WAEC Physics Past Questions & Answers

231.

Which of the following pairs of concepts constitutes complementary variables as expressess by the Heinsberg uncertainty principle?

Momentum and time

Energy and acceleration

Position and momentum

Velocity and Momentum

View answer & explanation

Correct answer is C

No explanation has been provided for this answer.

232.

The opposition offered to an alternating current in a C-R series circuit is known as

impedance

resistance

reactance

friction

View answer & explanation

Correct answer is A

No explanation has been provided for this answer.

233.

The diagram above illustrates two charged bodies, X and Y separated by a distance r. Each body has charge q and mass m. Which of the following statements about the electrostatic force $F_{e}$ and gravitational force $F_{g}$ acting between the bodies is correct?

$F_{e}$ is attraction while $F_{g}$ is repulsion

Both $F_{e}$ and $F_{g}$ are that of repulsion

Both $F_{e}$ and $F_{g}$ is that of attraction

$F_{e}$ is repulsion while $F_{g}$ is attraction

View answer & explanation

Correct answer is D

electrostatic force can be repulsive or attractive and gravitational force is only attractive.

234.

The main reason why a.c. is transmitted at very high voltage and low current is that

flows faster through the conductor

can be stepped down

is cheaper to generate

reduces heat loss

View answer & explanation

Correct answer is D

No explanation has been provided for this answer.

235.

A battery of e.m.f 3.0V is connected across a potentiometer wire AB of length 10 $\Omega m^{-1}$ as illustrated in the diagram above. If the jockey J makes contact at the 30cm mark, determine the resistance of length AJ and voltage across AJ.

3.0 $\Omega$ ; 0.9V

3.0 $\Omega$ ; 3.0V

3.3 $\Omega$ ; 0.3V

3.3 $\Omega$ ; 9.0V

View answer & explanation

Correct answer is A

E = 3.0V; $\frac{R}{l} = 10\Omega m^{-1}$

l = 30cm = 0.3m; R = $10\Omega m^{-1} \times 0.3m = 3\Omega$

$\frac{E}{V} = \frac{l_{1}}{l_{2}} \implies \frac{3}{V} = \frac{1}{0.3}$

= $\frac{3\times 0.3}{1} = 0.9V$

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