Paper I — Q1
(a) A particle moving in a central force field describes the path r = ke^αθ, where k and α are constants. If the mass of the…
A particle moving in a central force field describes the path r = ke^αθ, where k and α are constants. If the mass of the particle is m, find the law of force. 10 marks
A capillary tube having 1·0 mm diameter, 20 cm in length is fitted horizontally to a vessel in which alcohol is kept fully up to the neck. Density of alcohol is 8 × 10² kg/m³. The depth of the centre of the capillary tube below the surface of alcohol is 40 cm. Find the amount of alcohol that will flow out of the capillary tube in 10 minutes. Coefficient of viscosity of alcohol is 0·0012 Ns/m². 10 marks
An observer detects two explosions, one that occurs near him at a certain time and another that occurs 2·0 ms later 100 km away. Another observer finds that the two explosions occur at the same place. What time interval separates the explosions to the second observer? 10 marks
A thin film of petrol of thickness 9 × 10⁻⁶ cm is viewed at an angle 30° to the normal. Find the wavelength(s) of light in visible spectrum which can be viewed in the reflected light. The refractive index of the film μ = 1·35. 10 marks
A mass m is suspended by two springs having force constants k₁ and k₂ as shown in the figure. The mass m is displaced vertically downward and then released. If at any instant t, the displacement of the mass m is x, then show that the motion of the mass is simple harmonic motion having frequency
f = 1/(2π) √[1/m (k₁k₂)/(k₁+k₂)] 10 marks
हिंदी में प्रश्न पढ़ें
एक कण का पथ केंद्रीय बल क्षेत्र में r = ke^αθ से वर्णित किया जाता है, जहाँ कि k और α नियतांक हैं । यदि कण का द्रव्यमान m है, तो बल का नियम ज्ञात कीजिए । (10 अंक)
एक 1·0 mm व्यास और 20 cm लंबाई वाली केशिकीय नलिका एक बर्तन, जिसमें 8 × 10² kg/m³ घनत्व का एल्कोहॉल भरा है, से क्षैतिज दिशा में जुड़ी है । केशिकीय नलिका के केंद्र की गहराई एल्कोहॉल के पृष्ठ से 40 cm नीचे है । 10 मिनट में केशिकीय नलिका से बहने वाले एल्कोहॉल की मात्रा ज्ञात कीजिए । एल्कोहॉल का श्यानता गुणांक 0·0012 Ns/m² है । (10 अंक)
एक प्रेक्षक दो विस्फोट देखता है, पहला जो कि उसके पास किसी समय पर होता है तथा दूसरा जो कि 2·0 ms बाद 100 km दूर होता है । एक दूसरा प्रेक्षक पाता है कि दोनों विस्फोट एक ही स्थान पर होते हैं । दूसरे प्रेक्षक के लिए विस्फोटों का समय अंतराल क्या होगा? (10 अंक)
9 × 10⁻⁶ cm पतले पेट्रोल की परत (फिल्म) को लम्बवत् दिशा से 30° कोण पर देखा जाता है । परावर्तित प्रकाश के उन तरंगदैर्घ्य(यों) को ज्ञात कीजिए जो कि दृश्य वर्णक्रम (स्पेक्ट्रम) में आते हैं । पेट्रोल की परत (फिल्म) का अपवर्तनांक μ = 1·35 है । (10 अंक)
चित्र में दर्शाए अनुसार एक द्रव्यमान m, दो स्प्रिंगों जिनका बल नियतांक k₁ और k₂ है, से लटकाया गया है । द्रव्यमान m को उर्ध्वाधर दिशा में नीचे की तरफ थोड़ा-सा विस्थापित करके छोड़ दिया जाता है । यदि किसी समय t पर द्रव्यमान m का विस्थापन x हो, तो दिखाइए कि द्रव्यमान की गति सरल आवर्त गति है जिसकी आवृत्ति
f = 1/(2π) √[1/m (k₁k₂/(k₁+k₂))] है । (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 vertical mass-spring system suspended from a fixed horizontal ceiling with hatching. A helical spring with spring constant k1 hangs vertically from the ceiling. Connected in series directly below it is a second helical spring with spring constant k2. Suspended from the bottom of the second spring is a rectangular mass block, labeled m to its right.
(e) A vertical mass-spring system suspended from a fixed horizontal ceiling indicated by hatching. Connected to the ceiling and hanging vertically is a first helical spring labelled k1. Connected directly in series below the first spring is a second helical spring labelled k2. Suspended from the bottom of the second spring is a rectangular mass labelled m to its right.
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) Law of force for the orbit r = ke^(αθ)
Use Binet's equation for a central force, with u = 1/r and h = r²(dθ/dt) a constant (twice the areal velocity):
F(r) = −m h² u² (d²u/dθ² + u)
Here u = (1/k)e^(−αθ), so du/dθ = −αu and d²u/dθ² = α²u. Then
F = −m h² u² (α²u + u) = −m h²(1 + α²)u³ = −m h²(1 + α²)/r³.
With angular momentum L = mh this is F = −L²(1 + α²)/(m r³). The force is attractive and varies as the inverse cube of distance (valid for h ≠ 0).
(b) Flow of alcohol through a capillary
Use Poiseuille's formula for a horizontal tube, with the driving pressure given by the head of alcohol:
Δp = ρgh = 800 × 9.8 × 0.40 = 3136 Pa Q = πR⁴Δp/(8ηL), R = 5.0×10⁻⁴ m, L = 0.20 m, η = 0.0012 N s/m² Q = π(6.25×10⁻¹⁴)(3136)/(8 × 0.0012 × 0.20) = 3.21×10⁻⁷ m³/s
Check on laminar flow: mean speed = Q/πR² ≈ 0.41 m/s, so Reynolds number = ρvd/η ≈ 270, well below 2000.
In 10 min (600 s): V = Qt = 1.92×10⁻⁴ m³ = 192 cm³, and mass M = ρV ≈ 0.154 kg ≈ 154 g. End effects and surface tension are neglected; using g = 10 m/s² gives about 157 g.
(c) Time interval in the second frame
In frame S, Δx = 100 km = 1.0×10⁵ m and Δt = 2.0×10⁻³ s. In frame S′ the events happen at the same place, Δx′ = 0, so by the Lorentz transformation Δx′ = γ(Δx − vΔt) = 0 and
v = Δx/Δt = 5.0×10⁷ m/s, so v/c = 1/6.
Since the events are at one place in S′, the interval there is the proper time:
Δt′ = Δt/γ = Δt√(1 − v²/c²) = 2.0 ms × √(35)/6 = 1.97 ms.
(d) Thin petrol film: visible wavelengths in reflected light
Assume the film is in air, so the reflection at the top surface has a phase change of π and that at the bottom has none. Reflected light is bright when
2μt cos r = (n + ½)λ, n = 0, 1, 2, …
Snell's law: sin r = sin 30°/1.35 = 0.370, so r = 21.74° and cos r = 0.9289.
2μt cos r = 2 × 1.35 × 9×10⁻⁸ m × 0.9289 = 2.257×10⁻⁷ m = 225.7 nm
λ = 225.7/(n + ½) nm = 451.4 nm (n = 0), 150.5 nm (n = 1), 90.3 nm (n = 2) …
Only 451 nm lies in the visible range (400-700 nm). The film therefore appears blue in reflected light.
(e) Two springs in series: SHM frequency
For springs in series the same tension acts in each. Let the mass be displaced down by x from equilibrium and let ΔT be the extra tension. The extensions are y₁ = ΔT/k₁ and y₂ = ΔT/k₂, so
x = y₁ + y₂ = ΔT(k₁ + k₂)/(k₁k₂), hence ΔT = [k₁k₂/(k₁ + k₂)] x.
The restoring force on the mass gives
m d²x/dt² = −[k₁k₂/(k₁ + k₂)] x.
This has the form d²x/dt² + ω²x = 0, so the motion is simple harmonic with ω² = k₁k₂/[m(k₁ + k₂)], i.e.
f = ω/2π = (1/2π)√[(1/m)·k₁k₂/(k₁ + k₂)]
How this answer will be evaluated
Framework: Principle > Setup and diagram > Derivation > Result and limiting case
What earns full marks: Complete 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 law of force for a particle in a central field describing r = ke^(αθ). 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Uses Binet's equation F = -m h² u²(u'' + u)
- Computes u, du/dθ = -αu and d²u/dθ² = α²u correctly
- Result F = -m h²(1+α²)/r³: attractive inverse-cube force
Loses marks
- Wrong sign or missing (1+α²) factor
- Using a force law without derivation
Earns more
- Expresses in terms of L = mh
- States h ≠ 0 condition
Extra mark
- Notes attractive sign
- (b) Find the mass of alcohol flowing out of a horizontal capillary in 10 minutes. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Poiseuille formula Q = πR⁴Δp/(8ηL)
- Δp = ρgh = 3136 Pa with R = 0.5 mm, L = 0.2 m
- Q = 3.21e-7 m³/s, V ≈ 192 cm³, M ≈ 154 g in 600 s
Loses marks
- Using diameter in place of radius
- Unit or arithmetic slip in R⁴
Earns more
- Converts units correctly
- Checks laminar flow (Re ≈ 270)
Extra mark
- Mentions neglected end effects
- (c) Find the time interval between the explosions in the frame where they occur at the same place. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- v = Δx/Δt = 5×10⁷ m/s from Δx' = 0
- Recognises interval is proper time: Δt' = Δt/γ
- Δt' = 1.97 ms
Loses marks
- Using time dilation backwards (Δt' = γΔt)
- Wrong v
Earns more
- v/c = 1/6 computed
- Uses Lorentz transformation
Extra mark
- Notes Δt' is shorter than Δt
- (d) Find the visible wavelengths seen in reflected light from the petrol film. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Condition 2μt cos r = (n+½)λ with π phase change at top
- r = 21.74° from Snell's law, 2μt cos r = 225.7 nm
- Only 451 nm visible
Loses marks
- Ignoring refraction angle
- Using the wrong phase condition without reason
Earns more
- Lists 150.5 nm and 90.3 nm as outside visible range
- States film-in-air assumption
Extra mark
- Notes dark condition gives no visible wavelength
- (e) Show that the mass on two springs in series performs SHM and find its frequency. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Same tension in both springs; x = ΔT(1/k₁+1/k₂)
- Equivalent constant k₁k₂/(k₁+k₂); equation m x'' = -k_eq x
- f = (1/2π)√[k₁k₂/(m(k₁+k₂))]
Loses marks
- Adding spring constants as in parallel
- Frequency formula without derivation
Earns more
- Extensions y₁ = ΔT/k₁ and y₂ = ΔT/k₂
- States SHM standard form
Extra mark
- Notes ideal massless springs
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