Question

5. The sides of a small rectangular box are measured to be 1.8 +/-0.01 cm long, 2.05 +/-0.02 cm long. and 3.1 +/-0.01 cm long. Calculate the box's volume and approximate uncertainty in cubic centimeters. (1 point)

11.4+/ - 0.2c * m ^ 2 * 3

11.4+/ - 0.21c * m ^ A * 3

11+/ - 0.21c * m ^ 3

11 1 + i - 0.2c * m ^ 3

Answer 1

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. \( 1.8 \times 2.05 = 3.69 \, \text{cm}^2 \)
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:

\[
\boxed{11 \pm 0.2 \, \text{cm}^3}
\]

This format aligns with the choices presented in your question.