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.
Bernoulli's Principle: In a steady, incompressible, non-viscous flow, the total energy per unit volume remains constant along a streamline.
Derivation: Consider fluid flow through a pipe of varying cross-section. Using work-energy theorem:
\( P_1A_1\Delta x_1 - P_2A_2\Delta x_2 = \frac{1}{2}\rho v_2^2\Delta m + \rho g h_2\Delta m - \frac{1}{2}\rho v_1^2\Delta m - \rho g h_1\Delta m \)
Since \( \Delta m = \rho A_1\Delta x_1 = \rho A_2\Delta x_2 \), dividing by volume:
Thus, \( P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant} \).
Given: \( \rho = 1000 \, \text{kg/m}^3 \), \( A_P = 4\times10^{-3} \, \text{m}^2 \), \( A_Q = 8\times10^{-3} \, \text{m}^2 \), \( h_P = 2 \, \text{m} \), \( h_Q = 5 \, \text{m} \), \( v_P = 1 \, \text{m/s} \).
From continuity: \( A_P v_P = A_Q v_Q \)
Work done by gravity per unit volume = \( \rho g (h_Q - h_P) \)
Using Bernoulli between P and Q:
Work done by pressure per unit volume = \( P_P - P_Q = \frac{1}{2}\rho(v_Q^2 - v_P^2) + \rho g(h_Q - h_P) \)
Capillaries in series with radii ratio 1:2, length L each. Total pressure difference = 1 m of water = \( \rho g h = 1000\times9.8\times1 = 9800 \, \text{Pa} \).
For streamline flow, flow rate Q is same through both.
Let \( r_1 = r \), \( r_2 = 2r \). Then:
Total: \( \Delta P_1 + \Delta P_2 = \Delta P_1 + \frac{\Delta P_1}{16} = \frac{17}{16}\Delta P_1 = 9800 \)
For sphere falling in viscous fluid, forces: weight (down), buoyancy (up), viscous drag (up).
Buoyancy = \( \frac{4}{3}\pi R^3 \rho g \)
Stokes' drag = \( 6\pi\eta R v \)
At terminal velocity \( v_T \), net force = 0:
Solving for \( v_T \):
The given formula \( V_T = 2R^2 (\rho_0 - \rho) g \) appears to have omitted the \( 9\eta \) denominator or assumed \( 9\eta = 1 \).
Tank height = 5 m, orifice at bottom. Using Torricelli's theorem:
Given: Steel and copper wires similar (same L, A), stretched by equal forces F. Total elongation = 1 cm = 0.01 m.
Young's modulus: \( Y_s = 2.0\times10^{11} \, \text{N/m}^2 \), \( Y_c = 1.3\times10^{11} \, \text{N/m}^2 \).
Elongation: \( \Delta L = \frac{FL}{AY} \)
Ratio: \( \frac{\Delta L_s}{\Delta L_c} = \frac{Y_c}{Y_s} = \frac{1.3}{2.0} = 0.65 \)
Let \( \Delta L_s = 0.65x \), \( \Delta L_c = x \). Total: \( 0.65x + x = 0.01 \) ⇒ \( 1.65x = 0.01 \)
Given: L = 0.2 m, A = 1 cm² = 10⁻⁴ m², F = 49 N, Y = 1.0×10¹¹ N/m².
Stress = \( \frac{F}{A} = \frac{49}{10^{-4}} = 4.9\times10^5 \, \text{Pa} \)
Strain = \( \frac{\text{Stress}}{Y} = \frac{4.9\times10^5}{1.0\times10^{11}} = 4.9\times10^{-6} \)
Energy stored per unit volume = \( \frac{1}{2} \times \text{Stress} \times \text{Strain} \)
Volume = \( A L = 10^{-4} \times 0.2 = 2\times10^{-5} \, \text{m}^3 \)
Total energy = \( u \times \text{Volume} = 1.2005 \times 2\times10^{-5} \approx 2.40\times10^{-5} \, \text{J} \)
Consider a soap bubble of radius R, surface tension T. Bubble has two surfaces (inner and outer).
Force due to surface tension on hemisphere = \( 2 \times (2\pi R T) = 4\pi R T \) (2 surfaces)
Pressure difference \( \Delta P \) exerts force = \( \Delta P \times \pi R^2 \)
At equilibrium: \( \Delta P \cdot \pi R^2 = 4\pi R T \)
Thus, excess pressure inside soap bubble = \( \frac{4T}{R} \).
Two charges Q each, mass m, suspended by strings length L. At equilibrium:
Forces: Tension T, weight mg, Coulomb repulsion \( F = \frac{Q^2}{4\pi\epsilon_0 x^2} \) where x = separation.
Vertical: \( T\cos\theta = mg \)
Horizontal: \( T\sin\theta = \frac{Q^2}{4\pi\epsilon_0 x^2} \)
Also, \( \sin\theta = \frac{x/2}{L} = \frac{x}{2L} \)
Dividing: \( \tan\theta = \frac{Q^2}{4\pi\epsilon_0 mg x^2} \)
For small θ, \( \tan\theta \approx \sin\theta = \frac{x}{2L} \)
Thus: \( x = \left( \frac{Q^2 L}{2\pi\epsilon_0 mg} \right)^{1/3} \)
Also, from geometry: \( \theta = \frac{x}{2L} \) ⇒ \( \theta = \frac{1}{2L} \left( \frac{Q^2 L}{2\pi\epsilon_0 mg} \right)^{1/3} \)
The given expression \( \theta = \sqrt{\frac{Q^2}{16\pi\epsilon_0 mg L^2}} \) appears different; may be approximate form.
Energy stored in capacitor: \( U = \frac{1}{2} C V^2 \)
This energy heats block: \( U = m c \Delta T \)
Given: V = 200 V, m = 0.1 kg, c = 2.5×10² J/kg·K (assuming 2.5×10², not 2.5×10⁻⁷), ΔT = 0.4 K.
\( 20000 C = 10 \) ⇒ \( C = 5\times10^{-4} \, \text{F} = 500 \, \mu\text{F} \)
Capacitors: \( C_1 = 20 \, \mu\text{F} \) at 500 V, \( C_2 = 10 \, \mu\text{F} \) at 200 V.
Charges: \( Q_1 = C_1 V_1 = 20\times10^{-6} \times 500 = 0.01 \, \text{C} \)
\( Q_2 = C_2 V_2 = 10\times10^{-6} \times 200 = 0.002 \, \text{C} \)
Total charge = 0.012 C
Total capacitance when parallel = \( 20 + 10 = 30 \, \mu\text{F} = 3\times10^{-5} \, \text{F} \)
Common potential: \( V = \frac{\text{Total charge}}{\text{Total capacitance}} = \frac{0.012}{3\times10^{-5}} = 400 \, \text{V} \)
Biot-Savart law: \( dB = \frac{\mu_0 I dl \sin\theta}{4\pi r^2} \)
For circular coil, at center, \( \theta = 90^\circ \), \( \sin\theta = 1 \), and integral over circle gives circumference \( 2\pi r \).
For one turn: \( B = \frac{\mu_0 I}{2r} \)
For N turns: \( B = \frac{\mu_0 N I}{2r} \)
Given: N = 300, r = 0.1 m, I = 8 A, \( \mu_0 = 4\pi\times10^{-7} \)
Given: H = 360 A/m, B = 0.6 T
(i) Permeability \( \mu = \frac{B}{H} = \frac{0.6}{360} = 1.667\times10^{-3} \, \text{T·m/A} \)
(ii) \( \mu = \mu_0 \mu_r \), \( \mu_0 = 4\pi\times10^{-7} \)
Susceptibility \( \chi_m = \mu_r - 1 \approx 1325 \)
For solenoid: B = \( \mu_0 \mu_r n I \), where n = N/l
Magnetic flux through one turn: \( \phi = B A = \mu_0 \mu_r n I A \)
Total flux linkage: \( N\phi = \mu_0 \mu_r n N I A = \mu_0 \mu_r \frac{N}{l} N I A \)
Inductance L = \( \frac{\Psi}{I} = \frac{\mu_0 \mu_r N^2 A}{l} \)
Thus proved.
Young's double slit: d = 0.125 mm = 1.25×10⁻⁴ m, λ = 4500 Ã… = 4.5×10⁻⁷ m, D = 1 m.
Position of m-th bright fringe: \( y_m = \frac{m \lambda D}{d} \)
Second bright fringe (m=2): \( y_2 = \frac{2 \times 4.5\times10^{-7} \times 1}{1.25\times10^{-4}} = 7.2\times10^{-3} \, \text{m} \)
Separation between both sides: \( 2y_2 = 2 \times 7.2\times10^{-3} = 0.0144 \, \text{m} \)
Diffraction grating: λ = 589 nm = 5.89×10⁻⁷ m, angle between first orders = 34.2°, so θ₁ = 17.1°.
Grating equation: \( d\sin\theta = m\lambda \)
For m=1: \( d = \frac{\lambda}{\sin\theta_1} = \frac{5.89\times10^{-7}}{\sin 17.1^\circ} = \frac{5.89\times10^{-7}}{0.2940} \approx 2.003\times10^{-6} \, \text{m} \)
Lines per meter: \( \frac{1}{d} = \frac{1}{2.003\times10^{-6}} \approx 4.99\times10^5 \, \text{lines/m} \)
Lines per mm: \( 4.99\times10^5 / 1000 \approx 499 \, \text{lines/mm} \)
Maximum order: \( m_{\max} \leq \frac{d}{\lambda} = \frac{2.003\times10^{-6}}{5.89\times10^{-7}} \approx 3.4 \)
So visible orders: m = 0, ±1, ±2, ±3 (7 orders total).
Sonometer frequency: \( f \propto \frac{1}{L} \sqrt{T} \)
When tension increased by 21%, T' = 1.21T, frequency increases by 51 Hz.
Let original frequency = f. New frequency = f + 51 = \( f \times \sqrt{1.21} = 1.1 f \)
So 1.1f = f + 51 ⇒ 0.1f = 51 ⇒ f = 510 Hz.
Now if length increased by 10%, L' = 1.1L.
New frequency f'' = \( \frac{f}{1.1} = \frac{510}{1.1} \approx 463.6 \, \text{Hz} \)
Kinetic energy: \( \frac{1}{2}mv^2 = eV \) ⇒ \( v = \sqrt{\frac{2eV}{m}} \)
De Broglie wavelength: \( \lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2meV}} \)
For electron: \( m = 9.1\times10^{-31} \, \text{kg} \), \( e = 1.6\times10^{-19} \, \text{C} \), \( h = 6.63\times10^{-34} \, \text{Js} \)
In Ångströms: \( \lambda = \frac{12.27}{\sqrt{V}} \, \text{Å} \)
For V = 60 V: \( \lambda = \frac{12.27}{\sqrt{60}} = \frac{12.27}{7.746} \approx 1.584 \, \text{Ã…} \)
This corresponds to X-ray region of electromagnetic spectrum.
Reaction: \( ^{235}_{92}U + n → ^{141}_{56}Ba + ^{92}_{36}Kr + 3n \)
Mass defect:
Final: 140.91 u + 91.94 u + 3×1.01 u = 140.91 + 91.94 + 3.03 = 235.88 u
Mass defect Δm = 236.05 - 235.88 = 0.17 u
Energy per fission = 0.17 × 931 MeV = 158.27 MeV
Number of atoms in 10 kg of U-235:
Total energy = 158.27 MeV × 2.56×10²⁵ = 4.05×10²⁷ MeV
In joules: 4.05×10²⁷ × 1.6×10⁻¹³ ≈ 6.48×10¹⁴ J
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