Categories > Mechanical Properties of Solids > Young’s Modulus > An aluminum rod (Young's modulus $=7\times {{10}^{9}}\,N/{{m}^{2}})$ has a breaking strain of $0.2%$. The minimum cross-sectional area of the rod in order to support a load of ${{10}^{4}}$ Newton's is

Mechanical Properties of Solids

Question: An aluminum rod (Young's modulus $=7\times {{10}^{9}}\,N/{{m}^{2}})$ has a breaking strain of $0.2%$. The minimum cross-sectional area of the rod in order to support a load of ${{10}^{4}}$ Newton's is


1) $1\times {{10}^{-2}}\,{{m}^{2}}$
2) $1.4\times {{10}^{-3}}\,{{m}^{2}}$
3) $3.5\times {{10}^{-3}}\,{{m}^{2}}$
4) $7.1\times {{10}^{-4}}\,{{m}^{2}}$
Solution: Explanation: No Explanation
Young’s Modulus Breaking Stress

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