The internal and external diameters of a hollow cylinder are measured with the help of a Vernier callipers . Their values are `4.23 +- 0.01 cm and 3.87 +- 0.01 cm`, respectively . The thickness of the wall of the cylinder is

Options

A. `0.36 +- 0.02 cm`
B. `0.18 +- 0.02 cm`
C. `0.36 +- 0.01 cm`
D. `0.18 +- 0.01 cm`

Explanation

To find the thickness of the wall of a hollow cylinder given the internal and external diameters, we can follow these steps:

### Step 1: Identify the given values
- Internal diameter (d1) = 4.23 cm ± 0.01 cm
- External diameter (d2) = 3.87 cm ± 0.01 cm

### Step 2: Calculate the internal and external radii
The radius is half of the diameter. Therefore:
- Internal radius (r1) = d1 / 2 = 4.23 cm / 2 = 2.115 cm
- External radius (r2) = d2 / 2 = 3.87 cm / 2 = 1.935 cm

### Step 3: Calculate the thickness of the wall
The thickness (t) of the wall can be calculated using the formula:
\[ t = r1 - r2 \]
Substituting the values:
\[ t = 2.115 cm - 1.935 cm = 0.18 cm \]

### Step 4: Calculate the uncertainty in thickness
To find the uncertainty in the thickness, we need to consider the uncertainties in the diameters. The uncertainty in the thickness (Δt) can be calculated as:
\[ Δt = Δd1 + Δd2 \]
Where:
- Δd1 = 0.01 cm (uncertainty in internal diameter)
- Δd2 = 0.01 cm (uncertainty in external diameter)

Thus:
\[ Δt = 0.01 cm + 0.01 cm = 0.02 cm \]

### Step 5: Combine the results
The final result for the thickness of the wall of the cylinder is:
\[ t = 0.18 cm ± 0.02 cm \]

### Final Answer
The thickness of the wall of the cylinder is \( 0.18 \, \text{cm} \pm 0.02 \, \text{cm} \).

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