1)
TABULATE br /> i:1,2,3,4,5,

L(cm):80.00,70.00,60.00,50.00,40.00

t1(s):17.87,16.22,16.09,14.08,13.28

t2(2):17.56,16.46,15.09,14.77,13.36

t(s)mean

:17.715,16.340,15.590,14.425,13.320

T(t/10)s:1.7715,1.6340,1.5590,1.4425,1.3320

LogT(s)
.2483,0.2133,0.1928,0.1591,0.1212

Log L(cm):1.9031,1.8451,1.7782,1.6990,1.6021

Log T*10^-2(s):24.83,21.33,19.28,15.91,12.12

Log L*10^-1(cm)
:19.031,18.451,17.782,16.990,16.021

SLOPE(s)=(LogT2*10^-2 – LogT1*10^-2)(/Log L2
*10^-1 – LogL1*10^-1)
=(21.33-12.12(s))*10^-2/(18.50-16.02(cm)
*10^-1)
=3.714*10^-1
=0.3714cm^-1

1axi)
i)ensured supports of pendula were rigged
ii)avoided parallax error on meter rule/stop watch

1bi)
simple harmonic motion is the motion of a body
whose acceleration is always direct towards a fixed
point and is proportional to the displacement from
the fixed points

1bii)
T=1.2secs
Log 1.2=0.079
0079 shown on graph with corresponding Log L
read L correctly determined

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

3a)
TABULATE br /> x(cm):10,20,30,40,50,60

V(v) .65,0.75,1.00,1.20,1.45,1.55

I(A) .20,0.30,0.35,0.40,0.45,0.55

LogV(v):-0.187,-0.125,0.000,0.079,0.161,0.190
LogI(ohm)
:-0.699,-0.523,-0.456,-0.396,-0.391,-0.360
SLOPE(s):=( LogI2-LOgI1)/(LogV2-LogV1)
=-0.2-(-0.7)/0.3-(-0.2)
=0.5/0.5
=1AV^-1

3axi)
i)ensured clean/tight terminals
ii)open key when reading was taken

3bi)
i)brightness of the bulb increases
ii)voltage/current through the bulb increases

3bii)
i)diode
ii)transistor

3bi)As × increases the Brightness of the bulb
increases, voltage and current through the bulb
increases hence brightness increases.

3bii)Electrical device that do not obey ohm’s law
are.. br /> ¡)Diode
¡¡) capacitor
¡¡¡)transistor
¡v)thermistor
V)thyristor

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