# Engineering Physics Questions and Answers – Elastic Limit

This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Elastic Limit”.

1. Amorphous solids do not melt at a sharp temperature, rather these have softening range.
a) True
b) False

Explanation: All bonds in an amorphous solid are not equally strong. When then solid is heated, weaker bonds gets ruptured at the lowest temperature and the stronger ones at higher temperatures. So the solid first softens and then finally melts.

2. Which is more elastic?
a) Water
b) Air
c) Solid
d) Crystal

Explanation: Water is more elastic than anything else. Air can be compressed easily whereas water is incompressible and bulk modulus is reciprocal of compressibility.

3. The following wires are made of the same material. Which of these will have the largest extension, when the same tension is applied?
a) Length = 50cm, diameter = 0.5mm
b) Length = 100cm, diameter = 1mm
c) Length = 200cm, diameter = 2mm
d) Length = 300cm, diameter = 3mm

Explanation: ∆l=F/A l/Y=4F/(πD2) l/Y
In Length = 50cm, diameter = 0.5mm, l/D2 = 50/0.052 = 2×104 cm-1
In Length = 100cm, diameter = 1mm, l/D2 = 100/0.12 = 104 cm-1
In Length = 200cm, diameter = 2mm, l/D2 = 200/0.22 = 5×103 cm-1
In Length = 300cm, diameter = 3mm, l/D2 = 300/0.32 = 3.3×103 cm-1
Hence ∆l is maximum in Length = 50cm, diameter = 0.5mm.

4. There are two wires of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension the diameter of first wire, then the ratio of extension produced in the wire by applying same load will be _____________
a) 1:1
b) 2:1
c) 1:2
d) 4:1

Explanation: Extension,
∆l=F/A l/Y=F/(πr2) l/Y
For the two wires F, l and Y are same, so
(∆l1)/(∆l2)=((r2)2)/((r1)2)=(2/1)2 4:1.

5. A cube is subjected to a uniform volume compression. If the side of the cube decreases by 2%, the bulk strain is ___________
a) 0.02
b) 0.03
c) 0.04
d) 0.06

Explanation: V=l3
∆V/V=3∆l/l=3(2/10)=0.06.
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6. Energy stored in stretching a string per unit volume is ___________
a) 1/2×stress×strain
b) stress×strain
c) Y(Strain)2
d) 1/2 Y(Stress)2

Explanation: Energy stored per unit volume= Work done per unit volume
Energy stored per unit volume = 1/2×Stress×strain.

7. According to Hooke’s law of elasticity, if stress is increased, the ratio of stress to strain?
a) Increased
b) Decreased
c) Becomes zero
d) Remains constant

Explanation: According to Hooke’s law,
Stress/Strain=constant.

8. If in a wire of Young’s modulus Y, longitudinal strain is produced, then the value of potential energy stored in its unit volume will be ___________
a) YX2
b) 0.5Y2 X
c) 2YX2
d) 0.5YX2

Explanation: Potential energy stored per unit volume,
u=1/2×Stress×Strain=1/2 (Y×Strain)×Strain
u=0.5YX2.

9. A stretched rubber has ___________
a) Increased kinetic energy
b) Increased potential energy
c) Decreased kinetic energy
d) Decreased potential energy

Explanation: The work done in stretching the rubber is stored as its potential energy.

10. The breaking stress of a wire depends upon ___________
a) Length of the wire
c) Material of the wire
d) Shape of the cross-section

Explanation: The stress at which rupture of the wire occurs is called its breaking stress. Its value depends on the material of the wire.

11. Which of the following affects the elasticity of a substance?
a) Hammering and annealing
b) Change in temperature
c) Impurity in substance
d) All of the mentioned

Explanation: All the factors change in temperature, hammering and annealing and impurity in substance affects the elasticity of a substance.

12. The diameter of brass rod is 4mm. Young’s modulus of brass is 9×109 N/m2. The force required to stretch 0.1% of its length is ___________
a) 360πN
b) 36N
c) 36π×105 N
d) 144π×103N

Explanation: ∆l/l=0.1/100
F=YA∆l/l=(Y×πr2×∆l)/l
F=(9×109×π×(2×10(-3))2×0.1)/100 N=360πN.

13. A substance breaks down by a stress of (106 N)/m2. If the density of the material of the wire is 3×(103) kg/m3, then the length of the wire of the substance which will break under its own weight when suspended vertically will be ___________
a) 66.6m
b) 60.0m
c) 33.3m
d) 30.3m

Explanation: Breaking stress = mg/area=Alρg/A=lρg
l=Stress/ρg=106/(3×103×10)=33.3m.

Sanfoundry Global Education & Learning Series – Engineering Physics.

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