To calculate the uncertainty in the area of a circle based on the uncertainty in its diameter, we can use the propagation of uncertainty formula. The formula for the area $A$ of a circle is:

$A = \pi \times \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4}$

Uncertainty in Area ($\Delta A$):

When uncertainty in diameter ($D$) is given, the uncertainty in area ($A$) can be calculated using the following propagation of error formula:

$\frac{\Delta A}{A} = 2 \times \frac{\Delta D}{D}$

Where:

• $\Delta A$ is the uncertainty in area.
• $\Delta D$ is the uncertainty in diameter.
• $A$ is the calculated area.
• $D$ is the measured diameter.

Now let's apply this formula to each object.

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Object 1:

• Diameter $D = 9.7 \, \text{cm}$, Uncertainty in diameter $\Delta D = 0.1 \, \text{cm}$
• Calculated area $A = 73.90 \, \text{cm}^2$

$\frac{\Delta A}{73.90} = 2 \times \frac{0.1}{9.7} = 0.020619$ $\Delta A = 0.020619 \times 73.90 = 1.524 \, \text{cm}^2$

So, the area with uncertainty is:

$A = 73.90 \pm 1.52 \, \text{cm}^2$

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Object 2:

• Diameter $D = 6.7 \, \text{cm}$, Uncertainty in diameter $\Delta D = 0.1 \, \text{cm}$
• Calculated area $A = 35.26 \, \text{cm}^2$

$\frac{\Delta A}{35.26} = 2 \times \frac{0.1}{6.7} = 0.029851$ $\Delta A = 0.029851 \times 35.26 = 1.053 \, \text{cm}^2$

So, the area with uncertainty is:

$A = 35.26 \pm 1.05 \, \text{cm}^2$

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Object 3:

• Diameter $D = 6.9 \, \text{cm}$, Uncertainty in diameter $\Delta D = 0.1 \, \text{cm}$
• Calculated area $A = 37.39 \, \text{cm}^2$

$\frac{\Delta A}{37.39} = 2 \times \frac{0.1}{6.9} = 0.028986$ $\Delta A = 0.028986 \times 37.39 = 1.084 \, \text{cm}^2$

So, the area with uncertainty is:

$A = 37.39 \pm 1.08 \, \text{cm}^2$

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Object 4:

• Diameter $D = 5.0 \, \text{cm}$, Uncertainty in diameter $\Delta D = 0.1 \, \text{cm}$
• Calculated area $A = 19.63 \, \text{cm}^2$

$\frac{\Delta A}{19.63} = 2 \times \frac{0.1}{5.0} = 0.04$ $\Delta A = 0.04 \times 19.63 = 0.785 \, \text{cm}^2$

So, the area with uncertainty is:

$A = 19.63 \pm 0.79 \, \text{cm}^2$

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Object 5:

• Diameter $D = 16.0 \, \text{cm}$, Uncertainty in diameter $\Delta D = 0.1 \, \text{cm}$
• Calculated area $A = 201.06 \, \text{cm}^2$

$\frac{\Delta A}{201.06} = 2 \times \frac{0.1}{16.0} = 0.0125$ $\Delta A = 0.0125 \times 201.06 = 2.513 \, \text{cm}^2$

So, the area with uncertainty is:

$A = 201.06 \pm 2.51 \, \text{cm}^2$

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Final Results:

1. Object 1: $A = 73.90 \pm 1.52 \, \text{cm}^2$
2. Object 2: $A = 35.26 \pm 1.05 \, \text{cm}^2$
3. Object 3: $A = 37.39 \pm 1.08 \, \text{cm}^2$
4. Object 4: $A = 19.63 \pm 0.79 \, \text{cm}^2$
5. Object 5: $A = 201.06 \pm 2.51 \, \text{cm}^2$

Would you like further details or have any questions?

Related Questions:

1. How do we propagate uncertainties for different types of formulas (e.g., addition vs multiplication)?
2. How would the uncertainty change if the diameter measurements were more precise?
3. How does the uncertainty in circumference relate to the uncertainty in diameter?
4. What would happen if we used a different uncertainty value for the diameter?
5. How does changing the shape (e.g., to a square) affect uncertainty propagation?

Tip: When propagating uncertainties, always ensure the uncertainty in each measured quantity is expressed with the same number of significant figures as the uncertainty itself.