2017/18 NECO PHYSICS PRACTICAL ANSWERS BY COOLBOY

2ai)
fo=15cm
2av)
a=60.00cm
b=20.00cm
Hence L=a/b=60.00/20.00
L=3
2avi)
TABULATE
S/N:1,2,3,4,5
b(cm):20.00,25.00,35.00,40.00
a(cm):60.00,37.50,30.00,26.25,24.00
L=a/b:3.00,1.50,1.00,0.75,0.60
2avii)
Slope=Change in L/Change in a
=(3-0.25)/(60-18.6)
=2.75/41.4
=0.006642
2aviii)
S^-1=1/S
=(1/0.0066425cm)
S=15.05
S=15cm

2bi)
he characteristics of imaged formed are :
i)It is virtual
ii) It is enlarged or magnified i.e. twice or two times bigger as the object(m=2)

2bII) The concave mirror ,mounted in its holder ,is moved to and fro in front of the 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

3a)
Vo=0.35V
TABULATE
S/N:1,2,3,4,5,6
R(ohm):1.0,2.0,3.0,4.0,5.0,6.0
V(v):1.40,1.65,1.80,2.00,2.10,2.15
R^-1(ohm^-1):1.000,0.500,0.333,0.200,0.167
V^-1(v^-1):0.714,0.606,0.556,0.500,0.476,0.465
DRAW THE GRAPH of V^-1(v^-1) against R^-1
3aiii)
Vo=0.35V
TABULATE
S/N:1,2,3,4,5,6
R(ohm):1.0,2.0,3.0,4.0,5.0,6.0
V(v):1.40,1.65,1.80,2.00,2.10,2.15
R^-1(ohm^-1):1.000,0.500,0.333,0.200,0.167
V^-1(v^-1):0.714,0.606,0.556,0.500,0.476,0.465
3aiv)
DRAW THE GRAPH of V^-1(v^-1) against R^-1
3av)
Slope=change inV^-1/Change inR^-1
=(7.6-4.3)/(0.84-0.10)
3.3/0.74
=4.460
3avi)
Intercept=0.385
3avii)
I=0.385
I^-1=(0.385)^-1
=2.597