Mechanical Properties of Solids

Question: To double the length of a iron wire having $0.5\,c{{m}^{2}}$ area of cross-section, the required force will be $(Y={{10}^{12}}\,dyne/c{{m}^{2}})$



1) $1.0\times {{10}^{-7}}N$
2) $1.0\times {{10}^{7}}N$
3) $0.5\times {{10}^{-7}}N$
4) $0.5\times {{10}^{12}}$dyne
Solution: Explanation: No Explanation
Young’s Modulus Breaking Stress

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