PAPER 3
SECTION A
QUESTION 2
INTERPRETATION OF GRAPHS
1
PN TAN SHIM YU PAPER 3 - SECTION A - Q2 7/19/2012
2
1
d
. A student carried out an experiment to
investigate the relationship between the
resistance, R, and diameter, d, for a
constantan wire, and determine the
resistivity, ρ, of constantan. The student
used five constantan wires with different
diameters and fixed the length,l, of each
wire as 2x103 mm.
The results of the experiment are
shown in the graph R against
2
PN TAN SHIM YU PAPER 3 - SECTION A - Q2 7/19/2012
Relationship between variables
a
b
A) a is inversely proportional to b
B) a is directly proportional to b
C) a increases linearly with b
a
b
A) a is inversely proportional to b
B) a is directly proportional to b
C) a increases linearly with b
3
PN TAN SHIM YU PAPER 3 - SECTION A - Q2 7/19/2012
A) a is inversely proportional to b
B) a is directly proportional to b
C) a increases linearly with b
A) a is inversely proportional to b
B) a is directly proportional to 1/b
C) a increases linearly with 1/b
a
b
a
1/b
Relationship between variables
4
PN TAN SHIM YU PAPER 3 - SECTION A - Q2 7/19/2012
A) a is inversely proportional to b
B) a is directly proportional to b
C) a is linearly increasing with b
A) a is inversely proportional to 1/b
B) a is directly proportional to 1/b
C) a is linearly increasing with 1/b
a increases with b
a
b
a
1/b
Relationship between variables
(a) R is directly proportional to 1/d2
OR R is inversely proportional to d2
5
PN TAN SHIM YU PAPER 3 - SECTION A - Q2 7/19/2012
1/d2 = 0.41mm-2
1/d = 0.64 mm- 1
d = 1.56 mm
m = 0.73/0.6
= 1.22 Ωmm2
ρ = 0. 786(1.22 Ωmm2)
2 x 103 mm
= 4.79 x 10-4 Ωmm
R = ρ l
0.786 d2
= 4.79 x 10-4 Ωmm x 500 mm
0.786 (2.5mm)2
= 4.88 x 10-2 Ω
0.5
0.41
6
PN TAN SHIM YU PAPER 3 - SECTION A - Q2 7/19/2012
0.73
0.1
0.08 0.6 (0.6 – 0.08)
(0.73 – 0.1)
Gradient
= (0.73 – 0.1)
(0.6 – 0.08)
= 0.63
0.52
= 1.211Ωmm2
PN TAN SHIM YU PAPER 3 - SECTION A - Q2
7
7/19/2012
(1.16- 1.28)
l
Rd 2
786.0
l
m786.0
2
1
d
2
1
d
(c) The resistivity, ρ, of constantan is given by the formula:
If m = Rd2, then
where m in is the gradient of the graph of R against
(i) Calculate the gradient, m, of the graph of R against
Show on the graph how you determined m.
l
m786.0
mml 3102
(ii) Using the formula,
, calculate the resistivity, ρ of constantan.
Use
[3 marks]
[2 marks]
2 gantian betul
berdasarkan graf
3 unit yg betul
1
2 unit betul
1.22
1.22
4.78 x 10-4 Ωmm
PN TAN SHIM YU PAPER 3 - SECTION A - Q2
8
7/19/2012
l
Rd 2
786.0
(d) Another constantan wire has a diameter of 2.5 mm and a length of 500 mm.
Using the formula
and the value of in (c)(ii), calculate the resistance of the wire.
R = ............................................................
[2 marks]
1 gantian
2 dgn unit betul
4.78 x 10-4 Ωmm =
4.78 x 10-4 Ωmm
0.0487
PN TAN SHIM YU PAPER 3 - SECTION A - Q2
9
7/19/2012
(e) State one precaution that should be taken during this experiment.
1. Securely fasten the connecting wire
2. Avoid parallax error when taking readings of current and
potential difference by making sure that the image of the
needle in the mirror overlaps with the needle
..............................................................................................................................
[1 mark]
PN TAN SHIM YU PAPER 3 - SECTION A - Q2
10
7/19/2012
A student carries out an experiment to investigate the relationship between the
potential difference , V , across a filament bulb and the current , I , flowing through it.
The results of the experiment is shown in the graph of V against I.
(a) Based on the graph, determine the value of V when I = 0.2 A. Show on
the graph , how you determine the value of V.
V = 2.92 V ± 0.02 [2 marks]
(b) Ohm’s Law states that the potential difference across a conductor is
directly proportional to the current flowing through the conductor.
Based on the statement above m determine the range of the current that
obeys the Ohm’ s Law.
Show on the graph how you determine the range of the current.
Accepted Value = 0.00A - 0.12 A [2 marks]
(c ) The gradient of the graph represents resistance. Show how resistance varies with
current.
When I increases, resistance increases.
…………………………………….............................................. [1 mark]
PN TAN SHIM YU PAPER 3 - SECTION A - Q2 11
7/19/2012
V = 2.92V
0 A – 0.14A
m = 4.08 – 1.00
0.256 – 0.112
= 21.39 VA -1
P = I2m
= (0.2A)2 x 21.39 VA -1
= 0.86 VA
= 0.86 W
PN TAN SHIM YU PAPER 3 - SECTION A - Q2
12
7/19/2012
0.256 0.112
(d) The electric power , P, used by the bulb is given by the formula
P = I2m, where m is the gradient of the graph.
(i) Calculate the gradient , m , when I = 0.2 A
Show on the graph how you determine m. [4 marks]
4
3
(ii) Calculate the value of P when the current I = 0.2 A.
correct substitution - P= ( 0.2 )2 (22.6)
Answer with correct unit = 0.904 W
(0.85 - 0.95) [2 marks]
(e) State on precaution that should be taken to improve the results of this
experiment. 1. Ensure all connections in the circuit are tight
No leakage of current / short circuit.
2. Avoid parallax error when taking readings of current and
potential difference by making sure that the image of the
needle in the mirror overlaps with the needle
...............................................................................................................................
[1 mark]
PN TAN SHIM YU PAPER 3 - SECTION A - Q2 13
7/19/2012
22.6Ω ( 21.47 - 23.73)
SELAMAT MENDUDUKI
SPM !
SEMOGA MENDAPAT
KEPUTUSAN FIZIK YANG
CEMERLANG
PN TAN SHIM YU PAPER 3 - SECTION A - Q2
14
7/19/2012
Top Related