Skip to main content

Material Science - short answer type questions from AMIE exams (Winter 2021)

Answer the following in brief. (20 marks)

Discuss Scanning electron microscopy.

The scanning electron microscope (SEM) uses a focused beam of high-energy electrons to generate a variety of signals at the surface of solid specimens. The signals that derive from electron-sample interactions reveal information about the sample including external morphology (texture), chemical composition, and crystalline structure and orientation of materials making up the sample.
 

Differentiate between SEM and FESEM.

The difference between a FESEM and an SEM lies in the electron generation system. As a source of electrons, the FESEM uses a field emission gun that provides extremely focused high and low-energy electron beams, which greatly improves spatial resolution and enables work to be carried out at very low potentials . This helps to minimise the charging effect on non-conductive specimens and to avoid damage to electron beam-sensitive samples.

What is AFM?

The atomic force microscope (AFM) is a type of scanning probe microscope whose primary roles include measuring properties such as magnetism, height, and friction. The resolution is measured in a nanometer, which is much more accurate and effective than the optical diffraction limit.

 

Explain XRD.

X-ray diffraction analysis (XRD) is a technique used in materials science to determine the crystallographic structure of a material. XRD works by irradiating a material with incident X-rays and then measuring the intensities and scattering angles of the X-rays that leave the material. 


The primary use of XRD analysis is the identification of materials based on their diffraction pattern.  As well as phase identification, XRD also yields information on how the actual structure deviates from the ideal one, owing to internal stresses and defects.
 

Discuss EDS analysis.

EDS can be used to determine which chemical elements are present in a sample and can be used to estimate their relative abundance. EDS also helps to measure the multi-layer coating thickness of metallic coatings and analysis of various alloys.
 

How the dislocations are detected?

  • Transmission electron microscopy can be used to observe dislocations within the microstructure of the material.
  • Field ion microscopy and atom probe techniques offer methods of producing much higher magnifications and permit the observation of dislocations at an atomic level.

What is the atomic packing factor of the BCC structure?

Explain the importance of determining packing efficiency.

Packing efficiency = (Total volume of spheres/volume of cube) x 100

Packing Efficiency is important as:
  • Packing Efficiency represents the solid structure of the object.
  • It displays different properties of solids like consistency, density, and isotropy.
  • With the help of Packing Efficiency, different attributes of solid structures can be derived.

Discuss in brief the age-hardening process.

Age hardening, also known as precipitation hardening, is a type of heat treatment that is used to impart strength to metals and their alloys. It is called precipitation hardening as it makes use of solid impurities or precipitates for the strengthening process. The metal is aged by either heating it or keeping it stored at lower temperatures so that precipitates are formed. 

The process of age hardening is executed in a sequence of three steps. 
  • First, the metal is treated with a solution at high temperatures. 
  • The next step is rapid cooling across the solvus line so that the solubility limit is exceeded. The result is a super-saturated solid solution that remains in a metastable state. The lowering of temperatures prevents diffusion. 
  • Finally, the supersaturated solution is heated to an intermediate temperature in order to induce precipitation. The metal is maintained in this state for some time.

Discuss in brief the role of the shot peening process in property enhancement.

Shot peening is a cold work process used to impart compressive residual stresses onto the surface of a component, which results in modified mechanical properties. The shot peening process is used to add strength and reduce the stress profile of components.

Shot peening works by striking a surface with a shot (round metallic, glass or ceramic particle) with enough force to generate plastic deformation. When a group of shots impact the surface they generate multiple indentations, resulting in the component being encased by a compressively stressed layer on the metal surface.

The main advantage of shot peening is to extend the service life of a component by creating an induced compressive stress layer to increase resistance to fatigue (including corrosion fatigue, stress corrosion and cavitation erosion) while also helping to resist the development and propagation of cracks. The creation of compression stresses that resists metal fatigue help prevent the propagation of cracks through the material.
---

The study material for AMIE/Junior Engineer exams is available at https://amiestudycircle.com

Comments

Popular posts from this blog

Mechanics of Fluids (Solved Numerical Problems)

Numerical The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks) (AMIE Summer 2023) Solution Surface tension, σ = 0.0725 N/m Pressure intensity, P = 0.02 N/m 2 P = 4σ/d Hence, the Diameter of the dropd = 4 x 0.0725/200 = 1.45 mm Numerical Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks) (AMIE Summer 2023) Answer: 0.0125 N/m Numerical The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (AMIE Summer 2023, 7 marks) Answer: 0.725 N/cm 2   Numerical An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at t...

Design of Electrical Systems (Solved Numerical Problems)

Important note There is something wrong with this question paper. It seems that instead of "Design of Electrical Systems" the IEI has given problems from "Electrical Machines". You should raise a complaint to director_eea@ieindia.org in this regard. Numerical A 120 V DC shunt motor draws a current of 200A. The armature resistance is 0.02 ohms and the shunt field resistance is 30 ohms. Find back emf. If the lap wound armature has 90 slots with 4 conductors per slots, at what speed will the motor run when flux per pole is 0.04 Wb?​ (AMIE Summer 2023, 8 marks) Solution The back EMF (E b ) of a DC motor can be calculated using the formula: E b = V - I a R a   Given: V = 120 V I a = 200 A R a = 0.02 ohms Substituting the values into the formula: E b = 120 − 200 × 0.02 = 120 − 4​ = 116 V Now, let's calculate the speed (N) at which the motor will run using the given flux per pole (φ p ). The formula to calculate the speed of a DC motor is: N = 60×E b /(P×φ p ) Wh...

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\) Total Power \({P_{total}} = \rho A{V_1}^3/2\) Power density \(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\) Maximum power density \(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\) Assuming eff...