Complete waec physics pratical question/answers

Below is a real and comfirmed
waec real physics practical
questions and answers.
OVER VIEW
THEPHYSIC
PRACTICAL
QUESTION
ELECTRICITY
Connect the circuit
at shown in the
diagram above:-
Close the key k.
read and record the
emitters reading
Io when the clip is
not
constant wire. Open
the key with clip
making constant
with the constant
wire when
l=80cm. close the
key read and record
the ammeter and
I.Repeat the
procedure with
l=70,60,50,40 and
30 respectively in
each case read and
record ammeter
reading I. evaluate
I-1 .
Tabulate your
readings.
plot a graph of I on
the vertical axis
against I-1 on the
horizontal axis
starting from the
origin (0.0)
Determine the
slope S of the graph
and its intercept C
Evaluate P=C/S.
using your graph
determine the
current
I when I=55cm.
State two
precautions taken
to ensure accurate
result Explain what
is mean by
potential difference
between two point
in an electric circuit.
State two factor on
which the
resistance of a wire
depend.
The resistivity of a
given wire of cross
sectional area
0.7mm2 is 4.9x10-4
ohms
Calculate the
resistance of two
length of a wire
LIGHT
I. Trace the outline
ABCD of glass block
on a sheet of paper
as shown above.
II. Remove the
block mark a
position very close
to A
draw the normal at
the point N.
III. Draw the
incident ray
such that the angle
of
incidence i=20 o
IV. Fix two pins at
the point P and Q on
the incidence ray V.
Replace the black
and fix two other
pins at R and S
such the pins
appear to be at P
and Q when viewed
through side D.C of
the
glass block. VI.
Remove the block
and
join the points at R
and S
producing the line
to meet DC at F
VII. Join NF
VIII. Draw the
normal EF, at F and
produce it to meet
the
force AB at G
IX. Extend PG to
meet the
normal EF at H.
X. Measure and
record in
the table below NF
and NH Repeat the
experiment
i=30 o,40 o, 50o,
and
60o in each case,
measure
NF and read your
readings in
the table below.
Plot a graph of NF
on the vertical axis
and NH on the
horizontal axis.
Determine the
slope of the graph.
State two
precautions to
obtain accurate
result
State Snells law of
refraction
[b]2013 PHYSICS
PRACTICAL SOLUTION
SET
QUESTION 2:
(a.) Measure and
record the thickness
of the glass block
provided.
-Trace the outline
ABCD of theglass
block on the sheet of
paper as shown.
Remove the block and
draw Normal at N.
-Draw an incident ray
such that the angle of
incidence,i=25degree.
Fix two pins at points
P and Q onthe incident
ray.
Replace the glass block
and fix two otherpins
at point R and Y such
that the pins appear
to be in a straight line
with the images of
the pins at P and
Qwhen viewed
through the side DC of
the glass block.
Remove the block and
join the points at R
and Y producing the
line to meet DC and X.
Join NX and measure
its length L.
-Evaluate L^-2 and
Sin^2i.
Repeat the
experiment for
i=35degree,
45degree,55 degree
and 65degree. In each
case determine the
corresponding values
of L, L^-2 and Sin^i.
Tabulate your
readings.
Plot a graph of L^- on
the vertical axis and
Sin^2i on the
horizontal axis,
starting both axes
from the origin.
Determine the slope
(s) of the graph and
the intercept(I) on the
vertical axis.
Evaluate the
expression: K=(I/S)
^1/2.
State two precautions
taken to ensure
accurate results.
(i.) Using your graph
deduce the value of L
when i=0degree.
(ii.) State Snell's law of
refraction and explain
why refraction occurs
at the boundary
between two media.
ANSWER:
Table of values/
Observation:
Width of glass block
b=6.5cm.
S/N: 1,2,3,4,5.
P(degree):
25.00,35.00,45. ­
00,55.00,65.00.
L(cm): 6.80,7.00,7.30,
7.80,8.20.
L^2(cm^2):
46.24,49.00,53.
29,60.84,67.24.
L^-2(cm^-2):
0.0216,0.0204,0 .0188,0.0
0149.
SinPdegree:
0.04226,0.5736,
0.7071,0.8191,0.9063.
Sin^2Pdegree:
0.1786,0.3290,0 .5000,0.6
NOTE THE FOLLOWING:
*Draw a table for the
above values.
*Note that ^ means
Raise to Power. Eg:
2^-1 means Two raise
to power minus One.
*Note that COMMA(,)
in the above table
means NEXT LINE. Eg:
2,3,4,5 means dat 3 is
under 2, 4 is under 3
and 5 is under 4 in a
table form.
Therefore, S/N has
1,2,3,4,5under it, in a
table as shown in the
table of values above.
i.e 1 is under S/N and 2
is under 1 and 3 is
under 2 and4 is under
3 and so on.....
Slope= DL^-2(cm^-2)/
DSin^2i =
0.0136-0.0224/ 0.96-0.1
= 0.0088/0.86 = -0.
01023(cm^-2)
K=(1/5)^1/2 = (0.234/
0.01023)^1/2 =
1.512cm,=1.5cm.
Deduction intercept on
vertical axis = 0.
0234(cm2).
PRECAUTIONS:
(i.) I avoided error of
parallax.
(ii.) I ensured that the
object pins and the
image pins wereerect
and in a straight line.
(bi) When i = OL^-2 =
0.0236cm^-2 = L^-2 =
42.37cm^2.
Hence L=6.5cm.
- QUESTION 3 -
(ELECTRICITY):
You are provided with
a constantan wire, a
2-ohm standard
resistor, an
accumulator E, an
ameter(A), a key(K)
and other necessary
apparatus.
(i.) Measure and record
the e.m.f of the
accumulator provided.
(ii.) Connect a circuit as
shown in the diagram
above.
(iii.) Close the Key, read
and record the ameter
readings to when the
crocodile clip is not in
contact with the
constantan wire.
(iv.) Open the key with
the key making
contact with the wire,
when L=90cm. Close
the key. Read and
record the ammeter
reading I, Evaluate
I^-1.
(v.) Repeat the
procedure
forL=80,70,60 and
50cm.
(vi.) In each case, read
and record the
ammeter reading and
Evaluate I^-1(A^-1).
Tabulate your
readings.
(vii.) Plot a graph, L on
the vertical axis and
I^-1 on the horizontal
axis.
(viii.) Determine the
slope of the graph and
its intercept,c on the
vertical axis.
(ix) Evaluate K = c/S.
(x) Using your graph,
determine the current
(i) whenL = 55cm.
(xi.) State two
precautions taken to
ensure accurate result
SOLUTIONS:
TABLE OF VALUES:
S/N: 1,2,3,4,5.
Io(A): 0.650,0.700,0.7
50,0.800,0.850.
L(cm): 90.00,80.00,70. ­
00,60.00,50.00.
I^-1(A^- 1):
0.011,0.013,0.0
14,0.017,0.020.
I(A): 0.70,0.75,0.80,
0.85,0.90.
NOTE THE FOLLOWING:
*Draw a table for the
above values.
*Note that ^ means
Raise to Power. Eg:
2^-1 means Two raise
to power minus One.
*Note that COMMA(,)
in the above table
means NEXT LINE. Eg:
2,3,4,5 means dat 3 is
under 2, 4 is under 3
and 5 is under 4 in a
table form.
Therefore, S/N has
1,2,3,4,5under it, in a
table as shown in the
table of values above.
i.e 1 is under S/N and 2
is under 1 and 3 is
under 2 and4 is under
3 and so on.....
(viii.) Slope(S) = DI/
DI^-1 = 0.65/0.24
= 27.1.
Intercept (c) on the
vertical axis =
0.43amps.
(ix.) K= c/S = 0.43/27.1
= 0.016.
(x.) The value of the
current I when L =
55cm
= 0.018cm.
(xi.) Precautions:
(a.) I made sure the
key was removed
when readings
werenot being taken.
(b.) I ensured that the
terminals were clean.