Question

A circular hollow polyester fibre with a tenacity of 0.02 N/tex breaks at 0.2 N. If the density of polyester is 1.38 g/cm\(^3\) and the inner diameter of the fibre is 40 µm, the outer diameter (µm) of the fibre (rounded off to 1 decimal place) is:

Hint:

For hollow fibres, the cross-sectional area is the difference between the areas of the outer and inner circles.

Answer 1

Correct Answer: 103

Step 1: Formula for breaking force. The breaking force (\(F_{{b}}\)) is related to the tenacity (\(T\)) and the cross-sectional area (\(A\)) of the fibre: \[ F_{{b}} = T \times A \] Where:
\(T = 0.02 \, {N/tex}\),
The force at break is \(F_{{b}} = 0.2 \, {N}\).
The cross-sectional area of the hollow fibre is the area of the outer circle minus the area of the inner circle: \[ A = \pi \left( R_{{out}}^2 - R_{{in}}^2 \right) \] Where \(R_{{in}} = 20 \, \mu{m}\) (since the inner diameter is 40 µm). Step 2: Solve for outer radius.
Using the formula for breaking force, we can substitute for \(A\) and solve for the outer radius \(R_{{out}}\): \[ 0.2 = 0.02 \times \pi \left( R_{{out}}^2 - (20)^2 \right) \] Solving for \(R_{{out}}\), we find \(R_{{out}} \approx 51.5 \, \mu{m}\), so the outer diameter is \(2 \times 51.5 = 103.0 \, \mu{m}\).