Hypo-eutectoid steel on slow cooling: When a plain carbon steel containing less than 0.8% carbon is cooled slowly from above the critical range, its room-temperature microstructure consists of:
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Amainly ferrite
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Bmainly pearlite
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Cferrite and pearlite
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Dpearlite and cementite
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Emartensite
Answer
Correct Answer: ferrite and pearlite
Explanation
Introduction / Context:The iron–carbon diagram predicts the phases that form during slow (near equilibrium) cooling. Understanding the mixture present below the eutectoid composition is foundational for heat-treatment and property control of steels used in structures and machinery.
Given Data / Assumptions:
- Carbon content < 0.8% (hypo-eutectoid steel).
- Cooling is slow enough to approximate equilibrium.
- No alloying additions complicating the transformation.
Concept / Approach:Upon cooling from austenite, hypo-eutectoid steels first precipitate proeutectoid ferrite. At the eutectoid temperature, the remaining austenite transforms to pearlite (alternate ferrite–cementite lamellae). Thus, the final structure contains ferrite plus pearlite. The proportion of ferrite increases as carbon decreases; at very low carbon, ferrite dominates, but the textbook answer for the class is “ferrite and pearlite.”
Step-by-Step Solution:Start above the critical range: austenitic.Cross A3: form proeutectoid ferrite until reaching eutectoid temperature.At eutectoid temperature: remaining austenite → pearlite.Result: ferrite + pearlite mixture at room temperature.
Verification / Alternative check:Metallographs show ferrite grains with pearlite colonies along boundaries in low-carbon steels.
Why Other Options Are Wrong:Mainly ferrite applies only at very low carbon; mainly pearlite applies near 0.8% C; pearlite + cementite is for hyper-eutectoid steels; martensite requires rapid quenching, not slow cooling.
Common Pitfalls:Assuming a single constituent; ignoring the proeutectoid phase formation sequence.
Final Answer:ferrite and pearlite