a mass of 10kg is suspended from the end of a steel rod of length 2m and radius 1mm. what is the elongation of the rod beyond its original length (Take E= 200 × 109^9 N/m²).​

ΔL = 3.12 x 10⁻⁴ m = 0.312 mm

Explanation:

First, we will find the stress applied by the mass on the rod:

where,

σ = stress = ?

F = Force applied by the mass = weight = mg = (10 kg)(9.81 m/s²) = 98.1 N

A = cross-sectional area = πr² = π(1 mm)² = π(0.001 m)² = 3.14 x 10⁻⁶ m²

Therefore,

σ = 3.12 x 10⁷ Pa

Now, we will find the value of strain:

∈ = 1.56 x 10⁻⁴

Now, the change in length can be given by the formula of strain:

where,

L = Original Length = 2 m

ΔL = Elongation = ?

Therefore,

ΔL = (1.56 x 10⁻⁴)(2 m)

ΔL = 3.12 x 10⁻⁴ m = 0.312 mm

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