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 theorem is based on the principle of conservation of energy for a flowing fluid. It states that in a steady, incompressible, and non-viscous flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline.
According to the continuity equation:
When the cross-sectional area decreases at a constriction, the velocity must increase to maintain the same flow rate. From Bernoulli's principle:
An increase in kinetic energy (velocity) leads to a decrease in pressure energy to keep the total energy constant.
Due to the Bernoulli effect: The water flowing between the ships moves faster than on the outer sides, creating lower pressure between them. The higher pressure on the outer sides pushes the ships together, resulting in an attractive force.
Using Poiseuille's equation for laminar flow through a capillary:
Where \( \Delta P = \rho g h \), \( Q = 1.5 \times 10^{-6} \, \text{m}^3/\text{s} \), \( r = 0.001 \, \text{m} \), \( \eta = 1.0 \times 10^{-3} \, \text{N·s/m}^2 \), \( L = 0.2 \, \text{m} \), \( \rho = 1000 \, \text{kg/m}^3 \).
Substituting values:
Solving for \( h \):
Using Torricelli's theorem:
Drainage rate:
For the stream area \( A_s = \frac{1}{2} A \), continuity gives \( A v = A_s v_s \) ⇒ \( v_s = 2v \).
Energy equation from hole to point below:
Substitute \( v_s = 2v \), \( v = \sqrt{2gD} \):
This indicates the area contraction occurs at the hole itself (vena contracta effect).
Sound requires a medium to propagate. The moon has no atmosphere, so sound waves cannot travel. Communication must be via electromagnetic waves (e.g., radio).
Doppler shift formula:
Given: \( \Delta f = 4400 \, \text{Hz} \), \( f = 5.0 \times 10^6 \, \text{Hz} \), \( c = 1500 \, \text{m/s} \), \( \theta = 30^\circ \)
Solve for \( v \):
Volume flow rate:
Grating spacing:
Grating equation: \( \sin\theta = \frac{m \lambda}{d} \)
For \( \lambda_1 = 5.890 \times 10^{-7} \, \text{m} \):
For \( \lambda_2 = 5.896 \times 10^{-7} \, \text{m} \):
Angular separation:
Given: \( r = 1 \times 10^{-3} \, \text{m} \), \( h = 0.05 \, \text{m} \), \( \rho = 1200 \, \text{kg/m}^3 \), \( \gamma = 0.05 \, \text{N/m} \), \( P_0 = 1.013 \times 10^5 \, \text{Pa} \)
Pressure due to depth:
Excess pressure inside bubble due to surface tension:
Total pressure inside bubble:
Using the ideal gas law: \( P = n k T \)
Where: \( n = 5 \times 10^6 \, \text{molecules/m}^3 \), \( k = 1.38 \times 10^{-23} \, \text{J/K} \), \( T = 3 \, \text{K} \)
Given: \( m = 45 \, \text{kg} \), \( r = 0.5 \times 10^{-2} \, \text{m} \), \( \theta = 15^\circ \), \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \)
Weight supported by each wire:
Cross-sectional area:
Stress:
Strain:
At the midpoint, fields due to each charge are opposite. Let \( q_1 = 1.5 \times 10^{-6} \, \text{C} \), \( q_2 = 2.5 \times 10^{-6} \, \text{C} \), distance to midpoint = 0.15 m.
Net field:
Time constant:
At \( t = 4 \, \text{s} = \tau \):
From part (i): \( V_C = 4.5 \times 10^6 \, \text{V} \), \( V_D = 7.5 \times 10^6 \, \text{V} \)
Work done:
Faraday's law: \( \mathcal{E} = -\frac{d\phi}{dt} \)
At \( t = 10 \, \text{s} \):
Induced current:
For a toroid:
Where: \( N = 1200 \), \( A = 12.0 \times 10^{-4} \, \text{m}^2 \), \( r = 0.15 \, \text{m} \), \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \)
Initial flux:
If field removed: \( \Phi_f = 0 \) ⇒ \( \Delta\Phi = 6 \times 10^{-3} \, \text{Wb} \)
If field reversed: \( \Phi_f = -6 \times 10^{-3} \, \text{Wb} \) ⇒ \( \Delta\Phi = 12 \times 10^{-3} \, \text{Wb} \)
Induced charge: \( Q = \frac{N |\Delta\Phi|}{R} \)
Number of atoms in 1 g:
Decay constant:
Activity:
Closest approach distance \( r_0 \) from energy conservation:
Using \( R = R_0 A^{1/3} \) with \( R_0 = 1.2 \times 10^{-15} \, \text{m} \), \( A = 23 \):
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