1998 - WAEC Physics Past Questions and Answers - page 5
41
Calculate the resistance of the filament of lamp rated at 240V 60W
A
240\(\Omega\)
B
360\(\Omega\)
C
960\(\Omega\)
D
144\(\Omega\)
42
The diagram above illustrates a freely suspended bar magnet NS hanging from a point in a horizontal ceiling. The thread used for suspending the magnet and the axis of the magnet are as indicated on the diagram. The angle marked \(\theta\) is called the angle os
A
depression
B
declination
C
variation
D
elevation
correct option: e
Users' Answers & Comments43
Lenz's law of electromagnet states that
A
the induced charge is constant for a fixed change of force
B
a force is exerted on a current-carrying conductor in a magnetic field
C
the magnitude of the induced e.m.f in a circuit is proportional to the rate of change of the number of lines of force linking the circuit
D
the induced current in a conductor is in such a direction as to oppose the change producing it
correct option: d
Users' Answers & Comments44
A proton of charge 1.6 x 1019C is projected into a uniform magnetic field of flux density with a constant speed of 1.6 x 106ms-1, calculate the magnitude of the force exerted on it by the field
A
0.0N
B
2.0 x 10-21N
C
1.3 x 10-17N
D
5.1 x 10-14N
45
The direction of the magnetic field at a point in the vicinity of a bar magnetic is
A
along the line joining the point to the neutral point
B
always away from the south pole of the magnet
C
opposite the direction of theresultant field at that point
D
always towards the north pole of the magnet
correct option: e
Users' Answers & Comments46
The diagram below illustrates an a.c.source of 50V(r.m.s.), \(\frac{100}{\pi}\)Hz connected in series with an inductor of inductor of inductance L and a resistor of resistance R. The current in the circuit is 2A and the p.d across L and R are 30V and 40V respectively. Calculate the power factor of the circuit
A
1.33
B
1.25
C
0.80
D
0.75
correct option: e
cos\(\theta = \frac{R}{Z}\)
R = \(\frac{V}{I} = \frac{40}{2} = 20\)
Z = R2 + \(^X_L\) = 202 + 152 = 300
cos \(\theta = \frac{R}{Z} = \frac{20}{300} = 0.06\)
Users' Answers & CommentsR = \(\frac{V}{I} = \frac{40}{2} = 20\)
Z = R2 + \(^X_L\) = 202 + 152 = 300
cos \(\theta = \frac{R}{Z} = \frac{20}{300} = 0.06\)
47
The diagram below illustrates an a.c.source of 50V(r.m.s.), \(\frac{100}{\pi}\)Hz connected in series with an inductor of inductor of inductance L and a resistor of resistance R. The current in the circuit is 2A and the p.d across L and R are 30V and 40V respectively. Calculate the average power dissipated in the circuit
A
100W
B
80W
C
60W
D
20W
48
Use the following data to determine the length L of a wire when a fore of 30N is applied, assuming Hooke's law is obeyed
\(\begin{array}{c|c}\text{Force applied/N} & 0 & 5 & 10 & 30\ \hline \text{length of wire/mm} & 500.0 & 500.0 & 501.0 & L \end{array}\)
\(\begin{array}{c|c}\text{Force applied/N} & 0 & 5 & 10 & 30\ \hline \text{length of wire/mm} & 500.0 & 500.0 & 501.0 & L \end{array}\)
A
3.0mm
B
3.5mm
C
503.00mm
D
503.5mm
correct option: c
F \(\alpha\) e
F = ke
K = \(\frac{F}{e} = \frac{5}{0.5} = 10\)
when F = 30, e = \(\frac{F}{K} = \frac{30}{10} = 3\)
L = 500 + 3 = 503.00
Users' Answers & CommentsF = ke
K = \(\frac{F}{e} = \frac{5}{0.5} = 10\)
when F = 30, e = \(\frac{F}{K} = \frac{30}{10} = 3\)
L = 500 + 3 = 503.00
49
Which of the following explains the concave meniscus of water in a clean glass tube? the
A
adhension between water and glass molecules is greater than the sdhension between water molecules
B
cohension between water molecules is greater then the adhension between glass and water molecule
C
molecules of water near the glass move faster than the molecules at the centre of the tube
D
molecules of water at the water-air boundary are often attracted to the centre of the tube
correct option: a
Users' Answers & Comments50
An electron is accelerated from rest through a potential difference of 70kV in a vacuum. Calculate the maximum speed acquired by the electron (electronic charge = -1.6 x 10-19; mass of an electron = 9.1 x 10-31kg)
A
3.00 x 108ms-1
B
2.46 x 108ms-1
C
1.57 x 108ms-1
D
1.32 x 108ms-1
correct option: c
ev = \(\frac{1}{2}\)mv2
v2 = \(\frac{ev}{\frac{1}{2}m}\)
V = \(\sqrt{\frac{ev}{\frac{1}{2m}}} = \sqrt{\frac{1.6 \times 10^{-19} \times 70 \times 10^3}{\frac{1}{2} \times 9.1 \times 10^{-31}}}\)
= \(\sqrt{246.2 \times 10^{14}}\) = 1.57 x 108
Users' Answers & Commentsv2 = \(\frac{ev}{\frac{1}{2}m}\)
V = \(\sqrt{\frac{ev}{\frac{1}{2m}}} = \sqrt{\frac{1.6 \times 10^{-19} \times 70 \times 10^3}{\frac{1}{2} \times 9.1 \times 10^{-31}}}\)
= \(\sqrt{246.2 \times 10^{14}}\) = 1.57 x 108