WAEC SSCE 2016 EXPO: PHYSICS PRACTICAL Questions & Answers

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(PHYSICS PRACTICAL)
1i)
Centre of gravity =50.5
TABULATE
S/N|A(cm)| B(cm)
1 |10.0 | 51.90
2 |15.0 | 53.50
3 |20.0 | 53.80
4 |25.0 | 54.90
5 |30.0 | 55.80
1a) Slope = Change in Y/Change in X
= Y2 – Y1 / X2 – X1
= 51.50 – 50.00/25.00 – 20.00
= 1.50/5.00
= 0.30
Ki = (1 – 2S / S)100
= (1 – 2 x 0.30 / 0.3)100
= 1.33
K2 = 2C/S – 160
= 2×44.50/0.3 – 160
= 136.67
(1xii)
-I avoided the effect of draught during the experiment
-I avoided error due to parallax on the metre rule
-I balanced the metre rule horizontally on the knife edge before taking my readings
(1bi)
Moment of a force about a point is the product of the force and its perpendicular distance from the point to the line of action of the force. i.e moment=force*time and its unit is Nm
(1bii)
-Sum of forces acting in one direction must equal to the forces acting in another direction.
-The algebraic sum of moment acting about a point must be equal to zero
(3a)
TABULATE
S/N|E(v)|V(v)|I(A)|V^-1(v^-1)|I^-1(A^-1)
1 |1.50|0.80|0.40| 1.250 |2.500
2 |3.00|1.40|0.70| 0.714 |1.429
3 |4.50|2.00|1.00| 0.500 |1.000
4 |6.00|2.50|1.20| 0.400 |0.833
5 |7.50|3.20|1.50| 0.313 |0.667
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Precautions:
-I tightened all the wires tightly together
-I ensured clean terminals
-I avoided zero error on the ammeter used
(3bi)
-By using electric cells
-By using electrostatic method
(3bii)
For parallel connection=1/R=1/R+1/R 2Rt=R
Rt=R/2ohm
Total:(R/2)+2
4:(R/2)+2
4:(R+4)/2
8=R+4
8-4=R

R=4ohm

(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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