2018 Neco Physics Practical Questions and Answers | Physics Practical Expo

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PHYSICS PRACTICAL ANSWERS


(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




Tuesday 5th June 2018
Practical � Physics � 10:00am � 12:45pm


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Last edited by Emmi on Tue Jun 05, 2018 10:18 am; edited 2 times in total

Boss Emmi,  I believe in you sir

Sir,  thanks for your help, please we need physics early