The result should have the same number of significant figures as the measurement with the least number of significant figures.
| Dimension | Value | Significant Figures |
|---|---|---|
| Length (\(l\)) | \(4.234 \, \text{m}\) | 4 |
| Breadth (\(b\)) | \(1.005 \, \text{m}\) | 4 |
| Thickness (\(t\)) | \(2.01 \, \text{cm} = 0.02101 \, \text{m}\) | 3 |
Limiting Factor: Thickness has 3 significant figures
Formula: \( \text{Area} = l \times b \)
\[ 4.234 \times 1.005 = 4.25487 \, \text{m}^2 \]
Raw precision: 5 sig figs (4.25487)
Limit by measurements: Both have 4 sig figs → Round to 4 sig figs
Formula: \( \text{Volume} = l \times b \times t \)
\[ 4.234 \times 1.005 \times 0.02101 = 0.089402 \, \text{m}^3 \]
Raw precision: 5 sig figs (0.089402)
Limit by measurements: Thickness (3 sig figs) → Round to 3 sig figs
| Calculation | Raw Result | Limiting Sig Figs | Rounded Result |
|---|---|---|---|
| Area (\(l \times b\)) | \(4.25487 \, \text{m}^2\) | 4 (both dimensions) | \(4.255 \, \text{m}^2\) |
| Volume (\(l \times b \times t\)) | \(0.089402 \, \text{m}^3\) | 3 (thickness) | \(0.0894 \, \text{m}^3\) |
Exam Answer Format:
Area = \( 4.255 \, \text{m}^2 \)
Volume = \( 0.0894 \, \text{m}^3 \)
Mass = \( (28 \pm 0.01) \, \text{g} \), Volume = \( (5 \pm 0.1) \, \text{cm}^3 \). What is the percentage error in density?