Answer :
Certainly! Let's proceed step by step to find the thickness of the wall of the hollow cylinder along with its uncertainty.
### Step 1: Understanding the problem
We are given:
- The external diameter of the cylinder [tex]\( D_\text{ext} = 4.23 \)[/tex] cm with an uncertainty of [tex]\( \pm 0.01 \)[/tex] cm.
- The internal diameter of the cylinder [tex]\( D_\text{int} = 3.89 \)[/tex] cm with an uncertainty of [tex]\( \pm 0.01 \)[/tex] cm.
We need to determine the thickness of the wall of the cylinder and the associated uncertainty.
### Step 2: Calculate the wall thickness
The thickness of the wall [tex]\( t \)[/tex] can be calculated using the formula:
[tex]\[ t = \frac{D_\text{ext} - D_\text{int}}{2} \][/tex]
Substituting the given values:
[tex]\[ t = \frac{4.23 \, \text{cm} - 3.89 \, \text{cm}}{2} \][/tex]
[tex]\[ t = \frac{0.34 \, \text{cm}}{2} \][/tex]
[tex]\[ t = 0.17 \, \text{cm} \][/tex]
So, the thickness of the wall is [tex]\( 0.17 \, \text{cm} \)[/tex].
### Step 3: Calculate the uncertainty in the wall thickness
The uncertainty in the thickness, [tex]\( \Delta t \)[/tex], of the wall can be found using the propagation of uncertainties. The formula for the propagation of uncertainties for subtraction and division is:
For subtraction:
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = \sqrt{(\Delta D_\text{ext})^2 + (\Delta D_\text{int})^2} \][/tex]
where [tex]\( \Delta D_\text{ext} \)[/tex] and [tex]\( \Delta D_\text{int} \)[/tex] are the uncertainties of the external and internal diameters, respectively.
Given that both uncertainties are [tex]\( \pm 0.01 \, \text{cm} \)[/tex]:
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = \sqrt{(0.01 \, \text{cm})^2 + (0.01 \, \text{cm})^2} \][/tex]
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = \sqrt{0.0001 \, \text{cm}^2 + 0.0001 \, \text{cm}^2} \][/tex]
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = \sqrt{0.0002 \, \text{cm}^2} \][/tex]
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = 0.014142 \, \text{cm} \][/tex]
Since the thickness is calculated by dividing the result by 2:
[tex]\[ \Delta t = \frac{\Delta (D_\text{ext} - D_\text{int})}{2} \][/tex]
[tex]\[ \Delta t = \frac{0.014142 \, \text{cm}}{2} \][/tex]
[tex]\[ \Delta t = 0.007071 \, \text{cm} \][/tex]
### Step 4: Match the calculated values with given options
The calculated wall thickness and uncertainty are:
- Thickness [tex]\( t = 0.17 \, \text{cm} \)[/tex]
- Uncertainty [tex]\( \Delta t = 0.007071 \, \text{cm} \approx 0.01 \, \text{cm} \)[/tex] (when rounding to two decimal places)
Among the given options, the one that best matches our calculated values is:
(3) [tex]\((0.17 \pm 0.01) \, \text{cm}\)[/tex]
### Final Answer:
The thickness of the wall of the cylinder is [tex]\((0.17 \pm 0.01) \, \text{cm}\)[/tex], which corresponds to option (3).
### Step 1: Understanding the problem
We are given:
- The external diameter of the cylinder [tex]\( D_\text{ext} = 4.23 \)[/tex] cm with an uncertainty of [tex]\( \pm 0.01 \)[/tex] cm.
- The internal diameter of the cylinder [tex]\( D_\text{int} = 3.89 \)[/tex] cm with an uncertainty of [tex]\( \pm 0.01 \)[/tex] cm.
We need to determine the thickness of the wall of the cylinder and the associated uncertainty.
### Step 2: Calculate the wall thickness
The thickness of the wall [tex]\( t \)[/tex] can be calculated using the formula:
[tex]\[ t = \frac{D_\text{ext} - D_\text{int}}{2} \][/tex]
Substituting the given values:
[tex]\[ t = \frac{4.23 \, \text{cm} - 3.89 \, \text{cm}}{2} \][/tex]
[tex]\[ t = \frac{0.34 \, \text{cm}}{2} \][/tex]
[tex]\[ t = 0.17 \, \text{cm} \][/tex]
So, the thickness of the wall is [tex]\( 0.17 \, \text{cm} \)[/tex].
### Step 3: Calculate the uncertainty in the wall thickness
The uncertainty in the thickness, [tex]\( \Delta t \)[/tex], of the wall can be found using the propagation of uncertainties. The formula for the propagation of uncertainties for subtraction and division is:
For subtraction:
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = \sqrt{(\Delta D_\text{ext})^2 + (\Delta D_\text{int})^2} \][/tex]
where [tex]\( \Delta D_\text{ext} \)[/tex] and [tex]\( \Delta D_\text{int} \)[/tex] are the uncertainties of the external and internal diameters, respectively.
Given that both uncertainties are [tex]\( \pm 0.01 \, \text{cm} \)[/tex]:
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = \sqrt{(0.01 \, \text{cm})^2 + (0.01 \, \text{cm})^2} \][/tex]
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = \sqrt{0.0001 \, \text{cm}^2 + 0.0001 \, \text{cm}^2} \][/tex]
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = \sqrt{0.0002 \, \text{cm}^2} \][/tex]
[tex]\[ \Delta (D_\text{ext} - D_\text{int}) = 0.014142 \, \text{cm} \][/tex]
Since the thickness is calculated by dividing the result by 2:
[tex]\[ \Delta t = \frac{\Delta (D_\text{ext} - D_\text{int})}{2} \][/tex]
[tex]\[ \Delta t = \frac{0.014142 \, \text{cm}}{2} \][/tex]
[tex]\[ \Delta t = 0.007071 \, \text{cm} \][/tex]
### Step 4: Match the calculated values with given options
The calculated wall thickness and uncertainty are:
- Thickness [tex]\( t = 0.17 \, \text{cm} \)[/tex]
- Uncertainty [tex]\( \Delta t = 0.007071 \, \text{cm} \approx 0.01 \, \text{cm} \)[/tex] (when rounding to two decimal places)
Among the given options, the one that best matches our calculated values is:
(3) [tex]\((0.17 \pm 0.01) \, \text{cm}\)[/tex]
### Final Answer:
The thickness of the wall of the cylinder is [tex]\((0.17 \pm 0.01) \, \text{cm}\)[/tex], which corresponds to option (3).
