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A metallic wire of length L, area of cro...

A metallic wire of length L, area of cross-section A is suspended by attaching some wieght to it. If `alpha` is the longitudinal strain and Y is Young's modulus, find the ratio between elastic potential energy and the elastic energy density.

Text Solution

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Elastic potential energy
` = 1/2 xx stress xx strai n xx volume`
`= 1/2 Y xx (strai n)^(2) xx volumw`
Elastic energy density
`=("elastic potential energy")/("volume of wire")`
`=1/2 Y xx (strai n)^(2)`
`:. ("elasticity pontential energy" )/("elastic energy density" )`
= volume of the wire = AL .
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Knowledge Check

  • A metallic wire is suspended by attaching some weight to it. If alpha is the longitudinal strain and Y is Young's modulus, then the ratio of elastic potential energy to the energy density is equal to

    A
    stress on the wire
    B
    volume of the wire
    C
    strain in the wire
    D
    change in the volume of the wire

  • The elastic energy per unit volume is terms of longitudinal strain sigma and Young's modulus Y is

    A
    `(Ysigma^2)/(2)`
    B
    `(1)/(2)Ysigma`
    C
    `2Ysigma^2`
    D
    `2Ysigma`

  • A metal wire of length L, area of cross-section A and Young's modulus Y behaves as a spring of spring constant k.

    A
    `k = YA//L`
    B
    `k = 2YA//L`
    C
    `k = YA//2L`
    D
    `k = YL//A`

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