Mechanical Properties of Solids

Question: The area of cross section of a steel wire $(Y=2.0\times {{10}^{11}}N/{{m}^{2}})$is $0.1\ c{{m}^{2}}$. The force required to double its length will be

1) $2\times {{10}^{12}}N$

2) $2\times {{10}^{11}}N$

3) $2\times {{10}^{10}}N$

4) $2\times {{10}^{6}}N$

Solution:

Explanation: No Explanation

Young’s Modulus Breaking Stress

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