Paper I — Q7
(a) A region 1, z < 0, has a dielectric material with εᵣ = 3·2 and a region 2, z > 0 has a dielectric material with εᵣ = 2·0. Let…
A region 1, z < 0, has a dielectric material with εᵣ = 3·2 and a region 2, z > 0 has a dielectric material with εᵣ = 2·0. Let the displacement vector in the region 1 be, D⃗₁ = – 30 aₓ + 50 aᵧ + 70 aᵤ nCm⁻². Assume the interface charge density is zero. Find in the region 2, the D⃗₂ and P⃗₂, where P⃗₂ is the electric polarization vector in the region 2. 20 marks
Calculate the skin depth of electromagnetic waves of 1 MHz incident on a good conductor having σ = 5·8 × 10⁷ Sm⁻¹. Assume that inside the conductor μ = μ₀ = 4π × 10⁻⁷ Hm⁻¹. 10 marks
The spectral composition of solar radiation is similar to that of a black body radiator whose maximum emission corresponds to the wavelength 0·48 μm. Find the mass lost by the Sun every second due to radiation. Evaluate the time interval during which the mass of the Sun reduces by 1 per cent.
Given : Stefan Boltzmann constant = 5·669 × 10⁻⁸ W m⁻² K⁻⁴, radius of the Sun = 6·957 × 10⁸ m, surface temperature of the Sun = 5772 K and mass of the Sun is 1·9885 × 10³⁰ kg. 20 marks
हिंदी में प्रश्न पढ़ें
क्षेत्र 1, z < 0 परावैद्युत पदार्थ εᵣ = 3·2 का बना है और क्षेत्र 2, z > 0 परावैद्युत पदार्थ εᵣ = 2·0 का है । मान लीजिए कि विस्थापन सदिश क्षेत्र 1 में D⃗₁ = – 30 aₓ + 50 aᵧ + 70 aᵤ nCm⁻² है । मान लीजिए कि अंतरापृष्ठ आवेश घनत्व शून्य है । क्षेत्र 2 में D⃗₂ और P⃗₂ ज्ञात कीजिए, जहाँ P⃗₂ क्षेत्र 2 में वैद्युत ध्रुवण सदिश है । (20 अंक)
एक सुचालक, जिसका σ = 5·8 × 10⁷ Sm⁻¹ है, पर 1 MHz की विद्युत-चुंबकीय तरंगें आपतित होती हैं । इस सुचालक के लिए त्वचा गहराई की गणना कीजिए । मान लीजिए कि सुचालक के अंदर μ = μ₀ = 4π × 10⁻⁷ Hm⁻¹ है । (10 अंक)
सौर विकिरण का वर्णक्रम (स्पेक्ट्रल) संयोजन एक कृष्णिका विकिरक के समान है जिसके अधिकतम उत्सर्जन का तरंगदैर्ध्य 0·48 μm है । विकिरण के कारण सूर्य की द्रव्यमान क्षति प्रति सेकंड ज्ञात कीजिए । उस समय अंतराल की गणना कीजिए जिसमें सूर्य का द्रव्यमान 1% घट जाता है ।
दिया गया है : स्टीफन बोल्ट्ज़मान नियतांक = 5·669 × 10⁻⁸ W m⁻² K⁻⁴, सूर्य की त्रिज्या = 6·957 × 10⁸ m, सूर्य के पृष्ठ का ताप = 5772 K और सूर्य का द्रव्यमान 1·9885 × 10³⁰ kg है । (20 अंक)
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) D₂ and P₂ at a dielectric interface
The interface is the plane z = 0, so the normal is a_z. With zero free surface charge, the boundary conditions are: the normal component of D is continuous, and the tangential component of E is continuous.
Normal: D₂z = D₁z = 70 nC/m².
Tangential: E₂t = E₁t, i.e. D₂t/ε₂ = D₁t/ε₁, so D₂t = (ε_r2/ε_r1) D₁t = (2.0/3.2) D₁t = 0.625 D₁t.
D₂x = 0.625 × (−30) = −18.75 nC/m², D₂y = 0.625 × 50 = 31.25 nC/m².
D⃗₂ = (−18.75 a_x + 31.25 a_y + 70 a_z) nC/m².
Polarisation: from D = ε₀E + P and D = ε₀ε_r E, P = D(1 − 1/ε_r). In region 2, ε_r = 2.0, so P⃗₂ = D⃗₂/2:
P⃗₂ = (−9.375 a_x + 15.625 a_y + 35 a_z) nC/m².
(b) Skin depth of a good conductor
For a good conductor (σ ≫ ωε), the fields fall to 1/e of the surface value over the skin depth
δ = √(2/(ωμσ)) = 1/√(πfμσ).
Here f = 10⁶ Hz, μ = μ₀ = 4π×10⁻⁷ H/m and σ = 5.8×10⁷ S/m:
πfμσ = π × 10⁶ × 4π×10⁻⁷ × 5.8×10⁷ = 2.29×10⁸ m⁻², so δ = 1/√(2.29×10⁸) = 6.6×10⁻⁵ m ≈ 66 μm.
(The conducting condition holds, since σ/ωε₀ ≈ 10¹² ≫ 1.)
(c) Mass lost by the Sun per second and the time for a 1 % loss
Treating the Sun as a black body of radius R and temperature T = 5772 K, its radiated power is
P = 4πR²σT⁴ = 4π(6.957×10⁸)² × 5.669×10⁻⁸ × (5772)⁴ = 3.83×10²⁶ W.
By E = mc², the mass lost per second is
dm/dt = P/c² = 3.83×10²⁶/(3.0×10⁸)² ≈ 4.3×10⁹ kg/s.
(With c = 2.998×10⁸ m/s this is 4.26×10⁹ kg/s.)
A loss of 1 % means ΔM = 0.01 × 1.9885×10³⁰ = 1.99×10²⁸ kg. At constant luminosity,
t = ΔM/(dm/dt) = 1.99×10²⁸/4.26×10⁹ = 4.7×10¹⁸ s ≈ 1.5×10¹¹ years.
The given λ_max = 0.48 μm is not needed, because T is given; Wien's law with it would give about 6040 K, which differs from 5772 K because the Sun is not an exact black body.
How this answer will be evaluated
What earns full marks: Correct application of boundary conditions, formulas, and unit consistency with clear steps.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Find D₂ and P₂ in region 2 using the interface conditions. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Normal D continuous: D₂z = 70 nC/m²
- Tangential E continuous: D₂t = (ε_r2/ε_r1)D₁t = 0.625 D₁t
- D₂ = (-18.75, 31.25, 70) nC/m² and P₂ = D₂/2 = (-9.375, 15.625, 35) nC/m²
Loses marks
- Making tangential D continuous instead of E
- Using P = ε₀E with wrong ε_r
Earns more
- P = D(1 - 1/ε_r) derivation
- Zero surface charge assumption stated
Extra mark
- Identifies interface plane z = 0 and normal a_z
- (b) Calculate the skin depth in a good conductor at 1 MHz. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- δ = √(2/ωμσ) = 1/√(πfμσ)
- Substitution of f = 10⁶, μ₀, σ = 5.8×10⁷
- δ ≈ 6.6×10⁻⁵ m (66 μm)
Loses marks
- Using f in place of ω without 2π
- Unit slip
Earns more
- Checks good-conductor condition σ ≫ ωε
- Defines skin depth as fall to 1/e
Extra mark
- Notes that depth scales as 1/√f
- (c) Find the Sun's mass loss per second and the time for it to lose 1 % of its mass. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- P = 4πR²σT⁴ ≈ 3.83×10²⁶ W
- dm/dt = P/c² ≈ 4.3×10⁹ kg/s
- t = 0.01M/(dm/dt) ≈ 4.7×10¹⁸ s ≈ 1.5×10¹¹ years
Loses marks
- Using Wien-derived T without noting the given T
- Missing c² in mass loss
Earns more
- Uses given T = 5772 K
- Converts seconds to years
Extra mark
- Notes Wien's law with 0.48 μm gives about 6040 K, so λ_max is not needed
What "Calculate" is asking you to do
Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.
Structure that answers it
Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units
Where marks are lost
Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.
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