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Offline Mr. Success

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2015 waec gce practical physics answer for free
« on: September 09, 2015, 05:58:49 pm »
(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.

(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


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