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A mild steel wire of length 2L and cross-sectional areaA is stretched, well within elastic limit, horizontally between two pillars, A mass m is suspended from the midpoint of the wire. Strain in the wire is |
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Answer» `x^2/(2L^2)`  CHANGE in LENGTH , `triangleL=AC-AO` (From `triangleACO` ) `=[L^2+x^2]^(1//2) -L=L[1+1/2 x^2/L^2]-L=x^2/(2L)` `therefore` Longitudinal STRAIN = `(DELTAL)/L = (x^2//2L)/L=x^2/(2L^2)` |
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