UNIVERSITI TEKNOLOGI MARA
PHYSICS 094
LAB REPORT
EXPERIMENT 1: BASIC PHYSICAL MEASUREMENT
PI080S08 (Group 2)
PREPARED FOR:
DR. NUR ASYIKIN BINTI AHMAD NAZRI
PREPARED BY:
NO
NAME
STUDENT ID
1.
ADAM KAMAL EEHAN BIN KAMAL AZNIZAM
2022872288
2.
ALIAH MAISARAH BINTI ROSLI
2022872628
3.
SHARIFAH NORALYA SYAHIRAH BINTI SYED
MUHAMMAD ARIFF AL-ATTAS
2022871364
4.
NUR ALIAH SAFIYYAH BINTI BADLISHAH
2022871324
5.
ASMA ATHIRAH BINTI MD AZAM AKMAL
2022871966
6.
MUHAMMAD DANIAL BIN RIDUWAN
2022871094
DATE OF SUBMISSION
6 SEPTEMBER 2022
1
INTRODUCTION
A physical quantity is a physical property of a material or system that can be quantified by
measurement. Length, mass, and time are some examples of basic physical quantities that need to
be written in system International Unit (SI). The SI units of these quantity are the meter (m) for
length, the kilogram, (kg) for the mass and the second(s) is the unit of time. These physical
quantities are measured by using measuring tools such as meter ruler for length. Each device gives
a limited number of significant figures in the reading due to the size of the smallest division on the
scale. In general, the problem statement of this experiment is to measure the length (l), width (w)
and height (h) of the object using a vernier caliper. Then, use a micrometer screw gauge to measure
thickness (t) and diameter (d) of an object. Also, triple beam mass balances are used to measure
the mass (m), while the measuring tape is used to measure the circumference of an object. Before
starting an experiment, it is important for us to come up with the hypothesis first. In order to
confirm the theory, we need to make sure the data collection is in line with the hypothesis that we
made earlier. By comparing the result of accurate measurements to the prediction of the theory,
we can confirm the theory is correct.
OBJECTIVE OF THE EXPERIMENT
a) To measure length using vernier calliper and micrometre screw gauge.
b) To weigh object by using triple beam balance.
c) To calculate uncertainties for measurements made using laboratory instruments.
d) To determine the density of the objects.
e) To identify the significant of the gradient of the graph and calculate its percentage error.
THEORY
It is very important to get the accurate and precise reading in any measurements. Accuracy is when
the reading is near to the true value while precise is when the reading is very reproducible or in
other words that particular reading is not much differ to the previous reading. The imperfection in
all measurements is called uncertainty. When taking measurement, it is necessary to write its
uncertainty based on the measurement device. For example, when using ruler, the smallest reading
is 0.1cm therefore its uncertainty is 0.1cm and thus a reading of 2.2 cm from a ruler can be written
as (2.2±0.1) cm. For single measurement, the smallest scale of the measurement device is taken as
the uncertainty while for repeated measurements, analysis by calculating deviation and standard
deviation is necessary.
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MATERIALS APPARATUS
Vernier caliper, micrometer screw gauge, triple beam mass balance, π set and objects to be
measured.
METHODOLOGY
A. Measurement using vernier caliper and micrometer screw gauge
1. The length l, width w and height h of a rectangular objects were measured by using
vernier caliper. All the readings were recorded respectively in Table 1.1 and 1.2.
2. The thickness t and diameter d of a cylindrical objects were measured by using
micrometer screw gauge. All the readings were recorded respectively in Table 1.3 and
Table 1.4.
3. Steps 1 and 2 were repeated to obtain three sets of reading. The average reading was
calculated by using the appropriate formula.
B. Measurement of mass using triple beam mass balance
1. The mass m of the rectangular objects were measured by using triple mass balance. All
the reading were recorded respectively in Table 1.1 and Table 1.2.
2. The mass m of the cylindrical objects were measured by using triple beam mass balance.
All the reading were recorded respectively in Table 1.3 and Table 1.4.
3. Steps 1 and 2 were repeated to obtain three sets of readings. The average reading was
calculated by using the appropriate formula.
C. Determine value of by using set
1. The folded end of the measuring tape was slipped into the slot on the side. The smallest
disk was marked as M and the bigger disk as N, O and P.
2. The tap was wrapped once around the disk so that it overlaps the zero-line marker.
3. The circumference C was measured and the reading was recorded in Table 1.5. Three sets
of reading were obtained and the average value was calculated by using the appropriate
formula.
4. The diameter d was measured along the line marked on the face of the disk. The reading
was recorded in Table 1.5. Three sets of reading were obtained and the average value was
calculated by using appropriate formula.
5. Step 1 to 4 was repeated for disk N, O and P.
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RESULTS
A. Rectangular Objects
Length, l (m)
(10−2 )
Table 1.1
Object W
Width, w (m)
(10−2 )
Height, h (m)
(10−2 )
Mass, m (kg)
Reading 1, 𝑎1
2.00
2.00
5.690
3.35
Reading 2, 𝑎2
2.00
2.00
5.690
3.36
Reading 3, 𝑎3
2.00
2.00
5.692
3.33
Average reading, 𝑎
2.00
2.00
5.691
3.35
Deviation, |𝑎𝑛 − 𝑎|
0
0
0
0
0
0
0.001
0.001
0.001
0
0.01
0.01
0
0
0.00122
0.0158
Measurement
Standard deviation,
σ
(10−2 )
4
Measurement
Length, l (m)
(10−2 )
Table 1.2
Object X
Width, w (m)
(10−2 )
Reading 1, 𝑎1
3.570
1.922
1.922
3.80
Reading 2, 𝑎2
3.564
1.920
1.920
3.80
Reading 3, 𝑎3
3.570
1.920
1.920
3.78
Average reading, 𝑎
3.568
1.920
1.920
3.80
Deviation, |𝑎𝑛 − 𝑎|
0.002
0.004
0.002
0.002
0
0
0.002
0
0
0
0
0.02
0.00346
0.0014
0.0014
0.014
Standard deviation,
σ
Height, h (m)
(10−2 )
Mass, m (kg)
(10−2 )
5
B. Cylindrical Objects
Table 1.3
Object Y
Measurement
Thickness, l (m)
(10−3 )
Diameter, d (m)
(10−2 )
Mass, m (kg)
(10−3 )
Reading 1, 𝑎1
9.85
2.22
8.40
Reading 2, 𝑎2
9.87
2.23
8.50
Reading 3, 𝑎3
9.82
2.22
8.60
Average reading, 𝑎
9.85
2.22
8.50
Deviation, |𝑎𝑛 − 𝑎|
0
0.02
0.03
0
0.01
0
0.10
0
0.10
Standard deviation, 𝜎
0.0255
0.0071
0.10
6
Measurement
Table 1.4
Object Z
Thickness, l (m)
Diameter, d (m)
(10−2 )
(10−2 )
Mass, m (kg)
(10−3 )
Reading 1𝑎1
1.506
2.219
9.80
Reading 2, 𝑎2
1.505
2.220
9.90
Reading 3, 𝑎3
1.508
2.218
9.70
Average reading, 𝑎
1.506
2.219
9.80
Deviation, |𝑎𝑛 − 𝑎|
0
0.001
0.002
0
0.001
0.001
0
0.10
0.10
Standard deviation, 𝜎
0.00158
0.001
0.10
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C. Disk M, N, O and P
Table 1.5
N
O
Circumference, C (m)
Disk
Measurement
M
Reading 1, 𝑎1
0.162
0.186
0.263
0.298
Reading 2, 𝑎2
0.162
0.185
0.263
0.297
Reading 3, 𝑎3
0.161
0.185
0.264
0.298
Average reading, 𝑎
0.162
0.185
0.263
0.298
Measurement
P
Diameter, d (m)
(10−2 )
Reading 1, 𝑎1
3.940
5.850
8.352
9.412
Reading 2, 𝑎2
3.942
5.852
8.354
9.412
Reading 3, 𝑎3
3.940
5.852
8.352
9.412
Average reading, 𝑎
3.940
5.851
8.353
9.412
8
ANALYSIS
A. Rectangular Object
i) Density for object W:
V=lxwxh
= (2.00x10−2 )(2.00x10−2 )(5.691x10−2 )
V = 2.2764 x 10−5 𝑚3
𝜌=
𝑚
𝑉
3.35 ×10−2
= 2.2764 ×10−5
𝜌 = 1.471 x 103 kg 𝑚 −3
∴ Density for object W is 1.471 x 103 kg 𝑚 3
ii) Uncertainty of density for object W:
∆𝑙
ΔV = ( 𝑙 +
∆𝑤
𝑤
+
∆ℎ
ℎ
)𝑉
0
0
0.001×10−2
= (2.00 𝑥 10−2 + 2.00 𝑥 10−2 + 5.691×10−2 )
ΔV = 0.26 𝑚3
∆𝑚
Δ𝜌 = ( 𝑚 +
∆𝑉
𝑉
)𝜌
0
0.26
= (3.35×10−2 + 1.471×103 ) (1.471 × 103 )
Δ𝜌 = 0.26 kg 𝑚 −3
∴ Uncertainty of density for object W is (1.471 x 103 ± 0.26) kg 𝑚 −3
iii) Density for object X:
V=lxwxh
= (3.568 x 10−2 )(1.920 x 10−2 )(1.920 x 10−2 )
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= 1.315 x 10−5 𝑚3
𝜌 =
𝑚
𝑉
3.80×10−2
= 1.315×10−5
= 2.889 x 103 𝑘𝑔 𝑚−3
iv) Uncertainty of density for object X:
∆𝑙
ΔV = ( 𝑙 +
=(
∆𝑤
𝑤
+
∆ℎ
ℎ
0.00346×10−2
3.568×10−2
)𝑉
+
0.0014×10−2
1.920×10−2
+
1.920×10−2
0.0014×10−2
) (1.315 × 10−5 )
ΔV = 0.018 𝑚3
∆𝑚
Δ𝜌 = ( 𝑚 +
∆𝑉
𝑉
)𝜌
0.014×10−2
0.018
= ( 3.80 ×10−2 + 1.315×10−5) (2.889 × 103 )
Δ𝜌 = 3.955 x 106 kg 𝑚−3
∴ Uncertainty of density for object X is (2.889 x 103 ± 3.955 x 106 ) kg 𝑚 −3
B. Cylindrical Object
i) Density for object Y:
V = π𝑟 2 𝑙
= π(
2.22×10−2
2
2
) (9.85 × 10−2 )
V = 3.813 x 10−6 𝑚3
𝜌=
𝑚
𝑉
8.500×10−3
= 3.813×10−6
𝜌 = 2.229 x 103 kg 𝑚 −3
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∴ Density for object W is 2.229 x 103 kg 𝑚 −3
ii) Uncertainty of density for object Y:
∆𝑚
Δ𝜌 = ( 𝑚 + 2
∆𝑑
𝑑
∆𝑙
+ 𝑙 )𝜌
0.10×10−3
= (8.50 ×10−3 +
2(0.0071×10−2 )
2.22 ×10−2
+
0.0255×10−3
9.85×10−3
) (2.229 × 103 )
Δ𝜌 = 46.25 kg 𝑚−3
∴ Uncertainty of density for object X is 46.25 kg 𝑚 −3
iii) Density for object Z:
V = π𝑟 2 𝑙
= π(
2.219×10−2
2
2
) (1.506 × 10−2 )
V = 5.824 x 10−6 𝑚3
𝜌=
=
𝑚
𝑉
9.80×10−3
5.824×10−6
𝜌 = 1.682 x 10−3 kg 𝑚 −3
∴ Density for object W is 1.682 x 10−3 kg 𝑚 −3
iv) Uncertainty of density for object Z:
∆𝑚
Δ𝜌 = ( 𝑚 + 2
∆𝑑
𝑑
0.10×10−3
∆𝑙
+ 𝑙 )𝜌
= (9.80×10−3 +
2(0.001×10 −2)
2.219×10 −2
+
0.00158×10−2
1.506×10−2
) (1.682 × 10−3 )
Δ𝜌 = 0.0111 kg 𝑚−3
∴ Uncertainty of density for object X is 0.0111 kg 𝑚 −3
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C. Disk
12
i) Gradient from the circumference vs diameter graph
0.30 − 0.18
Gradient, m = 0.09 − 0.05
=3
ii) Percentage error by comparing the actual value to the value from the graph
Percentage Error =
=
(ℎ𝑖𝑔ℎ𝑒𝑠𝑡 − 𝑙𝑜𝑤𝑒𝑠𝑡)
ℎ𝑖𝑔ℎ𝑒𝑠𝑡
(0.9 − 0.5)
0.9
× 100%
× 100%
= 44.4%
DISCUSSION
We will be using a vernier calliper and a micrometre screw gauge to do measurements and
calculating the uncertainty for three types of objects which are;
a) Rectangular Objects
b) Cylindrical Objects
c) Disks
In each formula, there are multiple variables which varies every time we measure it, this is called
uncertainty. As we measured it three times, there were changes or in some measurements no
changes.
Using the formula ;
Measured quantity = Average quantity ± Standard Deviation
Rectangular Objects
Object W
Measurement
Length , l (m)
Width , w (m)
Height , h (m)
Measured quantity, q
2.00 x 10−2 ± 0.000
2.00 x 10−2 ± 0.000
5.69 x 10−2 ± 0.001
Object X
Measurement
Length , l (m)
Width , w (m)
Height , h (m)
Measured quantity, q
3.68 x 10−2 ± 0.001
1.92 x 10−2 ± 0.000
1.92 x 10−2 ± 0.000
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Cylindrical Objects
Object Y
Measurement
Thickness, l (m)
Diameter, d (m)
Measured quantity
9.85 x 10−3 ± 0.0255 x 10−3
2.22 x 10−2 ± 0.0071 x 10−2
Object Z
Measurement
Thickness, l (m)
Diameter, d (m)
Measured quantity
1.506 x 10−2 ± 0.00158 x 10−2
2.219 x 10−2 ± 0.00100 x 10−2
Disks
Disk M
Measurement
Circumference, c (m)
Diameter, d (m)
Measured quantity
0.162
3.940
Disk N
Measurement
Circumference, c (m)
Diameter, d (m)
Measured quantity
0.185
5.851
Disk O
Measurement
Circumference, c (m)
Diameter, d (m)
Measured quantity
0.263
8.353
Disk P
Measurement
Circumference, c (m)
Diameter, d (m)
Measured quantity
0.298
9.412
Every measurement has its own error, which varies on every experiment or situation. Some
common errors are known as zero error, random error and parallax error.
The most common error known when making measurements is zero error. This happens
when the 0 value is not lined up correctly in the beginning therefore making it a must for making
adjustments such as adding or subtracting small values to correct the error. For instance positive
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zero error is when the 0 value goes further than 0 thus having to subtract by how much it is
extended and vice versa for negative zero error.
CONCLUSION
As a conclusion, we could consider our experiment as a successful experiment as we got
to measure length, width and height. From the result that we get, we conclude that vernier caliper
is the most suitable tools to get an accurate measurement. This is because vernier caliper have 2
scale which the main one has one decimal place and the next one has two decimal places. By
measuring the length, width and height, we could calculate the volume, density and uncertainty of
density. We can calculate the volume by using the formula length x width x height and calculate
the density with the formula p = m/v.
REFERENCES
1. https://en.wikipedia.org/wiki/Physical_quantity
2. (1) New Message! (sage-advices.com)
3. Author. (2020, June 27). Is water displacement more accurate than calculating volume?
Short. Retrieved September 5, 2022, from https://short-facts.com/is-water-displacementmore-accurate-than-calculating-volume/
POST-LAB QUESTIONS
1. Why is it better to take more than one reading for the same measured quantity compared
to one reading?
- To get accurate and precise reading in any measurement. The more readings you have,
the more accurate the result would be.
2. A measurement of a rectangular box is given as 1.98 ft x 1.98 ft x 2.84 ft. Determine the
surface area of the box in meter square.
-
Surface Area = 2(1.98)2 + 4(2.84 x 1.98)
= 30.336 ft x 0.3048
= 9.246 m^2
3. Give one different method that could be used to determine the density of object W, X, Y
and Z? Explain.
- Water displacement. The amount in which the water level increases, is the volume of the
dunked object. The change in water level equals the volume of the submerged object. The
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most accurate measuring tools to measure small amount of water is volumetric flasks,
burettes and pipetted with tolerances of less than ±0.02. The weight of the displaced fluid
can be found mathematically. The mass of the displaced fluid can be expressed in terms
of the density and its volume, m = ρV. The fluid displaced has a weight W = mg, where g
is acceleration due to gravity. Therefore, the weight of the displaced fluid can be
expressed as W = ρVg.
4. Is there any difference in measurement if we use ruler to find the diameter of a cylinder
instead of using vernier caliper.
- Yes. Vernier caliper can measure until 0.01mm, however ruler can only measure until
1mm.
5. How to eliminate zero error?
- zero error should be subtracted from the total reading.
6. From the experiment that you have conducted, is it necessary to find uncertainties?
Explain your answer.
- Yes, it is necessary because in every measurement with our human abilities, we are
bound to make some minor mistakes while measuring. Therefore, we should be fair in
taking the results by measuring it at least more than 2 times and take the average
compared to taking from only one measurement
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