Find the uncertainty values for the table Object	Measured Quantity	Calculated Quantities
	Diameter (cm)	Circumference (cm)	Area (cm2)
1	9.7±0.1	30.47±0.314	73.90±
2	6.7±0.1	21.05±0.314	35.26±
3	6.9±0.1	21.68±0.314	37.39±
4	5.0±0.1	15.71±0.314	19.63±
5	16.0±0.1	50.27±0.314	201.06±


To find the uncertainty values for the area of each object, we'll use error propagation rules. The area \(A\) of a circle is related to the diameter \(D\) by the formula:

\[
A = \pi \left( \frac{D}{2} \right)^2 = \frac{\pi D^2}{4}
\]

The uncertainty in the area, \( \Delta A \), can be determined by propagating the uncertainty in the diameter. For a function of the form \( A = kD^2 \) (where \(k = \frac{\pi}{4}\)), the uncertainty in \(A\) is given by:

\[
\Delta A = 2kD \cdot \Delta D = \frac{\pi D}{2} \cdot \Delta D
\]

### Step-by-Step Calculation:

For each object, we'll calculate the uncertainty in the area using the diameter and its uncertainty.

#### 1. Object 1: Diameter \(D = 9.7 \pm 0.1 \) cm
\[
\Delta A_1 = \frac{\pi \cdot 9.7}{2} \cdot 0.1 = 1.524 \, \text{cm}^2
\]
So, the area uncertainty for object 1 is \( \pm 1.52 \, \text{cm}^2 \).

#### 2. Object 2: Diameter \(D = 6.7 \pm 0.1 \) cm
\[
\Delta A_2 = \frac{\pi \cdot 6.7}{2} \cdot 0.1 = 1.052 \, \text{cm}^2
\]
So, the area uncertainty for object 2 is \( \pm 1.05 \, \text{cm}^2 \).

#### 3. Object 3: Diameter \(D = 6.9 \pm 0.1 \) cm
\[
\Delta A_3 = \frac{\pi \cdot 6.9}{2} \cdot 0.1 = 1.083 \, \text{cm}^2
\]
So, the area uncertainty for object 3 is \( \pm 1.08 \, \text{cm}^2 \).

#### 4. Object 4: Diameter \(D = 5.0 \pm 0.1 \) cm
\[
\Delta A_4 = \frac{\pi \cdot 5.0}{2} \cdot 0.1 = 0.785 \, \text{cm}^2
\]
So, the area uncertainty for object 4 is \( \pm 0.79 \, \text{cm}^2 \).

#### 5. Object 5: Diameter \(D = 16.0 \pm 0.1 \) cm
\[
\Delta A_5 = \frac{\pi \cdot 16.0}{2} \cdot 0.1 = 2.513 \, \text{cm}^2
\]
So, the area uncertainty for object 5 is \( \pm 2.51 \, \text{cm}^2 \).

### Final Table with Uncertainties in Area:

| Object | Diameter (cm)         | Circumference (cm)        | Area (cm²)        |
|--------|-----------------------|---------------------------|-------------------|
| 1      | \(9.7 \pm 0.1\)       | \(30.47 \pm 0.314\)       | \(73.90 \pm 1.52\)|
| 2      | \(6.7 \pm 0.1\)       | \(21.05 \pm 0.314\)       | \(35.26 \pm 1.05\)|
| 3      | \(6.9 \pm 0.1\)       | \(21.68 \pm 0.314\)       | \(37.39 \pm 1.08\)|
| 4      | \(5.0 \pm 0.1\)       | \(15.71 \pm 0.314\)       | \(19.63 \pm 0.79\)|
| 5      | \(16.0 \pm 0.1\)      | \(50.27 \pm 0.314\)       | \(201.06 \pm 2.51\)|

Would you like more details or have any questions?

Here are 5 related questions for deeper understanding:
1. How does the uncertainty in circumference relate to that in diameter?
2. Why is the uncertainty in area larger than that in diameter?
3. What role does the constant \(\pi\) play in propagating uncertainties?
4. How would a larger diameter uncertainty affect the final results?
5. Can the uncertainties in the measurements be minimized, and how?

**Tip:** When propagating uncertainties, always consider how each measured variable affects the final calculated result to avoid underestimating the total uncertainty.

Find the uncertainty values for the table Object Measured Quantity Calculated Quantities Diameter (cm) Circumference (cm) Area (cm2) 1 9.7±0.1 30.47±0.314 73.90± 2 6.7±0.1 21.05±0.314 35.26± 3 6.9±0.1 21.68±0.314 37.39± 4 5.0±0.1 15.71±0.314 19.63± 5 16.0±0.1 50.27±0.314 201.06±

Learn how to calculate the uncertainty in the area of circles using error propagation rules. This problem involves the geometry of circles and uncertainty in measurements, suitable for Grades 10-12. Includes step-by-step calculations for objects with different diameters and uncertainties.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation