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To find the thickness of the tape, we need to consider the volume of the tape in both its rolled and unrolled forms.

First, let's summarize the given information:

1. The tape length when unrolled: ( L = 100 ) meters or ( 10000 ) cm.
2. Outer diameter of the roll: ( D_\text{outer} = 10 ) cm, so the outer radius ( R_\text{outer} = 5 ) cm.
3. Inner diameter of the roll: ( D_\text{inner} = 5 ) cm, so the inner radius ( R_\text{inner} = 2.5 ) cm.

Step 1: Calculate the Volume of the Tape

The volume ( V ) of the tape can be found by calculating the volume of the cylindrical shell formed by the tape: [ V = \pi \left( R_\text{outer}^2 - R_\text{inner}^2 \right) \times \text{Width of the tape} ]

Let ( t ) be the thickness of the tape, then:

[ V = \pi (R_\text{outer}^2 - R_\text{inner}^2) \cdot t ]

Step 2: Relation Between Volume and Length When Unrolled

When the tape is unrolled, its volume is:

[ V = \text{Length} \times \text{Width} \times \text{Thickness} ]

Given the length ( L = 10000 ) cm and the tape width ( t ), we have:

[ V = 10000 \times \text{Width} \times \text{Thickness} ]

Since the volume remains the same when rolled or unrolled, we equate the two expressions for volume:

[ \pi (R_\text{outer}^2 - R_\text{inner}^2) \cdot t = 10000 \cdot \text{Width} \cdot t ]

Step 3: Solve for the Thickness

Divide both sides by ( t ):

[ \pi (R_\text{outer}^2 - R_\text{inner}^2) = 10000 \cdot \text{Width} ]

Given that ( \text{Width} ) appears on both sides, it cancels out:

[ t = \frac{\pi (R_\text{outer}^2 - R_\text{inner}^2)}{10000} ]

Substitute ( R_\text{outer} = 5 ) cm and ( R_\text{inner} = 2.5 ) cm:

[ t = \frac{\pi (5^2 - 2.5^2)}{10000} ] [ t = \frac{\pi (25 - 6.25)}{10000} ] [ t = \frac{\pi \cdot 18.75}{10000} ] [ t = \frac{18.75\pi}{10000} ] [ t \approx \frac{58.9055}{10000} ] [ t \approx 0.00589 \text{ cm} ]

Therefore, the thickness of the tape is approximately ( 0.00589 ) cm, or ( 0.0589 ) mm.

If you have any questions or need further details, feel free to ask!