Question 1 |

A steel sample with 1.5 wt. % carbon (no other alloying elements present) is
slowly cooled from 1100^{\circ}C to just below the eutectoid temperature (723^{\circ}C). A
part of the iron-cementite phase diagram is shown in the figure. The ratio of the
pro-eutectoid cementite content to the total cementite content in the
microstructure that develops just below the eutectoid temperature is ________.

(Rounded off to two decimal places)

(Rounded off to two decimal places)

0.54 | |

0.35 | |

0.85 | |

0.95 |

Question 1 Explanation:

Mass friction of proeutectoid cementite \mathrm{W}_{\mathrm{Fe}_{3}} C^{\prime}=\frac{1.5-0.8}{6.7-0.8}=0.1186

Mass fraction of total cementite

\mathrm{W}_{\mathrm{Fe}_{3}} \mathrm{C}=\frac{1.5-0.035}{6.7-0.035}=0.2197

So, \quad \frac{W_{F_{3}} C^{\prime}}{W_{F_{e 3}} C}=\frac{0.1186}{0.2197}=0.54

Question 2 |

The atomic radius of a hypothetical face-centered cubic (FCC) metal is (\sqrt{2}/10) nm. The atomic weight of the metal is 24.092 g/mol. Taking
Avogadros number to be 6.023 \times 10^{23} atoms/mol, the density of the metal is
_______ kg/m^3. (Answer in integer)

2400 | |

3500 | |

2500 | |

3200 |

Question 2 Explanation:

Data given:

\begin{aligned} r & =\frac{\sqrt{2}}{10} \mathrm{~nm} \\ M & =24.092 \mathrm{~kg} / \mathrm{mol} \\ \mathrm{A} & =6.023 \times 10^{13} \end{aligned}

For FCC structure a=2 \sqrt{2} r=2 \sqrt{2} \times \frac{\sqrt{2}}{10}=\frac{4}{10}

So,volume of unit cell =a^{3}=\left(\frac{4}{10}\right)^{3}=0.064 \mathrm{~nm}^{3}

\mathrm{Ne}=4 for \mathrm{FCC}

\begin{aligned} \rho & =\frac{N_{e} \times M}{A \times \text { Volume of unit cell }} \\ & =\frac{4 \times \frac{24.092}{1000}}{6.023 \times 10^{3} \times \frac{0.064}{\left(10^{-9}\right)^{3}}} \\ & =2500 \mathrm{~kg} / \mathrm{m}^{3} \end{aligned}

\begin{aligned} r & =\frac{\sqrt{2}}{10} \mathrm{~nm} \\ M & =24.092 \mathrm{~kg} / \mathrm{mol} \\ \mathrm{A} & =6.023 \times 10^{13} \end{aligned}

For FCC structure a=2 \sqrt{2} r=2 \sqrt{2} \times \frac{\sqrt{2}}{10}=\frac{4}{10}

So,volume of unit cell =a^{3}=\left(\frac{4}{10}\right)^{3}=0.064 \mathrm{~nm}^{3}

\mathrm{Ne}=4 for \mathrm{FCC}

\begin{aligned} \rho & =\frac{N_{e} \times M}{A \times \text { Volume of unit cell }} \\ & =\frac{4 \times \frac{24.092}{1000}}{6.023 \times 10^{3} \times \frac{0.064}{\left(10^{-9}\right)^{3}}} \\ & =2500 \mathrm{~kg} / \mathrm{m}^{3} \end{aligned}

Question 3 |

Fluidity of a molten alloy during sand casting depends
on its solidification range. The phase diagram of a
hypothetical binary alloy of components A and B
is shown in the figure with its eutectic composition
and temperature. All the lines in this phase diagram,
including the solidus and liquidus lines, are straight
lines. If this binary alloy with 15 weight % of B
is poured into a mould at a pouring temperature of
800^{\circ}C , then the solidification range is

400 \; ^{\circ}C | |

250 \; ^{\circ}C | |

800 \; ^{\circ}C | |

150 \; ^{\circ}C |

Question 3 Explanation:

Solidification range =A'B

\triangle ABC \text{ and } \triangle MB'C is similar

\begin{aligned} \frac{MA}{MB'}&=\frac{BC}{BC'}\\ \frac{700-T_A}{700-400}&=\frac{15}{30}\\ 700-T_A&=\frac{1}{2} \times 300\\ T_A&=550^{\circ}C \end{aligned}

Solidification range =T_A-T_B=550-400=150 ^{\circ}C

Question 4 |

In Fe-Fe_3C phase diagram, the eutectoid composition
is 0.8 weight % of carbon at 725 ^{\circ}C . The maximum
solubility of carbon in \alpha -ferrite phase is 0.025
weight % of carbon. A steel sample, having no other
alloying element except 0.5 weight % of carbon, is
slowly cooled from 1000 ^{\circ}C to room temperature.
The fraction of pro-eutectoid \alpha -ferrite in the above
steel sample at room temperature is

0.387 | |

0.864 | |

0.475 | |

0.775 |

Question 4 Explanation:

Fraction of pro eutectoid \alpha- Ferrite

=\frac{0.8-0.5}{0.8-0.025}=0.387

Question 5 |

Which of the following methods can improve the
fatigue strength of a circular mild steel (MS) shaft?

**MSQ**Enhancing surface finish | |

Shot peening of the shaft | |

Increasing relative humidity | |

Reducing relative humidity |

Question 5 Explanation:

Surface Treatments:

During machining operations,small scratches and grooves are invariably introduced into the workpiece surface by cutting tool action. These surface markings can limit the fatigue life. It has been observed that improving the surface finish by polishing will enhance fatigue life significantly.

One of the most effective methods of increasing fatigue performance is by imposing residual compressive stresses within a thin outer surface layer. Thus, a surface tensile stress of external origin will be partially nullified and reduced in magnitude by the residual compressive stress. The net effect is that the likelihood of crack formation and therefore of fatigue failure is reduced.

Residual compressive stresses are commonly introduced into ductile metals mechanically by localized plastic deformation within the outer surface region. Commercially, this is often accomplished by a process termed shot peening. Small, hard particles (shot) having diameters within the range of 0.1 to 1.0 mm are projected at high velocities onto the surface to be treated. The resulting deformation induces compressive stresses to a depth of between one-quarter and one-half of the shot diameter.

During machining operations,small scratches and grooves are invariably introduced into the workpiece surface by cutting tool action. These surface markings can limit the fatigue life. It has been observed that improving the surface finish by polishing will enhance fatigue life significantly.

One of the most effective methods of increasing fatigue performance is by imposing residual compressive stresses within a thin outer surface layer. Thus, a surface tensile stress of external origin will be partially nullified and reduced in magnitude by the residual compressive stress. The net effect is that the likelihood of crack formation and therefore of fatigue failure is reduced.

Residual compressive stresses are commonly introduced into ductile metals mechanically by localized plastic deformation within the outer surface region. Commercially, this is often accomplished by a process termed shot peening. Small, hard particles (shot) having diameters within the range of 0.1 to 1.0 mm are projected at high velocities onto the surface to be treated. The resulting deformation induces compressive stresses to a depth of between one-quarter and one-half of the shot diameter.

There are 5 questions to complete.

For q8, answers should be given as follows :

1. A&B

2.A&C

3. B&C

4. ONLY B

It’s Correct ans is A or D.

Q5 please check figure

kindly make that type of paper by ESE QUES.

Q.23 answer should be D. “Tempering”.