# NABTEB GCE 2017 Physics Practical Solution Questions and Answer – Nov/Dec Expo

## NABTEB GCE 2017 Physics (Alternative to Practical) Solution Questions And Answer – Nov/Dec Expo Runz.

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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NABTEB GCE 2017 Nov/Dec Physics Practical Answer and Solution to the questions. #### Derek

Editor at FindEB; I'm just another self-employed job writer fanatic. You can reach me via jayderek1@gmail.com

### 1 Comment

1. Anonymous
January 14, 2019

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