1994 - JAMB Physics Past Questions and Answers - page 3

t = 4s ; v = 150ms^-1

depth = X ; but 2X = vt

X = \(\frac {vt}{2}\) = \(\frac {1500 \times 4}{2}\)

X = 3000m

F \(\alpha\) \(\frac{1}{L}\)

\(\lambda\) = 4L = 4 X 45 = 180cm

F = \(\frac {V}{\lambda}\)

= \(\frac {360}{180 \times 10^{-2}}\)

= 200Hz

F = \(\frac {1}{2L}\) \(\sqrt\frac{T}{m}\)

T₂ = 4T₁ ; L₂ = 2L₁

\(\frac{2F_1L_1}{2F_2L_2}\) = \(\sqrt\frac{T_1}{T_2}\)

\(\frac{400\times L_1}{F \times L_1}\) = \(\sqrt\frac{T_1}{4T_1}\)

F = 400Hz

1.5 = \(\frac{6 \times 10^{14}}{F}\)

F = 4 x 10¹⁴Hz

V = -2u

F = \(\frac{uv}{u + v}\)

20 = \(\frac{uv - 2u}{u - 2u}\)

u = 10cm

1.5 = \(\frac{3}{\text {apparent displacement}}\)

Apparent depth = 2cm

displacement = 3 - 2

= 1cm