Skip to main content

Material Science (Mixed Bag)

1. Ductility of a material can be defined as
(a) ability to undergo large permanent deformations in compression
(b) ability to recover its original form
(c) ability to undergo large permanent deformations in tension
(d) all of the above
(e) none of the above.

2. Malleability of a material can be defined as
(a) ability to undergo large permanent deformations in compression
(b) ability to recover its original form
(c) ability to undergo large permanent deformations in tension
(d) all of the above
(e) none of the above.

3. In compression, a prism of brittle material will break
(a) by forming a bulge (l>) by shearing along oblique plane
(c) in direction perpendicular to application of load
(d) by crushing into thousands of pieces
(e) none of the above.

4. The ability of a material to resist softening at high temperature is known as
(a) creep
(b) hot tempering
(c) hot hardness
(d) fatigue
(e) super hardening.

5. Mild steel belongs to the following category
(a) low carbon steel
(b) medium carbon steel
(c) high carbon steel
(d) alloy steel
(e) special steel.

6. The ultimate tensile strength of low carbon steel by working at a high strain rate will (a) decrease
(b) increase
(c) remain constant
(d) first increase and then decrease
(e) first decrease and then increase.

7. Slow plastic defoliation of metals under a constant stress is known as
(a) creep
(b) fatigue
(c) endurance
(d) plastic deformation
(e) non-plastic deformation.

8. The ultimate tensile strength and yield strength of most of the metals, when temperature
falls from 0 to l00°C will
(a) increase
(b) decrease
(c) remain same
(d) first increase and then decrease
(e) show unpredictable behaviour.

9. The number of electrons in 1 cm3 of metal would be of the order of
(a) 1010
(b) 1010
(c) 1022
(d) 1040
(e) 1052

10. Stress relaxation is- the phenomenon
(a) in which parts are not loaded
(b) in which stress remains constant on increasing load
(c) in which deformation tends to loosen the joint and produces a stress reduced
(d) stress reduces on increasing load
(e) none of the above.

Answer:

    1. c    2. a    3. b    4. c    5. a
    6. b    7. a    8. a    9. c    10. c

---


Labels:

Comments

Post a Comment

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...
Post a Comment
Read more

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...
Post a Comment
Read more

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...
Post a Comment
Read more