Question

Pearlite in a steel is formed from

• A. Ferrite
• B. Martensite
• C. Cementite
• D. Austenite

Hint:

Remember the phase transformations on the Iron-Carbon diagram. Austenite is the parent phase at high temperatures. Upon cooling, it can transform into:

• Pearlite (Ferrite + Cementite) on slow cooling.

• Bainite on moderate cooling.

• Martensite on very fast cooling (quenching).

Answer 1

The Correct Option is D

Step 1: Set up the cooling path.
Start with a piece of eutectoid steel fully heated into the single phase austenite field, above $727^\circ C$, where all the carbon is dissolved uniformly in the FCC iron lattice.
Step 2: Follow what happens as it cools through the critical line.
As the temperature drops just below $727^\circ C$, the FCC austenite becomes thermodynamically unstable because FCC iron can no longer hold that much carbon in solution. The lattice must reorganize, and it does so cooperatively, carbon poor regions flip to BCC ferrite while carbon is simultaneously rejected and concentrated into thin plates of cementite next to them.
Step 3: Recognize the resulting structure.
Because ferrite and cementite grow side by side in this cooperative, diffusion controlled way, they end up as alternating lamellae, the characteristic layered structure we call pearlite. Both phases making up pearlite, ferrite and cementite, are simply the split products of the parent phase, not sources feeding into it.
Step 4: Conclusion.
Since pearlite is the direct decomposition product of the parent austenite at the eutectoid temperature, and neither ferrite, cementite, nor martensite could be the parent, the phase it forms from is austenite.
\[ \boxed{\text{Austenite}} \]