NECO 2017/2018 QUESTIONS AND ANSWERS

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

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