Young’s modulus of steel being
1.2 × 1012 N/m2 means that
a stress of 1.2 × 1012 N/m2
is required to produce a unit longitudinal strain
in a steel rod, provided the elastic limit is not exceeded.
In other words, steel is a very rigid material, and
a very large force is needed to produce even a small change in its length.
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2. Define Young’s modulus. What is its unit in SI?
Young’s modulus is defined as the ratio of
longitudinal stress to the corresponding
longitudinal strain produced in a body within the
elastic limit.
Mathematically,
Young’s modulus (Y) = Longitudinal stress / Longitudinal strain
The SI unit of Young’s modulus is
newton per square metre (N/m2),
which is also called the pascal (Pa).
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3. State Hooke’s law.
Hooke’s law states that, within the elastic limit of a material,
the strain produced is directly proportional to the applied stress.
Mathematically,
Stress ∝ Strain
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or
Stress = k × Strain,
where k is a constant depending on the nature of the material.
4. Draw a stress-strain curve for a metallic wire.
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5. Which one is more elastic, a piece of rubber or a piece of steel?
A piece of steel is more elastic than a piece of rubber.
This is because elasticity is measured by Young’s modulus.
Steel has a much higher Young’s modulus than rubber, which means steel
undergoes a smaller strain for the same applied stress.
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6. Define stress and strain.
Stress is defined as the internal restoring force per unit area
developed in a body when an external force is applied to it.
Stress = Force / Area
Strain is defined as the ratio of the change in dimension of a body
to its original dimension produced due to the applied force.
Strain = Change in dimension / Original dimension
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Elasticity – Important Questions and Numericals | Engineering Physics (Polytechnic 1st Year)
1. Two wires of same material have lengths in the ratio 1:2 and radii in the ratio 2:1. Find the ratio of the elongation produced in them when they are stretched by same amount of force.
Let the two wires be 1 and 2.
Given:
Elongation of a wire is given by:
ΔL = (F L) / (A Y)
Since A = πr2,
ΔL ∝ L / r2
Therefore,
ΔL1 / ΔL2
= (L1 / r12) ÷ (L2 / r22)
Substituting values:
ΔL1 / ΔL2
= (1 / 22) ÷ (2 / 12)
= (1 / 4) ÷ 2
= 1 / 8
Hence, the ratio of elongations is
ΔL1 : ΔL2 = 1 : 8.
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Lengths: L1 : L2 = 1 : 2
Radii: r1 : r2 = 2 : 1
Same material ⇒ same Young’s modulus (Y)
Same stretching force (F)
2. Find the tension in a steel wire of length 2 m and diameter 1 mm. The wire is stretched by 1 mm. Given that Young’s modulus,
Y = 2 × 1011 N m−2
Given:
Length of wire, L = 2 m
Area of cross-section,
A = πr2 = π(0.5 × 10−3)2
= 0.25π × 10−6 m2
Using the relation:
Y = (F L) / (A ΔL)
Therefore,
F = (Y A ΔL) / L
Substituting values:
F = (2 × 1011 × 0.25π × 10−6 × 1 × 10−3) / 2
F ≈ 78.5 N
Hence, the tension in the wire is approximately 78.5 N.
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Diameter, d = 1 mm = 1 × 10−3 m
Radius, r = d/2 = 0.5 × 10−3 m
Extension, ΔL = 1 mm = 1 × 10−3 m
Young’s modulus, Y = 2 × 1011 N m−2
3. A wire of length 10 m has a cross-sectional area of 1.25 × 10−4 m2. It is subjected to a load of 5 kg. If Young’s modulus of the material is
Y = 4 × 1010 N m−2, calculate the elongation produced in the wire.
Given:
Length of wire, L = 10 m
The elongation of a wire is given by:
ΔL = (F L) / (A Y)
Substituting the given values:
ΔL = (49 × 10) / (1.25 × 10−4 × 4 × 1010)
ΔL = 490 / (5 × 106)
ΔL = 9.8 × 10−5 m
Hence, the elongation produced in the wire is 9.8 × 10−5 m.
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Cross-sectional area, A = 1.25 × 10−4 m2
Load, m = 5 kg
Force applied, F = mg = 5 × 9.8 = 49 N
Young’s modulus, Y = 4 × 1010 N m−2
Thank you