# 2018/19 NECO GCE PHYSICS PRACTICAL RUNZ & ANSWERS – NOV/DEC EXPO

## 2018/19 NECO GCE PHYSICS PRACTICAL RUNZ & ANSWERS – NOV/DEC EXPO

###### NECO GCE 2018/2019 Physics practical Question and Answers Expo/Runz,NECO Gce 2018 physics Runz, NECO gce 2018 physics practical questions and answers Runs – nov/dec expo,physics PRACTICAL neco gce expo/runs physics practical nov/december sure Expo

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(1aiv)
In a tabular form
Under S/N
1.0, 2.0, 3.0, 4.0, 5.0

Under y(g)
8.00, 10.00, 12.00, 14.00, 16.00

[Under yR(g)= 12.5*y]
100.00, 125.00, 150.00, 175.00, 200.00

Under X(cm)
17.60, 14.00, 11.60, 10.00, 8.70

Under XR(cm) = x * 2
35.00, 28.00, 23.20, 20.00, 17.40

Under L = 70/X
2.000, 2.500, 3.000, 3.500, 4.00

DRAW ANOTHER TABLE OF y(g) and L(cm)
Under y(g)
100.00
125.00
150.00
175.00
200.00

Under L(cm)
2.00
2.50
3.00
3.50
4.00

(1av)
DRAW THE GRAPH

(1avi)
Slope(s) = Dy/DL =
y2 – y1/L2 – L1
Where y2 = 175g, y1 = 100g
L2 = 3.5cm, L1 = 2.00cm

Slope(s) = 175 – 100/3.5 – 2

(1avii)
(PICK TWO)
(i)I ensured mass did not touch/rest on the table
(ii)I avoided error due to parallax on metre rule
(iii)I noted zero error on metre rule
(iv)I avoided draught

(1bi)
The principal of moments state that if a body is in equilibrium,then the sum of the clockwise moments about any point on the body is equal to the sum of the anticlockwise moments about the same point

(1bii)
Draw the diagram

Taking moments about point B ie CWM = ACM
(FA*7) = 10*2
7F = 20
7F/7 = 20/7
F = 2.9N
= 75/1.5
= 50g/cm

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(3a)
In a tabular form
Under S/N
1, 2, 3, 4, 5, 6

Under R(ohms)
5.0, 10.0, 15.0, 20.0, 25.0, 30.0

Under X(cm)
33.00, 49.50, 60.00, 67.00, 71.00, 74.50

Under K = 100 -X(cm)
67.00, 50.50, 40.00, 33.00, 29.00, 25.50

Under Y = x/k
0.4925, 0.9802, 1.5000, 2.0303, 2.4483, 2.9216

(3avi)
DRAW THE GRAPH

(3avii)
Slope
R2 – R1/y2 -y1
= 20 – 5/2.0 – 0.5
= 15/1.5
=10.0 ohms

(3aviii)
(i) I must use a freshly charged accumulator.
(ii) I must take proper readings through my resistance box and potentiometer to avoid error due to parallax

(3bi)
Lost volt is the voltage lost due to the internal resistance of a cell. It is measured in volts.

(3bii)
E = I(R+r)
4 = I(2+0.5)
4 = 2.5I
I = 4/2.5
= 1.6A

Loodinq…..

Stay tuned!!!

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2018/19 NECO GCE PHYSICS PRACTICAL RUNZ & ANSWERS – NOV/DEC EXPO
2018/19 NECO GCE PHYSICS PRACTICAL RUNZ & ANSWERS – NOV/DEC EXPO
2018/19 NECO GCE PHYSICS PRACTICAL RUNZ & ANSWERS – NOV/DEC EXPO
2018/19 NECO GCE PHYSICS PRACTICAL RUNZ & ANSWERS – NOV/DEC EXPO
2018/19 NECO GCE PHYSICS PRACTICAL RUNZ & ANSWERS – NOV/DEC EXPO
2018/19 NECO GCE PHYSICS PRACTICAL RUNZ & ANSWERS – NOV/DEC EXPO