Practice portal › Mechanical Properties of Solids › Elastic Potential Energy
Asked in JEE Main 9th Jan 2nd Shift 2020 · Energy stored in a stretched wire
Given: the same length, the same material and the same load F for both wires; only the areas differ.
Idea: u=1/2(σ²)/Y with σ=F/A, so at fixed F and Y, u∝1/(A²)∝1/(d⁴).
Then (u₁)/(u₂)=((d₂)/(d₁))⁴=1/4, so ((d₁)/(d₂))⁴=4.
(d₁)/(d₂)=4^1/4=√2.
The ratio of diameters is √2:1.
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