Neco 2018 Physics Answers – May/June – Expo

PHYSICS PRACTICAL ANSWERS

CLICK HERE FOR NUNBER 2 GRAPH

(2a)
In a tabular form:

Under tita°:
75, 65, 55, 45, 35

Under MO(cm):
1.1, 2.0, 2.5, 3.4, 3.9

Under NO(cm):
6.2, 6.4, 6.6, 6.8, 7.2

Under H=MO/NO:
0.177, 0.313, 0.379, 0.500, .542

Under Costita:
0.2588, 0.4226, 0.5736, 0.7071, 0.8192

(2axiii)
From the graph
Slope, S =Δcostita/^ΔH
= 0.75-0.45/0.5-0.3
= 0.3/0.2
=1.5

(2axiv)
(i) I ensured both the object and the pins were in straight lines so as to avoid error due to parallax.
(ii) I made sure there was no air interference

2bi)
Snell’s law of refraction states that the ratio of the sine of angle of incidence to the sine of the angle of refraction is a constant for a given pair of media.
I.e Sini/sinr = Π
Where Π is known as refractive index.

(2bii)
Given refractive index of glass = 1.5
i.e aΠg = 1.5(from air to glass)
SinC/sin90 = gΠa
SinC/Sinn90 = 1/aΠg
SinC/1 = 1/1.5
SinC= 0.6667
C = sin^-1(0.6667)
Critical angle for glass C = 42°

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CLICK HERE FOR NUMBER 3 GRAPH

(3a)
(ii) Vo = 2.00v

(3av)
In a tabular form
Under S/N
1, 2, 3, 4, 5

Under R(ohms):
2, 3, 4, 5, 6

Under V(v):
2.10, 2.30, 2.40, 2.50, 2.60

Under R^-1:
0.500, 0.333, 0.250, 0.200, 0.167

Under V^-1(v^-1):
0.476, 0.435, 0.417, 0.400, 0.385

3vii Slope, = Δv^-1/ΔR-1
= 0.5 – 0.355/0.6 – 0
= 0.145/0.6
S = 0.242

Intercept, C = 0.355v^-1

(3aviii)
K = S/C
K = 0.242/0.355
K = 0.68

(3ix)
(i) I ensured tight connections.
(ii) I ensured clean terminals.

(3bi)
(i) Temperature of wire
(ii) Cross sectional area of wire
(iii) Length of wire
(iv) Nature of wire

(3bii)
Draw the diagram

Effective E.m.f = 2v(parallel connection)
Effective internal resistance
= r * r/r + r = r/2ohms

Current, I , 0.8A
External resistance R = 2A
Using,
E = I(R+r)
2 = 0.8(2+r/2)
2.5 = (2+r/2)
2.5 – 2 = r/2
0.5 = r/2
r = 0.5*2 = 1ohms

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