2002 - JAMB Physics Past Questions and Answers - page 3
it could be noted that polarization phenomenon is only exhibited by electromagnetic waves
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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
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The work done on this object is given by the area of the triangle, which is equal to 1/2 fx
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| 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
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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 & CommentsThe 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 |
| = | 1 |
| |||
| 2 |
1/2 10-3 5.0 x 10-4Users' Answers & Comments
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
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