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March 27, 2017

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When a wire of lenght 5m & radius 0.5mm is stretched by a load of 49N,the elongation produced in the wire is 0.1cm. Find the energy stored per unit volume of the wire.

  • PHYSICS - ,

    σ=F/A=F/πr²
    ε=ΔL/L

    Hook’s law for the wire:
    k ε= σ
    k= σ/ ε= F L /πr²ΔL

    E=k•ΔL²/2= F• L•ΔL²/2πr²ΔL=
    =F•L•ΔL/2πr²=49•5•10⁻³/2•π•25•10⁻⁸=1.56•10⁵ J
    E₀=E/V= E/πr²L = 1.56•10⁵/π•25•10⁻⁸•5=3.97•10¹⁰ J/m³

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