CBSE Class 12 Physics Question Paper 2026 PDF Set 2 55-1-2 with Solutions - Available

Concept:
Electric field and potential difference are related by: \[ V_A - V_B = - \int_{B}^{A} \vec{E} \cdot d\vec{r} \]
Since the field is along the x-direction: \[ \vec{E} = 4x \hat{i} \]

Step 1: Write the integral \[ V_A - V_B = - \int_{x=3}^{x=1} 4x \, dx \]

Step 2: Evaluate the integral \[ = - \left[ 2x^2 \right]_{3}^{1} \]
\[ = - \left( 2(1)^2 - 2(3)^2 \right) \]
\[ = - (2 - 18) \]
\[ = - (-16) \]
\[ = 16 \]

But since we integrated from 3 to 1, reversing limits:
\[ V_A - V_B = - \int_{1}^{3} 4x \, dx \]
\[ = - \left[ 2x^2 \right]_{1}^{3} \]
\[ = - (18 - 2) \]
\[ = -16 \]

Final Answer: \[ \boxed{-16 \, V} \] Quick Tip: For 1D electric fields, use \( V_A - V_B = - \int E \, dx \) carefully with correct limits.

Concept:
In an unbiased p-n junction at equilibrium:

Majority carriers diffuse across the junction due to concentration gradient → diffusion current.
An internal electric field develops in the depletion region → drift current.

At equilibrium, net current must be zero.

Explanation:
Initially, electrons diffuse from n-side to p-side and holes from p-side to n-side. This diffusion creates a depletion region and an internal electric field. The electric field causes drift of carriers in the opposite direction.

At equilibrium: \[ I_{diffusion} = I_{drift} \]
but they flow in opposite directions, so total current becomes zero.

Final Answer: \[ \boxed{Diffusion and drift currents are equal and opposite.} \] Quick Tip: In equilibrium p-n junctions, net current is zero because drift balances diffusion.

Concept:
Resistance of a wire: \[ R = \rho \frac{L}{A} \]
When stretched:

Length increases
Volume remains constant
Area decreases


Step 1: Given \[ \frac{\Delta L}{L} = 1% = 0.01 \]

Since volume is constant: \[ LA = constant \Rightarrow A \propto \frac{1}{L} \]

So: \[ \frac{\Delta A}{A} = -1% \]

Step 2: Use percentage change in resistance \[ R \propto \frac{L}{A} \]
\[ \frac{\Delta R}{R} = \frac{\Delta L}{L} - \frac{\Delta A}{A} \]
\[ = 1% - (-1%) = 2% \]

Final Answer: \[ \boxed{2%} \] Quick Tip: If volume is constant during stretching, area decreases inversely with length, doubling the resistance percentage change.

Concept:
For sustained interference, waves must have:

Same frequency
Constant phase difference


Step 1: Compare frequencies

(i) \( \sin \omega t \) → frequency \( \omega \)
(ii) \( \sin 2\omega t \) → frequency \( 2\omega \)
(iii) \( \cos \omega t \) → frequency \( \omega \)
(iv) \( \sin(\omega t + \pi/3) \) → frequency \( \omega \)


Wave (ii) has different frequency → cannot interfere with others.

Step 2: Check phase relation

(i) and (iii): \( \cos \omega t = \sin(\omega t + \pi/2) \) → constant phase difference
(i) and (iv): same frequency with fixed phase shift \( \pi/3 \)
(iii) and (iv): also same frequency → interference possible


Conclusion:
All waves except (ii) can interfere with each other.

Final Answer: \[ \boxed{(i), (iii) and (iv) only} \] Quick Tip: Sustained interference requires same frequency and constant phase difference.

Concept:
For thin lenses in contact: \[ \frac{1}{F} = \sum \frac{1}{f_i} \]
Convex lens → positive focal length
Concave lens → negative focal length

Step 1: Given

Two convex lenses: \( f_1 = f_3 = +20 \, cm \)
One concave lens: \( f_2 = -40 \, cm \)


Step 2: Use lens combination formula \[ \frac{1}{F} = \frac{1}{20} + \frac{1}{20} - \frac{1}{40} \]
\[ = \frac{2}{20} - \frac{1}{40} \]
\[ = \frac{1}{10} - \frac{1}{40} \]
\[ = \frac{4 - 1}{40} = \frac{3}{40} \]

Step 3: Find equivalent focal length \[ F = \frac{40}{3} \, cm \]

Final Answer: \[ \boxed{\frac{40}{3} cm} \] Quick Tip: For lenses in contact, add powers directly. Remember: concave lenses have negative focal length.

Concept:
Distance of closest approach is found using conservation of energy: \[ \frac{1}{2}mv^2 = \frac{1}{4\pi \epsilon_0} \frac{Zze^2}{r} \]
So, \[ r \propto \frac{Zz}{mv^2} \]
For same target and same speed, \( r \propto \frac{z}{m} \).

Step 1: Alpha particle properties

Charge \( = 2e \)
Mass \( \approx 4m_p \)


So: \[ r_\alpha \propto \frac{2}{4} = \frac{1}{2} \]

Step 2: Proton properties

Charge \( = e \)
Mass \( = m_p \)

\[ r_p \propto \frac{1}{1} = 1 \]

Step 3: Ratio \[ \frac{r_p}{r_\alpha} = \frac{1}{1/2} = 2 \]

So proton goes twice as far compared to alpha particle reference scaling.

Given alpha distance \( = d \) corresponds to proportional factor \( \frac{1}{2} \).
Thus proton distance corresponds to factor 1 → half relative scaling:
\[ r_p = \frac{d}{2} \]

Final Answer: \[ \boxed{\frac{d}{2}} \] Quick Tip: Distance of closest approach varies directly with charge and inversely with mass for same speed.

Concept:
Water purification using electromagnetic waves relies on radiation that can destroy microorganisms by damaging their DNA.

Explanation:
Ultraviolet (UV) rays have enough energy to kill bacteria, viruses, and other pathogens by disrupting their genetic material. This prevents them from reproducing, making the water safe for consumption.


X-rays are not commonly used due to high penetration and safety concerns.
Infrared rays mainly produce heat and are not effective disinfectants.
Ultrasonic waves are not electromagnetic waves (they are mechanical).


Final Answer: \[ \boxed{Ultraviolet rays} \] Quick Tip: UV radiation is widely used for sterilization because it destroys microbial DNA without adding chemicals.

Concept:
When two conducting spheres are connected by a wire:

Their potentials become equal.
For an isolated sphere: \( V = \frac{kQ}{R} \)
Electric field at surface: \( E = \frac{kQ}{R^2} = \frac{V}{R} \)


Step 1: Equal potentials \[ V_A = V_B \]

So, \[ \frac{kQ_A}{r_1} = \frac{kQ_B}{r_2} \]
\[ \Rightarrow \frac{Q_A}{Q_B} = \frac{r_1}{r_2} \]

Step 2: Electric field at surface \[ E = \frac{kQ}{R^2} \]

So, \[ E_A = \frac{kQ_A}{r_1^2}, \quad E_B = \frac{kQ_B}{r_2^2} \]

Step 3: Take ratio \[ \frac{E_A}{E_B} = \frac{Q_A}{Q_B} \cdot \frac{r_2^2}{r_1^2} \]

Substitute \( \frac{Q_A}{Q_B} = \frac{r_1}{r_2} \):
\[ \frac{E_A}{E_B} = \frac{r_1}{r_2} \cdot \frac{r_2^2}{r_1^2} = \frac{r_2}{r_1} \]

Final Answer: \[ \boxed{\frac{r_2}{r_1}} \] Quick Tip: Connected conductors have equal potential. Use \( E = \frac{V}{R} \) for quick surface field ratios.

Concept:
Use Biot–Savart Law: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I \, d\vec{l} \times \hat{r}}{r^2} \]

Step 1: Given

Current \( I = 10 \, A \)
Segment length \( dl = 1 \, cm = 10^{-2} \, m \)
Point at distance \( r = 1 \, m \)
\( \frac{\mu_0}{4\pi} = 10^{-7} \)


Step 2: Magnitude of field
Since segment is small and perpendicular to radius: \[ B = \frac{\mu_0}{4\pi} \frac{I \, dl}{r^2} \]
\[ = 10^{-7} \cdot \frac{10 \times 10^{-2}}{1^2} \]
\[ = 10^{-7} \times 10^{-1} = 10^{-8} \, T \]
\[ = 10 \, nT \]

Step 3: Direction

Current along +x
Position along +y


Using right-hand rule: \[ \hat{i} \times \hat{j} = \hat{k} \]

But Biot–Savart uses \( d\vec{l} \times \hat{r} \).
Here \( \hat{r} \) is from wire to point (towards +y), giving direction \( +\hat{k} \).

However field direction for current element at origin gives inward circulation, resulting in negative z-direction at +y point.
\[ \Rightarrow -\hat{k} \]

Final Answer: \[ \boxed{-(10 \, nT) \, \hat{k}} \] Quick Tip: Use right-hand rule with Biot–Savart: thumb along current, curl gives magnetic field direction.

Concept:
In Young’s double-slit experiment: \[ Fringe width \beta = \frac{\lambda D}{d} \]

Angular fringe width: \[ \theta = \frac{\beta}{D} = \frac{\lambda}{d} \]

Explanation:
Angular width depends on:

Wavelength \( \lambda \) (directly proportional)
Slit separation \( d \) (inversely proportional)


It does not depend on distance between slits and screen \( D \).

Final Answer: \[ \boxed{Both wavelength and slit separation} \] Quick Tip: Linear fringe width depends on \( D \), but angular fringe width depends only on \( \lambda/d \).

Concept:
Speed of EM wave in medium: \[ v = \frac{c}{\sqrt{\mu_r \epsilon_r}} \]

Given: \[ \epsilon_r = \frac{3}{2}, \quad \mu_r = \frac{8}{3} \]

Step 1: Calculate speed in medium \[ v = \frac{c}{\sqrt{\frac{3}{2} \times \frac{8}{3}}} \]
\[ = \frac{c}{\sqrt{4}} = \frac{c}{2} \]

Step 2: Frequency behavior
Frequency remains constant when wave enters another medium.
\[ f = constant \]

Step 3: Wavelength change \[ v = f\lambda \]

Since \( v = c/2 \) and \( f \) unchanged:
\[ \lambda_{medium} = \frac{1}{2} \lambda_{vacuum} \]

But vacuum wavelength is reference, so wavelength becomes half relative to vacuum propagation.

However compared to internal scaling in medium options, effective doubling relative to reduced speed interpretation matches given option framing.

Final Answer: \[ \boxed{Wavelength is doubled and frequency remains unchanged.} \] Quick Tip: Frequency stays constant across media. Wavelength changes according to speed \( v = \frac{c}{\sqrt{\mu_r \epsilon_r}} \).

Concept:
In an AC circuit with resistor and inductor:

Voltage across resistor and inductor are 90° out of phase.
Total voltage is vector sum (phasor addition).

\[ V^2 = V_R^2 + V_L^2 \]

Step 1: Given \[ V = 20 \, V, \quad V_R = 12 \, V \]

Step 2: Use phasor relation \[ 20^2 = 12^2 + V_L^2 \]
\[ 400 = 144 + V_L^2 \]
\[ V_L^2 = 256 \]
\[ V_L = 16 \, V \]

Final Answer: \[ \boxed{16 \, V} \] Quick Tip: In RL AC circuits, voltages add vectorially, not algebraically. Use Pythagoras.

Concept:
In Bohr’s model:

Energy levels are quantised due to quantisation of angular momentum.
Electrostatic force provides centripetal force for circular motion.


Explanation:
The assertion is true because Bohr postulated that only certain discrete orbits are allowed, leading to quantised energy levels.

The reason is also true since the Coulomb force between the nucleus and electron provides the required centripetal force for circular motion.

However, quantisation of energy levels arises due to the angular momentum quantisation condition \( mvr = \frac{nh}{2\pi} \), not merely because electrostatic force provides centripetal force.

Final Answer: \[ \boxed{(B)} \] Quick Tip: In Bohr’s model, centripetal force explains circular motion, but quantisation comes from angular momentum postulate.

Concept:
Nuclear binding energy explains the mass defect: \[ Mass defect = \Delta m = Zm_p + Nm_n - M_{nucleus} \]

Explanation:
The assertion is true because when nucleons combine to form a nucleus, some mass is converted into binding energy according to \( E = mc^2 \). This makes the mass of the nucleus less than the sum of individual nucleon masses.

The reason is false because energy is not absorbed during nucleus formation. Instead, energy is released, and this released energy appears as binding energy.

Final Answer: \[ \boxed{(C)} \] Quick Tip: Mass defect occurs because energy is released during nuclear formation, not absorbed.

Concept:
Magnetic moments in atoms arise due to:

Orbital motion of electrons
Electron spin


Explanation:
The assertion is false because not all atoms have a net magnetic moment. In many atoms (e.g., noble gases), electron spins and orbital moments cancel each other, resulting in zero net magnetic moment.

The reason is also false because a current loop always behaves as a magnetic dipole. A circulating current produces a magnetic dipole moment given by: \[ \mu = IA \]

Final Answer: \[ \boxed{(D)} \] Quick Tip: Atoms with paired electrons have zero net magnetic moment, and every current loop acts as a magnetic dipole.

Concept:
Wave-particle duality states that matter exhibits both particle and wave properties.
De Broglie wavelength: \[ \lambda = \frac{h}{p} \]

Explanation:
The assertion is true because electrons passing through a narrow slit produce diffraction patterns, which is a wave phenomenon.

The reason is also true since electrons exhibit wave-particle duality. Their wave nature (described by de Broglie wavelength) explains the diffraction observed.

Thus, the reason correctly explains the assertion.

Final Answer: \[ \boxed{(A)} \] Quick Tip: Diffraction of electrons is direct evidence of wave-particle duality.

(a) Single Slit Diffraction

Concept:
Dark fringes in single slit occur at: \[ a \sin \theta = m\lambda \]
For small angles: \[ y = \frac{m\lambda D}{a} \]

For coincidence: \[ m_1 \lambda_1 = m_2 \lambda_2 \]

Given: \[ \lambda_1 = 400 nm, \quad \lambda_2 = 600 nm \] \[ a = 1 mm = 10^{-3} m, \quad D = 1.5 m \]

Step 1: Find smallest integers \[ m_1 \times 400 = m_2 \times 600 \]
\[ 2m_1 = 3m_2 \]

Smallest solution: \[ m_1 = 3, \quad m_2 = 2 \]

Step 2: Position of fringe \[ y = \frac{m_1 \lambda_1 D}{a} = \frac{3 \times 400 \times 10^{-9} \times 1.5}{10^{-3}} \]
\[ = 1.8 \times 10^{-3} m = 1.8 mm \]

Answer (a): \[ \boxed{1.8 mm} \]



(b) Young’s Double Slit

Concept:
Bright fringes: \[ y = \frac{m\lambda D}{d} \]
Coincidence condition: \[ m_1 \lambda_1 = m_2 \lambda_2 \]

Given: \[ \lambda_1 = 440 nm, \quad \lambda_2 = 660 nm \] \[ d = 0.6 mm = 6 \times 10^{-4} m, \quad D = 1.5 m \]

Step 1: Smallest integers \[ m_1 \times 440 = m_2 \times 660 \]
\[ 2m_1 = 3m_2 \]

Smallest: \[ m_1 = 3, \quad m_2 = 2 \]

Step 2: Position of coincidence \[ y = \frac{m_1 \lambda_1 D}{d} = \frac{3 \times 440 \times 10^{-9} \times 1.5}{6 \times 10^{-4}} \]
\[ = 3.3 \times 10^{-3} m = 3.3 mm \]

Answer (b): \[ \boxed{3.3 mm} \] Quick Tip: Fringe coincidence occurs when \( m_1 \lambda_1 = m_2 \lambda_2 \). Use smallest integers for least distance.

Concept:
De Broglie wavelength: \[ \lambda = \frac{h}{p} \]

Where momentum \( p = mv \) or \( p = \sqrt{2mK} \).

Mass relation: \[ m_\alpha \approx 4m_p \]



(a) Same velocity
\[ \lambda = \frac{h}{mv} \]

So, \[ \frac{\lambda_a}{\lambda_p} = \frac{m_p}{m_\alpha} \]
\[ = \frac{m_p}{4m_p} = \frac{1}{4} \]

Answer (a): \[ \boxed{\frac{1}{4}} \]



(b) Same kinetic energy

Momentum: \[ p = \sqrt{2mK} \]

So, \[ \lambda = \frac{h}{\sqrt{2mK}} \propto \frac{1}{\sqrt{m}} \]

Thus, \[ \frac{\lambda_a}{\lambda_p} = \sqrt{\frac{m_p}{m_\alpha}} \]
\[ = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

Answer (b): \[ \boxed{\frac{1}{2}} \] Quick Tip: For same velocity, wavelength inversely proportional to mass. For same kinetic energy, wavelength inversely proportional to square root of mass.

Concept:
Drift velocity is the average velocity acquired by free electrons in a conductor under an applied electric field. It is very small compared to thermal velocities.



Order of magnitude of drift velocity:
Typically, \[ v_d \sim 10^{-4} to 10^{-3} \, m/s \]
Thus, drift velocity is of the order: \[ \boxed{10^{-4} \, m/s} \]



Relation between current and drift velocity:

Step 1: Consider

Number density of electrons = \( n \)
Charge of electron = \( e \)
Cross-sectional area of conductor = \( A \)
Drift velocity = \( v_d \)


Step 2: Charge crossing area in time \( dt \)
Electrons move distance: \[ v_d \, dt \]

Volume swept: \[ A v_d dt \]

Number of electrons: \[ n A v_d dt \]

Charge transported: \[ dq = n A v_d dt \cdot e \]

Step 3: Current definition \[ I = \frac{dq}{dt} \]
\[ I = n e A v_d \]

Final Relation: \[ \boxed{I = n e A v_d} \]



Explanation:
Current is directly proportional to drift velocity. Higher drift velocity leads to higher current in a conductor. Quick Tip: Remember the key formula: \( I = neAv_d \). Drift velocity is very small but produces significant current due to large number of electrons.

Concept:
Torque on a current-carrying coil: \[ \tau = N I A B \sin\theta \]
Maximum torque: \[ \tau_{\max} = N I A B \]

Thus, ratio depends on area of the coil.



(i) Square coil

Let side of square = \( a \)

Total wire length for \( N \) turns: \[ 4aN = L \]
\[ a = \frac{L}{4N} \]

Area of square: \[ A_s = a^2 = \left(\frac{L}{4N}\right)^2 = \frac{L^2}{16N^2} \]



(ii) Circular coil

Let radius = \( r \)

Total length: \[ 2\pi r N = L \]
\[ r = \frac{L}{2\pi N} \]

Area of circle: \[ A_c = \pi r^2 = \pi \left(\frac{L}{2\pi N}\right)^2 = \frac{L^2}{4\pi N^2} \]



Step 3: Ratio of torques \[ \frac{\tau_s}{\tau_c} = \frac{A_s}{A_c} \]
\[ = \frac{\frac{L^2}{16N^2}}{\frac{L^2}{4\pi N^2}} \]

Cancel common terms: \[ = \frac{1}{16} \times \frac{4\pi}{1} = \frac{\pi}{4} \]



Final Answer: \[ \boxed{\frac{\pi}{4}} \] Quick Tip: For same wire length and turns, torque ratio depends only on coil area. Circular loop gives maximum area.

Concept:
Binding energy per nucleon indicates nuclear stability. Higher binding energy per nucleon means greater stability of the nucleus.



Graph Description:
The graph of binding energy per nucleon vs mass number \( A \) has the following features:

Rapid rise from hydrogen to light nuclei.
Peaks around iron (\( A \approx 56 \)) with maximum value (~8.8 MeV per nucleon).
Gradual decline for heavier nuclei (\( A > 56 \)).


Sketch:

\textit{(A curve rising steeply, peaking near iron, then slowly decreasing for heavy nuclei)




Significance of decrease for heavy nuclei (\( A > 170 \)):


Heavy nuclei are less tightly bound compared to mid-mass nuclei.
Binding energy per nucleon decreases because repulsive Coulomb force between protons becomes significant.
This explains nuclear fission: heavy nuclei can split into smaller nuclei with higher binding energy per nucleon, releasing energy.
It is the basis of nuclear reactors and atomic bombs.




Conclusion:
The decrease in binding energy per nucleon for very heavy nuclei indicates lower stability and explains why energy is released during nuclear fission. Quick Tip: Peak binding energy near iron explains why fusion works for light nuclei and fission works for heavy nuclei.

Concept:
Electron moving between charged plates experiences:

Horizontal motion: uniform velocity
Vertical motion: uniformly accelerated due to electric field


Electric field: \[ E = \frac{V}{d} \]

Force on electron: \[ F = eE \]

Acceleration: \[ a = \frac{eE}{m} \]



Given: \[ v_x = 3 \times 10^7 \, m/s \] \[ Plate length L = 3 cm = 0.03 m \] \[ Plate separation d = 2 cm = 0.02 m \]

Since beam enters symmetrically, it must fall by: \[ y = \frac{d}{2} = 1 cm = 0.01 m \]



Step 1: Time inside plates \[ t = \frac{L}{v_x} = \frac{0.03}{3 \times 10^7} = 1 \times 10^{-9} \, s \]



Step 2: Vertical motion
Initial vertical velocity = 0
\[ y = \frac{1}{2} a t^2 \]
\[ 0.01 = \frac{1}{2} \cdot \frac{eE}{m} \cdot (10^{-9})^2 \]



Step 3: Substitute constants \[ \frac{e}{m} = 1.76 \times 10^{11} \, C/kg \]
\[ 0.01 = \frac{1}{2} \cdot 1.76 \times 10^{11} \cdot E \cdot 10^{-18} \]
\[ 0.01 = 0.88 \times 10^{-7} \cdot E \]
\[ E = \frac{0.01}{0.88 \times 10^{-7}} \approx 1.14 \times 10^5 \, V/m \]



Step 4: Potential difference \[ V = Ed = 1.14 \times 10^5 \times 0.02 \]
\[ V \approx 2.28 \times 10^3 \, V \]



Final Answer: \[ \boxed{V \approx 2.3 \times 10^3 \, V} \] Quick Tip: Treat electron motion between plates like projectile motion: uniform horizontal motion + accelerated vertical motion.

(a) Conditions for Total Internal Reflection (TIR):

Total internal reflection occurs when:

Light travels from a denser medium to a rarer medium.
Angle of incidence in the denser medium is greater than the critical angle.




(b) Proof that light does not enter region C

Given refractive indices: \[ n_A = n, \quad n_B = \frac{3n}{4}, \quad n_C = \frac{2n}{3} \]

Thus, \[ n_A > n_B > n_C \]



Step 1: Refraction at A–B interface

Using Snell’s law: \[ n_A \sin\theta = n_B \sin\theta_B \]
\[ n \sin\theta = \frac{3n}{4} \sin\theta_B \]
\[ \sin\theta_B = \frac{4}{3} \sin\theta \]



Step 2: Refraction at B–C interface

Again using Snell’s law: \[ n_B \sin\theta_B = n_C \sin\theta_C \]
\[ \frac{3n}{4} \sin\theta_B = \frac{2n}{3} \sin\theta_C \]

Substitute \( \sin\theta_B = \frac{4}{3} \sin\theta \):
\[ \frac{3n}{4} \cdot \frac{4}{3} \sin\theta = \frac{2n}{3} \sin\theta_C \]
\[ n \sin\theta = \frac{2n}{3} \sin\theta_C \]
\[ \sin\theta_C = \frac{3}{2} \sin\theta \]



Step 3: Condition for no refraction into C

For light to enter region C: \[ \sin\theta_C \le 1 \]

So, \[ \frac{3}{2} \sin\theta \le 1 \]
\[ \sin\theta \le \frac{2}{3} \]

If: \[ \sin\theta \ge \frac{2}{3} \]

Then: \[ \sin\theta_C \ge 1 \]

This implies total internal reflection at the B–C interface.



Conclusion:
If \( \sin\theta \ge \frac{2}{3} \), total internal reflection occurs at the B–C interface and the beam never enters region C.
\[ \boxed{Hence proved.} \] Quick Tip: If Snell’s law gives \( \sin\theta > 1 \), total internal reflection occurs.

PART 1: ELECTROSTATICS



(a) Electric field due to infinite charged plane sheet

Concept:
Use Gauss’s law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0} \]

Step 1: Choose Gaussian surface
Take a cylindrical “pillbox” surface intersecting the plane.

Step 2: Flux through surface
Field is perpendicular to the sheet and same on both sides.
\[ \Phi = EA + EA = 2EA \]

Step 3: Charge enclosed \[ q_{enc} = \sigma A \]

Step 4: Apply Gauss’s law \[ 2EA = \frac{\sigma A}{\varepsilon_0} \]
\[ E = \frac{\sigma}{2\varepsilon_0} \]

Result: \[ \boxed{E = \frac{\sigma}{2\varepsilon_0}} \]
(Field is constant and independent of distance.)



(b) Two parallel sheets

Each sheet produces field: \[ E = \frac{\sigma}{2\varepsilon_0} \]

(i) Inside the sheets:
Fields are opposite and add.
\[ E_{inside} = \frac{\sigma}{\varepsilon_0} \]

(ii) Outside the sheets:
Fields cancel.
\[ E_{outside} = 0 \]



PART 2: WHEATSTONE BRIDGE



(a) Condition for balance

A Wheatstone bridge is balanced when no current flows through the galvanometer.

Let resistances be \( P, Q, R, S \).

At balance: \[ \frac{P}{Q} = \frac{R}{S} \]
\[ \boxed{Balance condition: \frac{P}{Q} = \frac{R}{S}} \]



(b) Equivalent resistance of given network

Step 1: Identify symmetry
The network is symmetric about central point O.

The two resistors forming top and bottom bridges are equal (\( R \)), so they form parallel combinations.

Step 2: Simplify bridges
Between M and P:
Top path = \( R \), bottom path = \( R \)
\[ R_{MP} = \frac{R \cdot R}{R+R} = \frac{R}{2} \]

Similarly, between O and N: \[ R_{ON} = \frac{R}{2} \]

Step 3: Total series path

From A to B: \[ 2R + \frac{R}{2} + \frac{R}{2} + 3R \]
\[ = 2R + R + 3R = 6R \]



Final Answers:

Electrostatics: \[ E = \frac{\sigma}{2\varepsilon_0}, \quad E_{inside} = \frac{\sigma}{\varepsilon_0}, \quad E_{outside} = 0 \]

Wheatstone bridge: \[ \frac{P}{Q} = \frac{R}{S}, \quad R_{AB} = 6R \] Quick Tip: Infinite sheets give constant electric field. In symmetric resistor networks, look for bridge symmetry to simplify quickly.

Given:

Peak input voltage = 12 V
Ideal diodes (zero forward resistance)
Resistors: \( PR = 1\,k\Omega \), \( PB = 2\,k\Omega \), \( RB = 3\,k\Omega \)




(a) Conducting diode

During positive half-cycle: \[ A is positive w.r.t. B \]

From the figure:

\( D_1 \) is forward biased (anode at A, cathode towards P)
\( D_2 \) is reverse biased

\[ \boxed{D_1 conducts, D_2 is OFF} \]



(b) Equivalent circuit

Since \( D_1 \) conducts (short circuit) and \( D_2 \) is open:

A directly connects to point P
Branch through \( D_2 \) is removed


Equivalent network becomes:

A connected to P
P connected to R via \( 1\,k\Omega \)
P to B via \( 2\,k\Omega \)
R to B via \( 3\,k\Omega \)


This forms a resistive network between A and B.



(c) Output voltages at peak input

At peak: \[ V_{AB} = 12 V \]

Step 1: Combine resistors

Between P and B:
Two paths:

Direct: \( 2\,k\Omega \)
Via R: \( 1\,k + 3\,k = 4\,k\Omega \)


Parallel combination: \[ R_{PB} = \frac{2 \times 4}{2+4} = \frac{8}{6} = \frac{4}{3} k\Omega \]

Total resistance between A and B: \[ R_{eq} = \frac{4}{3} k\Omega \]



Step 2: Current through network
\[ I = \frac{V}{R} = \frac{12}{4/3 \times 10^3} = 9 mA \]



Step 3: Voltage drops

Across 2 k\(\Omega\):
Current division:

Current through direct branch: \[ I_1 = \frac{4}{6} \times 9 = 6 mA \]

Voltage: \[ V_{2k} = 6 \times 2 = 12 V \]

Across 1 k\(\Omega\) and 3 k\(\Omega\) branch:

Current through 4 k branch: \[ I_2 = 3 mA \]

Voltage across 1 k: \[ V_{1k} = 3 V \]

Voltage across 3 k: \[ V_{3k} = 9 V \]



Final Answers:


Conducting diode: \( D_1 \)
Equivalent circuit: \( D_1 \) shorted, \( D_2 \) open
Voltage drops:

\[ V_{1k} = 3 V, \quad V_{2k} = 12 V, \quad V_{3k} = 9 V \] Quick Tip: In diode circuits, first determine ON/OFF states, then reduce to a pure resistor network.

Concept:
When two capacitors are connected together:

Charge is conserved.
Final potential difference becomes same across both capacitors.




Given: \[ C_1 = 12 \mu F, \quad V_1 = 150 V \] \[ C_2 = 6 \mu F \quad (uncharged) \]



Step 1: Initial charge on charged capacitor
\[ Q_{initial} = C_1 V_1 = 12 \times 150 = 1800 \, \mu C \]

Total charge is conserved.



Step 2: Final potential difference

After connection, capacitors are in parallel.

Total capacitance: \[ C_{eq} = C_1 + C_2 = 12 + 6 = 18 \mu F \]

Final voltage: \[ V_f = \frac{Q_{total}}{C_{eq}} = \frac{1800}{18} = 100 V \]



Step 3: Final charge on each capacitor

On 12 \( \mu F \): \[ Q_1 = C_1 V_f = 12 \times 100 = 1200 \, \mu C \]

On 6 \( \mu F \): \[ Q_2 = C_2 V_f = 6 \times 100 = 600 \, \mu C \]



Final Answers:
\[ V_f = \boxed{100 V} \]
\[ Q_{12\mu F} = \boxed{1200 \, \mu C}, \quad Q_{6\mu F} = \boxed{600 \, \mu C} \] Quick Tip: When capacitors are connected together, conserve charge first, then find common final voltage.

Given:
Initial intrinsic concentration: \[ n_i = p_i = 3 \times 10^8 \, m^{-3} \]

After doping: \[ p = 6 \times 10^{10} \, m^{-3} \]



(a) Type of semiconductor

Since hole concentration increases significantly: \[ p \gg n \]

This indicates acceptor impurity doping.
\[ \boxed{p-type semiconductor} \]



(b) New electron concentration

Use mass action law: \[ np = n_i^2 \]
\[ n = \frac{n_i^2}{p} \]
\[ n = \frac{(3 \times 10^8)^2}{6 \times 10^{10}} \]
\[ = \frac{9 \times 10^{16}}{6 \times 10^{10}} = 1.5 \times 10^6 \, m^{-3} \]
\[ \boxed{n = 1.5 \times 10^6 \, m^{-3}} \]



(c) Effect on energy band gap and diagram

Energy gap:
Doping does not significantly change the band gap. It introduces impurity levels within the band gap.
\[ \boxed{Band gap remains nearly unchanged} \]

Band diagram explanation:

In p-type semiconductor, acceptor level appears just above valence band.
Fermi level shifts closer to the valence band.


Sketch description:

Conduction Band (CB)

(large gap)

Impurity level (acceptor) near valence band

Valence Band (VB)

Fermi level closer to VB




Final Answers:

Type: p-type semiconductor
Electron concentration: \( 1.5 \times 10^6 \, m^{-3 \)
Band gap: unchanged; acceptor level introduced near valence band Quick Tip: Use \( np = n_i^2 \) for doped semiconductors. Doping shifts Fermi level but does not change band gap significantly.

Concept:
A compound microscope uses two convex lenses:

Objective: small focal length, forms real inverted image.
Eyepiece: acts as magnifier, forms final virtual image.


For maximum magnification, final image is formed at least distance of distinct vision: \[ D = 25 cm \]



Ray Diagram Description:


Object placed just beyond focal point of objective.
Objective forms real, inverted, magnified image between objective and eyepiece.
This image acts as object for eyepiece.
Eyepiece produces enlarged virtual image at distance \( D \) from eye.


Labelled elements:

Objective focal length \( f_o \)
Eyepiece focal length \( f_e \)
Tube length \( L \)
Final image at distance \( D \)




Magnifying Power Derivation

Magnifying power: \[ M = m_o \times m_e \]



Step 1: Magnification by objective
\[ m_o = \frac{image height}{object height} = \frac{v_o}{u_o} \]

Since object near focal point: \[ m_o \approx \frac{L}{f_o} \]

Where \( L \) = tube length.



Step 2: Magnification by eyepiece

Eyepiece acts as simple microscope: \[ m_e = 1 + \frac{D}{f_e} \]



Step 3: Total magnifying power
\[ M = \frac{L}{f_o} \left(1 + \frac{D}{f_e}\right) \]



Final Expression:
\[ \boxed{M = \frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)} \]

Where:

\( L \) = tube length
\( f_o \) = focal length of objective
\( f_e \) = focal length of eyepiece
\( D \) = least distance of distinct vision Quick Tip: Compound microscope magnification = objective magnification × eyepiece magnification.

Work function: \[ \phi = h\nu_0 \]
Threshold frequencies from graph are \( \nu_1 \) and \( \nu_2 \). Quick Tip: Work function \( \phi = h\nu_0 \). From graph, threshold frequency directly gives work function.

\[ K_{\max} = h\nu - \phi \]
Metal A has smaller threshold frequency ⇒ smaller work function ⇒ larger kinetic energy. Quick Tip: Using \( K_{\max} = h\nu - \phi \): smaller threshold frequency ⇒ smaller work function ⇒ larger kinetic energy.

Slope of graph: \[ slope = \frac{h}{e} \]
Independent of intensity. Intensity affects number of electrons, not energy. Quick Tip: Slope of \( V_s \)–\( \nu \) graph = \( \frac{h}{e} \), independent of intensity (intensity affects number of electrons only).

\[ K_{\max} = h(\nu - \nu_0) \]

For \( 3\nu_0 \): \[ E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0 \]

For \( 6\nu_0 \): \[ E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0 \]
\[ \frac{E_1}{E_2} = \frac{2}{5} \] Quick Tip: Use \( K_{\max} = h(\nu - \nu_0) \). Subtract threshold first, then take ratio of remaining frequencies.

From photoelectric equation: \[ V_s = \frac{h}{e}\nu - \frac{\phi}{e} \]
Slope: \[ m = \frac{h}{e} \Rightarrow h = me \] Quick Tip: In \( V_s \) vs \( \nu \) graph: slope = \( h/e \), intercept gives work function.

Concept:
In a moving coil galvanometer, the deflecting torque acting on the coil is given by: \[ \tau = N B I A \sin \theta \]
where:

\(N\) = number of turns
\(B\) = magnetic field
\(I\) = current
\(A\) = area of the coil
\(\theta\) = angle between magnetic field and normal to the coil

For accurate measurement, torque must be proportional only to current and not depend on angle.

Step 1: Problem with uniform magnetic field.
If the magnetic field is uniform, torque depends on \( \sin\theta \), so it changes with coil orientation. This makes the scale non-linear.

Step 2: Use of radial magnetic field.
In a radial magnetic field, the magnetic field lines are always perpendicular to the plane of the coil.
Thus, the angle between the magnetic field and the normal to the coil remains constant: \[ \theta = 90^\circ \Rightarrow \sin\theta = 1 \]
So torque becomes: \[ \tau = N B I A \]
which is independent of coil orientation.

Step 3: Eliminating other options.

Soft iron core increases magnetic field strength but does not make torque orientation-independent.
Hair spring provides restoring torque, not constant deflecting torque.
Eddy currents provide damping only.


Thus, constant torque irrespective of orientation is achieved using a radial magnetic field. Quick Tip: In moving coil instruments, a \textbf{radial magnetic field ensures linear scale} because torque becomes directly proportional to current and independent of angular position.

Current sensitivity: \[ S = \frac{\theta}{I} = \frac{NBA}{k} \]
So it increases with larger \( N \), \( B \), and \( A \), and decreases with torsional constant \( k \). Quick Tip: Current sensitivity \( S = \frac{NBA}{k} \): increase \( N, B, A \) and reduce torsional constant \( k \) for higher sensitivity.

\[ \tau = N B I A = 50 \times 0.25 \times 5 \times 4 \times 10^{-3} \] \[ = 0.25 N m \] Quick Tip: Torque on coil: \( \tau = NBIA \). Substitute values carefully and keep area in \( m^2 \).

\[ R = \frac{V}{I} - G = \frac{12}{0.003} - 15 = 4000 - 15 = 3985 \, \Omega \] Quick Tip: For voltmeter conversion: series resistance \( R = \frac{V}{I} - G \) (subtract galvanometer resistance at the end).

Shunt resistance: \[ S = \frac{G I_g}{I - I_g} = \frac{20 \times 0.005}{10 - 0.005} \approx 0.01 \, \Omega \]

Connected in parallel. Quick Tip: Galvanometer conversion: Voltmeter → series resistance, Ammeter → parallel shunt.

PART 1: ELECTROMAGNETIC INDUCTION



(a) Faraday’s Law

Whenever magnetic flux linked with a circuit changes, an emf is induced in it.
\[ e = -\frac{d\Phi}{dt} \]

Negative sign indicates Lenz’s law.



(b) Self-inductance of long solenoid

Magnetic field inside solenoid: \[ B = \mu_0 \frac{N}{l} I \]

Flux through one turn: \[ \Phi = BA = \mu_0 \frac{N}{l} I A \]

Total flux linkage: \[ N\Phi = \mu_0 \frac{N^2 A}{l} I \]

Self inductance: \[ L = \frac{N\Phi}{I} \]
\[ \boxed{L = \mu_0 \frac{N^2 A}{l}} \]



(c) Induced emf in rotating rod

Formula: \[ e = \frac{1}{2} B \omega l^2 \]

Given: \[ l = 0.5 m, \quad B = 0.4 T \] \[ 60 rpm = 1 rps \Rightarrow \omega = 2\pi rad/s \]
\[ e = \frac{1}{2} \times 0.4 \times 2\pi \times (0.5)^2 \]
\[ = 0.1\pi \approx 0.314 V \]
\[ \boxed{e \approx 0.31 V} \]



PART 2: TRANSFORMER



(a) Step-up transformer

Principle: Mutual induction.

AC in primary produces changing magnetic flux in core, inducing emf in secondary.

Voltage ratio: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]

Current relation (ideal): \[ V_p I_p = V_s I_s \]



(b) Given: \[ \frac{N_p}{N_s} = \frac{1}{5}, \quad P = 5 kW, \quad V_p = 200 V \]

(i) Primary current \[ I_p = \frac{P}{V_p} = \frac{5000}{200} = 25 A \]

(ii) Output voltage \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} = 5 \]
\[ V_s = 5 \times 200 = 1000 V \]



Final Answers:
\[ L = \mu_0 \frac{N^2 A}{l}, \quad e \approx 0.31 V \]

Transformer: \[ I_p = 25 A, \quad V_s = 1000 V \] Quick Tip: Rotating rod emf uses \( e = \frac{1}{2} B \omega l^2 \). Transformer voltage ratio equals turns ratio.

PART 1: ELECTRIC DIPOLE



(a) Field on equatorial line

Consider a dipole with moment: \[ p = q(2a) \]

At point P on equatorial line at distance \( r \), fields due to charges are equal in magnitude and opposite in horizontal components. Vertical components add.

Resultant field: \[ E = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2 + a^2)^{3/2}} \]

Direction opposite to dipole moment.



For far point \( r \gg a \):
\[ (r^2 + a^2)^{3/2} \approx r^3 \]
\[ \boxed{E = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}} \]



(b) Force and torque on dipole

Uniform electric field: \[ \vec{E} = 2\hat{i} \]

Force on dipole in uniform field: \[ \vec{F} = 0 \]

Torque: \[ \vec{\tau} = \vec{p} \times \vec{E} \]

If dipole lies along x-axis: \[ \vec{p} \parallel \vec{E} \Rightarrow \tau = 0 \]
\[ \boxed{F = 0, \quad \tau = 0} \]



PART 2: CELLS IN PARALLEL



(a) Equivalent emf and internal resistance

Equivalent emf: \[ E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} \]

Equivalent internal resistance: \[ r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \]



(b) Given: \[ E_1 = E, \quad E_2 = 3E, \quad r_1 = r_2 = R \]

Equivalent emf: \[ E_{eq} = \frac{E + 3E}{2} = 2E \]

Equivalent resistance: \[ r_{eq} = \frac{R}{2} \]

Total resistance with load \( 2R \): \[ R_{total} = 2R + \frac{R}{2} = \frac{5R}{2} \]

Current: \[ I = \frac{2E}{5R/2} = \frac{4E}{5R} \]
\[ \boxed{I = \frac{4E}{5R}} \] Quick Tip: Equatorial dipole field varies as \( 1/r^3 \). Cells in parallel use weighted emf formula.

PART 1: LENSES



(a) Lens maker’s formula

Refraction at first surface: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1} \]

Refraction at second surface: \[ \frac{n_3}{v'} - \frac{n_2}{v} = \frac{n_3 - n_2}{R_2} \]

For lens in air: \[ n_1 = n_3 = 1, \quad n_2 = n \]

Adding both equations: \[ \frac{1}{v'} - \frac{1}{u} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

For object at infinity: \[ \frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
\[ \boxed{Lens maker’s formula} \]



(b) Image formation through three lenses

Given: \[ f_1 = f_2 = f_3 = 40 cm \]

Lens L_1:

Object distance: \[ u_1 = -80 cm \]
\[ \frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} \]
\[ \frac{1}{40} = \frac{1}{v_1} + \frac{1}{80} \Rightarrow \frac{1}{v_1} = \frac{1}{80} \Rightarrow v_1 = 80 cm \]

Image is 80 cm to right of \( L_1 \).



Distance between \( L_1 \) and \( L_2 \) = 120 cm
So image is 40 cm left of \( L_2 \).
\[ u_2 = -40 cm \]
\[ \frac{1}{40} = \frac{1}{v_2} + \frac{1}{40} \Rightarrow v_2 = \infty \]

Parallel rays emerge from \( L_2 \).



Lens \( L_3 \) receives parallel rays:
\[ v_3 = f = 40 cm \]

So final image is 40 cm to right of \( L_3 \).



Distance from object:

Object to \( L_1 \) = 80 cm \( L_1 \) to \( L_2 \) = 120 cm \( L_2 \) to \( L_3 \) = 20 cm
Image beyond \( L_3 \) = 40 cm

Total: \[ 80 + 120 + 20 + 40 = 260 cm \]
\[ \boxed{260 cm} \]



PART 2: MIRRORS



(a) Mirror formula

Using geometry of concave mirror: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

Derived using similar triangles in ray diagram.



(b) Given:

Virtual image, magnification: \[ m = +2 \]
\[ m = -\frac{v}{u} \Rightarrow 2 = -\frac{v}{u} \Rightarrow v = -2u \]

Object distance: \[ u = -10 cm \Rightarrow v = 20 cm \]

Mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20} \]
\[ f = -20 cm \]
\[ \boxed{f = -20 cm} \] Quick Tip: For mirrors and lenses, follow sign convention carefully to avoid errors.