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 approximate uncertainty in cubic centimeters.

Answer 1

To calculate the volume 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 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, we calculate the volume:

\[
V = 1.80 \times 2.05 \times 3.10
\]

Calculating step by step:

1. Calculate \( 1.80 \times 2.05 \):
\[
1.80 \times 2.05 = 3.69
\]

2. Calculate \( 3.69 \times 3.10 \):
\[
3.69 \times 3.10 = 11.439
\]

Thus, the volume of the box is:

\[
V \approx 11.439 \, \text{cm}^3
\]

Next, we calculate the uncertainty in volume using the formula for propagation of uncertainty for multiplication:

\[
\frac{\Delta V}{V} = \sqrt{\left(\frac{\Delta L}{L}\right)^2 + \left(\frac{\Delta W}{W}\right)^2 + \left(\frac{\Delta H}{H}\right)^2}
\]

Calculating each relative uncertainty:

1. For length \( L \):
\[
\frac{\Delta L}{L} = \frac{0.01}{1.80} \approx 0.00556
\]

2. For width \( W \):
\[
\frac{\Delta W}{W} = \frac{0.02}{2.05} \approx 0.00976
\]

3. For height \( H \):
\[
\frac{\Delta H}{H} = \frac{0.01}{3.10} \approx 0.00323
\]

Now we can compute the total relative uncertainty in volume:

\[
\frac{\Delta V}{V} = \sqrt{(0.00556)^2 + (0.00976)^2 + (0.00323)^2}
\]

Calculating each term:

- \( (0.00556)^2 \approx 0.0000309136 \)
- \( (0.00976)^2 \approx 0.0000956976 \)
- \( (0.00323)^2 \approx 0.0000104329 \)

Now summing these:

\[
\frac{\Delta V}{V} = \sqrt{0.0000309136 + 0.0000956976 + 0.0000104329} \approx \sqrt{0.0001360441} \approx 0.01165
\]

Multiply \( 0.01165 \) by the volume \( V \) to find \( \Delta V \):

\[
\Delta V = V \times 0.01165 \approx 11.439 \times 0.01165 \approx 0.1334
\]

Thus, rounding to two decimal places, the uncertainty in volume is approximately \( 0.13 \, \text{cm}^3 \).

Finally, we report the volume with its uncertainty:

\[
V = 11.44 \pm 0.13 \, \text{cm}^3
\]