Solution:
The classification of polymers into fibres is typically based on their ability to form long chains that exhibit strong intermolecular forces like hydrogen bonding, van der Waals forces, and dipole-dipole interactions. These forces must be strong enough to hold the chains together tightly, allowing the material to be drawn into fibers that are both strong and flexible.
Vulcanized rubber (Option A): Vulcanized rubber is not classified as a fibre. It is a thermoset elastomer, made by adding sulfur to raw rubber which creates crosslinked polymer chains. The resulting material is elastic and resilient, ideal for products like tires and other elastic materials, but not for fibers.
Buna-S (Option B): Also known as styrene-butadiene rubber, is a copolymer made up of styrene and butadiene. It has good abrasion resistance and is used for items like shoe soles and car tires. However, it does not form fibers and is not classified as a fibrous polymer.
Terylene (Option C): Terylene, also known as polyester or polyethylene terephthalate (PET), is a polymer that has strong intermolecular forces due to its aromatic rings and ester linkages. These forces allow the molecules to align closely with each other, thereby forming fibers. Hence, Terylene is classified as a fibre and is used in textiles, recording tapes, and other products requiring material in fiber form.
Polystyrene (Option D): Polystyrene is a thermoplastic polymer with a structure unsuitable for fiber formation. It is commonly used in products like foam packaging, disposable cups, and insulating material. It is not characterized by the same degree of intermolecular forces seen in fibrous materials.
Therefore, the correct answer to the question is Option C, Terylene, which is classified as a fiber due to its strong intermolecular forces allowing it to be drawn into fibers.
© examsnet.com