(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 particle is m, find the law of force. (10 marks)

(b) A capillary tube having 1·0 mm diameter, 20 cm in length is fitted hori

Question

(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 particle is m, find the law of force. (10 marks)

(b) 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)

(c) 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)

(d) 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)

(e)

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

f = 1/(2π) √[1/m (k₁k₂)/(k₁+k₂)] (10 marks)

UPSC Answer Check · Accepted Answer

(a) For a central orbit, use Binet’s formula:
F(r) = −m h² u² (d²u/dθ² + u), where u = 1/r and h = r² dθ/dt.
Given r = k e^(αθ), so
u = 1/r = (1/k)e^(−αθ).
Differentiate:
du/dθ = −αu,
d²u/dθ² = α²u.
Substitute in Binet’s formula:
F(r) = −m h² u²(α²u + u)
= −m h²(1 + α²)u³
= −m h²(1 + α²)/r³.
If L = m h is the angular momentum of the particle, then h = L/m, so
F(r) = −L²(1 + α²)/(m r³).
Thus the force is an attractive inverse-cube force:
F(r) = −m h²(1 + α²)/r³ = −L²(1 + α²)/(m r³).
This is valid for h ≠ 0, i.e. non-radial motion.

(b) Use Poiseuille’s formula for steady laminar flow through a horizontal capillary tube:
Q = πR⁴Δp/(8ηL).
Here
R = 1.0/2 mm = 0.5 mm = 5.0×10⁻⁴ m,
L = 20 cm = 0.20 m,
h = 40 cm = 0.40 m,
ρ = 8×10² kg/m³,
η = 0.0012 N s/m²,
t = 10 min = 600 s.
The driving pressure difference is the hydrostatic head:
Δp = ρgh = (8×10²)(9.8)(0.40) = 3136 Pa.
Now
R⁴ = (5.0×10⁻⁴)⁴ = 6.25×10⁻¹⁴ m⁴.
Therefore
Q = π(6.25×10⁻¹⁴)(3136) / [8(0.0012)(0.20)]
= π(1.96×10⁻¹⁰)/(0.00192)
= 3.206×10⁻⁷ m³/s.
Volume flowing out in 10 minutes:
V = Qt = (3.206×10⁻⁷)(600)
= 1.924×10⁻⁴ m³
= 192.4 cm³.
Mass flowing out:
M = ρV = (8×10²)(1.924×10⁻⁴)
= 0.1539 kg
= 153.9 g.
Hence the amount of alcohol flowing out is
V ≈ 1.92×10⁻⁴ m³ = 192 cm³, M ≈ 0.154 kg = 154 g.
This assumes steady laminar flow, no end effects, and neglects surface-tension effects.

(c) Let the first observer’s frame be S. Take event 1 at x₁ = 0, t₁ = 0. For event 2,
Δx = 100 km = 1.00×10⁵ m,
Δt = 2.0 ms = 2.0×10⁻³ s.
Let S′ be the observer’s frame in which the two explosions occur at the same place, so Δx′ = 0. By the Lorentz transformation,
Δx′ = γ(Δx − vΔt) = 0,
which gives
v = Δx/Δt = (1.00×10⁵)/(2.0×10⁻³)
= 5.0×10⁷ m/s.
Also,
Δt′ = γ(Δt − vΔx/c²).
Using v = Δx/Δt,
Δt′ = Δt/γ = Δt√(1 − v²/c²).
Now
v/c = (5.0×10⁷)/(3.0×10⁸) = 1/6.
Thus
Δt′ = (2.0×10⁻³)√(1 − 1/36)
= (2.0×10⁻³)(√35/6)
= (√35/3)×10⁻³ s
≈ 1.972×10⁻³ s.
So the time interval for the second observer is
Δt′ = (√35/3) ms ≈ 1.97 ms.

(d) For thin-film interference in reflected light, the optical path difference between rays reflected from the top and bottom surfaces is
Δ = 2μt cos r,
where r is the angle of refraction inside the film. The angle of incidence is i = 30°.
By Snell’s law,
sin i = μ sin r,
so
sin r = sin 30°/1.35 = 0.5/1.35 = 0.37037.
Hence
r ≈ 21.74°,
cos r = √(1 − sin²r)
= √(1 − 0.13717)
= 0.92888.
The film thickness is
t = 9×10⁻⁶ cm = 9×10⁻⁸ m.
Therefore
2μt cos r = 2(1.35)(9×10⁻⁸)(0.92888)
= 2.257×10⁻⁷ m.
For a petrol film in air or on a rarer medium, one reflection suffers an extra π phase change. The condition for constructive interference in reflected light is
2μt cos r = (n − 1/2)λ, n = 1, 2, 3, …
Thus
λ = 2μt cos r/(n − 1/2).
For n = 1:
λ = (2.257×10⁻⁷)/0.5 = 4.514×10⁻⁷ m = 451.4 nm.
For n = 2:
λ = (2.257×10⁻⁷)/1.5 = 1.505×10⁻⁷ m = 150.5 nm.
For n = 3:
λ = (2.257×10⁻⁷)/2.5 = 9.03×10⁻⁸ m = 90.3 nm.
Only 451.4 nm lies in the visible range 400–700 nm. Therefore the visible wavelength viewed in reflected light is
λ ≈ 451 nm (blue-violet).

(e) The two springs are in series, so the tension in both springs is the same at any instant. Let T₀ = mg be the equilibrium tension. Let T be the instantaneous tension when the mass is displaced downward by x from equilibrium.
For the mass,
m d²x/dt² = mg − T = −(T − T₀).
Let ΔT = T − T₀. The additional extensions of the two springs are
y₁ = ΔT/k₁, y₂ = ΔT/k₂.
Since the springs are in series, the total additional downward displacement of the mass is
x = y₁ + y₂ = ΔT(1/k₁ + 1/k₂)
= ΔT(k₁ + k₂)/(k₁k₂).
Hence
ΔT = (k₁k₂/(k₁ + k₂))x.
Substituting into the equation of motion,
m d²x/dt² = −(k₁k₂/(k₁ + k₂))x.
This is the standard simple-harmonic equation
d²x/dt² + ω²x = 0,
with
ω² = (k₁k₂)/[m(k₁ + k₂)].
Therefore
ω = √[(k₁k₂)/(m(k₁ + k₂))].
The frequency is
f = ω/(2π)
= 1/(2π) √[(k₁k₂)/(m(k₁ + k₂))]
= 1/(2π) √[1/m (k₁k₂)/(k₁ + k₂)].
Thus the motion is simple harmonic with frequency
f = 1/(2π) √[1/m (k₁k₂)/(k₁ + k₂)].
This assumes ideal massless springs obeying Hooke’s law and neglects damping.

(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…

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