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Physics 2021 Paper I 50 marks Calculate

Paper I — Q8

(a) (i) The melting point of tin is 232°C, its latent heat of fusion is 14 cal/g and the specific heat of solid and molten tin…

(a)
(i)

The melting point of tin is 232°C, its latent heat of fusion is 14 cal/g and the specific heat of solid and molten tin are 0·055 and 0·064 cal/g °C respectively. Calculate the change in entropy when 1·0 gm of tin is heated from 100°C to 300°C.

(ii)

Calculate the efficiency of an engine having compression ratio 13·8 and expansion ratio 6 and working on diesel cycle. Given γ = 1·4. (10+5 marks)

(b)
(i)

Write the expression for the Fermi-Dirac distribution. Plot the Fermi-Dirac distribution at T = 0 and for T₁ > T₂ > 0. Now from the plot propose two alternative definitions of the Fermi level.

(ii)

Calculate the probability of an electron occupying an energy level 0·02 eV above the Fermi level at T = 300 K. (15+5 marks)

(c)

Given an infinite line charge of charge density 2 nCm⁻¹ parallel to the y-axis and passing through the point (3, 0, 4) m and an infinite sheet of charge of charge density 4 nCm⁻² parallel to the x-y plane and passing through the point (0, 0, 6) m. Calculate the electric field intensity at the point (10, 10, 10) m. Assume free space. 15 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

टिन का गलनांक 232°C है, इसके संगलन की गुप्त ऊष्मा 14 cal/g है और ठोस टिन और गलित टिन की विशिष्ट ऊष्मा क्रमशः 0·055 और 0·064 cal/g °C हैं । 1·0 gm टिन को 100°C से 300°C तक गर्म करने में एन्ट्रॉपी में हुए परिवर्तन की गणना कीजिए ।

(ii)

एक इंजन की दक्षता की गणना कीजिए जिसका संपीडन अनुपात 13·8 है तथा प्रसार अनुपात 6 है । यह इंजन डीजल साइकिल पर काम करता है । दिया गया है γ = 1·4. (10+5 अंक)

(b)
(i)

फर्मी-डिराक वितरण के लिए व्यंजक लिखिए । T = 0 और T₁ > T₂ > 0 के लिए फर्मी-डिराक वितरण को आरेखित कीजिए । इस आरेखण से फर्मी स्तर की दो विकल्पित परिभाषाएँ प्रस्तावित कीजिए ।

(ii)

T = 300 K पर एक इलेक्ट्रॉन को फर्मी स्तर से 0·02 eV ऊपर ऊर्जा स्तर पर पाए जाने की प्रायिकता की गणना कीजिए । (15+5 अंक)

(c)

y-अक्ष के समांतर तथा बिंदु (3, 0, 4) m से जाने वाले एक अपरिमित रेखीय आवेश का आवेश घनत्व 2 nCm⁻¹ है तथा x-y तल के समांतर एवं बिंदु (0, 0, 6) m से जाने वाले एक अपरिमित आवेश तल (शीट) का आवेश घनत्व 4 nCm⁻² है । बिंदु (10, 10, 10) m पर वैद्युत क्षेत्र की तीव्रता की गणना कीजिए । मुक्त आकाश की स्थिति मान लीजिए । (15 अंक)

Q8 of the 2021 UPSC Mains Physics Paper I, as printed
The question as printed in the 2021 Physics paper
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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)(i) Entropy change of 1.0 g of tin heated from 100 °C to 300 °C

Entropy is a state function, so evaluate it along a reversible path with three stages: heating the solid to the melting point, melting, and heating the liquid. T₁ = 373.15 K, T_m = 505.15 K, T₂ = 573.15 K.

ΔS₁ = m c_s ln(T_m/T₁) = 0.055 × ln(505.15/373.15) = 0.0167 cal/K ΔS₂ = mL/T_m = 14/505.15 = 0.0277 cal/K ΔS₃ = m c_l ln(T₂/T_m) = 0.064 × ln(573.15/505.15) = 0.0081 cal/K

ΔS = 0.0525 cal/K = 0.0525 × 4.184 = 0.22 J/K (constant pressure, melting reversible).

(a)(ii) Efficiency of a Diesel engine

Compression ratio r = V₁/V₂ = 13.8, expansion ratio V₄/V₃ = 6, so the cut-off ratio is ρ = V₃/V₂ = 13.8/6 = 2.3. The air-standard Diesel efficiency is

η = 1 − [1/(γr^(γ−1))]·(ρ^γ − 1)/(ρ − 1).

With γ = 1.4: r^0.4 = 2.857 and ρ^1.4 = 3.209, so

η = 1 − (3.209 − 1)/(1.4 × 2.857 × 1.3) = 1 − 0.425 = 0.575, i.e. about 57.5 %.

(b)(i) Fermi-Dirac distribution and the Fermi level

f(E) = 1/[exp((E − E_F)/k_BT) + 1], the probability that a state of energy E is occupied.

Plot (E on the horizontal axis, f on the vertical axis): at T = 0 the curve is a step, f = 1 for E < E_F and f = 0 for E > E_F. For T > 0 the step is rounded over a width of a few k_BT about E_F, and every curve passes through f = ½ at E = E_F. The T₁ curve (higher temperature) is more spread out than the T₂ curve: it lies above it for E > E_F and below it for E < E_F.

Diagram: Fermi-Dirac distribution at T = 0 and at two temperatures T1 > T2 > 0

Two definitions of the Fermi level from the plot: (1) the energy at which the occupation probability is ½, at any T > 0; (2) the energy that separates completely filled from completely empty states at T = 0 (the highest occupied level).

(b)(ii) Occupation probability 0.02 eV above E_F at 300 K

k_BT = 8.617×10⁻⁵ × 300 = 0.02585 eV, and E − E_F = 0.02 eV.

f = 1/[exp(0.02/0.02585) + 1] = 1/(2.168 + 1) = 0.316, about 31.6 %.

(c) Field of a line charge and a sheet at (10, 10, 10) m

Use superposition.

Line charge λ = 2 nC/m, parallel to y and through (3, 0, 4): the perpendicular displacement to P is (7, 0, 6) m, with d² = 85 m². Then

E_line = λ r/(2πε₀d²) = [2×10⁻⁹ × 1.7975×10¹⁰/85](7 î + 6 k̂) = 0.423(7 î + 6 k̂) = (2.96 î + 2.54 k̂) V/m.

Sheet σ = 4 nC/m² at z = 6 m: the point has z = 10 m, above the sheet, so the field points in +z with magnitude

E_sheet = σ/2ε₀ = 4×10⁻⁹/(2 × 8.854×10⁻¹²) = 225.9 V/m.

Total: E = 2.96 î + 0 ĵ + 228.4 k̂ V/m, with |E| ≈ 228.4 V/m.

How this answer will be evaluated

Framework: Principle > Setup > Derivation > Result

What earns full marks: Rigorous derivation, correct units, clear diagrams, and physical interpretation.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a)(i) Calculate the entropy change of 1.0 g of tin heated from 100 °C to 300 °C. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Three reversible stages: solid heating, melting, liquid heating
    • ΔS = mc_s ln(T_m/T₁) + mL/T_m + mc_l ln(T₂/T_m) in kelvin
    • ΔS ≈ 0.0525 cal/K ≈ 0.22 J/K

    Loses marks

    • Omitting the latent-heat term
    • Using Celsius in logarithms

    Earns more

    • Individual stage values 0.0167, 0.0277, 0.0081 cal/K
    • Converts to kelvin

    Extra mark

    • Converts calories to joules
  2. (a)(ii) Calculate the efficiency of a Diesel engine with compression ratio 13.8, expansion ratio 6 and γ = 1.4. 5 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Cut-off ratio ρ = 13.8/6 = 2.3
    • η = 1 - [1/(γr^(γ-1))](ρ^γ-1)/(ρ-1)
    • η ≈ 57.5 %

    Loses marks

    • Using the Otto formula
    • Using 6 as cut-off ratio

    Earns more

    • r^0.4 = 2.857 and ρ^1.4 = 3.209
    • Uses air-standard cycle

    Extra mark

    • Compares with Otto cycle of same compression ratio
  3. (b)(i) Write the Fermi-Dirac distribution, describe its plot at T = 0 and T₁ > T₂ > 0, and give two definitions of the Fermi level. 15 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • f(E) = 1/[exp((E-E_F)/k_BT)+1]
    • Step at T = 0; rounded curves through f = ½ at E_F; higher T more spread
    • Two definitions: f = ½ at T>0; boundary between filled and empty at T = 0

    Loses marks

    • Writing the Maxwell-Boltzmann form
    • Saying f(E_F) = 1 at T > 0

    Earns more

    • Width of rounding of order k_BT
    • Correct relative position of T₁ and T₂ curves

    Extra mark

    • Links E_F to chemical potential
  4. (b)(ii) Calculate the probability of occupation 0.02 eV above the Fermi level at 300 K. 5 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • k_BT = 0.02585 eV
    • f = 1/[exp(0.02/0.02585)+1]
    • f ≈ 0.316

    Loses marks

    • Mixing joules and eV
    • Giving f = 0.5

    Earns more

    • Converts T = 300 K with k_B = 8.617e-5 eV/K

    Extra mark

    • Notes f < 0.5 above the Fermi level
  5. (c) Calculate the electric field at (10, 10, 10) m due to the line charge and the sheet. 15 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Line charge: perpendicular displacement (7, 0, 6), d² = 85; E = λr/(2πε₀d²)
    • Sheet: E = σ/2ε₀ = 225.9 V/m along +z above the sheet
    • Total E ≈ (2.96, 0, 228.4) V/m

    Loses marks

    • Using the distance from the origin or including the y offset in d
    • Wrong direction for the sheet field

    Earns more

    • Uses superposition
    • Component of line field values 2.96 and 2.54 V/m

    Extra mark

    • Gives magnitude 228.4 V/m
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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