}{dt} \]
This implies that \(\vec{L}\) is a constant with respect to time. The derivative of a constant is zero.
Therefore, \(\vec{L} = constant\).
This proves that the angular momentum of the system is conserved when no net external torque acts on it. Quick Tip: This is the rotational equivalent of the conservation of linear momentum. No external force means linear momentum is constant. No external torque means angular momentum is constant. Think of an ice skater spinning faster when she pulls her arms in.

Four advantages of Light Emitting Diodes (LEDs) are:

High Energy Efficiency: LEDs convert a much larger percentage of electrical energy directly into light energy compared to incandescent bulbs, which lose most of their energy as heat. This makes them highly efficient and reduces power consumption.
Long Lifespan: LEDs have a very long operational life, typically lasting tens of thousands of hours, which is significantly longer than traditional incandescent or fluorescent lamps.
Durability and Compact Size: Being solid-state devices, LEDs are very durable, resistant to shock and vibrations, and can be made very small, allowing for flexible and compact designs.
Fast Switching and Controllability: LEDs can be turned on and off almost instantaneously and can be easily dimmed and controlled, making them ideal for applications requiring rapid switching and dynamic lighting. Quick Tip: Remember the key benefits: They are {Cheap} to run (efficient), {Long-lasting}, {Tough} (durable), and {Fast} (instant on/off).

According to the Stefan-Boltzmann law, the power (P) or rate of energy radiated by a perfectly black body is given by: \[ P = \sigma A T^4 \]
where \(E = P \times t\) is the total energy radiated in time \(t\). So, \(E = \sigma A T^4 t\).
Given:

Stefan's constant, \(\sigma \approx 5.67 \times 10^{-8} \, W/m^2K^4\)
Surface area, \(A = 200 \, cm^2 = 200 \times 10^{-4} \, m^2 = 2 \times 10^{-2} \, m^2\)
Temperature, \(T = 127^{\circ}C = 127 + 273 = 400 \, K\)
Time, \(t = 1 \, minute = 60 \, s\)

Now, calculate the energy radiated (E): \[ E = (5.67 \times 10^{-8}) \times (2 \times 10^{-2}) \times (400)^4 \times 60 \] \[ E = (5.67 \times 10^{-8}) \times (2 \times 10^{-2}) \times (256 \times 10^8) \times 60 \] \[ E = 5.67 \times 2 \times 256 \times 10^{-2} \times 60 \] \[ E = 11.34 \times 256 \times 0.6 \] \[ E \approx 2903 \times 0.6 \] \[ E \approx 1741.8 \, J \]
The energy radiated in one minute is approximately 1742 Joules. Quick Tip: Always convert units before calculating! Area must be in \(m^2\), time in seconds, and most importantly, temperature must be in Kelvin (\(K = ^{\circ}C + 273\)) for the Stefan-Boltzmann law.

The mutual inductance (\(M\)) between two coils is related to their self-inductances (\(L_1\) and \(L_2\)) and the coefficient of coupling (\(k\)) by the formula: \[ M = k \sqrt{L_1 L_2} \]
Given:

Self-inductance of the first coil, \(L_1 = 60 \, mH = 60 \times 10^{-3} \, H\)
Self-inductance of the second coil, \(L_2 = 60 \, mH = 60 \times 10^{-3} \, H\)
Coefficient of coupling, \(k = 0.75\)

Substitute the values into the formula: \[ M = 0.75 \times \sqrt{(60 \times 10^{-3}) \times (60 \times 10^{-3})} \] \[ M = 0.75 \times \sqrt{(60 \times 10^{-3})^2} \] \[ M = 0.75 \times (60 \times 10^{-3}) \] \[ M = 45 \times 10^{-3} \, H \] \[ M = 45 \, mH \]
The mutual inductance between the coils is 45 mH. Quick Tip: The coefficient of coupling (\(k\)) tells you how much of the magnetic flux from one coil links with the other. \(k=1\) means perfect coupling, \(k=0\) means no coupling. The mutual inductance is just this fraction of the geometric mean of the self-inductances.

The phase difference (\(\phi\)) between the voltage and current in a series LCR circuit is given by the formula: \[ \tan \phi = \frac{X_L - X_C}{R} \]
Given:

Resistance, \(R = 3 \, \Omega\)
Inductive reactance, \(X_L = 8 \, \Omega\)
Capacitive reactance, \(X_C = 4 \, \Omega\)

Substituting the values: \[ \tan \phi = \frac{8 - 4}{3} = \frac{4}{3} \]
Therefore, the phase difference is: \[ \phi = \tan^{-1}\left(\frac{4}{3}\right) \]
Since \(X_L > X_C\), the circuit is inductive, and the voltage leads the current by an angle of \(\tan^{-1}(4/3)\), which is approximately \(53.13^{\circ}\). Quick Tip: Remember {ELI the ICE man}. In an {L} (inductive) circuit, Voltage ({E}) leads Current ({I}). In a {C} (capacitive) circuit, Current ({I}) leads Voltage ({E}). Since \(X_L > X_C\) here, the circuit acts inductive (ELI).

A potentiometer has several advantages over a voltmeter for measuring potential difference:

Greater Accuracy: A potentiometer is more accurate because it works on the null deflection method. At the balance point, it draws no current from the circuit whose potential difference is being measured. In contrast, a voltmeter always draws some current, which alters the potential difference it is intended to measure.
Measures EMF Accurately: Because it draws no current at balance, a potentiometer can measure the true electromotive force (e.m.f.) of a cell. A voltmeter measures the terminal potential difference, which is less than the e.m.f. when current is flowing.
High Sensitivity: The sensitivity of a potentiometer is very high and can be easily increased by increasing the length of the potentiometer wire.
Versatility: A potentiometer is a versatile instrument that can also be used to compare the e.m.f.s of two cells and to determine the internal resistance of a cell. Quick Tip: The main advantage is simple: a potentiometer measures voltage without touching the circuit's current, giving a true reading. A voltmeter has to "sip" a little current, which slightly changes the voltage it's trying to measure.

A full wave rectifier converts both halves of an AC input signal into a pulsating DC output. A common circuit uses a center-tapped transformer and two diodes.

% A labeled diagram should be inserted here showing the components.
% Since I cannot generate an image, here is a description of the diagram:
% 1. An AC input source is connected to the primary coil of a transformer.
% 2. The secondary coil is center-tapped. The top end is connected to the anode of a diode D1. The bottom end is connected to the anode of a diode D2.
% 3. The cathodes of both diodes D1 and D2 are connected together.
% 4. A load resistor (RL) is connected between the common cathode point and the center tap of the transformer.
% 5. The output voltage is measured across the load resistor RL.
% An input sine wave and a corresponding output full-wave rectified waveform (showing only positive peaks) should be drawn.

\textit{
\caption{Circuit Diagram of a Full Wave Rectifier

Components:

AC Source: Provides the input alternating voltage.
Center-tapped Transformer: Steps down the voltage and provides two AC signals that are 180° out of phase.
Diodes (D1, D2): Act as one-way gates, each allowing current to flow during one half-cycle of the input.
Load Resistor (RL): The output voltage is developed across this resistor. Quick Tip: Think of the two diodes working in shifts. For the positive half of the AC cycle, D1 works. For the negative half, D2 works. The center-tap ensures they both push the current in the same direction through the load resistor.

Given:

Mass of the body, \(m = 0.8 \, kg\)
Restoring force, \(F = 0.4 \, N\)
Displacement, \(x = 4 \, cm = 0.04 \, m\)

1. To find the Force Constant (k):
The restoring force in Simple Harmonic Motion (S.H.M.) is given by Hooke's Law, \(F = kx\) (considering magnitude). \[ k = \frac{F}{x} \] \[ k = \frac{0.4 \, N}{0.04 \, m} = 10 \, N/m \]
The force constant is 10 N/m.

2. To find the Period of S.H.M. (T):
The period of S.H.M. is given by the formula: \[ T = 2\pi\sqrt{\frac{m}{k}} \] \[ T = 2\pi\sqrt{\frac{0.8}{10}} = 2\pi\sqrt{0.08} = 2\pi\sqrt{\frac{8}{100}} \] \[ T = 2\pi\frac{\sqrt{8}}{10} = 2\pi\frac{2\sqrt{2}}{10} = \frac{2\sqrt{2}\pi}{5} \, s \]
Using \(\pi \approx 3.14\) and \(\sqrt{2} \approx 1.414\): \[ T \approx \frac{2 \times 1.414 \times 3.14}{5} \approx \frac{8.88}{5} \approx 1.776 \, s \]
The period of S.H.M. is approximately 1.78 s. Quick Tip: S.H.M. problems often have two steps: First, use given force and displacement to find the spring/force constant (\(k\)). Second, use \(k\) and the mass to find the period or frequency.

The work done (\(W\)) by a gas during an isothermal expansion is given by the formula: \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]
Given:

Number of moles, \(n = 0.5\)
Universal gas constant, \(R \approx 8.314 \, J/mol \cdot K\)
Temperature, \(T = 300 \, K\)
Initial volume, \(V_i = 2.0 \, L\)
Final volume, \(V_f = 6.0 \, L\)

Substitute the values into the formula: \[ W = 0.5 \times 8.314 \times 300 \times \ln\left(\frac{6.0}{2.0}\right) \] \[ W = 0.5 \times 8.314 \times 300 \times \ln(3) \]
Using the value \(\ln(3) \approx 1.0986\): \[ W = 1247.1 \times 1.0986 \] \[ W \approx 1370.16 \, J \]
The work done by the gas is approximately 1370 J. Quick Tip: For {isothermal} processes (constant temperature), the work done involves a natural logarithm (\(\ln\)) of the volume ratio. For {isobaric} processes (constant pressure), work is simply \(P \Delta V\).

When a bar magnet of magnetic moment \(M\) and moment of inertia \(I\) is suspended in a uniform magnetic field \(B\), and displaced by a small angle \(\theta\), it experiences a restoring torque \(\tau\).
The restoring torque is given by: \[ \tau = -MB\sin\theta \]
For small angular displacements (\(\theta\)), \(\sin\theta \approx \theta\). \[ \tau \approx -MB\theta \]
This torque is also related to the angular acceleration \(\alpha\) by the equation \(\tau = I\alpha\). \[ I\alpha = -MB\theta \] \[ \alpha = -\left(\frac{MB}{I}\right)\theta \]
This equation is of the form \(\alpha = -\omega^2\theta\), which is the condition for angular Simple Harmonic Motion (S.H.M.).
By comparing the two equations, we get the angular frequency \(\omega\): \[ \omega^2 = \frac{MB}{I} \quad \Rightarrow \quad \omega = \sqrt{\frac{MB}{I}} \]
The period of oscillation \(T\) is related to the angular frequency by \(T = \frac{2\pi}{\omega}\).
Substituting the expression for \(\omega\): \[ T = 2\pi\sqrt{\frac{I}{MB}} \]
This is the expression for the period of a bar magnet vibrating in a uniform magnetic field. Quick Tip: This derivation is very similar to that of a simple pendulum. Just as \(T = 2\pi\sqrt{L/g}\) depends on length and gravity, the magnetic pendulum's period depends on its rotational inertia (\(I\)) and the magnetic restoring force (\(MB\)).

For a thermodynamic system to be in thermodynamic equilibrium, it must satisfy three types of equilibrium simultaneously:

[label=(\alph*)]
Mechanical Equilibrium: A system is in mechanical equilibrium when there are no unbalanced forces acting on any part of the system or between the system and its surroundings. This means the pressure within the system is uniform and does not change with time.

Chemical Equilibrium: A system is in chemical equilibrium when its chemical composition is constant and there are no net chemical reactions occurring within it. The rate of forward reactions equals the rate of reverse reactions.

Thermal Equilibrium: A system is in thermal equilibrium when there is no net flow of heat energy between its parts or between the system and its surroundings. This implies that the temperature is uniform throughout the system and is the same as that of its surroundings. Quick Tip: Think of equilibrium as "no change." {Mechanical:} No change in pressure/motion. {Chemical:} No change in composition. {Thermal:} No change in temperature (no heat flow). A system is in full thermodynamic equilibrium only when all three are true.

Brewster's Law: This law states that for a particular angle of incidence, known as Brewster's angle or the polarizing angle (\(i_p\)), the reflected light from a transparent dielectric surface is completely plane-polarized. This occurs when the reflected ray and the refracted ray are perpendicular to each other.

Derivation:
Let unpolarized light be incident at Brewster's angle \(i_p\) on the surface separating two media (e.g., air and glass) with refractive indices \(n_1\) and \(n_2\) respectively. Let \(r\) be the angle of refraction.
According to Snell's law: \[ n_1 \sin(i_p) = n_2 \sin(r) \quad \cdots (1) \]
By the condition of Brewster's law, the reflected and refracted rays are perpendicular. From the geometry of reflection and refraction, this means: \[ i_p + r = 90^{\circ} \] \[ r = 90^{\circ} - i_p \]
Now, substitute this value of \(r\) into Snell's law (equation 1): \[ n_1 \sin(i_p) = n_2 \sin(90^{\circ} - i_p) \]
Since \(\sin(90^{\circ} - \theta) = \cos(\theta)\), the equation becomes: \[ n_1 \sin(i_p) = n_2 \cos(i_p) \]
Rearranging the terms to find the angle \(i_p\): \[ \frac{\sin(i_p)}{\cos(i_p)} = \frac{n_2}{n_1} \] \[ \tan(i_p) = \frac{n_2}{n_1} \]
If the first medium is air or vacuum (\(n_1 \approx 1\)) and the second medium has refractive index \(n\) (\(n_2 = n\)), the formula simplifies to: \[ \tan(i_p) = n \]
This is the formula for Brewster's angle. Quick Tip: The key to the derivation is the special condition: at Brewster's angle, the angle between the reflected and refracted rays is 90°. This simple geometric fact, when combined with Snell's law, leads directly to the \(\tan(i_p) = n\) formula.

Derivation of the Law of Radioactive Decay:
The law of radioactive decay states that the rate of disintegration of radioactive nuclei in a sample is directly proportional to the number of undecayed nuclei present at that instant.
Let \(N\) be the number of undecayed nuclei in a sample at time \(t\), and \(dN\) be the number of nuclei that decay in a small time interval \(dt\). The rate of decay is \( -dN/dt \).
According to the law: \[ -\frac{dN}{dt} \propto N \] \[ -\frac{dN}{dt} = \lambda N \]
where \(\lambda\) is the decay constant, a positive constant characteristic of the radioactive substance.
Rearranging the equation to separate the variables: \[ \frac{dN}{N} = -\lambda dt \]
To find the number of nuclei remaining after a time \(t\), we integrate this equation. Let \(N_0\) be the initial number of nuclei at time \(t=0\), and \(N\) be the number of nuclei at time \(t\). \[ \int_{N_0}^{N} \frac{dN}{N} = \int_{0}^{t} -\lambda dt \] \[ [\ln N]_{N_0}^{N} = -\lambda [t]_0^t \] \[ \ln N - \ln N_0 = -\lambda(t - 0) \] \[ \ln\left(\frac{N}{N_0}\right) = -\lambda t \]
Taking the exponential of both sides: \[ \frac{N}{N_0} = e^{-\lambda t} \] \[ N(t) = N_0 e^{-\lambda t} \]
This is the mathematical expression for the law of radioactive decay, which shows that the number of undecayed nuclei decreases exponentially with time.

Definition of Becquerel (Bq):
One becquerel is the unit of activity of a radioactive sample in the SI system. It is defined as one disintegration (or decay) per second. \[ 1 \, Bq = 1 \, decay/second \] Quick Tip: Radioactive decay is a first-order process, meaning the rate depends only on the amount of "reactant" (\(N\)) you have. This always leads to an exponential decay formula, just like in chemical kinetics or capacitor discharge.

When a bar magnet of magnetic moment \(M\) and moment of inertia \(I\) is suspended in a uniform magnetic field \(B\), and displaced by a small angle \(\theta\), it experiences a restoring torque \(\tau\).
The restoring torque is given by: \[ \tau = -MB\sin\theta \]
For small angular displacements (\(\theta\)), \(\sin\theta \approx \theta\). \[ \tau \approx -MB\theta \]
This torque is also related to the angular acceleration \(\alpha\) by the equation \(\tau = I\alpha\). \[ I\alpha = -MB\theta \] \[ \alpha = -\left(\frac{MB}{I}\right)\theta \]
This equation is of the form \(\alpha = -\omega^2\theta\), which is the condition for angular Simple Harmonic Motion (S.H.M.).
By comparing the two equations, we get the angular frequency \(\omega\): \[ \omega^2 = \frac{MB}{I} \quad \Rightarrow \quad \omega = \sqrt{\frac{MB}{I}} \]
The period of oscillation \(T\) is related to the angular frequency by \(T = \frac{2\pi}{\omega}\).
Substituting the expression for \(\omega\): \[ T = 2\pi\sqrt{\frac{I}{MB}} \]
This is the expression for the period of a bar magnet vibrating in a uniform magnetic field. Quick Tip: This derivation is very similar to that of a simple pendulum. Just as \(T = 2\pi\sqrt{L/g}\) depends on length and gravity, the magnetic pendulum's period depends on its rotational inertia (\(I\)) and the magnetic restoring force (\(MB\)).

For a thermodynamic system to be in thermodynamic equilibrium, it must satisfy three types of equilibrium simultaneously:

[label=(\alph*)]
Mechanical Equilibrium: A system is in mechanical equilibrium when there are no unbalanced forces acting on any part of the system or between the system and its surroundings. This means the pressure within the system is uniform and does not change with time.

Chemical Equilibrium: A system is in chemical equilibrium when its chemical composition is constant and there are no net chemical reactions occurring within it. The rate of forward reactions equals the rate of reverse reactions.

Thermal Equilibrium: A system is in thermal equilibrium when there is no net flow of heat energy between its parts or between the system and its surroundings. This implies that the temperature is uniform throughout the system and is the same as that of its surroundings. Quick Tip: Think of equilibrium as "no change." {Mechanical:} No change in pressure/motion. {Chemical:} No change in composition. {Thermal:} No change in temperature (no heat flow). A system is in full thermodynamic equilibrium only when all three are true.

Brewster's Law: This law states that for a particular angle of incidence, known as Brewster's angle or the polarizing angle (\(i_p\)), the reflected light from a transparent dielectric surface is completely plane-polarized. This occurs when the reflected ray and the refracted ray are perpendicular to each other.

Derivation:
Let unpolarized light be incident at Brewster's angle \(i_p\) on the surface separating two media (e.g., air and glass) with refractive indices \(n_1\) and \(n_2\) respectively. Let \(r\) be the angle of refraction.
According to Snell's law: \[ n_1 \sin(i_p) = n_2 \sin(r) \quad \cdots (1) \]
By the condition of Brewster's law, the reflected and refracted rays are perpendicular. From the geometry of reflection and refraction, this means: \[ i_p + r = 90^{\circ} \] \[ r = 90^{\circ} - i_p \]
Now, substitute this value of \(r\) into Snell's law (equation 1): \[ n_1 \sin(i_p) = n_2 \sin(90^{\circ} - i_p) \]
Since \(\sin(90^{\circ} - \theta) = \cos(\theta)\), the equation becomes: \[ n_1 \sin(i_p) = n_2 \cos(i_p) \]
Rearranging the terms to find the angle \(i_p\): \[ \frac{\sin(i_p)}{\cos(i_p)} = \frac{n_2}{n_1} \] \[ \tan(i_p) = \frac{n_2}{n_1} \]
If the first medium is air or vacuum (\(n_1 \approx 1\)) and the second medium has refractive index \(n\) (\(n_2 = n\)), the formula simplifies to: \[ \tan(i_p) = n \]
This is the formula for Brewster's angle. Quick Tip: The key to the derivation is the special condition: at Brewster's angle, the angle between the reflected and refracted rays is 90°. This simple geometric fact, when combined with Snell's law, leads directly to the \(\tan(i_p) = n\) formula.

Derivation of the Law of Radioactive Decay:
The law of radioactive decay states that the rate of disintegration of radioactive nuclei in a sample is directly proportional to the number of undecayed nuclei present at that instant.
Let \(N\) be the number of undecayed nuclei in a sample at time \(t\), and \(dN\) be the number of nuclei that decay in a small time interval \(dt\). The rate of decay is \( -dN/dt \).
According to the law: \[ -\frac{dN}{dt} \propto N \] \[ -\frac{dN}{dt} = \lambda N \]
where \(\lambda\) is the decay constant, a positive constant characteristic of the radioactive substance.
Rearranging the equation to separate the variables: \[ \frac{dN}{N} = -\lambda dt \]
To find the number of nuclei remaining after a time \(t\), we integrate this equation. Let \(N_0\) be the initial number of nuclei at time \(t=0\), and \(N\) be the number of nuclei at time \(t\). \[ \int_{N_0}^{N} \frac{dN}{N} = \int_{0}^{t} -\lambda dt \] \[ [\ln N]_{N_0}^{N} = -\lambda [t]_0^t \] \[ \ln N - \ln N_0 = -\lambda(t - 0) \] \[ \ln\left(\frac{N}{N_0}\right) = -\lambda t \]
Taking the exponential of both sides: \[ \frac{N}{N_0} = e^{-\lambda t} \] \[ N(t) = N_0 e^{-\lambda t} \]
This is the mathematical expression for the law of radioactive decay, which shows that the number of undecayed nuclei decreases exponentially with time.

Definition of Becquerel (Bq):
One becquerel is the unit of activity of a radioactive sample in the SI system. It is defined as one disintegration (or decay) per second. \[ 1 \, Bq = 1 \, decay/second \] Quick Tip: Radioactive decay is a first-order process, meaning the rate depends only on the amount of "reactant" (\(N\)) you have. This always leads to an exponential decay formula, just like in chemical kinetics or capacitor discharge.

Statement: Kirchhoff's law of heat radiation states that at a given temperature, the ratio of the emissive power (\(E\)) to the coefficient of absorption (\(a\)) is the same for all bodies and is equal to the emissive power of a perfectly black body (\(E_b\)) at that same temperature. \[ \frac{E}{a} = E_b \]
This implies that a good absorber of radiation is also a good emitter of radiation at a given temperature.

Proof (Thought Experiment):
Consider an ordinary body 'O' and a perfectly black body 'B' of the same surface area, suspended in a constant temperature enclosure. After some time, both bodies will reach thermal equilibrium with the enclosure, meaning they will all be at the same temperature.
Let \(E\) be the emissive power of the ordinary body and \(a\) be its coefficient of absorption.
Let \(E_b\) be the emissive power of the black body. By definition, its coefficient of absorption is \(a_b = 1\).
Let \(Q\) be the total radiant energy incident per unit area per unit time on each body from the enclosure.

For the ordinary body 'O' at thermal equilibrium:
Energy absorbed per unit area per unit time = \(aQ\).
Energy emitted per unit area per unit time = \(E\).
Since the temperature is constant, Energy absorbed = Energy emitted. \[ aQ = E \quad \cdots (1) \]

For the perfectly black body 'B' at thermal equilibrium:
Energy absorbed per unit area per unit time = \(a_b Q = 1 \cdot Q = Q\).
Energy emitted per unit area per unit time = \(E_b\).
Since the temperature is constant, Energy absorbed = Energy emitted. \[ Q = E_b \quad \cdots (2) \]

Now, substitute the value of \(Q\) from equation (2) into equation (1): \[ a(E_b) = E \] \[ \frac{E}{a} = E_b \]
This proves Kirchhoff's law of heat radiation. Quick Tip: Remember the simple concept: "Good absorbers are good emitters." A black T-shirt gets hotter in the sun (good absorber) and also cools down faster in the shade (good emitter) compared to a white one.

End correction (\(e\)) accounts for the fact that the antinode at the open end of a pipe is not formed exactly at the edge, but slightly outside it. The corrected length is \(L' = L + e\) for a closed pipe and \(L' = L + 2e\) for an open pipe.

Let \(n_1\) and \(n_2\) be the frequencies of two successive resonating modes.

(i) For a pipe open at both ends:
The fundamental frequency is \(n = \frac{v}{2L'} = \frac{v}{2(L+2e)}\).
Let the pipe resonate with a tuning fork of frequency \(n_1\) at length \(L_1\), and with a tuning fork of frequency \(n_2\) at length \(L_2\).
Then, \(v = 2n_1(L_1 + 2e)\) and \(v = 2n_2(L_2 + 2e)\). \[ 2n_1(L_1 + 2e) = 2n_2(L_2 + 2e) \] \[ n_1L_1 + 2n_1e = n_2L_2 + 2n_2e \] \[ 2e(n_1 - n_2) = n_2L_2 - n_1L_1 \] \[ e = \frac{n_2L_2 - n_1L_1}{2(n_1 - n_2)} \]

(ii) For a pipe closed at one end:
The fundamental frequency is \(n = \frac{v}{4L'} = \frac{v}{4(L+e)}\).
Let the pipe resonate with a tuning fork of frequency \(n\) at two successive lengths, \(L_1\) (for the fundamental mode) and \(L_2\) (for the third harmonic).
For the first resonance: \(v = 4n(L_1 + e)\).
For the second resonance (3rd harmonic): The corrected length is effectively \(\frac{3\lambda}{4}\), so \(v = \frac{4n(L_2+e)}{3}\) is not the way. Instead, we use the fact that the distance between two consecutive nodes is \(\lambda/2\).
Let the first resonating length be \(L_1\) and the second be \(L_2\).
Then \(L_1 + e = \frac{\lambda}{4}\) and \(L_2 + e = \frac{3\lambda}{4}\).
Subtracting the first equation from the second: \[ (L_2 + e) - (L_1 + e) = \frac{3\lambda}{4} - \frac{\lambda}{4} \] \[ L_2 - L_1 = \frac{2\lambda}{4} = \frac{\lambda}{2} \implies \lambda = 2(L_2 - L_1) \]
Substitute \(\lambda\) back into the first equation: \[ L_1 + e = \frac{2(L_2 - L_1)}{4} = \frac{L_2 - L_1}{2} \] \[ e = \frac{L_2 - L_1}{2} - L_1 = \frac{L_2 - 3L_1}{2} \]

A simpler experimental method uses two different frequencies \(n_1, n_2\) and one length L. Let's re-derive for two lengths.
From \(v = 4n_1(L_1+e)\) and \(v=4n_2(L_2+e)\): \[ n_1(L_1+e) = n_2(L_2+e) \implies n_1L_1 + n_1e = n_2L_2 + n_2e \] \[ e(n_1-n_2) = n_2L_2 - n_1L_1 \implies e = \frac{n_2L_2 - n_1L_1}{n_1 - n_2} \]
The second derivation is correct for closed pipes as well. Quick Tip: End correction is like adding a little extra "acoustic length" to the pipe. The key to finding it experimentally is to use two different resonance conditions (either different lengths or different frequencies) and solve the resulting simultaneous equations.

Consider a conducting bar of length \(L\) rotating with a constant angular velocity \(\omega\) about a pivot at one end. The rotation occurs in a uniform magnetic field \(B\) which is perpendicular to the plane of rotation.
Consider a small element of the bar of length \(dr\) at a distance \(r\) from the pivot.
The linear velocity of this element is \(v = r\omega\).
The motional e.m.f. induced in this small element \(dr\) is given by: \[ d\mathcal{E} = B v \, dr \]
Substituting \(v = r\omega\): \[ d\mathcal{E} = B (r\omega) \, dr \]
To find the total e.m.f. induced across the entire length of the bar, we integrate this expression from the pivot (\(r=0\)) to the other end (\(r=L\)): \[ \mathcal{E} = \int_{0}^{L} d\mathcal{E} = \int_{0}^{L} B\omega r \, dr \]
Since \(B\) and \(\omega\) are constant: \[ \mathcal{E} = B\omega \int_{0}^{L} r \, dr \] \[ \mathcal{E} = B\omega \left[ \frac{r^2}{2} \right]_0^L \] \[ \mathcal{E} = B\omega \left( \frac{L^2}{2} - 0 \right) \] \[ \mathcal{E} = \frac{1}{2} B \omega L^2 \]
This is the expression for the rotational e.m.f. induced in the conducting bar. Quick Tip: Think of the induced EMF as coming from the average velocity of the rod. The tip moves at \(v = L\omega\) and the pivot at \(v=0\). The average velocity is \(\frac{0 + L\omega}{2} = \frac{1}{2}L\omega\). So the EMF is just \(B \times L \times v_{avg} = B L (\frac{1}{2}L\omega) = \frac{1}{2}B\omega L^2\).

First, we need to find the velocity (\(v_n\)) and radius (\(r_n\)) of the electron in the second orbit (\(n=2\)) of a hydrogen atom.
The velocity of an electron in the \(n^{th}\) orbit is \(v_n = \frac{e^2}{2\epsilon_0 h n}\), and the radius is \(r_n = \frac{\epsilon_0 h^2 n^2}{\pi m e^2}\).
A simpler approach is to calculate the time period of one revolution.
The time period \(T_n\) of revolution in the \(n^{th}\) orbit is given by \(T_n = \frac{2\pi r_n}{v_n}\).
It can be shown that the angular frequency is \(\omega_n = \frac{\pi m e^4}{2\epsilon_0^2 h^3 n^3}\).
The frequency of revolution is \(f_n = \frac{\omega_n}{2\pi} = \frac{m e^4}{4\epsilon_0^2 h^3 n^3}\).
For the second orbit, \(n=2\). \(f_2 = \frac{(9.1 \times 10^{-31}) (1.6 \times 10^{-19})^4}{4 (8.85 \times 10^{-12})^2 (6.63 \times 10^{-34})^3 (2)^3}\)
This is complex. Let's use a simpler known relation. The speed of the electron in the first orbit is \(v_1 \approx c/137 \approx 2.19 \times 10^6 \, m/s\).
The speed in the \(n^{th}\) orbit is \(v_n = v_1/n\). So, \(v_2 = v_1/2 \approx 1.095 \times 10^6 \, m/s\).
The radius of the first orbit is the Bohr radius, \(r_1 \approx 0.529 \times 10^{-10} \, m\).
The radius of the \(n^{th}\) orbit is \(r_n = r_1 n^2\). So, \(r_2 = r_1 (2)^2 = 4r_1 \approx 2.116 \times 10^{-10} \, m\).
The time period of one revolution in the second orbit is: \[ T_2 = \frac{2\pi r_2}{v_2} = \frac{2 \pi (2.116 \times 10^{-10})}{1.095 \times 10^6} \approx 1.21 \times 10^{-15} \, s \]
The total time the electron stays in the orbit is \(t_{total} = 10^{-8} \, s\).
The number of revolutions is the total time divided by the time for one revolution: \[ Number of revolutions = \frac{t_{total}}{T_2} = \frac{10^{-8}}{1.21 \times 10^{-15}} \approx 8.26 \times 10^6 \]
The electron will make approximately \(8.3 \times 10^6\) revolutions. Quick Tip: To find revolutions, you need the frequency (revolutions per second). Calculate the time for one revolution (Period, \(T\)) by finding the orbit's circumference (\(2\pi r_n\)) and dividing by the electron's speed (\(v_n\)). Then, divide the total time by the period.

The work done (\(W\)) by a constant torque (\(\tau\)) is given by the formula: \[ W = \tau \cdot \theta \]
where \(\theta\) is the angular displacement in radians.
Given:

Torque, \(\tau = 2000 \, Nm\)
Number of revolutions = 4

First, convert the number of revolutions into radians for the angular displacement \(\theta\).
One revolution corresponds to an angle of \(2\pi\) radians. \[ \theta = 4 \, revolutions \times 2\pi \, \frac{rad}{revolution} = 8\pi \, rad \]
Now, calculate the work done: \[ W = 2000 \, Nm \times 8\pi \, rad = 16000\pi \, J \]
Using \(\pi \approx 3.14159\): \[ W \approx 16000 \times 3.14159 \approx 50265.48 \, J \]
The work done during the first 4 revolutions is \(16000\pi\) Joules, or approximately 50265 J. (Note: The mass and radius of the flywheel are not needed for this calculation as the torque is given directly). Quick Tip: This is the rotational version of Work = Force \(\times\) distance. Here, Work = Torque \(\times\) angular distance. Remember to always convert revolutions to radians (\(1 rev = 2\pi rad\)) before calculating.

Given:

Galvanometer resistance, \(R_g = 50 \, \Omega\)
Full scale deflection current, \(I_g = 2 \, mA = 2 \times 10^{-3} \, A\)


(i) Conversion into an Ammeter:
To convert a galvanometer into an ammeter of range \(I\), a small resistance called a shunt (\(S\)) is connected in parallel with it.
The required range is \(I = 0.5 \, A\).
The formula for the shunt resistance is: \[ S = \frac{I_g R_g}{I - I_g} \] \[ S = \frac{(2 \times 10^{-3}) \times 50}{0.5 - (2 \times 10^{-3})} = \frac{0.1}{0.5 - 0.002} = \frac{0.1}{0.498} \] \[ S \approx 0.2008 \, \Omega \]
A shunt resistance of approximately 0.201 \(\Omega\) is required.

(ii) Conversion into a Voltmeter:
To convert a galvanometer into a voltmeter of range \(V\), a large resistance (\(R_s\)) is connected in series with it.
The required range is \(V = 10 \, V\).
The formula for the series resistance is: \[ R_s = \frac{V}{I_g} - R_g \] \[ R_s = \frac{10}{2 \times 10^{-3}} - 50 = \frac{10000}{2} - 50 \] \[ R_s = 5000 - 50 = 4950 \, \Omega \]
A series resistance of 4950 \(\Omega\) is required. Quick Tip: {Ammeter}: Add a tiny resistor in {parallel} (a "shunt") to bypass most of the current. {Voltmeter}: Add a huge resistor in {series} to limit the current and drop most of the voltage.

First, we calculate the excess pressure (\(P_{excess}\)) inside the water drop due to surface tension. The formula for a liquid drop is: \[ P_{excess} = \frac{2T}{r} \]
We need to convert the given values to SI units.

Diameter = 0.6 mm, so radius \(r = 0.3 \, mm = 0.3 \times 10^{-3} \, m\)
Surface tension, \(T = 72 \, dyne/cm\).
Since \(1 \, dyne = 10^{-5} \, N\) and \(1 \, cm = 10^{-2} \, m\),
\(T = \frac{72 \times 10^{-5} \, N}{10^{-2} \, m} = 72 \times 10^{-3} \, N/m\)

Now, calculate the excess pressure: \[ P_{excess} = \frac{2 \times (72 \times 10^{-3})}{0.3 \times 10^{-3}} = \frac{144}{0.3} = 480 \, N/m^2 (or Pa) \]
The total pressure inside the drop (\(P_{inside}\)) is the sum of the atmospheric pressure (\(P_{atm}\)) and the excess pressure. \[ P_{inside} = P_{atm} + P_{excess} \] \[ P_{inside} = (1.013 \times 10^5) + 480 \] \[ P_{inside} = 101300 + 480 = 101780 \, N/m^2 \]
The pressure inside the water drop is \(1.0178 \times 10^5\) N/m\(^2\). Quick Tip: Remember the formula has a factor of 2 for a drop (\(2T/r\)) because there is one surface, but a factor of 4 for a soap bubble (\(4T/r\)) because it has two surfaces (inner and outer).

For a long solenoid, the magnetic field (\(B\)) at its centre is uniform and given by the formula: \[ B = \mu_0 n I \]
where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, T \cdot m/A\)), \(I\) is the current, and \(n\) is the number of turns per unit length.
Given:

Length, \(L = \pi \, m\)
Number of turns, \(N = 1000\)
Current, \(I = 5 \, A\)

First, calculate the number of turns per unit length (\(n\)): \[ n = \frac{N}{L} = \frac{1000}{\pi} \, turns/m \]
Now, calculate the magnetic field \(B\): \[ B = (4\pi \times 10^{-7}) \times \left(\frac{1000}{\pi}\right) \times 5 \]
The \(\pi\) terms cancel out: \[ B = 4 \times 10^{-7} \times 1000 \times 5 \] \[ B = 20 \times 10^{-4} \, T = 2 \times 10^{-3} \, T \]
The magnetic field at the centre of the solenoid is \(2 \times 10^{-3}\) Tesla. (Note: The diameter of the solenoid is not needed for this calculation, assuming it is a long solenoid where end effects are negligible). Quick Tip: The key to solenoid problems is calculating \(n\), the turn density (turns per meter). Once you have \(n\), the formula \(B = \mu_0 n I\) is straightforward. Don't get distracted by the diameter unless you need to check if the solenoid is 'long' (\(L \gg D\)).

Ferromagnetism is the property of certain materials (like iron, cobalt, and nickel) that causes them to be strongly attracted to an external magnetic field and to retain their magnetic properties even after the external field is removed, thus becoming permanent magnets.

Domain Theory of Ferromagnetism:
The domain theory explains this strong magnetic behavior.

Atomic Magnets: In ferromagnetic materials, individual atoms have a net magnetic moment due to the intrinsic spin of their electrons.
Formation of Domains: Due to a quantum mechanical interaction called exchange coupling, the magnetic moments of adjacent atoms align themselves parallel to each other in large groups. These small regions of spontaneous and uniform magnetization are called magnetic domains. Each domain acts like a tiny, powerful magnet.
Unmagnetized State: In an unmagnetized piece of ferromagnetic material, these domains are oriented randomly. Their magnetic fields point in all different directions, so the net magnetic field of the object is zero.
Magnetization Process: When an external magnetic field is applied, two processes occur:

Domain Growth: Domains that are already aligned (or nearly aligned) with the external field grow in size by moving their boundary walls and taking over adjacent domains that are less favorably aligned.
Domain Rotation: As the external field becomes stronger, the magnetic moments of entire domains that are not yet aligned with the field will rotate to align with it.

Saturated State: When all the domains are aligned with the external field, the material is said to be magnetically saturated. It has reached its maximum possible magnetization.
Permanent Magnetism: When the external field is removed, many of the domains remain locked in their aligned positions due to crystal defects and impurities. This retained magnetism is what makes the material a permanent magnet. Quick Tip: Think of domains as tiny compass needles. In a normal iron bar, they're all jumbled up, pointing in random directions. When you bring a strong magnet nearby, you're like a drill sergeant shouting "Attention!", causing all the compass needles to snap into alignment, creating one big, strong magnetic field.

In a series LCR circuit, the instantaneous voltage \(V\) and current \(I\) are given by: \[ V = V_0 \sin(\omega t) \] \[ I = I_0 \sin(\omega t - \phi) \]
where \(V_0\) and \(I_0\) are the peak voltage and current, \(\omega\) is the angular frequency, and \(\phi\) is the phase difference between the voltage and current.

The instantaneous power (\(P\)) dissipated in the circuit is the product of instantaneous voltage and current: \[ P = VI = [V_0 \sin(\omega t)][I_0 \sin(\omega t - \phi)] \] \[ P = V_0 I_0 \sin(\omega t)[\sin(\omega t)\cos(\phi) - \cos(\omega t)\sin(\phi)] \] \[ P = V_0 I_0 [\sin^2(\omega t)\cos(\phi) - \sin(\omega t)\cos(\omega t)\sin(\phi)] \]
To find the average power (\(P_{avg}\)) over one complete cycle (\(T\)), we need to find the average value of this expression.
We know the average values over one cycle are:

\(\langle \sin^2(\omega t) \rangle = \frac{1}{2}\)
\(\langle \sin(\omega t)\cos(\omega t) \rangle = \langle \frac{1}{2}\sin(2\omega t) \rangle = 0\)

Therefore, the average power is: \[ P_{avg} = V_0 I_0 [ \langle \sin^2(\omega t) \rangle \cos(\phi) - \langle \sin(\omega t)\cos(\omega t) \rangle \sin(\phi) ] \] \[ P_{avg} = V_0 I_0 [ \frac{1}{2} \cos(\phi) - (0) \sin(\phi) ] \] \[ P_{avg} = \frac{V_0 I_0}{2} \cos(\phi) \]
We can express this in terms of the root mean square (RMS) values, where \(V_{rms} = \frac{V_0}{\sqrt{2}}\) and \(I_{rms} = \frac{I_0}{\sqrt{2}}\). \[ P_{avg} = \left(\frac{V_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) \cos(\phi) \] \[ P_{avg} = V_{rms} I_{rms} \cos(\phi) \]
This is the expression for the average power dissipated in a series LCR circuit. The term \(\cos(\phi)\) is called the power factor. Quick Tip: Power is only dissipated in the resistor. The inductor and capacitor store and release energy but don't dissipate it over a full cycle. The \(\cos(\phi)\) factor, which equals \(R/Z\), essentially tells you what fraction of the total "apparent power" (\(V_{rms}I_{rms}\)) is actually doing work in the resistor.

Distinction between Interference and Diffraction:

\begin{tabular{|p{4.5cm|p{4.5cm|
\hline
Interference & Diffraction

\hline
It is the result of superposition of waves from two (or more) different coherent sources. & It is the result of superposition of secondary wavelets originating from different points of the same wavefront.

\hline
Fringes are of equal width. & Fringes are of unequal width; the central maximum is twice as wide as other secondary maxima.

\hline
The intensity of all bright fringes is the same. & The intensity of bright fringes decreases rapidly as we move away from the central maximum.

\hline
The dark fringes are perfectly dark, having zero intensity. & The dark fringes are not perfectly dark.

\hline
\end{tabular


Calculation:
The angular fringe separation (\(\theta\)) in a double-slit experiment is given by: \[ \theta = \frac{\lambda}{d} \]
where \(\lambda\) is the wavelength of light and \(d\) is the distance between the slits.

In air:

Wavelength, \(\lambda_{air} = 589 \, nm\)
Angular separation, \(\theta_{air} = 0.20^{\circ}\)

So, \(0.20^{\circ} = \frac{\lambda_{air}}{d}\).

When the apparatus is immersed in water:

The slit separation \(d\) remains the same.
The wavelength of light changes. The new wavelength in water is \(\lambda_{water}\).
Refractive index of water, \(n_w = 1.33\).

The relationship between wavelengths is \(\lambda_{water} = \frac{\lambda_{air}}{n_w}\).
The new angular fringe separation in water (\(\theta_{water}\)) will be: \[ \theta_{water} = \frac{\lambda_{water}}{d} = \frac{(\lambda_{air}/n_w)}{d} = \frac{1}{n_w} \left(\frac{\lambda_{air}}{d}\right) \]
Since we know \(\frac{\lambda_{air}}{d} = \theta_{air}\), we can substitute this into the equation: \[ \theta_{water} = \frac{\theta_{air}}{n_w} \] \[ \theta_{water} = \frac{0.20^{\circ}}{1.33} \approx 0.1503^{\circ} \]
The new angular fringe separation is approximately 0.15\(^{\circ}\). Quick Tip: When an optical experiment is moved from air into a medium like water, the only thing that changes is the wavelength of light (\(\lambda_{new} = \lambda_{air}/n\)). This makes the fringes closer together.

Einstein's Photoelectric Equation:
The equation is: \[ K_{max} = h\nu - \phi_0 \]
or \[ h\nu = \phi_0 + K_{max} \]

Physical Significance of Each Term:

\(h\nu\): This term represents the energy of the incident photon. According to Planck's quantum theory, light consists of discrete packets of energy called photons. The energy of each photon is proportional to its frequency (\(\nu\)), where \(h\) is Planck's constant. This is the total energy supplied to a single electron on the metal surface.
\(\phi_0\): This is the work function of the metal. It represents the minimum amount of energy required to just liberate an electron from the surface of the metal, overcoming the attractive forces holding it. The work function is a characteristic property of the material.
\(K_{max}\): This is the maximum kinetic energy of the emitted electron (photoelectron). It represents the excess energy from the photon that is converted into the electron's kinetic energy after it has escaped the surface. The kinetic energy is a maximum because some electrons deeper within the metal may lose energy through collisions before they escape.


Calculation:
We need to find the energy (\(E\)) of the incident photon.
Given:

Wavelength of light, \(\lambda = 4000 \, \AA = 4000 \times 10^{-10} \, m = 4 \times 10^{-7} \, m\)

Constants:

Planck's constant, \(h = 6.63 \times 10^{-34} \, J \cdot s\)
Speed of light, \(c = 3 \times 10^8 \, m/s\)

The energy of a photon is given by the formula: \[ E = h\nu = \frac{hc}{\lambda} \] \[ E = \frac{(6.63 \times 10^{-34} \, J \cdot s)(3 \times 10^8 \, m/s)}{4 \times 10^{-7} \, m} \] \[ E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} \, J \] \[ E = 4.9725 \times 10^{-19} \, J \]
The energy of the incident photon is \(4.97 \times 10^{-19}\) Joules. Quick Tip: Think of it as a transaction: The photon gives its energy (\(h\nu\)) to the electron. The electron pays a "tax" or "exit fee" (\(\phi_0\)) to leave the metal. The leftover money is its kinetic energy (\(K_{max}\)).

Four Uses of Van de Graaff Generator:

Nuclear Physics Research: It is used to accelerate charged particles (like protons, deuterons) to high energies. These high-energy particles are then used to bombard atomic nuclei to study nuclear reactions and structure.
Particle Accelerators: It often serves as the initial stage (injector) for larger particle accelerators.
Medical Applications: High-energy beams produced by the generator can be used to generate X-rays for radiation therapy in treating cancer.
Industrial Applications: The high-energy X-rays can be used for non-destructive testing of materials and welds, and the electron beams can be used for sterilization.


Calculation:
We need to calculate the displacement current (\(I_d\)).
Given:

Rate of change of electric field, \(\frac{dE}{dt} = 2 \times 10^{11} \, V/m \cdot s\)
Area of plates, \(A = 20 \, cm^2 = 20 \times 10^{-4} \, m^2\)

Constant:

Permittivity of free space, \(\epsilon_0 = 8.85 \times 10^{-12} \, F/m\)

The displacement current is given by Maxwell's equation: \[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \(\Phi_E\) is the electric flux. For a parallel plate capacitor with a uniform field, \(\Phi_E = E \cdot A\).
Since the area \(A\) is constant: \[ \frac{d\Phi_E}{dt} = A \frac{dE}{dt} \]
Substituting this into the formula for \(I_d\): \[ I_d = \epsilon_0 A \frac{dE}{dt} \] \[ I_d = (8.85 \times 10^{-12}) \times (20 \times 10^{-4}) \times (2 \times 10^{11}) \] \[ I_d = (8.85 \times 20 \times 2) \times 10^{(-12 - 4 + 11)} \] \[ I_d = 354 \times 10^{-5} \, A \] \[ I_d = 0.00354 \, A = 3.54 \, mA \]
The displacement current is 3.54 mA. Quick Tip: Maxwell's brilliant idea was that a changing electric field acts like a current. The displacement current isn't a flow of charges, but it creates a magnetic field just as a real current does. This insight completed the theory of electromagnetism.