2017 PHYSICS PRACTICAL
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(1)
TABULATE:
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):0.2483,0.2133,0.1928,0.1591,0.1212
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:
x(cm):10,20,30,40,50,60
V(v):0.65,0.75,1.00,1.20,1.45,1.55
I(A):0.20,0.30,0.35,0.40,0.45,0.55
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
2b)
bi)The boiling point of a liquid is the temperature
at which the vapor pressure of the
liquid equals the pressure surrounding the liquid
and the
liquid changes into a vapor.
bii) impurities increases the temperature range a
substance can exist as a liquid
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..:
¡)Diode
¡¡) capacitor
¡¡¡)transistor
¡v)thermistor
V)thyristor
PHYSICS PRACTICAL)
(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
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(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
(vii) |Slope(s)
= change in logI/change in logd
= (0.87-0.57)/(0.02-(-0.11))
= 0.3/0.13=2.3A
(vii)
-I will ensure the circuit is open when no readings are not taken
-i will ensure tight connection
bi)
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
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(4)
Ohm ’ s Law .
The current flow through a metallic conductor
is directly proportional to the potential
difference provided that temperature and other physical conditions are constant.
(4b)
Factors that determine resistance of conduction.
PL
Loading…
A
i – Nature Of the materials .
ii – Length of the conducting wire.
iii – Cross setional area .
iv – Temperature .
How Temperature affects
Conduction.
=======================
Answer => How Temperature affects Conduction Is that The more the temperature
will be the resistance . (If neccessary) .
(4c)
Since the lost volt have developed across the
cell when the supplied volt across the resistance onces will be affected
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To subscribe,call/WhatsApp-08170610547
(1)
TABULATE:
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):0.2483,0.2133,0.1928,0.1591,0.1212
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:
x(cm):10,20,30,40,50,60
V(v):0.65,0.75,1.00,1.20,1.45,1.55
I(A):0.20,0.30,0.35,0.40,0.45,0.55
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
2b)
bi)The boiling point of a liquid is the temperature
at which the vapor pressure of the
liquid equals the pressure surrounding the liquid
and the
liquid changes into a vapor.
bii) impurities increases the temperature range a
substance can exist as a liquid
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..:
¡)Diode
¡¡) capacitor
¡¡¡)transistor
¡v)thermistor
V)thyristor
PHYSICS PRACTICAL)
(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
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(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
(vii) |Slope(s)
= change in logI/change in logd
= (0.87-0.57)/(0.02-(-0.11))
= 0.3/0.13=2.3A
(vii)
-I will ensure the circuit is open when no readings are not taken
-i will ensure tight connection
bi)
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
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(4)
Ohm ’ s Law .
The current flow through a metallic conductor
is directly proportional to the potential
difference provided that temperature and other physical conditions are constant.
(4b)
Factors that determine resistance of conduction.
PL
Loading…
A
i – Nature Of the materials .
ii – Length of the conducting wire.
iii – Cross setional area .
iv – Temperature .
How Temperature affects
Conduction.
=======================
Answer => How Temperature affects Conduction Is that The more the temperature
will be the resistance . (If neccessary) .
(4c)
Since the lost volt have developed across the
cell when the supplied volt across the resistance onces will be affected
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