Paper I — Q4
(a) Find out the value of atomic packing factor for the FCC crystal structure. Calculate the radius of an iridium (Ir) atom…
Find out the value of atomic packing factor for the FCC crystal structure.
Calculate the radius of an iridium (Ir) atom, given that Ir has an FCC crystal structure, a density of 22.4 g/cm³ and an atomic weight of 192.2 g/mol.
[Avogadro's number (Nₐ) = 6.022 × 10²³ atoms/mol] 15 marks
An inscribed circular hole is made in a triangular lamina with each side 'a'. Find the area moment of inertia of this lamina about one of the sides. 15 marks
Four masses A, B, C and D revolve at equal radii and are equally spaced along a shaft. The mass B weighs 6 kg. Masses C and D make angles of 90° and 240° respectively with B in the same direction.
Find the magnitude of the masses A, C and D and the angular position of A, if the system is in complete balance. 20 marks
हिंदी में प्रश्न पढ़ें
FCC क्रिस्टल संरचना के लिए परमाणु संकुलन गुणक का मान ज्ञात कीजिए।
एक ईरिडियम (Ir) परमाणु की त्रिज्या की गणना कीजिए। दिया गया है कि Ir की संरचना एक FCC क्रिस्टल संरचना है, घनत्व 22.4 g/cm³ तथा परमाणु भार 192.2 g/mol है।
[आवोगाद्रो की संख्या (Nₐ) = 6.022 × 10²³ परमाणु/मोल है]। (15 अंक)
प्रत्येक भुजा 'a' वाले एक तिकोने पटल पर एक वृत्तीय छिद्र अन्तर्वृत (इन्स्क्राइब्ड) है। इस पटल का एक भुजा के सापेक्ष क्षेत्रफलीय जड़त्व आघूर्ण ज्ञात कीजिए। (15 अंक)
चार द्रव्यमान A, B, C तथा D समान त्रिज्या पर घूर्णन करते हैं और एक शाफ्ट पर बराबर दूरी पर हैं। द्रव्यमान B का भार 6 kg है। द्रव्यमान B से द्रव्यमान C तथा D क्रमशः 90° व 240° का कोण एक ही दिशा में बनाते हैं।
यदि तंत्र पूर्णतया संतुलन में है, तो द्रव्यमान A, C तथा D का परिमाण और A की कोणीय स्थिति ज्ञात कीजिए। (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) APF of FCC and radius of iridium
In FCC the atoms touch along the face diagonal, so 4R = a√2 and a = 2√2R. The cell has 4 atoms. Hence
APF = 4(4/3)πR³/(2√2R)³ = 16πR³/(3 × 16√2R³) = π/(3√2) = 0.7405.
For iridium, ρ = nM/(N_A a³) with n = 4:
a³ = 4 × 192.2/(22.4 × 6.022 × 10²³) = 5.699 × 10⁻²³ cm³, so a = 3.848 × 10⁻⁸ cm.
R = a/(2√2) = 3.848 × 10⁻⁸/2.828 = 1.361 × 10⁻⁸ cm = 0.136 nm.
(b) Moment of inertia of the lamina
Let the lamina be an equilateral triangle of side a with an inscribed circular hole, and take the axis along one side. The height is h = √3a/2.
Triangle: the width at distance y from the side is a(1 − y/h), so
I_tri = ∫₀ʰ y² a(1 − y/h) dy = ah³/12 = √3a⁴/32.
Circle: the incircle has r = a√3/6, and its centre is at distance r from the side. By the parallel axis theorem,
I_hole = πr⁴/4 + πr² · r² = 5πr⁴/4 = 5πa⁴/576, since r⁴ = a⁴/144.
Lamina: I = I_tri − I_hole = √3a⁴/32 − 5πa⁴/576 = (18√3 − 5π)a⁴/576 ≈ 0.02686a⁴.
(c) Complete balance of four masses
Let the planes A, B, C, D be 1 unit apart, so that, from A, the positions are 0, 1, 2, 3. Take B = 6 kg at 0°, C at 90° and D at 240°. All radii are equal, so mr can be replaced by m.
Couples about plane A: m_B(1)∠0° + m_C(2)∠90° + m_D(3)∠240° = 0.
Horizontal: 6 − 1.5m_D = 0, so m_D = 4 kg.
Vertical: 2m_C − 3m_D(√3/2) = 0, so m_C = 3√3 = 5.196 kg.
Forces: m_A∠θ + 6∠0° + m_C∠90° + m_D∠240° = 0.
Horizontal: m_A cos θ + 6 − 2 = 0, so m_A cos θ = −4.
Vertical: m_A sin θ + 5.196 − 3.464 = 0, so m_A sin θ = −√3.
Hence m_A = √(16 + 3) = √19 = 4.36 kg and tan θ = √3/4, with both components negative, so θ = 180° + 23.4° = 203.4° from B, measured in the same sense as C and D.
m_A = 4.36 kg at 203.4°, m_C = 5.20 kg, m_D = 4 kg.
How this answer will be evaluated
What earns full marks: Rigorous derivation with all steps, units, and physical interpretation.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Find the APF of FCC and the radius of an iridium atom 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- APF = π/(3√2) = 0.74 from a = 2√2R and 4 atoms per cell
- a³ = 4M/(ρN_A) giving a = 3.848 × 10⁻⁸ cm
- R = a/(2√2) = 0.136 nm
Loses marks
- Using n = 2 or a = 4R/√3
- Mixing cm and nm units
Earns more
- Atoms touching along the face diagonal
- Correct use of n = 4 and unit consistency
Extra mark
- Mentions that FCC is a close-packed structure
- (b) Find the area moment of inertia of the triangular lamina with an inscribed circular hole about one side 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- I_tri = √3a⁴/32 about one side
- I_hole = 5πr⁴/4 with r = a√3/6
- I = (18√3 − 5π)a⁴/576
Loses marks
- Taking the circle's centre at the wrong height
- Adding instead of subtracting the hole
Earns more
- Parallel axis theorem applied with the correct distance r
- Integration for the triangle shown
Extra mark
- Gives the numerical coefficient 0.0269a⁴
- (c) Find masses A, C, D and the angular position of A for complete balance 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Couple balance giving m_D = 4 kg and m_C = 5.196 kg
- Force balance giving m_A = √19 = 4.36 kg
- Angular position of A ≈ 203.4° from B
Loses marks
- Balancing forces only
- Wrong quadrant for the angle of A
Earns more
- Couples taken about a reference plane with correct distances 0, 1, 2, 3
- Components resolved with correct signs
Extra mark
- Verifies closure of the force and couple polygons
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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