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NABTEB GCE 2017 Physics Practical Expo Answer – Nov/Dec

NABTEB GCE Nov/Dec 2017 Physics practical Questions and Answer Expo

(1a)
Tabular form 
S/N=1| 2 |3 | 4 | 5 |
M(g)=23.0|29.0|35.0|37.0|40.0|45.0|
θ(°)=25.0|30.0|35.0|40.0
tan θ=0.4663|0.5774|0.7002|0.8391|1.000
Slope S =∆M / ∆tanθ
=50.0 – 35.0 / 1.00 – 0.70
=15 / 0.3 
=50.0
(1avii)
To comfims the mass needed to just move block b
(1viii)
(i)i would ensure that I avoided parallax error
(ii) I would ensure the mass did not rest on a table
(1bi)
(i)Friction opposes motion
(ii)The limiting frictional force is dependent of the nature of the surface in contact
(iii) The maximum force of static friction is comparative to the normal force
(1bii)
(i) By streamlining
(ii)By lubricating the surface
(iii)By polishing the surface
(1biii)
F=MgSinθ
where m=2kg
θ=30°
=2 × 10sin30°
=20 × 0.5
=10.0N
=========================================
(2a)
IN TABULAR FORM
S/N=| 1 I| 2 | 3 | 4 | 5 |
X(cm)=20.2|15.3|11.2|8.8|6.7|
Y(cm)=7.6|9.6|12.8|16.2|20.2|
X+Y(cm)=27.8|24.9|24.0|25.0|26.2|
XY(cm^3)=153.52|146.88|143.36|142.56|135.34|
(2av)
slope S = ∆(X+Y)cm / ∆XY(cm^3)
=23.0 – 5.0 /140 – 80
=0.3cm^-1
(2avi)
K = 1/S
=1/0.3
=3.333cm
(2avii)
(i)I would ensure avoided parallax error when reading the meter rule
(ii)I would ensure that sharp image is formed on the screen
(2bi)
K confirm the sum of the distance of the object and image formed
(2bii)
A real image is the one through which the ray of light pass. If a screen is placed at the position of real image, the image is seen on he screen While 
A virtual image is one whereby rays of light do not pass but which nevertheless visible to the eye
(2biii)
1/F= 1/U + 1/U
F=
U=25.0cm
V=40.0cm
1/F =1/25 + 1/40
1/F =40 + 5 / 200
1/F =200 / 45
=4.4cm
therefore focal length of the lens =4.4cm
=============================================
(3a)
IN TABULAR FORM
S/N =1 | 2 | 3 | 4 | 5 |
R(π)=1.0|2.0|3.0|4.0|5.0|
Y(cm)=16.5|25.5|30.0|33.0|36.0|
P=R^-1(π^-1)=1.000|0.500|0.333|0.143|0.091|
Q=100/1(cm^-1)=6.061|4.000|3.333|3.030|2.778|
(3avi)
slope S=∆P/∆Q
=0.76 – 0.20 / 5.00 – 2.80
=0.56 / 2.2
=0.2255cmΩ^-1
(3avii)
K = 1/S
= 1/0.255
=3.922Ω/cm
(3aviii)
(i)I would ensure tight connection of wires at the terminal 
(ii)I would ensure that the key is not closed when reading are not being taken to avoid cells being run down
(3bi)
Intercept C = – 0.5Ω^-1
therefore Z= 1/C
= 1/-0.5 
=-2.0Ω
significance:To confirmed of the resistance
(3bii)
(i)length of the wire 
(ii)Temeperature of wire. 
(iii)cross sectional area
(3biii)
Resistivity of a given wire is defined as the resistance of a unit length of the wire of a unit cross- sectional area
P =RA/L
A=cross sectional area
R=Electrical resistance 
L=length of the wire

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