Question

The sides of a small rectangular box are measured to be 1.80 ± 0.01 cm long, 2.05 ± 0.02 cm long, and 3.10 ± 0.01 cm long. Calculate the box's volume and appropximate uncertainty in cubic centimeters

Answer 1

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:

1. Calculate \( 1.80 \times 2.05 \):

\[
1.80 \times 2.05 = 3.69
\]

2. 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:

1. For \( l \):

\[
\frac{\Delta l}{l} = \frac{0.01}{1.80} \approx 0.00556
\]

2. For \( w \):

\[
\frac{\Delta w}{w} = \frac{0.02}{2.05} \approx 0.00976
\]

3. 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 = \( 0.13 \, \text{cm}^3 \)**.