2017 - WAEC Physics Past Questions and Answers - page 4
The inverse of the time required for a wave to complete one full cycle is called
The time required to complete one full cycle is the period; the inverse of the period gives the frequency.
Two bar magnets are placed near each other. Which of the following diagrams correctly illustrates the magnetic field pattern formed by the pair?
The force per unit charge experienced at a point in a field is the
Electric intensity at a point is defined as the force experienced per unit positive charge at a point placed in the electric field.
Which of the following objects is not a conductor of electricity?
A circuit is set up as shown in the diagram above. When the key is closed, the ammeter reading will be
The resistors are connected both in parallel and series. To calculate the total resistance, we have:
For the parallel connection: \(\frac{1}{R} = \frac{1}{1} + \frac{1}{1}\)
\(\frac{1}{R} = 2 \implies R = 0.5\Omega\)
For the series connection: \(R_{total} = (1 + 0.5)\Omega = 1.5\Omega\)
Recall, \(V=IR\)
\(3 = 1.5I \implies I = \frac{3}{1.5} =2A\)
Which of the following statements is/ are correct?
I. Gravitational potential due to a mass at a point a distance r away is proportional to \(r^{2}\)
II. Gravitational field strength is a vector quantity
III. Gravitational intensity is proportional to the weight of a unit mass.
The diagrams above illustrate electrical connections to the element of an electric pressing iron.In which of the connections is the fuse correctly fixed?
A lamp rated 100W, 240V is lit for 5hours. Calculate the cost of lighting the lamp if 1kWh of electrical energy cost N5.
Power = 100W; Voltage = 240V; Time = 5 hours
Total energy used = Power x time = 100 x 5 = 500wh
converting to kwh, we have
\(\frac{500}{1000} = 0.5kWh \times N5 = N2.50\)
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.
E = 3.0V; \(\frac{R}{l} = 10\Omega m^{-1}\)
l = 30cm = 0.3m; R = \(10\Omega m^{-1} \times 0.3m = 3\Omega\)
\(\frac{E}{V} = \frac{l_{1}}{l_{2}} \implies \frac{3}{V} = \frac{1}{0.3}\)
= \(\frac{3\times 0.3}{1} = 0.9V\)
The main reason why a.c. is transmitted at very high voltage and low current is that