WAEC - PHYSICS PRACTICAL ANSWER– May/June Expo Now Available

WAEC - PHYSICS PRACTICAL ANSWER– May/June Expo Now Available_

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(1aviii)
In a tabular form
Under y(cm)
20.0, 30.0, 40.0, 50.0, 60.0

Under t(s)
36.61, 33.66, 31.32, 28.62, 25.93

Under T = t/20(s)
1.8305, 1.6830, 1.5660, 1.4310, 1.2965

Under T²(s²)
3.3507, 2.8325, 2.4524, 2.0478, 1.6809

(1ax)
Slope s = ΔT²/Δy
S = 4 - 2.5/0 - 39
S = 1.5/-39
S = -0.0385

Intercept, C = 4.0

(1axi)
SR = c
R = c/s = 4/-0.0385 = -103.9

(1axii)
(i) I ensured that the retort stand was well clamped.
(ii) I avoided error due to parallax when reading the metre rule.

Q3)

(3ai)
E.m.f, E = 3.00v

(3avii)
In a tabular form
Under R(ohms)
2.0, 5.0, 10.0, 12.0, 15.0, 20.0

Under V(v)
0.86, 0.62, 0.43, 0.38, 0.32, 0.25

Under V-¹(v-¹)
1.1628, 1.6129, 2.3256, 2.6316, 3.1250, 4.0000

(3aix)
Slope s = ΔR/Δv-¹
=17.5 - 4.25/3.5 - 1.5
=13.25/2
=6.625

Intercept, C = -5.75

(3ax)
S = Roα
6.625 = 0.85(α)
α = 6.625/0.85 = 7.79

C = -(Ro + β)
= -5.75 = -(0.85 + β)
β = 5.75 - 0.85
β = 4.9

(3axi)
(i) I ensured tight connections.
(ii) I avoided error due to parallax when reading the voltmeter.

(3bi)
Vab = 12 - 8 = 4v
Rab = 4×5/4+5 = 20/9ohms
Iab = Vab/Rab = 4/20/9 = 4 ×9/20 = 9/5
=1.8A

Power = I²ab Rab
=1.8² * 20/9
= 7.2watts

(3bii)
Current in line = power/voltage = 3600/240
=15A
Circuit breaker remain closed because current in line is less than 20A

(xi)
1. I ensured that electric terminals were tightly connected to avoid partial contact.

2. I ensured that the key was open when the readings were not taken in other to avoid running down the cell or battery.

3. I avoided parallax error of the voltmeter when taking my reading.

Pick 2

(3ai)
E.m.f, E = 3.00v

(3avii)
In a tabular form
Under R(ohms)
2.0, 5.0, 10.0, 12.0, 15.0, 20.0

Under V(v)
0.86, 0.62, 0.43, 0.38, 0.32, 0.25

Under V-¹(v-¹)
1.1628, 1.6129, 2.3256, 2.6316, 3.1250, 4.0000

(3aix)
Slope s = ΔR/Δv-¹
=17.5 - 4.25/3.5 - 1.5
=13.25/2
=6.625

Intercept, C = -5.75

(3ax)
S = Roα
6.625 = 0.85(α)
α = 6.625/0.85 = 7.79

C = -(Ro + β)
= -5.75 = -(0.85 + β)
β = 5.75 - 0.85
β = 4.9

(3axi)
(i) I ensured tight connections.
(ii) I avoided error due to parallax when reading the voltmeter.

(3bi)
Vab = 12 - 8 = 4v
Rab = 4×5/4+5 = 20/9ohms
Iab = Vab/Rab = 4/20/9 = 4 ×9/20 = 9/5
=1.8A

Power = I²ab Rab
=1.8² * 20/9
= 7.2watts

(3bii)
Current in line = power/voltage = 3600/240
=15A
Circuit breaker remain closed because current in line is less than 20A

Except for otherwise cases✅

1a)
Tabulate
L(cm) |130.0|110.0|90.0|70.0|
50.0|
t^1(s) |.47.5|43.1|39.6|35.0|
29.0|
t^2(s) |47.1|43.5|39.4|34.6|
29.2|
t=t^1+t^2/2(s) |47.3|43.3|39.5|
39.29.1|
T(s) |2.365|2.165|1.975|1.740|
1.455|
T^2(s^1) |5.593|4.687|3.901|
3.028|2.117|
L=30(cm) |100.0|80.0|50.0|40.0|
20.0|
(1ix)
Graph.(scroll down)
(1x)
Slop (graph scroll down)
(1xi)
(i) K1=4x^2/s intercept C =1.32
=4/0.043*22/7*22/7=978.842
(ii) K2=C/S= 1.32 / 0.043=30.698
====================­
=============
(1xii)
(i) I avoid parallax error in the
metre rule and stop
(ii) I avoid conical oscillation
(1bi)
The period of oscillation of a
body is time taken by body to
complete one oscillation.
====================­
===============
(1bii)
If a body is falling under its own
weight alone the rate of change
of its velocity with time as a
result of earth gravitational pull is
known as the acceleration of free
fall due to gravity
====================­
===============
(3a)
Tabulate
S/N: 1,2,3,4,5,6
S(ohms) : 15,12,10,8,6,4
V(v): 0.45,0.52,0.57,0.75,­
0.86,1.10
1/v(v^-1) : 2.20,1.92,1.75,1.30,1.1­
6,0.90
(3vii)
Graph (scroll down to see the
graph)
(3viii)
Slop ( scroll down to see the
graph)
(3ix)
scroll down to see the graph)
====================­
=============
(3x)
(i) I avoided parallax error when
voltmeter
(ii) I ensured that the key was
been taking
====================­
=============
(3bi)
Resistivity is a measure of the
resistance of a given size of
material to electrical conduction.
It is measured in ohms - meter
(πm)
====================­
==============
(3bii)
Given that
E=1.5v, R=1π, r=2π
I =E/R+r=1.5/1+2
= 1.5/3= 0.5A
++++++++++++++++++++­
++++++++++++++
Completed

PHYSICS 3 SOLUTIONS
_________________________________
Q1.
determine the slope of the graph and the intercept c
- evaluate k1 = k1 = 4π2/s and k2 =

(Xiii)- state any two of the following precautions

- Avoided conical oscillations
- Ensured firm suspension of pendulum
- Ensured reading is repeated (shown on the table
- Avoided draught
-Corrected/noted for zero error of metre rule/ stop match/clock
- Avoided parallax error in reading metre rule/ stop clock

(b) (i) The period of oscillation of a body is the time taken for the body to make one complete to and from movement

(ii) When a body is falling under its weight (alone), the rate of increase of its velocity with time (as a result of the earth’s gravitational pull) is the acceleration of free fall due to gravity.
________________________________
Q2bi.
1) It produces erect image of an object in front of it .
2) The size of the image formed is smaller and therefore, has wide view of coverage.
3) all image are formed between pole and focus.

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