To double the length of a iron wire having $0.5\,c{m^2}$ area of cross-section, the required force will be $(Y = {10^{12}}\,dyne/c{m^2})$
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(d) If length of wire doubled then strain $= 1$
$Y = {\rm{stress}}$ ==> $F = Y \times A$$ = {10^{12}} \times 0.5$$ = 0.5 \times {10^{12}}dyne$
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