1a)
TABULATE:
S/N: 1, 2 , 3, 4, 5
L(cm): 90.00, 80.00, 70.00, 60.00, 50.00
T(s): 38.00, 36.00, 34.00, 31.00, 28.00
T = t/20(sqt): 1.90, 1.80, 1.70, 1.55, 1.40
√L(cm^1/2): 9.49, 8.94, 8.37, 7.73, 7.07
(1aix)
Slope(s) = Δvertical/Δhorizontal
=1.90 – 1.55/9.49 – 7.75
= 0.35/1.74
S = 0.201
(1ax)
g = 4π²/S²
= 4 ×(3.142)²/(0.201)²
= 977.4cm/s²
(1axi)
(i) I ensured the angle of oscillation is relatively small.
(ii) I ensured that no external force is added to the system of oscillation.
(iii) I ensured the oscillation is perfect
(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°

The concave mirror, mounted in its holder, is moved to and fro in front of them until a sharp image of the cross wire of the ray box is formed on the screen adjacent to the object. The distance between the mirror and the screen was measured as 30.1cm. Since the radius of the curvature r,=2fo ,then half this distance is the focal length of the mirror fo . Thus focal length was determined to be 15.05cm approximately 15cm
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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