# Mechanical Behaviour Questions and Answers – Deformation Behaviour in Ceramics and Polymers

This set of Mechanical Behaviour Questions and Answers for Experienced people focuses on “Deformation Behaviour in Ceramics and Polymers”.

1. Ceramics are characterized for their ___ shear strength and ___ ductility.
a) High, high
b) High, low
c) Low, high
d) Low, low

Explanation: Ceramics contain ionic and covalent bonds. These bonds give them high shear strength. They have low ductility.

2. Ceramics have low ___
a) Hardness
b) Compressive strength
c) Fracture strength
d) Notch sensitivity

Explanation: Ceramics have narrow dislocations. These have low fracture strength. These have high hardness, compressive strength and notch sensitivity.

3. There are ___ slip systems in NiO.
a) 4
b) 6
c) 8
d) 12

Explanation: Slip can occur in NiO on {1 1 0} planes and in <1 1 0> directions. There are only six slip systems of the type {1 1 0} <1 1 0> in NiO. Whereas slip systems in Ni are 12.

4. What is the magnitude of Burgers vector of dislocations in NiO?
a) 2 Å
b) 2.5 Å
c) 3 Å
d) 3.5 Å

Explanation: The magnitude of the Burgers vector of dislocations in ceramics is higher than metals. Its value is 3 Å for NiO. While it is 2.4 Å for Ni.

5. Stress-strain curve for semi crystalline thermoplastics can be divided into ___ regimes.
a) 2
b) 3
c) 4
d) 5

Explanation: Semi crystalline thermoplastics with unaligned molecules exhibit a different type of stress-strain behavior. Their stress-strain curve can be divided into 3 regimes.

6. Regime I in stress-strain curve for semi crystalline thermoplastics show ___ behavior.
a) Linear
b) Irregular
c) Fracture point
d) Yield phenomena

Explanation: In regime I, stress-strain curve comprises of linear portion. This consists of a low modulus. This represents the force required to overcome secondary intermolecular bonds.

7. Which polymer doesn’t obey Hooke’s law?
a) Thermoset plastic
b) Thermoplastic below Tg
c) Thermoplastic with oriented molecules
d) Semi crystalline thermoplastic

Explanation: Semi crystalline thermoplastics show plastic deformation and yield point. Thermoset plastics show the almost linear stress-strain curve. Thermoplastics below Tg and with oriented molecules obey Hooke’s law.

8. Regime I in a tensile curve for polymer comprises of ___ portion with ___ molecules.
a) Linear, low
b) Irregular, low
c) Non-linear, high
d) Linear, high

Explanation: In regime I, polymer shows a linear curve. According to its slope, it shows low modulus. This represents the stress required to overcome the intermolecular secondary bonds.

9. During the stage of work hardening ___ structure develops.
a) Equiaxed
b) Microfibrillar
c) Neck
d) Fine grained

Explanation: Work hardening begins during Regime III. It continues up to fracture point of a specimen. During this stage, microfibrillar structure is developed.

10. Above glass transition temperature, polymer is ___
a) Hard
b) Stiff
c) Brittle
d) Ductile

Explanation: Above glass transition temperature (Tg), polymer is soft and ductile. Below Tg, polymer is hard, stiff and brittle. This temperature depends on various parameters.

11. HDPE has a lower UTS than LDPE.
a) True
b) False

Explanation: HDPE is more crystalline than LDPE. Because of high crystallinity, Intermolecular bonding is stronger. Hence it shows higher UTS.

12. A tensile strength of a polymer may be equal, greater or less than the yield strength.
a) True
b) False

Explanation: UTS in case of polymers may be greater, equal to or less than the yield strength. As for PI (TP), it is equal (172 MPa). For Nylon 6,6 UTS is greater than yield strength.

Sanfoundry Global Education & Learning Series – Mechanical Behaviour & Testing of Materials.

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