CBSE Class 12 Physics Question Paper 2024 (Set 2- 55/4/2) Available- Download Solution PDF with Answer Key

Electrostatic Potential at the Midpoint of \( BC \)


Step 1: Formula for Electrostatic Potential

- The electrostatic potential at a point due to a charge \( q \) at distance \( r \) is:
\[ V = \frac{1}{4\pi \epsilon_0} \frac{q}{r} \]

Step 2: Contribution of Charges at \( B \) and \( C \)
- The midpoint of \( BC \) is equidistant from charges at \( B \) and \( C \).

- Distance of midpoint from \( B \) or \( C \):
\[ r_{B} = r_C = \frac{BC}{2} = \frac{6L}{2} = 3L \]

- Potential due to charge at \( B \):
\[ V_B = \frac{1}{4\pi \epsilon_0} \frac{q}{3L} \]

- Potential due to charge at \( C \):
\[ V_C = \frac{1}{4\pi \epsilon_0} \frac{q}{3L} \]

- Total contribution from \( B \) and \( C \):
\[ V_{BC} = 2 \times \frac{q}{12\pi \epsilon_0 L} = \frac{2q}{12\pi \epsilon_0 L} = \frac{q}{6\pi \epsilon_0 L} \]

Step 3: Contribution of Charge at \( A \)

- Distance from \( A \) to the midpoint of \( BC \) (Using Apollonius theorem):
\[ r_A = \sqrt{(5L)^2 - (3L)^2} = \sqrt{25L^2 - 9L^2} = \sqrt{16L^2} = 4L \]

- Potential due to charge at \( A \):
\[ V_A = \frac{1}{4\pi \epsilon_0} \frac{q}{4L} = \frac{q}{16\pi \epsilon_0 L} \]

Step 4: Total Electrostatic Potential at Midpoint \[ V = V_{BC} + V_A \]
\[ V = \frac{q}{6\pi \epsilon_0 L} + \frac{q}{16\pi \epsilon_0 L} \]
\[ V = \frac{16q + 6q}{96\pi \epsilon_0 L} \]
\[ V = \frac{22q}{96\pi \epsilon_0 L} = \frac{11q}{48\pi \epsilon_0 L} \]

Thus, the electrostatic potential at the midpoint of \( BC \) is:
\[ \frac{11q}{48\pi \epsilon_0 L} \]

which matches option (A).
Quick Tip: The electrostatic potential at a point due to multiple charges is the algebraic sum of the potentials due to individual charges.

Understanding Coulomb Force Variation


- According to Coulomb’s Law, the force between two charges is:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{|q_1 q_2|}{r^2} \]

- The slope of the \( F \) vs. \( \frac{1}{r^2} \) graph represents \( |q_1 q_2| \).

- Since \( q_2 \) has the least magnitude and is positive, comparing slopes gives:
\[ q_3 > q_1 > q_2 \]

Thus, the correct order is \( q_3 > q_1 > q_2 \), matching option (D).
Quick Tip: A steeper slope in the \( F \) vs. \( \frac{1}{r^2} \) graph means a greater product of charge magnitudes.

Magnetic Susceptibility of Diamagnetic Materials


- Magnetic susceptibility (\( \chi_m \)) measures how a material responds to an external magnetic field.

- For diamagnetic materials, \( \chi_m \) is:

- Negative (they repel magnetic fields).

- Small (weak effect compared to paramagnetic or ferromagnetic materials).
\[ \chi_m \approx -10^{-5} to -10^{-6} \]

Thus, the correct answer is (A) Small and negative.
Quick Tip: Diamagnetic materials like copper, bismuth, and gold have a small negative magnetic susceptibility and oppose external magnetic fields.

Magnetic Flux Linked with Loop B


Step 1: Magnetic Field at the Center of a Current-Carrying Loop

The magnetic field at the center of a circular loop carrying current \( I \) is:
\[ B = \frac{\mu_0 I}{2R} \]

where:

- \( \mu_0 \) = Permeability of free space,

- \( R \) = Radius of loop A.

Step 2: Magnetic Flux Through Loop B

The magnetic flux through loop B is given by:
\[ \Phi = B \cdot A \]

where:
- \( A \) is the area of loop B:
\[ A = \pi r^2 = \pi \left( \frac{R}{20} \right)^2 = \frac{\pi R^2}{400} \]

- Substituting \( B \) from Step 1:
\[ \Phi = \left( \frac{\mu_0 I}{2R} \right) \times \frac{\pi R^2}{400} \]
\[ \Phi = \frac{\mu_0 I \pi R^2}{800R} \]
\[ \Phi \propto R \]

Thus, the magnetic flux linked with loop B is proportional to \( R \), which matches option (A).
Quick Tip: For a concentric loop system, the magnetic flux linkage depends on both the field of the larger loop and the area of the smaller loop.

Motion of a Charged Particle in a Magnetic Field


- The Lorentz Force on a charged particle moving in a magnetic field is:
\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \]

- Given \( \mathbf{B} = B_0 \hat{i} \) and \( \mathbf{v} = v_x \hat{i} + v_y \hat{j} \), we calculate the cross-product:
\[ \mathbf{F} = q [(v_x \hat{i} + v_y \hat{j}) \times (B_0 \hat{i})] \]

- Since \( \hat{i} \times \hat{i} = 0 \) and \( \hat{j} \times \hat{i} = -\hat{k} \), we get:
\[ \mathbf{F} = -q v_y B_0 \hat{k} \]

- The force acts perpendicular to the velocity component in the \( yz \)-plane, causing circular motion in that plane.
- The velocity component along \( x \)-direction remains unchanged, leading to a helical path.

Thus, the correct answer is (C) helical path.
Quick Tip: A charged particle moves in a helical path when its velocity has a component parallel and perpendicular to the magnetic field.

Explanation of Photoelectric Effect


- According to Einstein’s Photoelectric Equation:
\[ K_{\max} = h f - \phi \]

where:

- \( h \) = Planck’s constant,

- \( f \) = Frequency of incident radiation,

- \( \phi \) = Work function of the metal.

- If \( f < f_0 \) (threshold frequency), no electrons are emitted.

- This confirms the quantum nature of light, supporting option (A).
Quick Tip: In the photoelectric effect, electrons are emitted only if the photon energy exceeds the work function of the metal.

Phase Relationship in Electromagnetic Waves


- In an electromagnetic wave, the electric field \( \mathbf{E} \) and the magnetic field \( \mathbf{B} \) are:

1. Perpendicular to each other.

2. Perpendicular to the direction of wave propagation.

3. In phase with each other.

- The wave equations are:
\[ E = E_0 \sin(kz - \omega t) \]
\[ B = B_0 \sin(kz - \omega t) \]

- Since both fields oscillate in phase, the phase difference is \( 0 \).

Thus, the correct answer is (A) \( 0 \).
Quick Tip: In an electromagnetic wave, the electric and magnetic fields are in phase and perpendicular to each other and the direction of wave propagation.

Bohr’s Model and Radius of Electron Orbits


Step 1: Formula for Bohr’s Radius

According to Bohr’s Model, the radius of an electron in the \( n^{th} \) orbit of a hydrogen atom is:
\[ r_n = \frac{4\pi \epsilon_0 \hbar^2}{m e^2} n^2 \]

where:

- \( \epsilon_0 \) = Permittivity of free space,

- \( \hbar \) = Reduced Planck’s constant,

- \( m \) = Mass of the electron,

- \( e \) = Charge of the electron.

Step 2: Understanding the Proportionality

- Since all constants remain unchanged, we get:
\[ r_n \propto n^2 \]

Step 3: Conclusion

- The radius of an orbit increases quadratically with \( n \).

- Thus, the correct answer is (C) \( r_n \propto n^2 \).
Quick Tip: In Bohr’s Model, orbit radius varies as \( n^2 \), while energy varies as \( \frac{1}{n^2} \).

RMS Current Calculation


Step 1: Given Data
- Peak voltage: \( V_0 = 282V \)
- Angular frequency: \( \omega = 100 \)
- Capacitance: \( C = 1 \mu F = 10^{-6} F \)

Step 2: Reactance of the Capacitor \[ X_C = \frac{1}{\omega C} = \frac{1}{(100) (10^{-6})} = 10^4 \Omega \]

Step 3: Peak Current Calculation
Using:
\[ I_0 = \frac{V_0}{X_C} \]
\[ I_0 = \frac{282}{10^4} = 0.0282 A = 28.2 mA \]

Step 4: RMS Current \[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{28.2}{1.41} = 20 mA \]

Thus, the correct answer is (B) 20 mA.
Quick Tip: For a capacitor in AC circuit, current leads voltage by \( 90^\circ \), and the capacitive reactance is \( X_C = \frac{1}{\omega C} \).

Finding the Galvanometer's Full-Scale Deflection Current


Step 1: Given Data
- Galvanometer resistance: \( R_g = 100 \Omega \)
- Shunt resistance: \( R_s = 0.1 \Omega \)
- Ammeter range: \( I = 1A \)

Step 2: Current Through the Galvanometer
Using the shunt formula:
\[ I_g = I \times \frac{R_s}{R_g + R_s} \]
\[ I_g = 1 \times \frac{0.1}{100 + 0.1} = 1 \times \frac{0.1}{100.1} \]
\[ I_g \approx 1 \times 10^{-3} A = 1 mA \]

Thus, the correct answer is (B) 1 mA.
Quick Tip: The shunt resistance allows most of the current to bypass the galvanometer, reducing the current through it to prevent damage.

Understanding the Balmer Series


Step 1: Understanding the Wavelength and Spacing

- The Balmer series consists of transitions where electrons fall to the \( n=2 \) level.
- The wavelengths are given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \]

where \( R_H \) is the Rydberg constant.

- As \( n \) increases, the term \( \frac{1}{n^2} \) gets smaller, causing wavelengths to decrease.

Step 2: Effect on Spectral Lines

- The spectral lines become closer together as \( n \) increases.

- The intensity of higher-order transitions decreases, making them weaker.

Thus, the correct answer is (D) Closer together and weaker in intensity.
Quick Tip: In the Balmer series, higher energy transitions have shorter wavelengths, weaker intensity, and spectral lines appear closer together.

Maximum Induced EMF in the Coil


Step 1: Faraday’s Law of Electromagnetic Induction

According to Faraday’s Law, the induced EMF is given by:
\[ \mathcal{E} = -N \frac{d\Phi}{dt} \]

where \( \Phi \) is the magnetic flux:
\[ \Phi = B A = B_0 A \cos\left(\frac{2\pi t}{T}\right) \]

where:

- \( A \) is the area of the coil,

- \( B_0 \) is the maximum magnetic field.

Step 2: Calculating the Induced EMF

Differentiating \( \Phi \) with respect to \( t \):
\[ \mathcal{E} = -N A B_0 \frac{d}{dt} \left[\cos\left(\frac{2\pi t}{T}\right)\right] \]
\[ \mathcal{E} = N A B_0 \left(\frac{2\pi}{T}\right) \sin\left(\frac{2\pi t}{T}\right) \]
\[ \mathcal{E} = \mathcal{E}_0 \sin\left(\frac{2\pi t}{T}\right) \]

where \( \mathcal{E}_0 = \frac{2\pi N A B_0}{T} \) is the peak EMF.

Step 3: Finding Maximum EMF

- The EMF is maximum when \( \sin\left(\frac{2\pi t}{T}\right) = 1 \), which occurs at:
\[ \frac{2\pi t}{T} = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots \]

- Solving for \( t \):
\[ t = \frac{T}{4}, \frac{3T}{4}, \frac{5T}{4}, \dots = \frac{nT}{4}, \quad n = 1, 3, 5, \dots \]

Thus, the correct answer is (B) \( t = \frac{nT}{4} \).
Quick Tip: The EMF in a coil is maximum when the rate of change of flux is maximum, which occurs when \( B \) passes through zero.

Understanding Interference and Energy Conservation


Step 1: Explanation of Interference and Diffraction

- In interference and diffraction, light waves superpose, leading to constructive and destructive interference.

- This results in bright fringes (high intensity) and dark fringes (low intensity).

- However, the total energy remains conserved, as energy is redistributed, not lost.

Step 2: Evaluating Assertion and Reason

- The Assertion (A) is correct because light energy indeed redistributes, creating bright and dark fringes.

- The Reason (R) is false because energy is conserved, not lost.

Step 3: Conclusion

- Since Assertion (A) is true but Reason (R) is false, the correct answer is (C).
Quick Tip: In interference and diffraction, energy is not lost but redistributed. The total energy remains constant.

Understanding Electron Drift in Conductors


Step 1: Understanding Drift Velocity

- In a conductor, free electrons move randomly in all directions due to thermal motion.

- When an electric field is applied, electrons acquire a small net velocity called drift velocity \( v_d \).

- However, the overall movement of electrons still includes their random thermal motion.

Step 2: Evaluating Assertion (A)

- Assertion (A) is correct because electrons continue their random motion, but on average, they experience a net drift.

Step 3: Evaluating Reason (R)

- Reason (R) is also correct because drift velocity is a small velocity superposed on the random thermal velocities of electrons.

Step 4: Conclusion

- Since both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A), the correct answer is (A).
Quick Tip: In a conductor, electrons move randomly due to thermal energy, but when an electric field is applied, they acquire a small drift velocity superposed on their random motion.

Understanding the Photoelectric Effect in Zinc


Step 1: Understanding the Photoelectric Effect

- The photoelectric effect occurs when light of sufficient energy (\( E = h f \)) is incident on a metal, causing electron emission.

- The work function (\( \phi \)) is the minimum energy required to eject an electron.

- For zinc, the work function is 4.3 eV.

Step 2: Energy of Yellow Light

- The wavelength of yellow light is approximately 570 nm.

- Using the energy equation:
\[ E = \frac{hc}{\lambda} \]
\[ E = \frac{(6.63 \times 10^{-34}) (3 \times 10^8)}{570 \times 10^{-9}} \]
\[ E \approx 2.18 eV \]

Step 3: Evaluating Assertion (A)

- Since yellow light has only 2.18 eV, it cannot eject electrons from zinc (work function = 4.3 eV).

- Thus, Assertion (A) is false.

Step 4: Evaluating Reason (R)

- The Reason (R) is also false because the energy of yellow light is less than the work function of zinc, not more.

Step 5: Conclusion

- Since both Assertion (A) and Reason (R) are false, the correct answer is (D).
Quick Tip: For photoelectric emission, the photon energy must be greater than the work function of the metal. Zinc requires ultraviolet light for electron emission.

Temperature Coefficient of Resistance and Charge Carriers


Step 1: Understanding the Temperature Coefficient of Resistance

- The temperature coefficient of resistance (\( \alpha \)) determines how resistance changes with temperature:

- For metals, \( \alpha \) is positive because resistance increases with temperature due to increased collisions between electrons.

- For p-type semiconductors, \( \alpha \) is negative because resistance decreases with temperature as more charge carriers become available.

Step 2: Evaluating Assertion (A)

- Assertion (A) is true since metals have a positive coefficient and p-type semiconductors have a negative coefficient.

Step 3: Evaluating Reason (R)

- The Reason (R) is true, as:

- In metals, charge carriers are negatively charged electrons.

- In p-type semiconductors, charge carriers are holes, which behave as positively charged entities.

Step 4: Evaluating the Explanation

- While Reason (R) is correct, it does not explain why the temperature coefficient is positive for metals and negative for p-type semiconductors.

- The real reason is that in metals, increased temperature increases electron scattering, raising resistance, while in semiconductors, higher temperatures generate more charge carriers, lowering resistance.

Step 5: Conclusion

- Since both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A), the correct answer is (B).
Quick Tip: For metals, resistance increases with temperature due to higher electron scattering. For semiconductors, resistance decreases as more charge carriers are generated at higher temperatures.

Refraction at Spherical Surface from Denser to Rarer Medium


Step 1: Refraction Formula at Spherical Surface

The refraction formula at a spherical surface from a denser medium to a rarer medium is given by:
\[ \frac{n_1}{v} - \frac{n_2}{u} = \frac{n_2 - n_1}{R} \]

where:

- \( n_1 \) is the refractive index of the rarer medium,

- \( n_2 \) is the refractive index of the denser medium,

- \( v \) is the image distance,

- \( u \) is the object distance,

- \( R \) is the radius of curvature of the spherical surface.

Step 2: Substituting Known Values

We are given the following values:

- \( u = -20 \, cm \) (since the object is inside the sphere),

- \( R = -40 \, cm \) (the radius of the spherical surface),

- \( n_1 = 1 \) (refractive index of the rarer medium, i.e., air),

- \( n_2 = 1.5 \) (refractive index of the denser medium, i.e., glass).

Substitute these values into the refraction formula:
\[ \frac{1}{v} - \frac{1.5}{-20} = \frac{1.5 - 1}{-40} \]

Step 3: Solving for the Image Distance \( v \)

Simplify the equation:
\[ \frac{1}{v} + \frac{1.5}{20} = \frac{0.5}{40} \]
\[ \frac{1}{v} + \frac{0.075}{1} = \frac{0.0125}{1} \]
\[ \frac{1}{v} = \frac{0.0125}{1} - \frac{0.075}{1} = -0.0625 \]

Thus:
\[ v = -16 \, cm \]

Step 4: Nature of the Image

The negative sign indicates that the image is formed on the same side as the object. Therefore, the image is virtual.

Thus, the nature of the image is virtual and the position is at \( -16 \, cm \).

\boxed{\text{Image position: -16 \, \text{cm Quick Tip: For a bubble inside a denser medium, refraction at a curved surface follows the lens-maker's formula to find the image position.

Calculation of Focal Lengths in a Refracting Telescope


Step 1: Understanding the Formula

In a refracting telescope, the magnifying power (\( M \)) in normal adjustment is given by:
\[ M = \frac{f_o}{f_e} \]

where:

- \( f_o \) = Focal length of the objective lens,

- \( f_e \) = Focal length of the eyepiece lens.

The total length of the telescope in normal adjustment is:
\[ L = f_o + f_e \]

Given:
- \( L = 1.00 \) m = 100 cm,
- \( M = 19 \).

Step 2: Expressing \( f_o \) and \( f_e \)

Rearranging the magnification equation:
\[ f_o = M f_e \]

Substituting in the length equation:
\[ M f_e + f_e = L \]
\[ 19 f_e + f_e = 100 \]
\[ 20 f_e = 100 \]
\[ f_e = \frac{100}{20} = 5 cm \]
\[ f_o = M f_e = 19 \times 5 = 95 cm \]

Step 3: Conclusion

- Focal length of the objective lens: 95 cm.

- Focal length of the eyepiece lens: 5 cm.
Quick Tip: For a refracting telescope, the objective lens has a large focal length, while the eyepiece lens has a small focal length to achieve high magnification.

N/A Quick Tip: For particles with the same kinetic energy, the de Broglie wavelength decreases with increasing mass.

Finding the Refractive Index of the Medium




Step 1: Understanding the Critical Condition

- Since the emergent ray grazes the adjacent face, the angle of refraction at the emerging face is 90°.

- Using Snell’s law at the second interface:
\[ \mu_{glass} \sin r = \mu_{medium} \sin 90^\circ \]
\[ \mu_{glass} \sin r = \mu_{medium} \]

Step 2: Finding Angle \( r \)

- The prism is equilateral, so the angle of the prism:
\[ A = 60^\circ \]

- The relation between the angle of incidence and refraction inside the prism is:
\[ \sin i = \mu_{glass} \sin r \]

- Since the ray enters normally, \( i = 0^\circ \), and applying prism formula:
\[ r = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \]

Step 3: Finding \( \mu_{medium} \) \[ \mu_{medium} = \mu_{glass} \sin 30^\circ \]
\[ \mu_{medium} = \mu_{glass} \times \frac{1}{2} \]
\[ \mu_{medium} = \frac{\mu_{glass}}{2} \]

Step 4: Conclusion
Thus, the refractive index of the medium is:
\[ \mu_{medium} = \frac{\mu_{glass}}{2} \] Quick Tip: When light just grazes the face of a prism, the critical angle condition applies. Use Snell’s law to find the refractive index of the medium.

Calculating Resistance and Resistivity Ratios


Step 1: Resistance Formula

The resistance of a wire is given by:
\[ R = \frac{\rho L}{A} \]

where:

- \( \rho \) = resistivity,

- \( L \) = length of the wire,

- \( A = \pi r^2 \) = cross-sectional area.

Given:

- Length ratio: \( L_A : L_B = 1:2 \),

- Radius ratio: \( r_A : r_B = 2:1 \),

- Cross-sectional area ratio:
\[ A_A : A_B = \pi (2r)^2 : \pi (r)^2 = 4:1 \]

Step 2: Ratio of Resistances \[ \frac{R_A}{R_B} = \frac{\rho_A L_A / A_A}{\rho_B L_B / A_B} \]
\[ = \frac{\rho_A (1) / 4}{\rho_B (2) / 1} \]
\[ = \frac{\rho_A}{\rho_B} \times \frac{1}{4} \times \frac{1}{2} \]
\[ = \frac{\rho_A}{\rho_B} \times \frac{1}{8} \]

Thus,
\[ \frac{R_A}{R_B} = \frac{\rho_A}{8 \rho_B} \]

From the I-V graph, the slope represents resistance \( R \). Using the given graph values,
\[ \frac{R_A}{R_B} = 2 \]

Thus,
\[ 2 = \frac{\rho_A}{8 \rho_B} \]

Step 3: Ratio of Resistivities
Rearranging:
\[ \rho_A = 16 \rho_B \]
\[ \frac{\rho_A}{\rho_B} = 16 \]

Step 4: Conclusion

- Ratio of resistances: \( \frac{R_A}{R_B} = 2 \)

- Ratio of resistivities: \( \frac{\rho_A}{\rho_B} = 16 \)
Quick Tip: Resistance depends on length, area, and resistivity. The resistivity of a material is independent of its dimensions and only depends on the nature of the material.

Understanding p-n Junction Diode Biasing


- A p-n junction diode allows current flow in one direction (forward bias) and blocks current in the opposite direction (reverse bias).


(i) Forward Biasing:


- The p-side (anode) is connected to the positive terminal of the battery, and the n-side (cathode) is connected to the negative terminal.

- This reduces the depletion region width, allowing current to flow.


Circuit Diagram for Forward Biasing:


I-V Characteristics in Forward Bias:


- The current increases exponentially after the threshold voltage (0.7V for silicon, 0.3V for germanium).



(ii) Reverse Biasing:


- The p-side is connected to the negative terminal, and the n-side is connected to the positive terminal.

- This increases the depletion region width, preventing current flow (except for a small leakage current).


Circuit Diagram for Reverse Biasing:



I-V Characteristics in Reverse Bias:


- A small leakage current flows until the breakdown voltage is reached.




Quick Tip: In forward bias, current flows easily after the threshold voltage. In reverse bias, only a small leakage current exists until breakdown.

Geiger-Marsden Experiment (Rutherford Scattering)


- The experiment measured the number of alpha particles (\( N \)) scattered at different angles (\( \theta \)).

- The observed scattering pattern led to significant conclusions about atomic structure.


Graph: Variation of \( N \) with \( \theta \)




Conclusions from the Graph

1. Most alpha particles pass undeflected, meaning the atom is mostly empty space.

2. Few particles are scattered at large angles, implying a small, dense, positively charged nucleus.

Expression for Distance of Closest Approach

The distance of closest approach (\( r_0 \)) is the minimum separation between the alpha particle and the nucleus before it stops and reverses.

- At the point of closest approach, the initial kinetic energy of the alpha particle is converted into electrostatic potential energy:
\[ \frac{1}{2} m v^2 = \frac{1}{4\pi\epsilon_0} \frac{Z e \cdot 2e}{r_0} \]

Solving for \( r_0 \):
\[ r_0 = \frac{1}{4\pi\epsilon_0} \frac{2 Z e^2}{\frac{1}{2} m v^2} \]
\[ r_0 = \frac{4 \pi \epsilon_0 \cdot 2 Z e^2}{m v^2} \]

Thus, the distance of closest approach is:
\[ r_0 = \frac{2 Z e^2}{4 \pi \epsilon_0 \cdot \frac{1}{2} m v^2} \]

This represents the minimum distance between the alpha particle and the nucleus before repulsion stops its motion.
Quick Tip: Rutherford’s experiment showed that atoms have a small dense nucleus and are mostly empty space.

Charge Density and Potential in a Conducting Shell


Step 1: Understanding the Charge Distribution

- The conducting shell redistributes its charge to maintain electrostatic equilibrium.

- The inner surface of the shell acquires a charge of \( -Q \) to neutralize the field inside the conductor.

- The outer surface must therefore carry a charge:
\[ q_{outer} = q + Q \]

Step 2: Surface Charge Density on Outer Surface
The surface charge density \( \sigma \) is given by:
\[ \sigma = \frac{Charge on outer surface}{Surface area of the sphere} \]
\[ \sigma = \frac{q + Q}{4\pi R^2} \]

Thus, the charge density on the outer surface is:
\[ \boxed{\sigma = \frac{q + Q}{4\pi R^2}} \]

Step 3: Potential at \( R/2 \) from the Center

- Inside a conducting shell, the potential is uniform and equal to the potential at the surface.

- The potential at the surface is given by:
\[ V = \frac{1}{4\pi \epsilon_0} \left( \frac{Q}{R} + \frac{q}{R} \right) \]
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q + q}{R} \]

Since the entire region inside the shell has the same potential, the potential at \( R/2 \) is the same as at the surface:
\[ \boxed{V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q + q}{R}} \]

Step 4: Conclusion

- Charge density on the outer surface: \( \frac{q + Q}{4\pi R^2} \)

- Potential at \( R/2 \): \( \frac{1}{4\pi \epsilon_0} \cdot \frac{Q + q}{R} \)
Quick Tip: For a conducting shell, the electric field inside is zero, and the potential remains constant throughout the interior.

Nuclear Fission vs. Nuclear Fusion

\[ \begin{array}{|c|c|c|} \hline \textbf{Property} & \textbf{Nuclear Fission} & \textbf{Nuclear Fusion}
\hline Definition & A nucleus splits into smaller nuclei. & Two nuclei combine to a heavier nucleus.
\hline Energy & Large amt per fission event. & Extremely large energy release.
\hline Example & \( ^{235U + n \rightarrow ^{92}Kr + ^{141}Ba + 3n + Energy \)} & \( ^2H + ^3H \rightarrow ^4He + n + \text{Energy \)}
\hline Occurrence & In nuclear reactors and atomic bombs. & In stars like the Sun.
\hline Requirement & Critical mass of fuel and a neutron. & Extremely high temperature and pressure.
\hline \end{array} \]


Thus, nuclear fission and fusion differ in process, energy release, and applications.
Quick Tip: Nuclear fission is used in reactors, while nuclear fusion powers stars and promises future clean energy.

Calculation of Energy Released


- The number of atoms in 1 g of \( ^{239}Pu \) is given by:
\[ N = \frac{Mass of sample}{Atomic mass} \times Avogadro's number \]
\[ N = \frac{1}{239} \times 6.022 \times 10^{23} \]
\[ N \approx 2.52 \times 10^{21} \]

- Since each fission releases 180 MeV, the total energy released is:
\[ E = N \times Energy per fission \]
\[ E = (2.52 \times 10^{21}) \times (180) \]
\[ E \approx 4.54 \times 10^{23} MeV \]

Thus, the total energy released is \( 4.54 \times 10^{23} \) MeV.
Quick Tip: Energy released in nuclear fission is enormous compared to chemical reactions, making nuclear power highly efficient.

Magnetic Field Due to Current-Carrying Wires


Step 1: Magnetic Field Due to a Long Straight Wire

The magnetic field at a distance \( r \) from an infinitely long straight wire carrying current \( I \) is given by Ampère’s Law:
\[ B = \frac{\mu_0 I}{2\pi r} \]

where:

- \( \mu_0 = 4\pi \times 10^{-7} \) Tm/A is the permeability of free space,

- \( I \) = Current in the wire,

- \( r \) = Perpendicular distance from the wire.

Using the right-hand rule, the direction of the field is determined.

Step 2: Magnetic Field at Point \( P_1(2m, 2m) \)

Step 2.1: Contribution from the 3A Current Wire

Let the distance from the wire to \( P_1 \) be \( r_1 \).
\[ B_1 = \frac{\mu_0 \times 3}{2\pi r_1} \]

Step 2.2: Contribution from the 5A Current Wire

Let the distance from the wire to \( P_1 \) be \( r_2 \).
\[ B_2 = \frac{\mu_0 \times 5}{2\pi r_2} \]

Step 2.3: Net Magnetic Field at \( P_1 \)

Using vector addition, find the resultant magnetic field:
\[ B_{net, P_1} = \sqrt{B_1^2 + B_2^2 + 2 B_1 B_2 \cos \theta} \]

where \( \theta \) is the angle between the field vectors.

Step 3: Magnetic Field at Point \( P_2(-1m, 1m) \)

- Repeat the same process for \( P_2 \), considering different distances.

Final Answer:

After calculating the values,
\[ B_{net, P_1} = (calculated value in Tesla) \]
\[ B_{net, P_2} = (calculated value in Tesla) \]

Directions: Use the right-hand rule to determine field directions.
Quick Tip: The magnetic field due to a straight current-carrying wire follows a circular pattern, determined by the right-hand rule.

Understanding Displacement Current and Continuity in a Charging Capacitor


Step 1: Definition of Displacement Current

- The displacement current (\( I_d \)) is a term introduced by James Clerk Maxwell to explain how a changing electric field can produce a magnetic field, similar to a conduction current.

- It is given by:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]

where:

- \( \epsilon_0 \) = Permittivity of free space,

- \( \Phi_E \) = Electric flux through a given surface.
\[ \begin{array}{|c|c|c|} \hline \textbf{Property} & \textbf{Conduction Current} (I_c) & \textbf{Displacement Current} (I_d)
\hline Define & Move of free electrons in conductor. & Changing electric field- dielectric medium.
\hline Formula & I_c = \sigma E (Ohm’s law) & I_d = \epsilon_0 \frac{d\Phi_E}{dt}
\hline Presence & in Conductors. & in capacitor gaps, insulators, vacuum.
\hline Physical Flow & Involves actual charge carriers. & No actual flow, only changing electric field.
\hline \end{array} \]

Step 3: Continuity of Current in a Charging Capacitor

- Consider a capacitor connected to a DC source.

- During charging, conduction current flows in the wires.

- Inside the capacitor, no free electrons move across the gap, but electric flux builds up.
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]

- Maxwell’s Equation states that displacement current maintains continuity, ensuring a continuous magnetic field around the circuit.

- Thus, even though no charge flows through the dielectric, the circuit remains complete due to displacement current.

Step 4: Conclusion

- Displacement current is required to modify Ampère’s Law and explain how a capacitor circuit remains continuous even in the absence of conduction current in the capacitor gap.
Quick Tip: The displacement current allows capacitor circuits to obey Kirchhoff’s laws by ensuring a continuous current flow in the circuit, even though no actual charge moves across the capacitor plates.

Applying Kirchhoff’s Laws


- Using Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL), we analyze the given circuit.

- Let the potential at node \( M \) be \( V_M \) and node \( B \) be \( V_B \).

- Applying KVL in loop \( ANBMA \), considering the given resistances and batteries:
\[ 6E - 3R I_1 - 4R I_{BM} = 0 \]
\[ I_1 = \frac{6E - 4R I_{BM}}{3R} \]

- Similarly, applying Kirchhoff’s rule in other loops and solving the equations, we determine the current in the BM branch.

Thus, after solving, the current in branch BM is:
\[ I_{BM} = \frac{6E}{13R} \]

Thus, the required current in branch BM is \( \frac{6E}{13R} \).
Quick Tip: Use Kirchhoff's laws to analyze complex circuits: KCL for junctions and KVL for loops.

Definition of Mutual Inductance


- Mutual inductance (\( M \)) is the property of two coils in which a change in current in one coil induces an electromotive force (emf) in the other due to electromagnetic induction.

- The induced emf in the secondary coil is given by:
\[ \mathcal{E}_2 = - M \frac{dI_1}{dt} \]

where \( I_1 \) is the current in the primary coil.

- The SI unit of mutual inductance is Henry (H), where 1 Henry = 1 Weber per Ampere (1 H = 1 Wb/A).
Quick Tip: Mutual inductance quantifies the ability of one coil to induce emf in another due to changing magnetic flux.

Derivation of Mutual Inductance


- Consider two long coaxial solenoids, one inside the other.

- The primary solenoid (inner solenoid) has:

- \( N_1 \) turns

- Radius \( r_1 \)

- Length \( l \)

- Current \( I_1 \) flowing through it.

- The magnetic field inside the primary solenoid is given by:
\[ B_1 = \frac{\mu_0 N_1 I_1}{l} \]

- The magnetic flux linked with the secondary solenoid (outer solenoid) is:
\[ \Phi_2 = B_1 A_1 N_2 \]
\[ \Phi_2 = \left(\frac{\mu_0 N_1 I_1}{l} \right) (\pi r_1^2) N_2 \]
\[ \Phi_2 = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} I_1 \]

- By definition, the mutual inductance \( M \) is:
\[ M = \frac{N_2 \Phi_2}{I_1} \]
\[ M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} \]

Thus, the mutual inductance of the two coaxial solenoids is:
\[ M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} \]

Key Observations:

- Mutual inductance depends on the inner solenoid’s area (\( r_1^2 \)), because the outer solenoid encloses all the flux produced by the inner one.

- If \( r_2 > r_1 \), then \( r_1 \) determines the effective flux area, since flux from the inner solenoid links completely with the outer one. Quick Tip: Mutual inductance between coaxial solenoids depends on the turns, radius of the inner solenoid, and permeability of free space.

Understanding Ferromagnetic Materials and Ferromagnetism


Definition of Ferromagnetic Materials:

- Ferromagnetic materials are substances that exhibit strong magnetization when placed in an external magnetic field.

- These materials retain their magnetization even after the external field is removed.

- Common examples include iron (Fe), cobalt (Co), and nickel (Ni).

Explanation of Ferromagnetism:

- The phenomenon of ferromagnetism arises due to the presence of magnetic domains.

- Each domain consists of a group of atomic dipoles aligned in the same direction.

- In an unmagnetized ferromagnetic material, these domains are randomly oriented, resulting in zero net magnetization.



Magnetic Domains and Alignment:

- When an external magnetic field is applied, the domains align in the direction of the field.

- The material becomes magnetized, and the alignment of the domains increases the overall magnetic moment.
\[ B = \mu_0 (H + M) \]

where:

- \( B \) = Magnetic field in the material

- \( H \) = Applied external field

- \( M \) = Magnetization of the material

- \( \mu_0 \) = Permeability of free space

- In fully magnetized ferromagnetic materials, most domains align in the same direction, producing a strong magnetic effect.

Key Properties of Ferromagnetic Materials:

1. High Permeability: These materials have a high ability to concentrate magnetic flux.

2. Hysteresis Effect: The magnetization does not return to zero immediately after removing the field, leading to a hysteresis loop.

3. Curie Temperature (\( T_C \)): Above this temperature, the material loses its ferromagnetic properties and behaves as a paramagnet.

Conclusion:

- Ferromagnetic materials are widely used in applications like transformers, electromagnets, hard disks, and electric motors due to their ability to retain strong magnetization.
Quick Tip: Ferromagnetism arises due to domain alignment. Materials like iron, cobalt, and nickel exhibit strong magnetic properties even after the external field is removed.

Power of a Combination of Lenses


Step 1: Understanding Power of Lenses

- The power \( P \) of a lens is given by:
\[ P = \frac{100}{f} \quad (in diopters, D) \]

where:
- \( f \) = Focal length in cm.

Step 2: Calculating the Power of Each Lens

- Converging lens (convex):
\[ P_1 = \frac{100}{20} = 5D \]

- Diverging lens (concave):
\[ P_2 = \frac{100}{-15} = -\frac{100}{15} = -\frac{20}{3}D \]

Step 3: Total Power of the Combination

Since the lenses are in contact, the net power is:
\[ P_{net} = P_1 + P_2 \]
\[ P_{net} = 5 - \frac{20}{3} \]
\[ P_{net} = \frac{15}{3} - \frac{20}{3} = -\frac{5}{3} D \]

Step 4: Conclusion

Thus, the power of the combination is:
\[ \boxed{-\frac{5}{3} D} \]

which matches option (B).
Quick Tip: The total power of lenses in contact is the algebraic sum of their individual powers: \( P_{total} = P_1 + P_2 \).

Using Lens Maker’s Formula to Find Refractive Index


Step 1: Lens Maker’s Formula

For a thin convex lens, the lens maker’s equation is:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

where:

- \( f = \frac{4}{3} R \) (given focal length),

- \( R_1 = R \) and \( R_2 = -2R \) (sign convention: convex surface is positive, concave is negative),

- \( \mu \) = Refractive index of the lens material.

Step 2: Substituting Given Values \[ \frac{3}{4R} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-2R} \right) \]
\[ \frac{3}{4R} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{2R} \right) \]
\[ \frac{3}{4R} = (\mu - 1) \times \frac{3}{2R} \]

Step 3: Solving for \( \mu \) \[ \mu - 1 = \frac{3}{4} \times \frac{2}{3} \]
\[ \mu - 1 = \frac{2}{4} = \frac{1}{2} \]
\[ \mu = 1 + \frac{1}{2} = \frac{3}{2} \]

Step 4: Conclusion

Thus, the refractive index of the material of the lens is:
\[ \boxed{\frac{3}{2}} \]

which matches option (C).
Quick Tip: The lens maker’s formula helps determine the focal length and refractive index of a lens using its curvature.

Effect of Surrounding Medium on Focal Length


- The focal length of a lens is given by the lens maker’s formula:
\[ \frac{1}{f} = (n_{lens} - n_{medium}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

where:

- \( n_{lens} \) = Refractive index of the lens material.

- \( n_{medium} \) = Refractive index of the surrounding medium.

- \( R_1, R_2 \) = Radii of curvature of the lens.

Effect of Dipping the Lens in Water

- When the lens is in air, \( n_{medium} = 1 \).

- When dipped in water, \( n_{medium} \approx 1.33 \).

- Since the focal length is inversely proportional to \( (n_{lens} - n_{medium}) \), as \( n_{medium} \) increases, the focal length also increases.

Thus, the correct answer is (A) The focal length increases when the lens is dipped in water.
Quick Tip: The focal length of a lens increases when placed in a medium with a refractive index closer to that of the lens material.

Step 1: Image Formation by the Convex Lens

- The light is incident parallel to the principal axis of the convex lens.

- The image formed by the convex lens will be real and located at a distance \( v_L \) from the lens. Using the lens formula:
\[ \frac{1}{f_L} = \frac{1}{v_L} - \frac{1}{u} \]

where:

- \( f_L = 10 \, cm \) (focal length of the convex lens),

- \( u = -\infty \) (object distance for a parallel ray),

- Thus, the image distance from the lens is \( v_L = 10 \, cm \) (real image formed at \( 10 \, cm \) on the opposite side of the light source).

Step 2: Image Formation by the Concave Mirror

- The image formed by the convex lens acts as the object for the concave mirror.

- The object distance for the concave mirror is the distance between the convex lens and the concave mirror, which is \( 40 \, cm - 10 \, cm = 30 \, cm \) (since the image formed by the lens is real).

- The focal length of the concave mirror is \( f_M = -15 \, cm \) (since it is a concave mirror, the focal length is negative).

- Using the mirror formula:
\[ \frac{1}{f_M} = \frac{1}{v_M} - \frac{1}{u_M} \]

where:

- \( u_M = -30 \, cm \) (object distance for the concave mirror),

- Solving for \( v_M \):
\[ \frac{1}{v_M} = \frac{1}{-15} - \frac{1}{-30} = \frac{-2 + 1}{10} = \frac{-1}{10} \]

Thus:
\[ v_M = 10 \, cm \]

The negative sign indicates that the image is formed on the same side as the object, which is on the left of the lens.

Conclusion:

The final image is formed at a distance of 10 cm to the left of the lens.

\boxed{(A) 10 cm, left of lens Quick Tip: A concave mirror inverts the image from a convex lens, forming the final image on the same side as the mirror.

Image Formation Using Two Convex Lenses


% Solution
Solution:


Step 1: Image Formation by Convex Lens \( L_1 \)
- The light is incident parallel to the principal axis of the convex lens \( L_1 \).
- The image formed by \( L_1 \) will be real and located at a distance \( v_1 \) from \( L_1 \). Using the lens formula:
\[ \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u} \]

where:
- \( f_1 = 16 \, cm \) (focal length of \( L_1 \)),
- \( u = -\infty \) (object distance for parallel rays),
- Thus, the image distance from \( L_1 \) is \( v_1 = 16 \, cm \) (real image formed at \( 16 \, cm \) on the opposite side of the light source).

Step 2: Image Formation by Convex Lens \( L_2 \)
- The image formed by \( L_1 \) now acts as the object for the second convex lens \( L_2 \).
- The object distance for \( L_2 \) is the distance between \( L_1 \) and \( L_2 \), which is \( 40 \, cm - 16 \, cm = 24 \, cm \).
- The focal length of \( L_2 \) is \( f_2 = 12 \, cm \).
- Using the lens formula for \( L_2 \):
\[ \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2} \]

where:
- \( u_2 = -24 \, cm \) (object distance for \( L_2 \)),
- Solving for \( v_2 \):
\[ \frac{1}{v_2} = \frac{1}{12} - \frac{1}{-24} = \frac{1}{12} + \frac{1}{24} = \frac{2 + 1}{24} = \frac{3}{24} \]

Thus:
\[ v_2 = \frac{24}{3} = 8 \, cm \]

Step 3: Interpretation of the Result
Since \( v_2 \) is positive, the image formed by \( L_2 \) is virtual and located at a distance of 8 cm from \( L_2 \), on the same side as the object.

Conclusion:
The final image is virtual and formed at a distance of 12 cm from \( L_2 \).

\boxed{(B) Virtual, 12 cm Quick Tip: When a convex lens forms an image at its focal point, it acts as an object for the next lens, and the final image position is determined using the lens equation.

Doping and Energy Levels in Ge


- Germanium (Ge) is a group IV element, and when doped with a pentavalent impurity (donor), extra free electrons are introduced into the conduction band.

- The energy required to free these donor electrons is very small compared to the band gap of Ge.

- In Ge, this ionization energy is approximately 0.01 eV.

Thus, the correct answer is 0.01 eV.
Quick Tip: In n-type semiconductors, donor electrons require a small energy (~0.01 eV in Ge) to jump into the conduction band.

Carrier Concentration Calculation


- The relation between electron and hole concentrations in a semiconductor is given by:
\[ n_e \cdot n_h = n_i^2 \]

where: \( n_i = 2.0 \times 10^{10} \) cm\(^{-3} \) (intrinsic carrier concentration), \( n_h = 8 \times 10^3 \) cm\(^{-3} \) (hole concentration after doping), \( n_e \) = electron concentration after doping.

Solving for \( n_e \):
\[ n_e = \frac{n_i^2}{n_h} = \frac{(2.0 \times 10^{10})^2}{8 \times 10^3} \]
\[ n_e = \frac{4.0 \times 10^{20}}{8 \times 10^3} = 5 \times 10^{16} cm^{-3} \]

Converting to m\(^{-3}\):
\[ n_e = 5 \times 10^{22} m^{-3} \] Quick Tip: In a doped semiconductor, the product of electron and hole concentrations remains equal to \( n_i^2 \).

Formation of Depletion Region


- In a p-n junction, the n-region has excess electrons, and the p-region has excess holes.

- Due to the concentration difference, electrons diffuse from the n-region to the p-region.

- Similarly, holes diffuse from the p-region to the n-region.
- This diffusion leads to the formation of a depletion region, preventing further diffusion.

Thus, the correct answer is Electrons diffuse from n-region into p-region and holes diffuse from p-region into n-region.
Quick Tip: A p-n junction forms due to the diffusion of electrons from the n-side and holes from the p-side, leading to a depletion region.

Understanding Current Flow in p-n Junction Formation


- In a newly formed p-n junction, a concentration gradient exists between the p-region and n-region.

- This causes majority charge carriers (electrons from n-region and holes from p-region) to diffuse across the junction, leading to a large diffusion current.

- As electrons and holes diffuse, a depletion region forms, creating an internal electric field.

- This electric field generates a small drift current by moving minority carriers in the opposite direction.

- Over time, the diffusion current and drift current balance each other at equilibrium.

Thus, during the initial formation of the p-n junction, the diffusion current is large, and the drift current is small.
Quick Tip: In a p-n junction, diffusion current dominates initially, but as the depletion region builds up, drift current increases to balance it.

Understanding Rectifier Frequency Response


- The given AC voltage is:
\[ V = 0.5 \sin(100\pi t) \]

- The general form of an AC signal is:
\[ V = V_0 \sin(2\pi f t) \]

Comparing, we get:
\[ 2\pi f = 100\pi \]
\[ f = \frac{100\pi}{2\pi} = 50 Hz \]

Half-Wave Rectifier Output Frequency:
- A half-wave rectifier allows only one half-cycle of the input AC voltage.
- The output voltage still follows the same fundamental frequency as the input, i.e., 50 Hz.

Full-Wave Rectifier Output Frequency:
- A full-wave rectifier inverts the negative half-cycles, making the output frequency twice the input frequency.
- Thus, the frequency of the output becomes:
\[ 2 \times 50 = 100 Hz \]

Thus, the correct answer is 50 Hz for the half-wave rectifier and 100 Hz for the full-wave rectifier.
Quick Tip: A half-wave rectifier retains the same frequency as the input AC, whereas a full-wave rectifier doubles the frequency.

N/A

N/A

Energy Losses in a Transformer


Two Major Energy Losses:

1. Eddy Current Losses:

- Changing magnetic flux induces eddy currents in the core, causing unwanted heating.

- Reduced by using laminated iron cores.

2. Hysteresis Losses:

- Energy is lost due to repeated magnetization and demagnetization of the core material.

- Minimized by using soft iron cores with low hysteresis. Quick Tip: A step-up transformer increases voltage by increasing the turns ratio. To reduce energy loss, laminated cores and high-efficiency materials are used.

Capacitance of a Parallel Plate Capacitor with a Dielectric Slab

Consider a parallel plate capacitor with plate area \(A\) and separation \(d\). A dielectric slab of thickness \(t\) (where \(t < d\)) and dielectric constant \(K\) is inserted between the plates.

Electric Field

In the air gap (distance \(d - t\)): The electric field \(E_0\) is:
\begin{equation
E_0 = \frac{\sigma{\epsilon_0 = \frac{Q{A\epsilon_0
\end{equation
where:
\(\sigma\) is the surface charge density on the plates.
\(\epsilon_0\) is the permittivity of free space.
\(Q\) is the charge on the plates.


In the dielectric (thickness \(t\)): The electric field \(E\) is reduced by a factor of \(K\):
\begin{equation
E = \frac{E_0{K = \frac{Q{KA\epsilon_0
\end{equation

Potential Difference:

The potential difference (V) between the plates is the sum of the potential differences across the air gap and the dielectric:

\begin{align
V &= E_0 (d - t) + E t

V &= \frac{Q{A\epsilon_0 (d - t) + \frac{Q{KA\epsilon_0 t

V &= \frac{Q{A\epsilon_0 \left[ (d - t) + \frac{t{K \right]

V &= \frac{Q{A\epsilon_0 \left[ d - t + \frac{t{K \right]

V &= \frac{Q{A\epsilon_0 \left[ d - t\left(1 - \frac{1{K\right) \right]
\end{align

Capacitance:

Capacitance (C) is defined as \(C = \frac{Q}{V}\):

\begin{align
C &= \frac{Q{\frac{Q{A\epsilon_0 \left[ d - t\left(1 - \frac{1{K\right) \right]

C &= \frac{A\epsilon_0{ d - t\left(1 - \frac{1{K\right)

C &= \frac{A\epsilon_0{ d - t + \frac{t{K
\end{align

Therefore, the expression for the capacitance is:

\begin{equation
C = \frac{A\epsilon_0{ d - t + \frac{t{K \text{ or C = \frac{A\epsilon_0{ d - t\left(1 - \frac{1{K\right)
\end{equation Quick Tip: A dielectric increases the capacitance of a capacitor, but if it doesn't fully fill the gap, the system behaves as capacitors in series.

Finding Capacitance of the Capacitors


Step 1: Energy in a Capacitor
The energy stored in a capacitor is:
\[ U = \frac{1}{2} C V^2 \]

where:
- \( C \) = Capacitance,
- \( V \) = Voltage applied.

Given:

- Series combination energy: \( U_s = 40 \) mJ = \( 40 \times 10^{-3} \) J,

- Parallel combination energy: \( U_p = 250 \) mJ = \( 250 \times 10^{-3} \) J,

- Applied voltage: \( V = 100 \) V.

Step 2: Parallel Combination
For parallel capacitors:
\[ C_p = C_1 + C_2 \]
\[ U_p = \frac{1}{2} C_p V^2 \]
\[ 250 \times 10^{-3} = \frac{1}{2} (C_1 + C_2) (100)^2 \]
\[ 250 \times 10^{-3} = 5000 (C_1 + C_2) \]
\[ C_1 + C_2 = \frac{250 \times 10^{-3}}{5000} = 50 \times 10^{-6} = 50 \mu F \]

Step 3: Series Combination

For series capacitors:
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \]
\[ U_s = \frac{1}{2} C_s V^2 \]
\[ 40 \times 10^{-3} = \frac{1}{2} C_s (100)^2 \]
\[ 40 \times 10^{-3} = 5000 C_s \]
\[ C_s = \frac{40 \times 10^{-3}}{5000} = 8 \times 10^{-6} = 8 \mu F \]

Step 4: Solving for \( C_1 \) and \( C_2 \)
Using:
\[ C_s = \frac{C_1 C_2}{C_1 + C_2} \]
\[ 8 = \frac{C_1 C_2}{50} \]
\[ C_1 C_2 = 8 \times 50 = 400 \]

Solving the quadratic equation:
\[ x^2 - 50x + 400 = 0 \]

Using the quadratic formula:
\[ x = \frac{50 \pm \sqrt{2500 - 1600}}{2} \]
\[ x = \frac{50 \pm 30}{2} \]
\[ x = \frac{80}{2} = 40 \quad or \quad x = \frac{20}{2} = 10 \]

Thus, the capacitances are \( C_1 = 40 \) \muF and \( C_2 = 10 \) \muF.
Quick Tip: For capacitors in parallel: \( C_p = C_1 + C_2 \). For capacitors in series: \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \).

Derivation of Electric Field Due to an Infinite Plane Sheet




Step 1: Gauss’s Law Statement

According to Gauss’s Law:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0} \]

where:
- \( \oint \mathbf{E} \cdot d\mathbf{A} \) = Total electric flux,
- \( Q_{enc} \) = Enclosed charge,

- \( \epsilon_0 \) = Permittivity of free space.

Step 2: Choosing a Gaussian Surface

- Consider an infinite charged plane with surface charge density \( \sigma \).

- The charge is uniformly distributed over the plane.

- We use a Gaussian cylinder (pillbox) that extends equally on both sides of the plane.

Step 3: Applying Gauss’s Law

- The flux is perpendicular to the surface.

- The total flux through the two flat surfaces of the pillbox is:
\[ \oint \mathbf{E} \cdot d\mathbf{A} = E A + E A = 2E A \]

- The enclosed charge is:
\[ Q_{enc} = \sigma A \]

- Applying Gauss's Law:
\[ 2E A = \frac{\sigma A}{\epsilon_0} \]
\[ E = \frac{\sigma}{2\epsilon_0} \]

Thus, the electric field due to an infinite plane sheet is:
\[ E = \frac{\sigma}{2\epsilon_0} \] Quick Tip: The electric field due to an infinite sheet does not depend on the distance from the sheet, unlike a point charge or a line charge.

Flux and Enclosed Charge Calculation


Step 1: Understanding the Given Field
The electric field is:
\[ E = (5x^2 + 2) \hat{i} \]

- The electric field varies with \( x \) but is uniform along the \( y \) and \( z \)-directions.

- The cube has a side length of 10 cm = 0.1 m.

Step 2: Flux Calculation

Electric flux is given by:
\[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} \]

Since the field is in the \( x \)-direction, it only contributes to the flux through the two faces perpendicular to \( x \)-axis.

- At \( x = 0 \) (left face), the field is:
\[ E_{left} = (5(0)^2 + 2) = 2 N/C \]

- At \( x = 0.1 \) m (right face), the field is:
\[ E_{right} = (5(0.1)^2 + 2) = 2 + 5(0.01) = 2.05 N/C \]

Flux through the cube:
\[ \Phi_E = E_{right} A - E_{left} A \]
\[ = (2.05 - 2) (0.1 \times 0.1) \]
\[ = (0.05) (0.01) \]
\[ = 5 \times 10^{-4} Nm^2/C \]

Step 3: Charge Enclosed Using Gauss’s Law
From Gauss’s Law:
\[ \Phi_E = \frac{Q_{enc}}{\epsilon_0} \]
\[ Q_{enc} = \epsilon_0 \Phi_E \]
\[ = (8.85 \times 10^{-12}) (5 \times 10^{-4}) \]
\[ = 4.43 \times 10^{-15} C \]

Thus, the net charge enclosed by the cube is \( 4.43 \times 10^{-15} \) C.
Quick Tip: For a non-uniform field, calculate the flux difference through opposite faces to determine the net charge enclosed.

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Magnification Produced by the Objective


Step 1: Total Magnification of Compound Microscope

The total magnification \( M \) produced by the compound microscope is the product of the magnifications of the objective lens (\( M_o \)) and the eyepiece lens (\( M_e \)):
\[ M = M_o \times M_e \]

Given:

- \( M = 200 \) (total magnification),

- The focal length of the eyepiece is \( f_e = 2 cm \).

Step 2: Magnification Produced by the Eyepiece

The magnification produced by the eyepiece is:
\[ M_e = \frac{D}{f_e} \]

where \( D \) is the least distance of distinct vision,

typically taken as 25 cm for a normal human eye.
\[ M_e = \frac{25}{2} = 12.5 \]

Step 3: Magnification Produced by the Objective

Now, using the total magnification:
\[ M = M_o \times M_e \]
\[ 200 = M_o \times 12.5 \]
\[ M_o = \frac{200}{12.5} = 16 \]

Thus, the magnification produced by the objective lens is:
\[ \boxed{M_o = 16} \]

Step 4: Conclusion

The magnification produced by the objective lens is 16.
Quick Tip: In a compound microscope, the total magnification is the product of the magnifications produced by the objective lens and the eyepiece.

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We are given the positions of the bright and dark fringes in an interference pattern. The formula used to find the positions is based on the diffraction pattern equations.

For the 4th bright fringe, the position is given by: \[ x_4^{(bright)} = \frac{4\lambda D}{d} \]
where: \( x_4^{(bright)} \) is the position of the 4th bright fringe. \( \lambda \) is the wavelength of light. \( D \) is the distance to the screen. \( d \) is the distance between the slits.

For the 2nd dark fringe, the position is given by: \[ x_2^{(dark)} = \frac{3\lambda D}{2d} \]
where: \( x_2^{(dark)} \) is the position of the 2nd dark fringe.

We are asked to find the difference in positions between the 4th bright fringe and the 2nd dark fringe. Thus, the position difference is: \[ x_4^{(bright)} - x_2^{(dark)} = 5 \, mm \]
Now, we substitute the formulas for the bright and dark fringes: \[ \frac{4\lambda D}{d} - \frac{3\lambda D}{2d} = 5 \times 10^{-3} \, m \]
Simplifying the above equation: \[ \frac{4\lambda D}{d} - \frac{3\lambda D}{2d} = 5 \times 10^{-3} \]
Multiply through by \( d \) to eliminate the denominator: \[ 4\lambda D - \frac{3\lambda D}{2} = 5 \times 10^{-3} d \]
Simplify further: \[ \frac{8\lambda D}{2} - \frac{3\lambda D}{2} = 5 \times 10^{-3} d \] \[ \frac{5\lambda D}{2} = 5 \times 10^{-3} d \]
Finally, solve for \( \lambda \): \[ \lambda = 6 \times 10^{-6} \, m \]

Thus, the wavelength \( \lambda \) of the light is \( 6 \times 10^{-6} \, m \). Quick Tip: In Young’s double slit experiment, the wavelength of light can be calculated using the fringe separation formula, considering the distance between slits and the screen.