– NABTEB Physics Questions –
NABTEB Physics Questions is out now on our website. In this article, I will show you past NABTEB Physics random repeated questions for free. All you need to do is to stay focused and follow this guide.
About NABTEB
The National Business and Technical Examinations Board were founded in 1992 to domesticate craft level examinations previously conducted by , Pittman’s, and the throughout the United Kingdom in conformity with the principles of the National Policy on Education.
Its formation was the culmination of a 15-year evolutionary process in which FOUR Government Panels were established at various periods to assess the place and structure of public examinations in our educational system.
Mission
Develop as a globally recognized assessment body for Craftsmen and Technicians.
Vision
NABTEB is a vision-led, mission-driven public institution with the following vision: To be a Globally Recognised Assessment and Certification Body Preparing Candidates for the Workplace and Academic/Professional Excellence.
The management of the National Business and Technical Examinations Board (NABTEB) is yet to release the 2022 November/December NBC/NTC and ANBC/ANTC examinations results.
Candidates who intend to take part in the 2022 NABTEB can learn how to check their results online.
NABTEB is one of the examination bodies set up by the Federal Government in 1992 to reduce the burden of conducting examinations, which involves a lot of technical and trade-related practicals.
The board conducts the (NTC), National Business (NBC) and their respective advance level examinations ( and ).
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NABTEB Physics Questions and Answers
PHYSICS OBJ ANSWERS
1-10: CCACDDADAC
11-20: CCABDDDDDC
21-30: CDADABBBDD
31-40: CACDACDAAC
41-50: CABDACABBA
PHYSICS THEORY ANSWERS
(1a)
Draw the velocity time graph
(i) Acceleration = slope of line OA = AE/OE
(ii) Retardation = slope of line BC = BD/DC
(iii) Total distances covered in the area of the trapezium OA = 1/2(AB+OC)AE
(1b)
Average Velocity = v+u/2; where v = final velocity
u = Initial velocity.
Displacement S = Average velocity × time
ie S = (v+u/2)t
But v = u+at(Newton first equation)
S = (u+at+u/2)t
S = 2ut + at²/2
S = ut + 1/2at²
(1c)
Total distance covered = 10km
Time = 5mins = 5/60hour
Velocity = 10/5/60 = 10×60/5 = 600/5 = 120kmh-¹
120kmh-¹ = 120×1000/60×60 = 33.33m/s
Draw the velocity time graph
(i) Maximum speed = 33.33m/s
(ii) Retardation = 0 – 33.33/300-270
= 33.36/30
= -1m/s²
=======================
(2a)
The boiling point of a liquid is the temperature at which its saturated vapour pressure (s.v.p) is equal to the atmospheric pressure.
(2b)
The instrument is a J-shaped tube with the end of the shorter arm sealed. This is shown below
Draw the diagram
The J-shaped tube is filled with Mercury such that the entire short arm is filled even the closed-end A. A drop of the liquid is introduced into the space on top of the mercury in the shorter arm of the tube.
As heat is applied, the liquid evaporates and the pressure exerted by the vapour depresses the mercury level slightly. Further application of heat cause more vapour to rise and this increases the pressure on the mercury.
The pressure exerted by the vapour is known as saturated vapour pressure S.V.P. At the temperature where the s.v.p is equal to the atmospheric pressure, the level of mercury in the shorter arm and mercury level in the longer arm will be equal.
The temperature recorded at this point is the boiling point of the liquid.
(2c)
Mass of copper = 400g = 0.4kg
Temperature of copper = 1000°c
Mass of liquid = 300g = 0.3kg
Temperature of liquid = 26°c
Temperature of water and copper = 108°c
Let the power of water that will boil away be m
Heat lost by copper in cooling from 1000°c to 108°c
= 0.4×4×10²×(1000 – 108)
= 0.4×4×10² × 892
= 1.4272 × 10^5J
Heat gained by water at 26° to reach atm pressure 108°
= 0.3×4.2×10³×(108 – 26)
= 0.3×4.2×10² × 82
= 1.0332 × 10^5J
Heat gained by the mass “m” of water to boil away = m × 2.26 × 10^6
Assuming no heat is gained or lost to the surrounding
Total heat lost by copper = total heat gained by the water.
1.0332×10^5 + 2.26×10^6m = 1.4272×10^5
2.26×10^6m=(1.4272-1.0332)×10^5
2.26×10^6m = 0.394×10^5
m = 0.394×10^5/2.26×10^6
= 0.0174kg
= 17.4g
=======================
(3a)
The principal focus(F) of a curved mirror(concave) is the point on the principal axis to which incident rays parallel to the principal axis and close to it’s converge after reflection.
Draw the diagram
(3aii)
The focal length is the distance measured from the principal focus to the pole of the concave mirror. The distance FP in the diagram above is the focal length of the concave mirror.
(3bi)
Draw the diagram
The candle is placed at the centre of the curvature of the concave mirror. A carbon paper receiving the reflected ray at the centre of curvature where the image of the candle is formed will be ignited.
(3bii)
Draw the diagram
First, the arrangement is as shown above. The object O is placed between the pole of the concave mirror and the principal focus of the mirror.
The image formed will be erect but virtual. For the image to be captured on the screen, another concave mirror is placed at the centre of the curvature of the first concave mirror.
Another, the image formed in the first mirror becomes an object to the 2nd concave mirror. The reflected image can then be captured on the screen.
(3c)
Height of object = 50cm
Distance of object = 1m
Length of camera = 20cm
Solution
F = 20cm, u = 1m = 100cm
Using mirror formula, 1/f = 1/u + 1/v
= 1/20 = 1/100 + 1/v
1/v = 1/20 – 1/100 = 4/100
V = 100/4 = 25cm
Magnification = v/u = 25/100 = 0.25
(i) Height of image is calculated using the magnification of the image
ie height of image/height of object = magnification
Height of image = height of object × magnification
= 50 × 0.25 = 12.5cm
(ii) Magnification = image distance/object distance = v/u
= 25/100 = 0.25
=======================
(4ai)
The capacitance of a capacitor is defined as the ratio of the charge Q on either plates or conductors to the potential difference v between them.
C = Q/V,
C = capacitance of a capacitor.
Q = charge.
V = potential difference.
(4aii)
(i) It depends on the nature of the dielectric substance.
(ii) The area between the plates.
(iii) The distance between the plates
(4aiii)
Workdone in charging a capacitor W is given as
W = 1/2v × q = 1/2qv
But c = q/v
Therefore v = q/c
W = 1/2qv = 1/2q × q/c
= 1/2 q²/c
Also, q = cv
W = 1/2qv = 1/2cv.v = 1/2cv²
The work done in charging a capacitor is the energy stored in the capacitor.
W = 1/2qv = 1/2 q²/c = 1/2cv²
(4b)
Draw the diagram
(i)
Total capacitance 1/c = 1/5 + 1/7 = 12/35
c = 35/12uF
Charge q on both capacitor q = cv
= 35/12 ×120 = 350×10^-6 or 3.5×10^-4C
Charge on 5uF capacitor q = 5×10^-6 × 120
= 6.0×10^-4C
Charge on 7uF capacitor
q = 7×10^-6 × 12.
= 8.40×10^-4C
(ii)
P.d across 5uF capacitor = 3.5×10^-4/5×10^-6
= 0.70×10² = 70v
P.d across 7uF capacitor = 3.5×10^-4/7×10^-6
= 0.50×10² = 50v
(iii)
E = W = 1/2cv² = 1/2qv
= 1/2 × 3.5 × 10^-4 × 120
= 2.10×10^-2J
= 0.021J
=======================
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DISCLAIMER! These are not real NABTEB Physics Questions, but likely repeated questions over the years to help candidates understand the nature of their examinations. Ensure to note every question provided on this page.
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