2007 - JAMB Physics Past Questions and Answers - page 5

Workdone = product of the charge Q, and the p.d V.

i.e Work done = QV
= 600 * 10^-6 *2.1 * 10³

=1.8j

\( R = \frac{1}{a} p \implies p = \frac{R.a}{1} = \frac{0.1 \times 3.14}{2} \left ( \frac{10^{-3}}{2} \right)^2 \\
\text{Therefore } p = \frac{0.1 \times 3.14 \times 10^{-6}}{2 \times 4} \\
= 0.3925 \times 10^{-7} \text{ OR } -3.930 \times 10 ^{-8}Ωm \)

In general, Power = IV; \( \implies 40KW = IV \\ \text{Therefore } 40000 = 1 \times 800\\
\implies I = \frac{40000}{800} = 50A,\text{tune the current through} \)

Resistor = 50A
power loss= I²R = 50² x 2
= 2500 x 2 = 5.0 x 10³W

since the wires are parallel to each other, the forces which they exert on each other are equal and opposite, since they must agree with newton's law of action an reaction. Hence the force exerted by the wire is equally
5 x 10^-5

Lenz's lew of electrical magnetic induction is usefully referred to as the law of conservation of energy

Power of the lamp = 60W = \( \frac{v^2}{R} \) and the voltage rate = 120V

\( \implies R = \frac{V^2}{60} = \frac{120 \times 120}{60} = 240Ω \)

Now, if the resistance of the lamp is 240Ω the fuse wire that should ensure current does not exceed this value should be rated 200Ω.

80 days = 4 half lives (i.e 4 x 20dqs = 80dq)

But the fraction always left by any radioactive material after 4 half lives is

\( \left ( \frac{1}{2} \right)^2 = \frac{1}{16} \)