Question

5. 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 . (1 point ) 11.4+/-0.21 cm^ ^ 3; 11.4 + 1 - 0.2c * m ^ 3; 11+/-0.21 cm^ ^ 3; 11 - 0.2c * m ^ 3

Answer 1

To calculate the volume of a rectangular box and its uncertainty, we need to follow these steps:

1. **Calculate the Volume**:
The volume \( V \) of a rectangular box is given by the formula:
\[
V = l \times w \times h
\]
where \( l \), \( w \), and \( h \) are the lengths of the sides of the box.

2. **Given Measurements**:
- Length \( l = 1.80 \pm 0.01 \) cm
- Width \( w = 2.05 \pm 0.02 \) cm
- Height \( h = 3.10 \pm 0.01 \) cm

3. **Calculate the Volume**:
\[
V = 1.80 \, \text{cm} \times 2.05 \, \text{cm} \times 3.10 \, \text{cm}
\]

Performing the multiplication:
\[
V = 1.80 \times 2.05 \times 3.10 \approx 11.3 \, \text{cm}^3
\]

4. **Calculate the Uncertainty**:
The uncertainty in volume can be calculated using the formula for the uncertainty of multiplication:
\[
\frac{\Delta V}{V} = \sqrt{\left(\frac{\Delta l}{l}\right)^2 + \left(\frac{\Delta w}{w}\right)^2 + \left(\frac{\Delta h}{h}\right)^2}
\]

Where \( \Delta l \), \( \Delta w \), and \( \Delta h \) are the uncertainties in the lengths.

Substituting the values:
- \( \Delta l = 0.01 \, \text{cm} \)
- \( \Delta w = 0.02 \, \text{cm} \)
- \( \Delta h = 0.01 \, \text{cm} \)

We first calculate the relative uncertainties:
\[
\frac{\Delta l}{l} = \frac{0.01}{1.80} \approx 0.00556
\]
\[
\frac{\Delta w}{w} = \frac{0.02}{2.05} \approx 0.00976
\]
\[
\frac{\Delta h}{h} = \frac{0.01}{3.10} \approx 0.00323
\]

Now substitute into the total relative uncertainty formula:
\[
\frac{\Delta V}{V} = \sqrt{(0.00556)^2 + (0.00976)^2 + (0.00323)^2}
\]
\[
\frac{\Delta V}{V} \approx \sqrt{0.0000309136 + 0.0000952976 + 0.0000104289}
\]
\[
\frac{\Delta V}{V} \approx \sqrt{0.0001366401} \approx 0.0117
\]

Now, calculate \( \Delta V \):
\[
\Delta V \approx V \times 0.0117 \approx 11.3 \times 0.0117 \approx 0.132
\]

5. **Final Result**:
Rounding \( \Delta V \) gives approximately \( 0.13 \, \text{cm}^3 \). Thus, the volume of the box with uncertainty is:
\[
V \approx 11.3 \pm 0.13 \, \text{cm}^3
\]

Therefore, the closest answer would be:
- \(11.3 \pm 0.13 \, \text{cm}^3\).

In your options, none aligns directly with this calculation. The initial volume estimate provided in the options as \( 11.4 \pm 0.21 \, \text{cm}^3 \) could be considered as a rounding that is close to the values calculated if we adjust our assumptions about error margins, making \( 11.4 \pm 0.21 \, \text{cm}^3 \) the most reasonable option.