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. 2 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. 3 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 7 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 ) 9 = 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 10 ∴ 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 11 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 13 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 14 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 15 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 16