Physics 2023 Jamb Examination

Diffraction is the bending or spreading out of waves as they pass through an aperture (opening) or around the edge of an obstacle. The extent of the spreading depends on the relationship between the wavelength and the size of the opening. Since the diagram illustrates waves expanding into the 'shadow' region after passing through a gap, it depicts diffraction.

Electrical conduction in gases involves ionizing gas molecules to allow current flow.
1. Identification of gases: Each gas emits a unique spectrum when conducting electricity (discharge tubes).
2. Fluorescent tubes: These rely on mercury vapor conduction to produce UV light.
3. CRO/TV tubes: These use electron beams (cathode rays) in a vacuum or low-pressure gas environment.
Photocells, however, typically rely on the photoelectric effect in solids or vacuums rather than gas conduction. Therefore, i, ii, and iv are correct applications.

Based on a standard static equilibrium setup where \( T_2 \) is the diagonal cord at 38° to the horizontal and \( T_1 \) is horizontal:
1. Vertical balance: \( T_2 \sin(38^\circ) = mg \)\[ T_2 = \frac{220 \times 9.8}{\sin(38^\circ)} = \frac{2156}{0.6157} \approx 3502 \, \text{N} \]2. Horizontal balance: \( T_1 = T_2 \cos(38^\circ) \)\[ T_1 = 3502 \times \cos(38^\circ) = 3502 \times 0.7880 \approx 2760 \, \text{N} \]Thus, \( T_1 = 2760 \, \text{N} \) and \( T_2 = 3502 \, \text{N} \).

The horizontal range (\( R \)) of a projectile is given by:\[ R = \frac{u^2 \sin(2\theta)}{g} \]Given:
\( u = 75 \, \text{ms}^{-1} \)
\( \theta = 22^\circ \Rightarrow 2\theta = 44^\circ \)
\( g = 10 \, \text{ms}^{-2} \)
Calculation:\[ R = \frac{75^2 \times \sin(44^\circ)}{10} \] \[ R = \frac{5625 \times 0.6947}{10} = 390.7 \approx 391 \, \text{m} \]

Relative Humidity (R.H.) is the ratio of the actual partial pressure of water vapor to the saturated vapor pressure (SVP) at that same temperature, expressed as a percentage.\[ \text{R.H.} = \frac{\text{Partial Pressure}}{\text{SVP}} \times 100 \] \[ \text{R.H.} = \frac{38.8}{55.3} \times 100 \approx 70.16\% \] The closest value is 70%.

Absolute pressure is the sum of atmospheric pressure and the hydrostatic pressure (gauge pressure) exerted by the liquid column. \[ P_{\text{abs}} = P_{\text{atm}} + h\rho g \]1. Hydrostatic pressure: \( 32.8 \times 1000 \times 9.8 = 321440 \, \text{Pa} = 321.44 \, \text{kPa} \)
2. Total pressure: \( 101.3 \, \text{kPa} + 321.44 \, \text{kPa} = 422.74 \, \text{kPa} \).

The terminal potential difference (\( V \)) is the voltage available to the external circuit.
First, find the total current (\( I \)) using Ohm's law for the whole circuit:\[ I = \frac{E}{R + r} = \frac{24.0}{5.0 + 1.0} = \frac{24}{6} = 4 \, \text{A} \]Now, find the terminal p.d.:\[ V = IR = 4 \times 5.0 = 20.0 \, \text{V} \](Alternatively: \( V = E - Ir = 24 - (4 \times 1) = 20 \, \text{V} \)).

1. Find initial velocity (\( u \)) for the first interval:\[ s = ut + 0.5at^2 \Rightarrow 250 = u(10) + 0.5(4)(100) \] \[ 250 = 10u + 200 \Rightarrow 10u = 50 \Rightarrow u = 5 \, \text{ms}^{-1} \]2. Find total distance (\( S_{20} \)) covered in 20 seconds:\[ S_{20} = (5 \times 20) + 0.5(4)(20^2) = 100 + 800 = 900 \, \text{m} \]3. Distance in the next 10s = Total distance - First 10s distance:\[ 900 - 250 = 650 \, \text{m} \].

The relationship between the decay constant (\( \lambda \)) and the half-life (\( T_{1/2} \)) is given by:

n\[ \lambda = \frac{\ln(2)}{T_{1/2}} \approx \frac{0.693}{T_{1/2}} \] \[ \lambda = \frac{0.6931}{12} \approx 0.05775 \, \text{day}^{-1} \].

An intrinsic semiconductor is a pure semiconductor without any significant dopant species present. In these materials, the only charge carriers are those generated by thermal excitation. When an electron gains enough energy to jump from the valence band to the conduction band, it leaves behind a 'hole' in the valence band. Therefore, for every free electron generated, exactly one hole is created, making their numbers equal.