20 Physics -- Elasticity

A cylindrical of copper wire and a cylindrical steel wire, each of length 1.5 m and diameter 2 mm, are joined at one end to form a composite wire 3 m long. The wire is loaded until its length becomes 3.003 m. Calculate the strains in copper and steel wires and the force applied to the w...

A cylindrical of copper wire and a cylindrical steel wire, each of length 1.5 m and diameter 2 mm, are joined at one end to form a composite wire 3 m long. The wire is loaded until its length becomes 3.003 m. Calculate the strains in copper and steel wires and the force applied to the wire. (Young modulus for copper = 1.2 × 10¹¹ N/m² , Young modulus for steel = 2.0 × 10¹¹ N/m² .)

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Solution:
The length of a copper wire, LC = 1.5 m;
the length of a steel wire, LS = 1.5 m;
the diameter of the wires, d = 2 mm = 2 × 10^–3 m;
the composite length, L = 3 m;
the extended composite length, L' = 3.003 m;
the total extension, l = L' – L = 3.003 – 3 = 0.003 m;
the extension of the copper wire, l_C = ?;
the extension of the steel wire, l_S = ?

l_C + l_S = 0.003 … (1)

For the copper wire, Young modulus,

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If the oscillation is simple harmonic motion then we have,

Inserting given data, we get the spring constant approximately equal to 79N/m.

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When they fly apart symmetrically relative to the initial motion direction with the angle of divergence,
From the conservation of momentum, along the horizontal and vertical direction

and m1v1sin(θ/2)=m2v2sin(θ/2)

or, m1v1=m2v2 ...(2)

Now, from conservation of kinetic energy,

1/2 m1 v1²+ 0 =1/2 m1 v1²+1/2 m2 v2² ...(3)

From (1) and (2),

m1 u1=cos(θ/2)(m1v1+m1v1m2m2)=2m1v1cos(θ/2)

So, u1=2v1cos(θ/2)...(4)

From (2), (3) and (4)

4m1cos²(θ/2) v1²=m1v1²+m2m1²v1²/m2²

or, 4cos²(θ/2)=1+m1/m2

or,...

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Explain which one is more elastic–rubber or steel.

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Solution:
The Young modulus of a wire, Y = 12 1010 N/m2 ;
the diameter of the wire, d = 0.56 mm = 0.56 10^–3 mm
The cross-section area of the wire,

The strain = one-tenth of 1 percent

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Solution:
The radius of a sphere, r = 10 cm = 0.10 m;
the increase in pressure, p = 5 10⁸N/m²;
the bulk modulus, B = 3.14 10¹¹N/m²;
the change in volume, V = ?

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Solution:
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Solution:
The extension, l = ?;
the force applied, F = 30 N;
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the diameter of the wire, d = 3 mm = 3 10^–3m
The cross-section area of the wire,

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Solution:
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We have,
the weight of the wire = mg = (volume density)g
or F = LAρg

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Solution:

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Y = FL/Ae

e = FL/AY

Further, F = mg and centre of mass of the rod is at the centre, so L' = L/2

Therefore, e = mgL/2AY

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When a wire is bent back and forth, heat is generated due to the area of the elastic hysteresis and frictional force. Hence it becomes hot.

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Note of chapter 8

Does the temperature affect elasticity?

As the temperature increases, the interatomic forces of attraction become weaker. For a given stress, a large strain is produced when temperature increased.

Hence, the modulus of elasticity(stress/strain) decreases with increasing temperature.

So YES, temperature does affect the elasticity of the body, i.e body becomes more plastic in nature at higher temperature.

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(a)
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(d)The...

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