2021 - JAMB Physics Past Questions and Answers - page 4

\(\Delta\) = \(\frac{\bigtriangleup\theta}{\delta}

= \frac{80-20}{0.02}\)

= 3.0 x 10\(^{3}\)km\(^{-1}\)

mc\(\Delta\theta\) = m\(\lambda\)f

2 x 400 (100 - 0) = m(3.3 x 10\(^{5}\))

m = \(\frac{2 \times 400 \times 100}{3.3 \times 10^{5}}

= \frac{8}{33}\)kg

y = A sin (wt + 0)

y = 0.25 x 10\(^{-3}\)sin (500t - 0.025x)

w = 5.00 x 10\(^{2}\)rads\(^{-1}\)

When sound travels from hot air to cold air or from cold air to hot air it will refract, the molecules of air are moving faster and the vibrations of the sound wave can therefore be transmitted faster. Hence, the speed of sound is greater in hot air than it is in cold air

f\(_{0}\) = 40H\(_{3}\)

1st overtone = 2f\(_{0}\) = 80H\(_{3}\)

2nd overtone = 3f\(_{0}\) = 120H\(_{3}\)

3rd overtone = 4f\(_{0}\) = 160H\(_{3}\)

4th overtone = 5f\(_{0}\) = 200H\(_{3}\)

Hence, the number of overtones between 40H\(_{3}\) and 180H\(_{3}\) = 3

object distance (u) = image distance (v)

u = 3m

u|v| = 3 + 3 = 6m

f = \(\frac{m}{4}\)

= \(\frac{d}{4}\)

= \(\frac{20}{4}\)

= 5cm

\(\frac{\text {sin i}}{\text {sin r}}\) = \(\frac{\text {velocity in air}}{\text {velocity in glass}}\)

\(\frac{\text {sin i}}{\text {sin 30}}\) = \(\frac{3 \times 10^{8}}{2 \times 10^{8}}\)

sin i = \(\frac{3}{4}\)

= 0.75