JAMB Physics 2019 Paper

Concept:A siren produces a note because each hole lets a puff of air through once per revolution, so the frequency equals the number of holes times the number of revolutions per second.Then use the wave equation $v = f \lambda$ to find the wavelength.Explanation:Convert the siren rotation from rev/min to rev/s:$132\ \text{rev/min} = \frac{132}{60}\ \text{rev/s} = 2.2\ \text{rev/s}$.The ring has $200$ holes, so the frequency is:$f = 200 \times 2.2 = 440\ \text{Hz}$.Now apply $v = f\lambda$, where $v = 350\ \text{m/s}$:$\lambda = \frac{v}{f} = \frac{350}{440} = 0.795\ \text{m}$.This matches option A.Answer:The note produced has frequency $440\ \text{Hz}$ and wavelength $0.795\ \text{m}$. Therefore, the correct answer is A.