Instructions: This paper consists of six (6) questions. Answer any five (5) questions. Each question carries twenty (20) marks. Mathematical tables and non-programmable calculators may be used.
Static Pressure (P): The actual pressure at a point in a fluid at rest or in motion. It's the pressure measured by a sensor moving with the fluid.
Dynamic Pressure: Pressure due to fluid motion = \( \frac{1}{2} \rho v^2 \), where ρ is density and v is velocity.
Total Pressure (Stagnation Pressure): Sum of static and dynamic pressures: \( P_{\text{total}} = P + \frac{1}{2} \rho v^2 + \rho gh \).
Given: Static pressure \( P = 4.8 \times 10^4 \, \text{Pa} \), Total pressure \( P_{\text{total}} = 4.9 \times 10^4 \, \text{Pa} \), Area \( A = 120 \, \text{cm}^2 = 0.012 \, \text{m}^2 \), ρ = 1000 kg/m³.
Dynamic pressure = \( P_{\text{total}} - P = 4.9\times10^4 - 4.8\times10^4 = 1000 \, \text{Pa} \)
Volume flow rate \( Q = A v = 0.012 \times 1.414 \approx 0.0170 \, \text{m}^3/\text{s} \)
Given: Capillary length L = 1 mm = 0.001 m, radius r = \( 2 \times 10^{-6} \, \text{m} \), velocity v = \( 0.66 \times 10^{-3} \, \text{m/s} \), η = \( 4.0 \times 10^{-3} \, \text{Pa·s} \).
Using Poiseuille's law for flow rate: \( Q = \frac{\pi r^4 \Delta P}{8 \eta L} \)
Also \( Q = A v = \pi r^2 v \)
Tank: radius = 0.9 m, initial water depth = 3 m, orifice at bottom, ground distance = 6 m.
Velocity at orifice (Torricelli): \( v_0 = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 3} = \sqrt{58.8} \approx 7.67 \, \text{m/s} \)
This is horizontal velocity. Vertical motion: falls 6 m from tank bottom.
Vertical velocity at ground: \( v_y = gt = 9.8 \times 1.106 \approx 10.84 \, \text{m/s} \)
Resultant speed: \( v = \sqrt{v_0^2 + v_y^2} = \sqrt{7.67^2 + 10.84^2} \approx \sqrt{58.8 + 117.5} \approx \sqrt{176.3} \approx 13.28 \, \text{m/s} \)
Tank cross-sectional area \( A_{\text{tank}} = \pi r^2 = \pi (0.9)^2 \approx 2.5447 \, \text{m}^2 \)
Orifice area \( A_{\text{orifice}} = 6.3 \, \text{cm}^2 = 6.3 \times 10^{-4} \, \text{m}^2 \)
Time to empty cylindrical tank with orifice at bottom:
Initially height \( H_1 = 3 \, \text{m} \), finally \( H_2 = 0 \).
Given: Ultrasound frequency f₀ = 5.0 MHz = \( 5.0 \times 10^6 \, \text{Hz} \), angle θ = 30°, Doppler shift Δf = 4.4 kHz = 4400 Hz, speed of ultrasound c = 1.5 km/s = 1500 m/s, vessel diameter D = 1.0 mm = 0.001 m.
Doppler formula: \( \Delta f = \frac{2 f_0 v \cos\theta}{c} \)
Volume flow rate: \( Q = A v = \pi \left( \frac{D}{2} \right)^2 v = \pi (0.0005)^2 \times 0.762 \)
Grating: 6000 lines/cm = 600,000 lines/m, so grating spacing \( d = \frac{1}{600,000} \approx 1.667 \times 10^{-6} \, \text{m} \)
Wavelengths: λ₁ = 5890 Å = \( 5.89 \times 10^{-7} \, \text{m} \), λ₂ = 5896 Å = \( 5.896 \times 10^{-7} \, \text{m} \)
Grating equation: \( \sin\theta = \frac{m\lambda}{d} \), m = 1 (first order)
Angular separation Δθ = \( \theta_2 - \theta_1 \approx 0.02^\circ \)
Two bubbles radii: r₁ = 0.02 m, r₂ = 0.04 m, surface tension T = 0.07 N/m.
Excess pressure inside bubble 1: \( P_1 - P_0 = \frac{4T}{r_1} \)
Excess pressure inside bubble 2: \( P_2 - P_0 = \frac{4T}{r_2} \)
Since \( r_2 > r_1 \), \( P_1 > P_2 \). When they join, air flows from higher pressure (smaller bubble) to lower pressure (larger bubble).
The common interface will be concave toward the smaller bubble (higher pressure).
Pressure difference across common film: \( P_1 - P_2 = 4T \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \)
For the film: \( P_1 - P_2 = \frac{4T}{r} \) where r is radius of curvature of common film.
The film is concave toward the smaller bubble (sense of curvature).
Two rods: each L₀ = 1.0 m, same A, joined to make 2.0 m rod.
Rod 1: α₁ = \( 2.0 \times 10^{-5} \, /^\circ\text{C} \), Y₁ = \( 10^{10} \, \text{N/m}^2 \)
Rod 2: Values not given? Assume same or different? Problem incomplete.
If both rods have same α and Y, thermal expansion ΔL = L₀ α ΔT for each.
To prevent expansion, compressive stress σ = Y α ΔT needed.
Without specific data for rod 2, assume similar calculation.
Two charges: q₁ = q₂ = 1.5 μC = \( 1.5\times10^{-6} \, \text{C} \), separation d = 30 cm = 0.3 m.
(i) At midpoint: equal distances r = 0.15 m, equal magnitude fields oppose.
Since charges same sign, fields cancel: \( E_{\text{net}} = 0 \).
(ii) Point 10 cm from midpoint in perpendicular plane.
Distance from each charge: \( r = \sqrt{(0.15)^2 + (0.1)^2} = \sqrt{0.0225 + 0.01} = \sqrt{0.0325} \approx 0.1803 \, \text{m} \)
Field magnitude: \( E = \frac{9\times10^9 \times 1.5\times10^{-6}}{(0.1803)^2} \approx \frac{13500}{0.0325} \approx 4.154\times10^5 \, \text{N/C} \)
Vertical components cancel, horizontal components add: Each field makes angle φ where \( \cos\phi = \frac{0.15}{0.1803} \approx 0.832 \).
C = 8 μF = \( 8\times10^{-6} \, \text{F} \), R = 0.5 MΩ = \( 5\times10^5 \, \Omega \), V = 200 V.
(i) Initial current: \( I_0 = \frac{V}{R} = \frac{200}{5\times10^5} = 4.0\times10^{-4} \, \text{A} = 0.4 \, \text{mA} \)
(ii) At t = 4 s: time constant τ = RC = \( 5\times10^5 \times 8\times10^{-6} = 4 \, \text{s} \)
Current: \( I = I_0 e^{-t/\tau} = 0.4 \times e^{-1} \approx 0.4 \times 0.3679 \approx 0.147 \, \text{mA} \)
Voltage across capacitor: \( V_C = V(1 - e^{-t/\tau}) = 200(1 - e^{-1}) \approx 200 \times 0.6321 \approx 126.4 \, \text{V} \)
From diagram: A (+200 μC), B (-100 μC), distances: AD = 20 cm, DC = 60 cm, CB = 20 cm (total AB = 100 cm).
Point C: 20 cm from B, 80 cm from A.
Point D: 20 cm from A, 80 cm from B.
Potential at C: \( V_C = \frac{k q_A}{0.8} + \frac{k q_B}{0.2} = 9\times10^9 \left( \frac{200\times10^{-6}}{0.8} + \frac{-100\times10^{-6}}{0.2} \right) \)
Potential at D: \( V_D = \frac{k q_A}{0.2} + \frac{k q_B}{0.8} = 9\times10^9 \left( \frac{200\times10^{-6}}{0.2} + \frac{-100\times10^{-6}}{0.8} \right) \)
Work to move q = +500 μC from C to D:
Square side = 2L, current I. Distance from each side to center = L.
Field due to one straight wire at perpendicular distance L: \( B = \frac{\mu_0 I}{4\pi L} (\sin\theta_1 + \sin\theta_2) \)
For square, each side: θ₁ = θ₂ = 45°, so sin 45° = √2/2 ≈ 0.7071.
Total from 4 sides: \( B_{\text{square}} = 4 \times \frac{\mu_0 I}{4\pi L} \times 1.4142 = \frac{\mu_0 I}{\pi L} \times 1.4142 \)
If reshaped into circle: circumference = 8L, so radius \( r = \frac{8L}{2\pi} = \frac{4L}{\pi} \)
Field at center of circle: \( B_{\text{circle}} = \frac{\mu_0 I}{2r} = \frac{\mu_0 I}{2 \times (4L/\pi)} = \frac{\mu_0 I \pi}{8L} \)
Compare: \( B_{\text{square}} = \frac{\mu_0 I \times 1.4142}{\pi L} \approx 0.450 \frac{\mu_0 I}{L} \)
\( B_{\text{circle}} = \frac{\mu_0 I \pi}{8L} \approx 0.393 \frac{\mu_0 I}{L} \)
Thus field decreases when reshaped to circle.
Circular coil: N = 120 turns, R = 0.18 m, I = 3 A, point on axis at distance x = R from center.
Field on axis: \( B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} \)
With x = R: \( B = \frac{\mu_0 N I R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 N I R^2}{2(2R^2)^{3/2}} = \frac{\mu_0 N I R^2}{2 \times 2^{3/2} R^3} = \frac{\mu_0 N I}{4\sqrt{2} R} \)
Solenoid: diameter = 0.04 m ⇒ radius r = 0.02 m, length l = 0.6 m, N = 4000 turns.
Cross-sectional area A = πr² = π(0.02)² ≈ 0.0012566 m²
Inductance (air core): \( L = \frac{\mu_0 N^2 A}{l} = \frac{4\pi\times10^{-7} \times (4000)^2 \times 0.0012566}{0.6} \)
With iron core μ_r = 4000: \( L_{\text{iron}} = \mu_r L_{\text{air}} = 4000 \times 0.04212 \approx 168.5 \, \text{H} \)
⁴⁰K decays to ⁴⁰Ar, half-life T_{1/2} = 1.37×10⁹ years.
Ratio K:Ar = 1:7 ⇒ For every 1 K atom, 7 Ar atoms (from decayed K).
Original K atoms = Current K + Ar = 1 + 7 = 8 atoms.
Fraction remaining = 1/8 = 0.125.
Decay law: \( N = N_0 e^{-\lambda t} \), λ = ln2 / T_{1/2}
Half-life = 1 hour. Decay constant λ = ln2 / 1 = 0.6931 per hour.
60% decay ⇒ 40% remaining: N/N₀ = 0.4.
Deuterium fusion: ²H + ²H → ³He + n (or similar). Given masses:
²H = 2.015 a.m.u, ³He = 3.017 a.m.u, n = 1.009 a.m.u.
Mass defect per fusion: 2 × 2.015 - (3.017 + 1.009) = 4.030 - 4.026 = 0.004 a.m.u.
Energy per fusion = 0.004 × 931.5 MeV = 3.726 MeV
Number of deuterium atoms in 1 kg: \( N = \frac{1000}{0.002015} \times 6.02\times10^{23} \approx 2.988\times10^{26} \)
Each fusion uses 2 deuterium nuclei, so number of fusions = N/2 ≈ 1.494×10²⁶
Total energy = 1.494×10²⁶ × 3.726 MeV = 5.566×10²⁶ MeV
In joules: 5.566×10²⁶ × 1.6×10⁻¹³ ≈ 8.906×10¹³ J
50% usable ⇒ 4.453×10¹³ J available.
Power = 1 MW = 10⁶ J/s. Time = Energy/Power = 4.453×10¹³ / 10⁶ = 4.453×10⁷ s
Days = 4.453×10⁷ / (24×3600) ≈ 515.4 days
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