GATE ME 2023 Set 1 — Question 52

The given steel sample has 1.5 wt.% carbon, which is a hypereutectoid steel since its carbon content is greater than the eutectoid composition of 0.8 wt.% C.When this steel is slowly cooled, it undergoes the following transformations:We need to find the ratio of the mass fraction of pro-eutectoid cementite to the total cementite. We will use the lever rule on the provided phase diagram.From the diagram:

• Overall carbon composition, $C_0 = 1.5$ wt.%
• Carbon composition of ferrite ($\alpha$) at 723°C, $C_\alpha = 0.035$ wt.%
• Carbon composition of austenite ($\gamma$) at 723°C, $C_\gamma = 0.8$ wt.%
• Carbon composition of cementite (Fe3C), $C_{\text{Fe}_3\text{C}} = 6.7$ wt.%

Step 1: Calculate the mass fraction of total cementite (W_{\text{total_cem}})
This is calculated just below the eutectoid temperature (723 °C), where the phases are ferrite ($\alpha$) and cementite (Fe3C).
Using the lever rule:W_{\text{total_cem}} = \frac{C_0 - C_\alpha}{C_{\text{Fe}_3\text{C}} - C_\alpha} = \frac{1.5 - 0.035}{6.7 - 0.035} = \frac{1.465}{6.665} \approx 0.2198 Step 2: Calculate the mass fraction of pro-eutectoid cementite (W_{\text{pro_cem}})
This is the cementite that forms above the eutectoid temperature. We calculate its mass fraction by applying the lever rule just above 723 °C, where the phases are austenite ($\gamma$) and pro-eutectoid cementite (Fe3C).W_{\text{pro_cem}} = \frac{C_0 - C_\gamma}{C_{\text{Fe}_3\text{C}} - C_\gamma} = \frac{1.5 - 0.8}{6.7 - 0.8} = \frac{0.7}{5.9} \approx 0.1186 Step 3: Calculate the required ratio
The ratio of pro-eutectoid cementite to total cementite is:\text{Ratio} = \frac{W_{\text{pro_cem}}}{W_{\text{total_cem}}} = \frac{0.1186}{0.2198} \approx 0.53958 Rounding off to two decimal places, the ratio is 0.54.The official answer range is 0.53 to 0.55.