JAMB Physics Past Questions & Answers - Page 83

u = , H = 20, g = 10

From equation of motion under gravity

H = ut ± \(\frac{1}{2}\)gt², It is falling so g is +ve

H = 0 + \(\frac{1}{2}\)gt²

H = \(\frac{1}{2}\)gt²

2H = gt²

\(\frac{2H}{g}\) = t²

t = \(\frac{\sqrt{2H}}{g}\)

= \(\frac{\sqrt {2 \times 20}}{10}\)

= √4

= 2s

V = IR from ohms law

I = \(\frac{V}{R}\)

Since the resistance are in parallel

\(\frac{1}{Reff}\) = \(\frac{1}{8}\) + \(\frac{1}{12}\) + \(\frac{1}{6}\)

= \(\frac{2 + 3 + 6}{36}\) = \(\frac{11}{36}\)

Reff = \(\frac{36}{11}\) = 3.27

I = I₁ + I₂ + I₃

= \(\frac{V}{Reff}\)= \(\frac{12}{3.27}\)

= 3.669A

I = 3.7A

Seven 40W lamps = 7 × 40 = 280W

Five 80W lamps = 5 × 80 = 400W

Total power = 280W + 400W

= 680W

= \(\frac{680KW}{1000}\) = 0.68KW

The kilowatt hour = 0.68 KW × 12 hrs

= 8.16 KWh

Cost is N 7 per KWh

So the cost of running them = 8.16KWh × N 7/KWh

= 57.12

For a refractive index () = \(\frac{\sin\frac{1}{2} (A + D)}{\sin\frac{1}{2}A}\)

D = angle of minimum deviation

A = refractive angle of the prism

1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin\frac{1}{2} \times 60}\)

1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin 30}\)

Sin \(\frac{1}{2}\) (60 + D) = 1.5 * Sin 30

Sin \(\frac{1}{2}\) (60 + D) = 0.75

\(\frac{1}{2}\) (60 + D) = Sin-1 (0.75)

but Sin-1 (0.75) = 49o

\(\frac{1}{2}\) (60 + D) = 49

60 + D = 2 * 49 = 98o

D = 98o - 60o

D = 38o

When silicon is doped with a trivalent atom. A p-type semi-conductor will be formed when silicon is doped with a pentavalent atom such as arsenic, a n-type semi-conductor will be formed
Phosphorous is a pentavalent element, so it will produce a o -type semi-conductor