Physics 2019 Jamb Examination

Physics 2019 Jamb Examination
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Question 101

Calculate the velocity ratio of a screw jack of pitch 0.2cm if the length of the tommy bar is 23cm
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The velocity ratio (V.R.) of a screw jack is given by the distance moved by the effort divided by the distance moved by the load in one full turn. The effort moves in a circle of radius \( L \), while the load moves up by one pitch \( p \): \[ V.R. = \frac{2\pi L}{p} \] Given: \( L = 23 \; \text{cm} \) \( p = 0.2 \; \text{cm} \) \[ V.R. = \frac{2\pi \times 23}{0.2} = \frac{46\pi}{0.2} = 230\pi \]
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Question 102

The Earth's magnetic equator passes through Jos in Nigeria. At Jos, the
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The 'angle of dip' (or inclination) is the angle the Earth's magnetic field lines make with the horizontal. At the magnetic poles, the field is vertical (dip = 90°). At the magnetic equator, the field lines are perfectly horizontal, so the angle of dip is 0°.
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Question 103

A car moving at 20m/s with its horn blowing (f = 1200Hz) is chasing another car going at 15m/s. What is the apparent frequency of the horn as heard by the driver being chased?
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This involves the Doppler Effect. The formula for frequency heard (\( f' \)) when the source (s) and observer (o) are moving in the same direction is: \[ f' = f \left( \frac{v - v_o}{v - v_s} \right) \] Where: \( v = 340 \; \text{m/s} \) (speed of sound) \( v_o = 15 \; \text{m/s} \) (observer, moving away from source) \( v_s = 20 \; \text{m/s} \) (source, moving toward observer) \( f = 1200 \; \text{Hz} \) \[ f' = 1200 \left( \frac{340 - 15}{340 - 20} \right) = 1200 \left( \frac{325}{320} \right) \] \[ f' = 1200 \times 1.015625 = 1218.75 \; \text{Hz} \approx 1.22 \; \text{KHz} \]
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Physics 2019 Jamb Examination
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