To calculate the volume \( V \) of the 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 of the box.
Given the measurements:
- Length \( l = 1.80 , \text{cm} \) with an uncertainty of \( \Delta l = 0.01 , \text{cm} \)
- Width \( w = 2.05 , \text{cm} \) with an uncertainty of \( \Delta w = 0.02 , \text{cm} \)
- Height \( h = 3.10 , \text{cm} \) with an uncertainty of \( \Delta h = 0.01 , \text{cm} \)
First, calculate the volume:
\[ V = 1.80 , \text{cm} \times 2.05 , \text{cm} \times 3.10 , \text{cm} \]
Calculating step by step:
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Calculate \( 1.80 \times 2.05 \):
\[ 1.80 \times 2.05 = 3.69 \]
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Now calculate \( 3.69 \times 3.10 \):
\[ 3.69 \times 3.10 = 11.439 \]
Thus, the volume \( V \approx 11.44 , \text{cm}^3 \).
Next, we need to calculate the uncertainty in volume using the formula for propagation of uncertainty for multiplication:
\[ \frac{\Delta V}{V} = \sqrt{\left( \frac{\Delta l}{l} \right)^2 + \left( \frac{\Delta w}{w} \right)^2 + \left( \frac{\Delta h}{h} \right)^2} \]
Substituting the values:
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For \( l \):
\[ \frac{\Delta l}{l} = \frac{0.01}{1.80} \approx 0.00556 \]
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For \( w \):
\[ \frac{\Delta w}{w} = \frac{0.02}{2.05} \approx 0.00976 \]
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For \( h \):
\[ \frac{\Delta h}{h} = \frac{0.01}{3.10} \approx 0.00323 \]
Now, calculating the combined uncertainty:
\[ \frac{\Delta V}{V} = \sqrt{(0.00556)^2 + (0.00976)^2 + (0.00323)^2} \]
Calculating each term inside the square root:
- \( (0.00556)^2 \approx 0.0000309136 \)
- \( (0.00976)^2 \approx 0.0000952576 \)
- \( (0.00323)^2 \approx 0.0000104329 \)
Summing these up:
\[ 0.0000309136 + 0.0000952576 + 0.0000104329 \approx 0.0001366041 \]
Taking the square root:
\[ \sqrt{0.0001366041} \approx 0.01169 \]
Now multiply by the volume to get the absolute uncertainty in volume:
\[ \Delta V \approx V \times 0.01169 \approx 11.44 \times 0.01169 \approx 0.134 \]
Rounding this value to two decimal places:
Thus, the approximate volume of the box and its uncertainty is:
\[ V \approx 11.44 \pm 0.13 , \text{cm}^3 \]
Final answer:
Volume = \( 11.44 , \text{cm}^3 \) with uncertainty = \( 0.13 , \text{cm}^3 \).