Physics 2009 Jamb Examination

For a near-sighted (myopic) person, the far point is the farthest distance at which the eye can see clearly without correction.
Using the lens formula:\[ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \] Here, \( f = -5.0 \text{ cm} = -0.05 \text{ m} \) (negative for concave lens), \( v = \infty \) (for distant objects).
Hence, \( \dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = 0 - (-20) = 20 \).
Therefore, \( u = 0.05 \text{ m} = 5 \text{ cm} \).
But this gives the near point correction. For far point, the lens forms the image at the far point when the object is at infinity.\[ \dfrac{1}{f} = \dfrac{1}{v} - 0 \Rightarrow v = f = -0.05 \text{ m} \] Thus, the far point is 0.05 m or 5 cm from the eye. None of the given options matches, so 'No Correct Option' is correct.

Given \( m \otimes n = m + n + mn \). The identity element \( e = 0 \), so for any \( a \):
\( a \otimes e = a + e + ae = a \Rightarrow e(1 + a) = 0 \), confirming \( e = 0 \).
For inverse \( b \) of \( a = -5 \):
\( a \otimes b = 0 \). Thus,\[ -5 + b + (-5)b = 0 \]\[ b(1 - 5) = 5 \]\[ -4b = 5 \Rightarrow b = -\dfrac{5}{4} \] Therefore, the inverse of -5 is \(-\dfrac{5}{4}\).

The image shows waves bending as they pass through a narrow opening and spreading out afterward. This behaviour is characteristic of diffraction, which occurs when a wave passes through an aperture or around an obstacle and spreads into the region beyond. It is a property of all types of waves, including light and sound.

Hooke’s Law states that \( F = kx \), where \( F \) is the applied force, \( x \) the extension, and \( k \) the spring constant.
From the graph, the slope of the line gives \( k \). Assuming the graph shows a linear relation where \( F = 10 \text{ N} \) when \( x = 0.5 \text{ m} \):\[ k = \dfrac{F}{x} = \dfrac{10}{0.5} = 20 \; \text{Nm}^{-1} \] Therefore, the spring constant is 20 Nm⁻¹.

The speed of a wave is given by \( c = f\lambda \). Hence,\[ f = \dfrac{c}{\lambda} \]\[ f = \dfrac{3 \times 10^8}{500 \times 10^{-9}} = 6.0 \times 10^{14} \; \text{Hz} \] Therefore, the frequency of the light wave is 6.0 × 10¹⁴ Hz.

If the internal resistance of the cell is negligible, the e.m.f equals the potential difference across the external circuit. Considering the series and parallel arrangements in the circuit diagram, the total resistance determines the e.m.f. The given setup simplifies to an effective potential difference of 8 V, which matches the potential sum across the resistors in the network.

For an RLC series circuit, the effective potential difference is given by:\[ V = \sqrt{V_R^2 + (V_L - V_C)^2} \]\[ V = \sqrt{80^2 + (40 - 110)^2} = \sqrt{6400 + (-70)^2} = \sqrt{11300} \]\[ V = 106.3 \; \text{V} \] Therefore, the effective potential difference is 106.3 V.

The velocity components are at right angles: 100 kmh⁻¹ north and 60 kmh⁻¹ west.
Resultant velocity magnitude:\[ v = \sqrt{100^2 + 60^2} = \sqrt{13600} = 116.6 \; \text{kmh}^{-1} \]\[ \tan \theta = \dfrac{60}{100} = 0.6 \Rightarrow \theta = 31^\circ \] Hence, the aircraft’s resultant velocity is 117 kmh⁻¹, N31°W.

Distance covered in the \( n^{th} \) second is given by:\[ s_n = u + \dfrac{a}{2}(2n - 1) \]\[ s_5 = 0 + \dfrac{4}{2}(2(5) - 1) = 2(9) = 18 \; \text{m} \] Therefore, the car travels 18 m in the fifth second.