CBSE Class 12 Physics Set 3 - (55/4/3) Question Paper 2026 with Solution PDF

\ Concept:

• Photoelectric emission occurs from a metal surface if and only if the energy of incident photon \(E\) is greater than or equal to the work function \(\Phi_0\) of that metal.
• The energy of an incident photon of wavelength \(\lambda\) is given by \(E = \frac{hc}{\lambda}\).

Step 1: Calculate the energy of incident photon
Given wavelength of incident radiation \(\lambda = 331\text{ nm} = 331 \times 10^{-9}\text{ m}\).
Substitute Planck’s constant \(h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}\) and speed of light \(c = 3 \times 10^8\text{ m/s}\):
\[ E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}}\text{ J} \]
Convert energy from Joules to electron-volts (\(1\text{ eV} = 1.6 \times 10^{-19}\text{ J}\)):
\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9} \times 1.6 \times 10^{-19}}\text{ eV} \]
\[ E = \frac{1.989 \times 10^{-25}}{5.296 \times 10^{-26}}\text{ eV} \approx 3.755\text{ eV} \]

Step 2: Compare photon energy with work functions
The incident photon energy is \(E \approx 3.75\text{ eV}\).
Comparing \(E\) with the given work functions (\(\Phi_0\)):
For Sodium (Na): \(\Phi_0 = 1.92\text{ eV} < 3.75\text{ eV}\) \(\implies\) Photoelectric emission occurs.
For Potassium (K): \(\Phi_0 = 2.15\text{ eV} < 3.75\text{ eV}\) \(\implies\) Photoelectric emission occurs.
For Calcium (Ca): \(\Phi_0 = 3.20\text{ eV} < 3.75\text{ eV}\) \(\implies\) Photoelectric emission occurs.
For Molybdenum (Mo): \(\Phi_0 = 4.17\text{ eV} > 3.75\text{ eV}\) \(\implies\) Photoelectric emission DOES NOT occur.

Step 3: Conclusion
Among the four metals, only Molybdenum (Mo) has a work function higher than the incident photon energy. Therefore, only Mo will not show photoelectric emission, corresponding to option (B).

\ Concept:

• For any electromagnetic wave propagating in a medium, the ratio of the electric field amplitude \(E_0\) to the magnetic field amplitude \(B_0\) equals the speed of the electromagnetic wave \(v\) in that medium.
• The speed of light in a medium of refractive index \(n\) is given by \(v = \frac{c}{n}\).

Step 1: Formula for velocity of electromagnetic wave
The ratio of electric field amplitude to magnetic field amplitude is:
\[ \frac{E_0}{B_0} = v \]
The velocity \(v\) in a medium is related to the vacuum speed of light \(c\) and refractive index \(n\) by:
\[ v = \frac{c}{n} \]

Step 2: Calculation
Given \(c = 3 \times 10^8\text{ m/s}\) and refractive index of glass \(n = 1.5\):
\[ \frac{E_0}{B_0} = \frac{3 \times 10^8}{1.5} \]
\[ \frac{E_0}{B_0} = 2 \times 10^8\text{ m/s} \]

Step 3: Conclusion
The ratio of amplitude of electric field to magnetic field in glass is \(2 \times 10^8\text{ ms}^{-1}\), which corresponds to option (B).

\ Concept:

• The average value of a sinusoidal alternating voltage over one complete full cycle is strictly zero due to positive and negative half-cycles canceling each other out.
• The effective or root-mean-square (rms) value of voltage is given by \(V_{rms} = \frac{V_0}{\sqrt{2}}\), where \(V_0\) is the peak voltage amplitude.

Step 1: Determine the average voltage
Given sinusoidal voltage \(v = 14 \sin(314t)\text{ V}\).
Comparing with standard expression \(v = V_0 \sin(\omega t)\), peak voltage \(V_0 = 14\text{ V}\).
The average value of a symmetrical AC voltage over a full cycle is:
\[ V_{avg} = 0\text{ V} \]

Step 2: Calculate the effective (rms) voltage
The effective value is the rms voltage:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]
Substitute \(V_0 = 14\text{ V}\) and \(\sqrt{2} \approx 1.414\):
\[ V_{rms} = \frac{14}{\sqrt{2}} = 14 \times 0.707 \approx 9.9\text{ V} \approx 10\text{ V} \]

Step 3: Conclusion
The average value over a full cycle is \(0\text{ V}\) and effective value is \(10\text{ V}\), which corresponds to option (C).

\ Concept:

• In a p-n junction diode, the depletion region is depleted of mobile charge carriers and consists only of immobile donor and acceptor ions.
• Applying reverse bias increases the width of the depletion layer and raises the potential barrier height.

Step 1: Understanding reverse bias resistance
When a reverse bias voltage is applied across a p-n junction diode, the majority charge carriers are pulled away from the junction.
This widens the depletion layer, rendering it virtually free of mobile charge carriers.
Consequently, the resistance of the depletion region becomes extremely high (order of megaohms) compared to the low resistance of the neutral p-type and n-type bulk regions.

Step 2: Voltage distribution across regions
In a series electrical circuit, voltage drop is proportional to resistance (\(V = I R\)).
Because almost all the total resistance of a reverse-biased diode resides in the depletion region, virtually the entire external applied voltage drops across the depletion region.

Step 3: Conclusion
The applied reverse voltage drops almost entirely across the depletion region, corresponding to option (C).

Concept:

• When a charged particle moves perpendicularly through a uniform magnetic field \(\vec{B}\), magnetic Lorentz force provides the necessary centripetal force: \(q v B = \frac{m v^2}{r}\).
• The radius of the resulting circular trajectory is \(r = \frac{m v}{q B}\).

Step 1: Express radius in terms of mass-to-charge ratio
The trajectory radius is:
\[ r = \frac{m v}{q B} \]
Since velocity \(v\) and magnetic field \(B\) are identical for both particles:
\[ r \propto \frac{m}{q} \implies \frac{m}{q} = r \left(\frac{B}{v}\right) \]

Step 2: Compare radii from the figure
Observing the given circular trajectories from the figure:
The path of particle 1 has a larger radius of curvature than particle 2:
\[ r_1 > r_2 \]

Step 3: Derive inequality
Substitute \(r_1 > r_2\) into the radius relation:
\[ \frac{m_1 v}{q_1 B} > \frac{m_2 v}{q_2 B} \]
Canceling common non-zero terms \(v\) and \(B\):
\[ \frac{m_1}{q_1} > \frac{m_2}{q_2} \]
Rearranging terms by cross-multiplying \(m_2\) and \(q_1\):
\[ \frac{m_1}{m_2} > \frac{q_1}{q_2} \]

Step 4: Conclusion
The ratio of masses satisfies \(\frac{m_1}{m_2} > \frac{q_1}{q_2}\), corresponding to option (A).

Concept:

• The spectral series of hydrogen atom transitions are categorized based on lower orbit quantum number \(n_1\).
• Lyman series (\(n_1 = 1\)) lies in the Ultraviolet (UV) region.
• Balmer series (\(n_1 = 2\)) lies in the Visible region.
• Paschen (\(n_1 = 3\)), Brackett (\(n_1 = 4\)), and Pfund (\(n_1 = 5\)) series lie in the Infrared (IR) region.

Step 1: Identify transition quantum numbers
For Paschen series, electron transitions occur from outer orbits \(n_2 = 4, 5, 6, \dots\) down to lower orbit \(n_1 = 3\).

Step 2: Calculate wavelength range
Using Rydberg formula \(\frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n_2^2}\right)\):
For shortest wavelength (\(n_2 = \infty\)): \(\lambda_{min} = \frac{9}{R} \approx 820\text{ nm}\).
For longest wavelength (\(n_2 = 4\)): \(\lambda_{max} = \frac{144}{7R} \approx 1875\text{ nm}\).
Since visible spectrum ranges from \(400\text{ nm}\) to \(700\text{ nm}\), wavelengths above \(700\text{ nm}\) belong to the Infrared region.

Step 3: Conclusion
Paschen series lies entirely in the infrared region, corresponding to option (A).

\ Concept:

• The de Broglie wavelength of a particle of mass \(m\) and kinetic energy \(K\) is \(\lambda = \frac{h}{\sqrt{2mK}}\).

Step 1: Relate new wavelength to initial wavelength
Let initial kinetic energy be \(K_1 = K\) and initial wavelength be \(\lambda_1 = \frac{h}{\sqrt{2mK}}\).
New kinetic energy \(K_2 = 4K\).
New de Broglie wavelength \(\lambda_2\) is:
\[ \lambda_2 = \frac{h}{\sqrt{2m(4K)}} = \frac{h}{2\sqrt{2mK}} = \frac{\lambda_1}{2} = 0.5 \lambda_1 \]

Step 2: Calculate percentage change
Percentage change in wavelength is:
\[ \text{Percentage change} = \left( \frac{\lambda_2 - \lambda_1}{\lambda_1} \right) \times 100\% \]
\[ \text{Percentage change} = \left( \frac{0.5\lambda_1 - \lambda_1}{\lambda_1} \right) \times 100\% = -0.5 \times 100\% = -50\% \]
The negative sign indicates a decrease of \(50\%\).

Step 3: Conclusion
The de Broglie wavelength decreases by \(50\%\) of its initial value, corresponding to option (D).

\ Concept:

• Resistivity of a metallic conductor is given by \(\rho = \frac{m}{n e^2 \tau}\), where \(m\) is electron mass, \(e\) is electron charge, \(n\) is free electron concentration, and \(\tau\) is relaxation time.

Step 1: Behavior of free electron density \(n\) in metals
In metals, free electron density \(n\) is extremely high (\(\approx 10^{28}\text{ m}^{-3}\)) and determined by metallic bonding.
Raising the temperature does not appreciably alter \(n\) because thermal energy is insufficient to liberate significant additional free electrons.

Step 2: Behavior of relaxation time \(\tau\) with temperature
As temperature rises, thermal energy increases the amplitude of vibration of lattice ions.
Free electrons collide much more frequently with vibrating ions.
The average time interval between consecutive collisions (relaxation time \(\tau\)) decreases significantly (\(\tau \downarrow\)).

Step 3: Effect on resistivity
From \(\rho = \frac{m}{n e^2 \tau}\), since \(n\) remains almost constant while \(\tau\) decreases, resistivity \(\rho\) increases proportionally (\(\rho \uparrow\)).

Step 4: Conclusion
Resistivity increases because \(\tau\) decreases with temperature while \(n\) does not change appreciably, corresponding to option (C).

\ Concept:

• Mutual inductance \(M\) represents the magnetic coupling between two coils, defined by \(\Phi_2 = M I_1\).
• \(M\) is purely a geometric constant determined by physical construction and arrangement of the coils.

Step 1: Analyze dependence of mutual inductance
Mutual inductance \(M\) depends on:
1. Number of turns \(N_1\) and \(N_2\) in the two coils.
2. Cross-sectional areas \(A_1\) and \(A_2\) and lengths of the coils.
3. Relative distance/separation between coils.
4. Relative spatial orientation and alignment of their axes (coupling coefficient \(K\)).
5. Magnetic permeability \(\mu\) of core material inside coils.

Step 2: Eliminate incorrect options
\(M\) is independent of current \(I\), rate of change of current \(\frac{dI}{dt}\), or induced electromotive force \(e\).
These quantities determine induced voltage (\(e = -M \frac{dI}{dt}\)), but do not affect the fundamental geometric constant \(M\) itself.

Step 3: Conclusion
Mutual inductance depends on relative position and orientation of the coils, corresponding to option (B).

\ Concept:

• Mobility \(\mu\) of a charge carrier is defined as the magnitude of drift velocity per unit applied electric field: \(\mu = \frac{v_d}{E}\).
• In terms of relaxation time \(\tau\) and carrier mass \(m\): \(\mu = \frac{e \tau}{m}\).

Step 1: Analyze temperature dependence of mobility
As temperature increases, thermal vibrations of lattice ions intensify.
Collision frequency of electrons increases, leading to a decrease in average relaxation time \(\tau\).
From \(\mu = \frac{e \tau}{m}\), a decrease in relaxation time \(\tau\) causes mobility \(\mu\) to decrease.

Step 2: Evaluate other statements
- Mobility is independent of applied electric field \(E\) in low-field regime (\(\mu = \text{constant}\)).
- Mobility depends inversely on carrier mass \(m\) (\(\mu \propto \frac{1}{m}\)).
- Mobility decreases (not increases) with temperature rise in metals.

Step 3: Conclusion
The statement "Mobility decreases with increase in temperature" is true, corresponding to option (B).

\ Concept:

• A galvanometer is converted into a voltmeter of a desired voltage range by connecting a high resistance in series with the galvanometer coil. \
• The total potential difference measured by the voltmeter is the product of full-scale deflection current \(I_g\) and the total series resistance \((G + R)\). \
• The fundamental relation governing the voltmeter conversion is given by \(V = I_g(G + R)\).

Step 1: Formulate the equation for the first case
In the initial configuration, the series resistor is \(R_1 = 250\ \Omega\) and the maximum measurable voltage range is \(V_1 = V\).
Let \(I_g\) be the current required for full-scale deflection of the galvanometer.
Applying Ohm’s law across the series combination:
\[ V = I_g(G + 250) \quad \text{--- (Equation 1)} \]

Step 2: Formulate the equation for the second case
In the modified configuration, the series resistor is replaced by \(R_2 = 900\ \Omega\), and the voltage range increases to \(V_2 = 3V\).
Applying the same relation for the new range:
\[ 3V = I_g(G + 900) \quad \text{--- (Equation 2)} \]

Step 3: Solve the two equations simultaneously
Divide Equation 2 by Equation 1 to eliminate the full-scale current \(I_g\) and voltage \(V\):
\[ \frac{3V}{V} = \frac{I_g(G + 900)}{I_g(G + 250)} \]
Simplify the ratio:
\[ 3 = \frac{G + 900}{G + 250} \]
Cross-multiply to solve for the unknown galvanometer resistance \(G\):
\[ 3(G + 250) = G + 900 \]
\[ 3G + 750 = G + 900 \]
Rearrange the terms containing \(G\) to one side:
\[ 3G - G = 900 - 750 \]
\[ 2G = 150 \]
\[ G = \frac{150}{2} = 75\ \Omega \]

Step 4: Conclusion
The internal resistance of the galvanometer coil is \(75\ \Omega\), which matches option (D).

\ Concept:

• According to Einstein’s photoelectric equation, the maximum kinetic energy of photoelectrons emitted from a metal surface is given by \(K_{\text{max}} = h\nu - \Phi_0\), where \(\Phi_0 = h\nu_0\) is the work function of the metal.
• The threshold frequency \(\nu_0\) represents the minimum frequency of incident radiation required to eject electrons from the metal surface.
• If the incident photon frequency \(\nu > \nu_0\), the excess energy appears as the maximum kinetic energy of the emitted photoelectrons.

Step 1: Calculate the maximum kinetic energy for metal A
For metal surface A, the threshold frequency is given as \(\nu_{0A} = \frac{\nu}{2}\).
Applying Einstein’s photoelectric equation:
\[ K_A = h\nu - h\nu_{0A} \]
Substitute \(\nu_{0A} = \frac{\nu}{2}\):
\[ K_A = h\nu - h\left(\frac{\nu}{2}\right) = h\nu\left(1 - \frac{1}{2}\right) = \frac{1}{2}h\nu \]

Step 2: Calculate the maximum kinetic energy for metal B
For metal surface B, the threshold frequency is given as \(\nu_{0B} = \frac{\nu}{3}\).
Applying Einstein’s photoelectric equation:
\[ K_B = h\nu - h\nu_{0B} \]
Substitute \(\nu_{0B} = \frac{\nu}{3}\):
\[ K_B = h\nu - h\left(\frac{\nu}{3}\right) = h\nu\left(1 - \frac{1}{3}\right) = \frac{2}{3}h\nu \]

Step 3: Determine the ratio of maximum kinetic energies
Taking the ratio of \(K_A\) to \(K_B\):
\[ \frac{K_A}{K_B} = \frac{\frac{1}{2}h\nu}{\frac{2}{3}h\nu} \]
Cancel the common factor \(h\nu\):
\[ \frac{K_A}{K_B} = \frac{1/2}{2/3} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} \]

Step 4: Conclusion
The ratio of the maximum kinetic energy of photoelectrons emitted from metal A to that from metal B is \(3 : 4\), which corresponds to option (B).

\ Concept:

• According to the Lens Maker’s Formula, the focal length \(f\) of a thin spherical lens is given by \(\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).
• The converging power of a lens is inversely proportional to its focal length, expressed as \(P = \frac{1}{f}\).
• The relative refractive index depends on both the material of the lens and the surrounding medium.

Step 1: Analyze the Assertion (A)
When a convex lens made of glass (\(\mu_g \approx 1.5\)) is placed in air (\(\mu_a = 1\)), the relative refractive index is \(\frac{\mu_g}{\mu_a} = 1.5\).
When the same lens is immersed in water (\(\mu_w \approx 1.33\)), the relative refractive index becomes:
\[ \mu_{\text{rel}} = \frac{\mu_g}{\mu_w} = \frac{1.5}{1.33} \approx 1.125 \]
Since \(\mu_{\text{rel}} - 1\) decreases significantly in water, the term \(\frac{1}{f}\) decreases, which means the focal length \(f\) increases (\(f_w \approx 4 f_a\)).
Since optical power \(P = \frac{1}{f}\), the converging power of the lens decreases when immersed in water.
Therefore, Assertion (A) is false.

Step 2: Analyze the Reason (R)
From the Lens Maker’s formula, the focal length of a lens depends on:
1. The refractive index of the material of the lens (\(\mu_{\text{lens}}\)).
2. The refractive index of the surrounding medium (\(\mu_{\text{medium}}\)).
3. The radii of curvature of its refracting faces (\(R_1\) and \(R_2\)).
4. The wavelength/color of incident light.
Thus, focal length does not depend only on the radii of curvature of the faces.
Therefore, Reason (R) is false.

Step 3: Conclusion
Since both Assertion (A) and Reason (R) are false statements, option (D) is the correct choice.

\ Concept:

• Electrical conductivity of an extrinsic semiconductor is determined by the concentration and mobility of charge carriers, given by \(\sigma = e(n_e \mu_e + n_h \mu_h)\).
• In an n-type semiconductor, majority carriers are electrons in the conduction band, while in a p-type semiconductor, majority carriers are holes in the valence band.
• Mobility refers to the magnitude of the drift velocity acquired by a charge carrier per unit applied electric field (\(\mu = \frac{v_d}{E}\)).

Step 1: Analyze the Assertion (A)
For semiconductors with identical doping levels (\(N_d \approx N_a\)), the majority carrier density in n-type semiconductor (\(n_e\)) is equal to that in p-type semiconductor (\(n_h\)).
The conductivity of an n-type semiconductor is dominated by electrons: \(\sigma_n \approx e n_e \mu_e = e N_d \mu_e\).
The conductivity of a p-type semiconductor is dominated by holes: \(\sigma_p \approx e n_h \mu_h = e N_a \mu_h\).
Because electron mobility \(\mu_e\) is substantially higher than hole mobility \(\mu_h\), the resulting conductivity \(\sigma_n\) of n-type semiconductor is greater than \(\sigma_p\) of p-type semiconductor.
Therefore, Assertion (A) is true.

Step 2: Analyze the Reason (R)
Electrons move freely in the conduction band where available energy states are largely vacant and inter-atomic binding is negligible.
Holes move in the valence band through a process of bound electrons jumping into adjacent vacancies, encountering significantly more resistance and scattering from the crystal lattice.
Consequently, the mobility of electrons (\(\mu_e\)) in the conduction band is inherently greater than the mobility of holes (\(\mu_h\)) in the valence band.
Therefore, Reason (R) is true and correctly explains why n-type semiconductors have higher conductivity.

Step 3: Conclusion
Both Assertion (A) and Reason (R) are correct, and Reason (R) provides the accurate physical justification for Assertion (A). Hence, option (A) is the correct choice.

\ Concept:

• In electrostatics and steady-state circuit analysis, the electrostatic field is a conservative vector field.
• The line integral of the electrostatic field along any closed loop is identically zero: \(\oint \vec{E} \cdot d\vec{l} = 0\).
• Kirchhoff’s Voltage Law (Loop Rule) is a direct manifestation of the principle of conservation of energy in an electrical circuit.

Step 1: Analyze the Assertion (A)
Kirchhoff’s loop rule states that the algebraic sum of changes in electric potential around any closed circuit loop containing resistors and voltage sources is zero:
\[ \sum \Delta V = 0 \]
The work done in moving a test charge \(q\) between two points with potential difference \(\Delta V\) is given by \(W = q\Delta V\).
For a unit charge (\(q = 1\text{ C}\)) completing a full traversal around a closed loop, the total change in potential is \(\Delta V_{\text{net}} = 0\).
Therefore, the total work done \(W = 1 \times 0 = 0\).
Thus, Assertion (A) is true.

Step 2: Analyze the Reason (R)
Electric potential \(V(\vec{r})\) is a single-valued scalar function of space (state function) determined entirely by the location of the point in the circuit.
When a charge traverses a complete closed path and returns to its initial starting location, the final electric potential is identical to the initial electric potential (\(V_f = V_i\)).
The net potential change is \(\Delta V = V_f - V_i = 0\), leading to zero net work done.
Therefore, Reason (R) is true and accurately explains why the net work done around any closed loop is zero.

Step 3: Conclusion
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). Hence, option (A) is the correct choice.

\ Concept:

• Ferromagnetic materials possess domain structures where atomic magnetic dipole moments are strongly aligned in parallel due to exchange coupling.
• As temperature rises, increased thermal agitation disrupts the spontaneous alignment of atomic dipoles within the domains.
• At temperatures exceeding the Curie temperature (\(T > T_C\)), the domain structure completely breaks down, and the substance transforms into a paramagnetic state. \

Step 1: Analyze the Assertion (A)
When a ferromagnetic material (such as iron, cobalt, or nickel) is heated above its characteristic Curie temperature \(T_C\), the thermal agitation overcomes the quantum exchange forces holding the magnetic domains together.
The domain structure is destroyed, and the substance exhibits paramagnetic behavior following the Curie-Weiss law: \(\chi = \frac{C}{T - T_C}\) for \(T > T_C\).
Therefore, Assertion (A) is true.

Step 2: Analyze the Reason (R)
The loss of spontaneous magnetization in a ferromagnetic material as temperature approaches \(T_C\) is a continuous (second-order) phase transition.
The spontaneous magnetization decreases gradually and continuously as temperature increases, reaching zero smoothly at \(T = T_C\) according to the power law \(M_s(T) \propto (T_C - T)^\beta\), where \(\beta\) is a critical exponent.
The disappearance of magnetization is therefore gradual over the temperature range and not an abrupt step discontinuity.
Therefore, Reason (R) is false.

Step 3: Conclusion
Assertion (A) is a true statement, whereas Reason (R) is false. Thus, option (C) is the correct choice.

\ Concept:

• In an intrinsic semiconductor, electron concentration equals hole concentration: \(n_e = n_h = n_i\).
• According to Mass Action Law for semiconductors in thermal equilibrium, \(n_e \cdot n_h = n_i^2\), regardless of doping level.

Step 1: Given parameters
Intrinsic carrier concentration \(n_i = 5 \times 10^8\text{ m}^{-3}\).
New electron concentration after doping \(n_e = 4 \times 10^{12}\text{ m}^{-3}\).

Step 2: Calculate new hole concentration
Using mass action law:
\[ n_e \cdot n_h = n_i^2 \]
\[ n_h = \frac{n_i^2}{n_e} \]
Substitute given values:
\[ n_h = \frac{(5 \times 10^8)^2}{4 \times 10^{12}} \]
\[ n_h = \frac{25 \times 10^{16}}{4 \times 10^{12}} \]
\[ n_h = 6.25 \times 10^4\text{ m}^{-3} \]

Step 3: Identify the type of semiconductor
Comparing electron concentration \(n_e\) and hole concentration \(n_h\):
\[ n_e = 4 \times 10^{12}\text{ m}^{-3} \quad \text{and} \quad n_h = 6.25 \times 10^4\text{ m}^{-3} \]
Since \(n_e \gg n_h\), electrons are majority charge carriers and holes are minority carriers.
Therefore, the doped semiconductor is an n-type semiconductor.

Step 4: Conclusion
The new hole concentration is \(6.25 \times 10^4\text{ m}^{-3}\) and the resulting doped material is an n-type semiconductor.

Concept:

• Electric dipole moment is \(p = q \times (2a)\).
• Potential energy of a dipole in a uniform electric field \(\vec{E}\) at angle \(\theta\) is \(U(\theta) = -p E \cos \theta\).
• Work done in rotating a dipole from angle \(\theta_1\) to \(\theta_2\) is \(W = U(\theta_2) - U(\theta_1) = -p E (\cos \theta_2 - \cos \theta_1)\).
• Stable equilibrium corresponds to \(\theta_1 = 0^\circ\) (\(\vec{p}\) parallel to \(\vec{E}\)).
• Unstable equilibrium corresponds to \(\theta_2 = 180^\circ\) (\(\vec{p}\) antiparallel to \(\vec{E}\)).

Step 1: Calculate electric dipole moment
Given charge magnitude \(q = 1\ \mu\text{C} = 1 \times 10^{-6}\text{ C}\).
Separation distance \(2a = 10\text{ cm} = 0.1\text{ m}\).
Electric field intensity \(E = 100\text{ N/C}\).
Dipole moment \(p\):
\[ p = q \times (2a) = (1 \times 10^{-6}\text{ C}) \times (0.1\text{ m}) = 10^{-7}\text{ C}\cdot\text{m} \]

Step 2: Formulate work done equation
For stable equilibrium position: \(\theta_1 = 0^\circ\).
For unstable equilibrium position: \(\theta_2 = 180^\circ\).
Work done \(W\):
\[ W = -p E (\cos \theta_2 - \cos \theta_1) \]
\[ W = -p E (\cos 180^\circ - \cos 0^\circ) \]
Substitute \(\cos 180^\circ = -1\) and \(\cos 0^\circ = 1\):
\[ W = -p E (-1 - 1) = 2 p E \]

Step 3: Substitute values and compute result
\[ W = 2 \times (10^{-7}\text{ C}\cdot\text{m}) \times (100\text{ N/C}) \]
\[ W = 2 \times 10^{-5}\text{ J} \]

Step 4: Conclusion
The work done in turning the dipole from stable to unstable equilibrium is \(2 \times 10^{-5}\text{ J}\).

Concept:

• Huygens’ Principle provides a geometrical construction to determine the position and shape of a wave front at a later instant from its known position at an earlier instant.

Step 1: Statement of Huygens’ Principle
Huygens’ Principle rests on two fundamental postulates:
1. Primary Wavefront as Secondary Sources: Every point on a given primary wavefront acts as a fresh source of secondary disturbance, emitting tiny spherical wavelets called secondary wavelets that spread out in all directions with the speed of light in that medium.
2. New Wavefront Construction: The tangential envelope or forward surface touching all these secondary wavelets in the forward direction at any subsequent time \(t\) gives the new position and shape of the wavefront at that instant.

Step 2: Justification for Absence of Backwave
According to Huygens’ original model, secondary wavelets radiate in all directions. This mathematically suggests the presence of a backward wavefront (backwave) traveling back towards the source.
To justify why backwaves do not physically exist, Huygens assumed that the amplitude/intensity of secondary wavelets is directional.
The amplitude of a secondary wavelet at an angle \(\theta\) with respect to the forward normal of the primary wavefront is proportional to the directivity factor (or obliquity factor):
\[ F(\theta) = \frac{1}{2}(1 + \cos \theta) \]

Step 3: Evaluating directivity factor
- For the forward direction (\(\theta = 0^\circ\)): \(F(0^\circ) = \frac{1}{2}(1 + \cos 0^\circ) = \frac{1}{2}(1 + 1) = 1\) (Maximum amplitude).
- For the backward direction (\(\theta = 180^\circ\)): \(F(180^\circ) = \frac{1}{2}(1 + \cos 180^\circ) = \frac{1}{2}(1 - 1) = 0\) (Zero amplitude).

Step 4: Conclusion
Since the obliquity factor \(\frac{1}{2}(1 + \cos \theta)\) vanishes completely at \(\theta = 180^\circ\), no secondary wavelets propagate backwards, accounting for the absence of backwave.

Concept:

• Single-slit diffraction produces a central bright maximum flanked by secondary minima and secondary maxima of rapidly decreasing intensity.
• Minima occur at angular positions \(\sin \theta = \pm \frac{n \lambda}{a} \approx \pm \frac{n \lambda}{a}\) for integer \(n = 1, 2, 3, \dots\).
• Secondary maxima occur approximately at \(\sin \theta \approx \pm \left(n + \frac{1}{2}\right) \frac{\lambda}{a}\).

Step 1: Key features of diffraction intensity curve
1. Central Maximum: Positioned at \(\theta = 0\) with maximum intensity \(I_0\). Its angular width is \(2\theta_0 = \frac{2\lambda}{a}\).
2. First Minima: Located at \(\theta = \pm \frac{\lambda}{a}\) with zero intensity (\(I = 0\)).
3. Secondary Maxima: Located near \(\theta = \pm \frac{3\lambda}{2a}, \pm \frac{5\lambda}{2a}, \dots\)
Intensity drops rapidly: \(I_1 \approx \frac{I_0}{22}\) (about \(4.5\%\) of \(I_0\)), \(I_2 \approx \frac{I_0}{61}\) (about \(1.6\%\) of \(I_0\)).

Step 2: Diagram depiction
The graph plots Intensity \(I\) along the vertical axis against angular position \(\theta\) (or path parameter \(\beta = \frac{\pi a \sin \theta}{\lambda}\)) along the horizontal axis.

Step 3: Conclusion
The intensity pattern consists of a central peak of maximum height \(I_0\) flanked symmetrically by secondary peaks whose heights decrease rapidly on either side.

Concept:

• The angular half-width of central maximum is \(\theta = \frac{\lambda}{a}\).
• The linear width \(\beta_0\) of the central maximum on a screen placed at distance \(D\) is \(\beta_0 = 2 y_1 = \frac{2 \lambda D}{a}\).

Step 1: Formula for linear width
The distance of the first minimum from the center of the screen is \(y_1 = \frac{\lambda D}{a}\).
The total linear width \(\beta_0\) of the central maximum extends between first minima on either side:
\[ \beta_0 = 2 y_1 = \frac{2 \lambda D}{a} \]

Step 2: Analyze effect of decreasing D
From the formula \(\beta_0 = \frac{2 \lambda D}{a}\), linear width \(\beta_0\) is directly proportional to slit-to-screen distance \(D\) (\(\beta_0 \propto D\)).
When the separation \(D\) between the slit and the screen is decreased, the linear width \(\beta_0\) of the central maximum decreases proportionally.

Step 3: Conclusion
Decreasing the distance \(D\) causes the central maximum on the screen to become narrower linearly. Note that angular width (\(2\theta = \frac{2\lambda}{a}\)) remains unchanged as it depends only on wavelength \(\lambda\) and slit width \(a\).

Concept:

• Lens Maker’s Formula in air: \(\frac{1}{f_a} = (\mu_g - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).
• Lens Maker’s Formula in a liquid of refractive index \(\mu_l\): \(\frac{1}{f_l} = \left(\frac{\mu_g}{\mu_l} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).

Step 1: Express curvature factor using air focal length
Given focal length in air \(f_a = +20\text{ cm}\), refractive index of glass lens \(\mu_g = 1.5\).
Using Lens Maker’s formula in air:
\[ \frac{1}{f_a} = (\mu_g - 1) K \]
where \(K = \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\) is the geometric curvature factor.
\[ \frac{1}{20} = (1.5 - 1) K = 0.5 K \]
\[ K = \frac{1}{20 \times 0.5} = \frac{1}{10}\text{ cm}^{-1} \]

Step 2: Calculate focal length in liquid
Given refractive index of liquid \(\mu_l = 1.25\).
Relative refractive index of lens with respect to liquid is \(\mu_{rel} = \frac{\mu_g}{\mu_l} = \frac{1.5}{1.25} = 1.2\).
Using Lens Maker’s formula in liquid:
\[ \frac{1}{f_l} = (\mu_{rel} - 1) K = (1.2 - 1) K \]
\[ \frac{1}{f_l} = 0.2 K \]
Substitute \(K = \frac{1}{10}\):
\[ \frac{1}{f_l} = 0.2 \times \frac{1}{10} = \frac{0.2}{10} = \frac{1}{50}\text{ cm}^{-1} \]
\[ f_l = +50\text{ cm} \]

Step 3: Determine nature of lens in liquid
Since \(f_l = +50\text{ cm}\) remains positive (\(\mu_g > \mu_l\)), the lens retains its original focal sign.
Hence, it continues to act as a converging (convex) lens.

Step 4: Conclusion
The focal length of the lens in liquid is \(+50\text{ cm}\) and its nature remains converging.

Concept:

• Nuclear fission and nuclear fusion are nuclear reactions that release massive energy by transforming nuclei into configurations with higher binding energy per nucleon (\(E_b/A\)).

Step 1: Tabulate key differences

• Definition:
  - Fission: Process in which a heavy unstable nucleus splits into two lighter medium-mass daughter nuclei upon absorbing a thermal neutron.
  - Fusion: Process in which two light nuclei combine together at extremely high temperature and pressure to form a heavier, more stable nucleus.
• Temperature Requirement:
  - Fission: Occurs readily at room temperature with slow (thermal) neutrons.
  - Fusion: Requires extremely high temperatures (\(\approx 10^7 - 10^8\text{ K}\)) to overcome strong electrostatic Coulomb repulsion between positively charged nuclei.
• Energy per unit mass:
  - Fission: Energy released per unit mass is relatively lower (\(\approx 0.85\text{ MeV/nucleon}\)).
  - Fusion: Energy released per unit mass is significantly higher (\(\approx 6.75\text{ MeV/nucleon}\)).
• Radioactive Waste:
  - Fission: Produces highly radioactive fission fragments and long-lived toxic nuclear waste.
  - Fusion: Produces mostly non-radioactive stable products (like Helium), generating negligible radioactive waste.

Step 2: Examples
1. Nuclear Fission Example:
Neutron-induced fission of Uranium-235:
\[ \text{}^{235}_{92}\text{U} + \text{}^{1}_{0}\text{n} \longrightarrow \text{}^{141}_{56}\text{Ba} + \text{}^{92}_{36}\text{Kr} + 3 \text{}^{1}_{0}\text{n} + Q\text{ (\approx 200 MeV)} \]

2. Nuclear Fusion Example:
Proton-Proton cycle or Deuterium-Tritium fusion reaction:
\[ \text{}^{2}_{1}\text{H} + \text{}^{3}_{1}\text{H} \longrightarrow \text{}^{4}_{2}\text{He} + \text{}^{1}_{0}\text{n} + Q\text{ (17.6 MeV)} \]

Step 3: Conclusion
Fission involves splitting heavy nuclei at normal temperatures, whereas fusion involves combining light nuclei at ultra-high thermonuclear temperatures.

Concept:

• A full-wave rectifier converts both positive and negative half-cycles of an alternating input voltage into unidirectional direct current (DC).
• It utilizes a center-tapped transformer and two p-n junction diodes operating in alternate conduction modes during opposite half-cycles.

Step 1: Circuit Diagram Construction
The circuit consists of a center-tapped secondary winding of a transformer, two p-n junction diodes (\(D_1\) and \(D_2\)), and a load resistor \(R_L\) connected between the junction of the diodes’ cathodes and the central tap of the transformer secondary.

Step 2: Working During Positive Half-Cycle
During the positive half-cycle of input AC voltage:
Terminal \(A\) of the secondary winding becomes positive with respect to the center-tap \(C\), while terminal \(B\) becomes negative.
Diode \(D_1\) becomes forward-biased and conducts current.
Diode \(D_2\) becomes reverse-biased and remains non-conducting.
Current flows through load resistor \(R_L\) in the direction from \(X\) to \(Y\).

Step 3: Working During Negative Half-Cycle
During the negative half-cycle of input AC voltage:
Terminal \(A\) becomes negative with respect to center-tap \(C\), while terminal \(B\) becomes positive.
Diode \(D_1\) becomes reverse-biased and ceases conduction.
Diode \(D_2\) becomes forward-biased and conducts current.
Current again flows through load resistor \(R_L\) in the exact same direction from \(X\) to \(Y\).

Step 4: Input and Output Waveforms
Since current flows through load \(R_L\) in the same direction during both half-cycles, a continuous pulsating DC voltage is obtained across \(R_L\).

Step 5: Conclusion
A full-wave rectifier conducts during both half-cycles of input AC, producing a pulsating DC output with twice the frequency of the input AC supply (\(f_{out} = 2 f_{in}\)).

Concept:

• De Broglie postulated that moving material particles exhibit wave-like properties under appropriate conditions, establishing the wave-particle duality of matter.
• The de Broglie relation connects particle properties (momentum \(p\) or mass \(m\) and velocity \(v\)) with wave properties (wavelength \(\lambda\)).

Step 1: Analyze the de Broglie Equation
The de Broglie wavelength \(\lambda\) of a moving matter particle is expressed as:
\[ \lambda = \frac{h}{p} = \frac{h}{m v} \]
where \(h\) is Planck’s universal constant.

Step 2: Demonstrate Dual Aspect in the Formula
- The left-hand side contains \(\lambda\) (wavelength), which is a fundamental characteristic property of a wave.
- The right-hand side contains \(p = mv\) (momentum), which is a fundamental characteristic property of a localized particle.
- Planck’s constant \(h\) acts as the bridge uniting these two seemingly contradictory aspects of nature into a single equation.

Step 3: Physical Implications
1. For macroscopic heavy objects (\(m\) is large), \(\lambda\) is extremely small (undetectably small), making particle nature predominant.
2. For microscopic subatomic particles like electrons (\(m\) is tiny), \(\lambda\) becomes comparable to interatomic spacings, making wave properties like diffraction directly observable (as verified in the Davisson-Germer experiment).

Step 4: Conclusion
The de Broglie relation explicitly demonstrates matter duality by showing that every moving particle possesses an associated wave whose wavelength is inversely proportional to its momentum.

Concept:

• Energy of an incident photon of wavelength \(\lambda\) is \(E = \frac{hc}{\lambda}\).
• Einstein’s photoelectric equation states \(K_{max} = E - \Phi_0\). For negligible work function (\(\Phi_0 \approx 0\)), maximum kinetic energy of emitted electron is \(K_{max} = E = \frac{hc}{\lambda}\).
• The de Broglie wavelength of an electron of mass \(m_e\) and kinetic energy \(K\) is \(\lambda_{dB} = \frac{h}{\sqrt{2 m_e K}}\).

Step 1: Determine maximum kinetic energy of photoelectrons
Given incident radiation wavelength = \(\lambda\).
Work function of photosensitive surface \(\Phi_0 \approx 0\).
Maximum kinetic energy \(K\) acquired by emitted electrons:
\[ K = \frac{hc}{\lambda} \]

Step 2: Express de Broglie wavelength in terms of kinetic energy
The de Broglie wavelength \(\lambda_{dB}\) of the photoelectrons is:
\[ \lambda_{dB} = \frac{h}{\sqrt{2 m_e K}} \]

Step 3: Substitute kinetic energy into the de Broglie relation
Substitute \(K = \frac{hc}{\lambda}\):
\[ \lambda_{dB} = \frac{h}{\sqrt{2 m_e \left(\frac{hc}{\lambda}\right)}} \]
\[ \lambda_{dB} = \frac{h}{\sqrt{\frac{2 m_e h c}{\lambda}}} = \sqrt{\frac{h^2 \lambda}{2 m_e h c}} \]
Simplifying \(h\) inside the square root:
\[ \lambda_{dB} = \sqrt{\frac{h \lambda}{2 m_e c}} \]

Step 4: Conclusion
The de Broglie wavelength of the most energetic photoelectrons emitted from the surface is \(\lambda_{dB} = \sqrt{\frac{h \lambda}{2 m_e c}}\).

Concept:

• Magnetic field at distance \(r\) from an infinitely long straight wire carrying current \(I\) is \(B = \frac{\mu_0 I}{2\pi r}\).
• Direction of magnetic field is determined by Right Hand Thumb Rule.
• Net magnetic field is the vector sum of individual magnetic fields produced by surrounding conductors.

Step 1: Determine magnetic field due to Conductor 2
Conductor 2 carries current \(I\) in the \(+x\) direction at distance \(d\) below conductor 1.
By Right Hand Thumb Rule, magnetic field \(\vec{B}_2\) at conductor 1 points out of the page (\(+z\) direction, along \(+\hat{k}\)):
\[ \vec{B}_2 = \frac{\mu_0 I}{2\pi d} \hat{k} \]

Step 2: Determine magnetic field due to Conductor 3
Conductor 3 carries current \(3I\) in the \(-x\) direction at distance \(2d\) below conductor 1.
By Right Hand Thumb Rule, magnetic field \(\vec{B}_3\) at conductor 1 points into the page (\(-z\) direction, along \(-\hat{k}\)):
\[ \vec{B}_3 = \frac{\mu_0 (3I)}{2\pi (2d)} (-\hat{k}) = -\frac{3 \mu_0 I}{4\pi d} \hat{k} \]

Step 3: Calculate net magnetic field at conductor 1
\[ \vec{B}_{net} = \vec{B}_2 + \vec{B}_3 \]
\[ \vec{B}_{net} = \frac{\mu_0 I}{2\pi d} \hat{k} - \frac{3 \mu_0 I}{4\pi d} \hat{k} \]
Take common denominator \(4\pi d\):
\[ \vec{B}_{net} = \left( \frac{2 \mu_0 I - 3 \mu_0 I}{4\pi d} \right) \hat{k} = -\frac{\mu_0 I}{4\pi d} \hat{k} \]

Step 4: Conclusion
Magnitude of net magnetic field at conductor 1 is \(B_{net} = \frac{\mu_0 I}{4\pi d}\), directed into the plane of the paper (along the negative z-axis, \(-\hat{k}\)).

Concept:

• Force per unit length between two parallel long straight currents \(I_1\) and \(I_2\) separated by distance \(r\) is \(f = \frac{\mu_0 I_1 I_2}{2\pi r}\).
• Parallel currents in the same direction attract each other; antiparallel currents in opposite directions repel each other.

Step 1: Force per unit length due to Conductor 2
Current in conductor 1 is \(2I\) along \(+x\). Current in conductor 2 is \(I\) along \(+x\).
Since currents flow in the same direction, force is attractive (towards conductor 2, i.e., downwards along \(-\hat{j}\)):
\[ \vec{f}_{12} = \frac{\mu_0 (2I)(I)}{2\pi d} (-\hat{j}) = -\frac{\mu_0 I^2}{\pi d} \hat{j} \]

Step 2: Force per unit length due to Conductor 3
Current in conductor 1 is \(2I\) along \(+x\). Current in conductor 3 is \(3I\) along \(-x\).
Since currents flow in opposite directions, force is repulsive (away from conductor 3, i.e., upwards along \(+\hat{j}\)):
\[ \vec{f}_{13} = \frac{\mu_0 (2I)(3I)}{2\pi (2d)} (+\hat{j}) = +\frac{6 \mu_0 I^2}{4\pi d} \hat{j} = +\frac{3 \mu_0 I^2}{2\pi d} \hat{j} \]

Step 3: Calculate net force per unit length on conductor 1
\[ \vec{f}_{net} = \vec{f}_{12} + \vec{f}_{13} \]
\[ \vec{f}_{net} = \left( \frac{3 \mu_0 I^2}{2\pi d} - \frac{\mu_0 I^2}{\pi d} \right) \hat{j} \]
Take common denominator \(2\pi d\):
\[ \vec{f}_{net} = \left( \frac{3 \mu_0 I^2 - 2 \mu_0 I^2}{2\pi d} \right) \hat{j} = +\frac{\mu_0 I^2}{2\pi d} \hat{j} \]

Step 4: Conclusion
Magnitude of net magnetic force per unit length on conductor 1 is \(f_{net} = \frac{\mu_0 I^2}{2\pi d}\), directed upwards along the positive y-axis (\(+\hat{j}\)).

Concept:

• According to Faraday’s Law of Electromagnetic Induction, induced electromotive force (emf) is equal to the negative rate of change of magnetic flux: \(e = -\frac{d\Phi}{dt}\).
• Magnetic flux linked with a loop of area \(A\) in a uniform magnetic field \(B\) perpendicular to loop plane is \(\Phi = B A\).

Step 1: Calculate magnetic flux linked with the loop
Area of rectangular loop \(A = l \times b\).
Time-varying magnetic field \(B(t) = B_0 \sin \omega t\).
Assuming magnetic field is perpendicular to the plane of the loop (\(\theta = 0^\circ\)):
\[ \Phi(t) = B(t) \cdot A = (B_0 \sin \omega t) (l b) = l b B_0 \sin \omega t \]

Step 2: Differentiate flux with respect to time
Apply Faraday’s law of induction:
\[ e = -\frac{d\Phi}{dt} = -\frac{d}{dt} \left( l b B_0 \sin \omega t \right) \]
Since \(l, b, B_0, \omega\) are constants:
\[ e = -l b B_0 \frac{d}{dt} (\sin \omega t) \]
\[ e = -l b B_0 \omega \cos \omega t \]

Step 3: Conclusion
The instantaneous induced emf in the rectangular loop is \(e(t) = -l b B_0 \omega \cos \omega t = e_0 \cos(\omega t + \pi)\), where peak emf \(e_0 = l b B_0 \omega\).

Concept:

• Instantaneous induced current is \(i(t) = \frac{e(t)}{R}\).
• Effective or root-mean-square (rms) current for a sinusoidal current \(i(t) = I_0 \cos \omega t\) is \(I_{rms} = \frac{I_0}{\sqrt{2}}\), where \(I_0\) is peak current amplitude.

Step 1: Determine instantaneous current
From Part (i), induced emf is \(e(t) = -l b B_0 \omega \cos \omega t\).
Resistance of the loop = \(R\).
Using Ohm’s law, instantaneous induced current \(i(t)\):
\[ i(t) = \frac{e(t)}{R} = -\frac{l b B_0 \omega}{R} \cos \omega t \]

Step 2: Identify peak current amplitude
The peak current amplitude \(I_0\) is:
\[ I_0 = \frac{l b B_0 \omega}{R} \]

Step 3: Calculate effective (rms) current
The effective (rms) current value \(I_{eff}\) is:
\[ I_{eff} = I_{rms} = \frac{I_0}{\sqrt{2}} \]
Substitute \(I_0 = \frac{l b B_0 \omega}{R}\):
\[ I_{eff} = \frac{l b B_0 \omega}{\sqrt{2} R} \]

Step 4: Conclusion
The effective value of current flowing through the loop is \(I_{eff} = \frac{l b B_0 \omega}{\sqrt{2} R}\).

Concept:

• X-rays are high-energy electromagnetic radiations with very short wavelengths ranging from approximately \(0.01\text{ nm}\) to \(10\text{ nm}\) (\(10^{-11}\text{ m}\) to \(10^{-8}\text{ m}\)).

Step 1: Definition of X-rays
X-rays are energetic, invisible electromagnetic waves located in the spectrum between ultraviolet rays and gamma rays, characterized by high frequencies (\(\approx 10^{16}\text{ Hz}\) to \(10^{19}\text{ Hz}\)) and strong penetrating power through matter.

Step 2: Production Mechanism
X-rays are produced in a Coolidge tube when high-speed energetic electrons emitted from a heated filament are accelerated through a high potential difference (\(\approx 10\text{ kV}\) to \(100\text{ kV}\)) and suddenly decelerated upon striking a heavy metal target (like Tungsten or Molybdenum) of high atomic number and high melting point.
The loss of kinetic energy of fast-moving electrons during sudden deceleration produces continuous X-rays (Bremsstrahlung or braking radiation), while inner-shell atomic transitions in the target produce characteristic X-rays.

Step 3: Two Uses of X-rays
1. Medical Diagnostics and Therapy: Used in radiography (X-ray imaging) to detect bone fractures, foreign metallic objects, dental cavities, and in radiation therapy to destroy malignant cancer tumors.
2. Industrial and Scientific Applications: Used in crystallography for analyzing crystal structures via X-ray diffraction, and in industrial quality control to inspect internal defects, cracks, and welds in metal castings.

Step 4: Conclusion
X-rays are short-wavelength electromagnetic waves generated by decelerating fast electrons against a metallic target, widely applied in medical imaging and structural crystal analysis.

Concept:

• Nuclides are classified based on atomic number \(Z\) (proton count), neutron number \(N\), and mass number \(A = Z + N\).

Step 1: Define Isotopes
Isotopes are nuclides belonging to the same chemical element that have the same atomic number (\(Z\)) but different mass numbers (\(A\)).
- They contain identical numbers of protons but different numbers of neutrons (\(N = A - Z\)).
- They possess identical chemical properties because chemical behavior is dictated by atomic number \(Z\).
- Example: Isotopes of Hydrogen: Protium (\(\text{}^{1}_{1}\text{H}\)), Deuterium (\(\text{}^{2}_{1}\text{H}\)), and Tritium (\(\text{}^{3}_{1}\text{H}\)). Alternatively, Carbon isotopes: \(\text{}^{12}_{6}\text{C}\) and \(\text{}^{14}_{6}\text{C}\).

Step 2: Define Isobars
Isobars are nuclides belonging to different chemical elements that have the same mass number (\(A\)) but different atomic numbers (\(Z\)).
- They contain different numbers of protons and neutrons, but their total nucleon sum (\(Z + N\)) is equal.
- They possess distinct chemical properties because their atomic numbers \(Z\) are different.
- Example: Argon (\(\text{}^{40}_{18}\text{Ar}\)) and Calcium (\(\text{}^{40}_{20}\text{Ca}\)). Alternatively, Carbon (\(\text{}^{14}_{6}\text{C}\)) and Nitrogen (\(\text{}^{14}_{7}\text{N}\)).

Step 3: Conclusion
Isotopes share the same \(Z\) but different \(A\) (same element), whereas Isobars share the same \(A\) but different \(Z\) (different elements).

Concept:

• \(1\text{ atomic mass unit (u)}\) is defined as \(\frac{1}{12}\text{th}\) of the mass of an unbound neutral Carbon-12 atom: \(1\text{ u} \approx 1.660539 \times 10^{-27}\text{ kg}\).
• According to Einstein’s mass-energy equivalence principle, \(E = m c^2\).
• \(1\text{ electron-volt (eV)} = 1.60218 \times 10^{-19}\text{ J}\).

Step 1: Calculate energy equivalent in Joules
Mass \(m = 1\text{ u} = 1.660539 \times 10^{-27}\text{ kg}\).
Speed of light \(c = 2.99792 \times 10^8\text{ m/s}\).
Apply Einstein’s equation \(E = m c^2\):
\[ E = (1.660539 \times 10^{-27}\text{ kg}) \times (2.99792 \times 10^8\text{ m/s})^2 \]
\[ E = 1.660539 \times 10^{-27} \times 8.98755 \times 10^{16}\text{ J} \]
\[ E \approx 1.49242 \times 10^{-10}\text{ J} \]

Step 2: Convert energy from Joules to eV
Since \(1\text{ eV} = 1.60218 \times 10^{-19}\text{ J}\):
\[ E\text{ (in eV)} = \frac{1.49242 \times 10^{-10}\text{ J}}{1.60218 \times 10^{-19}\text{ J/eV}} \]
\[ E\text{ (in eV)} \approx 931.5 \times 10^6\text{ eV} \]

Step 3: Convert to Mega electron-volts (MeV)
Since \(1\text{ MeV} = 10^6\text{ eV}\):
\[ E = 931.5\text{ MeV} \]

Step 4: Conclusion
The mass-energy equivalent relation is \(1\text{ u} \approx 931.5\text{ MeV} = 9.315 \times 10^8\text{ eV}\).

Concept:

• A physical quantity is defined as a vector if and only if it possesses both magnitude and direction, and strictly obeys the algebraic laws of vector addition (such as the triangle law or parallelogram law of vector addition).

Step 1: Role of the arrow in representing current
The arrow drawn along a conductor indicating current direction signifies the directional sense of flow of positive charge carriers (or direction opposite to electron drift).
It merely indicates flow sense along a 1-dimensional constrained conducting path, not a spatial vector direction in 3D space.

Step 2: Violation of vector addition laws
When two conducting wires carrying currents \(I_1\) and \(I_2\) meet at a circuit junction at an angle \(\theta\), the resultant current entering the third wire is simply the scalar algebraic sum:
\[ I_{net} = I_1 + I_2 \]
The net current does not depend on the geometric orientation angle \(\theta\) between the wires.
If current were a vector, the resultant would follow vector addition (\(I_{net} = \sqrt{I_1^2 + I_2^2 + 2 I_1 I_2 \cos \theta}\)), which is physically untrue for electrical currents.

Step 3: Conclusion
Because electric current obeys ordinary scalar algebra rather than vector addition rules, electric current is fundamentally a scalar quantity.

Concept:

• A Wheatstone bridge consists of four resistance arms \(P, Q, R, S\) forming a closed loop \(ABCD\), with a voltage source \(V\) connected across \(AC\) and a galvanometer \(G\) across \(BD\).
• Balance condition corresponds to zero current through the galvanometer (\(I_g = 0\)), implying \(V_B = V_D\).

Step 1: Circuit Diagram and Kirchhoff’s Current Law
Let current \(I\) from battery split at node \(A\) into \(I_1\) through arm \(AB\) (resistance \(P\)) and \(I_2\) through arm \(AD\) (resistance \(R\)).
At node \(B\), current \(I_g\) flows through galvanometer arm \(BD\) (resistance \(G\)). Current through arm \(BC\) (resistance \(Q\)) is \(I_1 - I_g\).
At node \(D\), current through arm \(DC\) (resistance \(S\)) is \(I_2 + I_g\).

Step 2: Apply Kirchhoff’s Voltage Law (KVL)
Apply KVL to closed loop \(ABDA\):
\[ -I_1 P - I_g G + I_2 R = 0 \implies I_1 P + I_g G = I_2 R \quad \text{--- (Equation 1)} \]
Apply KVL to closed loop \(BCDB\):
\[ -(I_1 - I_g) Q + (I_2 + I_g) S + I_g G = 0 \quad \text{--- (Equation 2)} \]

Step 3: Apply Balance Condition (\(I_g = 0\))
For balanced bridge, potential at node \(B\) equals potential at node \(D\) (\(V_B = V_D\)), so zero current passes through galvanometer (\(I_g = 0\)).
Substitute \(I_g = 0\) into Equation 1:
\[ I_1 P = I_2 R \quad \text{--- (Equation 3)} \]
Substitute \(I_g = 0\) into Equation 2:
\[ I_1 Q = I_2 S \quad \text{--- (Equation 4)} \]

Step 4: Divide Equation 3 by Equation 4
\[ \frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \]
Canceling non-zero current factors \(I_1\) and \(I_2\):
\[ \frac{P}{Q} = \frac{R}{S} \]

Step 5: Conclusion
The balance condition for a Wheatstone bridge is \(\frac{P}{Q} = \frac{R}{S}\).

Concept:

• Magnetic field at the common center of a circular coil of radius \(r_2\) and \(N_2\) turns carrying current \(I_2\) is \(B_2 = \frac{\mu_0 N_2 I_2}{2 r_2}\).
• Since radius of smaller coil \(r_1 \ll r_2\), magnetic field \(B_2\) is nearly uniform across the entire area \(A_1 = \pi r_1^2\) of the smaller coil.
• Total magnetic flux linked with smaller coil containing \(N_1\) turns is \(\Phi_1 = N_1 B_2 A_1\).

Step 1: Identify parameters
Smaller coil (Coil 1): \(r_1 = 0.5\text{ cm} = 5 \times 10^{-3}\text{ m}\), \(N_1 = 10\).
Larger coil (Coil 2): \(r_2 = 5\text{ cm} = 5 \times 10^{-2}\text{ m}\), \(N_2 = 50\), \(I_2 = 3\text{ A}\).

Step 2: Calculate magnetic field produced by larger coil
\[ B_2 = \frac{\mu_0 N_2 I_2}{2 r_2} \]
Substitute \(\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}\):
\[ B_2 = \frac{(4\pi \times 10^{-7}) \times 50 \times 3}{2 \times (5 \times 10^{-2})} \]
\[ B_2 = \frac{600 \pi \times 10^{-7}}{10^{-1}} = 6\pi \times 10^{-5}\text{ T} \]

Step 3: Calculate magnetic flux linked with smaller coil
Area of smaller coil \(A_1 = \pi r_1^2 = \pi (5 \times 10^{-3})^2 = 25\pi \times 10^{-6}\text{ m}^2\).
Total magnetic flux \(\Phi_1\):
\[ \Phi_1 = N_1 B_2 A_1 \]
\[ \Phi_1 = 10 \times (6\pi \times 10^{-5}) \times (25\pi \times 10^{-6}) \]
\[ \Phi_1 = 1500 \pi^2 \times 10^{-11}\text{ Wb} \]
Using \(\pi^2 \approx 9.87\):
\[ \Phi_1 = 1.5 \pi^2 \times 10^{-7}\text{ Wb} \approx 1.5 \times 9.87 \times 10^{-7}\text{ Wb} \approx 1.48 \times 10^{-6}\text{ Wb} \]

Step 4: Conclusion
The magnetic flux linked with the smaller coil is \(1.5 \pi^2 \times 10^{-7}\text{ Wb} \approx 1.48 \times 10^{-6}\text{ Wb}\).

Concept:

• Mutual inductance \(M\) between two coupled coils relates magnetic flux \(\Phi_1\) in coil 1 to current \(I_2\) in coil 2: \(\Phi_1 = M I_2 \implies M = \frac{\Phi_1}{I_2}\).
• Theoretical formula for concentric coaxial circular coils (\(r_1 \ll r_2\)) is \(M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2 r_2}\).

Step 1: Use flux-current ratio
From Part (a), flux linked with smaller coil when current \(I_2 = 3\text{ A}\) flows in larger coil is \(\Phi_1 = 1.5 \pi^2 \times 10^{-7}\text{ Wb}\).
Mutual inductance \(M\):
\[ M = \frac{\Phi_1}{I_2} \]
\[ M = \frac{1.5 \pi^2 \times 10^{-7}\text{ Wb}}{3\text{ A}} \]
\[ M = 0.5 \pi^2 \times 10^{-7}\text{ H} = 5 \pi^2 \times 10^{-8}\text{ H} \]

Step 2: Numerical Evaluation
Using \(\pi^2 \approx 9.87\):
\[ M = 0.5 \times 9.87 \times 10^{-7}\text{ H} \approx 4.93 \times 10^{-7}\text{ H} \]

Step 3: Verify using direct formula
\[ M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2 r_2} \]
\[ M = \frac{(4\pi \times 10^{-7}) \times 10 \times 50 \times \pi \times (5 \times 10^{-3})^2}{2 \times (5 \times 10^{-2})} \]
\[ M = \frac{2000 \pi^2 \times 10^{-7} \times 25 \times 10^{-6}}{10^{-1}} = \frac{50000 \pi^2 \times 10^{-13}}{10^{-1}} = 5 \pi^2 \times 10^{-8}\text{ H} \approx 4.93 \times 10^{-7}\text{ H} \]

Step 4: Conclusion
The mutual inductance between the two coaxial coils is \(5 \pi^2 \times 10^{-8}\text{ H} \approx 4.93 \times 10^{-7}\text{ H}\).

Concept:

• The magnifying power (or angular magnification) of a telescope is the ratio of the angle subtended at the eye by the image to the angle subtended by the object.
• In normal adjustment (when the final image is at infinity), the magnifying power \(m\) is given by \(m = \frac{f_o}{f_e}\).
• In near point adjustment (when the final image is at the least distance of distinct vision \(D\)), \(m = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)\).

Step 1: Analyze dependence on focal lengths
From the standard formulas mentioned in the concepts, it is mathematically evident that magnification heavily depends on both \(f_o\) (focal length of objective) and \(f_e\) (focal length of eyepiece).
An increase in the objective’s focal length directly increases magnification, while an increase in the eyepiece’s focal length decreases it.
Thus, options (B) and (C) are directly involved in the magnification formula.

Step 2: Analyze dependence on color of light
According to Lens Maker’s Formula, the focal length of a lens is inversely related to \((n - 1)\), where \(n\) is the refractive index of the material.
Cauchy’s equation tells us that the refractive index \(n\) varies strictly with the wavelength (or color) of light.
Therefore, a change in the color of light leads to a change in the refractive index, which in turn alters the focal lengths \(f_o\) and \(f_e\).
Since the focal lengths change based on color, the overall magnifying power is implicitly dependent on the color of light.

Step 3: Analyze dependence on apertures
The aperture of a lens refers to its effective diameter or the light-gathering area.
While a larger aperture for the objective lens dramatically improves the resolving power and the brightness of the resulting image, it completely fails to alter the geometric focal lengths.
Since the aperture size does not appear in the angular magnification equations at all, the magnifying power remains completely independent of the apertures.

Step 4: Conclusion
The magnifying power depends on focal lengths and color, but definitely not on the physical aperture sizes of the lenses.
Hence, the correct option is (D).

Concept:

• An astronomical telescope is designed to view extremely distant objects, necessitating specific lens configurations.
• To achieve high magnification, the formula \(m = \frac{f_o}{f_e}\) dictates that \(f_o \gg f_e\).
• The tube length of the telescope (\(L\)) represents the separation between the objective lens and the eyepiece.

Step 1: Evaluate Statement (A)
For a functional astronomical telescope, the objective must gather abundant light and the system must provide angular magnification.
This fundamentally requires the focal length of the objective (\(f_o\)) to be significantly greater than the focal length of the eyepiece (\(f_e\)).
Therefore, statement (A) is completely correct in its assertion.

Step 2: Evaluate Statements (B) and (D)
Using the fundamental relation for magnifying power in normal adjustment: \(m = \frac{f_o}{f_e}\).
It is mathematically obvious that magnification is directly proportional to \(f_o\). Increasing \(f_o\) will indeed increase the magnifying power, making statement (B) correct.
Conversely, magnification is inversely proportional to \(f_e\). Increasing \(f_e\) will effectively decrease the magnifying power, making statement (D) entirely correct.

Step 3: Evaluate Statement (C)
The physical distance between the two lenses is known as the tube length \(L\).
When the telescope is adjusted for normal vision (final image at infinity), the intermediate image forms exactly at the focal points of both lenses, giving \(L = f_o + f_e\).
When adjusted for near point vision (final image at distance \(D\)), the intermediate image forms closer to the eyepiece, yielding \(L = f_o + u_e\), where \(u_e < f_e\).
In both typical adjustments, the distance \(L\) is either exactly equal to or less than (\(f_o + f_e\)).
It is never more than (\(f_o + f_e\)).
Thus, statement (C) is the logically incorrect statement.

Step 4: Conclusion
Since the question asks for the incorrect statement, option (C) fits perfectly.

Concept:

• When a telescope is in "normal adjustment," it implies that the final image is being formed at optical infinity.
• This is the most relaxed viewing state for the human eye.
• To achieve this, the real image formed by the objective lens must fall precisely on the primary focal plane of the eyepiece.

Step 1: Identify the given parameters
From the text of the problem, the focal length of the objective lens is given as \(f_o = 80 \text{ cm}\).
The focal length of the eyepiece is given as \(f_e = 4 \text{ cm}\).

Step 2: Apply the normal adjustment condition
In normal adjustment, the principal focus of the objective lens perfectly coincides with the principal focus of the eyepiece.
The physical distance separating the optical centers of the two lenses is referred to as the length of the telescope tube, denoted by \(L\).
The geometric relationship for this specific alignment is given simply by the sum of their individual focal lengths:
\[ L = f_o + f_e \]

Step 3: Perform the final calculation
Substitute the given numerical values into the tube length equation:
\[ L = 80 \text{ cm} + 4 \text{ cm} \]
\[ L = 84 \text{ cm} \]
This means the lenses must be separated by exactly 84 cm for parallel light to emerge from the eyepiece.

Step 4: Conclusion
The computed distance is 84 cm, which corresponds exactly to option (A).

Concept:

• Magnifying power (or angular magnification) is a measure of how much larger the object appears when viewed through the telescope compared to the naked eye.
• For an astronomical telescope set to normal adjustment, the final image is projected at infinity.
• The standard formula relating magnifying power \(m\) to the lens parameters is \(m = \frac{f_o}{f_e}\).

Step 1: Extract the required parameters
Referring back to the data provided in part (iv)(a):
The objective lens focal length is given by \(f_o = 80 \text{ cm}\).
The eyepiece focal length is provided as \(f_e = 4 \text{ cm}\).

Step 2: Apply the magnification formula
For viewing an object at infinity with the final image also placed at infinity (normal adjustment), the magnification is solely determined by the ratio of the focal lengths.
Utilize the formula:
\[ m = \frac{f_o}{f_e} \]

Step 3: Execute the arithmetic
Substitute the known values into the ratio:
\[ m = \frac{80 \text{ cm}}{4 \text{ cm}} \]
Since both quantities are in centimeters, the units cancel out elegantly, yielding a dimensionless number.
\[ m = 20 \]
This calculated value means the telescope makes distant objects appear 20 times larger in angular terms.

Step 4: Conclusion
The calculated magnifying power is 20, which aligns precisely with option (D).

Concept:

• Manganin is a specialized alloy consisting primarily of copper, manganese, and nickel.
• A defining characteristic of manganin is its incredibly low temperature coefficient of resistivity (\(\alpha \approx 0\)).
• This means its electrical resistance remains practically constant even when subjected to significant temperature changes.
• Ohm’s law states that \(V \propto I\), provided the physical conditions (most notably temperature and resistance) remain constant.

Step 1: Analyze the physical scenario
When taking readings without ever switching off the circuit, a continuous steady current heavily flows through the resistor X.
According to Joule’s law of heating (\(H = I^2 R t\)), this continuous current inevitably causes the resistor wire to heat up and increase in temperature.
For standard metals like copper or iron, this rise in temperature would drastically increase their electrical resistance.

Step 2: Apply the properties of Manganin
However, the resistor X is explicitly specified to be made of manganin.
Because manganin has a temperature coefficient of resistivity that is nearly negligible, the generated Joule heating does not affect its resistance value.
The resistance \(R\) of the manganin wire will stay absolutely constant regardless of how long the current is kept flowing.

Step 3: Determine the correct V-I graph
Since the resistance \(R\) is strictly constant, the relationship between the potential difference \(V\) and the current \(I\) follows a perfect linear proportionality (\(V = IR\)).
In graphical terms, a direct proportionality is always represented by a perfectly straight line passing through the origin.
Graphs that curve upwards or downwards represent non-ohmic components (like filaments or semiconductors) where resistance changes with temperature.

Step 4: Conclusion
The V-I characteristic will be a perfectly straight, linear line originating from zero, which is depicted in option (A).

Concept:

• In experimental physics, the calculation of an unknown resistance relies on Ohm’s Law: \(X = \frac{V}{I}\).
• Whenever a quantity is derived from the division or multiplication of two measured variables, the maximum fractional (or relative) error in the derived quantity is the sum of the fractional errors of the individual variables.
• Mathematically, for \(Z = \frac{A}{B}\), the relative error is given by \(\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}\).

Step 1: Identify the measurement formula
The student determines the unknown resistance X by measuring the voltage drop \(V\) across it and the current \(I\) passing through the main circuit.
The fundamental equation used for the calculation is \(X = \frac{V}{I}\).
Both \(V\) and \(I\) are independently measured quantities subject to their own instrumental and observational errors, denoted as \(\Delta V\) and \(\Delta I\).

Step 2: Apply error propagation rules
According to the principles of error propagation for quotients, the fractional errors simply add up.
The maximum possible fractional error in the calculated value of resistance X is formulated as:
\[ \frac{\Delta X}{X} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]
This equation clearly demonstrates that the total error in X is contributed to by both the voltmeter’s error and the ammeter’s error.

Step 3: Interpret the given options
Option (A) and (B) wrongly isolate the error to only one instrument, ignoring the dual dependency of the calculation.
Option (D) incorrectly suggests dividing the errors, which violates mathematical error combination laws entirely.
Option (C) states the error is equal to the sum of the error in voltmeter reading and error in ammeter reading. In standard experimental contexts, this phrasing specifically implies the summation of their respective relative/fractional errors as shown in the derived formula.

Step 4: Conclusion
The cumulative error in resistance \(X\) is inherently the combined sum of the individual measurement errors from both meters.
Therefore, option (C) is the most scientifically accurate statement among the choices.

Concept:

• A rheostat is a variable resistor used to actively control the flow of electric current by manually changing the resistance in the circuit.
• The total resistance of a simple series circuit dictates the total current supplied by the voltage source (\(I = \frac{V_{total}}{R_{eq}}\)).
• The voltage across a fixed component in a series circuit is directly proportional to the current flowing through it (\(V = IR\)).

Step 1: Analyze the circuit diagram
Carefully observing the provided circuit diagram reveals the connections: The positive terminal of the battery connects to the ammeter, which then connects to the unknown fixed resistance X.
The other end of resistance X connects to the movable slider (wiper) of the rheostat.
The fixed upper end of the rheostat, labeled P, is directly connected to the negative terminal of the battery.
This configuration means that the active resistance of the rheostat currently included in the circuit is simply the portion of the wire spanning from the fixed point P to the slider’s current position.

Step 2: Determine the effect of moving the slider
The question states that the movable end (slider) is physically pushed towards point P.
As the slider gets closer to P, the physical length of the rheostat wire included in the current path dramatically shortens.
Since resistance is directly proportional to length (\(R = \rho \frac{l}{A}\)), a shorter length means the rheostat introduces significantly less resistance into the circuit.
Consequently, the total equivalent resistance of the entire series circuit (\(R_{eq} = X + R_{rheo}\)) effectively decreases.

Step 3: Determine the impact on meter readings
According to Ohm’s law applied to the entire circuit, total current is \(I = \frac{V_{battery}}{R_{eq}}\).
Because the total equivalent resistance \(R_{eq}\) has decreased while the battery voltage remains constant, the overall current \(I\) must increase.
Thus, the reading on the ammeter will visibly increase.
Next, consider the voltmeter, which is connected directly in parallel across the fixed resistor X.
The voltage drop across X is determined by \(V_X = I \cdot X\).
Since the resistance \(X\) is perfectly constant and the current \(I\) has just increased, the voltage drop \(V_X\) must identically increase.
Thus, the reading on the voltmeter also increases.

Step 4: Conclusion
Both the ammeter and the voltmeter will show increased readings when the slider is moved towards P.
This precisely matches the statement in option (B).

Concept:

• The relationship between macroscopic electric current \(I\) and microscopic drift velocity \(v_d\) is governed by the relation \(I = n e A v_d\).
• In a strict series combination, the exact same electric current \(I\) flows continuously through every segment, regardless of changes in cross-sectional area.
• Therefore, drift velocity is universally inversely proportional to the cross-sectional area for a constant current (\(v_d \propto \frac{1}{A}\)).

Step 1: Establish the mathematical relationship
Since the three varied segments AC, CD, and DB are connected sequentially in series, the steady state current \(I\) is identical in all three parts.
From the established formula \(I = n e A v_d\), we can rearrange it to isolate drift velocity:
\[ v_d = \frac{I}{n e A} \]
Assuming the wire is completely uniform in material, the free electron density \(n\) and elemental charge \(e\) are strictly constant.
Since the cross-section is circular, area \(A = \pi \cdot r_{segment}^2\).
This yields the powerful proportionality: \(v_d \propto \frac{1}{r_{segment}^2}\).

Step 2: Calculate relative drift velocities for each segment
Let’s analyze segment AC:
Radius \(r_{AC} = r\).
Drift velocity \(v_1 \propto \frac{1}{r^2}\).
Let’s analyze segment CD:
Radius \(r_{CD} = \frac{r}{3}\).
Drift velocity \(v_2 \propto \frac{1}{(r/3)^2} = \frac{9}{r^2}\).
This clearly indicates \(v_2 = 9 \cdot v_1\).
Let’s analyze segment DB:
Radius \(r_{DB} = \frac{r}{2}\).
Drift velocity \(v_3 \propto \frac{1}{(r/2)^2} = \frac{4}{r^2}\).
This clearly indicates \(v_3 = 4 \cdot v_1\).

Step 3: Compare and arrange the magnitudes
By comparing the calculated proportionality coefficients:
\(v_1\) has a factor of 1.
\(v_2\) has an enormous factor of 9.
\(v_3\) has a factor of 4.
Arranging them purely in strictly descending order of magnitude gives:
\(v_2 > v_3 > v_1\).

Step 4: Conclusion
The highest drift velocity occurs in the thinnest section, and the lowest in the thickest section.
This logical ordering matches option (B).

Concept:

• According to the microscopic form of Ohm’s Law, the electric field \(E\) inside a conductor is proportional to the current density \(J\).
• The explicit mathematical relation is \(E = \rho J\), where \(\rho\) is the electrical resistivity of the material.
• Current density \(J\) is strictly defined as the total current \(I\) divided by the cross-sectional area \(A\) (\(J = \frac{I}{A}\)).

Step 1: Derive the electric field proportionality
Because the varied segments are arranged in series, the total electric current \(I\) is universally constant across the entire wire structure.
Substitute the definition of current density into the microscopic Ohm’s law:
\[ E = \rho \cdot \left(\frac{I}{A}\right) \]
Since the wire is made of one uniform metal, the resistivity \(\rho\) is constant everywhere.
Because \(I\) is also constant, the internal electric field strictly depends on the area:
\[ E \propto \frac{1}{A} \]
Substituting the formula for circular area \(A = \pi \cdot r_{segment}^2\):
\[ E \propto \frac{1}{r_{segment}^2} \]

Step 2: Evaluate the field in each distinct segment
For the first segment AC:
Radius is \(r\).
So, \(E_1 \propto \frac{1}{r^2}\).
For the middle segment CD:
Radius is extremely narrow at \(\frac{r}{3}\).
So, \(E_2 \propto \frac{1}{(r/3)^2} = \frac{9}{r^2}\).
For the final segment DB:
Radius is \(\frac{r}{2}\).
So, \(E_3 \propto \frac{1}{(r/2)^2} = \frac{4}{r^2}\).

Step 3: Rank the electric fields
By comparing the numerical scaling factors computed above:
The field \(E_1\) acts as the baseline (factor of 1).
The field \(E_2\) is 9 times stronger than \(E_1\).
The field \(E_3\) is exactly 4 times stronger than \(E_1\).
Placing these systematically in descending order gives:
\(E_2 > E_3 > E_1\).

Step 4: Conclusion
The electric field is most intense in the narrowest part of the wire to drive the same amount of current through the restricted bottleneck.
This fully aligns with option (C).

Concept:

• In a series LCR (Inductor, Capacitor, Resistor) circuit, the overall opposition to AC current is called impedance (\(Z\)).
• Resonance occurs under specific conditions where the reactive components perfectly cancel each other’s effects out.
• Power dissipation heavily depends on the phase alignment between the applied voltage and the resulting current, described by the power factor.

Step 1: Define Resonant Frequency
(I) Resonant Frequency:
In an alternating current series LCR circuit, the resonant frequency is strictly defined as the specific driving frequency of the AC source at which the inductive reactance (\(X_L = \omega L\)) becomes exactly equal in magnitude to the capacitive reactance (\(X_C = \frac{1}{\omega C}\)).
Because these two reactances are exactly \(180^\circ\) out of phase, they completely cancel each other out (\(X_L - X_C = 0\)).
Consequently, the total impedance of the circuit drops to its absolute minimum possible value, which is simply the pure ohmic resistance (\(Z = R\)).
At this specific frequency, the circuit permits the maximum possible amplitude of current to flow.
Mathematically, the angular resonant frequency is expressed as \(\omega_r = \frac{1}{\sqrt{LC}}\), and the linear resonant frequency is \(f_r = \frac{1}{2\pi\sqrt{LC}}\).

Step 2: Define Power Factor
(II) Power Factor:
The power factor of an AC circuit is fundamentally defined as the cosine of the phase angle (\(\phi\)) that exists between the total applied voltage and the resulting circuit current.
It serves as a crucial indicator of what fraction of the total apparent power is actually converted into useful work (true power dissipation).
Mathematically, it is written as \(\cos\phi = \frac{R}{Z}\), representing the simple ratio of the true resistance \(R\) to the total impedance \(Z\) of the circuit.
A higher power factor means the circuit is more purely resistive and highly efficient at dissipating energy.

Step 3: Determine condition for maximum power dissipation
The average power dissipated in an AC circuit over a complete cycle is governed by the formula:
\[ P_{avg} = V_{rms} \cdot I_{rms} \cdot \cos\phi \]
To strictly maximize this average power \(P_{avg}\), the multiplicative power factor term \(\cos\phi\) must reach its maximum mathematical value.
The cosine function reaches its absolute maximum value of \(1\) when the phase angle is precisely zero (\(\phi = 0^\circ\)).
Therefore, the power dissipated in the circuit will be maximum strictly when the power factor equals \(1\).
This idealized condition naturally occurs at electrical resonance, where the entire circuit behaves exactly like a purely resistive circuit.

Concept:

• The total impedance \(Z\) of a series LCR circuit is calculated using phasor addition because voltage drops across L, C, and R are completely out of phase.
• The governing impedance formula is \(Z = \sqrt{R^2 + (X_L - X_C)^2}\).
• The RMS (root mean square) values are essential for AC power calculations, derived from peak values as \(V_{rms} = \frac{V_m}{\sqrt{2}}\).

Step 1: Extract and list the given parameters
From the problem text, we explicitly have:
Inductance \(L = \frac{5}{\pi} \text{ H}\).
Capacitance \(C = \frac{50}{\pi} \mu\text{F} = \frac{50}{\pi} \times 10^{-6} \text{ F}\).
Resistance \(R = 400 \, \Omega\).
The applied AC voltage equation is given as \(v = 140 \sin(100\pi t)\).
By directly comparing this to the standard waveform equation \(v = V_m \sin(\omega t)\), we identify:
Peak voltage \(V_m = 140 \text{ V}\).
Angular frequency \(\omega = 100\pi \text{ rad/s}\).

Step 2: Calculate the individual reactances
First, carefully calculate the inductive reactance (\(X_L\)):
\[ X_L = \omega L \]
\[ X_L = (100\pi) \cdot \left(\frac{5}{\pi}\right) \]
The \(\pi\) terms elegantly cancel out, leaving:
\[ X_L = 500 \, \Omega \]
Next, meticulously calculate the capacitive reactance (\(X_C\)):
\[ X_C = \frac{1}{\omega C} \]
\[ X_C = \frac{1}{(100\pi) \cdot \left(\frac{50}{\pi} \times 10^{-6}\right)} \]
Again, the \(\pi\) terms perfectly cancel out in the denominator:
\[ X_C = \frac{1}{5000 \times 10^{-6}} \]
\[ X_C = \frac{1}{0.005} = \frac{1000}{5} = 200 \, \Omega \]

Step 3: Calculate the total circuit impedance (I)
Now, substitute the computed reactances into the main impedance formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
\[ Z = \sqrt{(400)^2 + (500 - 200)^2} \]
\[ Z = \sqrt{160000 + (300)^2} \]
\[ Z = \sqrt{160000 + 90000} \]
\[ Z = \sqrt{250000} \]
\[ Z = 500 \, \Omega \]
The total impedance of the LCR circuit is strictly \(500 \, \Omega\).

Step 4: Calculate the RMS current (II)
To find the rms current, we first rapidly need the rms voltage.
\[ V_{rms} = \frac{V_m}{\sqrt{2}} \]
Using the given approximation \(\sqrt{2} = 1.4\):
\[ V_{rms} = \frac{140}{1.4} = 100 \text{ V} \]
Now, forcefully apply Ohm’s law for AC circuits to find the current:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
\[ I_{rms} = \frac{100}{500} \]
\[ I_{rms} = 0.2 \text{ A} \]
The rms value of current flowing effortlessly through the circuit is exactly \(0.2 \text{ A}\).

Concept:

• A transformer operates strictly on the fundamental principle of mutual induction.
• It effectively consists of two separate coils (primary and secondary) wrapped tightly around a common laminated magnetic core.
• A step-up transformer specifically increases the alternating voltage, which necessitates having far more turns in the secondary coil than in the primary coil.

Step 1: Labelled Diagram of a Step-up Transformer

The proper diagram must clearly display a laminated iron core forming a closed loop.
The primary coil is wound on one limb and contains a relatively small number of turns (\(N_p\)).
The secondary coil is deliberately wound on the opposite limb (or over the primary) and contains a much larger number of turns (\(N_s\)).
An AC voltage source is explicitly connected across the primary terminals, and the high voltage output is taken across the secondary terminals.
Crucial labels should prominently include: Laminated soft iron core, Primary coil (\(N_p\)), Secondary coil (\(N_s\)), Input AC voltage (\(V_p\)), and Output AC voltage (\(V_s\)).

Step 2: Derivation of the Voltage Ratio
When a continuously varying alternating voltage is applied directly to the primary coil, it aggressively drives an alternating current through it.
This alternating current generates a constantly changing magnetic flux within the highly permeable iron core.
Assuming an ideal scenario where there is absolute zero flux leakage, all the magnetic flux (\(\Phi\)) perfectly links with both the primary and the secondary coils simultaneously.
According to Faraday’s fundamental law of electromagnetic induction, the induced electromotive force (emf) in the primary coil is given precisely by:
\[ e_p = -N_p \frac{d\Phi}{dt} \quad \text{--- (Equation 1)} \]
Similarly, the induced emf in the extensively wound secondary coil is:
\[ e_s = -N_s \frac{d\Phi}{dt} \quad \text{--- (Equation 2)} \]

Step 3: Formulate the transformation ratio
By directly dividing Equation 2 by Equation 1, the time derivative of flux cancels out completely:
\[ \frac{e_s}{e_p} = \frac{-N_s \frac{d\Phi}{dt}}{-N_p \frac{d\Phi}{dt}} = \frac{N_s}{N_p} \]
For an ideal transformer with negligible coil resistances, the applied primary voltage \(V_p\) perfectly equals the back emf \(e_p\) (\(V_p = e_p\)), and the secondary terminal voltage \(V_s\) exactly equals the induced emf \(e_s\) (\(V_s = e_s\)).
Substituting these voltage relations directly yields the final ratio:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
This highly important relation proves that the ratio of secondary voltage to primary voltage is strictly equal to the ratio of their respective number of turns, known as the transformation ratio (\(k\)).

Concept:

• An ideal transformer is assumed to be \(100\%\) efficient, meaning absolutely zero power is lost during the transfer.
• Input electrical power perfectly equals output electrical power (\(P_{in} = P_{out}\)).
• Electrical power for AC circuits (assuming unity power factor for simple transformer problems) is \(P = V \cdot I\).
• The voltage elegantly scales strictly according to the turns ratio equation: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\).

Step 1: Identify the explicitly given parameters
Number of turns heavily wound in the primary coil: \(N_p = 100\).
Number of turns heavily wound in the secondary coil: \(N_s = 5000\).
The total input power rigorously supplied to the primary: \(P = 3.3 \text{ kW} = 3300 \text{ W}\).
The applied primary voltage: \(V_p = 220 \text{ V}\).

Step 2: Calculate the primary current (I)
The electrical power fundamentally delivered to the primary coil is completely determined by the product of primary voltage and primary current.
\[ P = V_p \cdot I_p \]
To find the primary current \(I_p\), we simply rearrange the formula:
\[ I_p = \frac{P}{V_p} \]
Substitute the known values heavily into the arranged equation:
\[ I_p = \frac{3300 \text{ W}}{220 \text{ V}} \]
Simplify the fraction carefully by removing a zero:
\[ I_p = \frac{330}{22} \]
Divide thoroughly to get the exact value:
\[ I_p = 15 \text{ A} \]
The current powerfully drawn by the primary coil is therefore precisely \(15 \text{ A}\).

Step 3: Calculate the output voltage (II)
To precisely find the output (secondary) voltage, we must employ the fundamental transformer turns ratio formula derived earlier.
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
We can dynamically rearrange this to solve specifically for the secondary voltage \(V_s\):
\[ V_s = V_p \times \left( \frac{N_s}{N_p} \right) \]
Insert all the designated given values directly into this relation:
\[ V_s = 220 \times \left( \frac{5000}{100} \right) \]
First, neatly resolve the bracketed turns ratio, which signifies a massive step-up factor:
\[ \frac{5000}{100} = 50 \]
Now, multiply this substantial factor by the primary voltage:
\[ V_s = 220 \times 50 \]
\[ V_s = 11000 \text{ V} \]
This enormous output can also be written efficiently as \(11 \text{ kV}\).
The final output voltage securely provided by the secondary coil is \(11000 \text{ V}\).

Concept:

• An equipotential surface is defined as a region in space where the electric potential \(V\) is completely uniform and constant at every single point.
• The relationship between the electric field vector \(\vec{E}\) and the change in electric potential \(dV\) is given by the fundamental calculus relation \(dV = -\vec{E} \cdot d\vec{l}\).
• Work done in moving a test charge along an equipotential surface must be exactly zero.

Step 1: Define the potential difference along the surface
By the very definition of an equipotential surface, the potential difference between any two infinitesimally close points on this specific surface is exactly zero.
Mathematically, this means \(dV = 0\).

Step 2: Apply the electric field and potential relation
The change in potential \(dV\) for a tiny displacement \(d\vec{l}\) along the surface is expressed using the dot product:
\[ dV = -\vec{E} \cdot d\vec{l} \]
Substitute \(dV = 0\) strictly into this equation:
\[ 0 = -\vec{E} \cdot d\vec{l} \]

Step 3: Expand the dot product mathematically
Expanding the vector dot product gives:
\[ E \cdot dl \cdot \cos\theta = 0 \]
Where \(\theta\) represents the exact angle between the electric field vector \(\vec{E}\) and the displacement vector \(d\vec{l}\) lying on the surface.
Since neither the electric field magnitude \(E\) is zero nor the displacement \(dl\) is zero (for a non-trivial movement), the only mathematical possibility remaining is:
\[ \cos\theta = 0 \]

Step 4: Conclusion
Solving for \(\theta\) yields:
\[ \theta = 90^\circ \]
This rigorously proves that the electric field must always be perfectly perpendicular (normal) to the equipotential surface at any given point to ensure no work is done moving charges along it.

Concept:

• A dielectric is fundamentally an insulating material containing bound charges that cannot move freely, but can shift slightly from their equilibrium positions.
• When exposed to an external field, these bound charges undergo physical polarization.
• Polarization creates an opposing internal electric field that strictly alters the net field present inside the material.

Step 1: Analyze the effect of the external field
When a non-conducting dielectric slab is actively placed inside a uniform external electric field \(\vec{E}_{ext}\), the external field strongly exerts forces on the microscopic molecules of the dielectric.
This forcefully stretches or actively rotates the molecules, causing their positive and negative charge centers to physically separate slightly.
This phenomenon is known scientifically as dielectric polarization.

Step 2: Identify the creation of the induced field
Due to this precise polarization, a net positive surface charge density securely accumulates on one face of the dielectric, while a net negative surface charge density builds up on the opposite face.
These newly induced surface charges naturally produce their own internal electric field, known strictly as the polarization field, denoted as \(\vec{E}_p\).

Step 3: Determine the net internal electric field
By fundamental laws of electrostatics, this induced polarization field \(\vec{E}_p\) points exactly in the opposite direction to the original applied external field \(\vec{E}_{ext}\).
According to the principle of superposition, the net electric field \(\vec{E}_{net}\) physically existing inside the dielectric material is the vector sum of both opposing fields:
\[ \vec{E}_{net} = \vec{E}_{ext} - \vec{E}_p \]

Step 4: Conclusion
Because the induced field \(\vec{E}_p\) aggressively subtracts from the applied field, the overall magnitude of the net internal electric field is significantly reduced.
Therefore, the electric field inside a polarized dielectric is always strictly less than the external electric field existing outside it.

Concept:

• The capacitance \(C\) of a parallel plate capacitor is fundamentally dependent on its precise geometric configuration.
• For an isolated charged capacitor, the total charge \(Q\) stored on the plates remains strictly conserved because there is no external path for the electrons to flow.
• The macroscopic relation between voltage, charge, and capacitance is universally given by \(V = \frac{Q}{C}\).

Step 1: Establish the geometric capacitance formula
The theoretical formula for the capacitance of an empty parallel plate capacitor is:
\[ C = \frac{\epsilon_0 A}{d} \]
Where:
\(A\) is the overlapping surface area of the plates.
\(d\) is the physical separation distance between the two metallic plates.
\(\epsilon_0\) is the permittivity of free space.

Step 2: Analyze the physical change
The problem specifically states that the plates are physically brought closer to each other.
This action means that the separation distance \(d\) actively decreases.
Looking directly at our established formula, the capacitance \(C\) is inversely proportional to the separation distance \(d\) (\(C \propto \frac{1}{d}\)).
Therefore, a mathematical decrease in \(d\) strictly causes a proportional increase in the overall capacitance \(C\).

Step 3: Apply the voltage-charge relation
Assume the capacitor is completely disconnected from any charging battery (it is isolated).
Under this highly specific condition, the stored charge \(Q\) is completely trapped and must remain strictly constant.
Now, substitute the changing variables into the voltage definition:
\[ V = \frac{Q}{C} \]

Step 4: Conclusion
Since the numerator (charge \(Q\)) is perfectly constant, and the denominator (capacitance \(C\)) has significantly increased due to the plates moving closer, the resulting fraction must mathematically decrease.
Consequently, the electric potential difference \(V\) strictly decreases.

Concept:

• The total electrostatic potential energy (\(U\)) of a discrete system of charges is precisely the total amount of external work that was originally done to assemble them from infinity to their current configuration.
• To completely "dissociate" the system, an external agent must do work exactly equal to the negative of this initial potential energy to tear them completely apart and push them back out to infinity.
• The potential energy for a continuous system of multiple charges is the algebraic sum of the potential energies of all unique pairs of charges.

Step 1: Formulate the total potential energy of the system
For a carefully defined system consisting of three point charges \(q_1\), \(q_2\), and \(q_3\) separated by pairwise distances \(r_{12}\), \(r_{23}\), and \(r_{13}\), the formula 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 2: Substitute the provided parameters
From the problem description, the charges are situated on the vertices of an equilateral triangle.
This guarantees that all pairwise separation distances are perfectly equal: \(r_{12} = r_{23} = r_{13} = a\).
The specific charges are defined as: \(q_1 = q\), \(q_2 = -4q\), and \(q_3 = 2q\).
Inserting these into the potential energy equation:
\[ U = \frac{1}{4\pi\epsilon_0} \left[ \frac{(q)(-4q)}{a} + \frac{(-4q)(2q)}{a} + \frac{(q)(2q)}{a} \right] \]

Step 3: Simplify the mathematical expression
Multiply the charges carefully keeping track of the negative signs:
\[ U = \frac{1}{4\pi\epsilon_0 a} \left[ -4q^2 - 8q^2 + 2q^2 \right] \]
Combine the algebraic terms strictly inside the bracket:
\[ U = \frac{1}{4\pi\epsilon_0 a} \left[ -10q^2 \right] \]
\[ U = -\frac{10q^2}{4\pi\epsilon_0 a} \]
This negative energy definitively indicates that the system is in an attractive, bound state.

Step 4: Calculate the final work done to dissociate
The total external work required to dissociate the system is completely given by the difference between the final potential energy (at infinity) and the initial potential energy.
Since the potential energy at infinite separation is exactly zero (\(U_f = 0\)):
\[ W_{dissociate} = U_f - U_i \]
\[ W_{dissociate} = 0 - \left( -\frac{10q^2}{4\pi\epsilon_0 a} \right) \]
\[ W_{dissociate} = \frac{10q^2}{4\pi\epsilon_0 a} \]

Concept:

• In a metallic conductor, electric current is physically carried by free electrons moving in a specific direction.
• While the actual physical translation of these electrons (drift velocity) is incredibly slow, the electrical signal propagates in an entirely different manner.
• The signal is established by an electromagnetic field that travels independently of the physical particles.

Step 1: Explain the nature of electron drift
It is entirely physically accurate that individual free electrons move at a microscopic crawl, generally a few millimeters per second (drift speed \(v_d\)).
This extreme slowness is due to the incredibly dense lattice of metal ions; the electrons suffer billions of chaotic, random collisions every single second, heavily retarding their forward progress.

Step 2: Explain the establishment of the electric field
However, an electric circuit does not operate by patiently waiting for one specific electron to travel all the way from the battery’s negative terminal to the positive terminal.
The absolute instant the circuit switch is mechanically closed, an electromagnetic field is established throughout the entire length of the conductor.
This invisible electric field propagates through the wire at nearly the absolute speed of light in a vacuum (\(c \approx 3 \times 10^8\) m/s).

Step 3: Connect the field to the macroscopic current
Because the electric field travels almost instantaneously, it reaches every single free electron physically present anywhere in the entire wire at practically the exact same moment.
This rapidly established field instantly exerts an electrostatic force (\(F = -eE\)) on all the free electrons simultaneously.
As a result, electrons everywhere in the circuit begin their slow drift at the exact same time.

Step 4: Conclusion
Since electrons at the far end of the circuit begin moving simultaneously with those near the battery, the macroscopic electric current is established universally the very instant the switch is closed, despite the extremely slow physical speed of the individual charge carriers.

Concept:

• Any real-world voltage supply (like a battery) consists of an ideal electromotive force (emf, \(E\)) strictly in series with an unavoidable internal resistance (\(r\)).
• When current flows out of the battery, a specific portion of the generated voltage is inevitably lost inside the battery itself due to this internal resistance.
• The macroscopic terminal voltage equation is given by \(V = E - Ir\).

Step 1: Analyze the maximum current formula
According to Ohm’s law applied to a complete circuit, the total current \(I\) drawn from a battery connected to an external load resistance \(R\) is strictly formulated as:
\[ I = \frac{E}{R + r} \]
To physically draw the absolute maximum theoretical current from this specific supply, we must short-circuit the terminals, making the external resistance absolutely zero (\(R = 0\)).
The maximum possible current equation then rigorously simplifies to:
\[ I_{max} = \frac{E}{r} \]

Step 2: Evaluate the specific given condition
The problem specifically states that we are utilizing a "low voltage supply."
This means the numerator in our equation, the electromotive force \(E\), is a remarkably small numerical value.
Simultaneously, the problem strictly requires this weak supply to deliver "high currents."
This mathematically demands that the overall resulting fraction (\(E/r\)) must be an exceptionally large number.

Step 3: Determine the necessary condition for internal resistance
In any fraction where the numerator is fixed to a small value, the only mathematically valid way to make the total quotient extremely large is to make the denominator incredibly tiny.
Therefore, the internal resistance \(r\) strictly situated in the denominator must be extremely low.

Step 4: Conclusion
If the internal resistance \(r\) were high, the substantial voltage drop strictly occurring inside the battery itself (\(V_{drop} = Ir\)) would completely consume the already low available emf \(E\).
Thus, to prevent massive internal voltage loss and successfully deliver a high current, a low voltage supply must inherently be constructed with a very low internal resistance.

Concept:

• Electrical resistance is fundamentally defined for any component as the mathematical ratio of voltage to current at any given instant.
• Ohm’s Law is a highly specific physical statement about the strict linearity of certain conducting materials under constant environmental conditions.
• A definition is universally true, while a physical law has specific strict boundaries and limitations.

Step 1: Define the equation \(V = IR\)
The mathematical equation \(V = IR\), or strictly rearranged as \(R = \frac{V}{I}\), is universally accepted as the fundamental macroscopic definition of electrical resistance.
This specific equation allows us to mathematically calculate the instantaneous resistance \(R\) of absolutely any electrical device, at any specific moment, regardless of how that device fundamentally operates.
It flawlessly applies to a simple copper wire, a complex semiconductor diode, a glowing vacuum tube, or a heated tungsten filament.

Step 2: Define strict Ohm’s Law
Ohm’s Law, as originally formulated by Georg Simon Ohm, is a much stricter, narrower physical postulate.
It explicitly asserts that the current \(I\) flowing through a specific conductor is directly and linearly proportional to the potential difference \(V\) firmly applied across its ends (\(V \propto I\)), strictly provided that all physical conditions (most notably temperature and mechanical strain) remain completely constant.
This strict proportionality implies that the resistance \(R\) must remain an absolutely constant value, completely independent of the varying applied voltage or current.

Step 3: Highlight the logical contradiction
There are thousands of commonly used electrical components (such as p-n junction diodes, thermistors, and transistors) known as non-ohmic devices.
For these non-ohmic devices, if you forcefully change the voltage \(V\), the resulting current \(I\) does not change linearly. Their resistance \(R\) actively fluctuates depending on the applied voltage.
These devices violently disobey Ohm’s Law.
However, even when a diode is flagrantly violating Ohm’s Law, you can still perfectly calculate its instantaneous resistance at any given exact voltage point using the formula \(R = \frac{V}{I}\).

Step 4: Conclusion
Because the equation \(V = IR\) remains universally valid and actively true even for devices where Ohm’s Law completely and utterly fails, it is logically impossible for \(V = IR\) to be the actual statement of Ohm’s Law itself.
It is merely the mathematical definition of resistance, whereas Ohm’s Law is the specific strict physical statement that \(R\) must remain constant.

Concept:

• When two non-ideal battery cells are connected completely in parallel, they form a single equivalent cell that can mathematically replace them.
• The total equivalent internal resistance follows the standard parallel resistor formula: \(\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}\).
• The equivalent electromotive force (emf) is a weighted average determined by the formula: \(E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}\).

Step 1: Identify parameters from the circuit diagram
Carefully inspecting the provided circuit diagram reveals the connections.
The upper cell has an emf \(E_1 = 12\text{ V}\) and an internal resistance \(r_1 = 1 \, \Omega\).
The lower cell has an emf \(E_2 = 6\text{ V}\) and an internal resistance \(r_2 = 0.5 \, \Omega\).
Crucially, the longer vertical lines of both battery symbols (representing the positive terminals) are facing the exact same direction, specifically towards terminal A.
Because they are perfectly aligned with matching polarities, we use standard positive addition in our formula.

Step 2: Calculate the equivalent internal resistance
Apply the standard formula for resistors situated in parallel:
\[ \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \]
Substitute the given resistance values:
\[ \frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{0.5} \]
Recognize that \(\frac{1}{0.5}\) is exactly equal to \(2\):
\[ \frac{1}{r_{eq}} = 1 + 2 = 3 \]
Invert the result to find \(r_{eq}\):
\[ r_{eq} = \frac{1}{3} \, \Omega \approx 0.33 \, \Omega \]

Step 3: Calculate the equivalent electromotive force (emf)
Utilize the established equivalent cell formula:
\[ E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_{eq}}} \]
Substitute the individual values rigorously:
\[ E_{eq} = \frac{\frac{12}{1} + \frac{6}{0.5}}{3} \]
Simplify the specific terms within the numerator:
\(\frac{12}{1} = 12\)
\(\frac{6}{0.5} = 12\)
Now insert these back into the main calculation:
\[ E_{eq} = \frac{12 + 12}{3} \]
\[ E_{eq} = \frac{24}{3} \]
\[ E_{eq} = 8\text{ V} \]
The equivalent cell operating between points A and B powerfully behaves precisely as an \(8\text{ V}\) battery with an internal resistance of \(\frac{1}{3} \, \Omega\).

Concept:

• The refractive index serves as an optical measure of how much a material actively slows down light passing through it.
• Vacuum represents the ultimate speed limit where light travels the absolute fastest.

Step 1: Provide the strict definition
The absolute refractive index (often simply called the refractive index) of a specific optical medium is strictly defined as the mathematical ratio of the speed of light propagating in a pure vacuum to the speed of light propagating in that specific medium.

Step 2: Provide the mathematical formula
Mathematically, it is perfectly expressed by the equation:
\[ n = \frac{c}{v} \]
Where:
\(n\) is the absolute refractive index of the designated medium.
\(c\) is the universal speed of light in a vacuum (approximately \(3 \times 10^8\) m/s).
\(v\) is the actual phase speed of light measured traveling inside the medium.
Because light strictly travels slower in any material medium than in a vacuum (\(v < c\)), the absolute refractive index is universally a dimensionless quantity always strictly greater than \(1\).

Concept:

• As light sequentially traverses the two slanted interfaces of a glass prism, it undergoes consecutive refractions, causing an overall angular deviation.
• The overall angle of deviation intricately depends on the exact angle of incidence.
• There is exactly one unique angle of incidence where this total deviation hits its absolute minimum value, creating high geometric symmetry.

Step 1: Establish the basic geometric prism equations
For an incoming light ray undergoing refraction through a solid prism, standard optical geometry dictates two fundamental relationships:
First, the refracting angle of the prism (\(A\)) is equal to the sum of the two internal angles of refraction (\(r_1\) and \(r_2\)):
\[ A = r_1 + r_2 \quad \text{--- (Equation 1)} \]
Second, the total angle of deviation (\(\delta\)) produced by the prism is strictly related to the angle of incidence (\(i\)), angle of emergence (\(e\)), and the prism angle:
\[ \delta = i + e - A \quad \text{--- (Equation 2)} \]

Step 2: Apply the specific condition for minimum deviation
Experimental observation proves that the deviation angle \(\delta\) rigorously reaches its absolute minimum value (\(\delta_m\)) exactly when the light ray travels perfectly symmetrically through the prism.
Under this highly symmetric condition:
The angle of incidence is exactly equal to the angle of emergence (\(i = e\)).
The internal angles of refraction at both faces are exactly equal (\(r_1 = r_2 = r\)).

Step 3: Solve for \(r\) and \(i\) in terms of \(A\) and \(\delta_m\)
Substitute the symmetric condition \(r_1 = r_2 = r\) directly into Equation 1:
\[ A = r + r = 2r \]
\[ r = \frac{A}{2} \quad \text{--- (Equation 3)} \]
Next, seamlessly substitute the conditions \(\delta = \delta_m\) and \(i = e\) strictly into Equation 2:
\[ \delta_m = i + i - A \]
\[ \delta_m = 2i - A \]
Rearrange to explicitly solve for the angle of incidence \(i\):
\[ 2i = A + \delta_m \]
\[ i = \frac{A + \delta_m}{2} \quad \text{--- (Equation 4)} \]

Step 4: Apply Snell’s Law to derive the final formula
According to Snell’s law applied at the very first interface, the refractive index (\(\mu\)) of the prism material relative to the surrounding air is given by:
\[ \mu = \frac{\sin i}{\sin r_1} \]
Since \(r_1 = r\) during minimum deviation, substitute Equations 3 and 4 directly into this law:
\[ \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \]
This gives the required expression for the refractive index accurately in terms of the measurable angles.

Concept:

• When a light ray strikes an optical boundary perfectly perpendicularly (normal incidence, \(i = 0^\circ\)), it proceeds entirely without deviation.
• At an internal boundary, if the angle of incidence strictly exceeds the critical angle (\(i > C\)), Total Internal Reflection (TIR) occurs rather than refraction.
• The critical angle formula for a glass-air interface is universally given by \(\sin C = \frac{1}{\mu}\).

Step 1: Calculate the critical angle of the prism
The prism is composed of glass with a given refractive index of \(\mu = 1.5\).
We first decisively calculate the critical angle (\(C\)) for the glass-air interface using the established formula:
\[ \sin C = \frac{1}{\mu} \]
\[ \sin C = \frac{1}{1.5} = \frac{2}{3} \approx 0.667 \]
Since \(\sin 41.8^\circ \approx 0.667\), the critical angle is firmly established as \(C \approx 41.8^\circ\).
Based on standard optical diagram conventions for such problems, we assume the prism ABC is an equilateral triangle, meaning all interior vertex angles are exactly \(A = B = C = 60^\circ\).

Step 2: Trace ray behavior at the first interface (BC)
The problem states that the incoming ray QP is incident exactly normally (perpendicularly) on the bottom face BC.
Because the angle of incidence is exactly \(0^\circ\) relative to the surface normal, the ray enters the dense glass perfectly undeviated.
It travels vertically straight upward through the solid interior of the prism.

Step 3: Trace ray behavior at the second interface (AB)
The vertically traveling ray then strikes the slanted interior face AB.
Using simple geometry inside the right-angled triangle formed by the vertical ray, the horizontal base BC, and the slanted face AB: the angle at vertex B is \(60^\circ\), the angle between the ray and the base is \(90^\circ\), therefore the angle between the vertical ray and the face AB is \(180^\circ - 90^\circ - 60^\circ = 30^\circ\).
The surface normal to face AB is drawn perfectly perpendicular (\(90^\circ\)) to it.
Thus, the angle of incidence \(i\) relative to this normal is mathematically \(90^\circ - 30^\circ = 60^\circ\).
We carefully compare this angle of incidence to our previously calculated critical angle:
\(i = 60^\circ\), while \(C = 41.8^\circ\).
Since \(i > C\), the ray is completely trapped and undergoes Total Internal Reflection (TIR) inside the prism.
According to the law of reflection, the ray reflects at an exact angle of \(60^\circ\) from the normal, which simultaneously means it makes a geometric angle of \(90^\circ - 60^\circ = 30^\circ\) with the face AB itself.

Step 4: Trace ray behavior at the third interface (AC)
The newly reflected ray travels straight across the upper portion of the prism towards the opposite slanted face AC.
We now examine the small geometric triangle formed by the top vertex A, the point of TIR on face AB, and the new point of incidence on face AC.
Inside this specific triangle:
The top vertex angle \(A\) is \(60^\circ\).
The angle between the reflected ray and face AB was just established as \(30^\circ\).
The sum of angles in any triangle must be \(180^\circ\). Therefore, the third angle (where the ray strikes face AC) must strictly be \(180^\circ - (60^\circ + 30^\circ) = 90^\circ\).
Because the ray strikes face AC perfectly at \(90^\circ\) (perpendicular to the actual surface), its angle of incidence relative to the surface normal is exactly \(0^\circ\).
Consequently, the ray completely emerges out of the prism from face AC perfectly undeviated, traveling in a straight line out into the air.

Concept:

• Wave optics provides two entirely complementary ways to visualize the propagation of light: discrete rays and continuous wavefronts.
• A wavefront focuses strictly on the phase characteristics of the physical wave.
• A ray focuses strictly on the geometric path and direction of energy transport.

Step 1: Provide a detailed definition of a Wavefront
A completely uniform source of light strictly emits waves in all available directions.
A wavefront is defined geometrically as the continuous locus (or surface) connecting all neighboring points in a medium that are being vibrated in the exact same physical phase at any given instant of time.
For example, if you connect all the precise wave crests expanding outward from a point source, that spherical surface constitutes a spherical wavefront. It represents the actual physical wave structure moving through space.

Step 2: Provide a detailed definition of a Ray
A ray of light is a purely idealized, imaginary line drawn with an arrow that points rigorously in the exact direction of the propagation of the wave’s energy.
By fundamental geometric necessity in an isotropic medium, a ray is always strictly drawn perpendicular (normal) to the expanding wavefront surface.
While a wavefront depicts a sweeping surface of identical timing, the ray provides a simple linear vector representing where that surface is heading.

Concept:

• Huygens principle boldly states that every single point on an advancing wavefront acts as a fresh source of secondary spherical wavelets.
• The new, updated wavefront at any later time is formed geometrically by taking the forward envelope (common tangent) of all these expanding secondary wavelets.

Step 1: Establish the geometric setup
Imagine a completely flat, plane wavefront named \(AB\) propagating steadily through a medium with wave speed \(v\).
This wavefront obliquely strikes a perfectly flat reflecting surface \(XY\) at an angle of incidence \(i\).
Point \(A\) of the wavefront firmly touches the reflecting mirror surface first. At this precise instant, point \(B\) is still physically distant from the mirror, separated by a distance \(BC\).

Step 2: Apply Huygens construction over time \(t\)
It precisely takes time \(t = \frac{BC}{v}\) for the trailing edge of the wavefront at point \(B\) to physically travel and strike the mirror surface at point \(C\).
During this exact same time interval \(t\), the secondary spherical wavelet originating from point \(A\) has been expanding rapidly outward into the same upper medium.
The radius of this newly generated expanding spherical wavelet from \(A\) is precisely \(AD = vt\).
Because both events happen in the exact same medium over the same time, it is an absolute geometric certainty that distance \(AD\) perfectly equals distance \(BC\) (\(AD = BC = vt\)).

Step 3: Construct the reflected wavefront
To completely trace the new reflected wavefront, we boldly draw a common tangent line from point \(C\) directly to the edge of the spherical wavelet at point \(D\).
The resulting straight line segment \(CD\) actively represents the new plane reflected wavefront.
The angle between this outgoing reflected wavefront \(CD\) and the mirror surface \(XY\) geometrically defines the angle of reflection \(r\).

Step 4: Prove the law of reflection mathematically
We carefully inspect the two right-angled triangles rigidly formed on the mirror surface: \(\Delta ABC\) and \(\Delta ADC\).
In these two distinct triangles:
Side \(AC\) is completely shared as the common hypotenuse (\(AC = AC\)).
Side \(BC\) perfectly equals side \(AD\) (\(BC = AD = vt\)), as established by uniform wave propagation.
The angle \(\angle ABC = 90^\circ\) (since \(AB\) is the incident wavefront and ray is normal).
The angle \(\angle ADC = 90^\circ\) (since \(CD\) is a tangent to the spherical wavelet at \(D\)).
By the RHS (Right-Angle Hypotenuse Side) congruence criterion, \(\Delta ABC \cong \Delta ADC\) perfectly.
Because the triangles are strictly congruent, their corresponding matching angles must be exactly equal.
Therefore, \(\angle BAC = \angle DCA\).
By geometric construction, \(\angle BAC = i\) and \(\angle DCA = r\).
This rigorously proves that the angle of incidence equals the angle of reflection (\(i = r\)), flawlessly verifying the universal law of reflection.

Concept:

• A wavefront physically tracks the speed of light.
• A convex lens is remarkably thicker strictly at its center and aggressively tapers to become thinner at its peripheral edges.
• Because the refractive index of solid glass is much higher than air, the wave travels noticeably slower through the glass.

Step 1: Describe the incident wavefront
Imagine a completely flat, perfectly vertical plane wavefront smoothly advancing through the air and making contact with the front face of a biconvex lens.
Because it is a flat plane wavefront, all portions of the wave (top, middle, bottom) strike the front surface of the lens at almost the exact same time.

Step 2: Analyze the optical path differences
As the entire wavefront begins to force its way through the dense glass structure, different portions experience vastly different physical thicknesses.
The absolute center portion of the wavefront is forced to travel through the thickest, densest chunk of glass. Since speed \(v = \frac{c}{\mu}\), it is severely slowed down for the longest duration, causing a massive phase delay.
Conversely, the upper and lower outer edges of the wavefront travel through the very thin, tapered edges of the lens. They exit the glass relatively quickly, experiencing far less delay.

Step 3: Depict the final emerging wavefront
Because the peripheral edges strictly travel faster and exit sooner, they powerfully surge forward ahead of the delayed central portion.
This differential delay aggressively curves the previously flat wavefront.
The emerging wavefront completely transforms from a flat plane into a deeply curved, spherical shape.
Because the center is dragging behind, the spherical curvature actively collapses inward, perfectly converging all the wave energy toward a single central focal point precisely on the principal axis.