PHYSICS REPORT 1
AIM: To obtain the percentage of uncertainty of the length, width and area of the laboratory table.
LIST OF APPARATUS: Long ruler.
PROCEDURES:
1. A long ruler is placed at the corner of the laboratory table. 2. The length and the width of the table are recorded. 3. Step 2 is repeated 7 more times to obtain an average reading. 4. The area of the laboratory table is calculated and recorded.
RESULTS:
EXPERIMENT | LENGTH(CM) | WIDTH(CM) | AREA(CM2) | 1 | 113.3 | 153.2 | 17357.56 | 2 | 114.6 | 152.9 | 17522.34 | 3 | 113.9 | 153.9 | 17483.04 | 4 | 113.6 | 153.6 | 17448.96 | 5 | 113.8 | 153.7 | 17491.06 | 6 | 113.8 | 153.8 | 17502.44 | 7 | 113.7 | 153.0 | 17396.10 | 8 | 113.4 | 153.2 | 17372.88 | TOTAL | 909.84 | 1227.30 | 139574.40 | MEAN | 113.73 | 153.41 | 17446.80 | STANDARD DEVIATION | 0.37 | 0.36 | - |
EXAMPLE1:
% UNCERTAINTY (LENGTH) = ±Standard deviation* Mean x 100 = ±0.37 * 113.73 x 100 = ±0.33%
% UNCERTAINTY (WIDTH) = ±0.36 * 153.41 x 100 = ±0.23%
AREA (AVERAGE) = WIDTH X LENGTH = 17446.80 CM2
UNCERTAINTY = 0.23 + 0.33 = ±0.56%
±0.56% OF 17446.80 = 97.7
Thus, the area is 17446.80 ±97.7 CM2. EXAMPLE1:
% UNCERTAINTY (LENGTH) = ±Standard deviation* Mean x 100 = ±0.37 * 113.73 x 100 = ±0.33%
% UNCERTAINTY (WIDTH) = ±0.36 * 153.41 x 100 = ±0.23%
AREA (AVERAGE) = WIDTH X LENGTH = 17446.80 CM2
UNCERTAINTY = 0.23 + 0.33 = ±0.56%
±0.56% OF 17446.80 = 97.7
Thus, the area is 17446.80 ±97.7 CM2.

DISCUSSION: 1. The percentage of uncertainty for length is about 0.33% difference from the original length. 2. The percentage of uncertainty for width is about 0.23% difference from the original width. 3. The differences between the readings are probably caused by the zero error in the ruler readings.
CONCLUSION:
The