VERIFIED WAEC GCE 2015/16 PHYSICS AANSWERS

(1a)
Tabulate.
Under s/n: 1,2,3,4,5
Under M(N): 140,120,110,84,66
Under tita(o): 24,32,38,51,62
Under Sin tita: 0.4067, 0.5299,0.6156,
0.7771,0.8829.
Note that 1cm =20N
(1aviii)

– I would ensure that the meter rules
balanced before the readings.

– I would avoid error due to parallax

(1bi)
i. Total forces in one direction are
equal to the forces in opposite
direction.

ii. The algebraic sum of the moment
of all forces about any point should
be zero.

(1bii.)
The moment of force at equilibrium
point o is equal to,
sum of clockwise moment = sum of
anti- clockwise moment
OC*M=CD*W.

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(3a)
(i)d1=1.40cm, d2=1.55cm,

d3=1.75cm, d4=2.1cm, d5=2.3cm

Real values of di
d1=1.4*0.5=0.7mm
d2=1.55*0.5=0.77mm
d3=1.75*0.5=0.875mm
d4=2.1*0.5=1.05mm
d5=2.3*0.5=1.15mm

(ii)Ia1=2.4A, Ia2=3.8A, Ia3=5.0A,
Ia4=7.4A, Ia5=11.6A
Ib1=2.4A, Ib2=3.6A, Ib3=5.2A,
Ib4=7.5A, Ib5=11.6A

(iii)I=(Ia+Ib)/2
I1=(2.4+2.4)/2=2.4A
I2=(3.8+3.6)/2=3.7A
I3=(5.0+5.2)/2=5.1A
I4=(7.4+7.5)/2=7.45A
I5=(11.6+11.6)/2=11.6A

(3aiv)
logd1=log0.7=-0.15mm
logd2=log0.775=0.11mm
logd3=log0.875=0.06mm
logd4=log1.05=0.02mm
logd5=log1.15=0.06mm
logI1=log2.4=0.38
logI2=log3.7=0.57
logI3=log5.1=0.71
logI4=log7.45=0.87
logI5=log11.6=1.06

(3av)
TABULATE
S/N; 1,2,,3,4,5
di(cm);1.40,1.55,1.75,2.10,2.30
di(mm);0.70,0.78,0.88,1.05,1.15
Ia(A);2.4,3.8,5.0,7.4,11.6
Ib(A);2.4,3.6,5.2,7.5,11.6
I(A);2.4,3.7,5.1,7.5,11.6
logd!(mm);-0.15,-0.11,-0.06,0.02,0.06
logI1;0.38,0.57,0.71,0.87,1.06

(3avi)
|Slope(s)=change in logI/change in
logd
=(0.87-0.57)/(0.02-(-0.11))
=0.3/0.13=2.3A

(3avii)
-I will ensure the circuit is open
when no readings are not taken

-i will ensure tight connection

(3bi)
diameter(d)=1.09mm

(ii)
R=eL/A
R=(5.0*10^-2*0.01)/(pie/4)*(0.001)^2
R=(5*10^-4)/(0.7855*1*10^-6)
=6.37*10^2ohm
P=I^2R
=10^2*6.37*10^2
=6.37*10^4W