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# NECO 2018 Physics Practical Answer – June/July Expo

NECO Physics Practical Solution Questions and Answer – June/July 2018 Expo Runz

ANSWERS HAS BEEN SENT TO SUBSCRIBERS SIBCE 5AM. ALWAYS SUBSCRIBE TO GET ANSWRR EARLIER.

(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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(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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