CBSE Class 12 Physics (Code 042) - 2024 Board Solved Question Paper

0°

45°

90°

180°

Official Verified Academic Outcome:

C: 90°

VidyaTrail Step-by-Step Pedagogical Working:

Electrostatic Torque Equation:
τ = p × E = p·E·sin(θ)

• For torque to be maximum, sin(θ) must achieve its peak value, which is 1.
• sin(θ) = 1 ==> θ = 90° (π/2 radians).

Physical Significance: When the dipole moment vector p is perpendicular to electric field lines E, restoring electrostatic couples exert maximum rotational turning torque.

VidyaTrail Examiner Insight:

💡 VidyaTrail Examiner Insight: Notice the difference: Torque is MAXIMUM at θ = 90° (τ = pE), but Potential Energy is MINIMUM at θ = 0° (U = -pE, stable equilibrium) and MAXIMUM at θ = 180° (U = +pE, unstable equilibrium).

CBSE Marking Guidelines Breakdown:

• 0.5 Mark: Stating torque formula τ = p E sin(θ)
• 0.5 Mark: Evaluating θ = 90° for maximum sine value

P = V_rms · I_rms

P = 0

P = V_rms · I_rms · sin(φ)

P = V_rms · I_rms · cos(φ)

Official Verified Academic Outcome:

D: P = V_rms · I_rms · cos(φ)

VidyaTrail Step-by-Step Pedagogical Working:

Instantaneous voltage and current equations in an AC circuit with phase difference φ:
v(t) = V₀ sin(ωt)
i(t) = I₀ sin(ωt - φ)

Average electrical power over one complete cycle T:
P_avg = (1/T) ∫ v(t) · i(t) dt
P_avg = (V₀ I₀ / 2) · cos(φ) = (V₀ / √2) · (I₀ / √2) · cos(φ)
P_avg = V_rms · I_rms · cos(φ)

where cos(φ) is termed the Power Factor.

VidyaTrail Examiner Insight:

💡 VidyaTrail Examiner Insight: In a purely inductive or purely capacitive circuit, φ = 90°, cos(90°) = 0, meaning average power consumed is ZERO (Wattless Current).

CBSE Marking Guidelines Breakdown:

• 1.0 Mark: Applying AC integration cycle average to deduce power factor equation.

The barrier height increases and depletion layer width increases.

The barrier height increases and depletion layer width decreases.

The barrier height and the depletion layer width both decrease.

The barrier height decreases and depletion layer width increases.

Official Verified Academic Outcome:

C: The barrier height and the depletion layer width both decrease.

VidyaTrail Step-by-Step Pedagogical Working:

Semiconductor Junction Dynamics under Forward Bias:
• The positive terminal of the external battery connects to the p-side and negative terminal to n-side.
• Applied external forward voltage opposes the built-in internal electric field (barrier field).
• Resultant effective barrier potential decreases from V₀ to (V₀ - V).
• Consequently, majority carriers overcome the reduced opposing barrier and push into the junction, significantly thinning the depletion layer width.

VidyaTrail Examiner Insight:

💡 VidyaTrail Examiner Insight: Under reverse bias, the external field aids the internal barrier, increasing both barrier potential (V₀ + V) and widening the depletion zone.

CBSE Marking Guidelines Breakdown:

• 1.0 Mark: Explaining depletion barrier reduction under opposing external electric bias.

Official Verified Academic Outcome:

EMF E = 6 V, Internal Resistance r = 0.5 Ω

VidyaTrail Step-by-Step Pedagogical Working:

Governing Terminal Voltage Equation for a Discharging Cell:
V = E - I·r ==> E = V + I·r

Setting up the simultaneous linear equations from given data:
• State 1: When I₁ = 2 A, V₁ = 5 V:
E = 5 + 2r ... (Equation 1)

• State 2: When I₂ = 4 A, V₂ = 4 V:
E = 4 + 4r ... (Equation 2)

Equating both expressions for EMF E:
5 + 2r = 4 + 4r
5 - 4 = 4r - 2r
1 = 2r ==> r = 1/2 = 0.5 Ω.

Substituting r = 0.5 Ω into Equation 1:
E = 5 + 2(0.5) = 5 + 1 = 6 V.

Conclusion: The electromotive force E = 6.0 V and cell internal resistance r = 0.5 Ω.

VidyaTrail Examiner Insight:

💡 VidyaTrail Examiner Insight: Notice that terminal potential difference V is strictly LESS than EMF E when discharging, but EQUALS E when open-circuited (I = 0).

CBSE Marking Guidelines Breakdown:

• 0.5 Mark: Setting up E = 5 + 2r
• 0.5 Mark: Setting up E = 4 + 4r
• 0.5 Mark: Solving for internal resistance r = 0.5 Ω
• 0.5 Mark: Solving for open-circuit EMF E = 6 V

Official Verified Academic Outcome:

Slit Width: a = 1.2 × 10⁻⁶ m (1.2 μm)

VidyaTrail Step-by-Step Pedagogical Working:

Diffraction Minima Condition:
For diffraction at a single slit of width 'a', the angular condition for minima is given by:
a · sin(θ) = n · λ

Given Parameters:
• Order of minimum: n = 1 (First minimum)
• Wavelength: λ = 600 nm = 600 × 10⁻⁹ m
• Angular diffraction angle: θ = 30° ==> sin(30°) = 0.5

Substituting values:
a · sin(30°) = 1 · (600 × 10⁻⁹ m)
a · (0.5) = 600 × 10⁻⁹ m
a = (600 × 10⁻⁹) / 0.5 = 1200 × 10⁻⁹ m = 1.2 × 10⁻⁶ m = 1.2 μm.

VidyaTrail Examiner Insight:

💡 VidyaTrail Examiner Insight: Beware the reversal! In Young's double-slit interference, d sin θ = nλ gives MAXIMA, but in single-slit diffraction, a sin θ = nλ gives MINIMA.

CBSE Marking Guidelines Breakdown:

• 0.5 Mark: Stating single-slit diffraction minima condition a sin θ = nλ
• 0.5 Mark: Setting n = 1 and sin(30°) = 0.5
• 1.0 Mark: Accurate numerical computation yielding a = 1.2 × 10⁻⁶ m

Official Verified Academic Outcome:

B_axis = [μ₀·I·R²] / [2(x² + R²)^(3/2)], B_center = μ₀·I / (2R)

VidyaTrail Step-by-Step Pedagogical Working:

Derivation using Biot-Savart Law:
1. Consider a circular loop of radius R lying in the y-z plane carrying steady current I. Let P be an axial point on the x-axis at distance x from center O.

2. Distance from any current element dl to point P is r = √(x² + R²).
Perpendicularity: Current element dl is perpendicular to position vector r, so |dl × r| = r · dl.

3. Elementary Field dB via Biot-Savart Law:
dB = (μ₀ / 4π) · [I · dl / r²] = (μ₀ / 4π) · [I · dl / (x² + R²)]

4. Resolution of Components:
• Resolving dB into axial component (dB_x = dB · cos θ) and perpendicular component (dB_perp = dB · sin θ).
• By symmetry of diametrically opposite elements, all perpendicular components cancel identically.
• Axial field: cos θ = R / r = R / √(x² + R²).

dB_x = dB · cos θ = (μ₀ I dl / 4π(x² + R²)) · (R / (x² + R²)^(1/2))
dB_x = [μ₀ I R / 4π(x² + R²)^(3/2)] · dl

5. Integration around full circumference (∫ dl = 2πR):
B_net = [μ₀ I R / 4π(x² + R²)^(3/2)] · (2πR)
B_axis = [μ₀ I R²] / [2(x² + R²)^(3/2)] î (directed along the axis).

Special Case: At the Center of the Loop (x = 0):
B_center = [μ₀ I R²] / [2(0 + R²)^(3/2)] = [μ₀ I R²] / [2R³] = μ₀ I / (2R) î.

VidyaTrail Examiner Insight:

💡 VidyaTrail Examiner Insight: Always draw a neat diagram indicating dl, distance r = √(x² + R²), and resolving dB into dB cos θ and dB sin θ. Missing the cancellation argument loses 1 full mark.

CBSE Marking Guidelines Breakdown:

• 0.5 Mark: Biot-Savart differential formulation dB = (μ₀/4π)(I dl / r²)
• 0.5 Mark: Component resolution and cancellation of transverse elements
• 1.0 Mark: Integration around circumference 2πR yielding B_axis
• 1.0 Mark: Evaluating limit x = 0 to establish B_center = μ₀ I / (2R)