2015 WAEC GCE NOV/DEC
EXAM DAY: THURSDAY
PAPER: PHYSICS 3 (Alternative to Practical Work)
(1a)Tabulate.Under s/n: 1,2,3,4,5Under M(N): 140,120,110,84,66Under tita(o): 24,32,38,51,62Under 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 momentOC*M=CD*W.
(3a)(i)d1=1.40cm, d2=1.55cm, d3=1.75cm, d4=2.1cm, d5=2.3cmReal values of did1=1.4*0.5=0.7mmd2=1.55*0.5=0.77mmd3=1.75*0.5=0.875mmd4=2.1*0.5=1.05mmd5=2.3*0.5=1.15mm
(ii)Ia1=2.4A, Ia2=3.8A, Ia3=5.0A, Ia4=7.4A, Ia5=11.6AIb1=2.4A, Ib2=3.6A, Ib3=5.2A, Ib4=7.5A, Ib5=11.6A
(iii)I=(Ia+Ib)/2I1=(2.4+2.4)/2=2.4AI2=(3.8+3.6)/2=3.7AI3=(5.0+5.2)/2=5.1AI4=(7.4+7.5)/2=7.45AI5=(11.6+11.6)/2=11.6A
(3aiv)logd1=log0.7=-0.15mmlogd2=log0.775=0.11mmlogd3=log0.875=0.06mmlogd4=log1.05=0.02mmlogd5=log1.15=0.06mmlogI1=log2.4=0.38logI2=log3.7=0.57logI3=log5.1=0.71logI4=log7.45=0.87logI5=log11.6=1.06
(3av)TABULATES/N; 1,2,,3,4,5di(cm);1.40,1.55,1.75,2.10,2.30di(mm);0.70,0.78,0.88,1.05,1.15Ia(A);2.4,3.8,5.0,7.4,11.6Ib(A);2.4,3.6,5.2,7.5,11.6I(A);2.4,3.7,5.1,7.5,11.6logd!(mm);-0.15,-0.11,-0.06,0.02,0.06logI1;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/AR=(5.0*10^-2*0.01)/(pie/4)*(0.001)^2R=(5*10^-4)/(0.7855*1*10^-6)=6.37*10^2ohmP=I^2R=10^2*6.37*10^2=6.37*10^4W\
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