Paper I — Q2
(a) A body of mass m at rest splits into two masses m₁ and m₂ by an explosion. After the split the bodies move with a total…
A body of mass m at rest splits into two masses m₁ and m₂ by an explosion. After the split the bodies move with a total kinetic energy T in opposite direction. Show that their relative speed is √(2Tm/m₁m₂). 15 marks
A light rod of length 100 cm is suspended from the ceiling, horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and its cross-section is 0·05 sq. cm and the other is of brass of cross-section 0·1 sq. cm. Find the position along the rod at which a weight may be hung to produce
Equal stresses in both the wires,
Equal strain in both the wires.
Young's modulus of elasticity of brass and steel are 1·0 × 10¹¹ N/m² and 2·0 × 10¹¹ N/m² respectively. 15 marks
Show that the phenomenon of Fraunhofer diffraction at two vertical slits is modulation of two terms viz. double slit interference and single slit diffraction. Obtain the condition for positions of maxima and minima. 20 marks
हिंदी में प्रश्न पढ़ें
एक द्रव्यमान m का पिंड, जो कि स्थिर अवस्था में है, एक विस्फोट के दौरान दो भागों m₁ और m₂ द्रव्यमानों के दो पिंडों में विभाजित हो जाता है । विभाजन के बाद दोनों पिंड कुल गतिज ऊर्जा T के साथ विपरीत दिशाओं में गति करते हैं । दिखाइए कि दोनों की आपेक्षिक चाल √(2Tm/m₁m₂) है । (15 अंक)
100 cm लम्बाई के एक हल्के छड़ को दोनों किनारों पर समान लम्बाई के दो ऊर्ध्वाधर तारों से बाँधकर छत से क्षैतिज दिशा में लटकाया गया है । एक तार इस्पात से बना है जिसका अनुप्रस्थ-परिच्छेद 0·05 sq. cm है तथा दूसरा तार ब्रास (पीतल) का बना है जिसका अनुप्रस्थ-परिच्छेद 0·1 sq. cm है । छड़ पर उस स्थान को ज्ञात कीजिए जहाँ पर एक भार को लटकाया जा सके जिससे कि दोनों तारों में
बराबर प्रतिबल,
बराबर विकृति
उत्पन्न किया जा सके ।
ब्रास (पीतल) और इस्पात के यंग के प्रत्यास्थता गुणांक क्रमशः 1·0 × 10¹¹ N/m² और 2·0 × 10¹¹ N/m² हैं । (15 अंक)
दिखाइए कि दो उर्ध्वाधर छिद्रों से होने वाले फ्राउनहोफर विवर्तन की परिघटना दो पदों, जैसे कि द्वि-छिरी से व्यतिकरण और एकल छिरी से विवर्तन का मॉडुलन होता है । अधिकतम और न्यूनतम मान की स्थितियों की शर्त प्राप्त 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) Relative speed of the fragments after an explosion
The body is at rest, so there is no net momentum and no external impulse. Let the fragments m₁ and m₂ move in opposite directions with speeds v₁ and v₂, and m = m₁ + m₂. Conservation of momentum gives m₁v₁ = m₂v₂ = p.
The total kinetic energy is
T = p²/(2m₁) + p²/(2m₂) = p²m/(2m₁m₂), so p² = 2Tm₁m₂/m.
Since the fragments move apart, the relative speed is the sum of the speeds:
v_rel = v₁ + v₂ = p/m₁ + p/m₂ = p m/(m₁m₂).
Squaring and substituting p²:
v_rel² = (2Tm₁m₂/m)·m²/(m₁m₂)² = 2Tm/(m₁m₂)
so v_rel = √(2Tm/(m₁m₂)). The result is non-relativistic.
(b) Position of the load on the rod supported by steel and brass wires
Let the load W hang at distance x from the steel end, with L = 100 cm. Vertical equilibrium and moments about the steel end give T_b = Wx/L and T_s = W(L − x)/L.
Data: A_s = 0.05 cm², A_b = 0.1 cm², Y_s = 2.0×10¹¹ N/m², Y_b = 1.0×10¹¹ N/m².
(i) Equal stress: T_s/A_s = T_b/A_b, so (L − x)/A_s = x/A_b and x = L·A_b/(A_s + A_b) = 100 × 0.1/0.15 = 66.7 cm from the steel end (33.3 cm from the brass end).
(ii) Equal strain: stress/Y must be equal, so T_s/(A_sY_s) = T_b/(A_bY_b), hence (L − x)/(A_sY_s) = x/(A_bY_b). Here A_sY_s = 5×10⁻⁶ × 2×10¹¹ = 1.0×10⁶ N and A_bY_b = 10⁻⁵ × 10¹¹ = 1.0×10⁶ N. These are equal, so x = L/2 = 50 cm, the midpoint. The wires are assumed to be within the elastic limit.
(c) Fraunhofer diffraction at a double slit
Let each slit have width a and the centre-to-centre separation be d. For a direction θ, the field of one slit is the sum of contributions across its width:
A₁ = A₀ ∫ e^(i(2π/λ)x sinθ) dx over (−a/2, a/2) = A₀ a (sinβ/β), where β = πa sinθ/λ.
The second slit is displaced by d, so it has an extra phase δ = 2πd sinθ/λ. The resultant is
A = A₁(1 + e^(iδ)) = 2A₁cos(δ/2)·e^(iδ/2).
Intensity: I = I₀ (sinβ/β)² cos²(δ/2).
The first factor is the single-slit diffraction envelope and the second is the double-slit interference term, so the interference fringes are modulated by the diffraction pattern.
Maxima: the interference maxima (principal) occur at cos²(δ/2) = 1, that is d sinθ = nλ, n = 0, ±1, ±2, …, with intensity limited by the envelope. For a screen at distance D, y_n = nλD/d.
Minima: (1) interference minima, cos(δ/2) = 0, give d sinθ = (n + ½)λ. (2) Diffraction minima, sinβ = 0 with β ≠ 0, give a sinθ = mλ, m = ±1, ±2, …
If an interference maximum coincides with a diffraction minimum (d sinθ = nλ and a sinθ = mλ), that order is missing; this happens when d/a = n/m.
How this answer will be evaluated
Framework: Principle > Setup and diagram > Derivation > Result and limiting case
What earns full marks: Complete derivations with correct algebra, 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.
- (a) Show that the relative speed of the two fragments after the explosion is √(2Tm/(m₁m₂)). 15 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Momentum conservation m₁v₁ = m₂v₂ = p
- Total KE T = p²m/(2m₁m₂)
- Relative speed v₁ + v₂ = pm/(m₁m₂), then substitution giving √(2Tm/m₁m₂)
Loses marks
- Writing the result without derivation
- Using v₁ - v₂ rather than v₁ + v₂
Earns more
- Notes no external impulse and rest initial state
- Algebraic steps shown
Extra mark
- States the non-relativistic assumption
- (b) Find the position of the load for equal stress and for equal strain in the two wires. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Moments and equilibrium: T_b = Wx/L, T_s = W(L-x)/L
- Equal stress: x = 66.7 cm from the steel end
- Equal strain: A_sY_s = A_bY_b, so x = 50 cm
Loses marks
- Ignoring Young's modulus in part (ii)
- Giving a position without naming the reference end
Earns more
- Unit conversion of areas
- Positions stated relative to a named end
Extra mark
- Elastic limit remark
- (c) Show that double-slit Fraunhofer diffraction is interference modulated by single-slit diffraction and obtain the maxima and minima conditions. 20 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Single-slit amplitude A₀a sinβ/β with β = πa sinθ/λ; phase δ = 2πd sinθ/λ
- I = I₀(sinβ/β)²cos²(δ/2) as product of envelope and interference term
- Maxima d sinθ = nλ; minima d sinθ = (n+½)λ and a sinθ = mλ
Loses marks
- Omitting the diffraction envelope
- Mixing the interference and diffraction minima conditions
Earns more
- Resultant amplitude 2A₁cos(δ/2)
- Missing orders d/a = n/m
- Screen positions y = nλD/d
Extra mark
- Sketch of fringes inside envelope described in words
What "Derive" is asking you to do
Reach the stated expression from a starting relation, justifying every step. The destination is printed in the question, so only the route earns marks, and the assumptions you work under are part of that route.
Structure that answers it
Assumptions and notation defined → starting relation or governing equation → each step with its justification → the required expression → limiting case or boundary check
Where marks are lost
Writing the standard result first and fitting three lines to it, which an examiner reads at a glance. Marks also go on assumptions left unstated — lossless medium, small amplitude, errors independent with zero mean — and on symbols used before they are defined, even when the question says usual notations.
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