A steel rod having radius r and length L gets stretched along its length by ΔL, when a force F is applied to it. If another rod made of the same material having radius 2r and length L is subjected to the same force F, then the elongation observed for the second rod is
4ΔL
2ΔL
ΔL /4
ΔL /2
Answer is Right!
Answer is Wrong!
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UPSC Geoscientist – 2023
Stress = F/A, where F is the force applied and A is the cross-sectional area.
Strain = ΔL/L, where ΔL is the elongation and L is the original length.
So, Y = (F/A) / (ΔL/L) = (F * L) / (A * ΔL).
Rearranging for elongation, ΔL = (F * L) / (A * Y).
The cross-sectional area of a rod with radius r is A = πr².
For the first rod: ΔL₁ = (F * L) / (πr² * Y).
For the second rod: Radius is 2r, length is L, material is the same (Y is the same), and the force is F.
The cross-sectional area of the second rod is A₂ = π(2r)² = 4πr².
The elongation for the second rod is ΔL₂ = (F * L) / (A₂ * Y) = (F * L) / (4πr² * Y).
Substituting the expression for ΔL₁ into the equation for ΔL₂:
ΔL₂ = (1/4) * [(F * L) / (πr² * Y)] = (1/4) * ΔL₁.
Thus, the elongation observed for the second rod is ΔL/4.