Question

Stress versus elongation profiles for different polymeric materials are shown in the figure. Choose the combination that best describes these profiles.

• A. 1–Nylon fibers; 2–Polyethylene; 3–Vulcanized rubber; 4–Polystyrene
• B. 1–Polyethylene; 2–Vulcanized rubber; 3–Polystyrene; 4–Nylon fibers
• C. 1–Polystyrene; 2–Nylon fibers; 3–Polyethylene; 4–Vulcanized rubber
• D. 1–Vulcanized rubber; 2–Polyethylene; 3–Nylon fibers; 4–Polystyrene

Hint:

Brittle polymers (e.g., polystyrene) fail early; elastomers (rubber) elongate a lot; semicrystalline polymers (PE) show ductility; fibers (Nylon) have high strength.

Answer 1

The Correct Option is C

Step 1: Profile 1 (highest stress at low elongation).
Nylon fibers are crystalline polyamides with hydrogen bonding between chains. They resist elongation strongly, showing high stress at small strain. $\Rightarrow$ Curve 1 = Nylon fibers.
Step 2: Profile 2 (moderate stress, ductile).
Polyethylene (especially HDPE) is semicrystalline and ductile, showing considerable elongation with moderate stress. $\Rightarrow$ Curve 2 = Polyethylene.
Step 3: Profile 3 (elastic, nonlinear, strain-hardening).
Vulcanized rubber exhibits large elongation, low initial stress, but then stress increases due to strain-induced crystallization. $\Rightarrow$ Curve 3 = Vulcanized rubber.
Step 4: Profile 4 (brittle fracture).
Polystyrene is amorphous and glassy; it breaks at low elongation with relatively low stress. $\Rightarrow$ Curve 4 = Polystyrene.

Step 5: Conclude.
Thus, the mapping is (1–Nylon fibers, 2–Polyethylene, 3–Vulcanized rubber, 4–Polystyrene), which matches option (A). Final Answer: \[ \boxed{\text{(A) 1–Nylon fibers; 2–Polyethylene; 3–Vulcanized rubber; 4–Polystyrene}} \]