# WAEC GCE 2017 Physics Practical Solutions Answer – Nov/Dec Expo

## WAEC GCE Physics Practical Questions and Answer Solutions – NOV/DEC 2017 Expo Runz.

COMPLETED
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
masses(m)- 50,53,55,58,60
depth(d)-5.50,5.80.6.00,6.30,6.60
ti(s)- 6.45,6.80,6.85,7.40,7.45
T=t/n(s)- 0.655,0.680,0.685,0.740,0.758
T^2(s^2)- 0.429,0.462,0.469,0.548,0.575
Li(g/cm)- 5.455,5.776,5.833,6.032,6.154
Mli(g)-30.0,33.5,35.0,38.0,40.0
Slope=dy/dx=33.5-30/0.462-0.429
=3.5/0.033
=106.06gls^2
1aviii)
k=4^-2^5/Q, Q=24.25/6
k=4*(3.42)*106.06/4.88
Q=4.88
k=858.23cm/s^2
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.

3a)
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 connect connection of positive to positive to negative to negative when connecting to circuit
3bi)
when temperature increases the molecules tends to vibrate more intense there by opposing the flow of current
3bii)
P=I^R
P=V^2/R
power=1000W
Voltage=200V
R=?
1000=200^2/R
R=40000/1000
R=40