2002 - JAMB Physics Past Questions and Answers - page 3
21
blowing air over a liquid aids evaporation by
A
increasing its surface area
B
decreasing its vapour pressure
C
increasing its temperature
D
decreasing its density
correct option: a
Users' Answers & Comments22
The phenomenon that makes sound persist when its source has been removed is known as
A
acoustic vibration
B
rarefaction
C
echo
D
reverberation
correct option: d
Users' Answers & Comments23
the pressure of 3 moles of an ideal gas at a temperature of 27°C having a volume of 10-3m3 is
A
2.49 x 105-2
B
[]
C
[]
D
[]
correct option: d
Users' Answers & Comments24
Vibration in a stretched spring cannot be polarized because they are
A
stationary waves
B
transverse waves
C
longitudinal waves
D
mechanical waves
correct option: d
it could be noted that polarization phenomenon is only exhibited by electromagnetic waves
Users' Answers & Comments25
A
7.5 x 10-3J
B
2.5 x 10-3J
C
1.5 x 10-2J
D
7.5 x 10-1J
correct option: d
Energy store is equal to the work done in stretching the wire from E to F given by the area of the shaded trapezium under the graph.
= 1/2(0.1 x 0.2) x (0.1 - 0.05)
= 1/2(0.3) x 0.05
= 0.15 x 0.05
= 0.075
= 7.5 x 10-1J
Users' Answers & Comments= 1/2(0.1 x 0.2) x (0.1 - 0.05)
= 1/2(0.3) x 0.05
= 0.15 x 0.05
= 0.075
= 7.5 x 10-1J
26
A
Fx J
B
Fx/2 J
C
Fx2 J
D
F/x J
correct option: b
The work done on this object is given by the area of the triangle, which is equal to 1/2 fx
Users' Answers & Comments27
A
√2/2
B
4/3
C
√2
D
3/2
correct option: c
| n = | sin((A+D)/2) |
| sin(A/2) |
where A = the refracting angle of the prism(60o)
D = angle of deviation of the ray (30o)
| n = | sin((60+30)/2) |
| sin(60/2) |
| = | sin 45 |
| sin 30 |
=
1/√2/1/2
√2/2/1/2
= √2/2 x 2/1 = √2
n = √2
28
A
6Ω
B
9Ω
C
12Ω
D
15Ω
correct option: d
Let the combined parallel resistance of 3 Ω and x = Q Ω
∴ Total resistance in the circuit = 2 + Q + 1.5 = 3.5 + Q Ω
Total current in the circuit = 2A
P.D a cross the circuit 12 volts
∴ From Ohm’s law, V = IR
12 = 2(3.5 + Q)
= 7.0 + 2Q
∴ 2Q = 12 – 7
= 5
∴ Q = 5/2 = 2.5 Ω
Thus from parallel formula: I/Q = 1/3 + 1/x
1/2.5 = 1/3 + 1/x
∴ 1/x = 1/2.5 - 1/3 = 3-2.5/7.5 = 0.5/7.5
X = 7.5/0.5 = 75/5 = 15 Ω
∴ X = 15 Ω
Users' Answers & Comments∴ Total resistance in the circuit = 2 + Q + 1.5 = 3.5 + Q Ω
Total current in the circuit = 2A
P.D a cross the circuit 12 volts
∴ From Ohm’s law, V = IR
12 = 2(3.5 + Q)
= 7.0 + 2Q
∴ 2Q = 12 – 7
= 5
∴ Q = 5/2 = 2.5 Ω
Thus from parallel formula: I/Q = 1/3 + 1/x
1/2.5 = 1/3 + 1/x
∴ 1/x = 1/2.5 - 1/3 = 3-2.5/7.5 = 0.5/7.5
X = 7.5/0.5 = 75/5 = 15 Ω
∴ X = 15 Ω
29
The energy stored in a capacitor of capacitance 10μF carrying a charge of 100μC
A
5 x 104 J
B
4 x 102 J
C
4 x 10-3 J
D
5 x 10-4 J
correct option: d
The energy stored in a given capacitor is given by
E = 1/2 CV2 = 1/2 QV = 1/2Q2/C
Where C = the capacitance of the capacitor
Q = charge carried by the capacitor
V = p.d across the plates of the capacitor thus from the data given, C = 10μF
= 10 x 10-6F
Q = 100μc = 100 x 10-6C
∴ Energy stored = 1/2 Q2/C
1/2 10-3 5.0 x 10-4
Users' Answers & CommentsE = 1/2 CV2 = 1/2 QV = 1/2Q2/C
Where C = the capacitance of the capacitor
Q = charge carried by the capacitor
V = p.d across the plates of the capacitor thus from the data given, C = 10μF
= 10 x 10-6F
Q = 100μc = 100 x 10-6C
∴ Energy stored = 1/2 Q2/C
| = | 1 |
| |||
| 2 |
| = | 1 |
| |||
| 2 |
1/2 10-3 5.0 x 10-4
30
When connected to a mains of 250V, the fuse rating in the plug of an electric device of 1 KW is
A
2 A
B
3 A
C
4 A
D
5 A
correct option: c
1 KW = 1000 watts, power P = IV, thus for a mains of 250 V, the maximum permissible Current I = 1/V = 1000/250 = 4 A
Thus the fuse rating in the plug should be 4 A
Users' Answers & CommentsThus the fuse rating in the plug should be 4 A