2002 - JAMB Physics Past Questions and Answers - page 3

it could be noted that polarization phenomenon is only exhibited by electromagnetic waves

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

The work done on this object is given by the area of the triangle, which is equal to ¹/₂ fx

Here's how to calculate the refractive index: 1. **Identify the angle of incidence and angle of refraction:** * The angle of incidence (i) is the angle between the incident ray and the normal, which is 45 degrees. * The angle of refraction (r) is the angle between the refracted ray and the normal, which is 30 degrees. 2. **Apply Snell's Law:** Snell's law states: n1 * sin(θ1) = n2 * sin(θ2), where: * n1 is the refractive index of the first medium (air, which is approximately 1). * θ1 is the angle of incidence (45 degrees). * n2 is

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

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	(100 x 10-6)2 10 x 10-6
2

=	1	100 x 10-12 10 x 10-6
2

¹/₂ 10^-3 5.0 x 10^-4

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