To calculate the volume \( V \) of a rectangular box, we use the formula:
\[ V = L \times W \times H \]
where \( L \) is the length, \( W \) is the width, and \( H \) is the height.
Given:
- Length \( L = 1.8 , \text{cm} \pm 0.01 , \text{cm} \)
- Width \( W = 2.05 , \text{cm} \pm 0.02 , \text{cm} \)
- Height \( H = 3.1 , \text{cm} \pm 0.01 , \text{cm} \)
First, calculate the volume:
\[ V = 1.8 , \text{cm} \times 2.05 , \text{cm} \times 3.1 , \text{cm} \]
Calculating step-by-step:
- \( 1.8 \times 2.05 = 3.69 , \text{cm}^2 \)
- Now multiply by the height: \[ 3.69 , \text{cm}^2 \times 3.1 , \text{cm} = 11.439 , \text{cm}^3 \]
Thus, the volume rounded to three significant figures is:
\[ V \approx 11.4 , \text{cm}^3 \]
Next, we need to calculate the uncertainty in volume. The relative uncertainties are derived from the individual measurements. The formula for the combined uncertainty in volume when multiplying gives:
\[ \frac{\Delta V}{V} = \sqrt{\left( \frac{\Delta L}{L} \right)^2 + \left( \frac{\Delta W}{W} \right)^2 + \left( \frac{\Delta H}{H} \right)^2} \]
Where:
- \( \Delta L = 0.01 , \text{cm} \), \( L = 1.8 , \text{cm} \)
- \( \Delta W = 0.02 , \text{cm} \), \( W = 2.05 , \text{cm} \)
- \( \Delta H = 0.01 , \text{cm} \), \( H = 3.1 , \text{cm} \)
Calculating the relative uncertainties:
\[ \frac{\Delta L}{L} = \frac{0.01}{1.8} \approx 0.00556 \]
\[ \frac{\Delta W}{W} = \frac{0.02}{2.05} \approx 0.00976 \]
\[ \frac{\Delta H}{H} = \frac{0.01}{3.1} \approx 0.00323 \]
Now calculate the total relative uncertainty:
\[ \text{Total relative uncertainty} = \sqrt{(0.00556)^2 + (0.00976)^2 + (0.00323)^2} \]
Calculating each squared term:
\[ (0.00556)^2 \approx 0.0000309, \] \[ (0.00976)^2 \approx 0.0000955, \] \[ (0.00323)^2 \approx 0.0000104 \]
Adding these gives:
\[ 0.0000309 + 0.0000955 + 0.0000104 \approx 0.0001368 \]
Taking the square root:
\[ \sqrt{0.0001368} \approx 0.0117 \]
Now, calculate the absolute uncertainty in volume:
\[ \Delta V = V \times \text{Total relative uncertainty} = 11.439 , \text{cm}^3 \times 0.0117 \approx 0.134 \]
Rounding this gives us:
\[ \Delta V \approx 0.14 , \text{cm}^3 \]
Finally, the volume of the box with its uncertainty is:
\[ V \approx 11.4 \pm 0.14 , \text{cm}^3 \]
Rounded to appropriate significant figures, we can express the volume more simply as:
\[ \boxed{11 \pm 0.2 , \text{cm}^3} \]
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