Important Posts

WAEC GCE 2017 - PHYSICS PRACTICAL ANSWERS

Get Waec Gce Expo Here

WAEC GCE - PHYSICS PRACTICAL ANSWERS

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


==========================
KEEP REFRESHING THIS PAGE
==========================


(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.

=====================================


(2bi)
[View 2bi drawing here]

(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

=====================================


(3.)

(3ai to 3a iv)
d1=0.009m
d2=0.005m

Table W1
s/n: 1,2,3,4,5,6
xn(cm) 3.06,3.50,3.70,4.30,4.50,4.80
xm(cm): 15.30,16.50,18.50,21.50,22.50,24.40
xm(cm)converted : 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

(3vii) 
GRAPH

(3viii)
slope
dy/dx=26.57-13.69/11.07-6.89
=12.88/4.18
=3.08

(3ix)
k=d2/d1*√s
=0.005/0.009*√3.08
=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




++++++++++++++++++++++++


Get Waec Gce Expo Here
THERE IS LOVE IN SHARING! :

No comments:

Post a Comment

Designed by NAIRACODED MEDIA.