Question

A eutectoid steel with 100% austenite is cooled from a temperature of 750°C to a room temperature of 35°C. Match the cooling methods with the transformed structures.
\[ \begin{array}{c|c} \textbf{Cooling method} & \textbf{Transformed structure} \\ \hline P\ \ \text{Water quenching} & 1\ \ \text{Coarse pearlite} \\ Q\ \ \text{Oil quenching} & 2\ \ \text{Fine pearlite} \\ R\ \ \text{Air cooling} & 3\ \ \text{Martensite} \\ S\ \ \text{Furnace cooling} & 4\ \ \text{Very fine pearlite} \\ \end{array} \]

• A. P-1, Q-2, R-3, S-4
• B. P-2, Q-3, R-4, S-1
• C. P-3, Q-4, R-2, S-1
• D. P-3, Q-4, R-1, S-2

Hint:

Remember: faster cooling produces finer or martensitic structures, while slower cooling produces coarser pearlite. Water → martensite, Oil → very fine pearlite, Air → fine pearlite, Furnace → coarse pearlite.

Answer 1

The Correct Option is C

To correctly match the cooling method with the transformed microstructure, we use the fundamental temperature–time–transformation (TTT) behaviour of eutectoid steel. Since the steel starts as 100% austenite at 750\,°C, the structure formed depends entirely on the cooling rate. Faster cooling produces harder, finer structures, whereas slower cooling produces coarser, softer structures.
1. Water quenching (P):
Water quenching provides extremely rapid cooling, fast enough to avoid diffusion-based transformations such as pearlite formation. The cooling rate crosses below the martensite start (M_s) temperature before the pearlite or bainite regions on the TTT curve can be entered. Thus the austenite transforms into martensite, a very hard and brittle phase formed by diffusionless shear transformation. Therefore, P corresponds to structure 3.
2. Oil quenching (Q):
Oil quenching has a slower cooling rate than water. The cooling is still fast enough to avoid coarse pearlite formation but not fast enough to fully bypass the nose of the TTT curve. As a result, the steel begins to transform into very fine pearlite. Fine microstructural spacing results from the rapid cooling, producing a harder and stronger pearlite. Hence, Q corresponds to structure 4.
3. Air cooling (R):
Cooling in air is significantly slower than both oil and water quenching. The steel cools through the pearlite-forming region of the TTT diagram at a moderate rate, providing sufficient time for diffusion to form pearlite with relatively small lamellar spacing. This produces fine pearlite, not martensite. Thus, R corresponds to structure 2.
4. Furnace cooling (S):
Furnace cooling is the slowest method, allowing maximum time for atomic diffusion. The steel remains for a long time in the high-temperature pearlite region of the TTT diagram, causing growth of pearlite with large interlamellar spacing. This leads to the formation of coarse pearlite, a softer and weaker structure compared to fine pearlite. Therefore, S corresponds to structure 1.
Putting all combinations together: \[ P-3,\ Q-4,\ R-2,\ S-1, \] which matches option (C).