# 2019 NABTEB GCE Physics Questions And Answers

National Examination Council, NABTEB GCE Physics Questions and Answers for 2019 NABTEB GCE Examination is now available, subscribe now and get the questions and already solved answers by Examrelief.

# 👑 2019 NABTEB GCE PHYSICS ANSWERS👑

11-20: CCABDDDDDC
31-40: CACDACDAAC
41-50: CABDACABBA

(1a)

(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

(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 that temperature at which it’s 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

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 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.

(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)

The candle is placed at the centre of 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)

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 curvature of the first concave mirror. 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 [https://i.imgur.com/7upz3iL.jpghttps://i.imgur.com/7upz3iL.jpg]

(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
=======================

(5ai)
Nuclear fission is the splitting up of the nucleus of a heavy element into two approximate equal parts with a release of a huge amount of energy and neutrons. The mass of the element before fission is slightly reduced. This is accounted for by the huge amount of nuclear energy releases. This energy is given by Albert Einstein equation
E = Δmc²
E = energy released,
Δm = The difference in mass
c = The velocity of light.

Nuclear fusion is the process in which two or more light nuclei combine or fuse to form a heavy nucleus with a release of a large amount of energy.
Element that undergoes nuclear fission is Uranium.
Element that undergoes fusion are isotopes of hydrogen such as denterium and Tritium.

(5aii)
(i) Fusion is more easily achieved with highest elements such as hydrogen. Nuclear repulsion is easily overcome as nuclei approach each other.
(ii) Fusion process is less dangerous as the products are non-radioactive.
(iii) Raw materials for fusion are readily available.

(5aiii)
Fusion has not been used to generate power because the amount of heat requires to bring two higher nuclei together is very high. The temperature is of the order of 10^6 – 10^8 degree Celsius. The material that can withstand such enormous amount of heat does not exist yet. This account for why fusion has not been used to generate power.

(5bi)
238U —> 4H + 2 0e
92 2 -1
+ 234Th + Energy
90

234Th —> 0e + 234Pa
90 -1 91
+ Energy.

(5bii)
(i) They are used in Medicine as radiotherapy
(ii) They are used in industries to study defects in metals.
(iii) They are used in agriculture as radioactive tracers and preservatives.

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