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2019/2020 Nabteb SSCE Physics Expo Questions and Answers Examination
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2019 NABTEB SSCE PHYSICS PRACTICAL ANSWERS
NABTEB PHYSICS PRACTICAL SOLUTIONS
(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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(2)
CLICK HERE FOR THE GRAPH
(2i)
Tabulate
S/N 1|2|3|4|5|
y(cm) 10.0|20.0|30.0|40.0|50.0|
X(cm) -30.0|60.0|30.0|24.0|21.4|
X^-1(cm^-1) -0.033|0.016|0.033|0.0416|0.046|
y^-1(cm^-1) 0.100|0.050|0.030|0.025|0.020|
(2avi)
I1 = 0.068cm^-1 , I2=0.066cm^-1
2vaii) m=1/2(I1+I2)=1/2(0.068+0.066)
=1/2*0.134=0.067
M=0.067cm^-1 , K=M^-1=1/0.067
=14.92
K=15cm
(2aviii)
– I ensured I avoided parallax error when taking readings of the meter rule by placing eye vertically above the scale
– I ensured the intensity of source of light was maintained
(2bi)
K=15cm means the focal length of the concave mirror is equal to 15cm
(2bii)
– To provide magnified image of the face
– For applying make-up or shaving
– It is used in motor vehicle head lights
(2biii)
– The image is formed on the screen
– The image is larger than the object
– Further away than the object position
– It is real and inverted
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(3)
CLICK HERE FOR THE GRAPH
(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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_COMPLETED!._
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