Neco Physic Practical QUESTION AND ANSWERS NOW 100% VERIFY

NECO Physics Practical Questions and Answer Solutions – May/June 2017 Expo Runz@ blackload

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
2aix)
PRECAUTIONS
-I ensured that all apparatus are in straight line
-I avoided error due to parallax when reading the metre rule
-I avoided zero error on the metre rule

2bi)
u=10cm
f=15cm
Using 1/v+1/u=1/f
1/f-1/u=1/v
1/v=1/15-1/10
1/v=(2-3)/30
1/v=-1/30
v=-30cm
The characteristics of image formed are:
-It is virtual
-It is enlarged and magnified ie twice or two times as big as the object m=2

2bii)
The concave mirror mounted in its holder is moved to and fro in front of the screen until a sharp image of the cross wire of the ray box is formed on the screen adjacent of the object.The distance between the mirror and the screen was measured as 30cm since the radius of curvature r=2fo then half is distance

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

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

Posted by Tblogger 5/6/2017