Paper I — Q5
(a) Given that the electric potential of a system of charges is V = 12/r² + 1/r³ volt. Calculate the electric field vector at the…
Given that the electric potential of a system of charges is V = 12/r² + 1/r³ volt. Calculate the electric field vector at the Cartesian point (4, 2, 3) m. 10 marks
Eight indistinguishable balls are to be arranged in six distinguishable boxes. Calculate the total number of ways in which the above can be done. 10 marks
A rod of length l is perpendicular to a uniform magnetic field B. The rod revolves at an angular speed ω about an axis passing through one end of the rod and parallel to the magnetic field B. Find the voltage induced across the rod's ends. 10 marks
Calculate the critical constants for CO₂ for which the Van der Waals constants are given by a = 0·0072 and b = 0·002. Also calculate the Boyle's temperature of CO₂. The unit of pressure is atmosphere and the unit of volume is that of a gm-mole of the gas at NTP. 10 marks
Consider the two branch parallel circuit shown in the diagram. Determine the resonant frequency of the circuit. 10 marks
हिंदी में प्रश्न पढ़ें
एक आवेशों के तंत्र का वैद्युत विभव V = 12/r² + 1/r³ वोल्ट से दिया गया है । कार्तीय निर्देशांक (4, 2, 3) m पर स्थित बिंदु के लिए वैद्युत क्षेत्र सदिश की गणना कीजिए । (10 अंक)
आठ अविभेदित गेंदों को छः विभेदित डिब्बों में रखना है । इस बात की गणना कीजिए कि उपर्युक्त कार्य को कुल कितनी विधियों से कर सकते हैं । (10 अंक)
l लंबाई की एक छड़ एकसमान चुंबकीय क्षेत्र B के लंबवत है । यह छड़ अपने एक सिरे से गुजरते हुए तथा चुंबकीय क्षेत्र B के समांतर अक्ष के चारों तरफ कोणीय चाल ω से घूर्णन कर रही है । छड़ के सिरों के आर-पार प्रेरित वोल्टता ज्ञात कीजिए । (10 अंक)
CO₂ के क्रांतिक नियतांक की गणना कीजिए जिसके लिए वान्डर वाल्स नियतांक a = 0·0072 और b = 0·002 दिया गया है । CO₂ के लिए बॉयल ताप की भी गणना कीजिए । दाब का मात्रक (इकाई) वायुमंडल (atmosphere) और आयतन का मात्रक (इकाई) NTP पर गैस के एक gm-mole के समान है । (10 अंक)
चित्र में दिए गए द्वि-शाखी समांतर परिपथ के लिए अनुनाद आवृत्ति की गणना कीजिए । (10 अंक)
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(e) A two-branch parallel circuit connected across two open input terminals on the left (an upper terminal line and a lower terminal line). Connected in parallel across the two lines are two vertical branches: the first branch consists of a resistor of resistance R_L connected in series above an inductor of inductance L; the second branch consists of a resistor of resistance R_C connected in series above a capacitor of capacitance C.
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) Electric field from the potential V = 12/r² + 1/r³
Since V depends only on r, E = −∇V = −(dV/dr) r̂, with dV/dr = −24/r³ − 3/r⁴. So
E = (24/r³ + 3/r⁴) r̂ = (24/r⁴ + 3/r⁵) r⃗.
At (4, 2, 3) m, r² = 29, r = 5.385 m, r⁴ = 841 and r⁵ = 4529. The factor is 24/841 + 3/4529 = 0.02920 V/m². Therefore
E⃗ = 0.0292(4 î + 2 ĵ + 3 k̂) = (0.117 î + 0.058 ĵ + 0.088 k̂) V/m,
with magnitude |E| = 24/r³ + 3/r⁴ = 0.157 V/m, directed radially outward.
(b) Eight identical balls in six distinct boxes
Only the number of balls in each box matters, so we count the non-negative integer solutions of x₁ + x₂ + … + x₆ = 8. By the stars-and-bars method the number is
C(8 + 6 − 1, 6 − 1) = C(13, 5) = (13 × 12 × 11 × 10 × 9)/5! = 1287.
(This assumes empty boxes are allowed. If every box had to hold at least one ball the count would be C(7, 5) = 21.)
(c) Emf across a rod rotating in a magnetic field
Take a small element dx at distance x from the axis through one end. Its speed is v = ωx, perpendicular to both the rod and B, so the motional emf of the element is dε = (v × B)·dl = Bωx dx. Integrating along the rod,
ε = ∫₀ˡ Bωx dx = ½Bωl².
Check by Faraday's law: in time dt the rod sweeps area ½l²ω dt, so ε = B dA/dt = ½Bωl². The end farther from the axis is at the higher or lower potential according to the sense of rotation.
(d) Critical constants and Boyle temperature of CO₂
For one mole, (P + a/V²)(V − b) = RT. At the critical point (∂P/∂V)_T = 0 and (∂²P/∂V²)_T = 0, giving
V_c = 3b, P_c = a/(27b²), T_c = 8a/(27Rb), and T_B = a/(Rb) (so T_B = 27T_c/8).
Units: P in atm and V in molar volumes at NTP, so R = PV/T = 1/273 atm·(volume unit)/K.
V_c = 3 × 0.002 = 0.006 volume unit (0.006 × 22.4 L = 0.134 L). P_c = 0.0072/(27 × 4×10⁻⁶) = 66.7 atm. T_c = 8 × 0.0072 × 273/(27 × 0.002) = 291.2 K. T_B = 0.0072 × 273/0.002 = 982.8 K.
(e) Resonant frequency of the parallel R-L and R-C circuit
Branch 1 has Z₁ = R_L + jωL and branch 2 has Z₂ = R_C − j/(ωC). Parallel resonance means the admittance Y = 1/Z₁ + 1/Z₂ is real, so its imaginary part (susceptance) vanishes:
−ωL/(R_L² + ω²L²) + ωC/(1 + ω²C²R_C²) = 0.
Cross-multiplying: L(1 + ω²C²R_C²) = C(R_L² + ω²L²), so L − CR_L² = ω²LC(L − CR_C²), and
ω_r = √[(L − CR_L²)/(LC(L − CR_C²))], f_r = ω_r/2π.
Equivalently ω_r² = (1/LC)(R_L² − L/C)/(R_C² − L/C). If R_C = 0, then ω_r² = (1/LC)(1 − R_L²C/L), and for small resistances f_r ≈ 1/(2π√LC). A real resonance needs the quantity under the root to be positive.
How this answer will be evaluated
Framework: Principle > Setup and diagram > Derivation > Result and limiting case
What earns full marks: Complete derivation with units, correct final values, physical interpretation
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Calculate the electric field vector at (4, 2, 3) m from the potential V = 12/r² + 1/r³. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- E = -∇V with V depending only on r: E = (24/r⁴ + 3/r⁵) r⃗
- At r² = 29: factor ≈ 0.0292 V/m²
- E ≈ (0.117, 0.058, 0.088) V/m, magnitude 0.157 V/m
Loses marks
- Missing negative sign in E = -∇V (wrong direction)
- Using r instead of r² = 29 at the point
Earns more
- Correct derivative dV/dr
- Radial outward direction
Extra mark
- Gives both vector and magnitude
- (b) Calculate the number of ways to arrange eight indistinguishable balls in six distinguishable boxes. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Counts solutions of x₁+…+x₆ = 8
- Stars and bars: C(13,5)
- 1287 ways
Loses marks
- Treating balls as distinguishable (6⁸)
- Using C(13,8) with no reasoning
Earns more
- Notes empty boxes allowed
- Evaluates C(13,5) correctly
Extra mark
- Gives 21 for the no-empty-box variant
- (c) Find the voltage induced across the ends of a rod rotating about one end in a uniform field B. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Element speed v = ωx perpendicular to B
- emf = ∫Bωx dx
- ε = ½Bωl²
Loses marks
- Using full speed ωl for the whole rod
- Factor ½ missing
Earns more
- Faraday's law area-swept check
- Polarity depends on sense of rotation
Extra mark
- Notes axis parallel to B
- (d) Calculate the critical constants and Boyle temperature of CO₂ from a = 0.0072 and b = 0.002. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- V_c = 3b, P_c = a/27b², T_c = 8a/27Rb, T_B = a/Rb
- R = 1/273 atm·unit/K from NTP units
- V_c = 0.006, P_c = 66.7 atm, T_c = 291.2 K, T_B = 982.8 K
Loses marks
- Using R = 0.0821 with the given units
- Arithmetic error in T_c or T_B
Earns more
- Derivation from the critical conditions
- T_B = 27T_c/8 relation
Extra mark
- Converts V_c to litres
- (e) Determine the resonant frequency of the parallel circuit with branches R_L-L and R_C-C. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Branch impedances Z₁ = R_L + jωL, Z₂ = R_C - j/ωC
- Resonance when susceptance of Y = 1/Z₁ + 1/Z₂ vanishes
- ω_r = √[(L - CR_L²)/(LC(L - CR_C²))]
Loses marks
- Assuming f = 1/2π√LC without derivation
- Setting only the imaginary parts of the impedances to zero
Earns more
- Intermediate susceptance expression
- Special case R_C = 0 or small resistance giving 1/2π√LC
Extra mark
- Notes the root must be real for resonance
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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