2018/2019 waec Physics practical alternative to expo and runz

The West African Examinations Council (WAEC) is an examination board that conducts the West African Senior School Certificate Examination, for University, polytechnic, colleges and Jamb entry examination in West Africa countries. … In a year, over three million candidates registered for the exams coordinated by WAEC.

Date of commencement of the waec Physics  alternative A and B papers 

The 2018 Waec Physics practical Alternative A expo and runz will take place on Wednesday, 4th April, 2018 from 08:30am to 11:15am (1st Set) and 11:40am to 02:25pm (2nd Set).

Physics waec Practical Alternative B will be written on Friday, 6th April, 2018 from 08:30am to 11:15am (1st Set) and 11:40am to 02:25pm (2nd Set)

The physics waec practical is majorly divided into 3 questions (consisting of Mechanics, Waves and Electricity) with a few other general knowledge calculations.

WAEC Physics Practical Alternative To Expo And Runz Answers.

(1)
TABULATE:
i:1,2,3,4,5,
L(cm):80.00,70.00,60.00,50.00,40.00
t1(s):17.87,16.22,16.09,14.08,13.28
t2(2):17.56,16.46,15.09,14.77,13.36
t(s)mean:17.715,16.340,15.590,14.425,13.320
T(t/10)s:1.7715,1.6340,1.5590,1.4425,1.3320
LogT(s):0.2483,0.2133,0.1928,0.1591,0.1212
Log L(cm):1.9031,1.8451,1.7782,1.6990,1.6021
Log T*10^-2(s):24.83,21.33,19.28,15.91,12.12
Log L*10^-1(cm):19.031,18.451,17.782,16.990,16.021
SLOPE(s)=(LogT2*10^-2 – LogT1*10^-2)(/Log L2 *10^-1 – LogL1*10^-1)
=(21.33-12.12(s))*10^-2/(18.50-16.02(cm)*10^-1)
=3.714*10^-1
=0.3714cm^-1

(1axi)
(i) I ensured supports of pendula were rigid
(ii) I avoided parallax error on the meter rule

(1bi)
Simple harmonic motion is the motion of a body whose acceleration is always direct towards a fixed point and is proportional to the displacement from the fixed points

(1bii)

T=1.2secs
Log 1.2=0.079
0079 shown on graph with corresponding Log L read L correctly determined

(1a)
TABULATE:
S/N: 1, 2 , 3, 4, 5
L(cm): 90.00, 80.00, 70.00, 60.00, 50.00
T(s): 38.00, 36.00, 34.00, 31.00, 28.00
T = t/20(sqt): 1.90, 1.80, 1.70, 1.55, 1.40
√L(cm^1/2): 9.49, 8.94, 8.37, 7.73, 7.07

(1aix)
Slope(s) = Δvertical/Δhorizontal
=1.90 – 1.55/9.49 – 7.75
= 0.35/1.74
S = 0.201

(1ax)
g = 4π²/S²
= 4 ×(3.142)²/(0.201)²
= 977.4cm/s²

(1axi)
(i) I ensured the angle of oscillation is relatively small.
(ii) I ensured that no external force is added to the system of oscillation.

(iii) I ensured the oscillation is perfect
==========================================
(3a)
TABULATE:
x(cm):10,20,30,40,50,60
V(v):0.65,0.75,1.00,1.20,1.45,1.55
I(A):0.20,0.30,0.35,0.40,0.45,0.55
LogV(v):-0.187,-0.125,0.000,0.079,0.161,0.190
LogI(ohm):-0.699,-0.523,-0.456,-0.396,-0.391,-0.360
SLOPE(s):=( LogI2-LOgI1)/(LogV2-LogV1)
=-0.2-(-0.7)/0.3-(-0.2)
=0.5/0.5
=1AV^-1

(3axi)
(i) I ensured neat and tight terminals
(ii) I opened the key when reading was taken

(3bi)
(i)The brightness of the bulb increases
(ii) The voltage and current through the bulb increases

(3bii)
(i) diode
(ii) transistor