Difficulty: Easy
Correct Answer: pearlite and cementite
Explanation:
Introduction / Context:
The iron–carbon equilibrium diagram predicts microstructures after very slow cooling. Understanding how hypo- and hyper-eutectoid compositions differ in resulting constituents is foundational to heat-treatment strategy and property prediction.
Given Data / Assumptions:
Concept / Approach:
At the eutectoid composition (~0.8% C), austenite transforms to pearlite. For hyper-eutectoid steels, proeutectoid cementite forms along austenite grain boundaries before the eutectoid reaction. At the eutectoid temperature, the remaining austenite transforms to pearlite. Thus, the final structure is pearlite plus networks or particles of cementite, not ferrite, because carbon is above eutectoid.
Step-by-Step Solution:
On cooling from austenite, precipitate proeutectoid cementite due to C > eutectoid.At 723°C, eutectoid reaction converts remaining austenite to pearlite.Final mixture at room temperature: pearlite + cementite.Hence, option (d) is correct.
Verification / Alternative check:
Metallographic images of hyper-eutectoid steels show cementite networks or particles decorating prior austenite boundaries with pearlite colonies inside grains.
Why Other Options Are Wrong:
(a) ignores proeutectoid cementite; (b) and (c) involve ferrite typical of hypo-eutectoid steels; (e) requires non-equilibrium cooling (bainite/martensite), not slow cooling.
Common Pitfalls:
Assuming ferrite appears in all plain-carbon steels; phase balance depends on carbon level relative to eutectoid composition.
Final Answer:
pearlite and cementite
Discussion & Comments