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Exam Time: Wednesday, 27th September, 2017
Physics 3 (Alternative to Practical)
9.30am – 12.15pm
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WAEC GCE PRACTICAL PHYSICS ANSWERS FROM EXAMHOT.COM
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1ai)
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
(1aii)
Real values of ti(secs)
t1=6.45
t2=6.80
t3=6.85
t4=7.40
t5=7.45
(1aiii)
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
(1aiv)
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
T=t/n(s):0.645,0.680,0.685,0.740,0.745 where n=10
li(g/cm):5.4545,5.6034,5.7500,6.0317,6.0606
T^2(S^2):0.4160,0.4624,0.4692,0.5476,0.5550
(1aix)
(i) I avoided parallax error in reading clock/weight balance
(ii) I ensue repeated readings are taken to avoid random error
(1bi)
(i) weight
(ii) upthrust
(iii) 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.
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(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
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3a)
CLICK HERE FOR NO3 GRAPH
Table 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 conect connection of positive to positive to negetive to negetive 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
Loadinq…….
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