NABTEB Physics Practical Questions and Answers – 2019 May/June Expo

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NABTEB Physics Practical Questions and Answers

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👑 2019 NABTEB GCE PHYSICS PRACTICAL ANSWERS👑

100% VERIFIED PHYSICS PRACTICAL ANSWERS

(1a)
Initial position of the pointer on the metre rule
Xo = 37.20cm

Table of values
In a tabular form

S/N: 1.0, 2.0, 3.0, 4.0, 5.0
M(g): 70.00, 90.00, 110.00, 130.00, 150.00
X1(cm): 39.80, 40.00, 42.00, 43.10, 44.20
e = X1 – X0(cm): 2.60, 3.70, 4.80, 5.90, 7.00

(1aviii)
(i) I ensured that I avoided parallax error when taking readings on the metre rule.
(ii) I ensured firm suspension of the scale pan before and after leading the masses.

(1bi)
Hooke’s law states that the extension experienced by an elastic material is directly proportional to the force applied, provided the elastic limit is not exceeded

(1bii)
Pls deduce from the graph.

(1biii)
1/2mv² = Fe
M = 5g = 0.005kg
V = ?
F = 70N
e = 7cm = 0.07m
1/2 × 0.005 × v² = 70 ×0.07
V² = 70×0.07×2/0.005 = 1960
V = √1960 = 44.272 = 44.27m

Instantaneous velocity of the stone when released = 44.27m
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(2ai)
The focal length of the concave mirror provided is approximately equal to 15.0cm

Table of value

In a tabular form
S/N: 1.0, 2.0, 3.0, 4.0, 5.0

y(cm): 10.00, 20.00, 30.00, 40.00, 50.00

x(cm): 75.00, 60.00, 30.00, 24.00, 21.50

1/x(cm-¹): 0.0133, 0.0167, 0.0333, 0.0417, 0.0465

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1/y(cm-¹): 0.100, 0.050, 0.033, 0.025, 0.020

(2avi)
Plot a graph

(2aviii)
(i) I ensured that the image of the cross wires were sharp on the screen before taking my readings.
(ii) I ensured that I avoided parallax error when taking readings on the metre rule by placing my eye vertically above the scale.

(2bi)
K is the focal length of the mirror

(2bii)
(i) They are used as shaving mirrors.
(ii) They are used as car driving mirrors.
(iii) They are used as dentist mirrors.

(2biii)
(i) The image is magnified (ie larger than the object)
(ii) The image is real.
(iii) The image is inverted.
================================

(3i)
EMF=2.0V
Tabulate
K(Ω) | I(A) | I^-1(A^-1)
1.0 | 0.44 | 2.573
2.0 | 0.34 | 2.941
3.0 | 0.28 | 3.571
4.0 | 0.24 | 4.167
5.0 | 0.22 | 4.545

(3avii)
Slope (s) =ΔI^-1(A^-1)/ΔR(Ω) =4.85-2.60/4.90-1.50
=2.25/3.40 =0.662A^-1Ω^-1
Intercept (c) =2.65A^-1
3aviii) K=E*C=2*2.65=5.3A^-1V

(3aix)
– I ensured that I tightened the connection
– I avoided parallax error in reading ammeter/voltmeter

(3bi)
The value of K gives the overall resistance R(Ω)

(3bii)
Terminal p.d is the potential difference a cell in a circuit unit. When no current is flowing the terminal p.d is equal to the e.m.f of the cell. If a current is flowing then the terminal p.d of the cell will be lower than it’s EMF as some loss of energy takes place due to its internal resistance

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