# 2019 JAN/FEB WAEC GCE Physics Questions And Answers

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# 👑 2019 WAEC GCE PHYSICS ANSWERS👑

1-10: BBABDACBBA
11-20: ABCDABCDBA
21-30: BACABCDCBA
31-40: BCCDDDDBBA
41-50: BACDCDAACA

(1a)
The yield point of an elastic material is defined and said to be reached when a stretched wire does not return to its original position.

(1b)
N/m²

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(2)
P – Filament Envelop
Q – Filament
R – Pin

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(3a)
A semiconductor is defined as a solid substance that has a conductivity between that of an insulator and that of most metals, either due to the addition of an impurity or because of temperature effects. Devices made of semiconductors, notably silicon, are essential components of most electronic circuits.

(3b)
(i) Sulphur
(ii) Carbon
(iii) Silicon

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(4)
(i) Photoelectricity
(ii) Momentum
(iii) Radiation of energy from hot bodies.

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(5a)
The airbus is a projectile

(5b)
It is a projectile because it must have taken off at an angle to the horizontal and then traced a parabolic path.

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(6)
Energy is ability to do work
Energy = mgh

In S.I Units,
Energy = kg(m/s²)m
= kgm²/s²

In dimension form,
Energy = f x d
= Ma x d
= Mass x acceleration x distance
MLT-² * L
Energy = ML²T-²

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(8ai)
Upthrust is define as less weight at a body partially or wholly immersed in a liquid

(8aii) set belts helps to check the effect at Newton first law. When an external force acts as a body moving in the uniform motion, seat belts helps to oppose the body willingness to continue in that state at uniform motion.

(8aiii)
– a satellite circling around the earth
– the moon circling the earth
– the planets moving around the sun

(8bi)
Given: mass, m = 300g (0.3kg)
Initial velocity, u = 10m/s
G = 10m/s
Finial velocity, v = 0
Using V² = U² – 2gh
0² = 10² – 2(10)h
0 = 100 – 20h
20h = 100
H = 100/20
Maximum height, h = 5m, potential energy, P.E = mgh
= 0.3 x 10 x 5 = 155

(8bii)
On return journey, u = om/s
V = ?,
h = 5m,
g = 10m/s²
using: V² = u² + 2gh
v² = 0² + 2(10 x 5)
v² = 0 + 100
v² = 100
v = √100
v = 10m/s
this its velocity just before it strikes the ground K.E = ½ mv²
= ½ x 0.3 x 10²
K.E = ½ x 0.3 x 100
K.e = 15J

(8c)
DRAW THE DIAGRAM
At the height, the body has potential energy only as it falls, it has both potential and kinetic energy, just before it touches the ground it has only K.E

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(9ai)
(i) change in state
(ii) thermonic emission
(iii) chemical charge
(iv) change in temperature

(9aii)
It means that 3.34 x 10⁵ joules at heat is required to melt a unit mass at ice

(9b)
Heat supplied = heat gained + heat gained by stem by water ice
Let the mass of steam be Ms
(msls + ms cw ∆Ø) = mw cw∆Ø + Mi Li + Mi Cw ∆Ø
(ms x 2.3 x 10⁶) + (ms x 4200 x (100-35) = {0.3 x 4200 + (35 – 350} + 0.02 x 3.34 x 10⁵) + (0.02 x 4200 x 35)
2300000Ms + 273,000Ms = 0 + 6680 + 2940 + 2573000Ms = 9620
Ms = 0.0037kg
Total mass of water in container = 0.3 + 0.02 + 0.0037 = 0.3237kg
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(10a)
(i) Length of the string
(ii) Tension in the string

(10bi)
P – Pulley
Q – Pivot

(10bii)
(i) It is used in producing sounds.
(ii) It determines the frequency of sound produced by a plucked string

(10c)
(i) Fundamental note : This is the lowest frequency but the strongest audible frequency.

(ii) Overtone: These are the next higher frequencies or modes of vibration above a fundamental that a body can emit.

(iii) Harmonics is the higher frequency which are integral or whole number multiples of the fundamental frequency can also be produced in the strings

(10d)
L =?
Frequency f1 = 350Hz
L+15 = ?, f2 = 280Hz
Given that F*1/l,
f = k/L
F1L1 = F2L2
350L1 = 280(L+15)
350L1 = 280L+4200
350L1-280L = 4200
70L = 4200
L = 105cm

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