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## 2019 WAEC GCE Physics Practical Answers

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2019 WAEC GCE Physics Practical Answers by : 12:00 am On September 26, 2017

Wednesday, 27th September, 2017
Physics 3 (Alternative to Practical)
9.30am – 12.15pm

1i)
Real values of masses(msi)g
msi=50.0
ms2=52.5
ms3=54.5
ms4=58.0
ms5=60.0
Real values of di(cm)
d1=5.50
d2=5.80
d3=6.00
d4=6.30
d5=6.60
1ii)
Real values of ti(secs)
t1=6.45
t2=6.80
t3=6.85
t4=7.40
t5=7.45
1iii)
Real values of mli(g)
ml1=>msi-mso=50-20=30
ml2=>ms2-mso=52.5-20=32.5
ml3=>ms3-mso=54.5-20=34.5
ml4=>ms4-mso=58.0-20=38.0
ml5=>ms5-mso=60.0-20=40.0
1)TABULATE
i:1,2,3,4,5
msi(g):50.0,52.5,54.5,58.0,60.0
di(cm):5.50,5.80,6.00,6.30,6.60
ti(s):6.45,6.80,6.85,7.40,7.45
mli(g):30.0,32.5,34.5,38.0,40.0
1ix)
the two precautions are:
1.avoid parallax error in reading clock/weight balance
2. I ensue repeated readings are taken to avoid random error
3. I avoid draught or wind when performing the experiment
1bi)
the forces are weight, upthrust and viscous force / drag or liquid friction
1bii)
the amplitude of oscillation of the loaded test tube decreases with time because of the viscous drag(force) which opposes the motion.

2i)
TABULATE:
Exp:1,2,3,4,5
hi(cm)Real value:1.80,2.20,2.27,3.10,3.60
hi(cm)Converted:9.00,11.00,13.50,15.50,18.00
yi(cm)Real value:5.30,3.40,2.70,2.40,1.70
yi(cm)Converted:26.50,17.00,13.50,12.00,8.50
hi^-1:0.111,0.90,0.074,0.065,0.056
yi^-1:0.038,0.059,0.074,0.083,0.118

2bii)
Given :
f = 10cm
u = 20cm
v = ?
Using:
1/f = 1/u + 1/v
1/10 = 1/20 + 1/v
Multiplying through with 20v, we have:
2v = v + 20
2v – v = 20
v = 20cm
Magnification, m = v/u
m = 20/20
m = 1

3a)
TABULATE W1
s/n:1,2,3,4,5,6
xn(cm)/d1=0.009m:3.06,3.50,3.70,4.30,4.50,4.80
xm(cm)converted:15.30,16.50,18.50,21.50,22.50,24.40
xm(cm)converted d2=0.005m:84.70,83.50,81.50,78.50,77.50,75.60
Ri(n):11.07,10.12,8.81,7.30,6.89,620

TABULATE W2
s/n:1,2,3,4,5,6
xmi(cm):1.40,1.65,1.80,2.20,2.55,2.94
converted/xmi(cm):7.00,8.25,9.00,11.00,12.75,14.70
xni(cm):93.00,91.75,91.00,89.00,87.25,85.3
R2i(n):26.57,22.24,20.22,16.18,13.69,11.61

SLOPE:
dy/dx=26.57-13.69/11.07-6.89
=12.88/4.18
=3.08
3ix)
k=d2/d1*squareroot5
=0.005/0.009*squareroot13.08
=0.56*1.75
=0.98
3x)
i)i would ensure that the terminal are well tightened to avoid partial contact
ii)i would obey the connect connection of positive to positive to negative to negative when connecting to circuit

3bi)
The resistance of a metallic conductor increases with increase in temperature of metallic conductor increases the velocity of free electrons increases which cause resistance in the path of free electrons
3bii)
p=I^2R
P=V^2/R
P=1000W
V=200v
R=?
1000=200^2/R
R=40000/1000
R=40ohms

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