CBSE Class 12 Physics Set 2 - (55/2/2) Question Paper 2026 with Solution PDF

Concept:

The electrostatic potential energy of a system of multiple point charges is the total work done in bringing these charges from infinity to their respective positions in the assembly.

For a system of three point charges \(q_1\), \(q_2\), and \(q_3\) separated by distances \(r_{12}\), \(r_{23}\), and \(r_{13}\) respectively, the total potential energy \(U\) is given by the algebraic sum of the potential energies of all unique pairs.

The foundational formula used is \(U = \frac{1}{4\pi\epsilon_0} \left[ \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \right]\).



Step 1: {\color{redIdentify the charges and their coordinates

Let the three charges be defined as follows:
\(q_1 = +q\) located at position \(x_1 = -1 m\).
\(q_2 = -2q\) located at position \(x_2 = 0 m\).
\(q_3 = +q\) located at position \(x_3 = 2 m\).


Step 2: {\color{redCalculate the separation distances between each pair

The distance between \(q_1\) and \(q_2\) is \(r_{12} = |x_2 - x_1| = |0 - (-1)| = 1 m\).

The distance between \(q_2\) and \(q_3\) is \(r_{23} = |x_3 - x_2| = |2 - 0| = 2 m\).

The distance between \(q_1\) and \(q_3\) is \(r_{13} = |x_3 - x_1| = |2 - (-1)| = 3 m\).


Step 3: {\color{redApply the potential energy formula for the system

Substitute the charges and their calculated distances into the potential energy equation:
\[ U = \frac{1}{4\pi\epsilon_0} \left[ \frac{(+q)(-2q)}{1} + \frac{(-2q)(+q)}{2} + \frac{(+q)(+q)}{3} \right] \]

Carefully compute the products in the numerators:
\[ U = \frac{1}{4\pi\epsilon_0} \left[ \frac{-2q^2}{1} + \frac{-2q^2}{2} + \frac{q^2}{3} \right] \]

Simplify the fractions inside the bracket:
\[ U = \frac{1}{4\pi\epsilon_0} \left[ -2q^2 - q^2 + \frac{q^2}{3} \right] \]

Combine the integer terms:
\[ U = \frac{1}{4\pi\epsilon_0} \left[ -3q^2 + \frac{q^2}{3} \right] \]


Step 4: {\color{redFind the final simplified expression

Find a common denominator to add the terms inside the bracket:
\[ -3q^2 + \frac{q^2}{3} = \frac{-9q^2 + q^2}{3} = \frac{-8q^2}{3} \]

Now substitute this back into the total energy expression:
\[ U = \frac{1}{4\pi\epsilon_0} \left[ \frac{-8q^2}{3} \right] \]

Cancel out the common factor of \(4\) in the numerator and the denominator:
\[ U = -\frac{2q^2}{3\pi\epsilon_0} \]


Step 5: {\color{redConclusion

The calculated potential energy perfectly matches option (B). The negative sign indicates that the system is bound and energy would need to be supplied to break the charges apart to infinity.
Quick Tip: Always draw a quick 1D axis and plot the points to avoid silly mistakes when calculating the relative distances between charges.

Concept:

The photoelectric effect describes the emission of electrons when light hits a photosensitive material.

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by \(K_{max} = \frac{hc}{\lambda} - \Phi\), where \(\lambda\) is the wavelength of incident light and \(\Phi\) is the work function.

The stopping potential \(V_0\) is directly proportional to this maximum kinetic energy via the relation \(eV_0 = K_{max}\).

Therefore, a smaller wavelength \(\lambda\) precisely corresponds to a higher incident photon energy, which inevitably leads to a higher stopping potential.



Step 1: {\color{redAnalyze the given wavelengths in different spectral regions

The problem states that radiations \(\lambda_1\) and \(\lambda_2\) strictly belong to the Ultraviolet (UV) spectrum.

The radiation \(\lambda_3\) belongs entirely to the visible light spectrum.

From the standard electromagnetic spectrum, we universally know that UV light possesses much shorter wavelengths than any visible light.

Therefore, we can establish the fundamental inequality: \(\lambda_{UV} < \lambda_{visible}\).

This strictly implies that both \(\lambda_1\) and \(\lambda_2\) are distinctly smaller than \(\lambda_3\).


Step 2: {\color{redOrder the wavelengths from smallest to largest

The problem explicitly provides a crucial condition between the two UV wavelengths: \(\lambda_2 > \lambda_1\).

Combining this given fact with our spectral knowledge from Step 1, we can write a complete, uninterrupted inequality for all three wavelengths:
\[ \lambda_1 < \lambda_2 < \lambda_3 \]


Step 3: {\color{redRelate wavelengths to incident photon energies

The energy of an individual photon is mathematically given by \(E = \frac{hc}{\lambda}\).

Because energy is inversely proportional to wavelength, the inequality order flawlessly reverses for the photon energies.

Thus, the corresponding energies of the incident radiations follow the strict order:
\[ E_1 > E_2 > E_3 \]


Step 4: {\color{redRelate photon energies to stopping potentials

Einstein's equation \(eV_0 = E - \Phi\) securely connects photon energy to stopping potential.

Since the work function \(\Phi\) remains perfectly constant for the same given photosensitive surface in all three cases, the stopping potential \(V_0\) will simply scale directly with the incident photon energy \(E\).

A higher photon energy definitively demands a correspondingly higher stopping potential to completely halt the fastest emitted photoelectrons.

Consequently, the stopping potentials must inherently follow the exact same ordering as the photon energies:
\[ V_1 > V_2 > V_3 \]


Step 5: {\color{redConclusion

Comparing our rigorously derived relationship with the given choices, we find that option (C) perfectly matches our result.
Quick Tip: Always remember the golden rule of the electromagnetic spectrum: Higher Frequency \(\implies\) Shorter Wavelength \(\implies\) Higher Energy \(\implies\) Higher Stopping Potential.

Concept:

The Bohr model successfully applies classical mechanics and early quantum quantization rules to deeply describe the hydrogen atom.

The electron revolves in well-defined circular orbits around the central positive nucleus, with the electrostatic Coulomb force securely providing the requisite centripetal force.

The absolute radius of the \(n\)-th allowed Bohr orbit is firmly established to be directly proportional to the square of the principal quantum number \(n\) (\(r_n \propto n^2\)).

The electrostatic potential energy of the electron-nucleus system depends strictly on their mutual separation distance.



Step 1: {\color{redState the formula for electrostatic potential energy

For a hydrogen atom, the central nucleus has a charge of \(+e\) (one proton) and the revolving electron has a charge of \(-e\).

The electrostatic potential energy \(U_n\) of this dynamic system at a separation distance \(r_n\) is mathematically formulated as:
\[ U_n = \frac{1}{4\pi\epsilon_0} \frac{(+e)(-e)}{r_n} = -\frac{1}{4\pi\epsilon_0} \frac{e^2}{r_n} \]

Here, it is undeniably evident that the potential energy \(U_n\) is strictly inversely proportional to the orbital radius \(r_n\).


Step 2: {\color{redRelate the orbital radius to the principal quantum number

From Bohr's rigid quantization condition for angular momentum (\(mvr = n\frac{h}{2\pi}\)), the explicitly derived formula for the radius of the \(n\)-th orbit is:
\[ r_n = \frac{\epsilon_0 h^2 n^2}{\pi m e^2} \]

From this fundamental expression, we can clearly isolate the crucial proportionality:
\[ r_n \propto n^2 \]


Step 3: {\color{redDetermine the variation of potential energy with 'n'

We now systematically substitute this proportional relationship for \(r_n\) directly back into the potential energy equation.

Since \(U_n \propto \frac{1}{r_n}\) and \(r_n \propto n^2\), combining these two mathematically yields:
\[ U_n \propto \frac{1}{n^2} \]

This elegantly demonstrates that the absolute magnitude of the potential energy diminishes according to the inverse square of the principal quantum number.


Step 4: {\color{redConclusion

The potential energy firmly varies as \(1/n^2\). This precisely aligns with option (A). It is also worth noting that the total energy and kinetic energy also scale exactly with \(1/n^2\) in the Bohr model.
Quick Tip: In the Bohr model, practically all energy parameters (Kinetic Energy, Potential Energy, Total Energy) for an electron strictly vary proportionally as \(1/n^2\).

Concept:

When two solid conductive metal spheres are physically brought into electrical contact (or seamlessly connected via a conductive wire), charge will immediately flow aggressively between them.

This rapid redistribution of charge firmly obeys the universal law of conservation of charge, completely ensuring no net charge is ever lost or created.

The transient flow of electrons strictly ceases only when the entire connected system achieves perfect electrostatic equilibrium, meaning both spheres securely attain the exact same common electrostatic potential.

The total capacitance of this newly combined system determines the final numerical value of this shared common potential.



Step 1: {\color{redAnalyze Option (A) - Charge Conservation

The foundational principle of electrostatics dictates that for any isolated system, the net total electrical charge remains forever constant.

When the two spheres make physical contact, they effectively merge into a single isolated conductive system.

Thus, the total initial charge perfectly equals the total final charge: \(Q_{total} = q_1 + q_2\).

Therefore, statement (A) is absolutely correct.


Step 2: {\color{redAnalyze Option (B) - Common Potential

By fundamental definition, a conductive material in stable equilibrium constitutes a complete equipotential volume.

When the two distinct spheres connect, they form one unified conductor. Charge will dynamically shift until the potential difference strictly becomes zero.

Hence, both spheres must inevitably reach the exact same common potential \(V\).

Therefore, statement (B) is absolutely correct.


Step 3: {\color{redAnalyze Option (C) - Derivation of Common Potential

We assume the spheres are sufficiently far apart such that their mutual electrostatic influence is negligible, acting purely as parallel capacitors.

The intrinsic capacitance of an isolated spherical conductor of radius \(r\) is formally given by \(C = 4\pi\epsilon_0 r\).

The total capacitance of the combined connected system is the sum of their individual capacitances:
\[ C_{total} = C_1 + C_2 = 4\pi\epsilon_0 r_1 + 4\pi\epsilon_0 r_2 = 4\pi\epsilon_0 (r_1 + r_2) \]

The final common potential \(V\) is mathematically derived by dividing the conserved total charge by the total system capacitance:
\[ V = \frac{Q_{total}}{C_{total}} = \frac{q_1 + q_2}{4\pi\epsilon_0 (r_1 + r_2)} \]

Therefore, statement (C) is mathematically correct.


Step 4: {\color{redAnalyze Option (D) - Dimensional Analysis

Let's rigorously examine the algebraic expression presented in statement (D):
\[ V' = \frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2) (r_1 + r_2)}{r_1 r_2} \]

In the numerator, we have physical dimensions of \([Charge] \times [Length]\).

In the denominator, we have \([Length]^2\).

This simplifies dimensionally to \([Charge] / [Length]\), which is standard for potential. Let us check the full expression.

The correct potential derived in Step 3 is \(\frac{1}{4\pi\epsilon_0} \frac{Q}{R}\), which represents \([Charge] / [Length]\).

The expression in (D) evaluates to \(\frac{1}{4\pi\epsilon_0} \frac{Q \cdot L}{L^2} = \frac{1}{4\pi\epsilon_0} \frac{Q}{L}\). Dimensionally it is not obviously wrong, but mathematically it is completely incorrect based on our rigorous derivation in Step 3.

Since there can be only one correct specific mathematical expression for the final common potential, and (C) is the proven correct one, (D) must undeniably be the incorrect statement.


Step 5: {\color{redConclusion

The question explicitly asks to identify the statement that is NOT correct. Based on our detailed derivation, statement (D) provides an entirely flawed formula for the final system potential.
Quick Tip: When tackling "Which is NOT correct" questions, always systematically verify the basic physical conservation laws (like charge and energy) first, as they quickly eliminate the definitely true statements.

Concept:

When an excited electron in a hydrogen atom securely drops from a higher energy orbit (\(n_i\)) to a lower energy orbit (\(n_f\)), it strictly emits a single photon.

The exact energy of this uniquely emitted photon is absolutely equal to the numerical energy difference between those two specific stationary states, governed by \(E_{photon} = E_i - E_f\).

The mathematical frequency of the emitted photon directly scales with this energy gap according to Planck's fundamental equation: \(E_{photon} = h\nu\).

The energy of an electron in the \(n\)-th Bohr orbit is effectively written as \(E_n = \frac{-13.6 eV}{n^2}\).



Step 1: {\color{redAnalyze the first transition (\(n=4 \rightarrow n=1\))

The energy gap for an electron jumping deeply from the fourth orbit down to the ground state is:
\[ \Delta E_1 = E_4 - E_1 \]

Substitute the standard Bohr energy level formula:
\[ \Delta E_1 = \left(\frac{-13.6}{4^2}\right) - \left(\frac{-13.6}{1^2}\right) \]
\[ \Delta E_1 = 13.6 \times \left( \frac{1}{1^2} - \frac{1}{4^2} \right) eV \]

Calculate the fractional part carefully:
\[ \Delta E_1 = 13.6 \times \left( 1 - \frac{1}{16} \right) = 13.6 \times \left( \frac{15}{16} \right) eV \]

The frequency \(\nu\) of the emitted photon is linked directly to this energy gap:
\[ h\nu = 13.6 \times \left( \frac{15}{16} \right) \]


Step 2: {\color{redAnalyze the second transition (\(n=4 \rightarrow n=2\))

Now, consider a different scenario where the electron drops from the fourth orbit only down to the second orbit (Balmer series).

The new energy gap is calculated similarly:
\[ \Delta E_2 = E_4 - E_2 \]
\[ \Delta E_2 = 13.6 \times \left( \frac{1}{2^2} - \frac{1}{4^2} \right) eV \]

Calculate the fractional part carefully:
\[ \Delta E_2 = 13.6 \times \left( \frac{1}{4} - \frac{1}{16} \right) \]

Find a common denominator to complete the strict subtraction:
\[ \Delta E_2 = 13.6 \times \left( \frac{4}{16} - \frac{1}{16} \right) = 13.6 \times \left( \frac{3}{16} \right) eV \]

Let the unknown frequency of this newly emitted photon be \(\nu'\). According to Planck's equation:
\[ h\nu' = 13.6 \times \left( \frac{3}{16} \right) \]


Step 3: {\color{redCalculate the ratio of the frequencies

To rigorously find the mathematical relationship between the two frequencies, we divide the second equation strictly by the first equation:
\[ \frac{h\nu'}{h\nu} = \frac{13.6 \times \left( \frac{3}{16} \right)}{13.6 \times \left( \frac{15}{16} \right)} \]

The physical constants (\(h\)) and the \(13.6 eV\) scaling factor elegantly cancel out from both the numerator and the denominator:
\[ \frac{\nu'}{\nu} = \frac{\frac{3}{16}}{\frac{15}{16}} \]

The common denominator of \(16\) also nicely cancels out entirely:
\[ \frac{\nu'}{\nu} = \frac{3}{15} \]

Simplify this final basic fraction:
\[ \frac{\nu'}{\nu} = \frac{1}{5} \implies \nu' = \frac{\nu}{5} \]


Step 4: {\color{redConclusion

The new frequency \(\nu'\) mathematically evaluates exactly to \(\frac{\nu}{5}\). This directly corresponds to option (C).
Quick Tip: Using the Rydberg formula structure \(\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\) is frequently the fastest and most foolproof algebraic method to compare spectral transition energies without computing absolute numerical eV values.

Concept:

Faraday's Law of Electromagnetic Induction dictates that a changing magnetic flux seamlessly cutting through a wire loop will absolutely induce an electromotive force (EMF) across it.

However, for this induced EMF to physically drive an actual induced electrical current, the loop must constitute a completely closed conductive electrical circuit.

Lenz's Law further states that any induced current will flow in a specific direction such that its own generated magnetic field vigorously opposes the physical change in flux that fundamentally caused it.

This magnetic opposition is what typically causes a falling magnet to perceptibly slow down when dropping through solid metallic rings or tubes.



Step 1: {\color{redAnalyze the physical geometry of the provided loop

The problem explicitly states and visually illustrates that the small loop possesses a "small cut".

This seemingly minor detail is profoundly important. A cut physically breaks the continuous electrical pathway.

Because the conductive pathway is broken, the loop inherently forms an "open circuit" with functionally infinite electrical resistance.


Step 2: {\color{redEvaluate the electromagnetic induction effects

As the permanent magnet physically falls downwards due to gravity, the magnetic flux penetrating the area of the open loop genuinely changes over time.

Consequently, according to Faraday's law, a real electromotive force (EMF) is definitively induced strictly across the two ends of the small cut.

However, because the electrical circuit is permanently open, absolutely no induced electrical current can flow through the wire loop (\(I = \frac{EMF}{\infty} = 0\)).


Step 3: {\color{redDetermine the consequence of zero induced current

Lenz's law relies entirely on the presence of an induced current to generate a secondary opposing magnetic field.

Since there is zero induced current circulating in the open loop, there is absolutely no induced magnetic field created by the loop.

Consequently, the falling magnet experiences zero upward opposing electromagnetic force.

The only physical force acting on the magnet throughout its entire journey is the constant downward pull of Earth's gravity.


Step 4: {\color{redConclusion regarding kinematics

Because gravity is the sole force acting, the net force on the magnet strictly equals its weight (\(F_{net} = mg\)).

Using Newton's Second Law (\(F = ma\)), we see the downward acceleration of the magnet is precisely \(a = g\).

Since the gravitational acceleration \(g\) is practically a constant, the acceleration of the magnet remains completely uniform throughout its entire fall, regardless of its position relative to the cut loop.

This unyielding uniformity directly validates option (D) as the uniquely correct statement.
Quick Tip: Always aggressively check the physical continuity of any loop in electromagnetic induction problems; an open loop generates EMF but definitively prevents any opposing force from manifesting.

Concept:

The mathematical representation of a propagating electromagnetic wave involves a sinusoidal function typically in the standard form \(B = B_0 \sin(kx \pm \omega t)\).

Here, \(k\) definitively represents the angular wave number, and \(\omega\) represents the angular frequency.

The physical propagation speed \(v\) of any such wave is robustly determined by the mathematical ratio of its angular frequency to its wave number: \(v = \frac{\omega}{k}\).

For an electromagnetic wave strictly travelling through the vacuum of free space, this calculated speed must exactly equal the universal speed of light, \(c \approx 3 \times 10^8 m/s\).



Step 1: {\color{redExtract parameters and calculate speed for Wave I

From the given expression for \(B_1\), we identify the coefficients:

Angular wave number, \(k_1 = 0.7 \times 10^3 rad/m\).

Angular frequency, \(\omega_1 = 1.4 \times 10^{11} rad/s\).

Calculate the corresponding wave speed \(v_1\):
\[ v_1 = \frac{\omega_1}{k_1} = \frac{1.4 \times 10^{11}}{0.7 \times 10^3} \]
\[ v_1 = 2 \times 10^8 m/s \]

This speed is significantly less than \(c\), meaning it is traveling in a denser medium, not free space.


Step 2: {\color{redExtract parameters and calculate speed for Wave II

From the given expression for \(B_2\):
\(k_2 = 0.6 \times 10^3 rad/m\).
\(\omega_2 = 1.5 \times 10^{11} rad/s\).

Calculate the corresponding wave speed \(v_2\):
\[ v_2 = \frac{\omega_2}{k_2} = \frac{1.5 \times 10^{11}}{0.6 \times 10^3} \]
\[ v_2 = 2.5 \times 10^8 m/s \]

This is also clearly not equal to the definitive speed of light \(c\).


Step 3: {\color{redExtract parameters and calculate speed for Wave III

From the given expression for \(B_3\):
\(k_3 = 0.5 \times 10^3 rad/m\).
\(\omega_3 = 1.5 \times 10^{11} rad/s\).

Calculate the corresponding wave speed \(v_3\):
\[ v_3 = \frac{\omega_3}{k_3} = \frac{1.5 \times 10^{11}}{0.5 \times 10^3} \]
\[ v_3 = 3.0 \times 10^8 m/s \]

This meticulously matches the widely accepted speed of light in free space perfectly.


Step 4: {\color{redExtract parameters and calculate speed for Wave IV for completeness

From the given expression for \(B_4\):
\(k_4 = 0.2 \times 10^4 rad/m\).
\(\omega_4 = 4.8 \times 10^{11} rad/s\).

Calculate the corresponding wave speed \(v_4\):
\[ v_4 = \frac{\omega_4}{k_4} = \frac{4.8 \times 10^{11}}{0.2 \times 10^4} = 24 \times 10^7 = 2.4 \times 10^8 m/s \]

This is also incorrect for free space.


Step 5: {\color{redConclusion

Only Wave III possesses a calculated propagation velocity that exactly equals \(3 \times 10^8 m/s\). Therefore, only Wave III is actively travelling in free space. This solidly aligns with option (C).
Quick Tip: When visually inspecting wave equations, quickly glancing at the ratio of the coefficients of '\(t\)' over '\(x\)' immediately reveals the wave's phase velocity.

Concept:

Materials in nature are broadly classified into three distinct categories strictly based on their inherent magnetic properties and their interaction with external magnetic fields.

Diamagnetic materials (e.g., Copper, Lead, Water) create a weak opposing internal magnetic field. For them, the relative magnetic permeability is strictly slightly less than one (\(\mu_r < 1\)).

Paramagnetic materials (e.g., Aluminium, Platinum, Oxygen) create a very weak reinforcing internal magnetic field. For them, the relative magnetic permeability is strictly slightly greater than one (\(\mu_r > 1\), but close to 1).

Ferromagnetic materials (e.g., Iron, Cobalt, Nickel) possess large crystalline domains that aggressively align with external fields, generating massive internal reinforcement. For them, the relative magnetic permeability is phenomenally larger than one (\(\mu_r \gg 1\), often in the thousands).



Step 1: {\color{redAnalyze the given mathematical condition

The problem presents the specific mathematical inequality condition: \(\mu_r \gg 1\).

The "much greater than" symbol definitively isolates the required material class.

This specific condition is the textbook hallmark exclusively characterizing ferromagnetic materials.


Step 2: {\color{redEvaluate the provided substance options

We systematically examine the known magnetic classification of each option provided:

(A) Aluminium: It is a widely known paramagnetic material. Its permeability is only slightly greater than 1, failing the "much greater" test.

(B) Copper: It is a textbook diamagnetic material. Its permeability is physically less than 1.

(C) Lead: It is also strongly diamagnetic, meaning its permeability is distinctly less than 1.

(D) Nickel: It is one of the very few famous elements that are strongly ferromagnetic at room temperature (alongside Iron and Cobalt).


Step 3: {\color{redConclusion

Since Nickel is the only genuine ferromagnetic material securely listed among the choices, it uniquely satisfies the extreme condition \(\mu_r \gg 1\). This perfectly aligns with option (D).
Quick Tip: Memorize the "Iron Triad" (Iron, Cobalt, Nickel) as the three quintessential ferromagnetic elements encountered relentlessly in physics exam problems.

Concept:

The Biot-Savart Law mathematically calculates the exact magnetic field generated by an infinitesimally small current-carrying wire segment at any specific point in surrounding space.

The law is officially stated in its rigorous vector form as: \(d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3}\).

The definitive direction of the resulting magnetic field \(d\vec{B}\) is strictly dictated by the mathematical cross product of the current element vector \(d\vec{l}\) and the position vector \(\vec{r}\).

The cross product inherently guarantees that the resulting magnetic field vector is definitively perpendicular to both the current element vector and the position vector, adhering strictly to the right-hand rule.



Step 1: {\color{redIdentify the fundamental vector components from the problem

The conductive wire carries an active current flowing strictly along the positive x-axis.

Therefore, the infinitesimal current element vector \(d\vec{l}\) points squarely in the \(+\hat{i}\) direction:
\[ d\vec{l} = dx \hat{i} \]

The target observation point is located specifically at the spatial coordinates \((0, 40 cm, 0)\).

This point clearly lies entirely on the positive y-axis.

Therefore, the position vector \(\vec{r}\) extending directly from the origin to the observation point firmly points in the \(+\hat{j}\) direction:
\[ \vec{r} = 40 \hat{j} cm \]


Step 2: {\color{redExecute the required vector cross product

To confidently ascertain the specific direction of the generated magnetic field \(d\vec{B}\), we must carefully evaluate the core cross product \((d\vec{l} \times \vec{r})\).

We substitute the identified unit vector directions directly into the cross product:
\[ Direction of d\vec{B} \propto (\hat{i} \times \hat{j}) \]

According to the universally established standard cyclic rules for orthogonal Cartesian unit vectors:
\[ \hat{i} \times \hat{j} = \hat{k} \]


Step 3: {\color{redInterpret the mathematical result physically

The final unit vector \(\hat{k}\) definitively represents the positive z-axis in a standard 3D coordinate system.

Therefore, the infinitesimally small magnetic field generated strictly by that tiny 1 cm central segment points straight out along the positive z-axis.


Step 4: {\color{redConclusion

The generated magnetic field points robustly along the z-axis, which flawlessly matches option (C).
Quick Tip: For cross products, consistently picture the Right-Hand Rule: If your index finger points to the current (\(\hat{i}\)) and your middle finger points to the location (\(\hat{j}\)), your thumb automatically points to the magnetic field (\(\hat{k}\)).

Concept:

A sensitive galvanometer is effectively and safely converted into a much higher-range ammeter by connecting a very low resistance resistor, called a shunt, strictly in parallel with it.

Because the galvanometer coil and the shunt resistor are firmly connected in parallel, the electrical potential difference (voltage drop) across both distinct pathways must be absolutely identical.

This creates a neat division of the total incoming current \(I\): a very small, safe fraction \(I_g\) flows through the delicate galvanometer, while the massive remainder \((I - I_g)\) bypasses it securely through the rugged shunt.



Step 1: {\color{redExtract the known variables from the problem

The internal electrical resistance of the bare galvanometer coil is strictly \(R_g = 27 \Omega\).

The tiny parallel shunt resistance creatively used for the conversion is firmly \(S = 3 \Omega\).

The maximum target measurable range of the newly constructed ammeter is the total current \(I = 10 mA\).

The objective is to accurately find the specific current \(I_g\) that physically forces the galvanometer to show its maximum full-scale deflection.


Step 2: {\color{redEstablish the foundational parallel voltage equation

Since the two resistors are strictly parallel, we systematically equate their voltage drops using Ohm's Law (\(V = IR\)):
\[ V_{galvanometer} = V_{shunt} \]
\[ I_g \times R_g = (I - I_g) \times S \]


Step 3: {\color{redSubstitute the known values and solve the algebra

Insert the meticulously collected numerical values directly into our established equation:
\[ I_g \times 27 = (10 - I_g) \times 3 \]

We can cleanly simplify the math by immediately dividing both sides entirely by 3:
\[ I_g \times 9 = 10 - I_g \]

Carefully rearrange the resulting terms to properly group the unknown variable \(I_g\) onto the left side:
\[ 9 I_g + I_g = 10 \]

Combine the grouped terms:
\[ 10 I_g = 10 \]

Perform the final basic division to isolate \(I_g\):
\[ I_g = \frac{10}{10} = 1 mA \]


Step 4: {\color{redConclusion

The galvanometer coil itself requires precisely 1 mA of circulating current to safely reach its extreme full-scale deflection point. This elegantly and perfectly matches option (C).
Quick Tip: Notice that the shunt resistance (\(3 \Omega\)) is exactly \(1/9\)th of the galvanometer resistance (\(27 \Omega\)). This instantly means the shunt will greedily take \(9\) times more current than the galvanometer. If total current is \(10\) parts, galvanometer takes exactly \(1\) part.

Concept:

The rigorous Principle of Homogeneity of Dimensions constitutes a cornerstone of physics.

It fundamentally states that in any physically meaningful and correct mathematical equation, every single term being added, subtracted, or equated must possess the exact same dimensional units.

You absolutely cannot logically add apples to oranges; you cannot add volts to meters, nor can you subtract seconds from kilograms.



Step 1: {\color{redAnalyze the given mathematical equation

The provided time-dependent equation modeling the magnetic flux is:
\[ \phi = 5 At^2 + Bt - 2C \]

According to the problem statement, the unit of the resulting magnetic flux \(\phi\) on the left side is exclusively Webers (Wb).

The unit of the variable time \(t\) heavily featured on the right side is exclusively seconds (s).


Step 2: {\color{redApply the Principle of Homogeneity to the first term

Because the final outcome \(\phi\) is measured strictly in Webers, the entirety of the first mathematical term, \(5 At^2\), must also universally resolve to Webers.

The pure number 5 is completely dimensionless and irrelevant here.

Therefore, the dimensional unit of \((A \times t^2)\) must absolutely equal Wb.
\[ Unit of A \times (s)^2 = Wb \]

Algebraically isolating the unit for the unknown coefficient A yields:
\[ Unit of A = \frac{Wb}{s^2} = Wb s^{-2} \]


Step 3: {\color{redApply the Principle of Homogeneity to the second term

Following the exact same strict logical reasoning, the entirety of the second mathematical term, \(Bt\), must also definitively resolve to Webers.

Therefore, the dimensional unit of \((B \times t)\) must absolutely equal Wb.
\[ Unit of B \times (s) = Wb \]

Algebraically isolating the unit for the unknown coefficient B yields:
\[ Unit of B = \frac{Wb}{s} = Wb s^{-1} \]


Step 4: {\color{redConclusion

The rigorously derived SI units for the mathematical coefficients A and B are decisively Wb s\(^{-2}\) and Wb s\(^{-1}\) respectively. This perfectly and unquestionably matches option (C).
Quick Tip: Whenever an unknown physics coefficient is multiplied by time \(t^n\) to match a specific unit \(Y\), the unit of that coefficient is instantly and always \(Y \times s^{-n}\).

Concept:

In any alternating current (AC) circuit, a capacitor actively opposes the continuous flow of charge. This specific opposition is formally quantified as capacitive reactance, denoted as \(X_C\).

The explicit mathematical formula firmly defining capacitive reactance is \(X_C = \frac{1}{\omega C}\), where \(\omega\) is the driving angular frequency and \(C\) is the physical capacitance.

If we intentionally plot a mathematical graph with \(X_C\) safely on the y-axis and the reciprocal factor \((1/\omega)\) strictly on the x-axis, the resulting relation \(y = \left(\frac{1}{C}\right) x\) vividly forms a straight line passing smoothly through the origin.

The steepness, or mathematical slope \(m\), of this straight line is definitively equal to \(\frac{1}{C}\). Therefore, the slope is strictly inversely proportional to the actual capacitance.



Step 1: {\color{redExtract slopes from the given graph


From the provided image, we meticulously observe two distinctly sloped lines corresponding to the two capacitors.

The plotted line representing capacitor \(C_1\) creates an angle of exactly \(\theta_1 = 45^\circ\) directly with the positive x-axis.

The mathematical slope \(m_1\) for this first line is computed using the tangent function:
\[ m_1 = \tan(45^\circ) = 1 \]

The plotted line representing capacitor \(C_2\) creates an angle of exactly \(\theta_2 = 30^\circ\) directly with the positive x-axis.

The mathematical slope \(m_2\) for this second line is similarly computed:
\[ m_2 = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]


Step 2: {\color{redRelate the slopes back to capacitance

As robustly established in the concept section, the slope \(m\) of this specific type of graph is mathematically identical to \(\frac{1}{C}\).

Therefore, for the first capacitor:
\[ \frac{1}{C_1} = m_1 = 1 \implies C_1 = 1 \]

And for the second capacitor:
\[ \frac{1}{C_2} = m_2 = \frac{1}{\sqrt{3}} \implies C_2 = \sqrt{3} \]


Step 3: {\color{redCalculate the required final ratio

We are strictly asked to find the exact numerical value of the specific fractional ratio \(C_1 / C_2\).

We meticulously substitute the derived capacitance values into this fraction:
\[ \frac{C_1}{C_2} = \frac{1}{\sqrt{3}} \]


Step 4: {\color{redConclusion

The rigorously derived ratio of the two capacitances is exactly \(1 / \sqrt{3}\). This flawlessly and undeniably matches option (D).
Quick Tip: Always double-check the precise axis labels on physics graphs. A steeper slope here physically means a smaller capacitance because the slope represents \(1/C\), not \(C\). Intuition can easily be flipped if you don't check the math.

Concept:

The principle of linear superposition states that when multiple distinct waves simultaneously overlap in physical space, the resultant total displacement is the precise vector sum of their individual independent displacements.

If two coherent light waves interact and meet exactly out of phase (specifically, a peak flawlessly meets a trough), their opposing wave amplitudes aggressively cancel each other out entirely.

This highly specific phenomenon is officially termed destructive interference. The localized result of total destructive interference of visible light is visually complete darkness, representing zero net energy at that precise spatial spot.



Step 1: {\color{redEvaluate the Assertion (A)

The assertion boldly claims that "Light added to light can produce darkness."

In classical ray optics (particle theory), adding more light always increases brightness.

However, in wave optics, the phenomenon of interference proves that two light waves can indeed completely cancel each other out if they overlap with a phase difference of precisely \(\pi\), \(3\pi\), \(5\pi\), etc.

This cancellation genuinely produces observable dark fringes, such as those famously seen in Young's Double Slit Experiment.

Therefore, the Assertion (A) is absolutely and factually true.


Step 2: {\color{redEvaluate the Reason (R)

The reason clearly states that "When two coherent light waves interfere, there is darkness at position of destructive interference."

This sentence is the textbook, rigorous physical definition of how and why dark fringes aggressively form in any coherent optical interference pattern.

Therefore, the Reason (R) is also undeniably true.


Step 3: {\color{redDetermine if the Reason correctly explains the Assertion

The assertion is a counter-intuitive phenomenon (light + light = dark). The reason explicitly provides the exact scientific mechanism (destructive interference of coherent waves) that physically makes this strange phenomenon possible.

Because the reason actively and flawlessly explains exactly "why" the assertion is true, Reason (R) firmly constitutes the correct and complete explanation of Assertion (A).


Step 4: {\color{redConclusion

Since both statements are perfectly true and the reason provides the necessary correct physical explanation, the correct choice is definitely option (A).
Quick Tip: "Light + Light = Darkness" is the quintessential trademark signature exclusively of the wave nature of light, impossible to explain using classical Newtonian particle models.

Concept:

The rated power \(P\) of any standard electrical heating device explicitly designed to operate at a specific fixed voltage \(V\) is intrinsically inversely proportional to its internal physical resistance \(R\), governed by the formula \(P = \frac{V^2}{R}\).

When multiple electrical resistors are physically joined end-to-end in a series configuration, their total effective equivalent resistance strictly increases, combining as \(R_{eq} = R_1 + R_2\).

A higher overall system resistance connected to the exact same fixed voltage source will inevitably draw much less total current, thereby drastically reducing the total power consumed by the entire system.



Step 1: {\color{redEvaluate the Assertion (A)

Let the internal resistances of the two electric heaters be strictly \(R_1\) and \(R_2\).

Since their nominal rated powers at voltage \(V\) are \(P_1 = \frac{V^2}{R_1}\) and \(P_2 = \frac{V^2}{R_2}\).

When these two heaters are violently joined in a series circuit, their new total effective resistance becomes \(R_{eq} = R_1 + R_2\).

It is mathematically obvious that \(R_{eq}\) is strictly greater than \(R_1\) alone (\(R_{eq} > R_1\)).

The total combined power actually consumed by this new series combination when connected across the same source voltage \(V\) is:
\[ P_{series} = \frac{V^2}{R_{eq}} \]

Because the denominator has significantly increased (\(R_{eq} > R_1\)), the resulting overall fraction must inevitably decrease:
\[ \frac{V^2}{R_{eq}} < \frac{V^2}{R_1} \implies P_{series} < P_1 \]

Therefore, the power consumed by the series combination is undeniably less than that consumed by heater 1 operating alone. The Assertion (A) is absolutely true.


Step 2: {\color{redEvaluate the Reason (R)

The provided reason emphatically claims that the power consumed by an electrical device tightly connected to a fixed DC voltage source \(V\) is strictly proportional to its internal resistance.

The foundational mathematical formula governing this exact scenario is \(P = \frac{V^2}{R}\).

This formula clearly and undeniably demonstrates that, for any constant voltage source, the power consumed is strictly inversely proportional to the physical resistance (\(P \propto \frac{1}{R}\)), not directly proportional.

Therefore, the Reason (R) is fundamentally mathematically and physically false.


Step 3: {\color{redConclusion

Since the Assertion (A) is completely true but the corresponding Reason (R) contains a fatal physical error and is false, the correct choice is definitively option (C).
Quick Tip: Always aggressively verify whether a problem assumes a constant voltage (parallel/household circuits, use \(P = V^2/R\)) or a constant current (series circuits, use \(P = I^2 R\)) before jumping to proportionality conclusions about power and resistance.

Concept:

The photoelectric effect firmly relies on the quantum particle nature of light, where light comprises discrete energy packets called photons.

The optical intensity of a light beam physically corresponds directly to the sheer number of photons striking a specific target area per second. Increasing intensity vigorously increases the photon count, not their individual energy.

Assuming the constant frequency is safely above the threshold (\(\nu > \nu_0\)), more bombarding photons will cleanly eject more photoelectrons, directly and linearly increasing the total measured saturation photocurrent.

The stopping potential, conversely, depends strictly and only on the maximum kinetic energy of the ejected electrons, which is dictated exclusively by the incident photon frequency (energy), completely independent of intensity.



Step 1: {\color{redEvaluate the Assertion (A)

The assertion states that increasing the intensity of valid incident light (\(\nu > \nu_0\)) directly causes the resulting photocurrent to increase.

According to established quantum principles, higher light intensity essentially means a significantly larger number of photons are aggressively hitting the metal surface every second.

Since one incoming photon generally ejects exactly one electron, more impacting photons will invariably result in a higher number of emitted photoelectrons per second.

More moving electrons directly constitute a higher electrical current.

Therefore, the Assertion (A) is a scientifically true statement.


Step 2: {\color{redEvaluate the Reason (R)

The reason states that the stopping potential required for a given photosensitive surface physically increases when the frequency of the incident light is actively increased.

Einstein's robust photoelectric equation is elegantly stated as \(eV_0 = h\nu - \Phi\).

It is mathematically obvious from this linear equation that if the incident frequency \(\nu\) increases, the corresponding stopping potential \(V_0\) must also simultaneously and linearly increase.

Therefore, the Reason (R) is also a completely scientifically true statement.


Step 3: {\color{redDetermine if the Reason correctly explains the Assertion

The Assertion thoroughly discusses how changing light intensity directly manipulates the sheer \textit{number of flowing electrons (photocurrent).

The Reason, however, discusses an entirely separate phenomenon: how changing light \textit{frequency manipulates the maximum \textit{energy of those electrons (stopping potential).

While both physics statements are undeniably factual within the realm of the photoelectric effect, they are addressing two fundamentally completely unlinked parameters.

The reason spectacularly fails to provide any logical explanation whatsoever for why intensity impacts current.


Step 4: {\color{redConclusion

Both statements are individually perfectly true, but the reason is entirely disconnected from explaining the assertion. Thus, the correct choice is definitely option (B).
Quick Tip: Keep these rules permanently separated in your mind: Intensity controls only the \textbf{number (current), while Frequency controls only the \textbf{energy} (stopping potential).

Concept:

A standard semiconductor p-n junction inherently possesses a narrow depletion region strictly depleted of mobile charge carriers, maintaining a built-in potential barrier that fiercely opposes further natural diffusion.

The built-in electric field consistently points stubbornly from the n-region (which has exposed positive donor ions) towards the p-region (which has exposed negative acceptor ions).

Forward biasing is actively achieved by tightly connecting the external battery's positive terminal directly to the p-type region and the negative terminal to the n-type region.

This specific external battery arrangement forces an external electric field that directly and aggressively opposes the internal built-in field, fundamentally weakening it.



Step 1: {\color{redEvaluate the Assertion (A)

The assertion firmly claims that applying a forward bias actually increases the overall height of the internal barrier potential.

As detailed in the core concept, applying a forward bias inherently forces the majority carriers from both the p and n sides to aggressively rush towards the central junction.

This aggressive influx effectively shrinks the physical width of the depletion region.

Consequently, the net effective barrier potential is mathematically reduced from \(V_0\) down to \((V_0 - V_{applied})\).

Because the barrier height is actually drastically reduced, allowing current to flow easily, the Assertion (A) is entirely physically false.


Step 2: {\color{redEvaluate the Reason (R)

The reason claims that during forward biasing, the direction of the newly applied external voltage perfectly aligns in the exact same direction as the internal built-in potential.

Let's trace the fields carefully.

The internal built-in potential always directs from the n-side securely to the p-side.

In forward bias, the positive battery terminal connects to the p-side, and the negative to the n-side.

This rigorously creates an applied external electric field directing entirely from the p-side straight to the n-side.

Clearly, the applied field and the built-in field are pointing in perfectly opposite, anti-parallel directions, relentlessly fighting each other.

Therefore, the Reason (R) makes a fundamentally incorrect claim and is completely false.


Step 3: {\color{redConclusion

Both the Assertion and the Reason are riddled with fundamental physical errors regarding the basic operation of diodes. Therefore, both are entirely false, pointing strictly to option (D).
Quick Tip: Forward bias always means "fighting" the barrier (reducing its height to let current through). Reverse bias means "reinforcing" the barrier (making it taller to block current).

Concept:

When an extended object (like a long pencil) is placed perfectly horizontally along the principal axis of a spherical mirror, its final resulting image will inherently possess a specific measurable length.

This longitudinal image length is meticulously found by treating the two extreme physical ends of the object as two entirely separate point objects.

We robustly apply the standard spherical mirror formula \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) twice, independently, to locate the precise spatial positions of the images of both ends.

The absolute final length of the image is simply the mathematical magnitude of the spatial difference between these two computed image positions.



Step 1: {\color{redIdentify the mirror parameters using Cartesian sign convention

The optical device is explicitly a concave mirror. Therefore, its primary focal point lies in front of it, meaning its focal length is strictly negative.

Focal length, \(f = -20 cm\).

The entire object is placed physically in front of the reflecting surface, so all object distances are strictly negative.


Step 2: {\color{redCalculate the image position for the nearest end of the pencil

The nearest end of the 5 cm pencil is stated to be exactly 25 cm from the mirror's pole.

So, \(u_1 = -25 cm\).

Apply the mirror formula:
\[ \frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f} \]
\[ \frac{1}{v_1} + \frac{1}{-25} = \frac{1}{-20} \]

Rearrange to isolate the unknown image distance term:
\[ \frac{1}{v_1} = \frac{1}{25} - \frac{1}{20} \]

Find the lowest common denominator (100) to subtract the fractions smoothly:
\[ \frac{1}{v_1} = \frac{4 - 5}{100} = \frac{-1}{100} \]

Taking the reciprocal solidly yields the position of the first end's image:
\[ v_1 = -100 cm \]


Step 3: {\color{redCalculate the image position for the farthest end of the pencil

Since the pencil is exactly 5 cm long and lies completely along the axis starting from 25 cm, its other, farther end must logically be at a distance of \(25 + 5 = 30 cm\).

So, \(u_2 = -30 cm\).

Apply the mirror formula again independently:
\[ \frac{1}{v_2} + \frac{1}{-30} = \frac{1}{-20} \]

Rearrange to isolate the unknown image distance term:
\[ \frac{1}{v_2} = \frac{1}{30} - \frac{1}{20} \]

Find the lowest common denominator (60) to subtract the fractions smoothly:
\[ \frac{1}{v_2} = \frac{2 - 3}{60} = \frac{-1}{60} \]

Taking the reciprocal solidly yields the position of the second end's image:
\[ v_2 = -60 cm \]


Step 4: {\color{redDetermine the final longitudinal length of the image

Both ends of the pencil generate real images positioned at 100 cm and 60 cm in front of the mirror respectively.

The total physical length of this extended image is simply the absolute difference between these two distinct coordinates:
\[ Length of image = |v_1 - v_2| \]
\[ Length of image = |-100 - (-60)| = |-100 + 60| = |-40| \]
\[ Length of image = 40 cm \]


Step 5: {\color{redConclusion

The calculated length of the pencil's image is dramatically stretched out to 40 cm. Since both image coordinates are negative, it forms a completely real image lying entirely in front of the mirror.
Quick Tip: For objects placed completely horizontally along the principal axis (longitudinal magnification), never just use the transverse magnification formula \(m = -v/u\) directly on the length. Always rigorously find the two end points separately.

Concept:

In Young's Double Slit Experiment, different wavelengths of incident light will simultaneously produce their own entirely independent interference patterns on the shared observation screen.

Because the two distinct patterns possess drastically different intrinsic fringe widths, their respective bright maxima will mostly fall at entirely different spatial locations.

However, at certain specific distances from the center, a bright fringe of one wavelength will perfectly overlap and physically coincide with a bright fringe of the second wavelength.

This perfect coincidence mathematically occurs exactly when the spatial distance \(y\) from the central maximum is absolutely identical for both independent interference patterns.



Step 1: {\color{redList the provided experimental parameters strictly in SI units

First wavelength, \(\lambda_1 = 500 nm = 500 \times 10^{-9} m\).

Second wavelength, \(\lambda_2 = 600 nm = 600 \times 10^{-9} m\).

Screen distance, \(D = 1.8 m\).

Slit separation, \(d = 0.3 mm = 0.3 \times 10^{-3} m = 3 \times 10^{-4} m\).


Step 2: {\color{redEstablish the mathematical condition for coincidence

The generalized position for the \(n\)-th bright fringe measured directly from the central maximum is formally given by the established formula:
\[ y_n = n \frac{\lambda D}{d} \]

For perfect spatial coincidence to occur, the precise distance of the \(n_1\)-th bright fringe strictly of wavelength \(\lambda_1\) must exactly equal the precise distance of the \(n_2\)-th bright fringe strictly of wavelength \(\lambda_2\):
\[ y_{n_1} = y_{n_2} \]
\[ n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d} \]


Step 3: {\color{redDetermine the lowest possible integer ratio

Since the physical parameters \(D\) and \(d\) are identical for both setups, they elegantly cancel out entirely from both sides of the equation:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]

We gracefully rearrange this to find the strict ratio of the fringe orders:
\[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} \]

Substitute the given wavelength values to establish the numerical ratio:
\[ \frac{n_1}{n_2} = \frac{600 nm}{500 nm} = \frac{6}{5} \]

To rigorously find the *least* possible distance for the very first coincidence event, we absolutely must select the smallest possible integers that satisfy this exact fractional ratio. These are clearly:
\[ n_1 = 6 and n_2 = 5 \]

This physically implies that the 6th bright fringe of the shorter 500 nm light perfectly overlaps the 5th bright fringe of the longer 600 nm light.


Step 4: {\color{redCalculate the actual physical least distance

We can now calculate the explicit spatial distance \(y_{min}\) using either \(n_1\) or \(n_2\). We will deliberately use \(n_1\) for this calculation:
\[ y_{min} = n_1 \frac{\lambda_1 D}{d} \]

Carefully substitute all the rigorously converted SI values directly into the formula:
\[ y_{min} = 6 \times \frac{500 \times 10^{-9} \times 1.8}{0.3 \times 10^{-3}} \]

Let's aggressively simplify the numerical coefficients first:
\[ \frac{1.8}{0.3} = 6 \]

Now plug this factor squarely back into the calculation:
\[ y_{min} = 6 \times 500 \times 10^{-9} \times 6 \times 10^3 \]
\[ y_{min} = 36 \times 500 \times 10^{-6} \]
\[ y_{min} = 18000 \times 10^{-6} m \]

For drastically better readability and to match standard conventions, convert this final result into millimeters:
\[ y_{min} = 18 \times 10^{-3} m = 18 mm \]


Step 5: {\color{redConclusion

The very first spatial location proceeding outwards from the central maximum where the two bright interference fringes will perfectly overlap is exactly 18 mm away.
Quick Tip: The coincidence formula \(n_1 \lambda_1 = n_2 \lambda_2\) is extremely common in exams. Remember that the larger integer order always corresponds strictly to the shorter wavelength.

Concept:

For standard solid metallic conductors, the intrinsic electrical resistance predictably and consistently increases as the ambient temperature increases.

This specific thermal dependence over moderate temperature ranges is mathematically modeled by an established linear empirical approximation formula.

The formula is rigorously given as \(R_T = R_0 [1 + \alpha (T - T_0)]\), where \(R_T\) is the new resistance at the target temperature \(T\), \(R_0\) is the known baseline resistance at reference temperature \(T_0\), and \(\alpha\) is the material's specific temperature coefficient of resistance.



Step 1: {\color{redExtract and define all known parameters

The baseline reference temperature is explicitly provided as \(T_0 = 27^\circC\).

Let the initial resistance strictly at this reference temperature be denoted as \(R_0\).

The material's specific temperature coefficient is provided as \(\alpha = 2.0 \times 10^{-4} ^\circC^{-1}\).

The problem firmly dictates that the final resistance \(R_T\) at the unknown target temperature \(T\) must become exactly 20% more than the baseline resistance \(R_0\).

Mathematically, a 20% increase is cleanly formulated as:
\[ R_T = R_0 + 0.20 R_0 = 1.20 R_0 \]


Step 2: {\color{redSet up the temperature dependence equation

We systematically substitute our defined parameters directly into the standard linear resistance formula:
\[ R_T = R_0 [1 + \alpha (T - T_0)] \]
\[ 1.20 R_0 = R_0 [1 + \alpha (T - 27)] \]


Step 3: {\color{redSolve the algebra for the temperature change

Since the baseline resistance \(R_0\) appears as a strict multiplier on both sides of the equation, we can cleanly and immediately cancel it out completely:
\[ 1.20 = 1 + \alpha (T - 27) \]

Subtract 1 from both sides to elegantly isolate the term containing the temperature difference:
\[ 1.20 - 1 = \alpha (T - 27) \]
\[ 0.20 = \alpha (T - 27) \]

Now, carefully substitute the provided numerical value for the temperature coefficient \(\alpha\) into the equation:
\[ 0.20 = (2.0 \times 10^{-4}) \times (T - 27) \]

Vigorously rearrange the equation to solve for the explicit temperature difference \((T - 27)\):
\[ T - 27 = \frac{0.20}{2.0 \times 10^{-4}} \]


Step 4: {\color{redExecute the final numerical calculation

To effectively handle the messy scientific notation in the denominator, multiply both numerator and denominator aggressively by \(10^4\):
\[ T - 27 = \frac{0.20 \times 10^4}{2.0} \]
\[ T - 27 = \frac{2000}{2.0} \]
\[ T - 27 = 1000 \]

Finally, cleanly isolate the target temperature \(T\) by shifting the reference temperature across:
\[ T = 1000 + 27 \]
\[ T = 1027^\circC \]


Step 5: {\color{redConclusion

The conductor must be heated to a substantial temperature of exactly \(1027^\circC\) to experience a 20% increase in its electrical resistance.
Quick Tip: A "\(20%\) more" phrase immediately translates to a multiplier of \(1.20\). Be extremely careful with scientific notation division to avoid off-by-ten calculation errors.

Concept:

When a ray of light cleanly traverses a triangular optical prism, it undergoes refraction strictly at two distinct interfaces: upon entering the first face and upon exiting the second face.

The fundamental geometric relation deeply connecting the two internal angles of refraction (\(r_1\) and \(r_2\)) to the prism's characteristic refracting angle \(A\) is universally given by \(A = r_1 + r_2\).

The phrase "just suffers total internal reflection" acts as a critical physical trigger word, implying that the light ray hits the second internal boundary exactly at the critical angle, meaning the final angle of emergence is a grazing \(90^\circ\).

The critical angle \(C\) for any transparent medium located in air is rigorously defined by the sine relation: \(\sin C = \frac{1}{\mu}\).



Step 1: {\color{redIdentify the given optical parameters

The principal refracting angle of the triangular prism is explicitly \(A = 60^\circ\).

The absolute refractive index of the glass material is strictly \(\mu = \sqrt{2}\).

The emergent condition "just suffers total internal reflection" mathematically mandates that the internal angle of incidence squarely on the second face is exactly equal to the critical angle (\(r_2 = C\)).


Step 2: {\color{redCalculate the critical angle for the prism material

We systematically use the critical angle formula to find the exact value of \(C\):
\[ \sin C = \frac{1}{\mu} \]

Substitute the known refractive index securely into the equation:
\[ \sin C = \frac{1}{\sqrt{2}} \]

From foundational trigonometry, we universally recognize this specific sine value:
\[ C = 45^\circ \]

Therefore, the light ray strikes the second internal face at precisely \(r_2 = 45^\circ\).


Step 3: {\color{redCalculate the angle of refraction at the first face

We leverage the core geometric prism identity to meticulously find the first internal refraction angle \(r_1\):
\[ A = r_1 + r_2 \]

Substitute the known values for the prism angle and the derived second refraction angle:
\[ 60^\circ = r_1 + 45^\circ \]

Cleanly isolate \(r_1\):
\[ r_1 = 60^\circ - 45^\circ = 15^\circ \]

This indicates that the incident light ray bends internally to a \(15^\circ\) angle immediately after penetrating the first face.


Step 4: {\color{redCalculate the initial angle of incidence using Snell's Law

We rigorously apply Snell's Law specifically at the very first air-glass interface to discover the initial angle of incidence \(i_1\):
\[ \mu_1 \sin i_1 = \mu_2 \sin r_1 \]

Since the ray originates externally from ambient air, we establish \(\mu_1 = 1\), and the prism's index is \(\mu_2 = \sqrt{2}\):
\[ 1 \times \sin i_1 = \sqrt{2} \times \sin(15^\circ) \]

To strictly evaluate \(\sin(15^\circ)\) without a calculator, we expand it strategically using standard angle subtraction formulas:
\[ \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \]

Substitute the standard known trigonometric exact values:
\[ \sin(15^\circ) = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \]
\[ \sin(15^\circ) = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

Now, carefully substitute this exact expanded expression squarely back into our pending Snell's Law equation:
\[ \sin i_1 = \sqrt{2} \times \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right) \]

The \(\sqrt{2}\) terms wonderfully and cleanly cancel out completely:
\[ \sin i_1 = \frac{\sqrt{3} - 1}{2} \]


Step 5: {\color{redConclusion

We can cleanly express the final exact incident angle formally using the inverse sine function:
\[ i_1 = \sin^{-1}\left(\frac{\sqrt{3} - 1}{2}\right) \]

(For context, since \(\sqrt{3} \approx 1.732\), \(\sin i_1 \approx 0.366\), which rigorously corresponds to an angle of roughly \(21.47^\circ\)).
Quick Tip: The specific phrasing "just suffers TIR" or "grazes the emergent surface" invariably implies exactly one thing mathematically: \(r_2 = C\).

Concept:

Semiconductors uniquely operate fundamentally between conductors and insulators, playing the starring role in all modern electronics.

Intrinsic semiconductors represent the absolute purest, unadulterated form of the crystalline material (like pure Silicon or Germanium) found in nature, containing virtually zero chemical impurities.

Extrinsic semiconductors, on the other hand, are strictly man-made versions that have been intentionally and aggressively "doped" with minute, highly controlled quantities of specific chemical impurities (like Phosphorus or Boron) to drastically and purposely alter their baseline electrical properties.



Step 1: {\color{redDifference 1: Inherent Purity and Structural Composition

An intrinsic semiconductor is a perfectly pure semiconductor crystal totally free from any deliberately introduced foreign atoms or chemical defects.

An extrinsic semiconductor is a deliberately impure semiconductor. It is carefully manufactured by heavily doping a pure intrinsic semiconductor matrix with specifically chosen trivalent or pentavalent impurity atoms.


Step 2: {\color{redDifference 2: Density of Charge Carriers

In any intrinsic semiconductor, because every single broken covalent bond inherently generates exactly one free electron and one corresponding hole simultaneously, the thermal number density of free electrons (\(n_e\)) is always absolutely equal to the thermal number density of holes (\(n_h\)). Mathematically, \(n_e = n_h = n_i\).

In stark contrast, an extrinsic semiconductor exhibits a massive, intentional disparity. Depending entirely on the specific type of doping applied, one specific type of charge carrier vastly outnumbers the other. In n-type materials, electrons brutally dominate (\(n_e \gg n_h\)), whereas in p-type materials, holes aggressively dominate (\(n_h \gg n_e\)).


Step 3: {\color{redAdditional points for completeness

For added completeness, one could also validly mention Electrical Conductivity.

The intrinsic semiconductor possesses an exceptionally low baseline electrical conductivity at standard room temperature, heavily limiting its practical use in devices.

The extrinsic semiconductor boasts a tremendously high, artificially boosted electrical conductivity directly engineered by the sheer abundance of the added impurity dopants, making them essential for creating diodes and transistors.
Quick Tip: When a physics question simply asks for "two points of difference," always cleanly structure your answer in a clear two-column table or use distinct bullet points highlighting directly contrasting traits (e.g., Pure vs Impure, \(n_e=n_h\) vs \(n_e \neq n_h\)) to guarantee maximum marks.

Concept:

The de Broglie wavelength \(\lambda\) of a moving particle is intrinsically linked to its momentum \(p\) by the equation \(\lambda = \frac{h}{p}\), where \(h\) is Planck's constant.
The momentum of a particle can be expressed strictly in terms of its kinetic energy \(K\) and mass \(m\) using the classical relation \(p = \sqrt{2mK}\).
Combining these yields the functional wavelength formula: \(\lambda = \frac{h}{\sqrt{2mK}}\).


Step 1: {\color{redEstablish the formulas for both particles

For the alpha particle (subscript \(\alpha\)), the wavelength is:
\[ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha K_\alpha}} \]

For the proton (subscript \(p\)), the wavelength is:
\[ \lambda_p = \frac{h}{\sqrt{2m_p K_p}} \]


Step 2: {\color{redIdentify the given conditions and mass relationships

The problem explicitly states that both particles possess the exact same kinetic energy, so \(K_\alpha = K_p = K\).

We also know the fundamental mass relationship between an alpha particle (a helium nucleus) and a proton: the mass of an alpha particle is approximately four times the mass of a proton.
\[ m_\alpha = 4m_p \]


Step 3: {\color{redCalculate the required ratio

Divide the alpha particle equation by the proton equation to find the ratio:
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_\alpha K}}}{\frac{h}{\sqrt{2m_p K}}} \]

The Planck's constant \(h\), the factor of \(2\), and the identical kinetic energy \(K\) completely cancel out:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p}{m_\alpha}} \]

Substitute the mass relationship \(m_\alpha = 4m_p\) into the root:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p}{4m_p}} = \sqrt{\frac{1}{4}} \]
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{1}{2} \]


Step 4: {\color{redConclusion

The specific ratio of their de Broglie wavelengths is strictly 1:2. Quick Tip: When dealing with de Broglie wavelength ratios where Kinetic Energy is constant, the ratio simplifies cleanly to the inverse square root of their masses: \(\lambda \propto 1/\sqrt{m}\).

Concept:

When a charged particle of charge \(q\) is aggressively accelerated from rest across a specific potential difference \(V\), it naturally acquires a kinetic energy mathematically equal to \(K = qV\).
Substituting this energy formulation into the standard de Broglie relation \(\lambda = \frac{h}{\sqrt{2mK}}\) yields a new functional equation: \(\lambda = \frac{h}{\sqrt{2mqV}}\).


Step 1: {\color{redEstablish the formulas for both particles

For the alpha particle, the wavelength after acceleration is:
\[ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha q_\alpha V_\alpha}} \]

For the proton, the corresponding wavelength is:
\[ \lambda_p = \frac{h}{\sqrt{2m_p q_p V_p}} \]


Step 2: {\color{redIdentify the given conditions, mass, and charge relationships

The problem explicitly states they are accelerated through the identical potential difference, meaning \(V_\alpha = V_p = V\).

The mass of an alpha particle is four times that of a proton: \(m_\alpha = 4m_p\).

The physical charge of an alpha particle (two protons) is precisely twice the charge of a single proton: \(q_\alpha = 2q_p\).


Step 3: {\color{redCalculate the required ratio

Divide the alpha particle equation strictly by the proton equation to cleanly find the ratio:
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_\alpha q_\alpha V}}}{\frac{h}{\sqrt{2m_p q_p V}}} \]

The physical constants \(h\), \(2\), and the identical potential difference \(V\) completely cancel out:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}} \]

Substitute the known mass and charge relationships into the mathematical root:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p \cdot q_p}{(4m_p) \cdot (2q_p)}} \]

The mass and charge variables cancel out elegantly:
\[ \frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{1}{4 \times 2}} = \sqrt{\frac{1}{8}} \]
\[ \frac{\lambda_\alpha}{\lambda_p} = \frac{1}{2\sqrt{2}} \]


Step 4: {\color{redConclusion

The specific ratio of their de Broglie wavelengths after being accelerated through the same potential is exactly \(1:2\sqrt{2}\). Quick Tip: For particles accelerated through the same voltage, the wavelength relies purely on the inverse square root of the product of mass and charge: \(\lambda \propto 1/\sqrt{mq}\). Always memorize the alpha particle's properties (\(4m_p\), \(2q_p\)) as they are heavily tested.

Concept:

Mutual inductance fundamentally characterizes the electromagnetic interaction specifically existing between two distinct, magnetically coupled coils.
When a dynamically changing electric current passes strictly through one coil, it actively generates a changing magnetic flux that subsequently links with the neighboring second coil, directly inducing an electromotive force (EMF) inside it.


Step 1: {\color{redState the formal definition

The mutual inductance of a pair of physically coupled coils is formally defined as the total magnetic flux mathematically linked with one coil when a steady unit electric current (1 Ampere) actively flows through the other neighboring coil.

Alternatively, it can be rigorously defined based on Faraday's law: it is mathematically equal to the magnitude of the induced electromotive force (EMF) generated in the secondary coil when the active current running through the primary coil changes precisely at a unitary rate of \(1 Ampere per second\).

Mathematically, \(\Phi_{21} = M_{21} I_1\) or \(|e_2| = M_{21} \left| \frac{dI_1}{dt} \right|\).


Step 2: {\color{redProvide the SI unit

The standard SI unit for mutual inductance is the Henry (H).

It can also be dimensionally expressed as Weber per Ampere (\(Wb A^{-1}\)) or Volt-second per Ampere (\(V s A^{-1}\)).
Quick Tip: Definitions in physics board exams often carry marks split into the theoretical statement and the mathematical formula. Providing both ensures you definitively secure full marks.

Concept:

To calculate the mutual inductance \(M\) rigidly existing between any two inductive geometries, we pass a theoretical current \(I\) through the primary structure to find its uniform magnetic field.
Next, we mathematically compute the total magnetic flux \(\Phi\) actively linking the secondary structure due exclusively to this generated field.
The mutual inductance is then securely extracted using the fundamental proportionality equation: \(M = \frac{\Phi}{I}\).


Step 1: {\color{redDetermine the magnetic field of the primary solenoid

Let us assume an active current \(I\) flows continuously through the long outer solenoid.

The solenoid aggressively generates a largely uniform magnetic field strictly along its central axis, penetrating its hollow interior.

The magnitude of this magnetic field \(B\) is universally given by Ampere's Law formulation:
\[ B = \mu_0 n I \]

where \(\mu_0\) is the permeability of free space and \(n\) represents the number of turns strictly per unit length.


Step 2: {\color{redCalculate the magnetic flux linked with the inner loop

The smaller circular loop of radius \(r\) is situated directly inside the solenoid, sharing the exact same central axis.

Because their geometric axes completely coincide, the generated magnetic field \(B\) is strictly perpendicular to the flat area \(A\) of the inner loop.

The total cross-sectional area of the small inner loop is mathematically exactly:
\[ A = \pi r^2 \]

The total magnetic flux \(\Phi\) successfully penetrating through this single circular loop is the product of the field and its area:
\[ \Phi = B \cdot A = B A \cos(0^\circ) \]

Substitute the previously derived expression for the magnetic field:
\[ \Phi = (\mu_0 n I) \times (\pi r^2) \]
\[ \Phi = \mu_0 n \pi r^2 I \]


Step 3: {\color{redExtract the expression for Mutual Inductance

By fundamental definition, the mutual inductance \(M\) tightly relates the linked flux to the driving source current:
\[ \Phi = M I \]

Comparing our geometrically derived flux expression directly with this definitional formula strictly isolates \(M\):
\[ M = \frac{\Phi}{I} = \frac{\mu_0 n \pi r^2 I}{I} \]

The driving current \(I\) perfectly cancels out, leaving the final physical geometry-dependent expression:
\[ M = \mu_0 n \pi r^2 \]


Step 4: {\color{redConclusion

The formal expression for the mutual inductance seamlessly existing between the long solenoid and the tiny coaxial inner loop is solidly exactly \(\mu_0 n \pi r^2\). Quick Tip: Notice that the mutual inductance depends strictly on the area of the smaller inner loop (\(\pi r^2\)), not the larger outer solenoid. The magnetic field only links flux physically through the area where the secondary loop actually exists.

Concept:

A long, straight current-carrying wire universally generates a cylindrical magnetic field circulating around it, described exactly by Ampere's Circuital Law.
When a second parallel wire carrying its own current is placed firmly within this generated magnetic field, it aggressively experiences a magnetic Lorentz force.
The definitive direction of this mutual force is strictly governed by Fleming's Left-Hand Rule or the right-hand cross product rule.


Step 1: {\color{redCalculate the magnetic field created by Conductor A





Let conductor A strongly carry a steady current \(I_a\) flowing directly upwards.

Conductor B is placed parallel to A at a fixed perpendicular distance \(d\).

According to Ampere's Circuital Law, the magnitude of the magnetic field \(\vec{B}_a\) produced uniquely by conductor A precisely at the location of conductor B is:
\[ B_a = \frac{\mu_0 I_a}{2\pi d} \]

Using the Right-Hand Grip Rule, if current \(I_a\) flows upwards, the magnetic field lines aggressively circulate counter-clockwise. At the exact position of wire B (located to the right of A), this magnetic field vector \(\vec{B}_a\) points strictly perpendicular and into the plane of the paper/page.


Step 2: {\color{redDeduce the force acting on length L of Conductor B

Conductor B rigidly carries a steady current \(I_b\), also flowing directly upwards in the same direction.

A specific length \(L\) of this conductor firmly sitting inside the external magnetic field \(\vec{B}_a\) will continuously experience a magnetic force \(\vec{F}_{BA}\).

The foundational formula for this magnetic force is rigorously given by:
\[ \vec{F}_{BA} = I_b (\vec{L} \times \vec{B}_a) \]

Since the straight length vector \(\vec{L}\) (pointing upwards) and the magnetic field vector \(\vec{B}_a\) (pointing strictly inwards) are perfectly mutually perpendicular (\(\theta = 90^\circ\)), the scalar magnitude is:
\[ F_{BA} = I_b L B_a \sin(90^\circ) = I_b L B_a \]

Substitute the previously derived expression for \(B_a\) into the force equation:
\[ F_{BA} = I_b L \left( \frac{\mu_0 I_a}{2\pi d} \right) \]
\[ F_{BA} = \frac{\mu_0 I_a I_b L}{2\pi d} \]

Applying Fleming's Left-Hand Rule (Current up, Field strictly inwards), the resulting force \(F_{BA}\) on wire B points squarely towards the left, pulling directly towards conductor A. Thus, it is an attractive force.


Step 3: {\color{redExpression for force on Conductor A and Newton's Third Law

By applying the exact same rigorous logic in reverse, conductor B physically creates a magnetic field \(B_b = \frac{\mu_0 I_b}{2\pi d}\) pointing out of the page at the specific location of conductor A.

The force \(F_{AB}\) experienced by a length \(L\) of conductor A strictly due to this field is:
\[ F_{AB} = I_a L B_b = I_a L \left( \frac{\mu_0 I_b}{2\pi d} \right) = \frac{\mu_0 I_a I_b L}{2\pi d} \]

Applying Fleming's Left-Hand Rule again (Current up, Field strictly outwards), the resulting force \(F_{AB}\) on wire A points squarely towards the right, pulling directly towards conductor B.

Mathematically comparing the two derived vector forces, their absolute magnitudes are strictly identical:
\[ |F_{BA}| = |F_{AB}| = \frac{\mu_0 I_a I_b L}{2\pi d} \]

However, their physical directions are perfectly exactly opposite to each other. Conductor A pulls B to the left, while Conductor B pulls A to the right.

In strict vector notation:
\[ \vec{F}_{BA} = -\vec{F}_{AB} \]

This explicitly and conclusively proves that the mutual magnetic forces actively existing between the two parallel conductors rigorously obey Newton's Third Law of Motion (action and equal, opposite reaction).
Quick Tip: A useful mnemonic for parallel wires: "Like currents Attract, Unlike currents Repel." This behaves totally oppositely to standard electrostatic charges where "Like charges repel".

Concept:

The quantized energy of an electron stably revolving in the \(n\)-th orbit of a standard hydrogen atom is robustly given by the formula \(E_n = \frac{-13.6 eV}{n^2}\).
For any bound electron, its dynamic kinetic energy \(K\) is exactly equal strictly to the positive magnitude of its total energy: \(K = |E_n|\).
The de Broglie wavelength uniquely associated with this revolving electron can be rapidly calculated using the standard matter-wave relation \(\lambda = \frac{h}{\sqrt{2mK}}\).


Step 1: {\color{redIdentify the quantum state and calculate its energy

The problem specifically mentions the "second excited state".

In quantum mechanics, the ground state is \(n=1\), the first excited state is \(n=2\), and the second excited state strictly corresponds to \(n=3\).

The total energy of the electron specifically in this \(n=3\) state is:
\[ E_3 = \frac{-13.6}{3^2} eV = \frac{-13.6}{9} eV \]
\[ E_3 = -1.511 eV \]

The kinetic energy \(K\) is the absolute positive value of this total energy:
\[ K = 1.511 eV \]


Step 2: {\color{redConvert Kinetic Energy securely into Joules

To utilize standard SI physics formulas, we must absolutely convert this energy from electron-volts directly into standard Joules (\(1 eV = 1.6 \times 10^{-19} J\)):
\[ K = 1.511 \times 1.6 \times 10^{-19} J \]
\[ K \approx 2.4176 \times 10^{-19} J \]


Step 3: {\color{redCalculate the de Broglie wavelength

We meticulously apply the de Broglie wavelength formula directly:
\[ \lambda = \frac{h}{\sqrt{2m_e K}} \]

Substitute the provided standard constants and our newly calculated kinetic energy:
\[ \lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times (9 \times 10^{-31}) \times (2.4176 \times 10^{-19})}} \]

First, carefully simplify the massive term securely trapped inside the square root:
\[ 2m_e K = 18 \times 10^{-31} \times 2.4176 \times 10^{-19} \]
\[ 2m_e K = 43.5168 \times 10^{-50} \]

Now, firmly take the mathematical square root of this value:
\[ \sqrt{43.5168 \times 10^{-50}} \approx 6.5967 \times 10^{-25} kg m/s \]

Finally, cleanly divide Planck's constant by this calculated momentum:
\[ \lambda = \frac{6.6 \times 10^{-34}}{6.5967 \times 10^{-25}} \]
\[ \lambda \approx 1.0005 \times 10^{-9} m \]


Step 4: {\color{redConclusion

The de Broglie wavelength strongly associated with the electron safely residing in the second excited state is mathematically evaluated to be approximately \(1.0 \times 10^{-9} m\) (or exactly \(1 nm\)). Quick Tip: An alternate, brilliantly faster method heavily utilizes Bohr's quantization condition \(2\pi r = n\lambda\). If you remember the radius formula \(r_n = 0.53 \AA \times n^2\), you can cleanly bypass all heavy energy/joule conversions! \(\lambda = 2\pi (0.53 \times 3^2) / 3 \approx 9.99 \AA \approx 1 nm\).

Step 1: {\color{redIdentify the electromagnetic wave

The electromagnetic waves famously and widely used purely as a diagnostic tool in modern medicine to image internal bones and structures are **X-rays**.


Step 2: {\color{redState the standard wavelength range

The defining wavelength range strictly for X-rays generally lies safely between **\(10^{-8} m\) and \(10^{-13} m\)** (or \(10 nm\) down to \(0.001 nm\)).
Quick Tip: X-rays possess immense penetrating power precisely due to their very short wavelengths, effortlessly passing through soft tissue while being heavily absorbed by denser calcium-rich bones.

Step 1: {\color{redIdentify the electromagnetic wave

The electromagnetic waves extensively utilized in basic remote control switches specifically for household TV sets and appliances are **Infrared (IR) rays**.


Step 2: {\color{redState the standard wavelength range

The defining wavelength range for Infrared waves universally spans from roughly **\(1 mm\) down to \(700 nm\)** (or exactly \(10^{-3} m\) to \(7 \times 10^{-7} m\)).
Quick Tip: Infrared is often colloquially known strictly as "heat waves" because it is powerfully emitted by hot bodies and aggressively induces molecular vibrations leading to heat.

Step 1: {\color{redIdentify the electromagnetic wave

The specific electromagnetic waves purposely utilized in modern water purifiers to actively destroy bacteria and deadly microbes are **Ultraviolet (UV) rays**.


Step 2: {\color{redState the standard wavelength range

The recognized wavelength range rigorously assigned to Ultraviolet radiation safely falls between **\(400 nm\) and \(1 nm\)** (or exactly \(4 \times 10^{-7} m\) to \(10^{-9} m\)).
Quick Tip: The immense energy carried securely by short-wavelength UV-C rays is biologically sufficient to aggressively shatter the DNA of microorganisms, rendering them completely sterile.

Concept:

A structural system purely comprising two equal and entirely opposite point charges rigorously separated by a tiny distance is officially termed an electric dipole.
Electric field lines organically originate strictly from the positive charge and terminate cleanly on the negative charge, smoothly curving through surrounding space.
Equipotential surfaces inherently represent 3D geometric surfaces where the total electrical potential remains completely constant. They must always rigidly intersect electric field lines at perfect \(90^\circ\) right angles.


Step 1: {\color{redDescribe the Electric Field Lines



Draw a distinct positive charge (\(+q\)) securely on the left and a negative charge (\(-q\)) securely on the right.

Draw continuous, smooth curves confidently leaving the positive charge and securely entering the negative charge.

The field lines strictly along the direct axis connecting them travel straight across. The lines above and below actively bow outward gracefully.

Ensure that absolutely no two field lines ever physically cross each other, and indicate the proper direction strictly with arrows pointing precisely towards the negative charge.


Step 2: {\color{redDescribe the Equipotential Surfaces

Since the electric potential of a dipole is \(V = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r_1} - \frac{q}{r_2} \right)\), the absolute central point exactly midway between the charges strictly has zero potential.

Draw a large, flat vertical plane perfectly bisecting the central distance between the charges. This central equatorial plane is exactly a \(0V\) equipotential surface.

Closer to the charges, carefully draw dashed or dotted circular curves closely encircling each individual charge.

These encircling surfaces must visually bunch up closer together aggressively within the central region between the two charges where the field is distinctly stronger, and dynamically spread further apart on the extreme outer sides.
Quick Tip: When sketching diagrams manually in exams, strictly guarantee that every single intersection point between an equipotential dashed line and a solid electric field line visually appears completely perpendicular (a perfect \(90^\circ\) cross).

Concept:

By foundational definition, an equipotential surface is a continuous 3D geometric boundary across which the absolute electric potential is physically exactly constant at every single point.
Because there exists absolutely zero potential difference directly between any two arbitrary points safely lying on this surface (\(dV = 0\)), the rigorous physical work required to move a tiny test charge anywhere along it is mathematically zero.


Step 1: {\color{redEstablish the mathematical relation for work done

The physical work \(dW\) continuously done in moving a tiny test charge \(q\) over a microscopically small displacement \(d\vec{l}\) entirely along any equipotential surface is rigidly given by the standard dot product equation:
\[ dW = \vec{F} \cdot d\vec{l} = q(\vec{E} \cdot d\vec{l}) \]
\[ dW = q E dl \cos\theta \]

where \(\theta\) actively represents the explicit angle strictly existing between the electric field vector \(\vec{E}\) and the surface displacement vector \(d\vec{l}\).


Step 2: {\color{redApply the equipotential condition

As logically stated, the absolute potential difference \(dV\) across any equipotential surface is definitively zero.

Since work done is also deeply defined as \(dW = -q dV\), it robustly follows that:
\[ dW = -q(0) = 0 \]


Step 3: {\color{redDeduce the required angle

Equating our two independent work expressions yields:
\[ q E dl \cos\theta = 0 \]

Since the test charge \(q\) is non-zero, the local electric field \(E\) is non-zero, and the displacement \(dl\) is certainly non-zero, the purely mathematical burden of establishing this zero strictly falls entirely upon the cosine term:
\[ \cos\theta = 0 \]

The only viable physical angle that beautifully satisfies this stringent trigonometric condition is:
\[ \theta = 90^\circ \]


Step 4: {\color{redConclusion

This elegantly proves mathematically that the electric field vector \(\vec{E}\) must absolutely, without exception, be oriented strictly perpendicular (normal) to the equipotential surface at every single point upon it. Quick Tip: If the field were not flawlessly perpendicular, it would inherently possess a non-zero horizontal component physically running along the surface. This renegade component would actively exert force and do real work on charges, violently violating the defining \(dW=0\) condition.

Concept:

Isotopes are distinct chemical nuclides that share the exact same atomic number \(Z\) (identical number of positive protons) but purposely possess entirely different mass numbers \(A\) (different number of neutrons).
Isotones are completely different chemical nuclides that coincidentally contain the exact same physical number of internal neutrons \(N\), which is mathematically calculated actively as \(N = A - Z\).


Step 1: {\color{redIdentify and group the Isotopes

We meticulously examine the lower subscript index (atomic number \(Z\)) strictly provided for each individual nuclide in the list.

For \(^{12}_6C\), the atomic number is \(Z = 6\).

For \(^{14}_6C\), the atomic number is also \(Z = 6\).

Because they both rigorously share the exact same atomic number (\(Z=6\)), but definitively have vastly different top mass numbers (\(12\) vs \(14\)), they chemically belong to the same specific element.

Therefore, \(^{12}_6C\) and \(^{14}_6C\) solidly form a perfect pair of isotopes.


Step 2: {\color{redIdentify and group the Isotones

To firmly hunt for isotones, we must first mathematically calculate the hidden internal neutron count (\(N = A - Z\)) specifically for the remaining heavy nuclides.

For the heavy Mercury nuclide \(^{198}_{80}Hg\), the neutron count calculates to:
\[ N = 198 - 80 = 118 \]

For the heavy Gold nuclide \(^{197}_{79}Au\), the neutron count similarly calculates to:
\[ N = 197 - 79 = 118 \]

Because both highly distinct heavy nuclei amazingly share the exact same internal neutron count of exactly \(118\), they fit the definition perfectly.

Therefore, \(^{198}_{80}Hg\) and \(^{197}_{79}Au\) solidly form a perfect pair of isotones.
Quick Tip: A brilliant mnemonic trick: Isoto\textbf{p}es have the same number of \textbf{P}rotons. Isoto\textbf{n}es have the same number of \textbf{N}eutrons. Iso\textbf{b}ars have the same mass number (A, resembles \textbf{B}oth combined).

Concept:

Extensive scattering experiments have historically revealed that atomic nuclei are approximately spherical physical objects.
The physical volume of any specific nucleus is directly and strictly proportional to its total contained number of nucleons (which is its mass number \(A\)).
Consequently, the geometric radius \(R\) aggressively scales mathematically with the cube root of the given mass number \(A\).


Step 1: {\color{redState the dependency of nuclear size on mass number

Based securely on experimental observations, the physical radius \(R\) of any spherical nucleus is mathematically related to its mass number \(A\) strictly by the established empirical formula:
\[ R = R_0 A^{1/3} \]

where \(R_0\) is a universal empirical constant roughly equal to \(1.2 \times 10^{-15} m\) (or \(1.2 fm\)). This directly demonstrates that size (radius) grows precisely as the cube root of the atomic mass number.


Step 2: {\color{redFormulate the physical Volume and Mass of the nucleus

Assuming the stable nucleus is a perfectly rigid sphere, its physical spatial volume \(V\) is mathematically calculated as:
\[ V = \frac{4}{3} \pi R^3 \]

Substitute our previously defined radius dependency explicitly into this volume equation:
\[ V = \frac{4}{3} \pi (R_0 A^{1/3})^3 \]
\[ V = \frac{4}{3} \pi R_0^3 A \]

This vividly proves that nuclear volume \(V\) is strictly and linearly proportional to \(A\).

Next, the total physical mass \(M\) of the entire nucleus is approximately cleanly equal to the total mass number \(A\) explicitly multiplied by the average baseline mass of a single isolated nucleon (\(m \approx 1.66 \times 10^{-27} kg\)):
\[ M \approx A \cdot m \]


Step 3: {\color{redDerive the density expression to prove constancy

Nuclear density \(\rho\) is definitively calculated by strictly dividing the total nuclear mass by its total spatial volume:
\[ \rho = \frac{Mass}{Volume} = \frac{M}{V} \]

Substitute the robust expressions meticulously established in Step 2 securely into the density formula:
\[ \rho = \frac{A \cdot m}{\frac{4}{3} \pi R_0^3 A} \]

The mass number variable \(A\) wonderfully and completely cancels out from both the numerator and the massive denominator:
\[ \rho = \frac{m}{\frac{4}{3} \pi R_0^3} \]
\[ \rho = \frac{3m}{4 \pi R_0^3} \]


Step 4: {\color{redConclusion

In this final, rigorously derived expression, every single remaining term (\(m\), \(\pi\), \(R_0\)) is a completely fixed universal constant. Because the variable \(A\) has entirely vanished, it explicitly proves that the physical density of nuclear matter is a universal constant (roughly \(2.3 \times 10^{17} kg/m^3\)), completely independent of the mass number \(A\) for all atoms.
Quick Tip: The fact that nuclear density remains staggeringly constant strongly implies that nucleons act like incompressible fluid drops; adding more nucleons purely adds more volume without squishing the existing ones tighter.

Concept:

The Biot-Savart Law acts as the absolute foundational cornerstone for mathematically determining the magnetic field strictly generated by arbitrary steady current distributions.
To successfully tackle a large circular coil, we meticulously break the continuous ring into infinitesimally tiny current elements \(d\vec{l}\) and integrate their individual tiny magnetic contributions over the entire ring.
Geometric symmetry plays a massive role here: perpendicular field components elegantly and completely cancel each other out, leaving strictly only the axial field components to sum together perfectly.


Step 1: {\color{redSetup the geometry and apply Biot-Savart Law

Consider a large circular coil of radius \(r\) placed stably in the y-z plane, with its center perfectly locked at the origin \((0,0,0)\).

Let a steady current \(I\) flow continuously through it.

We aim to find the exact magnetic field \(\vec{B}\) strictly at a point \(P\) located directly on the x-axis at a distance \(x\) from the absolute center.

Consider a tiny, infinitesimal current element \(Id\vec{l}\) situated precisely at the topmost edge of the circular coil.

The spatial distance vector \(\vec{s}\) running straight from this top element down to the observation point \(P\) has a rigid magnitude derived cleanly from Pythagoras:
\[ s = \sqrt{r^2 + x^2} \]

According to the rigorous Biot-Savart law, the tiny magnitude of the magnetic field \(d\vec{B}\) generated uniquely by this tiny top element at point \(P\) is:
\[ dB = \frac{\mu_0}{4\pi} \frac{I dl \sin(90^\circ)}{s^2} = \frac{\mu_0}{4\pi} \frac{I dl}{(r^2 + x^2)} \]

(The angle is definitively \(90^\circ\) because the tangent vector \(d\vec{l}\) and the slant vector \(\vec{s}\) are fundamentally mutually perpendicular in 3D space).


Step 2: {\color{redResolve components based on structural symmetry

Using right-hand cross product rules, the generated vector \(d\vec{B}\) strictly points perpendicular to the slant distance \(\vec{s}\).

We meticulously resolve \(d\vec{B}\) into two distinct orthogonal components: a vertical perpendicular component (\(dB_\perp = dB \sin\phi\)) and a horizontal axial component (\(dB_\parallel = dB \cos\phi\)). Let the angle between \(d\vec{B}\) and the vertical axis be \(\phi\), which geometrically implies the angle between \(s\) and \(x\) is \(\phi\).

(Actually, let's use angle \(\theta\) between the axis \(x\) and the slant distance \(s\). Then the axial component along x is \(dB \sin\theta\)).

Let's use standard notation: let the angle between slant \(s\) and axis \(x\) be \(\alpha\). Then the axial component of the magnetic field is precisely \(dB \sin\alpha\).

From the built right triangle, we can cleanly extract the sine function:
\[ \sin\alpha = \frac{Opposite}{Hypotenuse} = \frac{r}{\sqrt{r^2 + x^2}} \]

Due to the perfect rotational symmetry of the physical coil, every single top element has a perfect diametrically opposite bottom element.

The vertical perpendicular components (\(dB \cos\alpha\)) from these opposite pairs perfectly and violently cancel each other out entirely (\(| \Sigma dB_\perp | = 0\)).

Only the horizontal axial components strictly pointing along the positive x-axis actually survive and add up.


Step 3: {\color{redIntegrate over the entire coil

The total net magnetic field \(B\) is the mathematical integral of strictly the surviving axial components:
\[ B = \int dB \sin\alpha \]

Substitute the previously established specific expressions meticulously into the integral:
\[ B = \int \left[ \frac{\mu_0}{4\pi} \frac{I dl}{(r^2 + x^2)} \right] \left( \frac{r}{\sqrt{r^2 + x^2}} \right) \]
\[ B = \frac{\mu_0 I r}{4\pi (r^2 + x^2)^{3/2}} \int dl \]

The mathematical integral \(\int dl\) simply and cleanly evaluates strictly to the total physical circumference of the large circular loop, which is exactly \(2\pi r\):
\[ B = \frac{\mu_0 I r}{4\pi (r^2 + x^2)^{3/2}} (2\pi r) \]

Cancel the constants securely:
\[ B = \frac{\mu_0 I r^2}{2 (r^2 + x^2)^{3/2}} \]


Step 4: {\color{redFinal expression for an N-turn coil

If the physical coil actually strictly consists of \(N\) tightly wound identical turns, each single turn contributes an identical amount of magnetic field.

Therefore, we simply and aggressively multiply the single-turn result directly by the integer \(N\):
\[ B_{axis} = \frac{\mu_0 N I r^2}{2 (r^2 + x^2)^{3/2}} \]

The final magnetic field vector points strictly along the central axial line. Quick Tip: To instantly check if your derived formula is correct, plug in \(x = 0\). The messy denominator simplifies to \((r^2)^{3/2} = r^3\), resulting cleanly in \(B = \mu_0 N I / 2r\), which is the famous and heavily memorized formula for the center of a loop!

Concept:

Physical quantities in physics are fundamentally strictly classified as either scalars (possessing only pure magnitude) or vectors (possessing both magnitude and spatial direction, while also obeying strict geometric addition rules).
While electric current is routinely drawn securely with directional arrows on circuit diagrams to trace the physical flow path of charge, this alone does not mathematically qualify it to be a vector quantity.


Step 1: {\color{redExplain the specific physical purpose of the drawn arrow

When we actively draw a specific arrow next to an electrical current flowing in a circuit diagram, we are purely indicating the designated physical direction of the continuous flow of positive electric charge (conventional current).

The arrow simply functions strictly as a bookkeeping tool to confidently track exactly where the physical charges are traveling through the restricted, 1-dimensional conductive wire paths.


Step 2: {\color{redExplain why it fundamentally fails vector classification

To be officially and mathematically classified as a true vector, a physical quantity must not only simply point in a direction, but it must absolutely obey the rigorous geometric laws of vector addition (such as the Parallelogram Law of Vector Addition or the Triangle Law).

Electric currents absolutely fail to obey these geometric laws.

If two completely independent current-carrying wires meet precisely at a junction node at a specific physical angle (e.g., \(90^\circ\)), the total exiting current is exclusively determined by simple, straightforward algebraic scalar addition (\(I_{total} = I_1 + I_2\)), exactly as dictated securely by Kirchhoff’s Current Law (KCL).

The physical angle existing between the intersecting wires has absolutely zero mathematical effect on the resulting total output current.

Because it aggressively defies standard vector addition geometry and merely adds like pure numbers, electric current is officially classified strictly as a scalar quantity (or more formally, a zero-rank tensor).
Quick Tip: Always aggressively highlight the phrase "fails to obey the parallelogram law of vector addition" when answering this specific question, as it is the absolute definitive marking scheme keyword examiners explicitly hunt for.

Concept:

Kirchhoff's Current Law (KCL) firmly states that the total electrical current dynamically entering any junction node strictly equals the total current exiting it.
Kirchhoff's Voltage Law (KVL) firmly states that the directed algebraic sum of all potential differences (voltage drops and gains) traversing completely around any closed physical loop is precisely zero.


Step 1: {\color{redDefine Currents and established Loops


Let the specific current flowing forcefully out of the 3V battery (driving from F upwards to A) be labeled as \(I_1\).

This current \(I_1\) securely reaches the top junction node B. Let it physically split into \(I_3\) flowing directly downwards securely through the central \(3 \Omega\) resistor (from B down to E), and \(I_2\) flowing squarely rightwards securely towards node C.

By rigorously applying KCL exactly at junction node B:
\[ I_1 = I_2 + I_3 \implies I_2 = I_1 - I_3 \]


Step 2: {\color{redApply KVL securely to the Left Loop

Consider the completely closed physical loop A-B-E-F-A and deliberately traverse it squarely in the clockwise direction.

The potential drops across the middle \(3 \Omega\) and bottom \(4 \Omega\) resistors, and cleanly rises across the 3V battery (moving precisely from short negative plate to long positive plate):
\[ -3 I_3 - 4 I_1 + 3 = 0 \]

Rearrange to cleanly isolate the constants:
\[ 4 I_1 + 3 I_3 = 3 \quad --- (Equation 1) \]


Step 3: {\color{redApply KVL securely to the Right Loop

Consider the completely closed physical loop B-C-D-E-B and deliberately traverse it squarely in the clockwise direction.

The path goes aggressively through the top 5V battery (from long positive plate to short negative plate, causing a strict drop of 5V), drops across the right \(2 \Omega\) resistor, and crucially gains potential securely across the central \(3 \Omega\) resistor (since we are traversing forcefully strictly against the assigned direction of \(I_3\)):
\[ -5 - 2 I_2 + 3 I_3 = 0 \]

Rearrange the terms:
\[ 2 I_2 - 3 I_3 = -5 \]

Substitute the previously defined KCL relation \(I_2 = I_1 - I_3\) safely into this equation:
\[ 2(I_1 - I_3) - 3 I_3 = -5 \]
\[ 2 I_1 - 2 I_3 - 3 I_3 = -5 \]
\[ 2 I_1 - 5 I_3 = -5 \quad --- (Equation 2) \]


Step 4: {\color{redSolve the resulting System of Linear Equations

We will systematically eliminate \(I_1\) to directly solve securely for \(I_3\).

Multiply Equation 2 entirely by exactly \(2\) to perfectly align the respective coefficients of \(I_1\):
\[ 4 I_1 - 10 I_3 = -10 \quad --- (Equation 3) \]

Now, methodically subtract Equation 3 directly from Equation 1:
\[ (4 I_1 + 3 I_3) - (4 I_1 - 10 I_3) = 3 - (-10) \]

The \(I_1\) terms completely cancel out:
\[ 3 I_3 + 10 I_3 = 3 + 10 \]
\[ 13 I_3 = 13 \]
\[ I_3 = 1 A \]


Step 5: {\color{redConclusion

The defined current variable \(I_3\) specifically physically represents the absolute current continuously flowing straight through the central \(3 \Omega\) resistor.

Therefore, the exact current flowing directly through the \(3 \Omega\) resistor is precisely 1 Ampere, effectively directed downwards from junction B to junction E. Quick Tip: When aggressively applying KVL, always remember that traversing forcefully against the assigned current arrow across a given resistor solidly results in a positive voltage gain (\(+IR\)), not a drop.

Concept:

The question is based on the well-known displacement method used in optics to determine the focal length of a convex lens in laboratory experiments.

When the distance between an object and a screen (denoted as \(D\)) is greater than or equal to \(4f\), there exist exactly two positions of the convex lens for which a sharp real image is formed on the screen.

This happens because of the principle of reversibility of light, which implies that the object and image distances can be interchanged (conjugate foci).



Step 1: {\color{redSetting up the lens equation

Let the object be placed at the origin, and the screen is at a distance \(D\).

Let the lens be placed at a distance \(x\) from the object.

According to the sign convention, the object distance is \(u = -x\).

Since the image forms on the screen, the image distance is \(v = +(D - x)\).

The thin lens formula is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]


Step 2: {\color{redFormulating the quadratic equation

Substitute the values of \(u\) and \(v\) into the lens formula:
\[ \frac{1}{D - x} - \frac{1}{-x} = \frac{1}{f} \]
\[ \frac{1}{D - x} + \frac{1}{x} = \frac{1}{f} \]

Taking the least common multiple (LCM) on the left-hand side:
\[ \frac{x + (D - x)}{x(D - x)} = \frac{1}{f} \]
\[ \frac{D}{Dx - x^2} = \frac{1}{f} \]

Cross-multiplying to rearrange into a standard quadratic equation form:
\[ Dx - x^2 = fD \]
\[ x^2 - Dx + fD = 0 \]


Step 3: {\color{redSolving for the roots and finding their difference

The above equation is a quadratic equation in \(x\), which means it will have two solutions (say \(x_1\) and \(x_2\)) representing the two lens positions.

Using the quadratic formula, the roots are:
\[ x = \frac{D \pm \sqrt{D^2 - 4fD}}{2} \]

The two positions of the lens are \(x_1 = \frac{D - \sqrt{D^2 - 4fD}}{2}\) and \(x_2 = \frac{D + \sqrt{D^2 - 4fD}}{2}\).

The distance between these two lens positions is given as \(d\), which is the difference between the roots:
\[ d = x_2 - x_1 \]
\[ d = \frac{D + \sqrt{D^2 - 4fD}}{2} - \frac{D - \sqrt{D^2 - 4fD}}{2} \]
\[ d = \frac{2\sqrt{D^2 - 4fD}}{2} = \sqrt{D^2 - 4fD} \]

Taking \(D\) completely common inside the square root to match the options:
\[ d = \sqrt{D(D - 4f)} \]


Step 4: {\color{redConclusion

The derived expression for the separation between the two lens positions is mathematically \(\sqrt{D(D - 4f)}\).

This exactly corresponds to option (A).
Quick Tip: Always remember that for the displacement method to work and provide real roots, the discriminant must be non-negative (\(D^2 - 4fD \ge 0\)).
This leads to the fundamental condition \(D \ge 4f\) for a real image to form on the screen in two positions.
If \(D = 4f\), the two positions coincide (\(d = 0\)), and if \(D < 4f\), no sharp image can be formed on the screen.

Concept:

The magnification \(m\) of a thin lens is defined as the ratio of the height of the image to the height of the object.

In terms of object distance \(u\) and image distance \(v\), the linear magnification is given by the formula \(m = \frac{v}{u}\).

When the absolute value of magnification \(|m| > 1\), the image is strictly enlarged.

When the absolute value of magnification \(|m| < 1\), the image is strictly reduced or diminished.

In the displacement method, the two lens positions are completely conjugate to each other, meaning \(u_1 = v_2\) and \(v_1 = u_2\).



Step 1: {\color{redAnalyzing the first lens position

When the student starts moving the lens away from the object towards the screen, the first position encountered is closer to the object.

Let the object distance be \(u_1\) and the image distance (distance to the screen) be \(v_1\).

Because the lens is physically closer to the object, we definitely have \(|u_1| < |v_1|\).

The magnification for this first position is \(m_1 = \frac{v_1}{u_1}\).

Since the numerator is geometrically larger than the denominator (\(|v_1| > |u_1|\)), the absolute value \(|m_1|\) is greater than 1.

Therefore, the image formed at the first position is highly enlarged compared to the original object.


Step 2: {\color{redAnalyzing the second lens position

As the student continues to move the lens further towards the screen, the second position is reached.

At this new position, the lens is now physically closer to the screen and much farther from the object.

Let the new object distance be \(u_2\) and the new image distance be \(v_2\).

Due to the principle of reversibility (conjugate property), \(u_2 = v_1\) and \(v_2 = u_1\).

Since we already established \(|v_1| > |u_1|\), it strictly means that \(|u_2| > |v_2|\).

The magnification for this second position is \(m_2 = \frac{v_2}{u_2}\).

Since the numerator is now geometrically smaller than the denominator (\(|v_2| < |u_2|\)), the absolute value \(|m_2|\) is less than 1.

Therefore, the image formed at the second position is significantly reduced (diminished) compared to the original object.


Step 3: {\color{redConclusion

Summarizing the sequence of observations as the lens moves away from the object:

The first position yields an enlarged image.

The second position yields a reduced image.

Hence, the sequence of images formed is "enlarged, reduced", which matches option (D).
Quick Tip: A highly useful relationship in the displacement method is that the product of the two magnifications is exactly unity (\(m_1 \times m_2 = 1\)).
Furthermore, if \(O\) is the true size of the object and \(I_1, I_2\) are the sizes of the two images, the actual size of the object is the geometric mean of the image sizes: \(O = \sqrt{I_1 \times I_2}\).

Concept:

This problem provides practical numerical values to apply the mathematical formula derived for the displacement method.

The total separation between the object and the screen is represented by the capital letter \(D\).

The total separation between the two conjugate positions of the convex lens is represented by the small letter \(d\).

The focal length \(f\) of the convex lens can be determined directly by rearranging the formula \(d = \sqrt{D^2 - 4fD}\).



Step 1: {\color{redIdentify the given quantities

From the text of the problem, the distance between the object and the screen is given as:
\(D = 80.00 cm\).

The distance separating the two sharp image positions of the lens is given as:
\(d = 20.00 cm\).


Step 2: {\color{redRearrange the displacement formula

We previously established that the distance between the two lens positions is:
\[ d = \sqrt{D^2 - 4fD} \]

To isolate the focal length \(f\), we first square both sides of the equation:
\[ d^2 = D^2 - 4fD \]

Next, we bring the term containing \(f\) to the left side and \(d^2\) to the right side:
\[ 4fD = D^2 - d^2 \]

Finally, we divide both sides entirely by \(4D\) to explicitly solve for \(f\):
\[ f = \frac{D^2 - d^2}{4D} \]


Step 3: {\color{redSubstitute values and calculate

Now, substitute the provided numerical values into the derived expression:
\[ f = \frac{(80.00)^2 - (20.00)^2}{4 \times 80.00} \]

Calculate the squares of the terms in the numerator:
\[ (80.00)^2 = 6400 \]
\[ (20.00)^2 = 400 \]

Calculate the denominator:
\[ 4 \times 80.00 = 320 \]

Substitute these computed values back into the fraction:
\[ f = \frac{6400 - 400}{320} \]
\[ f = \frac{6000}{320} \]

Simplify the fraction by completely dividing by \(10\) and then evaluating:
\[ f = \frac{600}{32} \]
\[ f = 18.75 cm \]


Step 4: {\color{redConclusion

The calculated focal length of the given convex lens is precisely \(18.75 cm\).

This matches option (B).
Quick Tip: To prevent heavy multiplication errors during manual calculation, you can smartly use the algebraic identity \(a^2 - b^2 = (a - b)(a + b)\) in the numerator.
For example, \((80^2 - 20^2) = (80 - 20)(80 + 20) = (60)(100) = 6000\).
This trick usually saves time and heavily minimizes calculation blunders during timed competitive exams.

Concept:

The question explores the fundamental physical condition required to perform the displacement method successfully.

For a convex lens to form a real image of an object on a fixed screen, the total distance between the object and the screen must not be arbitrarily small.

Mathematically, this condition is heavily derived from the discriminant of the quadratic equation connecting lens position and focal length.

The discriminant must be strictly non-negative (\(D^2 - 4fD \ge 0\)) to yield real physical positions for the lens.



Step 1: {\color{redDefine the mathematical constraint

As proven in earlier parts, the separation \(d\) between the two lens positions is given by \(d = \sqrt{D^2 - 4fD}\).

For \(d\) to represent a real, measurable physical distance (or zero), the expression purely inside the square root must be greater than or equal to zero.
\[ D^2 - 4fD \ge 0 \]

Since the distance \(D\) is a physically positive quantity (\(D > 0\)), we can safely divide the entire inequality by \(D\):
\[ D - 4f \ge 0 \]
\[ D \ge 4f \]

This means the absolute minimum distance between an object and its real image formed by a convex lens is exactly \(4f\).


Step 2: {\color{redApply the specific given values

The problem states that the focal length of the convex lens is \(f = 15 cm\).

Substitute this specific focal length into the derived constraint:
\[ D_{min} = 4 \times f \]
\[ D_{min} = 4 \times 15 cm \]
\[ D_{min} = 60 cm \]

Therefore, to find two distinct (or strictly coincident) positions of the lens that form a sharp image, the object-screen distance \(D\) must be at least \(60 cm\).


Step 3: {\color{redEvaluate the given options

We carefully check each option against our mathematically established condition \(D \ge 60 cm\):

(A) \(D = 45 cm\): Since \(45 < 60\), this is purely impossible.

(B) \(D = 50 cm\): Since \(50 < 60\), this is purely impossible.

(C) \(D = 55 cm\): Since \(55 < 60\), this is purely impossible.

(D) \(D = 65 cm\): Since \(65 > 60\), this fully satisfies the condition and allows for two distinct real lens positions.


Step 4: {\color{redConclusion

The only provided option that is mathematically greater than the minimum required distance of \(60 cm\) is \(65 cm\).

Thus, option (D) is the correct answer.
Quick Tip: A classic conceptual question often asked in vivas or exams is: "What happens exactly if \(D = 4f\)?"
In that extremely specific scenario, the discriminant becomes exactly zero, implying the two roots are identical (\(x_1 = x_2 = 2f\)).
The lens has only one valid position perfectly midway between the object and screen, and the magnification is exactly \(-1\) (same size, inverted).

Concept:

When two or more thin lenses are placed coaxially in strict physical contact, their individual powers algebraically add up to give the total equivalent power of the combination.

The total power \(P_{eq}\) is given by the formula \(P_{eq} = P_1 + P_2 + ...\)

The power of a single lens in Diopters (D) is calculated precisely as the reciprocal of its focal length measured in meters (\(P = \frac{1}{f(in m)}\)).

Proper sign convention must be stringently followed: convex lenses have positive focal length and power, whereas concave lenses have negative focal length and power.



Step 1: {\color{redCalculate the power of the first lens

The first lens is a convex lens, so its focal length is strictly positive.

Given focal length \(f_1 = +10 cm\).

We must convert this dimension perfectly into meters to properly calculate power in Diopters:
\[ f_1 = \frac{+10}{100} m = +0.1 m \]

Now, calculate the absolute power \(P_1\) of this first lens:
\[ P_1 = \frac{1}{f_1} = \frac{1}{+0.1} = +10 D \]


Step 2: {\color{redSet up the combination equation

The total equivalent power of the lens combination is given in the problem as:
\[ P_{eq} = +\frac{10}{3} D \]

Using the fundamental additive property of lens powers in contact:
\[ P_{eq} = P_1 + P_2 \]

Substitute the known values deeply into the equation:
\[ \frac{10}{3} = 10 + P_2 \]


Step 3: {\color{redSolve for the power and focal length of the unknown lens

Isolate \(P_2\) algebraically:
\[ P_2 = \frac{10}{3} - 10 \]

Take a common denominator to subtract correctly:
\[ P_2 = \frac{10 - 30}{3} = -\frac{20}{3} D \]

The strongly negative sign securely indicates that the second lens is fundamentally diverging (a concave lens).

Now, convert this resulting power back into a focal length \(f_2\) in meters:
\[ f_2 = \frac{1}{P_2} = \frac{1}{-\frac{20}{3}} m = -\frac{3}{20} m \]

Finally, convert this physical dimension back into centimeters for the final answer matching the options:
\[ f_2 = -\frac{3}{20} \times 100 cm \]
\[ f_2 = -3 \times 5 cm = -15 cm \]


Step 4: {\color{redConclusion

The focal length 'f' of the unknown second thin lens is heavily calculated to be \(-15 cm\).

This beautifully matches option (A).
Quick Tip: Always double-check units when dealing comprehensively with optical power.
A highly common mistake is using focal length directly in centimeters in the formula \(P = 1/f\), which yields drastically wrong power values.
Remember the alternative handy formula: \(P(D) = \frac{100}{f(in cm)}\) to skip the tedious meter conversion step entirely.

Concept:

Pure elemental semiconductors like Silicon (Si) and Germanium (Ge) belong structurally to Group 14 of the periodic table and essentially have exactly 4 valence electrons.

Doping is the intentional introduction of specific impurities into an intrinsic semiconductor for the purpose of intensely modulating its electrical properties.

A p-type (positive-type) semiconductor is strategically created by heavily doping the pure crystal with trivalent impurities (atoms possessing only 3 valence electrons).

A trivalent impurity atom forms exactly three covalent bonds with neighboring Si atoms, leaving a sharp vacancy or "hole" in the fourth bond, which practically acts as a positive charge carrier.



Step 1: {\color{redAnalyze the fundamental requirements

To successfully obtain a p-type semiconductor, we strictly need a trivalent dopant element from Group 13 of the periodic table.

Commonly utilized Group 13 elements include Boron (B), Aluminum (Al), Gallium (Ga), and Indium (In).


Step 2: {\color{redEvaluate the chemical nature of the given options

We carefully check the periodic group properties of all four options provided:

(A) Phosphorus (P): This is a prominent Group 15 element. It has precisely 5 valence electrons (pentavalent). Doping with it yields an n-type semiconductor.

(B) Arsenic (As): This is also a Group 15 pentavalent element. It similarly produces an n-type semiconductor.

(C) Boron (B): This is a defining Group 13 element. It has exactly 3 valence electrons (trivalent). When heavily substituted into a silicon lattice, it successfully generates holes, creating a p-type semiconductor.

(D) Antimony (Sb): This is yet another Group 15 pentavalent element. It unequivocally produces an n-type semiconductor.


Step 3: {\color{redConclusion

Based on the thorough chemical classification, Boron is the sole trivalent impurity critically listed in the options.

Therefore, heavily doping silicon with Boron will reliably produce the desired p-type semiconductor.

This makes option (C) correct.
Quick Tip: A very helpful mnemonic to rapidly memorize semiconductor dopants:
For p-type (Group 13, Trivalent, Acceptors): Remember "B A G I" (Boron, Aluminum, Gallium, Indium).
For n-type (Group 15, Pentavalent, Donors): Remember "P As Sb Bi" (Phosphorus, Arsenic, Antimony, Bismuth).

Concept:

Under conditions of absolute thermal equilibrium, regardless of the level of chemical doping, a foundational relationship exists between the concentrations of electrons and holes.

This strict relationship is universally known as the Mass Action Law for semiconductors.

The highly significant mathematical formulation states that the product of majority and minority carrier concentrations is perfectly equal to the square of the intrinsic carrier concentration.

Formula: \(n_e \cdot n_h = n_i^2\), where \(n_e\) is electron concentration, \(n_h\) is hole concentration, and \(n_i\) is intrinsic carrier concentration.



Step 1: {\color{redIdentify the provided physical parameters

The intrinsic carrier concentration parameter is given carefully as:
\(n_i = 1.5 \times 10^{16} m^{-3}\)

The given steady-state electron concentration is heavily defined as:
\(n_e = 5 \times 10^{22} m^{-3}\)

Notice importantly that \(n_e \gg n_i\), clearly indicating this is a strongly n-type extrinsic semiconductor.


Step 2: {\color{redApply the Mass Action Law equation

We aim to precisely determine the unknown hole concentration (\(n_h\)).

Rearranging the Mass Action Law heavily for \(n_h\):
\[ n_h = \frac{n_i^2}{n_e} \]

Substitute the provided heavy numerical values deeply into the numerator and denominator:
\[ n_h = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} \]


Step 3: {\color{redPerform the mathematical calculation carefully

First, meticulously compute the square of the complex intrinsic concentration in the numerator:
\[ (1.5 \times 10^{16})^2 = (1.5)^2 \times (10^{16})^2 \]
\[ = 2.25 \times 10^{32} m^{-6} \]

Now, divide this heavily by the given steady electron concentration:
\[ n_h = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} \]

Isolate the numerical fraction from the deep powers of ten:
\[ n_h = \left(\frac{2.25}{5}\right) \times 10^{(32 - 22)} \]
\[ n_h = 0.45 \times 10^{10} \]

Convert this raw result fully into proper standard scientific notation for final matching:
\[ n_h = 4.5 \times 10^{-1} \times 10^{10} = 4.5 \times 10^9 m^{-3} \]


Step 4: {\color{redConclusion

The heavily calculated minority hole concentration is exactly \(4.5 \times 10^9 m^{-3}\).

This strictly correlates with option (D).
Quick Tip: The Mass Action Law (\(n_e n_h = n_i^2\)) impressively guarantees that as you dope a semiconductor to heavily increase one carrier type (e.g., electrons), the concentration of the opposite carrier type (holes) severely drops.
This happens because the extremely abundant electrons rapidly recombine with the holes, substantially driving their steady-state number down to maintain thermal equilibrium.

Concept:

A deeply fundamental p-n junction features two vastly competing charge transport mechanisms: diffusion and drift.

Diffusion current is strictly driven by the extreme concentration gradient across the junction. Majority carriers (holes from p-side, electrons from n-side) diffuse rapidly into the opposite region.

Drift current is forcefully driven by the internally built-in electric field situated securely within the depletion region. It sweeps newly generated minority carriers across the junction.

During forward bias, an external voltage severely opposes the built-in potential, heavily shrinking the depletion region and strongly lowering the potential barrier height.



Step 1: {\color{redAnalyze the internal physics of forward bias

When a p-n diode is robustly forward-biased, the external electric field aggressively pushes majority carriers heavily towards the junction.

Because the barrier height is heavily reduced, a massive number of majority carriers easily acquire sufficient energy to cross over the junction.

This results in an exponentially massive surge in the diffusion current.

Simultaneously, the heavily opposing external field slightly decreases the already minuscule drift of minority carriers.

Thus, the total forward current is overwhelmingly dominated by the massive motion of majority carriers forcefully crossing the junction.


Step 2: {\color{redEvaluate the phrasing of the provided options

Let us meticulously scrutinize the given options based on semiconductor physics:

(C) suggests diffusion and drift are perfectly equal. This is purely the condition for an unbiased, open-circuit diode in thermal equilibrium, not forward bias.

(D) suggests current is strictly around 1 A. Forward currents in typical small-signal semiconductor diodes are securely in the heavily lower milliampere (mA) range, so this is generally false.

(B) suggests current is mainly due to drifting of minority carriers. This is heavily true only for reverse bias, where diffusion is completely stopped and only leakage drift occurs.

(A) claims current is mainly due to drifting of majority carriers. Physically speaking, the dominant mechanism for majority carriers crossing the lowered barrier is strictly termed "diffusion", not "drift". Drift is firmly associated with the electric field forcefully moving carriers.

However, in heavily simplified or poorly translated contexts (like translating the Hindi term 'अपवाह' loosely), "drifting" might inappropriately be used interchangeably as general "motion" or "flow".

Despite the heavily flawed usage of the strict word "drifting", the undeniable core concept intended by the examiner is that forward current is overwhelmingly driven by the massive flow of majority carriers.

Therefore, option (A) is definitely the most conceptually aligned intended answer among the highly flawed choices.


Step 3: {\color{redConclusion

While strictly speaking forward current is a "diffusion" current of majority carriers, option (A) correctly identifies that the current is heavily dominated by "majority carriers".

We securely select option (A) as the intended correct response.
Quick Tip: Always firmly remember the golden rule of p-n junctions:
Forward Bias = Huge current heavily dominated by Majority Carriers (physically via Diffusion).
Reverse Bias = Tiny leakage current heavily dominated by Minority Carriers (physically via Drift).
Even if exam terminology is slightly warped, firmly picking the option matching the correct carrier type (majority vs minority) is typically the safest strategy.

Concept:

The threshold voltage (also widely known as the knee voltage, cut-in voltage, or strongly built-in potential barrier) of a p-n junction diode is a highly critical parameter.

It firmly represents the absolute minimum forward bias voltage that must be externally applied to massively overcome the internal potential barrier of the heavily depleted region.

Below this specific threshold voltage, the heavily suppressed forward current is practically negligible.

Once the external voltage firmly exceeds this threshold, the diffusion current rises massively and exponentially.



Step 1: {\color{redMaterial dependency of threshold voltage

The intrinsic value of the cut-in voltage is heavily dependent on the fundamental energy bandgap of the specific semiconductor material used to fabricate the diode.

Materials with a heavily larger forbidden energy bandgap strictly require a much higher forward voltage to enable carriers to successfully cross the junction barrier.


Step 2: {\color{redComparing standard semiconductor materials

Germanium (Ge) strongly possesses a relatively small bandgap of approximately \(0.66 eV\) at room temperature.

Because of this physically smaller energy requirement, the threshold voltage for a standard Germanium diode is securely around \(0.2 V\) to \(0.3 V\).

Silicon (Si) strongly possesses a significantly larger bandgap of approximately \(1.12 eV\) at room temperature.

Consequently, establishing a sufficient forward current in Silicon heavily requires a substantially higher applied external potential.

The universally accepted, standard threshold voltage for a typical Silicon diode firmly resides at approximately \(0.7 V\).


Step 3: {\color{redEvaluating the options

Looking closely at the rigidly provided options:

(A) 0.2 V roughly corresponds to heavily doped Germanium or some Schottky diodes.

(C) 0.7 V precisely matches the universally accepted physical standard for Silicon p-n junction diodes.


Step 4: {\color{redConclusion

The heavily demanded threshold voltage for a standard Silicon diode is universally recognized as approximately \(0.7 V\).

This strictly makes option (C) the firmly correct choice.
Quick Tip: In heavy numerical circuit analysis problems involving diodes, if the diode is strictly stated to be 'Silicon' but no other parameter is explicitly given, you must always firmly assume a forward voltage drop of exactly \(0.7 V\).
If the problem strictly specifies an 'Ideal' diode, then heavily assume the voltage drop is exactly \(0 V\) when fully forward-biased.

Concept:

When a pentavalent elemental impurity (like Phosphorus or Arsenic) heavily dopes a Germanium (Ge) or Silicon (Si) crystal lattice, it forcefully replaces a host atom.

Four of the five valence electrons from the impurity atom form incredibly strong covalent bonds with the four immediately adjacent host atoms.

The critical fifth electron is left entirely unbound to any specific covalent bond. It continues to orbit the heavily localized positive ion core of the impurity atom.

This highly specific physical setup can be modeled extremely well using a heavily modified version of the Bohr model of the Hydrogen atom.



Step 1: {\color{redUnderstanding the modified Bohr model for semiconductors

In a pure vacuum, the ionization energy of a pristine Hydrogen atom is heavily established as \(13.6 eV\).

However, the weakly bound fifth electron in a doped semiconductor operates inside a dense crystalline medium.

This heavy crystalline environment drastically alters the physics in two major ways:

1. The electrostatic Coulomb force is heavily reduced by the large relative dielectric constant (\(\epsilon_r\)) of the host semiconductor crystal.

2. The electron's dynamic motion is heavily altered, so it operates with an effective mass (\(m^*\)) that is typically much lighter than the true free electron mass (\(m_e\)).

The severely modified ionization energy \(E_d\) for this localized donor electron is given by the formula:
\[ E_d = 13.6 eV \times \left( \frac{m^*}{m_e} \right) \times \left( \frac{1}{\epsilon_r^2} \right) \]


Step 2: {\color{redComparing the physical values for Silicon and Germanium

Silicon (Si) has a relative dielectric constant of roughly \(\epsilon_r \approx 11.7\).

Due to this moderate dielectric screening, the calculated ionization energy required to firmly free the fifth electron in Silicon is experimentally found to be approximately \(0.05 eV\).

Germanium (Ge) operates with a significantly larger relative dielectric constant of roughly \(\epsilon_r \approx 16\).

Because the dielectric constant is heavily squared in the denominator of our modified formula, the physically higher \(\epsilon_r\) of Ge causes a massive reduction in the binding energy.

Additionally, the effective mass of an electron in Germanium is also structurally different.

As a direct result of these heavy crystalline factors, the ionization energy for the weakly bound fifth electron in heavily doped Germanium plunges to approximately \(0.01 eV\).


Step 3: {\color{redConclusion

The heavily researched ionization energy for liberating the fifth electron in a pentavalent-doped Germanium crystal firmly sits at about \(0.01 eV\).

This strictly matches the numerical value provided in option (A).
Quick Tip: This incredibly tiny ionization energy (roughly 10 to 50 meV) is massively important for modern electronics because it perfectly matches the available thermal energy (\(k_B T \approx 0.026 eV\)) at absolute room temperature (300 K).
This physically guarantees that almost 100% of the deeply embedded donor impurity atoms are fully ionized at standard room temperature, freely providing heavy numbers of conduction electrons to the bulk material.

Concept:

In a series LCR circuit, an inductor \(L\), capacitor \(C\), and resistor \(R\) are connected in series across an alternating voltage source \(V = V_0 \sin(\omega t)\).

The current \(I\) is common to all three components at any instant.

Voltage across resistor \(V_R = I R\) is in phase with current \(I\).

Voltage across inductor \(V_L = I X_L\) leads current \(I\) by \(\frac{\pi}{2}\) radians, where \(X_L = \omega L\).

Voltage across capacitor \(V_C = I X_C\) lags behind current \(I\) by \(\frac{\pi}{2}\) radians, where \(X_C = \frac{1}{\omega C}\).



Step 1: {\color{redPhasor Diagram Construction

Let current phasor \(\vec{I}\) be drawn along the positive \(x\)-axis.

The potential difference across resistor \(\vec{V}_R\) is along the \(x\)-axis (in phase with \(\vec{I}\)).

The potential difference across inductor \(\vec{V}_L\) is along the positive \(y\)-axis (leading \(\vec{I}\) by \(90^\circ\)).

The potential difference across capacitor \(\vec{V}_C\) is along the negative \(y\)-axis (lagging \(\vec{I}\) by \(90^\circ\)).

Assuming \(V_L > V_C\), the resultant vector along the \(y\)-axis is \((V_L - V_C)\) pointing along the positive \(y\)-axis.





Step 2: {\color{redDerivation of Impedance Expression

Using vector addition for phasors, the total applied voltage \(V\) is given by the hypotenuse of the right-angled triangle formed by \(V_R\) and \((V_L - V_C)\):
\[ V^2 = V_R^2 + (V_L - V_C)^2 \]

Substitute \(V_R = I R\), \(V_L = I X_L\), and \(V_C = I X_C\):
\[ V^2 = (I R)^2 + (I X_L - I X_C)^2 \]
\[ V^2 = I^2 \left[ R^2 + (X_L - X_C)^2 \right] \]

Taking the square root on both sides:
\[ V = I \sqrt{R^2 + (X_L - X_C)^2} \]

The total effective opposition offered by the series LCR circuit to alternating current is called impedance \(Z\):
\[ Z = \frac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2} \]

Substituting \(X_L = \omega L\) and \(X_C = \frac{1}{\omega C}\):
\[ Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]


Step 3: {\color{redDerivation of Phase Difference

From the phasor diagram, if \(\phi\) is the phase angle between the total supply voltage \(V\) and current \(I\):
\[ \tan \phi = \frac{V_L - V_C}{V_R} = \frac{I X_L - I X_C}{I R} = \frac{X_L - X_C}{R} \]
\[ \phi = \tan^{-1} \left( \frac{\omega L - \frac{1}{\omega C}}{R} \right) \]


Step 4: {\color{redConclusion

The impedance of the series LCR circuit is \(Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}\) and the phase difference between voltage and current is \(\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)\).
Quick Tip: When \(X_L > X_C\), the circuit is predominantly inductive, and voltage leads current.
When \(X_C > X_L\), the circuit is predominantly capacitive, and voltage lags behind current.
At resonance (\(X_L = X_C\)), \(Z = R\) and \(\phi = 0^\circ\), meaning voltage and current are in phase.

Concept:

Impedance of an LCR series circuit is given by \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).

Average power consumed in an AC circuit is \(P_{avg} = V_{rms} I_{rms} \cos \phi\), where \(\cos \phi = \frac{R}{Z}\) is the power factor.

Current is said to be wattless if the power consumed in the AC circuit is zero despite current flowing through it.



Step 1: {\color{redCondition for Minimum Impedance

The impedance formula is \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).

Since \(R^2 \ge 0\) and \((X_L - X_C)^2 \ge 0\), the minimum possible value of \(Z\) occurs when the term \((X_L - X_C)^2\) becomes zero.

This condition is satisfied when inductive reactance equals capacitive reactance:
\[ X_L = X_C \implies \omega L = \frac{1}{\omega C} \]

This condition is known as electrical resonance.

Under resonance, minimum impedance is equal to resistance:
\[ Z_{min} = R \]


Step 2: {\color{redCondition for Wattless Current

Wattless current flows when the average power dissipation in the circuit is zero (\(P_{avg} = 0\)).

The average power dissipated is given by:
\[ P_{avg} = V_{rms} I_{rms} \cos \phi \]

For \(P_{avg} = 0\) while \(V_{rms} \neq 0\) and \(I_{rms} \neq 0\), we must have:
\[ \cos \phi = 0 \implies \phi = \frac{\pi}{2} or 90^\circ \]

From \(\tan \phi = \frac{X_L - X_C}{R}\), \(\phi = 90^\circ\) requires resistance \(R = 0\).

Therefore, wattless current flows in a purely inductive or purely capacitive circuit containing zero resistance (\(R = 0\)).


Step 3: {\color{redConclusion

(i) Impedance is minimum at resonance when \(X_L = X_C\) or \(\omega = \frac{1}{\sqrt{LC}}\).

(ii) Wattless current flows when the circuit resistance is zero (\(R = 0\)), making the phase difference \(\phi = \frac{\pi}{2}\).
Quick Tip: Wattless current is component \(I_{rms} \sin \phi\) which does no useful electrical work over a full cycle because power consumed is zero.
Real choke coils are used in AC circuits to reduce current without significant energy loss because \(R \approx 0\).

Concept:

An AC generator (or alternator) is an electrical device that converts mechanical energy into alternating electrical energy.

It works on the principle of Faraday's Law of Electromagnetic Induction.



Step 1: {\color{redPrinciple

When a closed armature coil is rotated rapidly in a uniform magnetic field, the magnetic flux linked with the coil continuously changes with time.

According to Faraday's law of electromagnetic induction, an electromotive force (emf) is induced in the coil, generating an alternating current in the external circuit.


Step 2: {\color{redConstruction

An AC generator consists of four main parts:

1. Armature (Coil): A rectangular coil \(ABCD\) consisting of a large number of turns of insulated copper wire wound over a soft iron core to increase magnetic field intensity.

2. Field Magnet: A strong electromagnet or permanent horse-shoe magnet providing a strong uniform magnetic field perpendicular to the axis of rotation.

3. Slip Rings: The two ends of the armature coil are connected to two metallic hollow rings \(R_1\) and \(R_2\), which rotate along with the coil.

4. Carbon Brushes: Two stationary flexible carbon blocks \(B_1\) and \(B_2\) remain in light sliding contact with slip rings \(R_1\) and \(R_2\) to conduct current to the external load circuit.





Step 3: {\color{redWorking

When the armature coil \(ABCD\) is rotated mechanically in the magnetic field with angular velocity \(\omega\):

During the first half-rotation, side \(AB\) moves upwards and side \(CD\) moves downwards.

By Fleming's Right-Hand Rule, induced current flows along \(ABCD\) through brush \(B_1\) to \(B_2\) in the outer circuit.

In the second half-rotation, side \(AB\) moves downwards and \(CD\) moves upwards.

The direction of induced current reverses, flowing along \(DCBA\) from \(B_2\) to \(B_1\) in the outer circuit.

Hence, the direction of current in the load resistor reverses periodically after every half rotation, generating alternating current.


Step 4: {\color{redConclusion

The continuous rotation of the armature coil in a magnetic field induces an alternating voltage across the slip rings, providing AC current to the connected external load.
Quick Tip: Slip rings are used in AC generators to maintain electrical connectivity while reversing current direction every half cycle.
Split rings (commutators) are used instead in DC generators to convert internal AC into unidirectional DC.

Concept:

Magnetic flux linked with a single turn of area \(A\) inclined at angle \(\theta\) to magnetic field \(\vec{B}\) is \(\Phi = B A \cos \theta\).

If the coil rotates with uniform angular speed \(\omega\), \(\theta = \omega t\).

By Faraday's Law, induced emf is \(e = -N \frac{d\Phi}{dt}\).



Step 1: {\color{redMagnetic Flux Expression

Let \(N\) = total number of turns in the rectangular coil, \(A\) = area of each turn, \(B\) = magnitude of magnetic field.

At time \(t = 0\), let the normal to the coil be parallel to \(\vec{B}\) (\(\theta = 0\)).

At time \(t\), the coil rotates through angle \(\theta = \omega t\).

The magnetic flux linked with all \(N\) turns of the coil is:
\[ \Phi = N (\vec{B} \cdot \vec{A}) = N B A \cos(\omega t) \]


Step 2: {\color{redApplying Faraday's Law

According to Faraday's law of electromagnetic induction, induced emf \(e\) is:
\[ e = -\frac{d\Phi}{dt} = -\frac{d}{dt} \left[ N B A \cos(\omega t) \right] \]

Differentiating \(\cos(\omega t)\) with respect to \(t\):
\[ \frac{d}{dt} [\cos(\omega t)] = -\omega \sin(\omega t) \]

Substitute back into the expression:
\[ e = -N B A \left( -\omega \sin(\omega t) \right) \]
\[ e = N B A \omega \sin(\omega t) \]


Step 3: {\color{redPeak Value and Final Relation

Let \(e_0 = N B A \omega\) be the maximum or peak value of induced emf.

Then the instantaneous induced emf equation becomes:
\[ e = e_0 \sin(\omega t) \]


Step 4: {\color{redConclusion

The induced emf varies sinusoidally with time according to \(e = N B A \omega \sin(\omega t) = e_0 \sin(\omega t)\).
Quick Tip: Peak emf \(e_0 = N B A \omega\) can be increased by increasing the number of turns \(N\), magnetic field strength \(B\), coil area \(A\), or speed of rotation \(\omega\).

Concept:

The induced emf in an AC generator is \(e = e_0 \sin(\omega t)\).

Angular frequency \(\omega\) is related to time period \(T\) by \(\omega = \frac{2\pi}{T}\).

Magnitude of induced emf \(|e|\) is maximum when \(|\sin(\omega t)| = 1\).



Step 1: {\color{redCondition for Maximum EMF

The induced emf expression is:
\[ e = e_0 \sin\left(\frac{2\pi}{T} t\right) \]

The magnitude of induced emf reaches maximum value \(|e| = e_0\) when:
\[ \sin\left(\frac{2\pi}{T} t\right) = \pm 1 \]


Step 2: {\color{redFinding Values of t within One Cycle (\(0 \le t \le T\))

For \(\sin \theta = +1\):
\[ \theta = \frac{\pi}{2} \implies \frac{2\pi}{T} t = \frac{\pi}{2} \implies t = \frac{T}{4} \]

For \(\sin \theta = -1\):
\[ \theta = \frac{3\pi}{2} \implies \frac{2\pi}{T} t = \frac{3\pi}{2} \implies t = \frac{3T}{4} \]


Step 3: {\color{redPhysical Significance

At \(t = \frac{T}{4}\) and \(t = \frac{3T}{4}\), the plane of the rotating armature coil is parallel to the magnetic field lines (\(\theta = 90^\circ\) and \(270^\circ\)).

In this orientation, rate of change of magnetic flux is maximum, producing peak induced emf.


Step 4: {\color{redConclusion

The induced emf of the generator reaches its maximum values at times \(t = \frac{T}{4}\) and \(t = \frac{3T}{4}\) during one full rotation.
Quick Tip: When plane of coil is perpendicular to magnetic field lines (\(\theta = 0, \pi\)), flux is maximum, but rate of change of flux is zero, so induced emf is zero (\(t = 0, \frac{T}{2}, T\)).

Concept:

Coherent sources are two independent or derived sources of light that emit light waves having the same frequency, same wavelength, and maintain a constant phase difference over time.

Resultant intensity at any point due to superposition of two waves is \(I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos \phi\).



Step 1: {\color{redDefinition of Coherent Sources

Two light sources are said to be coherent if they emit light waves of the same frequency, same wavelength, identical waveform, and have zero or a constant phase difference between them over time.


Step 2: {\color{redNeed for Coherent Sources for Stable Interference

The intensity at a point on the screen due to two overlapping light waves is given by:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos \phi \]

If sources are incoherent, the phase difference \(\phi\) changes randomly and extremely rapidly with time (on the order of \(10^{-8}\) s).

As a result, the time-average value of \(\cos \phi\) over human observation time becomes zero:
\[ \langle \cos \phi \rangle = 0 \implies I_{avg} = I_1 + I_2 \]

Thus, interference term vanishes, and uniform illumination is observed on the screen without distinct bright and dark fringes.

Coherent sources ensure that phase difference \(\phi\) remains constant with time at every point, giving a time-independent, stationary, and stable interference pattern with sharp bright and dark fringes.


Step 3: {\color{redGraph of Intensity Distribution

In Young's double-slit experiment using monochromatic light where both slits have equal intensity \(I_0\):

Maximum intensity at bright fringes: \(I_{max} = 4 I_0\) at path difference \(\Delta x = n\lambda\).

Minimum intensity at dark fringes: \(I_{min} = 0\) at path difference \(\Delta x = \left(n + \frac{1}{2}\right)\lambda\).

All bright fringes have equal maximum intensity \(4I_0\), and fringe width \(\beta = \frac{\lambda D}{d}\) is uniform.





Step 4: {\color{redConclusion

Coherent sources maintain a constant phase difference required for stable, non-time-varying constructive and destructive interference fringes on the screen.
Quick Tip: Two independent light sources (like two separate bulbs) can never be coherent because atomic emission processes in independent sources are completely random and uncorrelated.

Concept:

The relation between phase difference \(\phi\) and path difference \(\Delta x\) is \(\phi = \frac{2\pi}{\lambda} \cdot \Delta x\).

Resultant intensity of two coherent waves each of intensity \(I_0\) is \(I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right)\).



Step 1: {\color{redFormulas Used

Phase difference \(\phi\) corresponding to path difference \(\Delta x\):
\[ \phi = \left( \frac{2\pi}{\lambda} \right) \Delta x \]

Resultant intensity \(I\):
\[ I = I_0 + I_0 + 2\sqrt{I_0 I_0}\cos \phi = 2 I_0 (1 + \cos \phi) = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \]


Step 2: {\color{redCase (i): Path Difference \(\Delta x = \frac{\lambda}{4}\)

Calculate phase difference \(\phi_1\):
\[ \phi_1 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} radians \]

Calculate intensity \(I_1\):
\[ I_1 = 4 I_0 \cos^2\left( \frac{\pi/2}{2} \right) = 4 I_0 \cos^2\left(\frac{\pi}{4}\right) \]

Since \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\):
\[ I_1 = 4 I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = 4 I_0 \left(\frac{1}{2}\right) = 2 I_0 \]


Step 3: {\color{redCase (ii): Path Difference \(\Delta x = \frac{\lambda}{3}\)

Calculate phase difference \(\phi_2\):
\[ \phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} radians \]

Calculate intensity \(I_2\):
\[ I_2 = 4 I_0 \cos^2\left( \frac{2\pi/3}{2} \right) = 4 I_0 \cos^2\left(\frac{\pi}{3}\right) \]

Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\):
\[ I_2 = 4 I_0 \left(\frac{1}{2}\right)^2 = 4 I_0 \left(\frac{1}{4}\right) = I_0 \]

let's verify:
\(\cos\left(\frac{2\pi}{3}\right) = -1/2\).
\(I_2 = 2 I_0 (1 + \cos(2\pi/3)) = 2 I_0 (1 - 0.5) = 2 I_0 \times 0.5 = I_0\).

Or using \(4 I_0 \cos^2(\pi/3) = 4 I_0 (1/2)^2 = I_0\).



Step 4: {\color{redConclusion

(i) For path difference \(\frac{\lambda}{4}\), resultant intensity is \(2 I_0\).

(ii) For path difference \(\frac{\lambda}{3}\), resultant intensity is \(I_0\).
Quick Tip: Remember key intensity values for equal source intensities \(I_0\):
\(\Delta x = 0 \implies I = 4I_0\) (Central Maxima)
\(\Delta x = \lambda/4 \implies I = 2I_0\)
\(\Delta x = \lambda/3 \implies I = I_0\)
\(\Delta x = \lambda/2 \implies I = 0\) (Minima)

Concept:

An astronomical refracting telescope consists of two convex lenses: an objective lens of large focal length \(f_o\) and large aperture, and an eyepiece of small focal length \(f_e\) and small aperture.

In normal adjustment, final image is formed at infinity.



Step 1: {\color{redRay Diagram

Parallel rays from a distant object enter objective lens at an angle \(\alpha\).

Objective lens forms a real, inverted, and diminished image \(A'B'\) in its focal plane (\(f_o\)).

For final image to be at infinity, \(A'B'\) must lie exactly at the principal focus \(F_e\) of eyepiece lens.

Eyepiece refractor forms parallel rays emerging out at angle \(\beta\) into observer's eye.





Step 2: {\color{redDerivation of Magnifying Power

Magnifying power \(m\) of telescope is defined as ratio of angle \(\beta\) subtended at eye by final image to angle \(\alpha\) subtended by object at eye:
\[ m = \frac{\beta}{\alpha} \]

Since angles \(\alpha\) and \(\beta\) are very small:
\[ \alpha \approx \tan \alpha = \frac{A'B'}{O B'} = \frac{A'B'}{f_o} \]
\[ \beta \approx \tan \beta = \frac{A'B'}{E B'} = \frac{A'B'}{-f_e} \]

Here \(O B' = +f_o\) is focal length of objective, and \(E B' = -f_e\) is focal length of eyepiece (using Cartesian sign convention).


Step 3: {\color{redFinal Expression

Substitute expressions for \(\beta\) and \(\alpha\):
\[ m = \frac{\frac{A'B'}{-f_e}}{\frac{A'B'}{f_o}} = -\frac{f_o}{f_e} \]

Length of telescope tube in normal adjustment is \(L = f_o + f_e\).


Step 4: {\color{redConclusion

The magnifying power of a refracting telescope in normal adjustment is \(m = -\frac{f_o}{f_e}\). The negative sign indicates that final image is inverted with respect to object.
Quick Tip: To achieve large angular magnification \(m = -f_o / f_e\), the objective lens must have a very large focal length (\(f_o\)), while the eyepiece must have a small focal length (\(f_e\)).

Concept:

Objective lens collects light from faint, distant astronomical bodies.

Light gathering power is proportional to area of aperture (\(\propto D^2\)).

Resolving power is given by \(R.P. = \frac{D}{1.22 \lambda}\).



Step 1: {\color{redReason for Large Aperture Objective

1. Light Gathering Power: Distant stars/galaxies emit extremely faint light. A large aperture objective lens collects a large quantity of light rays, forming bright and clear images.

2. Resolving Power: Resolving power of telescope is directly proportional to diameter \(D\) of objective aperture (\(R.P. = \frac{D}{1.22 \lambda}\)). A larger aperture enables resolving close double stars or fine details of celestial bodies.


Step 2: {\color{redConclusion

Large objective aperture maximizes light collection and spatial resolution. Reflecting telescopes overcome chromatic aberration and structural weight constraints inherent to refracting lens systems.
Quick Tip: Paraboloidal mirrors are preferred in reflecting telescopes over spherical mirrors to eliminate spherical aberration completely without needing corrective lenses.

Concept:

Objective lens collects light from faint, distant astronomical bodies.

Light gathering power is proportional to area of aperture (\(\propto D^2\)).

Resolving power is given by \(R.P. = \frac{D}{1.22 \lambda}\).



Step 1: {\color{redAdvantages of Reflecting Telescope over Refracting Telescope

1. Absence of Chromatic Aberration: Reflecting telescopes use parabolic mirrors instead of lenses. Since reflection does not depend on wavelength, images are completely free from chromatic aberration.

2. High Light Efficiency and Easy Support: Large parabolic mirrors can be supported along their entire back surface rather than just at edges (unlike heavy thick lenses), avoiding mechanical sagging. Also, mirrors require surface polishing only, making them cheaper and lighter to fabricate with huge diameters.


Step 2: {\color{redConclusion

Large objective aperture maximizes light collection and spatial resolution. Reflecting telescopes overcome chromatic aberration and structural weight constraints inherent to refracting lens systems.
Quick Tip: Paraboloidal mirrors are preferred in reflecting telescopes over spherical mirrors to eliminate spherical aberration completely without needing corrective lenses.

Concept:

A parallel plate capacitor consists of two conducting parallel plates, each of area \(A\), separated by small distance \(d\).

Plates carry equal and opposite surface charge densities \(\sigma = +\frac{Q}{A}\) and \(-\sigma = -\frac{Q}{A}\).

Electric field in outer regions is zero, while uniform electric field exists in region between plates.



Step 1: {\color{redElectric Field Between Plates

Electric field due to a single infinite thin plane sheet of charge is \(E = \frac{\sigma}{2\varepsilon_0}\).

In the region between positively charged plate (\(+\sigma\)) and negatively charged plate (\(-\sigma\)):

Both electric fields point in the same direction (from positive to negative plate).

Total electric field \(E\):
\[ E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \]

Substitute surface charge density \(\sigma = \frac{Q}{A}\):
\[ E = \frac{Q}{\varepsilon_0 A} \]


Step 2: {\color{redPotential Difference Between Plates

Since electric field \(E\) is uniform over distance \(d\), potential difference \(V\) between plates is:
\[ V = E \cdot d = \left( \frac{Q}{\varepsilon_0 A} \right) d \]


Step 3: {\color{redCapacitance Calculation

Capacitance \(C\) is defined as ratio of total charge \(Q\) to potential difference \(V\):
\[ C = \frac{Q}{V} \]

Substitute expression for \(V\):
\[ C = \frac{Q}{\frac{Q d}{\varepsilon_0 A}} = \frac{\varepsilon_0 A}{d} \]


Step 4: {\color{redConclusion

The capacitance of a parallel plate capacitor filled with air is \(C = \frac{\varepsilon_0 A}{d}\).
Quick Tip: Capacitance depends only on geometric factors: directly proportional to area \(A\), inversely proportional to plate separation \(d\), and permittivity of medium. It is independent of total charge \(Q\) or potential \(V\).

Concept:

When capacitors remain connected to a DC battery of potential \(V\), potential difference across plates remains constant (\(V' = V\)).

Insertion of dielectric slab of dielectric constant \(K\) increases capacitance from \(C\) to \(C' = K C\).



Step 1: {\color{redEffect on Capacitance

Initial capacitances are \(C_1\) and \(C_2\).

When dielectric slabs are inserted, new capacitances become:
\[ C_1' = K C_1 \quad and \quad C_2' = K C_2 \]

Thus, capacitance of each capacitor increases by factor \(K\).


Step 2: {\color{red(i) Effect on Charge on Each Capacitor

Since battery remains connected, potential difference \(V\) across each capacitor remains unchanged.

Initial charges: \(Q_1 = C_1 V\) and \(Q_2 = C_2 V\).

New charges after dielectric insertion:
\[ Q_1' = C_1' V = (K C_1) V = K Q_1 \]
\[ Q_2' = C_2' V = (K C_2) V = K Q_2 \]

Therefore, charge on each capacitor increases \(K\) times.


Step 3: {\color{red(ii) Effect on Stored Energy

Initial energy stored in capacitors:
\[ U_1 = \frac{1}{2} C_1 V^2 \quad and \quad U_2 = \frac{1}{2} C_2 V^2 \]

New energy stored in capacitors:
\[ U_1' = \frac{1}{2} C_1' V^2 = \frac{1}{2} (K C_1) V^2 = K U_1 \]
\[ U_2' = \frac{1}{2} C_2' V^2 = \frac{1}{2} (K C_2) V^2 = K U_2 \]

Therefore, energy stored in each capacitor increases \(K\) times.


Step 4: {\color{redConclusion

When battery remains connected:

(i) Charge on each capacitor increases to \(K\) times its initial value.

(ii) Stored energy in each capacitor increases to \(K\) times its initial value.
Quick Tip: Always check battery connection status:
Battery CONNECTED \(\implies V\) remains constant \(\implies Q' = KQ\), \(U' = KU\).
Battery DISCONNECTED \(\implies Q\) remains constant \(\implies V' = V/K\), \(U' = U/K\).

Concept:

Free electrons inside conductor undergo frequent collisions with fixed heavy lattice ions.

In absence of electric field, average thermal velocity is zero (\(\vec{u}_{avg} = 0\)).

Electric field \(\vec{E}\) exerts force \(\vec{F} = -e \vec{E}\), producing acceleration \(\vec{a} = \frac{-e \vec{E}}{m}\).



Step 1: {\color{redAttainment of Time-Independent Average Drift Velocity

An applied electric field \(\vec{E}\) accelerates free electrons opposite to field lines.

However, electrons do not accelerate indefinitely because they continuously collide with vibrating lattice ions.

In each collision, electron loses its directed momentum and resets its direction randomly.

The average time elapsed between two successive collisions is relaxation time \(\tau\).

Velocity gained by \(i\)-th electron just before next collision:
\[ \vec{v}_i = \vec{u}_i + \vec{a} \tau_i \]

Averaging over all \(N\) free electrons:
\[ \vec{v}_d = \frac{1}{N} \sum \vec{v}_i = \frac{1}{N} \sum \vec{u}_i + \vec{a} \left( \frac{1}{N} \sum \tau_i \right) \]

Since initial thermal velocities are completely random, \(\frac{1}{N}\sum \vec{u}_i = 0\).

Defining average relaxation time \(\tau = \frac{1}{N}\sum \tau_i\):
\[ \vec{v}_d = \vec{a} \tau = -\frac{e \vec{E}}{m} \tau \]

Magnitude of drift velocity:
\[ v_d = \frac{e E \tau}{m} \]

Since \(e, E, m, \tau\) are constant at a given temperature, \(v_d\) is a steady, constant average drift velocity independent of time.


Step 2: {\color{redRelation Between Current and Drift Velocity

Consider cylindrical conductor of length \(L\), cross-sectional area \(A\), and free electron density \(n\).

Total volume of conductor \(V_{vol} = A \cdot L\).

Total number of free electrons in length \(L\): \(N_{total} = n A L\).

Total free charge in length \(L\): \(Q = (n A L) e\).

Time taken by electrons to drift through distance \(L\): \(t = \frac{L}{v_d}\).

Current \(I\) flowing through conductor:
\[ I = \frac{Q}{t} = \frac{n A L e}{\frac{L}{v_d}} = n e A v_d \]


Step 3: {\color{redConclusion

Electrons attain steady drift velocity \(v_d = \frac{e E \tau}{m}\) due to balance between acceleration and lattice collisions. The electric current is \(I = n e A v_d\).
Quick Tip: Drift velocity \(v_d\) is extremely small (around \(10^{-4} m/s\) or \(0.1 mm/s\)), whereas thermal velocity of electrons is very high (around \(10^5 m/s\)).

Concept:

Drift velocity \(v_d\) is the actual physical speed at which individual electrons drift along the wire length.

Electric field establishment throughout circuit occurs via electromagnetic wave propagation.



Step 1: {\color{redExplanation

When a circuit switch is closed, an electric field is established across every section of the conducting wire almost instantaneously.

The electric field propagates through the wire at the speed of electromagnetic waves in a medium, which is nearly equal to the speed of light \(c \approx 3 \times 10^8 m/s\).


Step 2: {\color{redMechanism

Electric current is not caused by a single electron traveling all the way from source to appliance.

Instead, free electrons are already present densely everywhere throughout the entire volume of conducting wire (\(n \approx 10^{28} m^{-3}\)).

As soon as the electric field reaches every point along the conductor, all free electrons across the entire circuit begin drifting simultaneously at their respective positions.


Step 3: {\color{redConclusion

Current is established almost instantly because the electric field travels through the wire at light speed, initiating drift motion of electrons throughout the entire loop at the same instant.
Quick Tip: Analogy: Think of a pipe completely packed full of marbles. Pushing one marble in at one end immediately pushes a marble out at the far end, even though individual marbles move slowly.

Concept:

When components are connected in series, the same current \(I\) flows through each of them.

Current expression: \(I = n e A v_d = n e (\pi r^2) v_d\).



Step 1: {\color{redGiven Data

Ratio of radii of two copper wires: \(\frac{r_1}{r_2} = \frac{3}{2}\).

Since both wires are made of copper, electron number density \(n\) is identical for both wires.

Since wires are connected in series across battery: \(I_1 = I_2 = I\).


Step 2: {\color{redRelating Drift Velocity to Radius

From current formula \(I = n e A v_d\):
\[ I = n e (\pi r^2) v_d \]

Since \(I\), \(n\), \(e\), and \(\pi\) are constants:
\[ r_1^2 v_{d1} = r_2^2 v_{d2} \]
\[ \frac{v_{d1}}{v_{d2}} = \left( \frac{r_2}{r_1} \right)^2 \]


Step 3: {\color{redCalculation

Substitute \(\frac{r_1}{r_2} = \frac{3}{2} \implies \frac{r_2}{r_1} = \frac{2}{3}\):
\[ \frac{v_{d1}}{v_{d2}} = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \]


Step 4: {\color{redConclusion

The ratio of drift velocities of electrons in the two copper wires is \(4 : 9\).
Quick Tip: In series connection, drift velocity is inversely proportional to cross-sectional area (\(v_d \propto \frac{1}{A} \propto \frac{1}{r^2}\)). Thinner wire has higher drift velocity!